Gordon James, Martin Liebeck Representations and Characters of Groups, Second Edition 2001

Representations and Characters of Groups Now in its second edition, this text provides a modern introduction to the representation theory of ®nite groups. The authors have revised the popular ®rst edition and added a considerable amount of new material. The theory is developed in terms of modules, since this is appropriate for more advanced work, but considerable emphasis is placed upon constructing characters. The character tables of many groups are given, including all groups of order less than 32, and all simple groups of order less than 1000. Among the applications covered are Burnside's pa q b theorem, the use of character theory in studying subgroup structure and permutation groups, and a description of how to use representation theory to investigate molecular vibration. Each chapter is accompanied by a variety of exercises, and full solutions to all the exercises are provided at the end of the book. This will be ideal as a text for a course in representation theory, and in view of the applications of the subject, will be of interest to mathematicians, chemists and physicists alike. R E P R E S E N TAT ION S AN D C H AR ACT E R S OF G ROU P S G O R D O N JA M E S a n d M A RT I N L I E B E C K Department of Mathematics, Imperial College, London Second Edition PUBLISHED BY CAMBRIDGE UNIVERSITY PRESS (VIRTUAL PUBLISHING) FOR AND ON BEHALF OF THE PRESS SYNDICATE OF THE UNIVERSITY OF CAMBRIDGE The Pitt Building, Trumpington Street, Cambridge CB2 IRP 40 West 20th Street, New York, NY 10011-4211, USA 477 Williamstown Road, Port Melbourne, VIC 3207, Australia http://www.cambridge.org © Cambridge University Press 1993, 2001 This edition © Cambridge University Press (Virtual Publishing) 2003 First published in printed format 1993 Second edition 2001 A catalogue record for the original printed book is available from the British Library and from the Library of Congress Original ISBN 0 521 81205 4 hardback Original ISBN 0 521 00392 X paperback ISBN 0 511 01700 6 virtual (netLibrary Edition) Contents Preface 1 Groups and homomorphisms 2 Vector spaces and linear transformations 3 Group representations 4 FG-modules 5 FG-submodules and reducibility 6 Group algebras 7 FG-homomorphisms 8 Maschke's Theorem 9 Schur's Lemma 10 Irreducible modules and the group algebra 11 More on the group algebra 12 Conjugacy classes 13 Characters 14 Inner products of characters 15 The number of irreducible characters 16 Character tables and orthogonality relations 17 Normal subgroups and lifted characters 18 Some elementary character tables 19 Tensor products 20 Restriction to a subgroup 21 Induced modules and characters 22 Algebraic integers 23 Real representations 24 Summary of properties of character tables 25 Characters of groups of order pq 26 Characters of some p-groups 27 Character table of the simple group of order 168 v page vii 1 14 30 38 49 53 61 70 78 89 95 104 117 133 152 159 168 179 188 210 224 244 263 283 288 298 311 vi Representations and characters of groups 28 Character table of GL(2, q) 322 29 Permutations and characters 337 30 Applications to group theory 348 31 Burnside's Theorem 361 32 An application of representation theory to molecular vibration 367 Solutions to exercises 397 Bibliography 454 Index 455 Preface We have attempted in this book to provide a leisurely introduction to the representation theory of groups. But why should this subject interest you? Representation theory is concerned with the ways of writing a group as a group of matrices. Not only is the theory beautiful in its own right, but it also provides one of the keys to a proper understanding of ®nite groups. For example, it is often vital to have a concrete description of a particular group; this is achieved by ®nding a representation of the group as a group of matrices. Moreover, by studying the different representations of the group, it is possible to prove results which lie outside the framework of representation theory. One simple example: all groups of order p2 (where p is a prime number) are abelian; this can be shown quickly using only group theory, but it is also a consequence of basic results about representations. More generally, all groups of order pa q b ( p and q primes) are soluble; this again is a statement purely about groups, but the best proof, due to Burnside, is an outstanding example of the use of representation theory. In fact, the range of applications of the theory extends far beyond the boundaries of pure mathematics, and includes theoretical physics and chemistry ± we describe one such application in the last chapter. The book is suitable for students who have taken ®rst undergraduate courses involving group theory and linear algebra. We have included two preliminary chapters which cover the necessary background material. The basic theory of representations is developed in Chapters 3±23, and our methods concentrate upon the use of modules; although this accords with the more modern style of algebra, in several instances our proofs differ from those found in other textbooks. The main results are elegant and surprising, but at ®rst sight they sometimes have an air of mystery vii viii Representations and characters of groups about them; we have chosen the approach which we believe to be the most transparent. We also emphasize the practical aspects of the subject, and the text is illustrated with a wealth of examples. A feature of the book is the wide variety of groups which we investigate in detail. By the end of Chapter 28, we have presented the character tables of all groups of order less than 32, of all p-groups of order at most p4 , and of all the simple groups of order less than 1000. Every chapter is accompanied by a set of Exercises, and the solutions to all of these are provided at the end of the book. We would like to thank Dr Hans Liebeck for his careful reading of our manuscript and the many helpful suggestions which he made. In this second edition, we have included two new chapters; one (Chapter 28) deals with the character tables of an in®nite series of groups, and the other (Chapter 29) covers aspects of the representation theory of permutation groups. We have also added a considerable amount of new material to Chapters 20, 23 and 30, and made minor amendments elsewhere. Preface to Second Edition eg ˆ ge ˆ g. Groups A group consists of a set G. In addition. (3) for all g in G. most of which you should   know already. which we shall use extensively to illustrate the later theory. h. written gh. such as dihedral groups and symmetric groups. h of G to form another element of G. ( gh)k ˆ g(hk). there exists an element gÀ1 in G such that gg À1 ˆ g À1 g ˆ eX We refer to the rule for combining elements of G as the product operation on G.1 Groups and homomorphisms This book is devoted to the study of an aspect of group theory. An elementary course on abstract algebra would normally cover all the material in the chapter. (2) there exists an element e in G such that for all g in G. and any book on basic group theory will supply you with further details. so we begin with a resume of facts about groups. we introduce several examples. k in G. One or two results which we shall use only infrequently are demoted to the exercises at the end of the chapter ± you can refer to the solutions if necessary. together with a rule for combining any two elements g. this rule must satisfy the following axioms: (1) for all g. 1 . with the usual multiplication of complex numbers. for two symmetries f and g. . then we call G a ®nite group. and denote by C the set of all complex numbers. is a group of order n. a2 . The product of an element g with itself. . These 2n rotations and re¯ections form a group under the product operation of composition (that is. is a group. the number of elements in G is called the order of G. g0 ˆ 1. The set of nth roots of unity in C. It is written as Cn and is called the cyclic group of order n. gÀ2 ˆ ( gÀ1 )2 . There are also n re¯ection symmetries: these are re¯ections in the n lines passing through O and a corner or the mid-point of a side of the polygon. and that every g in G has just one inverse. 1. under addition. Usually we write 1. a. and so on. (3) Let n be an integer with n > 3. rather than e. and an ˆ 1. Let A be a corner of the polygon. the product fg means `®rst do f. If a ˆ e2ðia n . (2) The set Z of all integers. It is elementary to see that G has just one identity element. for the identity element of G. and gÀ1 is an inverse of g in axiom (3). r1 . . This group is called the dihedral group of order 2n. then Cn ˆ f1. Write b for the re¯ection in the . a nÀ1 g. r nÀ1 where r k is the (clockwise) rotation about the centre O through an angle 2ðkan. similarly 3 g ˆ g 2 g. There are n rotation symmetries: these are r0 . X X X . and is written D2 n. . is written g 2 . and consider the rotation and re¯ection symmetries of a regular n-sided polygon. Also. the element e in axiom (2) is an identity element of G. If the number of elements in G is ®nite. gg. and is written |G|.1 Examples (1) Let n be a positive integer. then do g').2 Representations and characters of groups Axiom (1) states that the product operation is associative. under the product operation of composition. It is called the symmetric group of degree n. and the n re¯ections are b. ab. a nÀ1 (where 1 denotes the identity. b: an ˆ 1. a. X X X . forms a group. which we denote by In or just I. under matrix multiplication. b2 ˆ 1 and bÀ1 ab ˆ aÀ1 X These relations determine the product of any two elements of the group. While Cn and Z are abelian. and write a for the rotation r1 . a2 . We use the notation H < G to indicate that H is a subgroup of G. . The identity of GL(n. and is written Sn . F) is of course the identity matrix. n}. b2 ˆ 1. The order of Sn is n!. F). A group G is said to be abelian if gh ˆ hg for all g and h in G. and hence (a i b)(a j b) ˆ a i ba j b ˆ a i aÀ j bb ˆ a iÀ j X We summarize all this in the presentation D2 n ˆ ha. the set of all permutations of {1. The set of all invertible n 3 n matrices with entries in F. . is a group. It is an in®nite group. Check that an ˆ 1. . a2 b. For example. we have ba j ˆ aÀ j b (using the relation ba ˆ aÀ1 b). Subgroups Let G be a group. . bÀ1 ab ˆ aÀ1 iX (4) For n a positive integer. 2. A subset H of G is said to be a subgroup if H is itself a group under the product operation inherited from G. most of the other examples given above are non-abelian groups. D2 n is generated by a and b. Then the n rotations are 1. (5) Let F be either R (the set of real numbers) or C (the set of complex numbers).Groups and homomorphisms 3 line through O and A. which leaves the polygon ®xed). . X X X . a nÀ1 bX Thus all elements of D2 n are products of powers of a and b ± that is. This group is called the general linear group of degree n over F. and is denoted by GL(n. X X X . the group of invertible 2 3 2 matrices with entries in C. let r be the least positive integer such that g r ˆ 1.1 are cyclic. all elements of the form a i1 b j1 a i2 b j2 X X Xa i n b j n for some n.2 Examples (1) For every group G. and let     i 0 0 1 Aˆ . then r is equal to the number of elements in k gl ± indeed. biX Given any ®nite set S of elements of G. and (2) if h. h gi ˆ f1. the subgroup of G generated by S. We illustrate the construction in the next example. jk P Z for 1 < k < n. De®ne H to be the subset of G consisting of all elements which are products of powers of a and b ± that is. This construction gives a powerful method of ®nding new groups as subgroups of given groups. The subset h gi ˆ f g n : n P Zg is a subgroup of G. 1. g rÀ1 gX We call r the order of the element g. then k gl is ®nite.5 below. If gn ˆ 1 for some n > 1. C). g 2 . and again in Example 1.4 Representations and characters of groups It is easy to see that a subset H of a group G is a subgroup if and only if the following two conditions hold: (1) 1 P H. In this case. If G ˆ k gl for some g P G then we call G a cyclic group. b P G. Then H is a subgroup of G. k P H then hkÀ1 P H. we call H the subgroup generated by a and b. called the cyclic subgroup generated by g. where ik . such as general linear or symmetric groups. both {1} and G are subgroups of G. we can similarly de®ne hSi. Bˆ X 0 Ài À1 0 . g. (4) Let G ˆ GL(2. and write H ˆ ha. The groups Cn and Z in Examples 1. (3) Let G be a group and let a. (2) Let G be a group and g P G. The above three relations determine the product of any two elements of Q8 . hh9) for all g. h)( g9. we see that every element of H has the form A i B j for some integers i. . A2 ˆ B2 . h9) ˆ ( gg9. n and ®xes the other n À 2 numbers. j. called the direct product of G and H. . and is written Q8. The group H is called the quaternion group of order 8. called the alternating group of degree n. in fact j Hj ˆ 8. accordingly. Since the matrices Ai B j (0 < i < 3. 2. G 3 H is a group. and consider G 3 H ˆ f( g. so we have the presentation Q8 ˆ hA. the subgroup of G generated by A and B. h9 P H. BÀ1 AB ˆ AÀ1 X 5 Using the third relation. 0 < j < 1) are all distinct. With this product operation. BÀ1 AB ˆ AÀ1 iX (5) A transposition in the symmetric group Sn is a permutation which interchanges two of the numbers 1. Every permutation g in Sn can be expressed as a product of transpositions. . or they all have an odd number of transpositions. Check that A4 ˆ I. . and using the ®rst two relations. The subset An ˆ f g P Sn : g is an even permutationg is a subgroup of Sn .Groups and homomorphisms Put H ˆ kA. A2 ˆ B2 . Let G and H be groups. we can take 0 < i < 3 and 0 < j < 1. Direct products We describe a construction which produces a new group from given ones. g9 P G and all h. B: A4 ˆ I. It can be shown that either all such expressions for g have an even number of transpositions. . Hence H has at most eight elements. we call g an even or an odd permutation. Bl. h): g P G and h P HgX De®ne a product operation on G 3 H by ( g. g9 ) ˆ ( g 1 g9 . 3 Gr is also ®nite. X X X . . . for every h P H there exists g P G such that gW ˆ h). then a homomorphism from G to H is a function W: G 3 H which satis®es ( g1 g 2 )W ˆ ( g 1 W)( g 2 W) for all g 1 . and is written as WÀ1 . . |Gr |. 1. g r ): g i P Gi for 1 < i < rg. . the image of g under a function W is written as gW. not as W g. then G1 3 . we generally apply functions on the right ± that is. we mean that h ˆ gW. of order |G1 | .6 Representations and characters of groups More generally. An invertible function is also called a bijection. if G1 . g r )( g9 . 3 C2 (r factors) has order 2 r and all its nonidentity elements have order 2. By an expression W: g 3 h.3 Example The group C2 3 . Gr are groups. ( gW)ö ˆ g and (hö)W ˆ hX Then ö is called the inverse of W. X X X . A function W: G 3 H is invertible if there is a function ö: H 3 G such that for all g P G. . those functions from G to H which `preserve the group structure' ± the so-called homomorphisms ± are of particular importance. . Functions A function from one set G to another set H is a rule which assigns a unique element of H to each element of G. where g P G and h P H. . h P H. . g2 P G implies that g1 ˆ g2 ) and surjective (that is. A function W from G to H is invertible if and only if it is both injective (that is. g 2 P GX . g r g9 )X 1 r 1 r If all the groups Gi are ®nite. X X X . If G and H are groups. . . We often indicate that W is a function from G to H by the notation W: G 3 H. X X X . In this book. then the direct product G1 3 X X X 3 Gr is f( g 1 . Homomorphisms Given groups G and H. g1 W ˆ g2 W for g1 . with product operation de®ned by ( g1 . 0 < s < 1.4 Example Let G ˆ D2 n ˆ ka. and write the 2n elements of G in the form ai b j with 0 < i < n À 1. 0 < j < 1. so H  G. Suppose that 0 < r < n À 1. If there is an isomorphism W from G to H. Let H be any group. y À1 xy ˆ x À1 X We shall prove that the function W: G 3 H de®ned by W: a i b j 3 x i y j (0 < i < n À 1. WÀ1 is an isomorphism from H to G.4 in action. 0 < j < 1) is a homomorphism. and suppose that H contains elements x and y which satisfy x n ˆ y 2 ˆ 1. 1. we can also deduce that xr ys xt yu ˆ x i y j X Therefore. and so W is a homomorphism. 0 < j < 1. y ˆ (2 5)(3 4)X . 0 < u < 1. Moreover. bÀ1 ab ˆ aÀ1 X Since we have x n ˆ y 2 ˆ 1. Then ar bs at bu ˆ a i b j for some i. then G and H are said to be isomorphic. bÀ1 ab ˆ aÀ1 l. also. (at bu )W. 1. The following example displays a technique which can often be used to prove that certain functions are homomorphisms. 0 < t < n À 1. j with 0 < i < n À 1. y be the following permutations in G: x ˆ (1 2 3 4 5). We now demonstrate the technique of Example 1. i and j are determined by repeatedly using the relations an ˆ b2 ˆ 1. and we write G  H. y À1 xy ˆ x À1 . (ar bs at bu )W ˆ (a i b j )W ˆ x i y j ˆ x r y s x t y u ˆ (ar bs )W . b: an ˆ b2 ˆ 1.5 Example Let G ˆ S5 and let x.Groups and homomorphisms 7 An invertible homomorphism is called an isomorphism. 4. Suppose now that G is ®nite. and let Hx1 . every element of G is in precisely one of the cosets). . Thus. and so on.8 Representations and characters of groups (Here we adopt the usual cycle notation ± thus. y À1 xy ˆ x À1 X Let H be the subgroup kx. (1 2 3 4 5) denotes the permutation 1 3 2 3 3 3 4 3 5 3 1. the function h 3 hxi (h P H) is a bijection from H to Hxi . H ˆ kx. . we deduce that jGj ˆ rj HjX In particular. Since G ˆ Hx1 ‘ X X X ‘ Hxr . Using the above relations. Since W is invertible. 0 < j < 1) is a homomorphism. b: a5 ˆ b2 ˆ 1. For x in G. Hxr be all the distinct right cosets of H in G. . a group of order 10.) Check that x 5 ˆ y 2 ˆ 1. The distinct right cosets of H in G form a partition of G (that is. yl of G. we see that H ˆ fx i y j : 0 < i < 4. bÀ1 ab ˆ aÀ1 iX By Example 1. the subset Hx ˆ fhx: h P Hg of G is called a right coset of H in G. . it is an isomorphism. Cosets Let G be a group and let H be a subgroup of G. 0 < j < 1g. and so j Hxi j ˆ j Hj. yl  D10. and Hxi ’ Hxj is empty if i Tˆ j. Now recall that D10 ˆ ha. the function W: D10 3 H de®ned by W: a i b j 3 x i y j (0 < i < 4. we have . For all i. a . we have An v Sn . (3) Let G ˆ D8 ˆ ka. Nb. ka2 l ˆ {1. but the subgroup H ˆ kbl is not normal in G. b: a4 ˆ b2 ˆ 1. If n > 2 then there are just two right cosets of An in Sn . the sub-groups {1} and G are normal subgroups of G. Normal subgroups A subgroup N of a group G is said to be a normal subgroup of G if gÀ1 Ng ˆ N for all g P G (where gÀ1 Ng ˆ f gÀ1 ng: n P Ng). we see that GaN  C2 3 C2 . 9 The number r of distinct right cosets of H in G is called the index of H in G. a2 }. then j Hj divides |G|.Groups and homomorphisms 1.6 Lagrange's Theorem If G is a ®nite group and H is a subgroup of G. h P GX This makes GaN into a group. Then N v G and bÀ1 ab ˆ aÀ1 l and let Nˆ GaN ˆ fN . The subgroup kal is also normal in G. h P G. Thus jG: Hj ˆ jGjaj Hj when G is ®nite. we have fxy: x P Ng and y P Nhg ˆ NghX Hence we can de®ne a product operation on GaN by (Ng)(Nh) ˆ Ngh for all g. called the factor group of G by N.7 Examples (1) For every group G. and so Sn aAn  C2 . NabgX Since (Na)2 ˆ (Nb)2 ˆ (Nab)2 ˆ N. The importance of the condition gÀ1 Ng ˆ N (for all g P G) is that it can be used to show that for all g. Suppose that N v G and let GaN be the set of right cosets of N in G. and is written as jG: Hj. since b P H while aÀ1 ba ˆ a2 b P H. namely An ˆ f g P Sn : g eveng. Na. and An (1 2) ˆ f g P Sn : g oddgX Thus |Sn :An | ˆ 2. we write N v G to indicate that N is a normal subgroup of G. (2) For n > 1. 1. We de®ne the kernel of W by (1X8) Ker W ˆ f g P G: gW ˆ 1gX Then Ker W is a normal subgroup of G. If G is a ®nite group which is not simple.10 Theorem Suppose that G and H are groups and let W: G 3 H be a homomorphism. Then GaKer W  Im WX An isomorphism is given by the function Kg 3 gW where K ˆ Ker W. (This is analogous to the fact that every positive integer is built out of its prime factors. we eventually see that G is `built' out of a collection of simple groups. with p a prime number. is simple. Let G and H be groups and suppose that W: G 3 H is a homomorphism. For example. we relate normal subgroups and factor groups to homomorphisms. We shall give examples of non-abelian simple groups in later chapters ± the smallest one is A5 . the cyclic group Cp .10 Representations and characters of groups Simple groups A group G is said to be simple if G Tˆ {1} and the only normal subgroups of G are {1} and G. G is `built' out of these two smaller groups. Continuing this process with the smaller groups. then G has a normal subgroup N such that both N and G/N have smaller order than G. the image of W is (1X9) Im W ˆ f gW: g P Gg. 1. and in a sense. Also. The following result describes the way in which the kernel and image of W are related. ( g P G) . Kernels and images To conclude the chapter. simple groups are fundamental to the study of ®nite groups.) Thus. and Im W is a subgroup of H. 7(2) that Sn /An  C2 . g2 in G. . is a subgroup of H. D2 n ˆ ka. is a normal subgroup of G. The kernel. Show that if G is an abelian group which is simple. 11 is a homomorphism. The factor group GaN consists of the right cosets Ng ( g P G). We know from Example 1. Summary of Chapter 1 1. GL(n. . a2 ˆ b2 . 3 Gr . and that W: G 3 H is a surjective homomorphism. and the image. . 2. bÀ1 ab ˆ aÀ1 l. Suppose that G and H are groups. with multiplication (Ng)(Nh) ˆ NghX 3. Exercises for Chapter 1 1. Im W. W: g 3 À1. illustrating Theorem 1. . . A normal subgroup N of G is a subgroup such that gÀ1 Ng ˆ N for all g in G. then G is cyclic of prime order. C) ˆ the group of invertible n 3 n matrices over C. Examples of groups are Cn ˆ ka: an ˆ 1l. Show that either W is an isomorphism or H ˆ {1}. the direct product of the groups G1 . Gr . bÀ1 ab ˆ aÀ1 l. Sn ˆ the symmetric group of degree n. An ˆ the alternating group of degree n.10. . Ker W. b: an ˆ b2 ˆ 1. and for n > 2. if g is an even permutation. The factor group GaKer W is isomorphic to Im W. Q8 ˆ ka. b: a4 ˆ 1. G1 3 .11 Example The function W: Sn 3 C2 given by & 1.Groups and homomorphisms 1. . We have Ker W ˆ An . if g is an odd permutation. Im W ˆ C2 . 2. A homomorphism W: G 3 H is a function such that ( g1 g 2 )W ˆ ( g 1 W)( g 2 W) for all g1 . with G simple. ø: cr d s 3 x r y s (0 < r < 3. y ˆ (3 4). i 0 1 0 and let L be the subgroup hX . 6. . (c) If G is a ®nite cyclic group and x. (b) Let G be a ®nite cyclic group. is a homomorphism. Prove that D4 m  D2 m 3 C2 if m is odd. are homomorphisms. Show that both the functions ö: G 3 K and ø: H 3 K. and that G is not contained in An . Prove that f g P G: g n ˆ 1g is a cylic subgroup of G of order n. b: a4 ˆ b2 ˆ 1. (b) Let X.12 Representations and characters of groups 3. C). Show that just one of the functions ë: G 3 L and ì: H 3 L. and let n be a positive integer which divides |G|. 5. show that x is a power of y. d: c4 ˆ 1. Prove that G ’ An is a normal subgroup of G. Let G ˆ D8 ˆ ha. Y be the 2 3 2 matrices which are given by     0 i 0 À1 X ˆ . de®ned by ë: ar bs 3 X r Y s . de®ned by ö: ar bs 3 x r y s . and Ga (G ’ An )  C2 X 4. Prove that this homomorphism is an isomorphism. Find Ker ö and Ker ø. Y i of GL(2. 0 < s < 1). bÀ1 ab ˆ aÀ1 i. yl of S4 .Y ˆ . and H ˆ Q8 ˆ hc. c2 ˆ d 2 . ì: cr d s 3 X r Y s (0 < r < 3. Suppose that G is a subgroup of Sn . d À1 cd ˆ cÀ1 iX (a) Let x. and let K be the subgroup kx. 0 < s < 1). (a) Show that every subgroup of a cyclic group is cyclic. y be the permutations in S4 which are given by x ˆ (1 2). y are elements of G with the same order. Show that every group of even order contains an element of order 2. Prove that H v G. Bl has order 16. B has order 4.) 10. . Prove that every ®nite subgroup of this group is cyclic. is a group. (Hint: compare Q8 in Example 1. 8. C) such that A has order 8. 9. Suppose that H is a subgroup of G with |G: H| ˆ 2. Show that the set of non-zero complex numbers.Groups and homomorphisms 13 7. and B2 ˆ A4 and BÀ1 AB ˆ AÀ1 X Show that the group kA. Find elements A and B of GL(2. under the usual multiplication.2(4). and a rule for multiplying any element v of V by any element ë of F to form an element ëv of V. (The latter rule is called scalar multiplication. (2) (ë ‡ ì)v ˆ ëv ‡ ìv. (4) 1v ˆ v. (1) ë(u ‡ v) ˆ ëu ‡ ëv. here. (b) for all u. together with a rule for adding any two elements u. For reference purposes. these rules must satisfy: (2. and those of F are called scalars. we explain in detail how the results work. Most of the material will be familiar to you if you have taken a ®rst course on linear algebra. where we deal with projections.) Moreover. We write 0 for the identity element of the abelian group V under addition. v in V and all ë. Vector spaces Let F be either R (the set of real numbers) or C (the set of complex numbers). The elements of V are called vectors.2 Vector spaces and linear transformations An attractive feature of representation theory is that it combines two strands of mainstream mathematics. so we omit the proofs.1) (a) V is an abelian group under addition. 14 . namely group theory and linear algebra. v of V to form an element u ‡ v of V. we gather the results from linear algebra concerning vector spaces. A vector space over F is a set V. in case you have not come across projections before. linear transformations and matrices which we shall use later. (3) (ëì)v ˆ ë(ìv). ì in F. An exception occurs in the last section. A vector v in V is a linear combination of v1 . 1 n 1 n ë(x1 . we shall consider only vector spaces V which are ®nite-dimensional ± this means that V has a basis consisting of ®nitely many vectors. x2 . . . X X X . v1 . X X X . . . otherwise. as above. . . . . . Throughout this book. X X X . ë n in F. . we consider row vectors (x1 . (2) More generally. ë(x. ëxn )X Then F n is a vector space over F. . . . We denote the set of all such row vectors by F n. The vector space V is n-dimensional if dim V ˆ n. X X X . . X X X . and de®ne addition and scalar multiplication on F n by (x1 . . x9 ) ˆ (x1 ‡ x9 . v n are said to span V if every vector in V is a linear combination of v1 . It turns out that any two bases of V have the same number of vectors. . . The vectors v1 . . v n be vectors in a vector space V over F. . . v n if v ˆ ë1 v1 ‡ X X X ‡ ë n v n for some ë1 . ë y)X Then R2 is a vector space over R. . . Bases of vector spaces Let v1 . xn belong to F. y) where x and y are real numbers. X X X . not all of which are zero. . xn ‡ x9 ). . De®ne addition and scalar multiplication on R2 by (x. ë n in F. . . . . . The number of vectors in a basis of V is called the dimension of V and is written as dim V. v n are linearly independent. . xn ) ‡ (x9 . v n are linearly dependent if ë1 v1 ‡ X X X ‡ ë n v n ˆ 0 for some ë1 . . . y ‡ y9). The vectors v1 . v n . . . xn ) ˆ (ëx1 . . . y) ‡ (x9. x2 . for each positive integer n. . y9) ˆ (x ‡ x9. . If V ˆ {0} then dim V ˆ 0. .Vector spaces and linear transformations 15 2. xn ) where x1 .2 Examples (1) Let R2 denote the set of all ordered pairs (x. . v n form a basis of V if they span V and are linearly independent. y) ˆ (ëx. We say that v1 . . (2. 1) is a basis of V. . sp (u1 . v n of a vector space V. . . 0). (1. 0. . ur ) ˆ fë1 u1 ‡ X X X ‡ ë r ur : ë1 . . v k are linearly independent vectors in V. ur . . 0. 1. X X X . Except in the case where V ˆ {0}.3 Example Let V ˆ F n. v n in V such that v1 . so dim V ˆ n. . 0. and it is called the subspace spanned by u1 . . v P U then u ‡ v P U. . . ë n . the next result says that any linearly independent vectors can be extended to a basis. . . there are many bases of V. X X X . 0. . . . ë n in F. . v n form a basis of V. (1. . . . (2) if u. Another basis is (1.5). X X X . 2. ë r P FgX By (2. 1. sp (u1 . . each vector v in V can be written in a unique way as v ˆ ë1 v1 ‡ X X X ‡ ë n v n . (3) if ë P F and u P U then ëu P U. .6 Examples (1) {0} and V are subspaces of V. (0. (0. . (2) Let u1 . For a subset U of V to be a subspace. . . that is. Subspaces A subspace of a vector space V over F is a subset of V which is itself a vector space under the addition and scalar multiplication inherited from V. 0. . . Then (1. . .5) (1) 0 P U. . it is necessary and suf®cient that all the following conditions hold: (2. X X X . 0). 1. X X X . . Indeed. . 0). then there exist v k‡1 . X X X . .4) If v1 .16 Representations and characters of groups 2. X X X . . . . . . ur . . We de®ne sp (u1 . . X X X . . The vector v therefore determines the scalars ë1 . 0. 1)X Given a basis v1 . 0. . X X X . ur ) is a subspace of V. 0. with ë1 . ur be vectors in V. . X X X . . 0). . . ur ) to be the set of all linear combinations of u1 . 1. .4)). v n is a basis of V. . . (2. ‡ Ur is a subspace of V. . . . . . . . that u1 . We say that the sum U1 ‡ . . Direct sums of subspaces 17 If U1. .8 Examples (1) Suppose that v1 . . .9) Suppose that V ˆ U ‡ W. then the sum U1 ‡ . . (2) u1 . . . . only if U ˆ V. . . v k to a basis v1 . . . . ‡ Ur is a direct sum if every element of the sum can be written in a unique way as u1 ‡ .5). dim U ˆ dim V if and . Then V ˆ U1 È X X X È Un X (2) Let U be a subspace of V and let v1 . w1. . Then the following three conditions are equivalent: (1) V ˆ U È W. . . ‡ ur with ui P Ui for 1 < i < r. . . . . . If the sum is direct. ws is a basis of V. . Then V ˆ U È WX From this construction it follows that there are in®nitely many subspaces W with V ˆ U È W. .7) Suppose that U is a subspace of the vector space V . ws is a basis of W.4 if you have dif®culty with the proof. and for 1 < i < n. . . .3 and 2.4). . Ur are subspaces of a vector space V. (2. . . . let Ui be the subspace spanned by v i. . ur is a basis of U and that w1. and let W ˆ sp (v k‡1 . . You should consult the solutions to Exercises 2. then we write it as U1 È X X X È Ur X 2. . . . v n of V (see (2. (3) U ’ W ˆ {0}. . The next result is frequently useful when dealing with the direct sum of two subspaces. ur . v n ). v k be a basis of U.Vector spaces and linear transformations Notice that the following fact is a consequence of (2. unless U is {0} or V. . Then dim U < dim V Also. U1 ‡ . Extend v1. . . ‡ Ur is de®ned by U1 ‡ X X X ‡ Ur ˆ fu1 ‡ X X X ‡ ur : ui P Ui for 1 < i < rgX By (2. and let V ˆ f(u1 .10) Suppose that U. X X X . Ua. (2. X X X . u9 i in Ui (1 < i < r) and all ë in F. A linear transformation from V to W is a function W: V 3 W which satis®es (u ‡ v)W ˆ uW ‡ vW (ëv)W ˆ ë(vW) for all u. can be deduced immediately from the de®nition of a direct sum. . . Wb are subspaces of a vector space V. ur ): ui P U i for 1 < i < rgX De®ne addition and scalar multiplication on V as follows: for all ui . If. Ur be vector spaces over F. X X X . . 1 r 1 r ë(u1 . then it is immediate that V ˆ U1 È X X X È U 9X 9 r We call V the external direct sum of U1. Let U1. . we write V ˆ U1 È X X X È Ur X Linear transformations Let V and W be vector spaces over F. . . X X X . ëur )X With these de®nitions. W1. X X X . u9 ) ˆ (u1 ‡ u9 . v P V . involving the direct sum of several subspaces. then V ˆ U1 È X X X È Ua È W 1 È X X X È W b X We now introduce a construction for vector spaces which is analogous to the construction of direct products for groups. ur ‡ u9 ). ur ) ‡ (u9 . 0): ui P U i g i (where the ui is in the ith position). . ui . . for 1 < i < r. let (u1 . . U1. X X X . Ur. . and. V is a vector space over F. . . and for all ë P F and v P V X . . X X X . . and W ˆ W1 È X X X È Wb . If V ˆ U È W and also U ˆ U1 È X X X È Ua . X X X .18 Representations and characters of groups Our next result. . abusing notation slightly. . ur ) ˆ (ëu1 . we put U 9 ˆ f(0. W. . . then W is a linear transformation. it is easy to check that Ker W is a subspace of V and Im W is a subspace of W.5). Im W ˆ fvW: v P V gX Using (2. . . . . so a linear transformation preserves addition and scalar multiplication. given any basis v1 . ë n in F we have (ë1 v1 ‡ X X X ‡ ë n v n )W ˆ ë1 (v1 W) ‡ X X X ‡ ë n (v n W)X Thus. The kernel of W (written Ker W) and the image of W (written Im W) are de®ned as follows: (2X11) Ker W ˆ fv P V : vW ˆ 0g. Kernels and images Suppose that W: V 3 W is a linear transformation. Im W ˆ V X . . by specifying the values of ö on a basis of V. Their dimensions are connected by the following equation. and then saying `extend the action of ö to be linear'. . Furthermore. . then W is a linear transformation. and Ker W ˆ f0g. . Notice that if W: V 3 W is a linear transformation and v1. the linear transformation ö is given by (ë1 v1 ‡ X X X ‡ ë n v n )ö ˆ ë1 w1 ‡ X X X ‡ ë n wn X We sometimes construct a linear transformation ö: V 3 W in this way. and Ker W ˆ V . W is determined by its action on a basis. wn in W. Im W ˆ f0gX (2) If W: V 3 V is de®ned by vW ˆ 3v for all v P V.Vector spaces and linear transformations 19 Just as a group homomorphism preserves the group multiplication. v n of V and any n vectors w1. . . . v n is a basis of V. .13 Examples (1) If W: V 3 W is de®ned by vW ˆ 0 for all v P V. . . there is a unique linear transformation ö: V 3 W such that v i öˆ wi for all i. which is known as the Rank±Nullity Theorem: (2X12) dim V ˆ dim (Ker W) ‡ dim (Im W)X 2. then for ë1 . Suppose that W and ö are endomorphisms of V and ë P F. À1)). Wö and ëW are endomorphisms of V. v(ëW) ˆ ë(vW). we have Ker W ˆ sp ((7. We write W2 for WW. À3. (2) Ker W ˆ {0}. If there exists an invertible linear transformation from V to W. Then W ‡ ö. By applying (2. It turns out that the inverse of an invertible linear transformation is also a linear transformation (see Exercise 2. so dim (Ker W) ˆ 1 and dim (Im W) ˆ 2. Endomorphisms A linear transformation from a vector space V to itself is called an endomorphism of V.1). and hence W is invertible precisely when W is surjective and Ker W ˆ {0}. let V and W be vector spaces over F. then V and W are said to be isomorphic vector spaces.14) Let W be a linear transformation from V to itself. À y ‡ 3z) for all x. y. Im W ˆ R2 . z P R. Invertible linear transformations Again. Then the following three conditions are equivalent: (1) W is invertible. By also taking (2. (2.12).20 Representations and characters of groups (3) If W: R3 3 R2 is given by (x.2). then W is a linear transformation. v(Wö) ˆ (vW)ö. (3) Im W ˆ V. we see that isomorphic vector spaces have the same dimension. we obtain the next result (see Exercise 2. We de®ne the functions W ‡ ö. z)W ˆ (x ‡ 2 y ‡ z. A linear transformation W from V to W is injective if and only if Ker W ˆ {0}.7) into account. for all v P V. Wö and ëW from V to V by (2X15) v(W ‡ ö) ˆ vW ‡ vö. . y. y)(3W) ˆ (3x ‡ 3 y. 1) of V and B 9 is the basis . and let W be an endomorphism of V. . . then [W]B ˆ In for all bases B of V. (0. 2x À 10 y). and let W. À2x ‡ 4 y)X Then W and ö are endomorphisms of V. y)W2 ˆ (2x À y. v i W ˆ a i1 v1 ‡ X X X ‡ ain v n X 2. y)ö ˆ (x À 2 y. If W is an endomorphism of V.Vector spaces and linear transformations 2. (x. Suppose that v1 . where In denotes the n 3 n identity matrix. 0). (x. Àx ‡ 2 y). Note that Ker (W À ë1 V ) ˆ fv P V : vW ˆ ëvgX (2) Let V ˆ R2. for all ë P F. v n is a basis of V and call it B . and is denoted by [W]B . x À 2 y) of V. (x.17 De®nition The n 3 n matrix (aij ) is called the matrix of W relative to the basis B . (2) Let V ˆ R2 and let W be the endomorphism (x. y)(W ‡ ö) ˆ (2x À y. ö be the functions from V to V de®ned by (x. 3x À 6 y). and W ‡ ö. Wö. y)(Wö) ˆ (Àx ‡ 5 y. x À 2 y). y) 3 (x ‡ y. (x. If B is the basis (1. . Then there are scalars aij in F (1 < i < n. 2. Àx ‡ 5 y)X Matrices Let V be a vector space over F. then so is W À ë1 V . . 3W and W2 are given by (x.16 Examples (1) The identity function 1 V de®ned by 1 V : v 3 v for all v P V 21 is an endomorphism of V.18 Examples (1) If W ˆ 1 V (so that vW ˆ v for all v P V). y)W ˆ (x ‡ y. 1 < j < n) such that for all i. the matrix ëA is the m 3 n matrix over F obtained from A by multiplying all the entries by ë. AB ˆ 3 . 3A ˆ X À3 9 6 3 4 2 2 À13 À12 À5 The matrix of the sum or product of two endomorphisms (relative to some basis) is related to the matrices of the individual endomorphisms in the way you would expect: (2. À2  [W]B 9 ˆ  1 X À1 (1. and that W and ö are endomorphisms of V.22 Representations and characters of groups   1 . Bˆ 1 2 2 3  À4 X À1 2 . then we describe A as a matrix over F. j.20) Suppose that B is a basis of the vector space V. (1. Given an m 3 n matrix A ˆ (aij ) and an n 3 p matrix B ˆ (bij ). the product of two matrices is de®ned in a less transparent way. Then [W ‡ ö]B ˆ [W]B ‡ [ö]B . 0). their sum A ‡ B is the m 3 n matrix over F whose ij-entry is aij ‡ bij for all i. 1) of V. As you know. and [Wö]B ˆ [W]B [ö]B X . then [W]B ˆ 1 1 0 3 If we wish to indicate that the entries in a matrix A come from F.19 Example Let Aˆ Then A‡ Bˆ 2 BA ˆ  À1 3   2 0 . their product AB is the m 3 p matrix whose ij-entry is n ˆ kˆ1 aik bkj X 2. Given two m 3 n matrices A ˆ (aij ) and B ˆ (bij ) over F. 2 À1 5 À2 À4 3 0 3 3 . and for ë P F. for all scalars ë. Suppose that A is an n 3 n matrix over F. We concentrate on a particular way of doing this. (2. the matrix product vA also lies in V. given a basis of V. . y) ˆ (x ‡ 3 y. 2. The connection between invertible endomorphisms and invertible matrices is straightforward.22 Example Let Aˆ  1 3  À1 X 2 Then A gives us an endomorphism W of F 2. [ëW]B ˆ ë[W]B X 23 We showed you in (2. Àx ‡ 2 y)X 3 2 Invertible matrices An n 3 n matrix A is said to be invertible if there exists an n 3 n matrix B with AB ˆ BA ˆ In . . Then a necessary and suf®cient condition for A to be invertible is that det A Tˆ 0. Write det A for the determinant of A. v 3 vA (v P F n ) is an endomorphism of F n.21) If A is an n 3 n matrix over F then the function .Vector spaces and linear transformations Also. the vector space of row vectors (x1 . is unique. and this same matrix is used to describe the way in which the matrix of an . and let V ˆ F n.20): given a basis B of V. Then for all v in V. and follows from (2. Invertible matrices turn up when we relate two bases of a vector space. .17) how to get a matrix from an endomorphism of a vector space V. An invertible matrix converts one basis into another. if it exists. y)W ˆ (x. . It is easy enough to reverse this process and use a matrix to de®ne an endomorphism. where   1 À1 (x. The following remark is easily justi®ed. xn ) with each xi in F. an endomorphism W of V is invertible if and only if the matrix [W]B is invertible. Such a matrix B. it is called the inverse of A and is written as AÀ1 . The scalar ë is said to be an eigenvalue of W if vW ˆ ëv for some non-zero vector v in V. 0). The inverse of T is the change of basis matrix from B 9 to B . . . and let v9 . The precise meaning of these remarks is revealed in the de®nition (2.24) below. (1. where T is the change of basis matrix from B to B 9. and suppose that W is an endomorphism of V.24) If B and B 9 are bases of V and W is an endomorphism of V.25 Example Suppose that V ˆ R2. . 2. 2. then       1 0 0 1 1 0 1 1 [W]B ˆ ˆ T À1 [W]B 9 T X ˆ À1 1 3 À1 1 1 1 À2 Eigenvalues Let V be an n-dimensional vector space over F. v9 ˆ t i1 v1 ‡ X X X ‡ tin v n i for certain scalars tij . . Then     1 0 1 0 À1 Tˆ . x À 2 y) of V. 0). . (0. and is called the change of basis matrix from B to B 9. 1) and B 9 the basis (1. v n be a basis B of the vector space V. y) 3 (x ‡ y.24 Representations and characters of groups endomorphism depends upon the basis.23 De®nition Let v1 . (2. 1) of V. v9 1 n be a basis B 9 of V. .T ˆ X 1 1 À1 1 If W is the endomorphism W: (x. The n 3 n matrix T ˆ (tij ) is invertible. . then [W]B ˆ T À1 [W]B 9 T .18(2). Then for 1 < i < n.23) and the result (2. Let B be the basis (1. as in Example 2. . . Such a vector v is called an eigenvector of W. then   0 1 [W]B ˆ X À1 0 We have det ([W]B À ëI2 ) ˆ ë2 ‡ 1. i). and W has no eigenvalues in R. (1. x)X If B is the basis (1. Since every non-constant polynomial with coef®cients in C has a root in C. Corresponding eigenvectors are (1. Then W has an eigenvalue.27 Examples (1) Let V ˆ C2 and let W be the endomorphism of V which is given by (x. Thus we depend upon F being C in result (2. Note that if B 9 is the basis (1. x)X This time. (2. which occurs if and only if W À ë1 V is not invertible. V is a vector space over R. 2. y)W ˆ (À y. y)W ˆ (À y. For an n 3 n matrix A over F. The eigenvalues of A are those elements ë of F which satisfy det (A À ëI n ) ˆ 0X . Ài).Vector spaces and linear transformations 25 Now ë is an eigenvalue of W if and only if Ker (W À ë1 V ) Tˆ {0}. then the eigenvalues of W are those scalars ë in F which satisfy the equation det ([W]B À ëI n ) ˆ 0X Solving this equation involves ®nding the roots of a polynomial of degree n. the element ë of F is said to be an eigenvalue of A if vA ˆ ëv for some non-zero row vector v in F n. so i and Ài are the eigenvalues of W. Therefore. if B is a basis of V. i) of V. (0. then   i 0 [W]B 9 ˆ X 0 Ài (2) Let V ˆ R2 and let W again be the endomorphism which is given by (x. we deduce the following result. 0).26) Let V be a non-zero vector space over C. 1) of V. Ài) and (1. and let W be an endomorphism of V.26). w P W X Then ð is an endomorphism of V.26 Representations and characters of groups 2.29 Proposition Suppose that V ˆ U È W. that aii ˆ ë i for 1 < i < n. w9 in W. . Further. . Projections If a vector space V is a direct sum of two subspaces U and W. Let v and v9 belong to V. it follows that ð is a function on V. . we have Im ˆ U. We have (v ‡ v9)ð ˆ (u ‡ u9 ‡ w ‡ w9)ð ˆ u ‡ u9 ˆ (u ‡ w)ð ‡ (u9 ‡ w9)ð ˆ vð ‡ v9ðX Also.28 Example We say that an n 3 n matrix A ˆ (aij ) is diagonal if aij ˆ 0 for all i and j with i Tˆ j. w P W. Also. Im ð ˆ U . ë n . (ëv)ð ˆ (ëu ‡ ëw)ð ˆ ëu ˆ ë(vð)X Therefore. Clearly Im ð # U. . in addition. the eigenvalues are ë1 . and since uð ˆ u for all u in U. De®ne ð: V 3 V by (u ‡ w)ð ˆ u for all u P U . . Then v ˆ u ‡ w and v9 ˆ u9 ‡ w9 for some u. We often display such a matrix in the form H I ë1 0 f g FF Aˆd e F ën 0 which indicates. Ker ð ˆ W and ð2 ˆ ðX Proof Since every vector in V has a unique expression in the form u ‡ w with u P U. for ë in F. u9 in U and w. For this diagonal matrix A. then we can construct a special endomorphism of V which depends upon the expression V ˆ U È W: 2. ð is an endomorphism of V. and so ð2 ˆ ð.32 Proposition Suppose that ð is a projection of a vector space V. y) 3 (2x ‡ 2 y. j . we have v ˆ uð for some u P V. 2. 2. As v P Im ð. Finally. Thus Im ð ’ Ker ð ˆ f0g. 27 j 2. Therefore vð ˆ uð2 ˆ uð ˆ vX Since v P Ker ð. Àx À y) of R2 is a projection.29. and (2. it follows that v ˆ vð ˆ 0.30 De®nition An endomorphism ð of a vector space V which satis®es ð2 ˆ ð is called a projection of V.31 Example The endomorphism (x. Then V ˆ Im ð È Ker ðX Proof If v P V then v ˆ vð ‡ (v À vð). and so Ker ð ˆ W. and the second term v À vð lies in Ker ð. since (v À vð)ð ˆ vð À vð2 ˆ vð À vð ˆ 0X This establishes that V ˆ Im ð ‡ Ker ð. as in Proposition 2. Now suppose that v lies in Im ð ’ Ker ð.9) now shows that V ˆ Im ð È Ker ð. Clearly the ®rst term vð belongs to Im ð.Vector spaces and linear transformations (u ‡ w)ð ˆ 0 D u ˆ 0 D u ‡ w P W . (u ‡ w)ð2 ˆ uð ˆ u ˆ (u ‡ w)ð. We now show that every projection can be constructed using a direct sum. 2. F n is the set of row vectors (x1 . Show that if V and W are vector spaces and W: V 3 W is an invertible linear transformation then WÀ1 is a linear transformation. . A linear transformation W: V 3 W is invertible if and only if Ker W ˆ {0} and Im W ˆ W. . and dim V ˆ dim (Ker W) ‡ dim (Im W)X 4. where F ˆ C or R. A projection is an endomorphism ð of V which satis®es ð2 ˆ ð. F F F .31. there exists an invertible matrix T such that [W]B ˆ T À1 [W]B 9 TX 6. and an endomorphism W of V. y) 3 (2x ‡ 2 y. . Exercises for Chapter 2 1. there is a correspondence between the endomorphisms W of V and the n 3 n matrices [W]B over F. v in V and all ë in F. Eigenvalues ë of an endomorphism W satisfy vW ˆ ëv for some nonzero v in V. 3. x n ) with each xi in F. and dimF n ˆ n. . Ker W is a subspace of V and Im W is a subspace of W. For example.28 Representations and characters of groups 2. Ker ð ˆ f(x. . Given a basis B of the n-dimensional vector space V. All our vector spaces are ®nite-dimensional over F. Given two bases B and B 9 of V. ‡ ur (ui P Ui ). Also. 7. and every element v of V has a unique expression of the form v ˆ u1 ‡ . V ˆ U1 È . Àx): x P RgX Summary of Chapter 2 1. A linear transformation W: V 3 W satis®es (u ‡ v)W ˆ uW ‡ vW and (ëv)W ˆ ë(vW) for all u. Àx À y) is the projection of R2 which appears in Example 2. then Im ð ˆ f(2x.33 Example If ð: (x. Àx): x P Rg. 5. V ˆ U È W if and only if V ˆ U ‡ W and U ’ W ˆ {0}. È Ur if each Ui is a subspace of V. . . with all diagonal entries equal to 1 or 0.Vector spaces and linear transformations 29 2. Let U and W be subspaces of the vector space V. ur is a basis of U and w1. . ws is a basis of W. 4. . U2 and U3 such that V ˆ U1 ‡ U2 ‡ U3 and U1 ’ U 2 ˆ U1 ’ U3 ˆ U2 ’ U 3 ˆ f0g. . Prove that dim V ˆ dim U1 ‡ X X X ‡ dim U r X 7. (2) Ker W ˆ {0}. 8. w1. 9. . with all diagonal entries equal to ‡1 or À1. ur . (3) Im W ˆ V. Suppose that W is an endomorphism of the vector space V. Show that W is a projection if and only if there is a basis B of V such that [W]B is diagonal. (a) Let U1. Suppose that u1 . . . Give an example of a vector space V with endomorphisms W and ö such that V ˆ Im W È Ker W. . . Show that V ˆ U È W if and only if u1 . where U ˆ fv P V : vW ˆ vg. W ˆ fv P V : vW ˆ ÀvgX Deduce that V has a basis B such that [W]B is diagonal. Suppose that U1. . . . . . 5. with V ˆ U1 ‡ U2 ‡ U3. . Let V be a vector space and let W be an endomorphism of V. . Ur are subspaces of the vector space V. Suppose that W is an endomorphism of the vector space V and W2 ˆ 1 V . Show that the following are equivalent: (1) W is invertible. U2 and U3 be subspaces of a vector space V. . Show that V ˆ U1 È U2 È U3 D U 1 ’ (U 2 ‡ U3 ) ˆ U 2 ’ (U 1 ‡ U3 ) ˆ U 3 ’ (U 1 ‡ U2 ) ˆ f0gX (b) Give an example of a vector space V with three subspaces U1. Show that V ˆ U È W. È Ur. . 3. but V Tˆ U1 È U2 È U3. ws is a basis of V. but V Tˆ Im ö È Ker ö. 6. . and that V ˆ U1 È . . . Let U and W be subspaces of the vector space V. . Prove that V ˆ U È W if and only if V ˆ U ‡ W and U ’ W = {0}. for some n. Representations Let G be a group and let F be R or C. F). Recall from the ®rst chapter that GL (n. F). we have 1r ˆ I n . where In denotes the n 3 n identity matrix.3 Group representations A representation of a group G gives us a way of visualizing G as a group of matrices. and g À1 r ˆ ( gr)À1 for all g P G. We set out this idea in more detail. F) denotes the group of invertible n 3 n matrices with entries in F. The degree of r is the integer n. F). We also introduce the concept of equivalence of representations. then r is a representation if and only if ( gh)r ˆ ( gr)(hr) for all g. and consider the kernel of a representation. it follows that for every representation r: G 3 GL (n. To be precise. h P GX Since a representation is a homomorphism. and give some examples of representations. 30 . a representation is a homomorphism from G into a group of invertible matrices. 3. Thus if r is a function from G to GL (n.1 De®nition A representation of G over F is a homomorphism r from G to GL (n. Then ( gh)r ˆ I n ˆ I n I n ˆ ( gr)(hr) for all g. so r is a representation of G. .2 Examples (1) Let G be the dihedral group D8 ˆ ka. Bˆ À1 0 0 À1 and check that A4 ˆ B2 ˆ I. where In is the n 3 n identity matrix. De®ne r: G 3 GL (n. F) which is given by r: a i b j 3 A i B j (0 < i < 3. BÀ1 AB ˆ AÀ1 X It follows (see Example 1.4) that the function r: G 3 GL (2. b: a4 ˆ b2 ˆ 1. 0 < j < 1) is a representation of D8 over F. F) by gr ˆ I n for all g P G. The matrices gr for g in D8 are given in the following table: g gr  1  1 0 0 1 a  0 1 À1 0 a2  À1 0 0 À1 a3  0 À1 1 0    g gr  b  1 0 0 À1  ab  0 À1 À1 0 2  a b  À1 0 0 1 3  a b  0 1 1 0 (2) Let G be any group. bÀ1 ab ˆ aÀ1 l. This shows that every group has representations of arbitrarily large degree. The degree of r is 2. h P G.Group representations 31 3. as usual. Equivalent representations We now look at a way of converting a given representation into another one. De®ne the matrices A and B by     0 1 1 0 Aˆ . and consider the representation r of G which appears in Example 3. b: a4 ˆ b2 ˆ 1. T À1 (hr)T ˆ ( gó )(hó ). we simply de®ne gó ˆ T À1 ( gr)T Then for all g. (3) if r is equivalent to ó and ó is equivalent to ô. (2) if r is equivalent to ó then ó is equivalent to r.4 Examples (1) Let G ˆ D8 ˆ ka. h P G. equivalence of representations is an equivalence relation. a representation. bÀ1 ab ˆ aÀ1 l.3 De®nition Let r: G 3 GL (m. we have (see Exercise 3. and let T be an invertible n 3 n matrix over F. Thus ar ˆ A for all g P GX . We say that r is equivalent to ó if n ˆ m and there exists an invertible n 3 n matrix T such that for all g P G. F) be a representation. we have (T À1 AT )(T À1 BT ) ˆ T À1 (AB)TX We can use this observation to produce a new representation ó from r.32 Representations and characters of groups Let r: G 3 GL (n. 3. then r is equivalent to ô. F) be representations of G over F. F) and ó : G 3 GL (n. indeed.2(1). gó ˆ T À1 ( gr)TX Note that for all representations r.4): (1) r is equivalent to r. 3. ó and ô of G over F. Note that for all n 3 n matrices A and B. and so ó is. ( gh)ó ˆ T À1 (( gh)r)T ˆ T À1 (( gr)(hr))T ˆ T À1 ( gr)T . In other words. . we have     i 0 0 1 À1 À1 T AT ˆ . Hence r: 1 3 I. If   2 À3 . Bˆ 0 0   1 X Ài  1 Ài X 1 i  0 X À1 33 0 À1 Assume that F ˆ C. and de®ne 1 T ˆp 2 Then 1 i 1 T À1 ˆ p 2  In fact. T BT ˆ . aó ˆ 1ó ˆ 0 À1 0 1 and ó is equivalent to r. . a 3 A is a representation of G. T has been constructed so that T À1 AT is diagonal. (2) Let G ˆ C2 ˆ ka: a2 ˆ 1l and let  À5 Aˆ À2  12 X 5 Check that A2 ˆ I. 0 Ài 1 0 and so we obtain a representation ó of D8 for which     0 1 i 0 X . À1 1 0 and so we obtain a representation ó of G for which     1 0 1 0 . bó ˆ aó ˆ 1 0 0 Ài The representations r and ó are equivalent. where Aˆ    1 1 .Group representations and br ˆ B. Tˆ 1 À1 then T À1 AT ˆ   0 . 6 De®nition A representation r: G 3 GL (n. that is.7 Proposition A representation r of a ®nite group G is faithful if and only if Im r is isomorphic to G. F) which is de®ned by gr ˆ (1) for all g P G. is called the trivial representation of G.34 Representations and characters of groups There are two easily recognized situations where the only representation which is equivalent to r is r itself. and when gr ˆ In for all g in G. However. these are when the degree of r is 1. In agreement with De®nition 1. 3. Thus Ker r ˆ f g P G: gr ˆ I n gX Note that Ker r is a normal subgroup of G. F). this consists of the group elements g in G for which gr is the identity matrix. To put the de®nition another way. the trivial representation of G is the representation where every group element is sent to the 1 3 1 identity matrix. . Kernels of representations We conclude the chapter with a discussion of the kernel of a representation r: G 3 GL (n.8. there are usually lots of representations which are equivalent to r. It can happen that the kernel of a representation is the whole of G. F) is said to be faithful if Ker r ˆ {1}. Of particular interest are those representations whose kernel is just the identity subgroup. if the identity element of G is the only element g for which gr ˆ In . as is shown by the following de®nition. 3. 3.5 De®nition The representation r: G 3 GL (1. 2.10. Conversely. for some n. it follows that all representations which are equivalent to a faithful representation are faithful. and so |Ker r| ˆ 1. Representations r and ó of G are equivalent if and only if there exists an invertible matrix T such that for all g P G. (3) The trivial representation of a group G if faithful if and only if G ˆ {1}. then these two groups have the same (®nite) order. A representation is faithful if it is injective. F). Therefore. In Chapter 6 we shall show that every ®nite group has a faithful representation.8 Examples (1) The representation r of D8 given by j  i 1 0 0 1 i j (a b )r ˆ 0 À1 À1 0 as in Example 3. if G  Im r. The group generated by the matrices     0 1 1 0 and À1 0 0 À1 is therefore isomorphic to D8. The basic problem of representation theory is to discover and understand representations of ®nite groups. (2) Since T À1 AT ˆ In if and only if A ˆ In . r is faithful. since the identity is the only element g which satis®es gr ˆ I. j 3. gó ˆ T À1 ( gr)T X 3. that is.2(1) is faithful. Summary of Chapter 3 1. A representation of a group G is a homomorphism from G into GL(n.Group representations 35 Proof We know that Ker r v G and by Theorem 1. . if Ker r ˆ {1} then G  Im r. the factor group G/ Ker r is isomorphic to Im r. 3. Bˆ 0 1 0  . say G ˆ ka: am ˆ 1l. Dˆ X À 3a2 1a2 0 À1 Prove that each of the functions r k : G 3 GL (2. C. Suppose that G ˆ D2 n ˆ ka. C) (1 < j < 3). then ó is equivalent to r. r2 : ar 3 Br .  1 0   1 0 . given by . bÀ1 ab ˆ aÀ1 l. De®ne the matrices A. and ó is equivalent to ô. Which of these representations are faithful? 3. F) such that ar ˆ (1) and br ˆ (À1). 4). Let G ˆ D12 ˆ ka. b: a6 ˆ b2 ˆ 1. B. (3) if r is equivalent to ó. bÀ1 ab ˆ aÀ1 l. 1 0 0 eÀiða3     p 1a2 3a2 1 0 p Cˆ . D over C by     0 1 eiða3 0 Aˆ . and F ˆ R or C. Suppose that r. Let G be the cyclic group of order m. Suppose that A P GL (n. 4. Bˆ . de®ned by r1 : ar 3 Ar . C) (k ˆ 1. 2. Prove: (1) r is equivalent to r. Let Aˆ (0 < r < m À 1)X Show that r is a representation of G over C if and only if Am ˆ I. and de®ne r: G 3 GL (n. Show that there is a representation r: G 3 GL (1. is a representation of G over C. r3 : a r 3 C r (0 < r < 2).Cˆ  0 À1 1 À1  e2ðia3 and let G ˆ ka: a3 ˆ 1l  C3 . then r is equivalent to ô.36 Representations and characters of groups Exercises for Chapter 3 1. Show that each of the functions r j : G 3 GL (2. (2) if r is equivalent to ó. 5. ó and ô are representations of G over F. C) by r: ar 3 Ar 2. b: an ˆ b2 ˆ 1. C). Does it follow that gh ˆ hg? . r 4 : ar bs 3 C r D s (0 < r < 5. 7. 8. Give an example of a faithful representation of D8 of degree 3. r3 : ar bs 3 (ÀA) r Bs . Let r be a representation of the group G. 0 < s < 1). Prove that Ga Ker r is abelian. Suppose that g and h are elements of G such that ( gr)(hr) ˆ (hr)( gr). Which of these representations are faithful? Which are equivalent? 6. Suppose that r is a representation of G of degree 1. r2 : ar bs 3 A3 r (ÀB) s .Group representations r1 : ar bs 3 Ar Bs . 37 is a representation of G. For all v P V and g P G. . the fact that r is a homomorphism shows that v(( gh)r) ˆ v( gr)(hr) for all v P V and all g. the matrix product v( gr). is a row vector in V (since the product of a 1 3 n matrix with an n 3 n matrix is again a 1 3 n matrix). . since 1r is the identity matrix. Much of the material in the remainder of the book will be presented in terms of FG-modules. 38 . the vector space of all row vectors (ë1 . We now list some basic properties of the multiplication v( gr). . we have v(1r) ˆ v for all v P V. ë n ) with ë i P F. h P G. First. as there are several advantages to this approach to representation theory. .4 FG-modules We now introduce the concept of an FG-module. FG-modules Let G be a group and let F be R or C. Next. Suppose that r: G 3 GL (n. of the row vector v with the n 3 n matrix gr. Write V ˆ F n . and show that there is a close connection between FG-modules and representations of G over F. F) is a representation of G. Finally. the properties of matrix multiplication give (ëv)( gr) ˆ ë(v( gr)). v(br) ˆ (ë1 . ë2 ) P F 2 then. v1 ˆ v. ë P F and g. g P G) is de®ned. v(a3 r) ˆ (ë2 . and let r: G 3 GL (2. v( gh) ˆ (v g)h. F) be the representation of G over F given in Example 3. Àë1 )X Motivated by the above observations on the product v( gr). b: a4 ˆ b2 ˆ 1. ë P F and g P G. (u ‡ v) g ˆ ug ‡ v g. Then V is an FG-module if a multiplication v g (v P V. (v P V ) . the function v 3 vg is an endomorphism of V. ë1 ).2(1). (4) and (5) in the de®nition ensure that for all g P G. Note that conditions (1).FG-modules (u ‡ v)( gr) ˆ u( gr) ‡ v( gr) for all u.2 De®nition Let V be a vector space over F and let G be a group. for example. satisfying the following conditions for all u. bÀ1 ab ˆ aÀ1 l. 4.1 Example Let G ˆ D8 ˆ ka. (ëv) g ˆ ë(v g). v(ar) ˆ (Àë2 . Thus     1 0 0 1 X . v P V. 39 4. br ˆ ar ˆ 0 À1 À1 0 If v ˆ (ë1 . v P V. we now de®ne an FG-module. Àë2 ). We use the letters F and G in the name `FG-module' to indicate that V is a vector space over F and that G is the group from which we are taking the elements g to form the products v g (v P V). h P G: (1) (2) (3) (4) (5) v g P V. let [ g]B denote the matrix of the endomorphism v 3 v g of V. h P G. 0). 0.3 De®nition Let V be an FG-module. (0. and V ˆ F n. 0). Proof (1) We have already observed that for all u.40 Representations and characters of groups 4. F) is a representation of G over F. X X X . X X X . (0. there is a basis B of V such that (2) Assume that V is an FG-module and let B be a basis of V. X X X .4 Theorem (1) If r: G 3 GL(n. ( g P G) . (u ‡ v)( gr) ˆ u( gr) ‡ v( gr)X Therefore. (ëv)( gr) ˆ ë(v( gr)). 0. then V becomes an FG-module if we de®ne the multiplication v g by v g ˆ v( gr) gr ˆ [ g]B (v P V . if we let B be the basis (1. relative to the basis B . Then the function g 3 [ g]B is a representation of G over F. v(1r) ˆ v. 4. 1) of F n. 0. 1. 0. g P G)X for all g P GX Moreover. then gr ˆ [ g]B for all g P G. ë P F and g. 0. The connection between FG-modules and representations of G over F is revealed in the following basic result. X X X . g P GX Moreover. we have v( gr) P F n . and let B be a basis of V. For each g P G. v(( gh)r) ˆ (v( gr))(hr). F n becomes an FG-module if we de®ne v g ˆ v( gr) for all v P F n . v P F n. then the representation is just the representation r (see Theorem 4. v2 . bÀ1 ab ˆ aÀ1 l. it follows that [ gh]B ˆ [ g]B [h]B X In particular.FG-modules 41 (2) Let V be an FG-module with basis B . F) (where n ˆ dim V ). Since v( gh) ˆ (v g)h for all g. h P G and all v in the basis B of V. b: a4 ˆ 1. g 3 [ g]B v1 b ˆ v1 . then we have v1 a ˆ v2 .5 Examples (1) Let G ˆ D8 ˆ ka.   ˆ (0. In Example . V becomes an FG-module if we de®ne v g ˆ v( gr) (v P V . br ˆ X À1 0 0 À1 Write V ˆ F 2. a2 ˆ b2 .2(1). j Our next example illustrates part (1) of Theorem 4. (2) Let G ˆ Q8 ˆ ka. 4. Now v1 ˆ v for all v P V. v2 b ˆ Àv2 X ( g P G) If B denotes the basis v1 . 1) of V. g P G)X For instance. 1)X 0 (1. and hence is a representation of G over F.4(1) again).4. (0. bÀ1 ab ˆ aÀ1 l and let r be the representation of G over F given in Example 3. so     0 1 1 0 ar ˆ . By Theorem 4. v2 a ˆ Àv1 . b: a4 ˆ b2 ˆ 1. [1]B ˆ [ g]B [ g À1 ]B for all g P G. 0) À1 1 0 If v1 . Therefore each matrix [ g]B is invertible (with inverse [ gÀ1 ]B ). v2 is the basis (1. We have proved that the function g 3 [ g]B is a homomorphism from G to GL (n. 0).4(1). so [1]B is the identity matrix. 0)a ˆ (1. . . Shortly. we shall show you various ways of constructing FG-modules directly. v1 b ˆ v2 . v n of V and then extending the action to be linear on the whole of V.42 Representations and characters of groups 1. C) generated by     0 1 i 0 . To illustrate Theorem 4. v2 b ˆ Àv1 X Notice that in the above examples. For instance. that we have a multiplication v g for all v in V and g in G which (ë i P F) . We then obtain a CG-module with basis v1 . and then de®ne (ë1 v1 ‡ X X X ‡ ë n v n ) g to be ë1 (v1 g) ‡ X X X ‡ ë n (v n g)X As you might expect. (v1 ‡ 2v2 )ab ˆ v1 ab ‡ 2v2 ab ˆ v2 b À 2v1 b ˆ Àv2 À 2v1 X A similar remark holds for all FG-modules V: if v1 . . v1 b and v2 b determine v g for all v P V and g P G. . gr generate G. v n is a basis of V and g1 . . there are restrictions on how we may de®ne the vectors v i g. . F F F . in Example 4. . . v n is a basis of a vector space V over F Suppose . the vectors v1 a. and B ˆ Aˆ À1 0 0 Ài so we already have a representation of G over C. To do this. The next result will often be used to show that our chosen multiplication turns V into an FG-module. without using a representation.5(1). . v2 a ˆ Àiv2 . v2 such that v1 a ˆ iv1 . . we ®rst de®ne v i g for each i and each g in G. that is.2(4) we de®ned Q8 to be the subgroup of GL (2. 4. then the vectors v i g j (1 < i < n. v2 a. we turn a vector space V over F into an FG-module by specifying the action of group elements on a basis v1 . .6 Proposition Assume that v1 . 1 < j < r) determine v g for all v P V and g P G. .4(1) we must this time take F ˆ C. FG-modules 43 satis®es the following conditions for all i with 1 < i < n, for all g, h P G, and for all ë1 , F F F , ë n P F: (1) (2) (3) (4) v i g P V; v i ( gh) ˆ (v i g)h; vi 1 ˆ vi ; (ë1 v1 ‡ . . . ‡ ë n v n ) g ˆ ë1 (v1 g) ‡ . . . ‡ ë n (v n g). Then V is an FG-module. Proof It is clear from (3) and (4) that v1 ˆ v for all v P V. Conditions (1) and (4) ensure that for all g in G, the function v 3 v g (v P V) is an endomorphism of V. That is, vg P V, (ëv) g ˆ ë(v g), (u ‡ v) g ˆ ug ‡ v g, for all u, v P V, ë P F and g P G. Hence (4X7) (ë1 u1 ‡ X X X ‡ ë n u n )h ˆ ë1 (u1 h) ‡ X X X ‡ ë n (un h) for all ë1 , . . . , ë n P F, all u1 , . . . , un P V and all h P G. Now let v P V and g, h P G. Then v ˆ ë1 v1 ‡ . . . ‡ ë n v n for some ë1 , . . . , ë n P F, and v( gh) ˆ ë1 (v1 ( gh)) ‡ X X X ‡ ë n (v n ( gh)) ˆ (ë1 (v1 g) ‡ X X X ‡ ë n (v n g))h ˆ (v g)h by condition (4)X by condition (4) ˆ ë1 ((v1 g)h) ‡ X X X ‡ ë n ((v n g)h) by condition (2) by (4X7) We have now checked all the axioms which are required for V to be an FG-module. j Our next de®nitions translate the concepts of the trivial representation and a faithful representation into module terms. 4.8 De®nitions (1) The trivial FG-module is the 1-dimensional vector space V over F with vg ˆ v for all v P V , g P GX 44 Representations and characters of groups (2) An FG-module V is faithful if the identity element of G is the only element g for which v g ˆ v for all v P V X For instance, the FD8 -module which appears in Example 4.5(1) is faithful. Our next aim is to use Proposition 4.6 to construct faithful FGmodules for all subgroups of symmetric groups. Permutation modules Let G be a subgroup of Sn , so that G is a group of permutations of {1, . . . , n}. Let V be an n-dimensional vector space over F, with basis v1 , . . . , v n . For each i with 1 < i < n and each permutation g in G, de®ne v i g ˆ v ig X Then v i g P V and v i 1 ˆ v i . Also, for g, h in G, v i ( gh) ˆ v i( gh) ˆ v(ig) h ˆ (v i g)hX We now extend the action of each g linearly to the whole of V; that is, for all ë1 , . . . , ë n in F and g in G, we de®ne (ë1 v1 ‡ X X X ‡ ë n v n ) g ˆ ë1 (v1 g) ‡ X X X ‡ ë n (v n g)X Then V is an FG-module, by Proposition 4.6. 4.9 Example Let G ˆ S4 and let B denote the basis v1 , v2 , v3 , v4 of V. If g ˆ (1 2), then v1 g ˆ v2 , v2 g ˆ v1 , v3 g ˆ v3 , v4 g ˆ v4 X And if h ˆ (1 3 4), then v1 h ˆ v3 , v2 h ˆ v2 , v3 h ˆ v4 , v4 h ˆ v1 X We have [ g]B 0 f1 ˆf d0 0 H 1 0 0 0 0 0 1 0 H I 0 0 f0 0g g, [h]B ˆ f d0 0e 1 1 0 1 0 0 1 0 0 0 I 0 0g gX 1e 0 FG-modules 45 4.10 De®nition Let G be a subgroup of Sn . The FG-module V with basis v1 , . . . , v n such that v i g ˆ v ig for all i, and all g P G, is called the permutation module for G over F. We call v1 , . . . , v n the natural basis of V. Note that if we write B for the basis v1 , . . . , v n of the permutation module, then for all g in G, the matrix [ g]B has precisely one nonzero entry in each row and column, and this entry is 1. Such a matrix is called a permutation matrix. Since the only element of G which ®xes every v i is the identity, we see that the permutation module is a faithful FG-module. If you are aware of the fact that every group G of order n is isomorphic to a subgroup of Sn , then you should be able to see that G has a faithful FG-module of dimension n. We shall go into this in more detail in Chapter 6. 4.11 Example Let G ˆ C3 ˆ ka: a3 ˆ 1l. Then G is isomorphic to the cyclic subgroup of S3 which is generated by the permutation (1 2 3). This alerts us to the fact that if V is a 3-dimensional vector space over F, with basis v1 , v2 , v3 , then we may make V into an FG-module in which v1 1 ˆ v1 , v2 1 ˆ v2 , v3 1 ˆ v3 , v1 a ˆ v2 , v2 a ˆ v3 , v3 a ˆ v1 , v1 a2 ˆ v3 , v2 a2 ˆ v1 , v3 a2 ˆ v2 X Of course, we de®ne v g, for v an arbitrary vector in V and g ˆ 1, a or a2 , by (ë1 v1 ‡ ë2 v2 ‡ ë3 v3 ) g ˆ ë1 (v1 g) ‡ ë2 (v2 g) ‡ ë3 (v3 g) for all ë1 , ë2 , ë3 P F. Proposition 4.6 can be used to verify that V is an FG-module, but we have been motivated by the de®nition of permutation modules in our construction. FG-modules and equivalent representations We conclude the chapter with a discussion of the relationship between FG-modules and equivalent representations of G over F. An FG- 46 Representations and characters of groups module gives us many representations, all of the form g 3 [ g]B ( g P G) for some basis B of V. The next result shows that all these representations are equivalent to each other (see De®nition 3.3); and moreover, any two equivalent representations of G arise from some FG-module in this way. 4.12 Theorem Suppose that V is an FG-module with basis B , and let r be the representation of G over F de®ned by r: g 3 [ g]B ö: g 3 [ g]B 9 ( g P G)X ( g P G) (1) If B 9 is a basis of V, then the representation of G is equivalent to r. (2) If ó is a representation of G which is equivalent to r, then there is a basis B 0 of V such that ó : g 3 [ g]B 0 ( g P G)X Proof (1) Let T be the change of basis matrix from B to B 9 (see De®nition 2.23). Then by (2.24), for all g P G, we have [ g]B ˆ T À1 [ g]B 9 TX Therefore ö is equivalent to r. (2) Suppose that r and ó are equivalent representations of G. Then for some invertible matrix T, we have gr ˆ T À1 ( gó )T for all g P GX Let B 0 be the basis of V such that the change of basis matrix from B to B 0 is T. Then for all g P G, [ g]B ˆ T À1 [ g]B 0 T , and so gó ˆ [ g]B 0 . j 4.13 Example Again let G ˆ C3 ˆ ka: a3 ˆ 1l. There is a representation r of G which is given by  1r ˆ 1 0 FG-modules     0 1 0 À1 À1 2 ,a rˆ , ar ˆ X À1 À1 1 1 0  47 (To see this, simply note that (ar)2 ˆ a2 r and (ar)3 ˆ I; see Exercise 3.2.) If V is a 2-dimensional vector space over C, with basis v1 , v2 (which we call B ), then we can turn V into a CG-module as in Theorem 4.4(1) by de®ning v1 1 ˆ v1 , v2 1 ˆ v2 , v1 a ˆ v2 , v1 a2 ˆ Àv1 À v2 , v2 a2 ˆ v1 X  À1 X 0 v2 a ˆ Àv1 À v2 , We then have    1 0 0 [1]B ˆ , [a]B ˆ 0 1 À1  1 À1 2 , [a ]B ˆ À1 1 Now let u1 ˆ v1 and u2 ˆ v1 ‡ v2 . Then u1 , u2 is another basis of V, which we call B 9. Since u1 1 ˆ u 1 , u2 1 ˆ u 2 , u1 a ˆ Àu1 ‡ u2 , u2 a ˆ Àu1 , u1 a2 ˆ Àu2 , u2 a2 ˆ u1 À u2 , we obtain the representation ö: g 3 [ g]B 9 where       1 0 À1 1 0 À1 2 [1]B 9 ˆ , [a]B 9 ˆ , [a ]B 9 ˆ X 0 1 À1 0 1 À1 Note that if Tˆ then for all g in G, we have [ g]B ˆ T À1 [ g]B 9 T , and so r and ö are equivalent, in agreement with Theorem 4.12(1).   1 0 1 1 Summary of Chapter 4 1. An FG-module is a vector space over F, together with a multiplication by elements of G on the right. The multiplication satis®es properties (1)±(5) of De®nition 4.2. 48 Representations and characters of groups 2. There is a correspondence between representations of G over F and FG-modules, as follows. (a) Suppose that r: G 3 GL (n, F) is a representation of G. Then F n is an FG-module, if we de®ne v g ˆ v( gr) (v P F n , g P G)X (b) If V is an FG-module, with basis B , then r: g 3 [ g]B is a representation of G over F. 3. If G is a subgroup of Sn , then the permutation FG-module has basis v1 , F F F , v n , and v i g ˆ v ig for all i with 1 < i < n, and all g in G. Exercises for Chapter 4 1. Suppose that G ˆ S3 , and that V ˆ sp (v1 , v2 , v3 ) is the permutation module for G over C, as in De®nition 4.10. Let B 1 be the basis v1 , v2 , v3 of V and let B 2 be the basis v1 ‡ v2 ‡ v3 , v1 À v2 , v1 À v3 . Calculate the 3 3 3 matrices [ g]B 1 and [ g]B 2 for all g in S3 . What do you notice about the matices [ g]B 2 ? 2. Let G ˆ Sn and let V becomes an FG-module & v, vg ˆ Àv, be a vector space over F. Show that V if we de®ne, for all v in V, if g is an even permutation, if g is an odd permutation. 3. Let Q8 ˆ ka, b: a4 ˆ 1, b2 ˆ a2 , bÀ1 ab ˆ aÀ1 l, the quaternion group of order 8. Show that there is an RQ8 -module V of dimension 4 with basis v1 , v2 , v3 , v4 such that v1 a ˆ v 2 , v1 b ˆ v 3 , v2 a ˆ Àv1 , v2 b ˆ v4 , v3 a ˆ Àv4 , v4 a ˆ v3 , and v3 b ˆ Àv1 , v4 b ˆ Àv2 X 4. Let A be an n 3 n matrix and let B be a matrix obtained from A by permuting the rows. Show that there is an n 3 n permutation matrix P such that B ˆ PA. Find a similar result for a matrix obtained from A by permuting the columns. 5 FG-submodules and reducibility We begin the study of FG-modules by introducing the basic building blocks of the theory ± the irreducible FG-modules. First we require the notion of an FG-submodule of an FG-module. Throughout, G is a group and F is R or C. FG-submodules 5.1 De®nition Let V be an FG-module. A subset W of V is said to be an FGsubmodule of V if W is a subspace and wg P W for all w P W and all g P G. Thus an FG-submodule of V is a subspace which is also an FGmodule. 5.2 Examples (1) For every FG-module V, the zero subspace {0}, and V itself, are FG-submodules of V. (2) Let G ˆ C3 ˆ ka: a3 ˆ 1l, and let V be the 3-dimensional FGmodule de®ned in Example 4.11. Thus, V has basis v1 , v2 , v3 , and v1 1 ˆ v1 , v2 1 ˆ v2 , v3 1 ˆ v3 , v1 a ˆ v2 , v2 a ˆ v3 , v3 a ˆ v1 , v1 a2 ˆ v3 , v2 a2 ˆ v1 , v3 a2 ˆ v2 X Put w ˆ v1 ‡ v2 ‡ v3, and let W ˆ sp (w), the 1-dimensional subspace spanned by w. Since 49 50 Representations and characters of groups w1 ˆ wa ˆ wa2 ˆ w, W is an FG-submodule of V. However, sp (v1 ‡ v2 ) is not an FGsubmodule, since (v1 ‡ v2 )a ˆ v2 ‡ v3 P sp (v1 ‡ v2 )X a Irreducible FG-modules 5.3 De®nition An FG-module V is said to be irreducible if it is non-zero and it has no FG-submodules apart from {0} and V. If V has an FG-submodule W with W not equal to {0} or V, then V is reducible. Similarly, a representation r: G 3 GL (n, F) is irreducible if the corresponding FG-module F n given by v g ˆ v( gr) (v P F n , g P G) (see Theorem 4.4(1)) is irreducible; and r is reducible if F n is reducible. Suppose that V is a reducible FG-module, so that there is an FGsubmodule W with 0 , dim W , dim V. Take a basis B 1 of W and extend it to a basis B of V. Then for all g in G, the matrix [ g]B has the form H I Xg 0 d e (5X4) Yg Z g for some matrices Xg , Yg and Zg , where Xg is k 3 k (k ˆ dim W). A representation of degree n is reducible if and only if it is equivalent to a representation of the form (5.4), where X g is k 3 k and 0 , k , n. Notice that in (5.4), the functions g 3 Xg and g 3 Z g are representations of G: to see this, let g, h P G and multiply the matrices [ g]B and [h]B given by (5.4). Notice also that if V is reducible then dim V > 2. 5.5 Examples (1) Let G ˆ C3 ˆ ka: a3 ˆ 1l and let V be the 3-dimensional FGmodule with basis v1 , v2 , v3 such that FG-submodules and reducibility v1 a ˆ v2 , v2 a ˆ v3 , v3 a ˆ v1 , 51 as in Example 4.11. We saw in Example 5.2(2) that V is a reducible FG-module, and has an FG-submodule W ˆ sp (v1 ‡ v2 ‡ v3 ). Let B be the basis v1 ‡ v2 ‡ v3 , v1 , v2 of V. Then H H I I 1 0 0 1 0 0 f f g g [1]B ˆ f 0 1 0 g, [a]B ˆ f 0 0 1 g, d d e e 0 0 1 1 À1 À1 H I 1 0 0 f g [a2 ]B ˆ f 1 À1 À1 gX d e 0 1 0 This reducible representation gives us two other representations: at the `top left' we have the trivial representation and at the `bottom right' we have the representation which is given by       0 1 1 0 À1 À1 2 ,a 3 ,a3 X 13 À1 À1 0 1 1 0 (2) Let G ˆ D8 and let V ˆ F 2 be the 2-dimensional FG-module described in Example 4.5(1). Thus G ˆ ka, bl, and for all (ë, ì) P V we have (ë, ì)a ˆ (Àì, ë), (ë, ì)b ˆ (ë, Àì)X We claim that V is an irreducible FG-module. To see this, suppose that there is an FG-submodule U which is not equal to V. Then dim U < 1, so U ˆ sp ((á, â)) for some á, â P F. As U is an FGmodule, (á, â)b is a scalar multiple of (á, â), and hence either á ˆ 0 or ⠈ 0. Since (á, â)a is also a scalar multiple of (á, â), this forces á ˆ ⠈ 0, so U ˆ {0}. Consequently V is irreducible, as claimed. Summary of Chapter 5 1. If V is an FG-module, and W is a subspace of V which is itself an FG-module, then W is an FG-submodule of V. 2. The FG-module V is irreducible if it is non-zero and the only FGsubmodules are {0} and V. 5 are irreducible? 4. and let V ˆ F 2. cr ˆ 0 å À1 0 çÀ1 1 0 (c) For which values of å. (b) Suppose that å and ç are complex cube roots of unity. br ˆ . b. 3. and let G ˆ ka. Verify that V is an FG-module and ®nd all the FG-submodules of V. â)1 ˆ (á. ç is r irreducible? 5. Let G ˆ C13 . Which of the four representations of D12 de®ned in Exercise 3. Find a CG-module which is neither reducible nor irreducible. â) and (á. ç is r faithful? (d) For which values of å. Prove that there is a representation r of G over C such that       å 0 ç 0 0 1 ar ˆ X . b ˆ (4 5 6). de®ne (á. c ˆ (2 3)(4 5). cÀ1 ac ˆ aÀ1 and cÀ1 bc ˆ bÀ1 X Deduce that G has order 18. For (á. â)a ˆ (â. Let G ˆ C2 ˆ ka: a ˆ 1l. Prove that if r is reducible then ó is reducible. (a) Check that a3 ˆ b3 ˆ c2 ˆ 1. cl. 2. c P S6 by a ˆ (1 2 3). b.52 Representations and characters of groups Exercises for Chapter 5 1. Let r and ó be equivalent representations of the group G over F. 2 . â) P V. ab ˆ ba. á). De®ne the permutations a. and let F be R or C. After de®ning the group algebra of G. if uˆ n ˆ iˆ1 ë i g i and v ˆ n ˆ iˆ1 ìi g i are elements of FG. . . . then 53 . Take as the elements of FG all expressions of the form ë1 g 1 ‡ X X X ‡ ë n g n (all ë i P F)X The rules for addition and scalar multiplication in FG are the natural ones: namely.6 Group algebras The group algebra of a ®nite group G is a vector space of dimension |G| which also carries extra structure involving the product operation on G. group algebras are the source of all you need to know about representation theory. the ultimate goal of representation theory ± that of understanding all the representations of ®nite groups ± would be achieved if group algebras could be fully analysed. g n . g n as a basis. and ë P F. In particular. . We de®ne a vector space over F with g1 . In a sense. The group algebra of G Let G be a ®nite group whose elements are g1 . which will be explored in greater detail later on. . . Group algebras are therefore of great interest. . known as the regular representation of G. . we shall use it to construct an important faithful representation. and we call this vector space FG. .1 Example Let G ˆ C3 ˆ ka: a3 ˆ el. g n . we write e for the identity element of G.hPG ˆ where all ë g . ì h P F. .) The vector space CG contains u ˆ e À a ‡ 2a2 and v ˆ 1 e ‡ 5aX 2 We have u ‡ v ˆ 3 e ‡ 4a ‡ 2a2 . g n is called the natural basis of FG. FG is a vector space over F of dimension n. . v are the elements of CG which appear in Example 6. in this example.54 Representations and characters of groups u‡vˆ n n ˆ ˆ (ë i ‡ ì i ) g i and ëu ˆ (ëë i ) g i X iˆ1 iˆ1 With these rules. . ˆˆ (ë h ì hÀ1 g ) g gPG hPG 6. FG carries more structure than that of a vector space ± we can use the product operation on G to de®ne multiplication in FG as follows: 3 2ˆ 32ˆ ˆ ëg g ìh h ˆ ë g ì h ( gh) gPG hPG g. . with basis g1 . . (To avoid confusion with the element 1 of F. The basis g1 . 6. . 1 u ˆ 1 e À 1 a ‡ 2 a2 X 2 3 3 3 3 Sometimes we write elements of FG in the form ˆ ë g g (ë g P F)X gPG Now.2 Example If G ˆ C3 and u.1. then uv ˆ (e À a ‡ 2a2 )(1 e ‡ 5a) 2 ˆ ˆ 1 2e ‡ 5a À 1 a À 5a2 ‡ a2 ‡ 10a3 2 ‡ 9 a À 4a2 X 2 21 2e . . j .Group algebras 6. r(s ‡ t) ˆ rs ‡ rt. ì g .kPG ˆ r(st)X We leave the proofs of the other equations as easy exercises.3 De®nition The vector space FG. for all r. t P FG and ë P F: (1) (2) (3) (4) (5) (6) (7) rs P FG.h. Then (rs)t ˆ ˆ ˆ g. (r ‡ s)t ˆ rt ‡ st. namely the element 1e (where 1 is the identity of F and e is the identity of G). 6.h. is called the group algebra of G over F. with multiplication de®ned by 3 2ˆ 32ˆ ˆ ëg g ìh h ˆ ë g ì h ( gh) gPG hPG g. (2) Let ˆ ˆ ˆ rˆ ë g g. r1 ˆ 1r ˆ r. s ˆ ì g g. (ër)s ˆ ë(rs) ˆ r(ës). We write this element simply as 1. r0 ˆ 0r ˆ 0. gPG gPG gPG (ë g . s. The group algebra FG contains an identity for multiplication. ì h P F).hPG 55 (ë g . í g P F). r(st) ˆ (rs)t.kPG ë g ì h í k ( gh)k ë g ì h í k g(hk) ˆ g. t ˆ í g g.4 Proposition Multiplication in FG satis®es the following properties. Proof (1) It follows immediately from the de®nition of rs that rs P FG. 5 De®nition Let G be a ®nite group and F be R or C. ë P F and g. The vector space FG. so g ˆ 1. j 6.4. v1 ˆ v. Then 1 g ˆ 1. Note that the regular FG-module has dimension equal to |G|.7 Example Let G ˆ C3 ˆ ka: a3 ˆ el. v( gh) ˆ (v g)h.4 is called an algebra. so that V is a vector space of dimension n over F. any vector space equipped with a multiplication satisfying properties (1)±(7) of Proposition 6. The representation g 3 [ g]B obtained by taking B to be the natural basis of FG is called the regular representation of G over F.6 Proposition The regular FG-module is faithful. Therefore V is an FG-module. The elements of FG have the form . g P G). by parts (1). We shall be concerned only with group algebras. 6. (ëv) g ˆ ë(v g). h P G. (u ‡ v) g ˆ ug ‡ v g. Let V ˆ FG. respectively. Proof Suppose that g P G and v g ˆ v for all v P FG. (3). v P V. with the natural multiplication v g (v P FG. but it is worth pointing out that the axioms for an algebra mean that it is both a vector space and a ring. 6. where n ˆ |G|. we have vg P V. For all u. is called the regular FG-module.56 Representations and characters of groups In fact. The regular FG-module We now use the group algebra to de®ne an important FG-module. (2). and the result follows. (4) and (5) of Proposition 6. say € r ˆ gPG ì g g (ì g P F). a. (ë1 e ‡ ë2 a ‡ ë3 a2 )a2 ˆ ë2 e ‡ ë3 a ‡ ë1 a2 X (ë i P F)X 57 By taking matrices relative to the basis e. This is done in the following natural way. v2 r ˆ ëv1 ‡ ìv2 . Now.8 De®nition Suppose that V is an FG-module. we obtain the regular representation of G: H I H I H I 1 0 0 0 1 0 0 0 1 e 3 d 0 1 0 e.9 Examples (1) Let V be the permutation module for S4 . De®ne vr by ˆ vr ˆ ì g (v g)X gPG 6. (2v1 ‡ v2 )r ˆ ëv1 ‡ (2ë ‡ ì)v2 ‡ 2ìv3 X (ë. and that v P V and r P FG. together with a multiplication v g for v P V and g P G (and the multiplication satis®es various axioms). a2 3 d 1 0 0 eX 0 0 1 1 0 0 0 1 0 FG acts on an FG-module You will remember that an FG-module is a vector space over F. a 3 d 0 0 1 e.Group algebras ë1 e ‡ ë2 a ‡ ë3 a2 We have (ë1 e ‡ ë2 a ‡ ë3 a2 )e ˆ ë1 e ‡ ë2 a ‡ ë3 a2 . 6. ì P F) . as described in Example 4. it is sometimes helpful to extend the de®nition of the multiplication so that we have an element vr of V for all elements r in the group algebra FG. (ë1 e ‡ ë2 a ‡ ë3 a2 )a ˆ ë3 e ‡ ë1 a ‡ ë2 a2 . a2 of FG.9. If r ˆ ë(1 2) ‡ ì(1 3 4) then v1 r ˆ ëv1 (1 2) ‡ ìv1 (1 3 4) ˆ ëv2 ‡ ìv3 . given by De®nition 6. (6) j ˆ (vr)sX . v(rs) ˆ (vr)s. s P FG with ˆ ˆ rˆ ë g g. s P FG: (1) (2) (3) (4) (5) (6) (7) vr P V. then for all v P V and r P FG. the element vr is simply the product of v and r as elements of the group algebra. and we leave them to you. (5). v0 ˆ 0r ˆ 0.4. We shall give a proof of part (2).58 Representations and characters of groups (2) If V is the regular FG-module. v(r ‡ s) ˆ vr ‡ vs. Let v P V. 6. and let r.h 2ˆ g 3 32ˆ ë g (v g) ìh h h by (4).10 Proposition Suppose that V is an FG-module. v P V. s ˆ ì h hX gPG hPG Then 2ˆ 3 v(rs) ˆ v ë g ì h ( gh) g. (ëv)r ˆ ë(vr) ˆ v(ër). v1 ˆ v.h ˆ ˆ ˆ ˆ g. assuming the other parts.3.h ë g ì h (v( gh)) ë g ì h ((v g)h) by (4) and (6) ˆ g. (u ‡ v)r ˆ ur ‡ vr. Then the following properties hold for all u. all ë P F and all r. Proof All parts except (2) are straightforward. Compare the next result with Proposition 6. . where B is the basis g1 . g n }. b: a4 ˆ b2 ˆ 1. Suppose that G ˆ D8 ˆ ka. What is the matrix [W]B . Show that for every ®nite group G. Assume that G is a ®nite group. . with the natural multiplication v g (v P FG. r P FG such that vr ˆ 0. (c) Let W: CG 3 CG be the linear transformation sending v to vc for all v in CG. . The vector space FG. 3. and has a natural multiplication de®ned on it. Let G ˆ C2 . . . yx and x 2 . with |G| . and v0 ˆ 0 for all v P V . 3. Work out matrices for the regular representation of C2 3 C2 over F. (b) Deduce that c2 ˆ |G|c. g P G) is the regular FG-module. If V is an FG-module. 1. (a) Let x and y be the following elements of CG: x ˆ a ‡ 2a2 . The group algebra FG of G over F consists of all linear combinations of elements of G. For r and s in CG. y ˆ b ‡ ab À a2 X Calculate xy. (b) Let z ˆ b ‡ a2 b. 2. . 2. where the symbol 0 is used for the zero elements of V and FG. prove from the de®nition that 0r ˆ 0 for all r P FG. say G ˆ { g1 . . and write c €n for the element iˆ1 g i of CG. . (a) Prove that ch ˆ hc ˆ c for all h in G. The regular FG-module is faithful. does rs ˆ 0 imply that r ˆ 0 or s ˆ 0? 4. but neither v nor r is 0.Group algebras Summary of Chapter 6 59 1. gn of CG? 5. bÀ1 ab ˆ aÀ1 l. Exercises for Chapter 6 1. there exists an FG-module V and elements v P V. Show that zg ˆ gz for all g in G. Deduce that zr ˆ rz for all r in CG. . Suppose that G ˆ D6 ˆ ka. and let ù ˆ e2ðia3 . de®ned by W ˆ sp (1 ‡ ù2 a ‡ ùa2 . . Prove that the 2-dimensional subspace W of CG. is an irreducible CG-submodule of the regular CG-module.60 Representations and characters of groups 6. b: a3 ˆ b2 ˆ 1. b ‡ ù2 ab ‡ ùa2 b). bÀ1 ab ˆ aÀ1 l. 1 De®nition Let V and W be FG-modules. The analogous functions for FG-modules are called FG-homomorphisms. we have (vr)W ˆ (vW)r since (vr)W ˆ ˆ gPG ë g (v g)W ˆ ˆ gPG ë g (vW) g ˆ (vW)rX The next result shows that FG-homomorphisms give rise to FGsubmodules in a natural way. group homomorphisms and linear transformations. 61 . g P GX In other words. A function W: V 3 W is said to be an FG-homomorphism if W is a linear transformation and (v g)W ˆ (vW) g for all v P V . and we introduce these in this chapter. then for all v P V and r ˆ gPG ë g g P FG. submodule of W. Note that if G is a ®nite group and W: V 3 W is an FG-homomorph€ ism.2 Proposition Let V and W be FG-modules and let W: V 3 W be an FG-homomorphism. Then Ker W is an FG-submodule of V and Im W is an FG. respectively. FG-homomorphisms 7. the `structure-preserving' functions are. if W sends v to w then it sends v g to wg. 7.7 FG-homomorphisms For groups and vector spaces. Then W is a linear transformation. and 2 3 2 3 ˆ ˆ (vW) g ˆ ë i wg ˆ ë i wX Therefore W is an FG-homomorphism. we have 2 3 2 3 ˆ ˆ (v g)W ˆ ë i v ig W ˆ ë i w. so vg P Ker W. . Im W ˆ V. v n ) be the permutation module for G over F (see De®nition 4.62 Representations and characters of groups Proof First note that Ker W is a subspace of V and Im W is a subspace of W. Provided ë Tˆ 0. De®ne 2 n 3 n ˆ ˆ W: ëivi 3 ë i w (ë i P F)X iˆ1 iˆ1 Thus v i W ˆ w for all i. Im W ˆ {0}. Now let w P Im W. j 7. and Ker W ˆ V. Then W is an FG-homomorphism. Here. and de®ne W: V 3 V by vW ˆ ëv for all v P V. Therefore Ker W is an FG-submodule of V. and so Im W is an FG-submodule of W. since W is a linear transformation.10). Let V ˆ sp (v1 . . @ n A n ˆ ˆ Ker W ˆ ëi vi : ëi ˆ 0 . Let v P Ker W and g P G. We construct an FG-homomorphism W from V to W. and for all € v ˆ ë i v i P V and all g P G. so that w ˆ vW for some v P V. .3 Examples (1) If W: V 3 W is de®ned by vW ˆ 0 for all v P V.8). wg ˆ (vW) g ˆ (v g)W P Im W. . Then (v g)W ˆ (vW) g ˆ 0 g ˆ 0. (2) Let ë P F. iˆ1 iˆ1 Im W ˆ W X . For all g P G. (3) Suppose that G is a subgroup of S n . and let W ˆ sp (w) be the trivial FG-module (see De®nition 4. then W is an FGhomomorphism. we have Ker W ˆ {0}. j as W is an FG-homomorphism Suppose that W: V 3 W is an FG-isomorphism. ((wWÀ1 ) g)W ˆ ((wWÀ1 )W) g ˆ wg ˆ ((wg)WÀ1 )WX Hence (wWÀ1 ) g ˆ (wg)WÀ1 . (2) V is irreducible if and only if W is irreducible (since X is an FGsubmodule of V if and only if XW is an FG-submodule of W).5 Proposition If W: V 3 W is an FG-isomorphism. For w P W and g P G. then the inverse WÀ1 : W 3 V is also an FG-isomorphism. Proof Certainly WÀ1 is an invertible linear transformation. Ker W is an FG-submodule of the permutation module V. . We call a function W: V 3 W an FGisomorphism if W is an FG-homomorphism and W is invertible. . Then we may use W and WÀ1 to translate back and forth between the isomorphic FGmodules V and W. If there is such an FG-isomorphism. 7.FG-homomorphisms 63 By Proposition 7. In the next result. Isomorphic FG-modules 7. we check that if V  W then W  V. . . . . then we say that V and W are isomorphic FG-modules and write V  W. so we need only show that WÀ1 is an FG-homomorphism. . We list some examples below: (1) dim V ˆ dim W (since v1 .2. and prove that V and W share the same structural properties. as required. (3) V contains a trivial FG-submodule if and only if W contains a trivial FG-submodule (since X is a trivial FG-submodule of V if and only if XW is a trivial FG-submodule of W). . v n is a basis of V if and only if v1 W. v n W is a basis of W).4 De®nition Let V and W be FG-modules. . Let W be the invertible linear transformation from V to W for which v i W ˆ w i for all i. Let g P G. Then by Theorem 4. suppose that r and ó are equivalent. we deduce that (v i g)W ˆ (v i W) g for all i. we show that isomorphic FG-modules correspond to equivalent representations. . . v n is a basis B 1 of V and w1. . suppose ®rst that W is an FG-isomorphism from V to W. Since [ g]B 1 ˆ [ g]B 2 . Hence r and ó are equivalent. . Since (v i g)W ˆ (v i W) g for each i. suppose that v1 . and hence W is an FG-isomorphism. .6 Theorem Suppose that V is an FG-module with basis B . . This completes the proof of (7. v n W is a basis B 2 of W. . Proof We ®rst establish the following fact: (7. v n be a basis B 1 of V. Then V and W are isomorphic if and only if the representations r: g 3 [ g]B and ó : g 3 [ g]B 9 are equivalent.7). . Now assume that V and W are isomorphic FG-modules. we frequently disdain to distinguish between isomorphic FG-modules. we continue simply to emphasize the similarity between isomorphic FG-modules. though. .64 Representations and characters of groups Just as we often regard isomorphic groups as being identical. there is a basis B 0 of V such that gó ˆ [ g]B 0 for all g P G.7). . . By (7. Let g P G. . In the next result. . De®ne a representation ö of G by ö: g 3 [ g]B 1 . . Conversely. 7. .7) The FG-modules V and W are isomorphic if and only if there are a basis B 1 of V and a basis B 2 of W such that [ g]B 1 ˆ [ g]B 2 for all g P GX To see this. . then v1 W. there are a basis B 1 of V and a basis B 2 of W such that [ g]B 1 ˆ [ g]B 2 for all g P G. Conversely. and let v1 . For the moment. . Then by Theorem 4. it follows that [ g]B 1 ˆ [ g]B 2 . ö is equivalent to both r and ó.12(2).12(1). and W is an FGmodule with basis B 9. w n is a basis B 2 of W such that [ g]B 1 ˆ [ g]B 2 for all g P G. 11.4(1) we encountered two equivalent representations r and ó of G. and let W denote the regular FG-module. where     0 1 1 0 ar ˆ . we have [ g]B ˆ [ g]B 9 for all g P GX According to (7. v3 of V. Indeed. We have H H I I 1 0 0 0 1 0 f f g g [1]B 9 ˆ f 0 1 0 g. Then 1.8 Example Let G ˆ ka: a3 ˆ 1l.7). [a]B 9 ˆ f 0 0 1 g. a cyclic group of order 3. bó ˆ Ài 1  1 X 0 (ë i P F) .9 Example Let G ˆ D8 ˆ ka. a2 is a basis of W. v3 a ˆ v1 X Writing B for the basis v1 . v3 such that v1 a ˆ v2 . b: a4 ˆ b2 ˆ 1. bÀ1 ab ˆ aÀ1 l. Therefore V and W are isomorphic FG-modules.FG-homomorphisms 65 that is. call it B 9. d d e e 0 0 1 1 0 0 H 0 0 0 1 1 0 I f [a2 ]B 9 ˆ f 1 d 0 g 0 gX e Compare the FG-module V de®ned in Example 4. [ g]B 9 ˆ [ g]B 0 for all g P G. br ˆ À1 0 0 À1 and aó ˆ  i 0   0 0 . j 7. the FG-modules V and W are therefore isomorphic. v2 . 7. a. v2 a ˆ v3 . the function W: ë1 v1 ‡ ë2 v2 ‡ ë3 v3 3 ë1 1 ‡ ë2 a ‡ ë3 a2 is an FG-isomorphism from V to W.7). by (7. with basis v1 . In Example 3. v2 . let W be the CG-module with basis w1. w2 b ˆ w1 Thus. v2 for which v1 a ˆ v2 . . and B i is a basis of Ui. v2 of V and B 9 for the basis w1. w n is a basis B of V. v2 b ˆ Àv2 (see Example 4. . let W: V 3 W be the invertible linear transformation such that W: v1 3 w1 ‡ w2 . u m . (Compare Example 3. . and hence W is a CG-isomorphism from V to W.4(1). if V ˆ U1 È .6. w1. . . . H I 0 [ g]B 1 eX [ g]B ˆ d 0 [ g]B 2 More generally. then we can amalgamate B 1 . . Then by (2. . . v2 3 iw1 À iw2 X Then (v j a)W ˆ (v j W)a and (v j b)W ˆ (v j W)b for j ˆ 1. . since r and ó are equivalent. B r to . . if we write B for the basis v1 . where U and W are FG-submodules of V. and we show that these give rise to FG-homomorphisms.5(1)). in a similar way. and suppose that V ˆ U È W. and. and w1. . v2 a ˆ Àv1 . u m be a basis B 1 of U. . w n be a basis B 2 of W. Let V be an FG-module. . . . To verify this directly. then for all g P G we have r: g 3 [ g]B and ó : g 3 [ g]B 9 X According to Theorem 7. Let u1 . u1 . the CG-modules V and W are isomorphic.) Direct sums We conclude the chapter with a discussion of direct sums of FGmodules. and for g P G. w2 for which w1 a ˆ iw1 . .66 Representations and characters of groups Let V be the CG-module with basis v1 . . v1 b ˆ v1 . È Ur. w2 a ˆ Àiw2 . w2 of W. w1 b ˆ w2 . .9). a direct sum of FG-submodules Ui. 2. . . . 11 Proposition Let V be an FG-module. v ˆ u1 ‡ . ‡ u r (u j P U j for all j). ‡ ur for unique vectors ui P Ui. vð 2 ˆ ui ð i ˆ ui ˆ vð i . For v P V we have . and suppose that V ˆ U1 È X X X È U r where each Ui is an FG-submodule of V. since for v P V with v ˆ u1 ‡ . 7.FG-homomorphisms obtain a basis B of V. i We now present a technical result concerning sums of irreducible FG-modules which will be used at a later stage. i so ð2 ˆ ð i . 7. . Proof Clearly ð i is a linear transformation.12 Proposition Let V be an FG-module.30). and ð i is an FG-homomorphism. . and suppose that V ˆ U1 ‡ X X X ‡ Ur . and g P G. H [ g]B 1 f FF [ g]B ˆ d (7X10) F 67 I g eX 0 [ g]B r 0 The next result shows that direct sums give rise naturally to FGhomomorphisms. and is also a projection of V . . . we have (v g)ð i ˆ (u1 g ‡ X X X ‡ ur g)ð i ˆ ui g ˆ (vð i ) gX Also. j . Thus ð i is a projection (see De®nition 2. sum of some of the FG-submodules Ui. and we de®ne ð i : V 3 V (1 < i < r) by setting vð i ˆ ui X Then each ð i is an FG-homomorphism. and for g P G. where each Ui is an irreducible FG-submodule of V Then V is a direct . F F F . j Finally. Exercises for Chapter 7 1. . and Ui is irreducible. Ur } which has the properties that W 1 ‡ X X X ‡ W s is direct (iXeX equal to W 1 È X X X È W s ). If V and W are FG-modules and W: V 3 W is a linear transformation which satis®es (v g)W ˆ (vW) g for all v P V. v r ) g ˆ (v1 g. we have V ˆ W ˆ W 1 È F F F ÈW s . V r are FG-modules. as claimed. we remark that if V1 . Ur so that the sum of our chosen FG-submodules is direct. . v r g) for all v i P Vi (1 < i < r) and all g P G.68 Representations and characters of groups Proof The idea is to choose as many as we can of the FG-submodules U1. Then W ‡ Ui is not a direct sum. If Ui P Y this is clear. . but W 1 ‡ X X X ‡ W s ‡ U i is not direct. Isomorphic FG-modules correspond to equivalent representations. But W ’ Ui a is an FG-submodule of Ui. . as required. . g P G. then W is an FG-homomorphism. Since U i  W for all i with 1 < i < r. then we can make the external direct sum V1 È F F F ÈVr (see Chapter 2) into an FGmodule by de®ning (v1 . V and W be FG-modules. Ws } of {U1. Prove that Wö: U 3 W is an FG-homomorphism. and let W: U 3 V and ö: V 3 W be FG-homomorphisms. Let U. . and so Ui # W. . if Ui P Y X a Let W ˆ W1 ‡ X X X ‡ WsX We claim that Ui # W for all i. so assume that Ui P Y. choose a subset Y ˆ {W1. Summary of Chapter 7 1. Kernels and images of FG-homomorphisms are FG-modules. so W ’ Ui Tˆ {0}. 3. . 2. therefore W ’ Ui ˆ Ui. . X X X . . . . X X X . . To this end. (a) Show that the function W: á1 ‡ âx 3 (á À â)(1 À x) (á. Let G be the subgroup of S5 which is generated by (1 2 3 4 5). De®ne the FGsubmodules V0 and W0 of V and W as in Exercise 3. Prove that the permutation module for G over F is isomorphic to the regular FG-module. Prove that the subset V0 ˆ fv P V : v g ˆ v for all g P Gg is an FG-submodule of V.FG-homomorphisms 69 2. 3. Let G be the subgroup of S4 which is generated by (1 2) and (3 4). (c) Find a basis B of FG such that   2 0 [W]B ˆ X 0 0 . (b) Prove that W2 ˆ 2W. Suppose that V and W are isomorphic FG-modules. Prove that V0 and W0 are isomorphic FG-modules. Assume that V is an FG-module. Show that the function ˆ W: v 3 v g (v P V ) gPG is an FG-homomorphism from V to V0 . Is the permutation module for G over F isomorphic to the regular FG-module? 6. Let G ˆ C2 ˆ kx: x 2 ˆ 1l. Is it necessarily surjective? 4. â P F) is an FG-homomorphism from the regular FG-module to itself. 5. in fact. and let V be an FG-module. let F be R or C. for instance sp (v2 . we illustrate it with some examples. where as usual F ˆ R or C.2(2) below. A consequence of this theorem is that every FGmodule is a direct sum of irreducible FG-submodules. since ug ˆ u for all g P G. 8. v2 . v2 À 2v3 ). If U is an FG-submodule of V. Put u ˆ v1 ‡ v2 ‡ v3 and U ˆ sp (u)X Then U is an FG-submodule of V. (The assumption on F is important ± see Example 8. We shall ®nd this W in an 70 .) This essentially reduces representation theory to the study of irreducible FG-modules.2 Examples (1) Let G ˆ S3 and let V ˆ sp (v1 . then there is an FG-submodule W of V such that V ˆ U È WX Before proving Maschke's Theorem. v3 ) be the permutation module for G over F (see De®nition 4. only one FGsubmodule W of V with V ˆ U È W.1 Maschke's Theorem Let G be a ®nite group. There are many subspaces W of V such that V ˆ U È W.8 Maschke's Theorem We now come to our ®rst major result in representation theory. namely Maschke's Theorem.10). v3 ) and sp (v1 . Maschke's Theorem 8. But there is. . extend it to a basis v1 . The corresponding FG-module is V ˆ sp (v1 . . 1. where. (2) The conclusion of Maschke's Theorem can fail if F is not R or C. and take F to be the ®eld of integers modulo p. For v P V and x P G. .Maschke's Theorem 71 example after proving Maschke's Theorem (but you may like to look for it yourself now). U ˆ sp (v1 ) is an FG-submodule of V. p À 1) j 1 is a representation from G to GL (2.1 We are given U. . By Proposition 2. v1 a j ˆ v1 . We aim to modify the projection ö to create an FG-homomorphism from V to V with image U. To this end. v n ). as can easily be seen. v2 a j ˆ jv1 ‡ v2 X Clearly. let G ˆ C p ˆ ka: a p ˆ 1l. let p be a prime number. (vx)W ˆ 1 ˆ (vx) gö g À1 X jGj gPG . v2 ). de®ne W: V 3 V by (8X3) vW ˆ 1 ˆ v gö g À1 jGj gPG (v P V )X It is clear that W is an endomorphism of V and Im W # U. .) For v P V. But there is no FGsubmodule W such that V ˆ U È W. For example. v m of U. . and we de®ne ö: V 3 V by setting vö ˆ u. Choose any subspace W0 of V such that V ˆ U È W0X (There are many choices for W0 ± simply take a basis v1 . . . . . an FG-submodule of the FG-module V. . ö is a projection of V with kernel W0 and image U. Check that the function   1 0 j a 3 ( j ˆ 0. We show ®rst that W is an FG-homomorphism. for 0 < j < p À 1. X X X . v n of V.29. F). Proof of Maschke's Theorem 8. since U is the only 1-dimensional FG-submodule of V. we have v ˆ u ‡ w for unique vectors u P U and w P W 0. and let W0 ˆ sp (v m‡1 . . 4) shows that Im W ˆ U. 2. Consequently W2 ˆ W. we prove that W2 ˆ W. v2 ). we have ug P U. 1 ˆ 1 ˆ 1 ˆ uW ˆ (8X4) ugö g À1 ˆ (ug) g À1 ˆ u ˆ uX jGj gPG jGj gPG jGj gPG Now let v P V. Moreover. let W0 ˆ sp (v1 . v2 . v2 À v3 )X Ȁ É € (In fact.2. Using this. so does h ˆ xg.) . 3)X 3 The required FG-submodule W is then Ker W.3) is W: v i 3 1(v1 ‡ v2 ‡ v3 ) (i ˆ 1. Hence 1 ˆ (vx)W ˆ vhöhÀ1 x jGj hPG 2 3 1 ˆ ˆ vhöhÀ1 x jGj hPG ˆ (vW)xX Thus W is an FG-homomorphism.32. The projection ö onto U is given by ö: v1 3 0. g P G. First. j 8. Next.3(3). Let W ˆ Ker W.5 Example Let G ˆ S3 and let V ˆ sp (v1 . and so (ug)ö ˆ ug. so W ˆ sp (v1 À v2 . First note that for u P U. as claimed.4) we have (vW)W ˆ vW. the FG-submodule constructed in Example 7. v3 ) be the permutation module. so by (8. Then W is an FG-submodule of V by Proposition 7. We have now established that W: V 3 V is a projection and an FGhomomorphism. Then vW P U. This completes the proof of Maschke's Theorem. Then V ˆ U È W0 (but of course W0 is not an FG-submodule). v3 3 v1 ‡ v2 ‡ v3 X Check now that the FG-homomorphism W given by (8. as in Example 8. W ˆ ë i v i : ë i ˆ 0 . v2 3 0. with submodule U ˆ sp (v1 ‡ v2 ‡ v3 ).2(1).72 Representations and characters of groups As g runs over the elements of G. (8. and V ˆ U È W by Proposition 2. We use the proof of Maschke's Theorem to ®nd an FG-submodule W of V such that V ˆ U È W. then we get H I j 0 0 [ g]B 9 ˆ d 0 j j e. This example illustrates the matrix version of Maschke's Theorem: for an arbitrary ®nite group G. the matrix [ g]B has the form H I j 0 0 [ g]B ˆ d j j j eX j j j The zeros re¯ect the fact that U is an FG-submodule of V (see (5. v1 À v2 . v1 . 0 j j because sp (v1 À v2 . n. v2 À v3 ) is also an FG-submodule of V. v2 À v3 as a basis B 9. . suppose that r is a reducible representation of a ®nite group G over F of degree n.Maschke's Theorem 73 Note that if B is the basis v1 ‡ v2 ‡ v3 . if we can choose a basis B of an FGmodule V such that [ g]B has the form H I à 0 d e à à for all g P G (see (5. then for all g P G. v2 of V. Yg. k . where X g is k 3 k with 0 . Z g . g 3d Yg Zg for some matrices X g . To put this another way. If instead we use v1 ‡ v2 ‡ v3 . Then we know that r is equivalent to a representation of the form H I Xg 0 e ( g P G). then we can ®nd a basis B 9 such that [ g]B 9 has the form H I à 0 d e 0 à for all g P G.4)).4)). j Another useful consequence of Maschke's Theorem is the next proposition. If V is irreducible then the result holds.7 Theorem If G is a ®nite group and F ˆ R or C. we have. g 3d 0 Bg where A g is also a k 3 k matrix. Then V has an FG-submodule U not equal to {0} or V. Since dim U . Then by (2. dim V and dim W . Proof Let V be a non-zero FG-module.10). there is an FG-submodule W such that V ˆ U È W. a direct sum of irreducible FG-modules.74 Representations and characters of groups Maschke's Theorem asserts further that r is equivalent to a representation of the form H I Ag 0 e. Consequences of Maschke's Theorem We now use Maschke's Theorem to show that every non-zero FGmodule is a direct sum of irreducible FG-submodules. where each Ui and W j is an irreducible FG-module. where each Ui is an irreducible FG-submodule of V.6 De®nition An FG-module V is said to be completely reducible if V ˆ U1 È F F F ÈU r . so suppose that V is reducible. (By an irreducible FG-submodule. W ˆ W 1 È X X X È W s . by induction. . V ˆ U1 È X X X È Ur È W 1 È X X X È W s . By Maschke's Theorem. The result is true if dim V ˆ 1. we simply mean an FG-submodule which is an irreducible FG-module. U ˆ U1 È X X X È Ur . dim V. The proof goes by induction on dim V. 8.) 8. then every non-zero FG-module is completely reducible. since V is irreducible in this case. Maschke's Theorem says that for every FG-submodule U of an FGmodule V. Theorem 8. there is an FG-submodule W of V such that V ˆ U È W. v2 x ˆ Àv1 À v2 X (This is a CG-module. in order to understand FG-modules. If G ˆ C2 3 C2 .Maschke's Theorem 75 8. where v1 x ˆ v2 . Suppose that U is an FG-submodule of V. 1. Every non-zero FG-module V is a direct sum of irreducible FGmodules: V ˆ U1 È X X X È Ur X Exercises for Chapter 8 1. We begin our study of these in the next chapter. there is an FG-submodule W with V ˆ U È WX 2. Then the function ð: V 3 U which is de®ned by ð: u ‡ w 3 u (u P U . Proof By Maschke's Theorem. and let V be the 2-dimensional CGmodule with basis v1 .11.7 tells us that every non-zero FG-module is a direct sum of irreducible FG-modules. where F ˆ R or C and G is a ®nite group. Summary of Chapter 8 Assume that G is a ®nite group and F ˆ R or C. Let G ˆ kx: x 3 ˆ 1l  C3 . by Exercise 3. by Proposition 7. . Then there exists a surjective FG-homomorphism from V onto U. v2 . express the group algebra RG as a direct sum of 1-dimensional RG-submodules.8 Proposition Let V be an FG-module. we may concentrate upon the irreducible FG-modules. w P W ) j is an FG-homomorphism onto U.) Express V as a direct sum of irreducible CG-submodules.2. Thus. 2. 5. 2 n 3 n n ˆ ˆ ˆ ëi vi . . g P G. ] is a complex inner product. Find a group G.2(2). An alternative proof of Maschke's Theorem for CG-modules. a CG-module V and a CG-homomorphism W: V 3 V such that V Tˆ Ker W È Im W. and de®ne U c ˆ fv P V : [u. with the natural multiplication by elements of G (so that for v P V. v P V )X xPG (1) Verify that [ . the vector vg is just the product of the row vector v with the matrix g). . vg] ˆ [u. . Let G be a ®nite group and let r: G 3 GL (2. (This shows that Maschke's Theorem fails for in®nite groups ± compare Example 8. v] ˆ (ux. v] for all u.2) for the de®nition of a complex inner product): for ë i . Suppose that G is the in®nite group &  ' 1 0 : nPZ n 1 and let V be the CG-module C2 . v P V and g P GX (2) Suppose that U is a CG-submodule of V.5(2) and Exercises 5.3. 6. 5. vx) (u.4. ] on V by ˆ [u.6 in the light of this result.1. ì j P C. ì jv j ˆ ëi ìiX iˆ1 jˆ1 iˆ1 De®ne another complex inner product [ . C) be a representation of G.) 5. h in G such that the matrices gr and hr do not commute. Suppose that there are elements g. Let V be a CG-module with basis v1 .76 Representations and characters of groups 3. v] ˆ 0 for all u P U gX Show that U c is a CG-submodule of V. ) on V as follows (see (14. 4. Prove that r is irreducible. . Show that V is not completely reducible.) 6. v n and suppose that U is a CG-submodule of V. . (You may care to revisit Example 5. De®ne a complex inner product ( . which satis®es [ug. (Hint: it is a standard property of complex inner products that V ˆ U È U c for all subspaces U of V. there exists a faithful irreducible CG-module.Maschke's Theorem 77 (3) Deduce Maschke's Theorem. Prove that for every ®nite simple group G. .) 7. and so Ker (W À ë1 V ) Tˆ {0}. and we give an immediate application by determining all the irreducible representations of ®nite abelian groups.26). (1) If W: V 3 W is a CG-homomorphism. we have Im W ˆ W. and hence is a CG-isomorphism. Schur's Lemma concerns CG-modules rather than RG-modules.2. (2) If W: V 3 V is a CG-isomorphism.9 Schur's Lemma Schur's Lemma is a basic result concerning irreducible modules. and since much of the ensuing theory depends on it. we shall deal with CG-modules for the remainder of the book (except in Chapter 23). Thus Ker (W À ë1 V ) is a non-zero CG-submodule 78 . G denotes a ®nite group. Schur's Lemma is fundamental to representation theory. the endomorphism W has an eigenvalue ë P C.2.1 Schur's Lemma Let V and W be irreducible CG-modules. Though simple in both statement and proof. then either W is a CGisomorphism. Then Im W Tˆ {0}. (2) By (2. Throughout. Schur's Lemma 9. or vW ˆ 0 for all v P V. Ker W is a CG-submodule of V. Also by Proposition 7. and W is irreducible. Proof (1) Suppose that vW Tˆ 0 for some v P V. as Ker W Tˆ V and V is irreducible. then W is a scalar multiple of the identity endomorphism 1 V . Thus W is invertible. Ker W ˆ {0}. As Im W is a CG-submodule of W by Proposition 7. which is a contradiction. for all g P G . 9. W ˆ ë1 V . Proof Suppose that V is reducible. so that V has a CG-submodule U not equal to {0} or V. Then V is irreducible. and suppose that every CG-homomorphism from V to V is a scalar multiple of 1 V . Therefore v(W À ë1 V ) ˆ 0 That is. and is not a scalar multiple of 1 V . C) be a representation of G.4(1). for all v P V X 79 j 9.3 Corollary Let r: G 3 GL (n. g P G. j We next interpret Schur's Lemma and its converse in terms of representations. By Maschke's Theorem. there is a CGsubmodule W of V such that V ˆ U È WX Then the projection ð: V 3 V de®ned by (u ‡ w)ð ˆ u for all u P U. Hence V is irreducible. regard C n as a CG-module by de®ning v g ˆ v( gr) for all v P C n . w P W is a CG-homomorphism (see Proposition 7.2 Proposition Let V be a non-zero CG-module. Ker (W À ë1 V ) ˆ V. Let A be an n 3 n matrix over C. g P G. Part (2) of Schur's Lemma has the following converse.Schur's Lemma of V. as required. Proof As in Theorem 4.11). Then r is irreducible if and only if every n 3 n matrix A which satis®es ( gr)A ˆ A( gr) has the form A ˆ ëI n with ë P C. Since V is irreducible. The endomorphism v 3 vA of C n is a CG-homomorphism if and only if (v g)A ˆ (vA) g for all v P C n . bÀ1 ab ˆ aÀ1 l.1 and Proposition 9.80 Representations and characters of groups that is. Hence  Aˆ á 0 0 á  ˆ áIX Consequently r is irreducible. and then (br)A ˆ A(br) gives á ˆ ä.4 Examples (1) Let G ˆ C3 ˆ ka: a3 ˆ 1l. C) be the representation for which  ar ˆ 0 À1 1 À1  (see Exercise 3. b: a5 ˆ b2 ˆ 1. by Corollary 9. ùÀ1  br ˆ 0 1  1 X 0 commutes with both ar and br. and let r: G 3 GL (2. C) for which  ar ˆ Assume that the matrix  Aˆ á ã â ä  ù 0  0 . Corollary 9. and let ù ˆ e2ðia5 . Check that there is a representation r: G 3 GL (2. Since the matrix  0 À1 1 À1  commutes with all gr ( g P G).2).2. if and only if ( gr)A ˆ A( gr) for all g P GX The result now follows from Schur's Lemma 9. j 9. The fact that (ar)A ˆ A(ar) forces ⠈ 㠈 0. . (2) Let G ˆ D10 ˆ ka.3.3 implies that r is reducible. 3 C n r . Since G is abelian. .Schur's Lemma Representation theory of ®nite abelian groups 81 Let G be a ®nite abelian group. we deduce that dim V ˆ 1. with g in i ˆ 1 and g i g j ˆ g j g i for all i. .6. let c i be a generator for C n i . and for 1 < i < r.5 Proposition If G is a ®nite abelian group. . X X X . We shall not prove it here. . Let G ˆ C n1 3 . Write g i ˆ (1. Pick x P G. say ë x 1 V . By Theorem 9. C) be an irreducible representation of G (ci in ith position)X . then every irreducible CG-module has dimension 1. jX Now let r: G 3 GL (n. X X X . g r i. 1) Then G ˆ h g1 . Thus we have proved 9. but refer you to Chapter 9 of the book of J. B. Fraleigh listed in the Bibliography.1(2). By Schur's Lemma 9. v gx ˆ vxg for all g P G. this endomorphism is a scalar multiple of the identity 1 V . We shall determine the irreducible representations of all direct products C n1 3 C n2 3 X X X 3 C n r where n1 . n r are positive integers. 9. and hence the endomorphism v 3 vx of V is a CG-homomorphism. this covers the irreducible representations of all ®nite abelian groups. ci . Thus vx ˆ ë x v for all v P V X This implies that every subspace of V is a CG-submodule.6 Theorem Every ®nite abelian group is isomorphic to a direct product of cyclic groups. The next result is a major structure theorem for ®nite abelian groups. X X X . As V is irreducible. . . and let V be an irreducible CGmodule. g2 l are V1 . . ë r i determine r.82 Representations and characters of groups over C. there exists ë i P C such that g i r ˆ (ë i ) (where of course (ë i ) is a 1 3 1 matrix).ë r of G constructed above are irreducible and have degree 1. and every irreducible representation of G over C is equivalent to precisely one of them. . .ë r X Conversely. There are |G| of these representations. where ak rù j ˆ (ù jk ) (0 < k < n À 1)X (2) The four irreducible CG-modules for G ˆ C2 3 C2 ˆ k g1 . V2 . that is. . and no two of them are equivalent. . As g i has order n i . i r . and then (9X7) i i gr ˆ ( g 11 X X X g irr )r ˆ (ë11 X X Xë irr )X For a representation r of G satisfying (9. . V3 . n r such representations. . Also. write r ˆ rë1 . . v2 g 1 ˆ v2 . . . we have g ˆ g11 . . given any n i th roots of unity ë i (1 < i < r). 2. v4 g 1 ˆ Àv4 . v2 g 2 ˆ Àv2 . the values ë1 .XXX. There are n1 n2 . . . 9. The representations rë1 .8 Theorem Let G be the abelian group C n1 3 . . 3. . We have proved the following theorem. since for g P G. where Vi is a 1-dimensional space with basis v i (i ˆ 1. 4) and v1 g 1 ˆ v1 .9 Examples (1) Let G ˆ C n ˆ ka: a n ˆ 1l. so for 1 < i < r. we have ë in i ˆ 1. . 9. Then n ˆ 1 by Proposition 9. i r . The n irreducible representations of G over C are rù j (0 < j < n À 1). the function i i g 11 X X X g irr 3 (ë11 X X Xë irr ) is a representation of G. v3 g 1 ˆ Àv3 . and put ù ˆ e2ðia n . ë i is an n i th root of unity. . g irr for some integers i1 . v4 g 2 ˆ Àv4 X . v1 g 2 ˆ v2 .7) for all i1 . V4 . 3 C n r .XXX. .5. v3 g 2 ˆ v3 . it is easy to check that Z(CG) is a subspace of CG. the result follows from (9. . If g P G. by Proposition 9.7. V ˆ U1 È X X X È Ur . Then for each i. let u i be a vector spanning Ui. If g has order n. there exists an integer m i such that ui g ˆ ù m i u i X Thus if B is the basis u1 . As V is also a C H-module. written Z(CG). we shall see that Z(CG) plays a crucial role in . then there is a basis B of V such that the matrix [ g]B is diagonal. For abelian groups G. . and let V be a non-zero C H-module.10). a direct sum of irreducible C H-submodules Ui of V. Proof Let H ˆ k gl.11 Proposition Let G be a ®nite group and V a CG-module. the centre Z(CG) is the whole group algebra. By Theorem 8. The centre of the group algebra CG.Schur's Lemma Diagonalization 83 Let H ˆ k gl be a cyclic group of order n. is de®ned by Z(CG) ˆ fz P CG: zr ˆ rz for all r P CGgX Using (2. Put ù ˆ e2ðia n . . u r of V. then H m1 I ù 0 f g FF (9X10) [ g]B ˆ d eX F 0 ù mr The following useful result is an immediate consequence of this. For arbitrary groups G. 9.12 De®nition Let G be a ®nite group.5. Each Ui has dimension 1. then the entries on the diagonal of [ g]B are nth roots of unity. . 9.5). j Some further applications of Schur's Lemma Our next application concerns an important subspace of the group algebra CG. then ˆ h P Z(CG)X To see this. Then there exists ë P C such that vz ˆ ëv for all v P V X Proof For all r P CG and v P V. bÀ1 ab ˆ aÀ1 l.14 Proposition Let V be an irreducible CG-module. . its dimension is equal to the number of irreducible representations of G ± see Chapter 15). this CG-homomorphism is equal to ë1 V for some ë P C. Indeed. 9. 1 ‡ a ‡ a2 and 1 ‡ a ‡ a2 ‡ b ‡ ab ‡ a2 b lie in Z(CG).84 Representations and characters of groups the study of representations of G (for example. so the elements 1. j Some elements of the centre of CG are provided by the centre of G. b: a3 ˆ b2 ˆ 1. We shall see later that these elements in fact form a basis of Z(CG). write z ˆ h.13 Example € The elements 1 and gPG g lie in Z(CG). then {1}. and let z P Z(CG). For example. 9. which we now de®ne. kal and G are normal subgroups of G. ˆ ˆ g À1 zg ˆ g À1 hg ˆ h ˆ z. Consequently zr ˆ rz for all r P CG.1(2). and the result follows. if G ˆ D6 ˆ ka. By Schur's Lemma 9. We use Schur's Lemma to prove the following important property of the elements of Z(CG). hP H hP H hP H € hP H and so zg ˆ gz. we have vrz ˆ vzr. and hence the function v 3 vz is a CG-homomorphism from V to V. if H is any normal subgroup of G. Then for all g P G. 16 Proposition If there exists a faithful irreducible CG-module. the function z 3 ëz (z P Z(G)) is an injective homomorphism from Z(G) into the multiplicative group Cà of non-zero complex numbers.14. there is no faithful irreducible CG-module unless G is cyclic.17 Example If G is an abelian group. they . Indeed. For example. 9. In particular. there exists ë z P C such that vz ˆ ë z v for all v P V X Since V is faithful.6 that for every ®nite group G there is a faithful CG-module. Although we have seen in Proposition 6. Therefore Z(G)  {ë z : z P Z(G)}.Schur's Lemma 9. it is not necessarily the case that there is a faithful irreducible CG-module.9(2)). is cyclic (see Exercise 1. which. is de®ned by Z(G) ˆ fz P G: zg ˆ gz for all g P GgX 85 Clearly Z(G) is a normal subgroup of G. The irreducible representations of non-abelian groups are more dif®cult to construct than those of abelian groups. and hence by Proposition 9. then G ˆ Z(G).6. written Z(G). C2 3 C2 has no faithful irreducible representation (compare Example 9. If z P Z(G) then z lies in Z(CG). Proof Let V be a faithful irreducible CG-module.16 is false. 9. being a ®nite subgroup of Cà . we give an example of a group G such that Z(G) is cyclic but there exists no faithful irreducible CG-module. since in Exercise 25. and so by Proposition 9. j We remark that the converse of Proposition 9.15 De®nition The centre of G. then Z(G) is cyclic.16. the following result shows that the existence of a faithful irreducible CG-module imposes a strong restriction on the structure of G.7). and is a subset of Z(CG). let v i be a vector spanning V i . Hence G is abelian. we can write CG ˆ V1 È X X X È Vn . Proof By Theorem 8. For 1 < i < n.5. C3 and C2 3 C2 .18 Proposition Suppose that G is a ®nite group such that every irreducible CG-module has dimension 1. call it B . Since the representation g 3 [ g]B ( g P G) of G is faithful (see Proposition 6. 9. 3. The elements of Z(CG) act as scalar multiples of the identity on all irreducible CGmodules. Then G is abelian. For all x. The centre Z(CG) of the group algebra CG consists of those elements which commute with all elements of CG. All irreducible CG-modules for a ®nite abelian group G have dimension 1. the matrices [x]B and [ y]B are diagonal.86 Representations and characters of groups do not all have degree 1. y P G.6). . Also. . j Summary of Chapter 9 1. Exercises for Chapter 9 1. and there are precisely |G| of them.7. where each Vi is an irreducible CG-submodule of the regular CGmodule CG. Schur's Lemma states that every CG-homomorphism between irreducible CG-modules is either zero or a CG-isomorphism. we deduce that x and y commute. . . v n is a basis of CG. since we are assuming that all irreducible CG-modules have dimension 1. 2. . Then dim Vi ˆ 1 for all i. and hence they commute. Then v1 . as required. Write down the irreducible representations over C of the groups C2 . as is shown by the following converse to Proposition 9. the only CG-homomorphisms from an irreducible CG-module to itself are scalar multiples of the identity. irreducible. Can G have a faithful representation of degree less than r? 4. Let G ˆ C4 3 C4 . (a) Find a non-trivial irreducible representation r of G such that g2 r ˆ (1) for all g P G. bÀ1 ab ˆ aÀ1 l.14. (Compare Proposition 9. b: a3 ˆ b2 ˆ 1. Show that if V is an irreducible CG-module.Schur's Lemma 87 2. b ‡ ù2 ab ‡ ùa2 b) (see Exercise 6. bÀ1 ab ˆ aÀ1 l. then there exists ë P C such that 2ˆ 3 v g ˆ ëv for all v P V X gPG 6. (b) Prove that there is no irreducible representation ó of G such that gó ˆ (À1) for all elements g of order 2 in G. Write ù ˆ e2ðia3 . bó 4 À5 M( gr) ˆ ( gr)M for all g P G. . (b) Find ë P C such that w(a ‡ aÀ1 ) ˆ ëw for all w P W. . where   À5 6 ˆ X À4 5 5. b: a4 ˆ b2 ˆ 1. and let W be the irreducible CG-submodule of the regular CGmodule de®ned by W ˆ sp (1 ‡ ù2 a ‡ ùa2 . Let G ˆ D6 ˆ ka.6). br ˆ X À5 7 À4 5 Find all 2 3 2 matrices M such that Hence determine whether or not r is Do the same for the representation   5 À6 aó ˆ .) . Prove that G has a faithful representation of degree r. 3. Let G be the ®nite abelian group C n1 3 . (a) Show that a ‡ aÀ1 P Z(CG). 3 C n r . Suppose that G ˆ D8 ˆ ka. ó of G. Check that there is a representation r of G over C such that     À7 10 À5 6 ar ˆ . (c) C2 3 D8 . (b) D8 . Which of the following groups have a faithful irreducible representation? (a) C n (n a positive integer). . (d) C3 3 D8.88 Representations and characters of groups 7. . this is not really a practical way of ®nding the irreducible CG-modules. . However. it is suf®cient to decompose CG as a direct sum of irreducible CG-submodules. . We shall show in this chapter that every irreducible CG-module is isomorphic to one of the CG-modules U1. Consider CG as the regular CG-module. As a consequence. Irreducible submodules of CG We begin with another consequence of Maschke's Theorem. there are only ®nitely many non-isomorphic irreducible CG-modules (a result which has already been established for abelian groups in Theorem 9. unless G is a small group.8). we can write CG ˆ U1 È X X X È Ur where each Ui is an irreducible CG-module. Ur. Also.7. 10. Proof Since Ker W is a CG-submodule of V by Proposition 7. By Theorem 8.2.1 Proposition Let V and W be CG-modules and let W: V 3 W be a CG-homomorphism. to ®nd all irreducible CG-modules. there is by Maschke's Theorem a CG-submodule U of V such that V ˆ Ker W È U . in theory. De®ne a function W: U 3 Im W by uW ˆ uW (u P U )X 89 . Then there is a CG-submodule U of V such that V ˆ Ker W È U and U  Im W.10 Irreducible modules and the group algebra Let G be a ®nite group and CG be the group algebra of G over C. . Therefore U  Ui. so w ˆ vW for some v P V. Now ð i is a CG-homomorphism (see Proposition 7. . However. If U is any irreducible CG-submodule of V. since W is a CG-homomorphism. we have ð i Tˆ 0. with basis v1 . As U and Ui are irreducible.4 De®nitions (1) If V is a CG-module and U is an irreducible CG-module. 10. Schur's Lemma 9. . and write V ˆ U1 È X X X È Us . È Us (each Ui irreducible) without U being equal to any Ui. If u P Ker W then u P Ker W ’ U ˆ {0}. Write v ˆ k ‡ u with k P Ker W. Now let w P Im W. then we say that U is a composition factor of V if V has a CG-submodule which is isomorphic to U. . ‡ us for unique vectors ui P Ui (1 < i < s). a direct sum of irreducible CG-submodules Ui. . Then w ˆ vW ˆ kW ‡ uW ˆ uW ˆ uWX Therefore Im W ˆ Im W. then U  Ui for some i. Choosing i such that ui Tˆ 0 for some u P U. We have now established that W: U 3 Im W is an invertible CG-homomorphism. Clearly W is a CG-homomorphism. as the following example shows. as required. . u P U. such that v g ˆ v for all v P V and g P G. U ˆ sp (v1 ‡ v2 ) is an irreducible CG-submodule which is not equal to U1 or U2. and ð i Tˆ 0.3 Example Let G be any group and let V be a 2-dimensional CG-module. where U1 ˆ sp (v1 ) and U2 ˆ sp (v2 ) are irreducible CG-submodules.11). Thus U  Im W. we have u ˆ u1 ‡ . De®ne ð i : U 3 Ui by setting uð i ˆ ui . Proof For u P U. j Of course it can happen that U is an irreducible CG-submodule of U1 È .90 Representations and characters of groups We show that W is a CG-isomorphism from U to Im W. 10. hence Ker W ˆ {0}. 10. v2 .2 Proposition Let V be a CG-module. Then V ˆ U1 È U2 .1(1) implies that ð i is a CG-isomorphism. (rs)W ˆ w(rs) ˆ (wr)s ˆ (rW)sX By Proposition 10. which shows that every irreducible CG-module is a composition factor of the regular CG-module.6).7 Corollary If G is a ®nite group.5 Theorem Regard CG as the regular CG-module. so is U. and the result is proved. then W  Ui. since W is irreducible. By Proposition 10. there is a CG-submodule U of CG such that CG ˆ U È Ker W and U  Im W ˆ W X As W is irreducible.2 we have U  Ui for some i. since for r. We record this fact in the following corollary. then there are only ®nitely many non-isomorphic irreducible CG-modules. W is a CG-homomorphism. Then every irreducible CG-module is isomorphic to one of the CG-modules Ui.1. and choose a non-zero vector w P W. 10. Proof Let W be an irreducible CG-module.Irreducible modules and the group algebra 91 (2) Two CG-modules V and W are said to have a common composition factor if there is an irreducible CG-module which is a composition factor of both V and W.5 shows that there is a ®nite set of irreducible CGmodules such that every irreducible CG-module is isomorphic to one of them. and write CG ˆ U 1 È X X X È U r . . Observe that {wr: r P CG} is a CG-submodule of W. a direct sum of irreducible CG-submodules. s P CG. it follows that (10X6) W ˆ fwr: r P CGgX Now de®ne W: CG 3 W by rW ˆ wr (r P CG)X Clearly W is a linear transformation. Moreover. and Im W ˆ W by (10. We now come to the main result of the chapter. j Theorem 10. 10. v2 is a basis of CG. The irreducible representation of G corresponding to Ui is the representation rù i of Example 9. v1 . We decompose CG as a direct sum of irreducible CG-submodules. 10. 1. De®ne v0 . (2) Let G ˆ D6 ˆ ka. and hence CG ˆ U0 È U 1 È U2 . As in (1) above. v2 b ˆ w1 . w2 b ˆ v1 X . v1 ˆ 1 ‡ ù2 a ‡ ùa2 .9(1). v1 b ˆ w2 . v2 ˆ 1 ‡ ùa ‡ ù2 a2 . 2X Hence Ui is a CG-submodule of CG for i ˆ 0. to ®nd all the irreducible CG-modules we need only decompose the regular CG-module as a direct sum of irreducible CG-submodules. and write ù ˆ e2ðia3 . 2. note that v0 b ˆ w 0 . Then v1 a ˆ a ‡ ù2 a2 ‡ ù1 ˆ ùv1 . v2 P CG by v0 ˆ 1 ‡ a ‡ a2 . w0 ˆ bv0 w1 ˆ bv1 .92 Representations and characters of groups According to Theorem 10. b: a3 ˆ b2 ˆ 1. 1. a direct sum of irreducible CG-submodules Ui. however. U1 or U2. Next. bÀ1 ab ˆ aÀ1 l. By Theorem 10. 1.5. Let ù ˆ e2ðia3 and de®ne v0 ˆ 1 ‡ a ‡ a2 . 1. w1 b ˆ v2 . and similarly vi a ˆ ùi vi for i ˆ 0. v2 ˆ 1 ‡ ùa ‡ ù2 a2 . v1 . w0 b ˆ v0 . every irreducible CG-module is isomorphic to U0. 2. v1 ˆ 1 ‡ ù2 a ‡ ùa2 .8 Examples (1) Let G ˆ C3 ˆ ka: a3 ˆ 1l. w2 ˆ bv2 X (ˆ b ‡ ba ‡ ba2 ). this is not a practical method for studying CG-modules in general. 2.5. and so sp (v i ) and sp (wi ) are Ckal-modules. It is easy to check that v0 . and let Ui ˆ sp (v i ) for i ˆ 0. We now do this for a couple of examples. v i a ˆ ù i v i for i ˆ 0. r2 : a 3 (1). w2 is a basis of CG. w0 ). w2 ) and sp (v2 . Let G ˆ C4 . Correspondingly. But U3  U4 (there is a CG-isomorphism sending v1 3 w1 .5(2). Every irreducible CG-module occurs as a composition factor of the regular CG-module. w1.8(1). U2 and U3. w1 ) are irreducible. every irreducible representation of D6 over C is equivalent to precisely one of the following: r1 : a 3 (1).) . There are only ®nitely many non-isomorphic irreducible CGmodules. w0. and hence CG ˆ U1 È U2 È U3 È U4 .Irreducible modules and the group algebra 93 Therefore. a direct sum of irreducible CG-submodules. We conclude from Theorem 10.b3 X 1 0 0 ùÀ1 Summary of Chapter 10 1. namely U1. the CG-submodules U3 ˆ sp (v1 . b 3 (1). sp (v1 . w1 ) are Ckbl-modules. b 3 (À1). v1 .     0 1 ù 0 r3 : a 3 . (Hint: copy the method of Example 10. By the argument in Example 5. Now v0. Let G be a ®nite group. the other 1-dimensional Ui. sp (v0 .5 that there are exactly three nonisomorphic irreducible CG-modules. w0 ) is reducible. Express CG as a direct sum of irreducible CGsubmodules. v2 . Exercises for Chapter 10 1. Find a CG-submodule of CG which is isomorphic to the trivial CG-module. Is there only one such CGsubmodule? 2. However. 2. Note that U1 is the trivial CG-module. and hence are CG-submodules of CG. w2 3 v2 ). as U 1 ˆ sp(v0 ‡ w0 ) and U2 ˆ sp(v0 À w0 ) are CG-submodules. sp(v0 . and U1 is not isomorphic to U2. w2 ) and U4 ˆ sp (v2 . u2 b ˆ u2 .5(2). Suppose that V is a where U1 and U2 are CG-submodule U of isomorphic to both of non-zero CG-module such that V ˆ U1 È U2. but is them. isomorphic CG-modules. such that u2 a ˆ Àu2 . bÀ1 ab ˆ aÀ1 l. v2 and v1 a ˆ iv1 . Thus V has basis v1 . 5. and u3 b ˆ Àu3 X 4. b2 ˆ a2 . v1 b ˆ v2 . b: a4 ˆ b2 ˆ 1. b: a4 ˆ 1. v2 a ˆ Àiv2 . sp (u2 ) and sp (u3 ). . u3 a ˆ Àu3 . and let V be the CG-module given in Example 4.94 Representations and characters of groups 3. v2 b ˆ Àv1 X Show that V is irreducible. u1 b ˆ Àu1 X Find also 1-dimensional CG-submodules. Let G ˆ D8 ˆ ka. Use the method of Example 10. of CG such that u1 a ˆ u1 . sp (u1 ) say. and ®nd a CG-submodule of CG which is isomorphic to V. bÀ1 ab ˆ aÀ1 l. Let G ˆ Q8 ˆ ka. Show that there is a V which is not equal to U1 or U2.8(2) to ®nd all the irreducible representations of D8 over C. Find a 1-dimensional CG-submodule. 6. 9). In Theorem 10.9 is based on a study of the vector space of CG-homomorphisms from one CG-module to another. With these de®nitions. Our proof of Theorem 11. de®ne W ‡ ö and ëW by v(W ‡ ö) ˆ vW ‡ vö. ö P HomCG (V . W ) and ë P C.11 More on the group algebra We now go further into the structure of the group algebra CG of a ®nite group G.5 we proved that every irreducible CG-module U is isomorphic to one of the Ui. we write CG ˆ U 1 È X X X È U r . The question arises: how many of the Ui are isomorphic to U ? There is an elegant and signi®cant answer to this question: the number is precisely dim U (see Theorem 11. ëW P HomCG (V . a direct sum of irreducible CG-modules Ui. Then W ‡ ö. We begin our study of the vector space HomCG (V. W ). As in Chapter 10. W ) for the set of all CG-homomorphisms from V to W.1 De®nition Let V and W be CG-modules. W ) is a vector space over C. W) with an easy consequence of Schur's Lemma. We write HomCG (V . 95 . v(ëW) ˆ ë(vW) for all v P V . The space of CG-homomorphisms 11. De®ne addition and scalar multiplication on HomCG (V . W ) as follows: for W. it is easily checked that HomCG (V. Therefore X is a common composition factor of V and W. V1 . Then V and W have a common composition factor. and so HomCG (V . (2) dim (HomCG (V1 È V2 . W 1 . so by Schur's Lemma 9.4. W )) ˆ 0. W ) ˆ fëW: ë P Cg. W ).1(1). W ) Tˆ f0g. The key step is the following proposition. there exists ë P C such that öWÀ1 ˆ ë1 V X Then ö ˆ ëW. . W1 È W2 )) ˆ dim (HomCG (V. and suppose that HomCG (V .1(1) implies that XW  X. and let W: V 3 W be a CG-isomorphism. If ö P HomCG (V . Then (1) dim (HomCG (V. 11. then öWÀ1 is a CG-isomorphism from V to V.3 Proposition Let V and W be CG-modules. if V  W . Then & 1. Now suppose that V  W.96 Representations and characters of groups 11. a 1-dimensional space.1(2). recall the de®nition of a composition factor of a CG-module from 10. W1 )) ‡ dim (HomCG (V. W )). dim (HomCG (V . W2 )). V2 and W .4 Proposition Let V . by Maschke's Theorem. Then V ˆ Ker W È U for some non-zero CG-module U. W )) ˆ dim (HomCG (V1 . 11. if V T W X Proof If V T W then this is immediate from Schur's Lemma 9. W)) ‡ dim (HomCG (V2 . W 2 be CG-modules. Let X be an irreducible CG-submodule of U. Schur's Lemma 9. j For the next result. W ) in general. Since XW Tˆ {0}. Proof Let W be a non-zero element of HomCG (V .2 Proposition Suppose that V and W are irreducible CG-modules. j The next few results show how to calculate the dimension of HomCG (V . W ). W1 ) and HomCG (V. then vWð1 ˆ 0 and vWð2 ˆ 0 for all v P V. Therefore W ˆ 0. 2) lies in HomCG (V1 È V2 . W2 ) (see Exercise 7. If W P HomCG (V . the function ö: v 3 vö1 ‡ vö2 (v P V ) lies in HomCG (V. 2. W2 ). Wð2 ) (W P HomCG (V . 2). W 1 ) and Wð2 P HomCG (V. Now let h be the function from HomCG (V1 È V2 . Wi ) (i ˆ 1. W Vi is the function v i W Vi ˆ v i W (v i P Vi )X to Then W Vi P HomCG (V i . We have established that f is an invertible linear transformation from HomCG (V. Hence f is surjective. and (2) follows. W ) for i ˆ 1. 2). then Wð1 P HomCG (V . Given Clearly h is an injective linear HomCG (Vi .1). Hence h is surjective. W V2 ) (W P HomCG (V1 È V2 . j . We have shown that h is an invertible linear transformation. so Ker f ˆ {0} and f is injective. for all w1 P W 1 .More on the group algebra Proof (1) De®ne W 1 È W 2 3 W 2 by the functions ð1 : W1 È W2 3 W1 (w1 ‡ w2 )ð2 ˆ w2 and 97 ð2 : (w1 ‡ w2 )ð1 ˆ w1 . If W P Ker f. ö2 ) under h. w2 P W 2 . W1 È W2 ). the function ö: v1 ‡ v2 3 v1 ö1 ‡ v2 ö2 öi P (v i P Vi for i ˆ 1. 2) to be the restriction of W to V i . We now de®ne a function f from HomCG (V. W ) È HomCG (V2 . W1 È W2 ) to the (external) direct sum of HomCG (V. ö2 ). W ) (i ˆ 1. We show that f is invertible. and (1) follows.11. Consequently these two vector spaces have equal dimensions. W 1 È W 2 ))X Clearly f is a linear transformation. so vW ˆ vW(ð1 ‡ ð2 ) ˆ 0. ð1 and ð2 are CG-homomorphisms. and the image of ö under f is (ö1 . (2) For W P HomCG (V1 È V2 . Given ö i P HomCG (V. W1 È W2 ) to HomCG (V. W2 ) by f : W 3 (Wð1 . W ) HomCG (V1 . By Proposition 7. W ) and has image (ö1 . W ))X transformation. W ) which is given by h: W 3 (W V1 . W1 ) È HomCG (V. de®ne W Vi : Vi 3 W (i ˆ 1. that is. W ). W 1 È W 2 ). 1 < j < s). we can ®nd dim (HomCG (V.4. W )) in general. By an obvious induction using Proposition 11. È Ws )) ˆ s ˆ jˆ1 dim (HomCG (V. and using Proposition 11. . . . In the following corollary we single out the case where one of the CGmodules is irreducible. W1 È . (2) dim(HomCG (V1 È F F F ÈV r . Vi. 11.2. È Vr . U i ))X . Wj )). V )) ˆ s ˆ iˆ1 s ˆ iˆ1 dim (HomCG (Ui . W)). we have (11. Then the dimensions of HomCG (V . and dim (HomCG (W . . . W )) ˆ r ˆ iˆ1 dim (HomCG (Vi. W1 È . Proof By (11. . These in turn imply (3) dim (HomCG (V1 È .98 Representations and characters of groups Now suppose that we have CG-modules V. Let W be any irreducible CG-module.5). Wj (1 < i < r.6 Corollary Let V be a CG-module with V ˆ U1 È X X X È Us . By applying (3) when all Vi and Wj are irreducible.5) (1) dim (HomCG (V. W ) and HomCG (W . where each Ui is an irreducible CG-module. dim (HomCG (V . W )) ˆ dim (HomCG (W . W )). È Ws )) ˆ r s ˆˆ iˆ1 jˆ1 dim (HomCG (Vi. Wj )). W. V ) are both equal to the number of CG-modules Ui such that Ui  W. U )) ˆ dim U X Proof Let d ˆ dim U. & 99 1. . 11. s P CG. W )) ˆ dim (HomCG (W .More on the group algebra And by Proposition 11. . Then 1ö ˆ ë1 u1 ‡ X X X ‡ ë d u d for some ë i P C. U3 )) ˆ dim (HomCG (U3 . ud of U.5 to ®nd bases for these two vector spaces of CG-homomorphisms. .6. dim (HomCG (Ui . we saw in Example 10. Since ö is a CG-homomorphism. Suppose that ö P HomCG (CG.6.8(2) that CG ˆ U1 È U2 È U3 È U4 .2.7 Example For G ˆ D6. The next proposition investigates the space of CG-homomorphisms from the regular CG-module to any other CG-module.8 Proposition If U is a CG-module. Choose a basis u1 . then dim (HomCG (CG. if U i T W X j a direct sum of irreducible CG-submodules. (rs)ö i ˆ ui (rs) ˆ (ui r)s ˆ (rö i )sX We shall prove that ö1 . . Ui )) ˆ The result follows. . Thus by Corollary 11. if U i  W . it will give the main result of this chapter. with U3  U4 but U3 not isomorphic to U1 or U2. ö d is a basis of HomCG (CG. CG)) ˆ 2X You are asked in Exercise 11. U ). de®ne ö i : CG 3 U by rö i ˆ ui r (r P CG)X Then ö i P HomCG (CG. U) since for all r. . we have dim (HomCG (CG. . for all r P CG we have . . U ). 0. 11. When combined with Corollary 11. For 1 < i < d. U ). . . U2 are non-isomorphic 1-dimensional CG-modules.8. . and . U )). where U1. j 11. . Now assume that Therefore ö1 . . . . HomCG (CG. this is equal to the number of Ui with Ui  U. which therefore has dimension d. then the number of CG-modules Ui with Ui  U is equal to dim U. U ). j We now come to the main theorem of the chapter. ‡ ë d ö d . If U is any irreducible CG-module. dim U ˆ dim (HomCG (CG. and by Corollary 11.9 Theorem Suppose that CG ˆ U1 È X X X È Ur .100 Representations and characters of groups rö ˆ (1r)ö ˆ (1ö)r ˆ ë1 u1 r ‡ X X X ‡ ë d u d r ˆ r(ë1 ö1 ‡ X X X ‡ ë d ö d )X Hence ö ˆ ë1 ö1 ‡ . which tells us how often each irreducible CG-module occurs in the regular CGmodule. Hence ö1 . ö d span ë1 ö1 ‡ X X X ‡ ë d ö d ˆ 0 (ë i P C)X Evaluating both sides at the identity 1.6. . Proof By Proposition 11. a direct sum of irreducible CG-submodules.10 Example Recall again from Example 10. which forces ë i ˆ 0 for all i.8(2) that if G ˆ D6 then CG ˆ U1 È U 2 È U3 È U 4 . . ö d is a basis of HomCG (CG. 11. . we have 0 ˆ 1(ë1 ö1 ‡ X X X ‡ ë d ö d ) ˆ ë1 u1 ‡ X X X ‡ ë d u d . write d i ˆ dim V i .) 11. X X X . (By Corollary 10. V k form a complete set of non-isomorphic irreducible CG-modules if every irreducible CG-module is isomorphic to some V i . .9.9: U1 occurs once. .11 De®nition We say that the irreducible CG-modules V1 .12 Theorem Let V1 . V k are isomorphic. . dk be the dimensions of all the irreducible CG-modules. Then k ˆ (dim Vi )2 ˆ jGjX iˆ1 Proof Let CG ˆ U1 È .9 concerning the dimensions of all irreducible CG-modules. the result follows. and let d1. 11. U3 occurs twice. and no two of V1 .13 Example Let G be a group of order 8. 11. a direct sum of irreducible CG-submodules. U4 are isomorphic irreducible 2-dimensional CG-modules. for each i. È Ur. For 1 < i < k. dim U1 ˆ 1. By Theorem 11. . .7. Therefore dim CG ˆ dim U 1 ‡ X X X ‡ dim Ur ˆ k ˆ iˆ1 k ˆ iˆ1 d i (dim Vi ) ˆ d2X i j As dim CG ˆ |G|. for any ®nite group G there exists a complete set of non-isomorphic irreducible CG-modules.More on the group algebra 101 U3. X X X . dim U3 ˆ 2X We conclude the chapter with a signi®cant consequence of Theorem 11. k ˆ iˆ1 d 2 ˆ 8X i . . the number of CG-modules Uj with Uj  Vi is equal to di .12. This illustrates Theorem 11. By Theorem 11. X X X . U2 occurs once. dim U2 ˆ 1. V k form a complete set of non-isomorphic irreducible CG-modules. 1. CG). 2. dim (HomCG (V1 È X X X È V r . and the second when G ˆ D8 (see Exercise 10.5). 1. Let G be a ®nite group. Summary of Chapter 11 1. 1. If G is a group of order 12. then k ˆ (dim Vi )2 ˆ jGjX iˆ1 Exercises for Chapter 11 1. Let CG ˆ U1 È . combined with Theorem 11.102 Representations and characters of groups Observe that the trivial CG-module is irreducible of dimension 1. 4.3. . If G is a non-abelian group of order 6.) 3. 1. If V1 . U )) ˆ dim U . We shall see later that dim Vi divides |G| for all i. Find a basis for HomCG (CG. Then the number of Ui with Ui  U is equal to dim U. (Hint: use Exercise 5. . and so di ˆ 1 for some i. V k is a complete set of non-isomorphic irreducible CGmodules. 1. X X X . 1.12. . dim (HomCG (CG. what are the possible degrees of all the irreducible representations of G? Find the degrees of the irreducible representations of D12. W 1 È X X X È W s )) ˆ r s ˆˆ iˆ1 jˆ1 and dim (HomCG (Vi . dk are 1. and let U be any irreducible CG-module. . a direct sum of irreducible CG-modules. 1. . 1 1. . 3. 1. and this fact. 1. 2X Both these possibilities do occur: the ®rst holds when G is an abelian group (see Proposition 9. ®nd the dimensions of all the irreducible CG-modules. is quite a powerful tool in ®nding the dimensions of irreducible CG-modules. W j ))X 2. È Ur.4). Hence the possibilities for d1. . 10. and let V. 6. as de®ned in 4. Find a basis for HomCG (CG. U3 ) and a basis for HomCG (U3. as in Example 10. Let G ˆ D6 and let CG ˆ U1 È U2 È U3 È U4. Vi ))X €k Show that dim (HomCG (V . a direct sum of irreducible CG-modules. U) has dimension 1. Vi )) and ei ˆ dim (HomCG (W . W be arbitrary CG-modules. 5. show that HomCG (V. Suppose that G ˆ Sn and V is the n-dimensional permutation module for G over C. If U is the trivial CGmodule. W )) ˆ iˆ1 d i ei . CG). Assume that for 1 < i < k.More on the group algebra 103 4.8(2). X X X . d i ˆ dim (HomCG (V . . Let V1 . V k be a complete set of non-isomorphic irreducible CG-modules. After de®ning conjugacy classes. Throughout the chapter.12 Conjugacy classes We take a break from representation theory to discuss some topics in group theory which will be relevant in our further study of representations. We say that x is conjugate to y in G if y ˆ g À1 xg for some g P GX The set of all elements conjugate to x in G is x G ˆ f g À1 xg: g P Gg. At the end of the chapter we prove a result linking the conjugacy classes of a group to the structure of its group algebra.1 De®nition Let x. 12. Then there exist g. and is called the conjugacy class of x in G. then either x G ˆ y G or x G ’ y G is empty. and pick z P x G ’ y G. Our ®rst result shows that two distinct conjugacy classes have no elements in common. Proof Suppose that x G ’ y G is not empty. G is a ®nite group. h P G such that z ˆ gÀ1 xg ˆ hÀ1 yhX 104 . symmetric and alternating groups. y P G.2 Proposition If x. y P G. Conjugacy classes 12. we develop enough theory to determine the conjugacy classes of dihedral. b: a3 ˆ b2 ˆ 1. 12. a2 gX Also. ab. a2 g. so bG ˆ fb. we have aG ˆ fa. .3 is to observe that conjugacy is an equivalence relation. ‘ x G . fb.3 Corollary Every group is a union of conjugacy classes. and that the conjugacy classes are the equivalence classes. Another way of seeing this Corollary 12. ab. .5 Examples (1) For every group G. ab. a2 bgX Thus the conjugacy classes of G are f1g. xl representatives of the conjugacy classes of G. 1 G ˆ {1} is a conjugacy class of G. and bÀ1 ab ˆ a2 . and so xG ˆ yG . where k ˆ hgÀ1 . x G are l l distinct. . a2 . So a P x G A a ˆ bÀ1 xb A a ˆ cÀ1 yc AaP y X G 105 for some b P G where c ˆ kb A a ˆ bÀ1 k À1 ykb Therefore x G # y G. . . where the conjugacy classes x1 . . 12. . The elements of G are 1. and distinct conjugacy classes are disjoint. . Similarly y G # x G (using y ˆ kxkÀ1 ). a2 bgX . . then we call x1 . G is the union of its conjugacy classes and so we deduce immediately 12. Since gÀ1 ag is a or a2 for every g P G. . a.4 De®nition G G If G ˆ x1 ‘ . aÀi ba i ˆ aÀ2i b for all integers i. j Since every element x of G lies in the conjugacy class x G (as x ˆ 1À1 x1 with 1 P G). a2 b. bÀ1 ab ˆ aÀ1 l. fa. (2) Let G ˆ D6 ˆ ka.Conjugacy classes Hence x ˆ ghÀ1 yhgÀ1 ˆ kÀ1 yk. b. Suppose that x is conjugate to y in G.1). g P G. Let x have order m. y r ˆ gÀ1 x r g Tˆ 1. so that y ˆ gÀ1 xg for some g P G.) It is easy to check that CG (x) is a subgroup of G (Exercise 12. b P G. j Conjugacy class sizes The next theorem determines the sizes of the conjugacy classes in G in terms of certain subgroups which we now de®ne. Then the size of the conjugacy class x G is given by jx G j ˆ jG: CG (x)j ˆ jGjajCG (x)jX In particular. written CG (x). 12. 12. The next proposition is often useful when calculating conjugacy classes. |x G | divides |G|. kxl # CG (x) for all x P G.8 Theorem Let x P G. r . then x n is conjugate to y n in G for every integer n. we have gÀ1 abg ˆ ( g À1 ag)( g À1 bg)X Hence gÀ1 x n g ˆ ( gÀ1 xg) n . that is. Proof Observe that for a. CG (x) ˆ f g P G: xg ˆ gxgX (So also CG (x) ˆ { g P G: gÀ1 xg ˆ x}. Then y m ˆ gÀ1 x m g ˆ 1. and for 0 . and so x G ˆ {x}. 12.6 Proposition Let x. y P G.106 Representations and characters of groups (3) If G is abelian then gÀ1 xg ˆ x for all x. is the set of elements of G which commute with x. . so y also has order m. m. Then y n ˆ gÀ1 x n g and therefore x n is conjugate to y n in G. If x is conjugate to y in G. and x and y have the same order. The centralizer of x in G. Hence every conjugacy class of G consists of just one element.7 De®nition Let x P G. Observe that x P CG (x) and indeed. it is convenient to consider separately the cases where n is odd and where n is even.15. Thus G ˆ ha. h P G. bÀ1 ab ˆ aÀ1 iX In ®nding the conjugacy classes of G. we may de®ne an injective function f from x G to the set of right cosets of CG (x) in G by f : g À1 xg 3 CG (x) g ( g P G)X Clearly f is surjective. xl be representatives of the conjugacy classes of G. Let G ˆ D2 n. the dihedral group of order 2n. and both | Z(G)| and |x G | divide |G|. . proving that |x G | ˆ |G:CG (x)|. We have now proved all parts of the following result. Since CG (ai ) contains kal. b: an ˆ b2 ˆ 1. j Before summarizing our results on conjugacy classes. as de®ned in 9. (1) n odd First consider ai (1 < i < n À 1). we have g À1 xg ˆ hÀ1 xh D hg À1 x ˆ xhg À1 D hg À1 P CG (x) D CG (x) g ˆ CG (x)hX 107 By dint of this.10 The Class Equation Let x1 . Then ˆ jx G j. . Hence f is a bijection.8 by ®nding the conjugacy classes of all dihedral groups. i Conjugacy classes of dihedral groups We illustrate the use of Theorem 12. . jG: CG (ai )j < jG: haij ˆ 2X . where Z(G) is the centre of G. jGj ˆ j Z(G)j ‡ i xiP Z(G) a for all g P G where |x G | i ˆ |G:CG (xi )|. . we make the observation that (12X9) jx G j ˆ 1 D g À1 xg ˆ x D x P Z(G). 12.Conjugacy classes Proof Observe ®rst that for g. a j (ab)aÀ j ˆ a2 j‡1 bX It follows that bG ˆ fa2 j b: 0 < j < m À 1g. As bÀ1 am b ˆ aÀ m ˆ am . aÀ1 }. aÀi gX Next. Thus CG (b) ˆ f1. and so |(ai ) G | > 2. (2) n even Write n ˆ 2m. That is.12) The dihedral group D2 n (n even. |bG | ˆ n. the centralizer of am in G contains both a and b. and CG (ai ) ˆ hai. (ai ) G ˆ {ai . aÀ m‡1 }. As n is odd. Using Theorem 12. a j baÀ j ˆ a2 j b. aÀ1 }. b}. bG ˆ fb. {a2 j b: 0 < j < m À 1}. aÀ( nÀ1)a2 }. . {a m }. bG must consist of the remaining n elements of G. X X X . Therefore the conjugacy class of am in G is just fam g. For every integer j. Since all the elements ai have been accounted for. (ab) G ˆ fa2 j‡1 b: 0 < j < m À 1gX Hence (12. {b.8. . {a. no element ai or ai b (with 1 < i < n À 1) commutes with b. so {ai . {a2 j‡1 b: 0 < j < m À 1}. . aÀi } # (ai ) G . a nÀ1 b}. CG (b) contains {1. {a mÀ1 . {a. ab. . . ab.108 Representations and characters of groups Also bÀ1 ai b ˆ aÀi . . . . {a( nÀ1)a2 . we have 2 > jG: CG (ai )j ˆ j(ai ) G j > 2X Hence equality holds here. and as bÀ1 ai b ˆ aÀi . . aÀi } for 1 < i < m À 1.8. As in case (1). n ˆ 2m) has precisely m ‡ 3 conjugacy classes: {1}. and hence CG (am ) ˆ G. (ai ) G ˆ fai .11) The dihedral group D2 n (n odd) has precisely 1(n ‡ 3) con2 jugacy classes: {1}. bgX Therefore by Theorem 12. . . . . a nÀ1 bgX We have shown (12. ai Tˆ aÀi . for 1 < i < n and i P A. and so by (12. y of the same cycle-shape. . . ik ) in Sn . say x ˆ (a1 X X X a k 1 ) X X X (c1 X X X ck s ). for g P Sn we have (12. . ik ) g ˆ (i1 g i2 g . with k1 > k2 > . . and note that x and gÀ1 xg have the same cycle-shape. . given any two permutations x. Proof Write A ˆ {i1 . 1 k 1 k (products of disjoint cycles). Write x ˆ (a1 X X X ak 1 )(b1 X X X bk 2 ) X X X (c1 X X X ck s ). . . .Conjugacy classes Conjugacy classes of S n 109 We shall later need to know the conjugacy classes of the symmetric group Sn . Then gÀ1 xg is the k-cycle (i1 g i2 g . a ig( gÀ1 xg) ˆ ixg ˆ igX Hence gÀ1 (i1 i2 . y ˆ (a9 X X X a9 1 ) X X X (c9 X X X c9 s ). .14). ir g( g À1 xg) ˆ i r xg ˆ i r‡1 g (or i1 g if r ˆ k)X Also. ik }. . . > ks . and let g P Sn . ks ) the cycle-shape of x. . Now consider an arbitrary permutation x P Sn . ik g). . . . .13. . 12.13 Proposition Let x be a k-cycle (i1 i2 . . a product of disjoint cycles. exists g P Sn sending j . as required.14) g À1 xg ˆ g À1 (a1 X X X a k 1 ) gg À1 (b1 X X X b k 2 ) g X X X g À1 (c1 X X X ck s ) g ˆ (a1 g X X X a k 1 g)(b1 g X X X b k 2 g) X X X (c1 g X X X c k s g)X We call (k1 . there a1 3 a9 . . On the other hand. ck s 3 c9 s . . 1 k g À1 xg ˆ yX We have proved the following result. By Proposition 12. . For ir P A. . Our ®rst observation is simple but crucial. ik g). 3.110 Representations and characters of groups 12. 4}. (1 2). (1 3)(2 4). 2) and is f(1 2)(3 4). (1 2 3). (The notation … n † means the binomial coef®cient 2 r n3a(r3(n À r)3). (2 3)} {(1 2 3). (1 3). (1 3 2)} Cycle-shape (1) (2) (3) (2) The conjugacy class of (1 2)(3 4) in S4 consists of all the elements of cycle-shape (2. 2) and there are six 4-cycles. and so on.) The number of 3-cycles is 4 3 2 (4 for the choice of ®xed point and 2 because there are two 3-cycles ®xing a given point). The number of 2-cycles is equal to the number of pairs that can be chosen from {1.15 Theorem For x P Sn . the conjugacy class x Sn of x in Sn consists of all permutations in Sn which have the same cycle-shape as x. the conjugacy class representatives g. 12. Similarly. (1 4)(2 3)gX (3) There are precisely ®ve conjugacy classes of S4 .16 Examples (1) The conjugacy classes of S3 are Class {1} {(1 2). with representatives (see De®nition 12. 2. we simply count the number of 2-cycles.4): 1.8) are as follows: Representative Class size g | gG | |CG ( g)| 1 1 24 (1 2) 6 4 (1 2 3) (1 2)(3 4) 8 3 3 8 (1 2 3 4) 6 4 We check our arithmetic by noting that jS4 j ˆ 1 ‡ 6 ‡ 8 ‡ 3 ‡ 6X . (1 2 3 4)X To calculate the sizes of the conjugacy classes. there are three elements of cycle-shape (2. 3-cycles. Thus for G ˆ S4 . the conjugacy class sizes | gG | and the centralizer orders |CG ( g)| (obtained using Theorem 12. (1 2)(3 4). ÀÁ which is 4 ˆ 6. Proof (1) Assume that x commutes with an odd permutation g. Thus x Sn # x An . while x S3 ˆ {x. here x A3 ˆ fxg. If h is even then y P x An . and so x Sn ˆ x An . given by x A n ˆ f g À1 xg: g P An g. we have seen in Theorem 12. The next result determines precisely when x An and x Sn are equal. is of course contained in x S n . (2) Assume that x does not commute with any odd permutation. g 1 (1 2) | gG | 1 10 |CG ( g)| 120 12 (1 2 3) (1 2)(3 4) 20 15 6 8 (1 2 3 4) 30 4 111 (1 2 3)(4 5) (1 2 3 4 5) 20 24 6 5 Conjugacy classes of A n Given an even permutation x P A n . the corresponding table for G ˆ S5 is Rep. The conjugacy class x A n of x in A n . however. and if h is odd then gh P An and y ˆ hÀ1 xh ˆ hÀ1 g À1 xgh ˆ ( gh)À1 x( gh). (2) If x does not commute with any odd permutation in Sn then x Sn splits into two conjugacy classes in An of equal size. Let y P x S n . 1. For an easy example where equality does not hold.17 Proposition Let x P An with n . x A n might not be equal to x S n .Conjugacy classes (4) Similarly. then x Sn ˆ x An . so again y P x An . with representatives x and (1 2)À1 x(1 2). and what happens when equality fails. 12. (1) If x commutes with some odd permutation in Sn . consider x ˆ (1 2 3) P A3 . so that y ˆ hÀ1 xh for some h P Sn . x À1 }.15 that the conjugacy class x S n consists of all permutations in S n which have the same cycle-shape as x. Then CSn (x) ˆ CAn (x)X . (1 4)(2 3)gX However. (2. but (1 2 3 4 5) commutes with no odd permutation. Hence by Proposition 12.17. Thus the conjugacy classes of A4 are Representative Class size Centralizer order 1 1 12 (1 2)(3 4) 3 4 (1 2 3) 4 3 (1 3 2) 4 3 (as jAn j ˆ 1jSn j) 2 (2) We ®nd the conjugacy classes of A5 . j 12. 2) and (5). we observe that fhÀ1 xh: h is oddg ˆ ((1 2)À1 x(1 2)) An since every odd permutation has the form (1 2)a for some a P An . together with the permutations of cycle-shapes (2.112 Representations and characters of groups Hence by Theorem 12. so g is 1. an even permutation. Since (1 2)(3 4) commutes with the odd permutation (1 2).17 implies that (1 2)(3 4) A4 ˆ (1 2)(3 4) S4 ˆ f(1 2)(3 4). with representatives (1 2 3) and (1 2)À1 (1 2 3)(1 2) ˆ (1 3 2).18 Examples (1) We ®nd the conjugacy classes of A4 . the . (1 2 3) or (1 3 2). The elements of A4 are the identity.13. jx An j ˆ jAn : CAn (x)j ˆ 1jSn : CAn (x)j 2 ˆ 1jSn : CSn (x)j ˆ 1jx Sn jX 2 2 Next. (1 2 3) S4 splits into two conjugacy classes in A4 of size 4. Proposition 12.) Hence by Proposition 12. (Check this by using the argument in (1) above. The elements (1 2 3) and (2 3)(4 5) commute with the odd permutation (4 5). the 3-cycle (1 2 3) commutes with no odd permutation: for if gÀ1 (1 2 3) g ˆ (1 2 3) then (1 2 3) ˆ (1 g 2 g 3 g) by Proposition 12.17. The non-identity even permutations in S5 are those of cycle-shapes (3).8. (1 3)(2 4). Now x Sn ˆ fhÀ1 xh: h is eveng ‘ fhÀ1 xh: h is oddg ˆ x An ‘ ((1 2)À1 x(1 2)) An X Since |x An | ˆ 1|x Sn |. as we wished to show. the conjugacy classes x An and ((1 2)À1 x(1 2)) An must 2 be disjoint and of equal size. 2) and (3). (1 2)(3 4). Proof If H is a union of conjugacy classes. j 12. Then by Proposition 12. 3. we have gÀ1 hg P H. Thus H v G. Let H v S4 . Using Proposition 12. 6. g P G A g À1 hg P H. g P G. and so hG # H. 8. (1 2 3 4 5) and (1 2)À1 (1 2 3 4 5)(1 2) ˆ (1 3 4 5 2). these conjugacy classes have sizes 1. H is a union of conjugacy classes of S4 . Therefore ‘ Hˆ hG . (1 2 3).19. Conversely. 1 ‡ 3. we see that the class sizes and centralizer orders are as follows: Representative Class size Centralizer order 1 1 60 (1 2 3) 20 3 (1 2)(3 4) 15 4 (1 2 3 4 5) 12 5 (1 3 4 5 2) 12 5 Normal subgroups Normal subgroups are related to conjugacy classes by the following elementary result. 6. hP H and so H is a union of conjugacy classes of G. Since j Hj divides 24 by Lagrange's Theorem.16(3).19 Proposition Let H be a subgroup of G. so gÀ1 Hg # H.17(2).Conjugacy classes 113 conjugacy classes of A5 are represented by 1. Then H v G if and only if H is a union of conjugacy classes of G.20 Example We ®nd all the normal subgroups of S4 . As we saw in Example 12. then h P H. if H v G then for all h P H. 1 ‡ 8 ‡ 3 or 1 ‡ 6 ‡ 8 ‡ 3 ‡ 6X . 12. and 1 P H. there are just four possibilities: j Hj ˆ 1. (1 3)(2 4). . . . hÀ1 C i h ˆ r ˆ jˆ1 r ˆ jˆ1 yÀ1 gyj X j hÀ1 yÀ1 gyj hX j . . in the last case H ˆ S4 . we have H ˆ 1 S4 ‘ (1 2)(3 4) S4 ˆ f1. (1 4)(2 3)gX The centre of a group algebra In this ®nal section we link the conjugacy classes of the group G to the centre of the group algebra CG. de®ne ˆ Ci ˆ g P CGX gPCi The elements C1 . and in the third case H ˆ A4 . (1 2)(3 4). . We have now shown that S4 has exactly four normal subgroups: f1g. Let Ci consist of the r distinct conjugates yÀ1 gy1. C l form a basis of Z(CG). A4 and V4 ˆ f1. . (1 3)(2 4). . Proof First we show that each C i belongs to Z(CG). . we write it as V4 (V stands for `Viergruppe'. (1 4)(2 3)gX This is easily checked to be a subgroup of S4 . . . For 1 < i < l. 12.114 Representations and characters of groups In the ®rst case H ˆ {1}. S4 . Recall from De®nition 9. C l of CG are called class sums. yÀ1 gyr of an element g. In the case where j Hj ˆ 1 ‡ 3. . (1 2)(3 4). . 12.12 that the centre of CG is Z(CG) ˆ fz P CG: zr ˆ rz for all r P CGgX We know that Z(CG) is a subspace of the vector space CG. . .22 Proposition The class sums C1 . . so 1 r Ci ˆ For all h P G. meaning `four-group'). .21 De®nition Let C1 . There is a convenient basis for this subspace which can be described in terms of the conjugacy classes of G. Cl be the distinct conjugacy classes of G. the centralizer CG (x) is the set of . since j hÀ1 yÀ1 gyj h ˆ hÀ1 yÀ1 gyk h D yÀ1 gyj ˆ yÀ1 gyk X j k j k Hence r ˆ jˆ1 À1 hÀ1 yÀ1 gyj h ˆ C i . . 12. and distinct conjugacy classes are disjoint. (1 2) ‡ (1 3) ‡ (2 3). the coef®cient ë g of g is equal to the coef®cient ë hÀ1 gh of hÀ1 gh. That is to say. . ˆ ˆ ë g hÀ1 gh ˆ ë g gX gPG gPG So for every h P G. For an element x of a group G. It follows that r ˆ iˆ1 ë i C i where ë i is the j coef®cient ë gi for some gi P Ci . hence with all € hPG ë h h P CG. .23 Examples (1) From Example 12. a2 . (1 2 3) ‡ (1 3 2)X (2) From (12. . . C l are linearly independent: for if €l iˆ1 ë i C i ˆ 0 (ë i P C). b ‡ a2 b. Every group is a union of conjugacy classes. j and so h C i h ˆ C i .3. That is.12). we have rh ˆ hr. a ‡ a3 . a basis for Z(CD8 ) is 1. That is.Conjugacy classes 115 As j runs from 1 to r. Let r ˆ € À1 gPG ë g g P Z(CG). so h rh ˆ r. C i h ˆ hC i X Therefore each C i commutes with all h P G. . and so C i P Z(CG). . ab ‡ a3 bX Summary of Chapter 12 1. then all ë i ˆ 0 as the classes C1 . the function g 3 ë g is constant on €l conjugacy classes of G. . the elements hÀ1 yÀ1 gyj h run through Ci . It remains to show that C1 . a basis for Z(CS3 ) is 1. observe that C1 . Cl are pairwise disjoint by Corollary 12. Next. This completes the proof. . For h P G. . . .16(1). 2. C l span Z(CG). Show that A5 is a simple group. Give a basis of the centre of the group algebra CQ8. 5.20. It is a subgroup of G.10 to show that Z(G) Tˆ {1}. n (a) Prove that j(1 2) G j ˆ …2 † and ®nd CG ((1 2)). The class sums in CG form a basis for the centre of CG. 7. (a) Use the Class Equation 12. 3. Find the conjugacy classes of the quaternion group Q8. Prove that G has a conjugacy class of size p. and the number of elements in the conjugacy class x G is equal to |G:CG (x)|. Suppose that G is a group of order pn . (Hint: use the method of Example 12. Exercises for Chapter 12 1. The conjugacy classes of Sn correspond to the cycle-shapes of permutations in Sn . (There are 11 conjugacy classes in all.116 Representations and characters of groups elements of G which commute with x. Verify that your solution satis®es Theorem 12. 4.) 4. (c) Now let n ˆ 6. and let n be a positive integer. 3. Let p be a prime number. Let G ˆ Sn . . If G is a group and x P G. 2. Show that j(1 2 3)(4 5 6) G j ˆ 40 and j(1 2)(3 4)(5 6) G j ˆ 15.) 6. Let G be a ®nite group and suppose that g P G and z P Z(G).8. show that CG (x) is a subgroup of G which contains Z(G). and ®nd the sizes of the other conjugacy classes of S6 . (b) Suppose that n > 3 and that | Z(G)| ˆ p. n n (b) Show that j(1 2 3) G j ˆ 2…3 † and j(1 2)(3 4) G j ˆ 3…4 †. If x P An then x Sn ˆ x An if and only if x commutes with some odd permutation in Sn . Prove that the conjugacy classes gG and ( gz) G have the same size. What are the cycle-shapes of those permutations x P A6 for which x A6 Tˆ x S6 ? 5. Characters of representations have many remarkable properties.1 De®nition If A ˆ (aij ) is an n 3 n matrix. With each n 3 n matrix gr ( g P G) we associate the complex number given by adding all the diagonal entries of the matrix. and they are the fundamental tools for performing calculations in representation theory. since from the de®nition of a representation r: G 3 GL(n. For example. In this chapter we present some basic properties and examples. written tr A. Moreover. The theory of characters will occupy a considerable portion of the rest of the book. then the trace of A. C). whereas the character records just one number for each matrix. C) is a representation of the ®nite group G.13 Characters Suppose that r: G 3 GL(n. basic problems. we shall show later that two representations have the same character if and only if they are equivalent. it appears that we must keep track of all the n2 entries in each matrix gr. is given by tr A ˆ n ˆ iˆ1 aii X That is. 117 . the trace of A is the sum of the diagonal entries of A. These facts are surprising. The trace of a matrix 13. and call this number ÷( g). The function ÷: G 3 C is called the character of the representation r. can be resolved by doing some easy arithmetic with the character of the representation. such as deciding whether or not a given representation is irreducible. 2 Proposition Let A ˆ (aij ) and B ˆ (bij ) be n 3 n matrices. Therefore tr (A ‡ B) ˆ and tr (AB) ˆ For the last part. tr (T À1 AT ) ˆ tr ((T À1 A)T ) ˆ tr (T (T À1 A)) ˆ tr AX (by the second part ) j n n ˆ ˆ iˆ1 jˆ1 n ˆ iˆ1 (aii ‡ bii ) ˆ n ˆ iˆ1 aii ‡ n ˆ iˆ1 bii ˆ tr A ‡ tr B. that is. and tr (AB) ˆ tr (BA)X Moreover.3 De®nition Suppose that V is a CG-module with a basis B . aij bji ˆ n n ˆ ˆ jˆ1 iˆ1 bji aij ˆ tr (BA)X Notice that. the trace function is not multiplicative. since if B and B 9 are bases of V. Then the character of V is the function ÷: G 3 C de®ned by ÷( g) ˆ tr [ g]B ( g P G)X The character of V does not depend on the basis B .118 Representations and characters of groups 13. if T is an invertible n 3 n matrix. unlike the determinant function. then tr (T À1 AT ) ˆ tr AX Proof The ii-entry of A ‡ B is aii ‡ bii . Characters 13. Then tr (A ‡ B) ˆ tr A ‡ tr B. tr (AB) need not equal (tr A)(tr B). then [ g]B 9 ˆ T À1 [ g]B T . and the ii-entry of AB is €n jˆ1 aij bji. 2. and so by Proposition 13. we have tr [x]B ˆ tr [ y]B . . and let B be a basis of V. we write ÷( g) and not g÷. You will have noticed that we are writing characters as functions on the left. namely ÷( g) ˆ tr ( gr) ( g P G)X 13. Proof (1) Suppose that V and W are isomorphic CG-modules. Then [x]B ˆ [ g À1 yg]B ˆ [ g]À1 [ y]B [ g]B X B Hence by Proposition 13. C) to be the character ÷ of the corresponding CGmodule C n . (2) If x and y are conjugate elements of the group G. Then by (7. then ÷(x) ˆ ÷( y) for all characters ÷ of G. and ÷ is reducible if it is the character of a reducible CG-module.5(1) for representations is that equivalent representations have the same character. and so V and W have the same character. j The result corresponding to Proposition 13.4 De®nition We say that ÷ is a character of G if ÷ is the character of some CGmodule. there are a basis B 1 of V and a basis B 2 of W such that [ g]B 1 ˆ [ g]B 2 for all g P GX Consequently tr [ g]B 1 ˆ tr [ g]B 2 for all g P G. That is. Therefore ÷(x) ˆ ÷( y).5 Proposition (1) Isomorphic CG-modules have the same character. tr [ g]B 9 ˆ tr [ g]B for all g P GX Naturally enough. (2) Assume that x and y are conjugate elements of G.7). Let V be a CG-module. ÷ is an irreducible character of G if ÷ is the character of an irreducible CG-module. Further. 13.24)). where ÷ is the character of V. so that x ˆ gÀ1 yg for some g P G. we de®ne the character of a representation r: G 3 GL(n.Characters 119 for some invertible matrix T (see (2.2. The matrices [ g]B ( g P G) are given by Exercise 4. g [ g]B ÷( g) 1 (1 2) (1 3) H 1 d0 0 0 0 1 0e 0 1 3 I H 0 d1 0 1 0 0 0e 0 1 1 I H 0 0 d0 1 1 0 1 I 1 0e 0 .) g gr ÷( g) 1 1 0 0 1 2 a 0 1 À1 0 0 a2 À1 0 0 À1 À2 a3 0 À1 1 0 0         g gr ÷( g)  b 1 0 0 À1 0   ab 0 À1 0 À1 0   a2 b À1 0 0 1 0   a3 b 0 1 0 1 0  (2) Let G ˆS3 . Let ÷ be the character of this representation. We record these matrices. and let r: G 3 GL(2. b: a4 ˆ b2 ˆ 1.1. C) be the representation for which     1 0 0 1 .6 Examples (1) Let G ˆ D8 ˆ ka.2(1)). together with the character ÷ of V.120 Representations and characters of groups Later. where v i g ˆ v ig for 1 < i < 3 and all g P G.5(1): if two CG-modules have the same character. then they are isomorphic. gr and ÷( g) as g runs through G. and take V to be the 3-dimensional permutation module for G over C (see De®nition 4. 13. v3 .10). bÀ1 ab ˆ aÀ1 l. The following table records g. v2 . br ˆ ar ˆ 0 À1 À1 0 (see Example 3. thus B is the basis v1 . we shall prove an astonishing converse of Proposition 13. Let B be the natural basis of V. (We obtain ÷( g) by adding the two entries on the diagonal of gr. every character is constant on conjugacy classes of G. U2. (4) Let G ˆ D6 ˆ ka. ÷3 . ÷2 . bÀ1 ab ˆ aÀ1 l (so G  S3 ). In Example 10. it is much quicker to write down the single complex number ÷( g) for the group element g than to record the matrix which corresponds to g. This will become clear as the theory of characters develops. r2 . This re¯ects the fact that by Proposition 13. G has just three irreducible characters ÷1 . the character encapsulates a great deal of information about the representation. U3. the characters given take few distinct values. with values g ÷1 ( g) ÷2 ( g) ÷3 ( g) 1 1 1 1 a 1 ù ù2 a2 1 ù2 ù where ù ˆ e2ðia3 .8(2).8. . and they are as follows: g ÷1 ( g) ÷2 ( g) ÷3 ( g) 1 1 1 2 a 1 1 À1 a2 1 1 À1 b 1 À1 0 ab 1 À1 0 a2 b 1 À1 0 Notice that in all the above examples. Nevertheless. Moreover.5(2). r3 given in Example 10.Characters g [ g]B ÷( g) H (2 3) I H (1 2 3) I H (1 3 2) I 0 1 0 0e 1 0 0 121 1 0 0 d0 0 1e 0 1 0 1 0 1 d0 0 1 0 0 0 1e 0 0 d1 0 (3) Let G ˆ C3 ˆ ha: a3 ˆ 1 i. By Theorem 9. Thus if ÷ i is the character of Ui for 1 < i < 3. The values of these characters on the elements of G can be calculated from the corresponding representations r1 . then the irreducible characters of G are ÷1 . b: a3 ˆ b2 ˆ 1.8(2). we found a complete set of non-isomorphic irreducible CG-modules U1. ÷2 and ÷3 . We denote it by 1 G . (2) If V is any 1-dimensional CG-module.9 Proposition Let ÷ be the character of a CG-module V. Characters of degree 1 are called linear characters.6(1) we gave a character of D8 of degree 2.8(1)) is a linear character.8 gives all the irreducible characters of ®nite abelian groups. Observe that Theorem 9. Then for all v P V X . namely the trivial character. Finding all the irreducible characters is usually dif®cult.7 De®nition If ÷ is the character of the CG-module V.6(4) we saw that the irreducible characters of D6 have degrees 1. we therefore know at least one of the irreducible characters of G. they are all linear characters. (3) The character of the trivial CG-module (see De®nition 4. they are. in particular. The values of a character The next result gives information about the complex numbers ÷( g).122 Representations and characters of groups 13. Thus 1 G : g 3 1 for all g P GX Given any group G. 13. these are the only non-zero characters of G which are homomorphisms (see Exercise 13.6(2) we gave a character of S3 of degree 3. irreducible characters. 13. and in 13. then the dimension of V is called the degree of ÷. Suppose that g P G and g has order m. of course. in 13. then for each g P G there is a complex number ë g such that v g ˆ ë gv The character ÷ of V is given by ÷( g) ˆ ë g ( g P G) and ÷ has degree 1. In fact. called the trivial character of G. 1 and 2. where ÷ is a character of G and g P G.4). Every linear character of G is a homomorphism from G to the multiplicative group of non-zero complex numbers.8 Examples (1) In Example 13. Therefore ÷( g) ˆ ù1 ‡ X X X ‡ ù n . . 123 Proof (1) Let n ˆ dim V.Characters (1) (2) (3) (4) ÷(1) ˆ dim V.11 there is a basis B of V such that H I ù1 0 f g FF [ g]B ˆ d e F ùn 0 where each ù i is an mth root of unity. and so ÷(1) ˆ dim V. a sum of mth roots of unity. Also ÷( gÀ1 ) ˆ ÷( g) by (3). When the element g of G has order 2. and let B be a basis of V. and so ÷( g) ˆ ÷( g). Consequently ÷(1) ˆ tr [1]B ˆ tr I n ˆ n. since for all real W. (eiW )À1 ˆ eÀiW . ÷( g) is a real number if g is conjugate to gÀ1 . ÷( g) is a sum of mth roots of unity. that is. (2) By Proposition 9. ‡ ùÀ1 . Then the matrix [1]B of the identity element 1 relative to B is equal to In .5(2). Therefore ÷( g À1 ) ˆ ù1 ‡ X X X ‡ ù n ˆ ÷( g)X (4) If g is conjugate to gÀ1 then ÷( g) ˆ ÷( gÀ1 ) by Proposition 13. ÷( gÀ1 ) ˆ ÷( g). which is the complex conjugate of eiW . ÷( g) is j real. (3) We have H f [ gÀ1 ]B ˆ d ùÀ1 1 0 FF 0 F ùÀ1 n I g e and so ÷( gÀ1 ) ˆ ùÀ1 ‡ . the n 3 n identity matrix. Every complex mth root of unity ù 1 n satis®es ùÀ1 ˆ ù. . we can be much more speci®c about the possibilities for ÷( g): . 124 Representations and characters of groups 13. j÷( g)j ˆ ÷(1) D gr ˆ ëI n (2) Ker r ˆ { g P G: ÷( g) ˆ ÷(1)}. 13. C) be a representation of G. suppose that |÷( g)| ˆ ÷(1). showing that we can determine the kernel of a representation just from knowledge of its character.11 Theorem Let r: G 3 GL(n. and ÷(1) ˆ r ‡ sX Certainly then. we have ÷( g)  ÷(1) mod 2. there is a basis B of C n such that H I ù1 0 f g FF [ g]B ˆ d e F ùn 0 where each ù i is an mth root of unity. where n ˆ ÷(1) and each ù i is a square root of unity. and let g be an element of order 2 in G. Then each ù i is ‡1 or À1.9. By Proposition 9. Then (13X12) j÷( g)j ˆ jù1 ‡ X X X ‡ ù n j ˆ ÷(1) ˆ nX for some ë P CX . ÷( g) P Z. Conversely. (1) For g P G. If gr ˆ ëIn with ë P C. and s are À1. j Our next result gives the ®rst inkling of the importance of characters.10 Corollary Let ÷ be a character of G. so |÷( g)| ˆ n ˆ ÷(1). and let ÷ be the character of r. and ÷( g)  ÷(1) mod 2X Proof By Proposition 13. Then ÷( g) is an integer.11. we have ÷( g) ˆ ù1 ‡ X X X ‡ ù n . and ÷( g) ˆ në. so that ÷( g) ˆ r À s. and suppose that g has order m. Proof (1) Let g P G. and since r À s ˆ r ‡ s À 2s  r ‡ s mod 2. then ë is an mth root of unity. Suppose r of them are ‡1. Part (2) follows.14 Examples (1) According to Example 13. . and so ÷( g) ˆ n ˆ ÷(1). 13. .11(2). we have gr ˆ ëIn for some ë P C.11(2). consider the picture in the Argand diagram. we have jz1 ‡ X X X ‡ zn j < jz1 j ‡ X X X ‡ jzn j. . Thus H I ù1 0 f g FF [ g]B ˆ d e ˆ ù1 I n X F ù1 0 Hence for all bases B 9 of C n we have [ g]B 9 ˆ ù1 In . b: a3 ˆ b2 ˆ 1. we deduce from (13. ÷2 . We call ÷ a faithful character if Ker ÷ ˆ {1}. . . written Ker ÷. 125 with equality if and only if the arguments of z1 . .Characters Note now that for any complex numbers z1 . 13. the irreducible characters of the group G ˆ D6 ˆ ka. if r is a representation of G with character ÷.6(4). Conversely. zn . and so g P Ker r. and so gr ˆ ù1 In .12) that ù i ˆ ù j for all i. Then by (1). is de®ned by Ker ÷ ˆ f g P G: ÷( g) ˆ ÷(1)gX By Theorem 13. In particular. . Therefore gr ˆ In . . then the kernel of ÷. with the following values: . Ker ÷ v G. ÷3 . then Ker r ˆ Ker ÷. we de®ne the kernel of a character as follows. zn are all equal. (To see this. j Motivated by Theorem 13.13 De®nition If ÷ is a character of G. j. (2) If g P Ker r then gr ˆ In .) Since |ù i | ˆ 1 for all i. whence ë ˆ 1. This implies that ÷( g) ˆ ë÷(1). This completes the proof of (1). bÀ1 ab ˆ aÀ1 l are ÷1 . suppose that ÷( g) ˆ ÷(1). (2) Let G ˆ D8 ˆ ka. C) is a representation with character ÷. b: a4 ˆ b2 ˆ 1. and let ÷ be the character of G given in Example 13. For a character ÷ of G. then we de®ne A to be the n 3 n matrix (a ij ).126 g ÷1 ( g) ÷2 ( g) ÷3 ( g) 1 1 1 2 Representations and characters of groups a 1 1 À1 a2 1 1 À1 b 1 À1 0 ab 1 À1 0 a2 b 1 À1 0 Then Ker ÷1 ˆ G. de®ne ÷: G 3 C by ÷( g) ˆ ÷( g) ( g P G)X Thus the values of ÷ are the complex conjugates of the values of ÷. then so is ÷. In particular.6(1): g ÷( g) 1 2 a 0 a2 À2 a3 0 b 0 ab 0 a2 b 0 a3 b 0 Then Ker ÷ ˆ {1}. 13.11(1) implies that if r: G 3 GL(2. then a2 r ˆ ÀI. bÀ1 ab ˆ aÀ1 l. Observe that if A ˆ (aij ) and B ˆ (bij ) are n 3 n matrices over C. Ker ÷2 ˆ kal and Ker ÷3 ˆ {1}. ÷3 is a faithful irreducible character of D6. Thus ÷( g) ˆ tr ( gr) ( g P G)X If A ˆ (aij ) is an n 3 n matrix over C. And since |÷(a2 )| ˆ |À2| ˆ ÷(1). If ÷ is irreducible.15 Proposition Let ÷ be a character of G. then (13X16) (AB) ˆ A B. We next prove a result which is sometimes useful for constructing a new character from a given one. Proof Suppose that ÷ is the character of a representation r: G 3 GL(n. C). . Theorem 13. Then ÷ is a character of G. so ÷ is a faithful character. and suppose that V ˆ U1 È X X X È Ur . Hence ÷ is irreducible if and only if ÷ is irreducible. .Characters since the ij-entry of A B is n ˆ kˆ1 127 a ik b kj . €n which is equal to the complex conjugate of kˆ1 aik bkj .17 De®nition The regular character of G is the character of the regular CG-module. It follows from (13. 13. j The regular character 13. X X X . Since tr ( gr) ˆ tr ( gr) ˆ tr ( gr) ˆ ÷( g) ( g P G). . . We write the regular character as ÷reg . the ij-entry of AB. a direct sum of irreducible CG-modules Ui. Ur. C) de®ned by gr ˆ ( gr) is a representation of G. .18 Proposition Let V be a CG-module.19. the character of the representation r is ÷.19 Theorem Let V1 . we shall express the regular character in terms of the irreducible characters of G. First we need a preliminary result. Then ÷reg ˆ d 1 ÷1 ‡ X X X ‡ d k ÷ k X . . . Proof This is immediate from (7.16) that the function r: G 3 GL(n. V k be a complete set of non-isomorphic irreducible CG-modules (see De®nition 11. k let ÷ i be the character of Vi and di ˆ ÷ i (1). Then the character of V is equal to the sum of the characters of the CG-modules U1.11). . j ( g P G) 13. . In Theorem 13. and for i ˆ 1.10). It is clear that if r is reducible then r is reducible. in particular. j The values of ÷reg on the elements of G are easily described. .19 and Proposition 13. Now the result follows from Proposition 13.21 Example We illustrate Theorem 13. then ÷reg (1) ˆ jGj.128 Representations and characters of groups Proof By Theorem 11. Now let g P G with g Tˆ 1. .9(1). gn of CG. and let B be the basis g1 . . the irreducible characters of G are ÷1 .9. . . and are given in the next result. 13.18. ÷2 . ÷ 3 : g ÷1 ( g) ÷2 ( g) ÷3 ( g) 1 1 1 2 a 1 1 À1 a2 1 1 À1 b 1 À1 0 ab 1 À1 0 a2 b 1 À1 0 We calculate ÷1 ‡ ÷2 ‡ 2÷3 : (÷1 ‡ ÷2 ‡ 2÷3 )( g) 6 0 0 0 0 0 . gn be the elements of G. By Proposition 13.6(4). By Example 13. Then for 1 < i < n. where for each i there are di factors V i . the ii-entry is zero for all i. CG  (V1 È X X X È V1 ) È (V2 È X X X È V2 ) È X X X È (Vk È X X X È Vk ).20 for the group G ˆ D6 . . we have gi g ˆ gj for some j with j Tˆ i. Therefore the ith row of the matrix [ g]B has zeros in every place except column j. ÷reg (1) ˆ dim CG ˆ |G|. .20 Proposition If ÷reg is the regular character of G. . and ÷reg ( g) ˆ 0 if g Tˆ 1X Proof Let g1 . It follows that ÷reg ( g) ˆ tr [ g]B ˆ 0X j 13. (1 2 3 4)X The permutation character ð takes the values gi ð( gi ) 1 4 (1 2) 2 (1 2 3) 1 (1 2)(3 4) 0 (1 2 3 4) 0 . . with representatives 1. . . and we now describe this. (1 2).19. . there is an easy construction using the permutation module which produces a character of degree n. v n . . and is 1 if ig ˆ i.10). . G has ®ve conjugacy classes. . so that G is a group of permutations of {1.23 Example Let G ˆ S4 . . The permutation module V for G over C has basis v1 . v i g ˆ v ig (1 < i < n) (see De®nition 4. where for all g P G.20. (1 2)(3 4). and the value 0 on all non-identity elements of G. let fix ( g) ˆ fi: 1 < i < n and ig ˆ igX Then (13X22) ð( g) ˆ jfix( g)j ( g P G)X We call ð the permutation character of G. .Characters 129 This is the regular character of G. Then by Example 12. illustrating Proposition 13.16(3). by Theorem 13. Let B denote the basis v1 . Then the iientry in the matrix [ g]B is 0 if ig Tˆ i. (1 2 3). n}. 13. Permutation characters In the case where G is a subgroup of the symmetric group Sn . . . . Suppose that G is a subgroup of Sn . Therefore the character ð of the permutation module V is given by ð( g) ˆ (the number of i such that ig ˆ i)X For g P G. and it takes the value |G| on 1. v n . By Maschke's Theorem 8. 2. . A character is obtained from a representation by taking the trace of each matrix. a subgroup of S4 . . By Example 12. . and let u ˆ v1 ‡ X X X ‡ v n . Then ð ˆ 1 G ‡ í. . (1 2)(3 4). (1 2 3).1. so |®x( g)| ˆ 1 ‡ í( g) for all g P G. so the character of U is the trivial character 1 G (see Example 13. U is isomorphic to the trivial CG-module.24 Proposition Let G be a subgroup of Sn . . Indeed. (1 3 2)X The values of the character í of G are gi í( gi ) 1 3 (1 2)(3 4) À1 (1 2 3) 0 (1 3 2) 0 Summary of Chapter 13 1. there is a CG-submodule W of V such that V ˆ U È WX Let í be the character of W.130 Representations and characters of groups 13. Proof Let v1 . the conjugacy classes of G are represented by 1.18(1). so U is a CG-submodule of V.8(3)). and therefore í( g) ˆ jfix( g)j À 1 ( g P G)X j ( g P G) 13. Then the function í: G 3 C de®ned by í( g) ˆ jfix ( g)j À 1 is a character of G.25 Example Let G ˆ A4 . v n be the basis of the permutation module V as above. Characters are constant on conjugacy classes. and U ˆ sp (u)X Observe that ug ˆ u for all g P G. and all g P G. 5. Prove that the only non-zero characters of G which are homomorphisms are the linear characters. 131 4. Suppose that z P Z(G) and that z has order m. Prove that there exists an mth root of unity ë P C such that for all g P G. 5. The regular character ÷reg of G takes the value |G| on the identity and the value 0 on all other elements of G. 4. 6. b: a6 ˆ b2 ˆ 1. r2 be the representations of G for which     ù 0 0 1 ar1 ˆ . Write the regular character of C4 as a linear combination of these. 7. br2 ˆ 1 0  0 X À1 ( g P G) Find the characters of r1 and r2 . Find ÷(x) for x ˆ (1 2) and for x ˆ (1 6)(2 3 5).11. Prove that if ÷ is a faithful irreducible character of the group G. Exercises for Chapter 13 1. For all characters ÷ of G. then Z(G) ˆ { g P G: |÷( g)| ˆ ÷(1)}. Find all the irreducible characters of C4 . Isomorphic CG-modules have the same character. br1 ˆ (where ù ˆ e2ðia3 ) 0 ùÀ1 1 0 and ar2 ˆ  À1 0   0 1 . Let G ˆ D12 ˆ ka.Characters 3. Let ÷ be the character of the 7-dimensional permutation module for S7 . Find also Ker r1 and Ker r2 . and ÷( gÀ1 ) ˆ ÷( g). Assume that ÷ is an irreducible character of G. the complex number ÷( g) is a sum of roots of unity. check that your answers are consistent with Theorem 13. 3. . 2. and let r1 . bÀ1 ab ˆ aÀ1 l. If G is a subgroup of Sn . ÷(zg) ˆ ë÷( g)X 6. then the function í which is given by í( g) ˆ jfix( g)j À 1 is a character of G. The character of a representation determines the kernel of the representation. 8. Show that G has a normal subgroup of index 2.) 10. where k is an odd integer. (a) Show that ä: g 3 det ( gr) ( g P G) is a linear character of G. 9.132 Representations and characters of groups 7. (b) Prove that G/Ker ä is abelian. show that G has a normal subgroup of index 2. Let ÷ be a character of a group G. Show that either (1) ÷( g)  ÷(1) mod 4. Let r be a representation of the group G over C. . (c) Assume that ä( g) ˆ À1 for some g P G. Let g be a group of order 2k. Hint: use Exercise 7. and let g be an element of order 2 in G. or (2) G has a normal subgroup of index 2.10. (Compare Corollary 13. Prove that if x is a non-identity element of the group G. then ÷(x) Tˆ ÷(1) for some irreducible character ÷ of G. By considering the regular representation of G. Inner products The characters of a ®nite group G are certain functions from G to C. using characters. and ë P C.1 Example Let G ˆ C3 ˆ ka: a3 ˆ 1l.) 14. The set of all functions from G to C forms a vector space over C. Also.14 Inner products of characters We establish some signi®cant properties of characters in this chapter. ö are functions from G to C. and in particular we prove the striking result (Theorem 14.21) that if two CG-modules have the same character then they are isomorphic. if we adopt the natural rules for adding functions and multiplying functions by complex numbers. That is. then we de®ne W ‡ ö: G 3 C by (W ‡ ö)( g) ˆ W( g) ‡ ö( g) and we de®ne ëW: G 3 C by ëW( g) ˆ ë(W( g)) ( g P G)X ( g P G) (We write these functions on the left to agree with our notation for characters. The proofs rely on an inner product involving the characters of a group. if W. we describe a method for decomposing a given CG-module as a direct sum of CG-submodules. and we describe this ®rst. and suppose that W: G 3 C and ö: G 3 C are given by 133 . W(a) ˆ i.2) (a) kW. Notice that condition (a) implies that kW. Wi for all W. We now introduce an inner product on the vector space of all functions from G to C. The de®nition of an inner product on a vector space over C runs as follows. öi ˆ W( g)ö( g)X jGj gPG . öl ˆ ë1 kW1 . With every ordered pair of vectors W. ë2 P C and all vectors W1 . De®ne 1 ˆ hW. as in this example. ö. The vector space of all functions from G to C can be equipped with an inner product in a way which we shall describe shortly. (c) kW. ö. ö in the vector space.134 Representations and characters of groups 1 W ö 2 1 a i 1 a2 À1 1 This means that W(1) ˆ 2. 14. ë2 P C and all vectors W1 . 0 if W Tˆ 0. W2 i for all ë1 . öl which satis®es the following conditions: (14. W(a2 ) ˆ À1 and ö(1) ˆ ö(a) ˆ ö(a2 ) ˆ 1. öl for all ë1 .3 De®nition Suppose that W and ö are functions from G to C. W2 . there is associated a complex number kW. Wl . öl ˆ hö. Wl is always real. ö. and that conditions (a) and (b) give hö. W2 . This will be of basic importance in our study of characters. öl ‡ ë2 kW2 . ë1 è1 ‡ ë2 W2 i ˆ ë1 hö. Then W ‡ ö and 3W are given by 1 W ‡ö 3W 3 6 a 1‡i 3i a2 0 À3 We shall often think of functions from G to C as row vectors. W1 i ‡ ë2 hö. (b) kë1 W1 ‡ ë2 W2 . so k .Inner products of characters 135 It is transparent that the conditions of (14. 1 ‡ 1 . l is an inner product on the vector space of functions from G to C.5 Proposition Assume that G has exactly l conjugacy classes. i ‡ (À1) . . 1 ‡ i . ÷i ˆ ÷( g)ø( g À1 ).9(3). and this is a real numberX jGj gPG (2) h÷. 1) ˆ 1X 3 Inner products of characters We can exploit the fact that characters are constant on conjugacy classes to simplify slightly the calculation of the inner product of two characters. öi ˆ 1(2 .4 Example As in Example 14. suppose that G ˆ C3 ˆ ka: a3 ˆ 1l and that W and ö are given by 1 W ö 2 1 a i 1 a2 À1 1 Then hè. öi ˆ 1(1 . 2 ‡ i .1. 3 3 hè. Therefore 1 ˆ h÷. Let ÷ and ø be characters of G. èi ˆ 1(2 . 3 hö. 14. 1 À 1 . øi ˆ ÷( g)ø( g À1 )X jGj gPG . 14. 1) ˆ 1(1 ‡ i). . .2) hold. by Proposition 13. gl . øi ˆ l ˆ ÷( g i )ø( g i ) iˆ1 jCG ( g i )j X Proof (1) We have ø( g) ˆ ø( gÀ1 ) for all g P G. 1 ‡ 1 . . with representatives g1 . (À1)) ˆ 2. 1 ˆ (1) h÷. øi ˆ hø. g 3 ˆ (1 2 3). we also have 1 ˆ h÷. ˆ ÷( g)ø( g) ˆ j g G j÷( g i )ø( g i )X i gP g G i Now Gˆ l ‘ iˆ1 g G and j g G j ˆ jGjajCG ( g i )j.3 and Theorem 12. in fact. i i by Corollary 12.6 Example The alternating group A4 has four conjugacy classes. ÷l ˆ h÷. Since characters are constant on conjugacy classes.) (2) Recall that g G denotes the conjugacy class of G which contains i gi . øi. g 4 ˆ (1 3 2) (see Example 12. øi is real.8. g 2 ˆ (1 2)(3 4). Hence h÷.18(1)). øi ˆ l 1 ˆ 1 ˆˆ ÷( g)ø( g) ˆ ÷( g)ø( g) jGj gPG jGj iˆ1 G gP g i ˆ ˆ l ˆ j gGj iˆ1 l ˆ iˆ1 jGj i ÷( g i )ø( g i ) j 1 ÷( g i )ø( g i )X jCG ( g i )j 14. ÷iX jGj gPG Since kø. with representatives g 1 ˆ 1. We shall see in Chapter 18 that there are characters ÷ and ø of A4 which take the following values on the representatives gi : gi |CG ( gi )| ÷ ø g1 12 1 4 g2 4 1 0 g3 3 ù ù2 g4 3 ù2 ù . it follows that h÷. an integer. øi is.136 Representations and characters of groups Since { gÀ1 : g P G} ˆ G. (We shall prove later that h÷. øi ˆ ÷( g À1 )ø( g) ˆ hø. ù ‡ ‡ ˆ 0. 4 0 . øi ˆ 0.12) that the irreducible characters of G form an orthonormal set of vectors in the vector space of functions from G to C. Among other results. ÷l ˆ 1. and then let W2 be the sum of the remaining CG-modules U i . . for distinct irreducible characters ÷ and ø of G. 0 ù 2 . ù2 ù2 . . . we may take W1 to be the sum of those irreducible CG-submodules Ui which are isomorphic to a given irreducible CG-module.Inner products of characters (where ù ˆ e2ðia3 ). Using part (2) of Proposition 14.5. we have h÷. There are several ways of choosing CG-submodules W1 and W2 of CG such that CG ˆ W1 È W2 and W1 and W2 have no common composition factor (see De®nition 10.4). and to ®nd the inner products of ÷ and ø with the trivial character (which takes the value 1 on all elements of A4 ). Ur. For example. We ®rst look at the effect of applying the elements e1 and e2 of CG to W1 and W 2 . . We shall investigate some consequences of writing CG like this. . we temporarily adopt the following Hypothesis: 14. 0 ù . øi ˆ ‡ ‡ ˆ 2X ‡ 12 4 3 3 h÷. We are now going to pave the way to proving the key fact (Theorem 14. Recall from Chapter 10 that the regular CG-module is a direct sum of irreducible CG-submodules. that is. ‡ 12 4 3 3 4 . øi ˆ 137 We advise you to check also that k÷. Write 1 ˆ e1 ‡ e2 where e1 P W1 and e2 P W2. where W1 and W2 are CG-submodules which have no common composition factor. we shall derive a formula for e1 in terms of the character of W 1 . therefore. we have 1 . ù hø. 4 1 . ù2 ù . ÷i ˆ 1 and h÷.7 Hypothesis Let CG ˆ W 1 È W 2 . and that every irreducible CG-module is isomorphic to one of the CGmodules U1. say CG ˆ U 1 È X X X È U r . 7. We shall calculate the trace of W in two ways. e2 ˆ e2 and e1 e2 ˆ e2 e1 ˆ 0X 1 2 Proof In Proposition 14. Proposition Let ÷ be the character of the CG-module W1 which appears in Hypothesis 14.138 Representations and characters of groups 14.10. so every CG-homomorphism from W2 to W1 is zero. and this completes the proof. w2 e1 ˆ 0. j 14. take w1 ˆ e1 and w2 ˆ e2 . The function W: w 3 we1 x À1 (w P CG) is an endomorphism of CG.8 Proposition For all w1 P W1 and w2 P W2. we have w1 e1 ˆ w1 . we evaluate e1 .8. by Proposition 11. Therefore w1 w2 ˆ 0 for all w1 P W 1 . 14. Then 1 ˆ e1 ˆ ÷( g À1 ) gX jGj gPG Proof Let x P G. Now w1 ˆ w1 1 ˆ w1 (e1 ‡ e2 ) ˆ w1 e1 . j . w1 e2 ˆ 0. w2 P W 2 . Next. we have e2 ˆ e1 . In particular.9 Corollary For the elements e1 and e2 of CG which appear in Hypothesis 14.3. w2 e 2 ˆ w2 X Proof If w1 P W1 then the function w2 3 w1 w2 (w2 P W2 ) is clearly a CG-homomorphism from W2 to W 1 . and w2 ˆ w2 1 ˆ w2 (e1 ‡ e2 ) ˆ w2 e2 .7. Similarly w2 w1 ˆ 0. w1 e2 ˆ w2 e1 ˆ 0. But W2 and W1 have no common composition factor. we see that for all x P G. for w1 P W1 and w2 P W2 we have.3 of the multiplication in CG. By the de®nition of the character ÷ of W 1 .11 Corollary Let ÷ be the character of the CG-module W1 which appears in Hypothesis 14. the endomorphism w 3 wgx À1 (w P CG) of CG has trace 0 if g Tˆ x and has trace |G| if g ˆ x. in view of Proposition 14. € Hence. By Proposition 13.Inner products of characters 139 First. so e1 ˆ ˆ gPG ëg g for some ë g P C.20.7. w2 W ˆ w2 e1 x À1 ˆ 0X Thus W acts on W1 by w1 3 w1 x À1 and on W2 by w2 3 0. and of course the endomorphism w2 3 0 of W2 has trace 0. we deduce from Proposition 14. Therefore tr W ˆ ÷(x À1 )X Secondly. w1 W ˆ w1 e1 x À1 ˆ w1 x À1 . as W: w 3 w gPG ë g gx À1 . ÷iX ÷( g À1 )÷( g) ˆ jGj2 gPG jGj . ÷i ˆ ÷(1)X Proof Using the de®nition 6. Then h÷. the endomorphism w1 3 w1 x À1 of W1 has trace equal to ÷(x À1 ). e1 P CG. we have tr W ˆ ë x jGjX Comparing our two expressions for tr W. ë x ˆ ÷(x À1 )ajGjX Therefore e1 ˆ 1 ˆ ÷( g À1 ) gX jGj gPG j 14.8.10 that the coef®cient of 1 in e2 is 1 ˆ 1 1 h÷. say CG ˆ U1 È X X X È Ur . and obtain hm÷.140 Representations and characters of groups On the other hand. each of which has character ÷. Hence k÷. as required. with characters ÷ and ø. . ÷i ˆ 1. j We can now prove the main theorem concerning the inner product k .9 that CG is a direct sum of irreducible CG-submodules. l. and no composition factor of X is isomorphic to U. ÷i ˆ 1. and h÷. øi ˆ 0X Proof Recall from Theorem 11. let Y be the sum of those CG-submodules Ui of CG which are isomorphic to either U or V. and let Z be the sum of the remaining CG-submodules Ui. The character of W is m÷. Then h÷. let X be the sum of the remaining CG-submodules Ui. ÷l ˆ ÷(1). every composition factor of W is isomorphic to U. m÷i ˆ m÷(1)X As ÷(1) ˆ dim U ˆ m. and the 1 coef®cient of 1 in e1 is ÷(1)/|G|. Let m ˆ dim U. we know from Corollary 14. Next. respectively.11 to the character of W. W and X have no common composition factor. where the number of CG-submodules Ui which are isomorphic to U is dim U. this yields h÷. Then CG ˆ W È X X Moreover.9 that e2 ˆ e1 . since W is the direct sum of m CG-submodules. and de®ne W to be the sum of the m irreducible CG-submodules Ui which are isomorphic to U. In particular. Then CG ˆ Y È Z.12 Theorem Let U and V be non-isomorphic irreducible CG-modules. 14. We now apply Corollary 14. 13). j Applications of Theorem 14. øi ‡ hø. then by Theorem 14. If ÷ i is the character of Vi (1 < i < k). dk such that (14X14) V  (V1 È X X X È V1 ) È (V2 È X X X È V2 ) È X X X È (Vk È X X X È Vk ). By Theorem 8. Now let V be a CG-module. and hence k÷. ø(1) ˆ n. and ÷(1) ˆ m. so there are non-negative integers d1. X X X . .11). øi ˆ d i for 1 < i < k. we have (14X13) h÷ i . øi ˆ d 2 ‡ X X X ‡ d 2 X 1 k Summarizing. Each of these is isomorphic to some V i . øi ˆ 1. we have . . k÷. V k be a complete set of nonisomorphic irreducible CG-modules (see De®nition 11. ÷l. . . .12.5(1). V is equal to a direct sum of irreducible CG-submodules.Inner products of characters 141 and Y and Z have no common composition factor. . ÷i ‡ n2 hø.7. øl ˆ kø. øi ‡ hø. this implies that the irreducible characters ÷1 .11. j. øl ˆ 0. by the part of the theorem which we have already proved. ÷i ˆ hø. there are d i factors Vi X Therefore the character ø of V is given by (14X15) ø ˆ d 1 ÷1 ‡ X X X ‡ d k ÷ k X Using (14. ÷ k are all distinct. m÷(1) ‡ nø(1) ˆ hm÷ ‡ nø.12 Let G be a ®nite group. ÷ i i ˆ h÷ i . In particular. where for each i. Therefore h÷. we obtain from this (14X16) hø. . where ä ij is the Kronecker delta function (that is. øi ‡ mn(h÷. ÷i ˆ 0X By Proposition 14. The character of Y is m÷ ‡ nø. . ÷ j i ˆ ä ij for all i. and let V1 . and hø. ÷i)X Now h÷. By Corollary 14. ä ij is 1 if i ˆ j and is 0 if i Tˆ j). m÷ ‡ nøi ˆ m2 h÷. where n ˆ dim V. ÷1 i ˆ 3. taking the following values on the conjugacy class representatives 1.1 1. dk .142 Representations and characters of groups 14. ÷ k be the irreducible characters of G. By Example 13. d i ˆ hø. we know that ø(1) ˆ 3. ÷3 l ˆ 1. and d2X i 14. . øi ˆ k ˆ iˆ1 for 1 < i < k. ÷2 . If ø is any character of G. . .6(4) that the irreducible characters of S3  D6 are ÷1 .6(2). Moreover. ø(1 2) ˆ 1. ÷3 . .) A more substantial calculation along these lines is given in Example 15. ÷ i i hø. then ø ˆ d 1 ÷1 ‡ X X X ‡ d k ÷ k for some non-negative integers d1. . kø.17.17 Theorem Let ÷1 . (1 2 3): Now let ø be the character of the 3-dimensional permutation module gi |CS3 ( g i )| ÷1 ÷2 ÷3 1 6 1 1 2 (1 2) 2 1 À1 0 (1 2 3) 3 1 1 À1 for S3 .5(2). by Proposition 14. ø(1 2 3) ˆ 0X Therefore.18 Example Recall from Example 13.7. . ÷2 l ˆ 0 and kø. hø. . . (1 2).1 ‡ ‡ 0 ˆ 1X 6 2 Similarly. . ø ˆ ÷1 ‡ ÷3 X (This can of course be checked immediately by comparing the values of ø and ÷1 ‡ ÷3 on each conjugacy class representative. Thus by Theorem 14. .12. ÷l Tˆ 0. the constituents of ø are the irreducible characters ÷ i of G for which the integer di in the expression ø ˆ d1 ÷1 ‡ . øi ˆ d 2 ‡ X X X ‡ d 2 X 1 k It follows that one of the integers di is 1 and the rest are zero.21 Theorem Suppose that V and W are CG-modules. øl ˆ 1. It gives us a quick and effective method of determining whether or not a given CG-module is irreducible. The next result is another signi®cant consequence of Theorem 14. We say that ÷ is a constituent of ø if kø. Thus. Then by (14. Proof If V is irreducible then kø. j We are now in a position to prove the remarkable result that `a CGmodule is determined by its character'. and by (14. øl ˆ 1. øl ˆ 1 by Theorem 14. with characters ÷ and ø. Then V and W are isomorphic if and only if ÷ ˆ ø. respectively. We have ø ˆ d 1 ÷1 ‡ X X X ‡ d k ÷ k for some non-negative integers di.14).16). and so V is irreducible. for it means that many questions about CG-modules can be answered using character theory. Proof In Proposition 13. .20 Theorem Let V be a CG-module with character ø. It is this fact which motivates our study of characters in much of the rest of the book. assume that kø. It is the converse which is the substantial part of this theorem.Inner products of characters 143 We shall see many more applications of the important Theorem 14. 14.17.12. V  Vi for some i.19 De®nition Suppose that ø is a character of G. 14. Conversely. ‡ dk ÷ k is non-zero. 14. Then V is irreducible if and only if kø. . and that ÷ is an irreducible character of G. 1 ˆ hø.5 we proved the elementary fact that if V  W then ÷ ˆ ø. di (1 < i < k) such that V  (V1 È X X X È V1 ) È (V2 È X X X È V2 ) È X X X È (Vk È X X X È Vk ) with ci factors Vi for each i. . 2. ÷ k . the representations r2 and r3 are equivalent. By (14. suppose that ÷ ˆ ø. The next theorem is another consequence of Theorem 14. and let r1 . and hence V  W. r3 and r4 . Again let V1 . d i ˆ hø. r2 . ÷ i i (1 < i < k)X Since ÷ ˆ ø. j 14. . . r2 . Vk be a complete set of non-isomorphic irreducible CG-modules with characters ÷1 . ci ˆ h÷. r4 be the representations of G over C for which 2 3 2 3 ù 0 ù 0 . We know by (14. X X X . ar2 ˆ 0 ù 0 ùÀ1 2 3 2 3 0 1 1 ùÀ1 ar3 ˆ . . ar1 ˆ .12. and W  (V1 È X X X È V1 ) È (V2 È X X X È V2 ) È X X X È (Vk È X X X È Vk ) with di factors Vi for each i. ÷ i i. .16).144 Representations and characters of groups Thus. it follows that ci ˆ di for all i.22 Example Let G ˆ C3 ˆ ka: a3 ˆ 1l. The characters ø i of the representations r i (i ˆ 1. r3 .14) that there are non-negative integers ci .21. but there are no other equivalences among r1 . 3. ar4 ˆ À1 À1 0 ù (ù ˆ e2ðia3 ). 4) are 1 ø1 ø2 ø3 ø4 2 2 2 2 a 2ù À1 À1 1‡ù a2 2ù2 À1 À1 1 ‡ ù2 Hence by Theorem 14. By Proposition 11. and W  (V1 È X X X È V1 ) È (V2 È X X X È V2 ) È X X X È (Vk È X X X È Vk ) with di factors Vi for each i. j We now relate inner products of characters to the spaces of CGhomomorphisms which we constructed in Chapter 11. for any i. using (11. 14. respectively.Inner products of characters 145 14. j we have dim (HomCG (Vi . di (1 < i < k) such that V  (V1 È X X X È V1 ) È (V2 È X X X È V2 ) È X X X È (Vk È X X X È Vk ) with ci factors Vi for each i. . . ÷ k are linearly independent. . .5)(3) we see that dim (HomCG (V .2.24 Theorem Let V and W be CG-modules with characters ÷ and ø. W )) ˆ k ˆ iˆ1 ci d i X . W )) ˆ h÷. ÷ i i ˆ ë i X Therefore ÷1 . Proof Assume that ë1 ÷1 ‡ X X X ‡ ë k ÷ k ˆ 0 (ë i P C)X Then for all i. . . . ÷ k be the irreducible characters of G.23 Theorem Let ÷1 . Vj )) ˆ ä ij X Hence.14) that there are non-negative integers ci . Then ÷1 . . .13) we have 0 ˆ hë1 ÷1 ‡ X X X ‡ ë k ÷ k . using (14. Then dim (HomCG (V . ÷ k are linearly independent vectors in the vector space of all functions from G to C. . øiX Proof We know from (14. . . Since W2 and V1 have no common composition factor. Let V be any CG-module. then 2ˆ 3 À1 V ÷( g ) g gPG . and 1 ˆ e1 ‡ e2 with e1 P W 1 . ÷ˆ ci ÷ i and ø ˆ di÷i and so (14. and V is any CG-module. v2 e1 ˆ 0.146 Representations and characters of groups k ˆ iˆ1 k ˆ iˆ1 On the other hand.26 Proposition If ÷ is an irreducible character of G. e2 P W 2 . for all v1 P V1 and v2 P V2 we have v1 e1 ˆ v1 .13) implies that h÷. we deduce the stated results just as in the proof of Proposition 14. where the CG-modules W1 and W2 have no common composition factor.25 Proposition With the above notation. Decomposing CG-modules It is sometimes of practical importance to be able to decompose a given CG-module into a direct sum of CG-submodules. Once more we adopt Hypothesis 14. where every composition factor of V1 is a composition factor of W1 and every composition factor of V2 is a composition factor of W2. 14. v1 e2 ˆ 0. øi ˆ The result follows. and we now describe a process for doing this.7: CG ˆ W 1 È W 2 . We can write V ˆ V1 È V2 . j 14.8. v 2 e2 ˆ v 2 X k ˆ iˆ1 j ci d i X Proof If v1 P V1 then the function w2 3 v1 w2 (w2 P W2 ) is clearly a CG-homomorphism from W2 to V1. (2) For each irreducible character ÷ of G.28 Examples (1) Let G be any ®nite group and let V be any non-zero CG-module. Proposition 14. The procedure is as follows: (14. Then the character of W1 is m÷ where m ˆ ÷(1). the element e1 of W1 is given by m ˆ e1 ˆ ÷( g À1 ) gX jGj gPG Let V1 be the sum of those CG-submodules of V which have character ÷. so 2ˆ 3 À1 V1 ˆ V ÷( g ) g X gPG j Once the irreducible characters of our group G are known. The character of V÷ is a multiple of ÷. and let V÷ be the subspace of V spanned by these vectors. Some more complicated uses of the method can be found in Chapter 32. . we de®ne Vr ˆ fvr: v P V g). and let W2 be the sum of the remaining CG-submodules Ui.Inner products of characters 147 is equal to the sum of those CG-submodules of V which have character ÷ (where for r P CG.7. calculate the vectors € v i ( gPG ÷( gÀ1 ) g) for 1 < i < n.26. Proof Write CG ˆ U 1 È X X X È U r . Then Proposition 14. . .26 provides a useful practical tool for ®nding CG-submodules of a given CG-module V. by Theorem 11. v n of V.25 shows that Ve1 ˆ V1.9. Taking ÷ to be the trivial character of G in Proposition 14. . Let W1 be the sum of those CG-submodules Ui which have character ÷. and by Proposition 14. We illustrate this method with a couple of simple examples. (3) Then V is the direct sum of the CG-modules V÷ as ÷ runs over the irreducible characters of G.27) (1) Choose a basis v1 .10. Clearly we may omit the constant multiplier m/|G|. 14. Also W1 and W2 satisfy Hypothesis 14. we see that . a direct sum of irreducible CG-submodules Ui. v2 À v4 )X . . with basis v1 . . Then 2ˆ 3 V g ˆ sp (v1 ‡ X X X ‡ v n )X gPG Hence V has a unique trivial CG-submodule. v4 such that v i g ˆ v ig for all i and all g P G.148 Representations and characters of groups 2ˆ 3 V g gPG is the sum of all the trivial CG-submodules of V. Ve4 ˆ sp (v1 À v2 ‡ v3 À v4 ). v2 . . For example. . Ve2 ˆ 0.3(3)): 1 ÷1 ÷2 ÷3 ÷4 ÷5 1 1 1 1 2 a 1 1 À1 À1 0 a2 1 1 1 1 À2 a3 1 1 À1 À1 0 b 1 À1 1 À1 0 ab 1 À1 À1 1 0 a2 b 1 À1 1 À1 0 a3 b 1 À1 À1 1 0 Let V be the permutation module for G. ÷5 of D8 (see Example 16. Ve3 ˆ 0. let G ˆ Sn and let V be the permutation module. with basis v1 . Ve5 ˆ sp (v1 À v3 . e5 ˆ 1(1 À a2 ). v3 . . let ÷ i (1) ˆ ei ˆ ÷ i ( g À1 ) gX 8 gPG For example.5). v n such that v i g ˆ v ig for all i and all g P G. . Here is a list of the irreducible characters ÷1 . For 1 < i < 5. . (2) Let G be the subgroup of S4 which is generated by a ˆ (1 2 3 4) and b ˆ (1 2)(3 4)X Then G  D8 (compare Example 1. Then 2 Ve1 ˆ sp (v1 ‡ v2 ‡ v3 ‡ v4 ). . . ÷ j i ˆ ä ij for all i.Inner products of characters We have V ˆ Ve1 È Ve4 È Ve5 . . . i ei ej ˆ 0 for i Tˆ jX Compare these results with Corollary 14. and ø is any character. then ø ˆ d 1 ÷1 ‡ X X X ‡ d k ÷ k where d i ˆ hø. h÷ i . . . Every CG-module is determined by its character.27) does not in general enable us to write a given CG-module as a direct sum of irreducible CG-submodules (since V÷ is not in general irreducible). ÷ k are the irreducible characters of G. ÷ k of G form an orthonormal set. öi ˆ W( g)ö( g)X jGj gPG 2. Also. ø is irreducible if and only if kø. The inner product of two functions W. If ÷1 . ÷ i iX Each di is a non-negative integer. The irreducible characters ÷1 . respectively. ÷4 and ÷5 . . øl ˆ 1. Summary of Chapter 14 1. ö from G to C is given by 1 ˆ hW. 149 and so we have expressed V as a direct sum of irreducible CGsubmodules whose characters are ÷1 . Note that the procedure described in (14. . j. 4. that is. e2 ˆ ei for 1 < i < 5. . . 3. You might like to check that e1 ‡ X X X ‡ e5 ˆ 1.9. br1 ˆ .150 Representations and characters of groups Exercises for Chapter 14 1. Which of ÷ and ø is irreducible? 2. and that ÷( g) is a non-negative real number for all g in G. b: a4 ˆ 1. Prove that ÷ is reducible. r2 . gó ˆ T À1 ( gr)TX 4. Suppose that ÷ is a non-zero. øi. Let G ˆ Q8 ˆ ka. We shall see in Chapter 18 that G has characters ÷ and ø which take the following values on the conjugacy classes: Class representative |CG ( gi )| ÷ ø 1 24 3 3 (1 2) 4 À1 1 (1 2 3) 3 0 0 (1 2)(3 4) 8 3 À1 (1 2 3 4) 4 À1 À1 Calculate h÷. Let G ˆ S4 . and let r1 . i 0 1 0 2 3 2 3 À1 0 1 0 ar3 ˆ . 5. but r3 is not equivalent to r1 or r2. ÷i. øi and hø. b2 ˆ a2 . show that h÷reg . r3 be the representations of G over C for which 2 3 2 3 i 0 0 1 ar1 ˆ . and that for each g in G there is an invertible matrix Tg such that gó ˆ T À1 ( gr)Tg X g Prove that there is an invertible matrix T such that for all g in G. If ÷ is a character of G. Suppose that r and ó are representations of G. br3 ˆ X 0 1 0 À1 Show that r1 and r2 are equivalent. bÀ1 ab ˆ aÀ1 l. non-trivial character of G. br2 ˆ . ÷i ˆ ÷(1)X . 0 Ài À1 0 2 3 2 3 0 i 0 À1 ar2 ˆ . 3. h÷. 3 or 4? 8. . . If ð is the permutation character of Sn . ÷ k be the irreducible characters of the group G. and suppose that ø ˆ d 1 ÷1 ‡ X X Xd k ÷ k is a character of G.) 151 7. . Let ÷1 . øl ˆ 1.Inner products of characters 6. What can you say about the integers di in the cases kø. ÷( g) is an even integer. prove that hð. Suppose that ÷ is a character of G and that for every g P G. Does it follow that ÷ ˆ 2ö for some character ö? . . 1 S n i ˆ 1X (Hint: you may ®nd Exercise 11. 2.4 relevant. 152 . the theorem provides machinery for investigating characters which is used in the remainder of the book. Together with the material from Chapter 14. A basis of C is given by those functions which take the value 1 on precisely one conjugacy class and zero on all other classes.1 De®nition A class function on G is a function ø: G 3 C such that ø(x) ˆ ø( y) whenever x and y are conjugate elements of G (that is. ø is constant on conjugacy classes).5(2). then (15X2) dim C ˆ lX 15. G is as usual a ®nite group. The set C of all class functions on G is a subspace of the vector space of all functions from G to C. the characters of G are class functions on G.3 Theorem The number of irreducible characters of G is equal to the number of conjugacy classes of G. and to some consequences of this theorem.15 The number of irreducible characters We devote this chapter to the theorem which states that the number of irreducible characters of a ®nite group is equal to the number of conjugacy classes of the group. Thus. Throughout. By Proposition 13. Class functions 15. if l is the number of conjugacy classes of G. we consider the regular CG-module. X X X .The number of irreducible characters 153 Proof Let ÷1 . so (15. . Since Z(CG) has dimension l by Proposition 12. we deduce that l < k. By Theorem 14. vz ˆ ë i vX Hence wz ˆ ë i w for all w P W i . for each i there exists ë i P C such that for all v P Vi. ÷ k of G form a basis of the vector space of all class functions on G. X X X . . . they span a subspace of C of dimension k. j (1 < i < k)X 15. ÷ k are linearly independent. then øˆ where ë i ˆ kø. W i is isomorphic to a direct sum of copies of V i . . if ø is a class function. Now let z P Z(CG). the centre of CG. f iz ˆ ëi f i It follows that z ˆ 1z ˆ ( f 1 ‡ X X X ‡ f k )z ˆ f 1 z ‡ X X X ‡ f k z ˆ ë1 f 1 ‡ X X X ‡ ë k f k X This shows that Z(CG) is contained in the subspace of CG spanned by f 1 . . Proof Since ÷1 . we know from Theorem 8.2) implies that k < l. In order to prove the reverse inequality l < k. ÷ i l for 1 < i < k. By Proposition 9. ÷1 .23. This completes the proof that k ˆ l.22. dim C ˆ l.7 that CG ˆ W 1 È X X X È W k . we can write 1 ˆ f1 ‡ X X X ‡ fk with f i P W i for 1 < i < k. and let l be the number of conjugacy classes of G. By (15. V k is a complete set of non-isomorphic irreducible CG-modules. .2). . . X X X . and in particular. Since CG contains the identity element 1. . f k . where for each i. ÷ k are linearly independent elements of C. which is equal to k by k ˆ iˆ1 ëi÷i . . ÷ k be the irreducible characters of G.14. . . If V1 . Indeed.4 Corollary The irreducible characters ÷1 . ÷ k span C. whose k entries are the values of ÷ on the k conjugacy classes of G. . Then g is conjugate to gÀ1 if and only if ÷( g) is real for all characters ÷ of G.4 has the following useful consequence. by Proposition 13.9(3)). .3.13). j 15. .4 that there is a certain group G of order 12 which has exactly six conjugacy classes with representatives g1 . we regard a character ÷ of G as a row vector.5 Proposition Suppose that g. . The last part follows. and so g is conjugate to h. . . Then ø( g) ˆ ø(h) ˆ 1. the result follows immediately from Proposition 15. and so they form a basis of C. Proof Since ÷( g) is real if and only if ÷( g) ˆ ÷( gÀ1 ) (see Proposition j 13. 15. In particular. Proof If g is conjugate to h then ÷( g) ˆ ÷(h) for all characters ÷ of G. . and six irreducible characters ÷1 .6 Corollary Suppose that g P G. As in previous examples.5(2). ø( g) ˆ ø(h) for all class functions ø on G. ÷6 given as follows: . . Then g is conjugate to h if and only if ÷( g) ˆ ÷(h) for all characters ÷ of G. j Corollary 15. h P G. this is true for the class function ø which takes the value 1 on the conjugacy class of g and takes the value 0 elsewhere. Hence ÷1 . 15. Conversely.5.4. Then by Corollary 15. g6 (where g1 ˆ 1). .7 Example We shall see in Section 18. . .154 Representations and characters of groups Theorem 15. We conclude the chapter with an example illustrating some practical methods of expressing characters and class functions of a group as combinations of irreducible characters. using (14. suppose that ÷( g) ˆ ÷(h) for all characters ÷. . . they are the degrees of the ÷ i ). . . ÷6 : g1 ë ì 2 4 g2 À2 4 g3 À2 1 g4 2 1 g5 0 0 g6 0 0 .The number of irreducible characters gi |CG ( g i )| ÷1 ÷2 ÷3 ÷4 ÷5 ÷6 g1 12 1 1 1 1 2 2 g2 12 1 À1 1 À1 2 À2 g3 6 1 À1 1 À1 À1 1 g4 6 1 1 1 1 À1 À1 g5 4 1 i À1 Ài 0 0 155 g6 4 1 Ài À1 i 0 0 Suppose we are given characters ÷ and ø of G as follows: g1 ÷ ø 3 4 g2 À3 0 g3 0 0 g4 0 4 g5 i 0 g6 Ài 0 Then it is easy to spot that ÷ ˆ ÷2 ‡ ÷6 . The reason for this is that the required coef®cients are known to be non-negative integers. ì of G as combinations of ÷1 . ÷4 and ÷6 . ø ˆ ÷1 ‡ ÷2 ‡ ÷3 ‡ ÷4 X For example. We suggest that you use the `guesswork method' to express the following characters ë. the second entry in the row vector for ÷ is equal to minus the ®rst entry. The correct answer now comes quickly to mind. we see that ÷ must be a combination of ÷2 . it is not hard to use tactical guesswork to express ö as a combination of the irreducible characters. . In fact. Inspecting the values of the irreducible characters ÷ i . given any character ö of G whose degree is not large compared with the degrees of the ÷ i . and the entries in the column corresponding to g1 ˆ 1 are positive integers (indeed. 5(2). 1 (À1 ‡ 2i) . ÷2 i ˆ ‡ ‡ ‡ ‡ 12 12 6 6 4 (À1 À 2i) . 1 ‡ ‡ ‡ ‡ 12 12 6 6 4 (À1 À 2i) . (À1) ‡ ˆ 2. 4 11 . . ÷6 l ˆ 0. 1 5 . (À1) hö. (À1) 5 . kö. 4 hö. 1 À3 . 1 5 . (À1) (À3) . we calculate these inner products: 11 . 1 (À1 ‡ 2i) . l. 1 3 .156 Representations and characters of groups How do we cope with a class function or with a more dif®cult character. 1 3 . 1 À3 . 1 ‡ ˆ 1. ÷3 i ˆ ‡ ‡ ‡ ‡ 12 12 6 6 4 (À1 À 2i) . ÷5 l ˆ 2 and kö. 1 (À1 ‡ 2i) . 1 3 . The number of irreducible characters of a group is equal to the number of conjugacy classes of the group. ‡ 4 11 .4 that the coef®cients ë i in the expression ö ˆ ë1 ÷1 ‡ X X X ‡ ë6 ÷6 are given by ë i ˆ hö. (Ài) hö. Therefore ö ˆ ÷1 ‡ 3÷2 ‡ 2÷3 ‡ ÷4 ‡ 2÷5 X Summary of Chapter 15 1. We know from Corollary 15. i ˆ 3. ÷ i i (1 < i < 6)X Using Proposition 14. ÷1 i ˆ and similarly kö. ÷4 l ˆ 1. like the following one? g1 ö 11 g2 3 g3 À3 g4 5 g5 À1 ‡ 2i g6 À1 À 2i The answer is to use the inner product k . ø2 and ø3 be the class functions on S3 taking the following values: 1 ø1 ø2 ø3 1 0 0 (1 2) 0 1 0 (1 2 3) 0 0 1 Express ø1 . ÷ k of G form a basis of the vector space of all class functions on G. The irreducible characters ÷1 . ø2 and ø3 as linear combinations of the irreducible characters ÷1 . If ø is a class function. ÷ i iX Exercises for Chapter 15 1. . Let ø1 . . ÷2 and ÷3 of S3 . . The three irreducible characters of S3 are ÷1 . . ÷2 . ÷3 : 1 ÷1 ÷2 ÷3 1 1 2 (1 2) 1 À1 0 (1 2 3) 1 1 À1 Let ÷ be the class function on S3 with the following values: 1 ÷ 19 (1 2) À1 (1 2 3) À2 Express ÷ as a linear combination of ÷1 . Is ÷ a character of S3 ? 2.The number of irreducible characters 157 2. . ÷2 and ÷3 . then øˆ k ˆ iˆ1 ë i ÷ i where ë i ˆ hø. (Hint: show that Z(G) cannot have order 6. ÷6 . . Let G be a group of order 12. with conjugacy class representatives g1 . . .2. (a) Show that G cannot have exactly 9 conjugacy classes. ÷6 as in that example. . . . . Suppose that G is the group of order 12 in Example 15. . . .7. . 6 or 12 conjugacy classes.158 Representations and characters of groups 3.) (b) Using the solution to Exercise 11. prove that G has 4. . . Let ø be the class function on G taking the following values: g1 ø 6 g2 0 g3 3 g4 À3 g5 À1 À i g6 À1 ‡ i Express ø as a linear combination of ÷1 . g6 and irreducible characters ÷1 . Is ø a character of G? 4. Find groups G in which each of these possibilities is realized. 4). Note that in the character table. g k be representatives of the conjugacy classes of G. Much of the later material in the book will be devoted to understanding character tables. Character tables 16. j with 1 < i < k. The motivation for this is Theorem 14. which tells us that every CG-module is determined by its character. ÷ k be the irreducible characters of G and let g 1 . the numbering is arbitrary. It is therefore convenient to record all the values of all the irreducible characters of G in a square matrix.1 De®nition Let ÷1 . the rows are indexed by the irreducible characters of G and the columns are indexed by the conjugacy classes (or. It is usual to number the irreducible characters and conjugacy classes of G so that ÷1 ˆ 1 G . 159 . and the number of them is equal to the number of conjugacy classes of G. many of which are encapsulated in the orthogonality relations (Theorem 16. the identity element of G. Thus. The k 3 k matrix whose ij-entry is ÷ i ( g j ) (for all i. by conjugacy class representatives). the trivial character. Beyond this. and g1 ˆ 1. X X X .21. X X X . The entries in a character table are related to each other in subtle ways. many problems in representation theory can be solved by considering characters. This matrix is called the character table of G.16 Character tables and orthogonality relations The irreducible characters of a ®nite group G are class functions. in practice. 1 < j < k) is called the character table of G. and hence also the rows of the character table. bÀ1 ab ˆ aÀ1 l. b as representatives of the conjugacy classes of G. You found all the irreducible representations of G in Exercise 10. The irreducible characters of G are given in Example 13. a. the character table of C2 ˆ ha: a2 ˆ 1i is 1 ÷1 ÷2 1 1 a 1 À1 and the character table of C3 ˆ ka: a3 ˆ 1l is 1 ÷1 ÷2 ÷3 1 1 1 a 1 ù ù2 a2 1 ù2 ù (ù ˆ e2ðia3 ) (3) Let G ˆ D8 ˆ ka.3 Examples (1) Let G ˆ D6 ˆ ka. For example. We take 1. j 16.8. Proof This follows immediately from the fact that the irreducible characters of G. b: a3 ˆ b2 ˆ 1.4. and then the character table of G is 1 ÷1 ÷2 ÷3 1 1 2 a 1 1 À1 b 1 À1 0 (2) We can write down the character table of any ®nite abelian group using Theorem 9.23).6(4). are linearly independent (Theorem 14.2 Proposition The character table of G is an invertible matrix. bÀ1 ab ˆ aÀ1 l. The conjugacy classes .160 Representations and characters of groups 16. b: a4 ˆ b2 ˆ 1. a2 . and these are given by part (2) of our next result. The orthogonality relations We have already seen many uses for the relations (14. . s P {1. . h÷ r . Then the following relations hold for any r. Similar relations exist between the columns of the character table. . .12). Hence the character table of G is 1 ÷1 ÷2 ÷3 ÷4 ÷5 1 1 1 1 2 a2 1 1 1 1 À2 a 1 1 À1 À1 0 b 1 À1 1 À1 0 ab 1 À1 À1 1 0 The character tables of all dihedral groups will be found in Chapter 18. X X X . .4 Theorem Let ÷1 . (1) The row orthogonality relations: k ˆ ÷ r ( g i )÷ s ( g i ) iˆ1 jCG ( g i )j ˆ ä rs X (2) The column orthogonality relations: k ˆ iˆ1 ÷ i ( g r )÷ i ( g s ) ˆ ä rs jCG ( g r )jX . These relations can be expressed in terms of the rows of the character table. by writing them as k ˆ ÷ r ( g i )÷ s ( g i ) ˆ ä rs jCG ( g i )j iˆ1 (see Proposition 14. a. 16. among the irreducible characters ÷1 . ÷ k of G.5(2)). . . . and representatives are 1. and let g 1 . .Character tables and orthogonality relations 161 of G are given by (12. b. ÷ k be the irreducible characters of G. k}. ÷ s i ˆ ä rs . ab. . g k be representatives of the conjugacy classes of G.13). . . and ø s ( g) ˆ 0 otherwise. For 1 < s < k.3(1). Hence ëi ˆ Therefore ä rs ˆ ø s ( g r ) ˆ k ˆ iˆ1 1 ˆ ÷ i ( gs ) ø s ( g)÷ i ( g) ˆ X jGj gP g G jCG ( g s )j s ëi ÷i ( gr ) ˆ k ˆ ÷ i ( g r )÷ i ( g s ) iˆ1 jCG ( g s )j . ÷ i i ˆ 1 ˆ ø s ( g)÷ i ( g)X jGj gPG Now ø s ( g) ˆ 1 if g is conjugate to g s . by Theorem 12. . and this time we record the order of the centralizer C G ( g i ) next to each conjugacy class representative gi : gi |CG ( g i )| ÷1 ÷2 ÷3 1 6 1 1 2 a 3 1 1 À1 b 2 1 À1 0 .162 Representations and characters of groups Proof The row orthogonality relations have already been proved. so ë i ˆ hø s . We copy the character table of G from Example 16. also there are jGjajC G ( g s )j elements of G which are conjugate to g s . . . ø s is a linear combination of ÷1 . They are recorded here merely for comparison with the column relations.5 Examples We illustrate the column orthogonality relations. ÷ k . j and the column orthogonality relations follow. ÷ j i ˆ ä ij .4. say øs ˆ ëi ÷i (ë i P C)X We know that h÷ i . . (1) Let G ˆ D6.8. let ø s be the class function which satis®es ø s ( g r ) ˆ ä rs k ˆ iˆ1 (1 < r < k)X By Corollary 15. 16. 12). (À1) ˆ 3. By the column orthogonality relations with r ˆ s ˆ 1. 1 ‡ 1 . 1 . 1 ‡ (À1) . Hence the last entry in the ®rst column is 3. The entries in the ®rst column of the character table are the degrees of the irreducible characters. By considering the orthogonality relations between the ®rst column and columns 3 and 4. 1 .Character tables and orthogonality relations € Consider the sums 3 ÷ i ( g r )÷ i ( g s ) for various cases: iˆ1 r ˆ 1. the sum of the squares of these numbers is 12 (this also follows from Theorem 11. we obtain the complete character table as . We shall use the column orthogonality relations to determine the last row of the character table. the order of the centralizer of g r ) if r ˆ s. The column orthogonality relation 4 ˆ iˆ1 ÷ i ( g 1 )÷ i ( g 2 ) ˆ 0 gives 1 . r ˆ 1. s ˆ 2: s ˆ 2: s ˆ 3: 1 . 0 ˆ 0X 163 In each case. so they are positive integers. (2) Suppose we are given the following part of the character table of a group G of order 12 which has exactly four conjugacy classes: gi |CG ( g i )| ÷1 ÷2 ÷3 ÷4 g1 12 1 1 1 g2 4 1 1 1 g3 3 1 ù ù2 g4 3 1 ù2 ù (where ù ˆ e2ðia3 ). taking the products of the numbers which appear. (À1) ˆ 0. 1 ‡ 1 . and is the number at the top of the column (that is. Let x denote the number at the foot of the second column. 1 ‡ 2 . (À1) ‡ 2 . 1 ‡ 3x ˆ 0X Therefore x ˆ À1. The sum of the products is 0 if r Tˆ s. we read down columns r and s of the character table. r ˆ 2. 1 ‡ 1 . 1 ‡ 1 . 1 ‡ 1 . if g ˆ 1. 1 ‡ ù . ù2 ‡ 0 . ˆ d i ÷ i ( g) ˆ X iˆ1 0.164 follows: gi |CG ( g i )| ÷1 ÷2 ÷3 ÷4 Representations and characters of groups g1 12 1 1 1 3 g2 4 1 1 1 À1 g3 3 1 ù ù2 0 g4 3 1 ù2 ù 0 Notice that the orthogonality relations hold between all pairs of columns. (À1) ˆ 4. if g Tˆ 1. 0 ˆ 0X We shall see later that the character table which we have constructed here is that of A4 . if g Tˆ 1. By taking the complex conjugate of each side of this equation. 1 ‡ ù . although our calculation has used only those relations which involve the ®rst column. ù2 ‡ ù2 . For example. ÷ i ( g 3 )÷ i ( g 3 ) ˆ 1 .20 give V k ` jGj. ù ‡ 0 . Those column orthogonality relations which involve the ®rst column of the character table were proved in Chapter 13. since Theorem 13. k ˆ ÷ i (1)÷ i ( g) ˆ X iˆ1 0. if g ˆ 1. 1 ‡ (À1) . ù ‡ ù2 . 1 ‡ 1 . 1 ‡ 1 . ÷ i ( g 3 )÷ i ( g 4 ) ˆ 1 . 0 ˆ 3. we get V ` jGj. .19 and Proposition 13. where d i ˆ ÷ i (1). 4 ˆ iˆ1 4 ˆ iˆ1 4 ˆ iˆ1 ÷ i ( g 2 )÷ i ( g 2 ) ˆ 1 . On the other hand.Character tables and orthogonality relations 165 and these are just the column orthogonality relations which involve the ®rst column. ÷4 i ˆ ä i4 to obtain four equations in the four unknown values ÷4 ( g j ) (1 < j < 4). Now the rs-entry in M M t is k ˆ ÷ r ( g i )÷ s ( g i ) iˆ1 jCG ( g i )j ˆ ä rs . the row and column orthogonality relations encapsulate the same information. the rs-entry in M t M is k ˆ 1 ÷ i ( g r )÷ i ( g s ) ˆ ä rs . where we were given three of the four irreducible characters of G. we see that the row and column orthogonality relations are equivalent. it is a fact that the column orthogonality relations contain precisely the same information as the row orthogonality relations. More importantly. The character table of G is a k 3 k matrix. Since the properties M t M ˆ I and M M t ˆ I of a square matrix M are equivalent to each other. Indeed. Rows versus columns Notice that in Example 16. by the row orthogonality relations. so M M t ˆ I. jCG ( g r )j1a2 jCG ( g s )j1a2 iˆ1 by the column orthogonality relations. . by letting the ij-entry of M be ÷ i ( gj ) X jCG ( g j )j1a2 Let M t denote the transpose of the complex conjugate of M.5(2). We could have used the above argument to deduce the column orthogonality relations from the row ones. the equation M M t ˆ I is just another way of expressing the row orthogonality relations. Although the calculation with the column orthogonality relations was easier to perform. so M t M ˆ I. as we shall now show. we calculated the values of the last character one at a time using the column orthogonality relations. An alternative approach would have been to use the row orthogonality relations h÷ i . and we adjust the entries ÷ i ( g j ) in this matrix to obtain another k 3 k matrix M. . . . 2. with representatives g1 . 3. . . k ˆ iˆ1 ÷ i ( g r )÷ i ( g s ) ˆ ä rs jCG ( g r )jX Exercises for Chapter 16 1. . Summary of Chapter 16 Let G be a ®nite group with irreducible characters ÷1 . k ˆ ÷ r ( g i )÷ s ( g i ) iˆ1 jCG ( g i )j ˆ ä rs X 3. g5 . . . A certain group G of order 8 is known to have a total of ®ve conjugacy classes. . Write down the character table of C2 3 C2 . The column orthogonality relations state that for all r. . . and four linear characters ÷1 . . gk . . s. The row orthogonality relations state that for all r. There exists a group G of order 10 which has precisely four conjugacy classes. ÷ k and conjugacy class representatives g1 . ÷4 taking the following values: gi |CG ( g i )| ÷1 ÷2 ÷3 ÷4 g1 8 1 1 1 1 g2 8 1 1 1 1 g3 4 1 1 À1 À1 g4 4 1 À1 1 À1 g5 4 1 À1 À1 1 Find the complete character table of G. . 2. g4 . . . . . ÷2 as follows: . we can deduce exactly the same results using either set of relations. The character table of G is the k 3 k matrix with ij-entry ÷ i ( g j ). . and has irreducible characters ÷1 . 1. s.166 Representations and characters of groups so when we are working with character tables. with representatives g1 . . Find the complete character table of G. then on g4 ± use Corollary 13.10. (a) Find æ. . 5. k ‰ iˆ1 jCG ( g i )jX . Let G be a ®nite group with conjugacy class representatives g1 . ÷ k be the irreducible characters of G. (b) Find another column of the character table. X X X . . (Hint: ®rst ®nd the values of the remaining irreducible characters on g1 . Show that det C is either real or purely imaginary. Let ÷1 .) 4. Show that @ A k ˆ Z(G) ˆ g P G: ÷ i ( g)÷ i ( g) ˆ jGj X iˆ1 6. . . g k and character table C.Character tables and orthogonality relations gi |CG ( g i )| ÷1 ÷2 g1 10 1 2 g2 5 1 á g3 5 1 â g4 2 1 0 167 p p where á ˆ (À1 ‡ 5)a2 and ⠈ (À1 À 5)a2. and that jdet Cj2 ˆ Find Æ(det C) when G ˆ C3 . A certain group G has two columns of its character table as follows: gi |CG ( g i )| ÷1 ÷2 ÷3 ÷4 ÷5 g1 21 1 1 1 3 3 g2 7 1 1 1 æ æ where g1 ˆ 1 and æ P C. in a way which we shall describe. can be used to get new irreducible characters from a given irreducible character. The characters of GaN should therefore be easier to ®nd than the characters of G. in turn. it is easy to tell from the character table whether or not G is simple.e.7. De®ne ÷ ÷: G 3 C by ÷( g) ˆ ~(Ng) ( g P G)X ÷ Then ÷ is a character of G. Lifted characters We begin by constructing a character of G from a character of GaN . In the opposite direction. it is also true that the character table of G enables us to ®nd the normal subgroups of G. the characters of degree 1) are obtained by lifting the irreducible characters of GaN in the case where N is the derived subgroup of G.1 Proposition Assume that N v G. (The derived subgroup is de®ned below in De®nition 17.) The linear characters. In fact. in particular. and ÷ and ~ have the same degree. The linear characters of G (i. C) be a representation of GaN with 168 . and N Tˆ {1}. then the factor group GaN is smaller than G. we can use the characters of GaN to get some of the characters of G. 17.17 Normal subgroups and lifted characters If N is a normal subgroup of the ®nite group G. normal subgroups help us to ®nd characters of G. and let ~ be a character of GaN . ÷ ~ Proof Let r: GaN 3 GL (n. by a process which is known as lifting. Thus. C). C) is a representation of G with character ÷. then ÷ ~(N) ˆ ÷(1). ÷ 17. and ÷ is the lift of ~ to G. Now let ÷ be a character of G with N < Ker ÷. ÷(1) ˆ ~(N).3 Theorem Assume that N v G. r r ( g P G)X ( g P G) . ÷ ÷ so N < Ker ÷.2 De®nition If N v G and ~ is a character of GaN . Thus r is a representation of G. j 17. if k P N then ÷ ÷(k) ˆ ~(Nk) ˆ ~(N ) ˆ ÷(1). Irreducible characters of GaN correspond to irreducible characters of G which have N in their kernel. If g 1 . The function r: G 3 GL (n. Suppose that r: G 3 GL (n. so ÷ and ~ have the same ÷ ÷ degree. By associating each character of GaN with its lift to G. Also. then the character ÷ of G ÷ which is given by ÷( g) ˆ ~(Ng) ÷ is called the lift of ~ to G.Normal subgroups and lifted characters 169 character ~. Moreover. C) by (Ng)~ ˆ gr r Then for all g. C) which is given by the ÷ composition g 3 Ng 3 (Ng)~ r ( g P G) is a homomorphism from G to GL (n. The character ÷ of r satis®es ÷( g) ˆ tr ( gr) ˆ tr ((Ng)~) ˆ ~(Ng) r ÷ for all g P G. We may therefore de®ne a function r: GaN 3 GL (n. ~ Proof If ÷ is a character of GaN . and hence 2 2 ~ g1 r ˆ g2 r. g2 P G and Ng1 ˆ Ng2 then g1 gÀ1 P N. we obtain a bijective correspondence between the set of characters of GaN and the set of characters ÷ of G which satisfy N < Ker ÷. so ( g1 gÀ1 )r ˆ I. h P G we have ((Ng)(Nh))~ ˆ (Ngh)~ ˆ ( gh)r ˆ ( gr)(hr) r r ˆ ((Ng)~)((Nh)~). so that N v G (see Example 12. (1 3)(2 4). (1 4)(2 3)g.170 Representations and characters of groups ~ ~ so r is a representation of GaN . Hence ÷ is irreducible if and only if ~ is irreducible. We know from Example 16. let U be a subspace of C n . If we put a ˆ N(1 2 3) and b ˆ N(1 2) then GaN ˆ ha.4 Example Let G ˆ S4 and N ˆ V4 ˆ f1. 17. (1 2)(3 4). and note that u( gr) P U for all u P U D u(Ng)~ P U for all u P U X r Thus.3 enables us to write down as many irreducible characters of G as there are irreducible characters of GaN . To see this. then Theorem 17. ÷ j If we know the character table of GaN for some normal subgroup N of G. It remains to show that irreducible characters correspond to irreducible characters.3(1) that the character table of GaN is N ~1 ÷ ~2 ÷ ~3 ÷ 1 1 2 N (1 2) 1 À1 0 N(1 2 3) 1 1 À1 . bi and a3 ˆ b2 ˆ N . We have now established that the function which sends each character of GaN to its lift to G is a bijection between the set of characters of GaN and the set of characters of G which have N in their kernel.20). so GaN  D6 . The representation r is therefore irreducible if and ~ only if the representation r is irreducible. U is a CG-submodule of C n if and only if U is a C(GaN )submodule of C n . bÀ1 ab ˆ aÀ1 . If ~ is the character of r then ÷ ~(Ng) ˆ ÷( g) ( g P G)X ÷ ~ Thus ÷ is the lift of ÷. we note that ÷ ÷((1 2)(3 4)) ˆ ~(N ) since (1 2)(3 4) P N .5 Proposition If N v G then there exist irreducible characters ÷1 . First we shall show how to ®nd all the normal subgroups of G. so g ˆ 1 by Proposition 15.Normal subgroups and lifted characters To calculate the lift ÷ of a character ~ of GaN . X X X . ~3 are ÷ ÷ ÷ irreducible characters of GaN . ÷2 .5. ~2 . since Ker ÷ ˆ f g P G: ÷( g) ˆ ÷(1)g (see De®nition 13. 17. The following proposition shows that every normal subgroup arises in this way. ~3 are ÷1 . Hence the intersection of the kernels of all the irreducible characters of G is {1}. Finding normal subgroups The character table contains accessible information about the structure of a group. ~2 . ÷2 . any subgroup which is the intersection of the kernels of irreducible characters is a normal subgroup too.13). then ÷( g) ˆ ÷(1) for all characters ÷. as our next two propositions will demonstrate. ÷3 . Recall that we can easily locate the kernel of an irreducible character ÷ from the character table. ÷3 are irreducible characters of G. ÷ ÷((1 2 3 4)) ˆ ~(N (1 3)) since N (1 2 3 4) ˆ N (1 3)X ÷ Hence the lifts of ~1 . ÷ s of G such that Nˆ s ’ iˆ1 Ker ÷ i X Proof If g belongs to the kernel of each irreducible character of G. once the character table of G is known. since ~1 . Of course. Also Ker ÷ v G. . which are given by ÷ ÷ ÷ 1 ÷1 ÷2 ÷3 1 1 2 (1 2) 1 À1 0 (1 2 3) 1 1 À1 (1 2)(3 4) 1 1 2 171 (1 2 3 4) 1 À1 0 Then ÷1 . so Ker ÷ Tˆ {1}.5. Since ÷ is non-trivial and irreducible. Therefore if g P Ker ÷ i then Ng P Ker ~ i ˆ ÷ ÷ {N}. hence Ker ÷ Tˆ G. since the .11(2). If r is a representation of G with character ÷. let ÷ i be the lift to G of ~ i . Thus Ker ÷ is a normal subgroup of G which is not equal to {1} of G. suppose that G is not simple. then Ker ÷ ˆ Ker r by Theorem 13. If g P Ker ÷ i then ÷ ~ i (N ) ˆ ÷ i (1) ˆ ÷ i ( g) ˆ ~ i (Ng). We shall show how to ®nd all linear characters of any group G. ÷ is non-trivial. there is an irreducible character ÷ of G such that Ker ÷ is not {1} or G. and so g P N.6 Proposition The group G is not simple if and only if ÷( g) ˆ ÷(1) for some non-trivial irreducible character ÷ of G. Hence Nˆ s ’ iˆ1 Ker ÷ i X j It is particularly easy to tell from the character table of G whether or not G is simple: 17. ÷ ÷ „ „ and so Ng P Ker ~ i. . Ker r Tˆ G. and some nonidentity element g of G. Then g P Ker ÷. and taking 1 Tˆ g P Ker ÷. . ~ s be the irreducible characters of GaN . and so G is not simple. By the ÷ ÷ above observation. s ’ iˆ1 Ker ~ i ˆ fN gX ÷ For 1 < i < s. so that there is a normal subgroup N of G with N Tˆ {1} and N Tˆ G. . Proof Suppose there is a non-trivial irreducible character ÷ such that ÷( g) ˆ ÷(1) for some non-identity element g. As Ker ÷ Tˆ G.172 Representations and characters of groups Now let ~1 . j Linear characters Recall that a linear character of a group is a character of degree 1. Then by Proposition 17. Conversely. we have ÷( g) ˆ ÷(1). . h]. it is necessary to determine the derived subgroup of G. Clearly [ g. One step is provided by the following proposition. . which is de®ned in the following way.Normal subgroups and lifted characters 173 ®rst move in constructing the character table of G is often to write down the linear characters. for all g. 17. Therefore. h]: g. Thus G9 ˆ h[ g. 17. let G9 be the subgroup of G which is generated by all elements of the form g À1 hÀ1 gh ( g. We are going to show that G9 v G and that the linear characters of G are the lifts to G of the irreducible characters of GaG9. We abbreviate gÀ1 hÀ1 gh as [ g. Then ÷ is a homomorphism from G to the multiplicative group of non-zero complex numbers. j Next. h P G)X Then G9 is called the derived subgroup of G. h P GiX 17. h P G. h P G. h] is always an even permutation. then G9 < Ker ÷. h] ˆ (1 2 3). (2) Let G ˆ S3 . so G9 < A3 .8 Examples (1) If G is abelian then [ g. As a preliminary step.9 Proposition If ÷ is a linear character of G. so G9 ˆ {1}. we explore some group-theoretic properties of the derived subgroup. h] ˆ 1 for all g.7 De®nition For a group G. If g ˆ (1 2) and h ˆ (2 3) then [ g. Hence G9 ˆ h(1 2 3)i ˆ A3 . Proof Let ÷ be a linear character of G. ÷( g À1 hÀ1 gh) ˆ ÷( g)À1 ÷(h)À1 ÷( g)÷(h) ˆ 1X Hence G9 < Ker ÷. j It follows from Proposition 17. . h] and their inverses. 17. We have ghgÀ1 hÀ1 P N D Ngh ˆ Nhg D (Ng)(Nh) ˆ (Nh)(Ng)X Hence G9 < N if and only if GaN is abelian.11 Theorem The linear characters of G are precisely the lifts to G of the irreducible characters of GaG9.3 they are precisely the irreducible characters ÷ of G . GaG9 is abelian. (2) Let g.10 that G9 is the smallest normal subgroup of G with abelian factor group. . Theorem 9. (1) G9 v G. and by Theorem 17. h. all of degree 1. the number of distinct linear characters of G is equal to jGaG9j. . b. . Since GaG9 is abelian. to prove that G9 v G it is suf®cient by the ®rst sentence to prove that x À1 [ g.8 shows that GaG9 has exactly m irreducible characters ~1 . In particular. Since we have proved that G9 v G. Proof Let m ˆ jGaG9j. we have x À1 (ab)x ˆ (x À1 ax)(x À1 bx). (2) G9 < N if and only if GaN is abelian. ~ m . x À1 hx]X Therefore G9 v G. and x À1 aÀ1 x ˆ (x À1 ax)À1 X Now G9 consists of products of elements of the form [ g. ÷ m of these characters to G also have degree 1.10 Proposition Assume that N v G. In particular. x P G. h P G. Given the derived subgroup G9. we can obtain the linear characters of G by applying the next theorem. Proof (1) Note that for all a. Therefore. . . . x P G. and so divides |G|. h]x ˆ x À1 g À1 hÀ1 ghx ˆ (x À1 gx)À1 (x À1 hx)À1 (x À1 gx)(x À1 hx) ˆ [x À1 gx. .174 Representations and characters of groups 17. But x À1 [ g. h]x P G9 for all g. ÷ ÷ The lifts ÷1 . we deduce that GaG9 is abelian. if g P An X a À1. ÷2 . as the next result shows. we have G9 < A n by Proposition 17. 2). . a 3-cycle or an element of cycle-shape (2. which are given by ÷1 ˆ 1 Sn . h ˆ (2 3) and k ˆ (1 2)(3 4). Not only are the linear characters of G important in being irreducible characters. j 17. But every product of two transpositions is equal to the identity. so G9 ˆ {1} ˆ A n . h] ˆ (1 2 3). k] ˆ (1 4)(2 3)X Since G9 v G. An (1 2)g  C2 . From the last example.Normal subgroups and lifted characters 175 such that G9 < Ker ÷. We proved that S9 ˆ A3 in Example 3 17. We have now proved that G9 ˆ A n .12 Example Let G ˆ S n . ÷ m are therefore all the linear characters of G. but they can also be used to construct new irreducible characters from old. [h. . If g ˆ (1 2). the characters ÷1 . and A n consists of permutations. Therefore. all the elements in (1 2 3) G and (1 4)(2 3) G belong to G9. 2).10(2). each of which is the product of an even number of transpositions. then [ g.9. @ ÷2 ( g) ˆ 1. Therefore A n < G9. the group n n S n aS9 has two linear characters ~1 and ~2.8(2). . . we know that S9 ˆ An . if g P An .15. S n has exactly two linear characters ÷1 . 17. We shall show that G9 ˆ An . In view of Proposition 17. where ÷ ÷ n ~1 (An (1 2)) ˆ 1. As S n aA n  C2 . by Theorem 12. .11.13 Example We ®nd the linear characters of S n (n > 2). so we assume that n > 4. Since Sn aS9 ˆ fAn . If n ˆ 1 or 2 then S n is abelian. G9 contains all 3-cycles and all elements of cycle-shape (2. ÷ ~2 (An (1 2)) ˆ À1X ÷ Therefore by Theorem 17. C) by g(rë) ˆ ë( g)( gr) ( g P G)X Thus g(rë) is the matrix gr multiplied by the complex number ë( g). 3. De®ne rë: G 3 GL (n. if ÷ is irreducible. . the complex number ë( g) is a root of unity. The normal subgroups of G can be found from the character table of G. The matrix g(rë) has trace ë( g) tr ( gr). so ë( g)ë( g) ˆ 1.14 Proposition Suppose that ÷ is a character of G and ë is a linear character of G. ÷iX jGj gPG By Theorem 14. The character of G which corresponds to the character ~ ÷ of GaN is the lift of ~. Now for all g P G. ÷ëi ˆ ÷( g)ë( g)÷( g)ë( g) jGj gPG ˆ 1 ˆ ÷( g)÷( g) ˆ h÷. Proof Let r: G 3 GL (n. Then the product ÷ë. ÷ ÷ 2. Summary of Chapter 17 1. it follows that ÷ë is irreducible if and only if ÷ is irreducible. Therefore 1 ˆ h÷ë. Characters of GaN correspond to characters ÷ of G for which N < Ker ÷. de®ned by ÷ë( g) ˆ ÷( g)ë( g) ( g P G) is a character of G. and is given by ÷( g) ˆ ~(Ng) ( g P G). which is ë( g)÷( g). The linear characters of G are precisely the lifts to G of the irreducible characters of GaG9. then so is ÷ë. Hence rë is a representation of G with character ÷ë.20. C) be a representation with character ÷. j The general case of a product of two characters will be discussed in Chapter 19.176 Representations and characters of groups 17. Since r and ë are homomorphisms it follows easily that rë is a homomorphism. Moreover. with representatives g1 . (a) Find the ®ve conjugacy classes of G. Let a and b be the following permutations in S7 : a ˆ (1 2 3 4 5 6 7). . 4 or 12 linear characters. and has irreducible characters ÷. (b) Find the conjugacy classes of G. a3 1 1 À1 À1 0 b. (c) Find the character table of G. g6 (where g1 ˆ 1). . ö with values as follows: g1 ÷ ö 1 2 g2 Ài 0 g3 i 0 g4 1 À1 g5 À1 À1 g6 À1 2 Use Proposition 17.14 to complete the character table of G.3(3)). . 2. Show that every group of order 12 has 3. The character table of D8 is as shown (see Example 16. Compare your table with the character table of D8 (Example 16. Let G ˆ Q8 ˆ ka. 4. a2 b 1 À1 1 À1 0 ab. and construct all the linear characters of G. b: a4 ˆ 1. . What are the sizes of the conjugacy classes of G? 5. a3 b 1 À1 À1 1 0 . bÀ1 ab ˆ a2 X (a) Show that G has order 21. 3. b ˆ (2 3 5)(4 7 6)X Let G ˆ ka. bÀ1 ab ˆ aÀ1 l. and hence cannot be simple. (c) Complete the character table of G. Check that a7 ˆ b3 ˆ 1.3(3)): 1 ÷1 ÷2 ÷3 ÷4 ÷5 1 1 1 1 2 a2 1 1 1 1 À2 a. b2 ˆ a2 . A certain group G of order 12 has precisely six conjugacy classes.Normal subgroups and lifted characters Exercises for Chapter 17 177 1. bl. (b) Find G9. b3 å 0 0 ù2 (b) Find all the irreducible representations of U6 n. For n > 1. as in Proposition 17. then there is a representation of V8 n over C which sends     å 0 0 1 X a3 . 7. aÀ1 ba ˆ bÀ1 i has order 6n.178 Representations and characters of groups Express each normal subgroup of D8 as an intersection of kernels of irreducible characters. bÀ1 a ˆ aÀ1 bi has order 8n. then there is a representation of T4 n over C which sends     å 0 0 1 a3 . bÀ1 ab ˆ aÀ1 i has order 4n. Let n be an odd positive integer.) (a) Show that if å is any (2n)th root of unity in C. b: a2 n ˆ b3 ˆ 1. (a) Let ù ˆ e2ðia3 . Show that if å is any (2n)th root of unity in C.b3 0 Àå À1 À1 0 (b) Find all the irreducible representations of V8 n . ba ˆ aÀ1 bÀ1 . 8. (a) Show that if å is any nth root of unity in C. b: a2 n ˆ 1. . The group V8 n ˆ ha. the group U6 n ˆ ha. then there is a representation of U6n over C which sends     0 å ù 0 X a3 . (It is known as a dicyclic group.b3 X 0 å À1 ån 0 (b) Find all the irreducible representations of T4 n. an ˆ b2 .5. You are given that the group T4 n ˆ ha. 6. b: a2 n ˆ b4 ˆ 1. including the groups S4 and A4 .1 The group S4 In Example 17. By Proposition 17.18 Some elementary character tables We now illustrate the techniques we have presented so far by constructing the character tables of several groups.4. The values of ÷2 . 24 4 8 4 179 . which deals with the product of a character with a linear character. ÷2 . ÷4 i ˆ 9 1 1 1 ‡ ‡ ‡ ˆ 1. and all dihedral groups. we produced three irreducible characters ÷1 . the product ÷4 ÷2 is also a character of S4 .24. We shall now use Proposition 17. Let ÷4 be the character ÷4 ( g) ˆ jfix ( g)j À 1 ( g P S4 ) which is given in Proposition 13. 18. ÷3 of S4 by lifting characters of the factor group S4 aV4 .14.14. ÷4 and ÷4 ÷2 are as follows: gi |CG ( gi )| ÷2 ÷4 ÷4 ÷2 1 24 1 3 3 (1 2) 4 À1 1 À1 (1 2 3) 3 1 0 0 (1 2)(3 4) 8 1 À1 À1 (1 2 3 4) 4 À1 À1 1 Note that h÷4 . to complete the character table of S4 . The values of í are as follows: gi |CG ( gi )| í 1 12 3 (1 2)(3 4) 4 À1 (1 2 3) 3 0 (1 3 2) 3 0 Note that hí. 12 4 so í is an irreducible character of G of degree 3.24. The character ÷4 ÷2 is also irreducible. (1 3 2) (see Example 12.180 Representations and characters of groups so ÷4 is irreducible. we have now found the complete character table of S4 . and we have produced ®ve irreducible characters. It is not dif®cult to con®rm this by showing that . and G has four conjugacy classes. Thus jGaG9j ˆ 3 by Theorem 17. (1 2 3). Let í be the character of A4 given by Proposition 13. as shown.2 The group A4 Let G ˆ A4 . Character table of S4 gi |CG ( gi )| ÷1 ÷2 ÷3 ÷4 ÷5 1 24 1 1 2 3 3 (1 2) 4 1 À1 0 1 À1 (1 2 3) 3 1 1 À1 0 0 (1 2)(3 4) 8 1 1 2 À1 À1 (1 2 3 4) 4 1 À1 0 À1 1 18.11. Let ÷5 ˆ ÷4 ÷2 . Since S4 has ®ve conjugacy classes. either by using the same calculation or by quoting the result of Proposition 17. with representatives 1. Then |G| ˆ 12. and the sum of the squares of their degrees is 12. the alternating group of degree 4. íi ˆ 9 1 ‡ ˆ 1. Since G has four irreducible characters.14.18(1)). so that í( g) ˆ |®x ( g)| À 1 for all g P A4 . there must be exactly three linear characters of G. (1 2)(3 4). bÀ1 ab ˆ aÀ1 iX We shall derive the character table of G. ~2 . together with the character ÷4 ˆ í. (1 3)(2 4). give the complete character table of A4 : Character table of A4 gi |CG ( gi )| ÷1 ÷2 ÷3 ÷4 1 12 1 1 1 3 (1 2)(3 4) 4 1 1 1 À1 (1 2 3) 3 1 ù ù2 0 (1 3 2) 3 1 ù2 ù 0 18. (1 4)(2 3)gX 181 Now GaG9 ˆ fG9. . we obtain a representation r j of G for each j with 1 < j . Bj ˆ X 1 0 0 åÀ j Check that A n ˆ B2 ˆ I. For each integer j with 1 < j . Write å ˆ e2ðia n .Some elementary character tables G9 ˆ V4 ˆ f1. G9(1 2 3). na2. and the character table of GaG9 is G9 ~1 ÷ ~2 ÷ ~3 ÷ 1 1 1 G9(1 2 3) 1 ù ù2 G9(1 3 2) 1 ù2 ù ~ ÷ ÷ (where ù ˆ e2ðia3 ). (1 2)(3 4).3 The dihedral groups Let G be the dihedral group D2 n of order 2n. so that G ˆ ha. with n > 3. s P Z). ~3 to G. C) by (ar bs )r j ˆ (Aj ) r (Bj ) s (r. G9(1 3 2)g  C3. de®ne  j    0 1 å 0 Aj ˆ . na2. b: an ˆ b2 ˆ 1. The lifts of ÷1 . BÀ1 Aj Bj ˆ AÀ1 X j j j j It follows that by de®ning r j : G 3 GL(2. The (n À 1)a2 irreducible characters ø1 . we have proved that D9 n ˆ kal for n odd. If i and j are distinct integers with 1 < i . we obtain two linear characters ÷1 . ÷2 of G by lifting the irreducible characters of Gahai to G. Let ø j be the character of r j . Therefore there is no matrix T with ar i ˆ T À1 (ar j )T. far . These characters ÷1 and ÷2 are given by ÷1 ˆ 1 G and & 1 if g ˆ ar for some r. either by the proof of Example 5. na2 and 1 < j .182 Representations and characters of groups Each r j is an irreducible representation. na2. fas b: 0 < s < n À 1gX Thus there are (n ‡ 3)a2 conjugacy classes. We have now constructed distinct irreducible characters ø j of G.4.5(2) or by applying the result of Exercise 8. so ar i and ar j have different eigenvalues.11. then å i Tˆ å j and å i Tˆ å À j . in view of 2 Theorem 17. (Incidentally. Since kal v G and Gahai  C2.) The character table of D2 n (n odd) is therefore as follows (where å ˆ e2ðia n ): gi |CG ( gi )| ÷1 ÷2 øj (1 < j < (n À 1)/2) 1 2n 1 1 2 ar (1 < r < (n À 1)/2) n 1 1 ‡ å À jr b 2 1 À1 0 å jr . na2. Case 1: n odd By (12. one for each j which satis®es 1 < j . At this point it is convenient to consider separately the cases where n is odd and where n is even. ø( nÀ1)a2 each have degree 2. ÷2 ( g) ˆ À1 if g ˆ ar b for some rX We have now found all the irreducible characters of D2 n (n odd). and so r i and r j are not equivalent. there are two more to be found.11) the conjugacy classes of D2 n (n odd) are f1g. ø2 . aÀ r g(1 < r < (n À 1)a2). X X X . As G has (n ‡ 3)/2 irreducible characters in all. and their values appear in the following complete character table of D2 n (n even. ÷3 . are f1g.Some elementary character tables 183 Case 2: n even If n is even. aÀ r g(1 < r < m À 1). of which m À 1 are given by ø1 . ha2 ib. ha2 ia. gi |CG ( gi )| ÷1 ÷2 ÷3 ÷4 øj (1 < j < m À 1) 1 2n 1 1 1 1 2 am 2n 1 1 (À1) m (À1) m 2(À1) j a r (1 < r < m À 1) n 1 1 (À1) r (À1) r jr å ‡ å À jr b 4 1 À1 1 À1 0 ab 4 1 À1 À1 1 0 18. ø2 . say n ˆ 2m.4 Another group of order 12 We shall now describe a non-abelian group G of order 12 which is not isomorphic to either A4 or D12. fas b: s eveng. fam g. far . It is in fact known that every non-abelian group of order . ÷2 . ÷4 (and G9 ˆ ka2 l). ha2 iabg  C2 3 C2 X Therefore G has four linear characters ÷1 . they are easy to calculate. ø mÀ1 X To ®nd the remaining four irreducible characters. å ˆ e2ðia n ). Since these linear characters are the lifts of the irreducible characters of G/ka2 l.12). X X X . then the conjugacy classes of D2 n. n ˆ 2m. as supplied by (12. we ®rst note that ha2 i ˆ fa j : j eveng is a normal subgroup of G and Gaha2 i ˆ fha2 i. and we shall construct the character table of G. fas b: s oddgX Hence G has m ‡ 3 irreducible characters. ÷4 of G given below: . we have Gaha2 i  C4 . a5 b} Representative gi 1 a3 a a2 b ab |CG ( gi )| 12 12 6 6 4 4 Therefore G has six irreducible characters. The relations further imply that CG (a) ˆ hai. D12 or G. and Gaha2 i ˆ fha2 i. Since a has order 6 and b P kal. bÀ1 ab ˆ aÀ1 X It follows from these relations that every element of G has the form ar bs with 0 < r < 5. By lifting the irreducible characters of C4 to G. a2 . 0 < s < 1 as given above. a3 . b ˆ (1 7 4 10)(2 12 5 9)(3 11 6 8). ha2 ib. ha2 iabgX Since ka2 la ˆ ka2 lb2 . a3 ˆ b2 . and so |G| ˆ 12. and similar facts. Observe that ka2 l ˆ {1. ar b (0 < r < 5)X Check that a and b satisfy a6 ˆ 1. ÷3 . namely ar . b. ÷2 . Let a and b be the following permutations in S12 : a ˆ (1 2 3 4 5 6)(7 8 9 10 11 12). aÀ1 } {a2 . aÀ2 } {b. a2 b. CG (a3 ) ˆ G. and let G ˆ ka. a3 b. a3 bgX These. which are tabulated below: Conjugacy class {1} {a3 } {a. a4 b} {ab. we obtain the linear characters ÷1 . ha2 ia. CG (b) ˆ f1. but we shall not prove this result here. a subgroup of S12 .184 Representations and characters of groups 12 is isomorphic to A4 . bl. a4 } v G. a the group G has at least 12 elements. help us to ®nd the conjugacy classes of G. 2 2 á1 á2 ‡ â1 â2 ˆ 0X Since á1 . so we can solve them for á r and â r . we shall use the column orthogonality relations. the ®rst equation gives á1 ˆ â1 ˆ 2. â r taken by the last two irreducible characters ÷5 . 4 ‡ 2á3 À 2â3 ˆ 0. so they are positive integers. the column orthogonality relations 6 ˆ iˆ1 ÷ i ( g r )÷ i ( g 1 ) ˆ 0 and 6 ˆ iˆ1 ÷ i ( g r )÷ i ( g 2 ) ˆ 0 now give us two equations involving 2á r ‡ 2â r and 2á r À 2â r . â1 are positive integers. 2á5 À 2â5 ˆ 0.Some elementary character tables gi |CG ( gi )| ÷1 ÷2 ÷3 ÷4 ÷5 ÷6 1 12 1 1 1 1 á1 â1 a3 12 1 À1 1 À1 á2 â2 a 6 1 À1 1 À1 á3 â3 a2 6 1 1 1 1 á4 â4 b 4 1 i À1 Ài á5 â5 185 ab 4 1 Ài À1 i á6 â6 It remains to ®nd the values á r . we may take á2 ˆ 2 and â2 ˆ À2. For r . The other two equations then imply that á2 ˆ Àâ2 ˆ Æ2. 4 ‡ 2á4 ‡ 2â4 ˆ 0. ÷6 . Theorem 16. ÷6 .4(2). Explicitly: r ˆ 3: r ˆ 4: r ˆ 5: r ˆ 6: 2á3 ‡ 2â3 ˆ 0. 2á6 À 2â6 ˆ 0X .10. also a3 is an element of order 2. respectively. 2á6 ‡ 2â6 ˆ 0. Observe that á1 . 2. 1 1 4 ‡ á2 ‡ â2 ˆ 12. 2á4 À 2â4 ˆ 0. so á2 and â2 are integers by Corollary 13. Since we have not yet distinguished between ÷5 and ÷6 . By the column orthogonality relations applied to columns 1 and 2. 2á5 ‡ 2â5 ˆ 0. â1 are the degrees of ÷5 . we have 4 ‡ á2 ‡ â2 ˆ 12. For this. ) Summary of Chapter 18 In this chapter we gave the character tables of various groups. without constructing the corresponding CG-modules. Section 18. 2. â4 ˆ À1. .186 Hence Representations and characters of groups á3 ˆ À1. á6 ˆ 0. â3 ˆ 1. 1. 3.6. as follows. â5 ˆ 0.1: the group S4 .3: the dihedral groups. á5 ˆ 0. (In fact.2: the group A4 . á4 ˆ À1. This is typical of more advanced calculations. and illustrates the fact that it is usually much easier to construct an irreducible character of a group than to obtain an irreducible representation. â6 ˆ 0X The complete character table of G is therefore as follows: gi CG ( gi )| ÷1 ÷2 ÷3 ÷4 ÷5 ÷6 1 12 1 1 1 1 2 2 a3 12 1 À1 1 À1 2 À2 a 6 1 À1 1 À1 À1 1 a2 6 1 1 1 1 À1 À1 b 4 1 i À1 Ài 0 0 ab 4 1 Ài À1 i 0 0 We can deduce that G is not isomorphic to A4 or D12 from the fact that the character table of G is different from those of A4 and D12. It is instructive to note that we produced the last two irreducible characters of G by simply using the orthogonality relations. Section 18. Section 18. it is not hard to construct the representations of the above group G with characters ÷5 and ÷6 ± see Exercise 17. 6. (Hint: use the result of Exercise 17.5. b: a2 n ˆ 1.) 3. Find the character table of G. and show that all its entries are integers. aÀ1 ba ˆ bÀ1 l. b: a2n ˆ b4 ˆ 1. Write down explicitly the character table of D12. Find the values of ð on the elements of D8. and express ð as a sum of irreducible characters. Find the character table of G.Some elementary character tables Exercises for Chapter 18 187 1. Find the character table of G.) 4. Let G ˆ U6 n ˆ ka. ba ˆ aÀ1 bÀ1 . 5.8. It is a good idea to do the cases n odd and n even separately. Let G ˆ V8 n ˆ ha. .7. Let G ˆ T 4n ˆ ha. Use the character table to ®nd seven distinct normal subgroups of D12.1(3). an ˆ b2 . Regard D8 as a subgroup of S4 permuting the four corners of a square. as in Example 1. bÀ1 ab ˆ aÀ1 i. as in Exercise 17. with n odd. 2. as in Exercise 17. bÀ1 a ˆ aÀ1 bi. as in Exercise 17. (Hint: use Proposition 17. Let ð be the corresponding permutation character of D8. b: a2n ˆ b3 ˆ 1.6. Tensor product spaces Let V and W be vector spaces over C with bases v1 . we introduce a symbol v i  wj. and indeed. given those of G and H. then. and more generally ÷ 3 . but a little ingenuity is required in order to justify the conclusion that the product ÷ø is a character of G. . It is therefore straightforward to calculate the product. wn . and to put them together to form a new CG-module. . then the degrees of ÷. which has character ÷ø. We shall illustrate this by constructing the character tables of S5 and S6 . so we consider the character ÷ 2 . . If ÷ is not linear. 1 < j < n. ÷ 4 . ÷ 2 . At the end of the chapter. we apply tensor products in a different way. . The plan is to take CG-modules V and W with characters ÷ and ø respectively. Potentially. v m and w1 . products of characters provide a very good source of new characters from given ones. and it can be extended to include the product of any pair of characters ÷ and ø. For each i. . . X X X . and so on. and by taking successive powers of ÷ we obtain arbitrarily many new characters. j with 1 < i < m. called the tensor product of V and W. The value of the product ÷ø on an element g of G is simply ÷( g)ø( g).19 Tensor products The idea of multiplying a character of a group G by a linear character of G was introduced at the end of Chapter 17. increase. respectively. to ®nd all the irreducible characters of a direct product G 3 H. we have a chance of getting a large proportion of the character table of G from just one non-linear character ÷ of G. An important special case of the product ÷ø occurs when ÷ ˆ ø. . The tensor product space V  W is de®ned to be the mn-dimensional vector space over C with a basis given by 188 . . ë(v  w) ˆ ë ˆ i. w P W and ë P C. it is impossible to express v1  w1 ‡ v2  w2 in the form v  w. xa P V and y1. Then jˆ1 2 3 2 3 ˆ ˆ ˆ v  (ëw) ˆ ëi vi  ëì j w j ˆ ëë i ì j (v i  wj ).Tensor products fv i  wj : 1 < i < m. j ë i ì j (v i  wj ) ˆ ëë i ì j (v i  wj )X . ì j P C). j i. . . . . because this is not the case. 1 < j < ngX Thus V  W consists of all expressions of the form ˆ ë ij (v i  wj ) (ë ij P C)X €m For v P V and w P W with v ˆ iˆ1 ë i v i and (ë i . j ˆ ˆ i. then v  (ëw) ˆ (ëv)  w ˆ ë(v  w)X (2) If x1 . .1 Proposition (1) If v P V. . (2v1 À v2 )  (w1 ‡ w2 ) ˆ 2v1  w1 ‡ 2v1  w2 À v2  w1 À v2  w2 X Do not be misled by the notation into believing that every element of V  W has the form v  w. j 189 wˆ €n jˆ1 ì j wj For example. yb P W. j ëë i ì j (v i  wj ). we de®ne v  w P V  W by ˆ vwˆ ë i ì j (v i  wj )X i. For instance. j (ëv)  w ˆ 2 ˆ i 3 ëë i v i  2ˆ j 3 ì jw j ˆ i. 19. then 2 a 3 H b I ˆ ˆ ˆ xi  d yj e ˆ xi  yj X iˆ1 jˆ1 i. i j i. . j €m € Proof (1) Let v ˆ iˆ1 ë i v i and w ˆ n ì j w j. Proof Write vi ˆ m ˆ kˆ1 ë ik ek . 1 < j < n) give a basis of V  W.2 Proposition If e1 . w n. v m and w1 . .1. . The proof of part (2) is equally straightforward. . we have ˆ v i  wj ˆ ë ik ì jl (ek  f l )X k. . . 1 < j < ng give a basis of V  W. and we leave it as an exercise. em is a basis of V and f1 . . the next proposition shows that other bases of V and W work equally well. Tensor product modules We have introduced the tensor product of two vector spaces. wj ˆ n ˆ lˆ1 ì jl f l (ë ik . Let G be a ®nite group and let V and W be CG-modules with bases v1 . 19. X X X . Since elements ek  fl are v i  wj (1 < i < m.190 Representations and characters of groups Therefore v  (ëw) ˆ (ëv)  w ˆ ë(v  w). . . respectively. so we are now in a position to de®ne the tensor product of two CG-modules. fn is a basis of W. l Now the elements V  W . 1 < j < n) are a basis of the mn elements ek  fl (1 < k < m. . . The multiplication of v i  wj by an element of . then the elements in fei  f j : 1 < i < m. j Our construction of V  W depended upon choosing a basis of V and a basis of W at the beginning. ì jl P C)X Then by Proposition 19. . We know that the elements v i  wj (1 < i < m. it follows that the also a basis of V  W. 1 < l < n) V  W has dimension mn. . and hence span V  W. de®ne (v i  wj ) g ˆ v i g  w j g and. given in De®nition 19. . 19.4 Proposition For all v P V. which is then extended linearly to a multiplication on the whole of V  W. consider what happens when r is a scalar multiple of g. j €m ˆ ˆ ˆ 2 i.3 De®nition Let g P G. j for arbitrary complex numbers ë ij . 19. j i. more generally.Tensor products 191 G is de®ned in the following simple way. w P W and all g P G. makes the vector space V  W into a CGmodule. Then jˆ1 2ˆ 3 (v  w) g ˆ ë i ì j (v i  wj ) g by Proposition 19X1 iˆ1 i. For example. j. we have (v  w) g ˆ v g  wgX Proof Let v ˆ € ë i v i and w ˆ n ì j wj. j ë i ì j (v i g  wj g) 3 ëivi g  2ˆ j j ˆ i 3 ì j wj g by Proposition 19X1 ˆ v g  wgX You should be warned that (v  w)r Tˆ vr  wr for most elements r in CG.3. let 2ˆ 3 ˆ ë ij (v i  wj ) g ˆ ë ij (v i g  wj g) i. 19. For all i.5 Proposition The rule for multiplying an element of V  W by an element of G. and 2ˆ i. Hence. em of V and a basis f1 . 19. . . Then (v i  wj ) g ˆ v i g  wj g P V  W . (ei  f j ) g ˆ ei g  f j g ˆ ë i ì j (ei  f j ). . j We now calculate the character of V  W.6 Proposition Let V and W be CG-modules with characters ÷ and ø. . and by Proposition 19. . and g.192 Representations and characters of groups Proof Let 1 < i < m. Then the character of the CG-module V  W is the product character ÷ø. j by Proposition 19X4 3 ˆ ë ij (v i  wj ) g ˆ ë ij ((v i  wj ) g)X i. h P G.2. these vectors ei  fj form a basis of V  W. 1 < j < n. respectively. ø( g) ˆ ì jX Now for 1 < i < m and 1 < j < n. if ö is the character of V  W then .6 are ful®lled. where ÷ø( g) ˆ ÷( g)ø( g) for all g P GX Proof Let g P G. . fn of W such that ei g ˆ ë i e i (1 < i < m) and f j g ˆ ì j f j m ˆ iˆ1 n ˆ jˆ1 (1 < j < n) for some complex numbers ë i . j Therefore all the conditions of Proposition 4. Then ÷( g) ˆ ë i . . (v i  wj )( gh) ˆ v i ( gh)  wj ( gh) ˆ (v i g)h  (wj g)h ˆ (v i g  wj g)h ˆ ((v i  wj ) g)h. and V  W is a CG-module.11 we can choose a basis e1 . (v i  wj )1 ˆ v i  wj . . By Proposition 9. ì j . and ÷4 ÷4 ˆ ÷1 ‡ ÷3 ‡ ÷4 ‡ ÷5 X Powers of characters Corollary 19. the powers of ÷ carry a lot of information about the whole character table of G. j 19. as can be seen from Theorem 19.8 Example The character table of S4 was given in Section 18. and calculate ÷3 ÷4 and ÷4 ÷4 .10 below. An inductive proof using Corollary 19. . for every nonnegative integer n.7 Corollary The product of two characters of G is again a character of G. We reproduce it here.1.Tensor products 2 32 ˆ ˆ ˆ 3 ö( g) ˆ ëi ì j ˆ ëi ì j ˆ ÷( g)ø( g).7 shows that ÷ n is a character of G. j i j 193 as required. gi |CG ( gi )| ÷1 ÷2 ÷3 ÷4 ÷5 ÷3 ÷4 ÷4 ÷4 1 24 1 1 2 3 3 6 9 (1 2) 4 1 À1 0 1 À1 0 1 (1 2 3) 3 1 1 À1 0 0 0 0 (1 2)(3 4) 8 1 1 2 À1 À1 À2 1 (1 2 3 4) 4 1 À1 0 À1 1 0 1 We see that ÷3 ÷4 ˆ ÷4 ‡ ÷5 . When ÷ is a faithful character (that is. 19.7 shows that if ÷ is a character of G then so is ÷ 2 . the product of ÷ with itself. where ÷ 2 ˆ ÷÷. i. Ker ˆ {1}). More generally. we de®ne ÷ n by ÷ n ( g) ˆ (÷( g)) n 0 for all g P GX Thus ÷ ˆ 1 G . the term x1 must come from all rÀ2 the factors in the ®rst row. we shall need the following result concerning the so-called `Vandermonde matrix'.194 Representations and characters of groups In the course of the proof of Theorem 19. we must take x1 from the rÀ2 ®rst row of the matrix. . j . It follows that Ä is divisible by ‰ (xi À x j ) ˆ (x1 À x2 )(x1 À x3 ) X X X (x1 À xr ) i. On the other hand.10. Hence rÀ1 rÀ2 the coef®cient of x1 x2 . to obtain x1 x2 X X X x rÀ1 rÀ1 in the expansion of the determinant Ä. then the matrix rÀ1 I á2 X X X á1 1 rÀ1 á2 X X X á2 g g 2 g X XXX X e á2 r XXX á rÀ1 r ár 3 (x2 À x3 ) X X X (x2 À xr ) F F F 3 (x rÀ1 À xr )X rÀ1 rÀ2 Now the coef®cient of x1 x2 X X Xx rÀ1 in this product is 1: for in the rÀ1 way we have displayed the product. Suppose that x1 . . xr are indeterminates. and so on. . . We ®rst sketch a proof of this result. á r are distinct H 1 á1 f1 á 2 f Aˆf dX X 1 is invertible. so Ä ˆ 0. x2 from the second row.9) If á1 . and so on. . and consider H rÀ1 I 1 x1 x2 X X X x1 1 f 1 x x2 X X X x rÀ1 g 2 f g 2 2 Ä ˆ detf gX dX X X XXX X e 1 xr x2 r XXX x rÀ1 r If i Tˆ j and xi ˆ xj then two rows of the given matrix are equal. (19. x2 must come from all the factors in the rÀ1 rÀ2 second row. . j complex numbers. It follows that ‰ ÄˆÆ (xi À xj )X i. . . . x rÀ1 in Ä is Æ1. . 5.11 Examples (1) If G Tˆ {1} and ÷ is the regular character of G. â r )X Now A is invertible by (19.8.Tensor products 195 To obtain (19. . we know this already. and b Tˆ 0 since â1 Tˆ 0. and deduce that the matrix A is invertible since its determinant is non-zero. j 19. Then ÷( g) takes four different values. 19. G1 ˆ {1}. We must show that h÷ j . then ÷( g) takes just two different values (see Proposition 13.9). á r . . and let b be the row vector which is given by b ˆ (â1 . so that G1 ˆ Ker ÷. . For 1 < i < r. de®ne Gi ˆ f g P G: ÷( g) ˆ á i gX Take á1 ˆ ÷(1). øi Tˆ 0 for some j with 0 < j < r À 1. (2) Let G ˆ S4 . Then every irreducible character of G is a constituent of one of the powers ÷ 0 . . let ˆ âi ˆ ø( g). We have seen that ÷ 2 ˆ ÷1 ‡ ÷3 ‡ ÷4 ‡ ÷5 . ÷ 1 . so Theorem 19. . But the ( j ‡ 1)th entry in the row vector bA is equal to jGjh÷ j . and suppose that ÷( g) takes precisely r different values as g varies over all the elements of G. ÷ rÀ1 . Let ÷ ˆ ÷4 . øi Tˆ 0 for some j with 0 < j < r À 1.10 says that every irreducible character of G is a constituent of 1 G or ÷. we substitute á i for xi (1 < i < r). . as we wished to prove. gPG i and note that â1 ˆ ø(1) Tˆ 0. Proof Let the r values taken by ÷ be á1 .9). h÷ j . As ÷ is faithful. øi. Then for all j > 0. and refer to Example 19. and thus h÷ j .10 Theorem Let ÷ be a faithful character of G. . .20). øi ˆ r 1 ˆ 1 ˆ (÷( g)) j ø( g) ˆ (á i ) j â i X jGj gPG jGj iˆ1 Let A be the r 3 r matrix with ij-entry (á i ) jÀ1 . and for 1 < i < r. hence bA Tˆ 0. Now let ø be an irreducible character of G. by Theorem 10. X X X . . V  V ˆ S(V  V ) È A(V  V )X . Let V be a CG-module with character ÷. j and extending linearly ± that is. . v n be a basis of V. it is of some importance to be able to decompose powers of a character ÷ into sums of irreducible characters. ÷5 of G. in this case.12 Proposition The subspaces S(V  V) and A(V  V) are CG-submodules of V  V. j i. it is easy to see that S(V  V) and A(V  V) are subspaces of V  V (indeed. ÷2 i ˆ 1X and we ®nd that Thus ÷ 0 .10.6. as we shall see. w P V. ÷ 3 ) have between them as constituents all the irreducible characters ÷1 . and the subspace A(V  V ) is known as the antisymmetric part of V  V. A(V  V ) ˆ fx P V  V : xT ˆ Àxg. . . . Decomposing ÷ 2 In view of Theorem 19. we have (v  w)T ˆ w  vX Hence T is independent of the choice of basis. the square of ÷. ÷ 1 . j Check that for all v. and de®ne a linear transformation T: V  V 3 V  V by (v i  v j )T ˆ v j  v i for all i.10. Also. Let v1 . This special case is particularly useful in ®nding irreducible characters. the module V  V has character ÷ 2 . ÷ 2 . illustrating Theorem 19. By Proposition 19. 2ˆ 3 ˆ ë ij (v i  v j ) T ˆ ë ij (v j  v i )X i. Now de®ne subsets of V  V as follows: S(V  V ) ˆ fx P V  V : xT ˆ xg. We are going to provide a method for decomposing ÷ 2 . The subspace S(V  V) is called the symmetric part of V  V. they are eigenspaces of T). ÷ 3 (indeed. . . 19.196 Representations and characters of groups h÷ 3 . just ÷ 2 . . Since T is linear. Thus S(V  V) and A(V  V) are CG-submodules of V  V. Further. . j ë ij (v i g  v j g)T 3 ë ij (v i  v j ) gT X 2ˆ i. . Hence. (1) The vectors v i  v j ‡ v j  v i (1 < i < j < n) form a basis of S(V  V ). and ( yg)T ˆ ( yT ) g ˆ À yg. j Therefore T is a CG-homomorphism from V  V to itself. while the antisymmetric part of V  V contains all vectors of the form v  w À w  v. We now present bases of the symmetric and antisymmetric parts of V  V which consist of elements like these.Tensor products Proof For all ë ij P C and g P G. 1 2(x ‡ xT ) P S(V  V ) and 1 2(x À xT ) P j V  V ˆ S(V  V ) È A(V  V )X Note that the symmetric part of V  V contains all vectors which have the form v  w ‡ w  v with v. for x P S(V  V). . j i. A(V  V ). 19. (2) The vectors v i  v j À v j  v i (1 < i . v n be a basis of V. j < n) form a basis of A(V  V). Therefore. If x P S(V  V) ’ A(V  V) then x ˆ xT ˆ Àx. so xg P S(V  V) and yg P A(V  V).13 Proposition Let v1 . 2ˆ 3 ˆ ë ij (v i  v j ) Tg ˆ ë ij (v j g  v i g) i. w P V. The dimension of A(V  V ) is n(n À 1)a2. j 197 ˆ ˆ ˆ i. we have (xg)T ˆ (xT ) g ˆ xg. The dimension of S(V  V ) is n(n ‡ 1)a2. y P A(V  V) and g P G. so x ˆ 0. . . for all x P V we have x ˆ 1(x ‡ xT ) ‡ 1(x À xT )X 2 2 Since T 2 is the identity. By Proposition 19. . j i i. Hence dim S(V  V ) > n(n ‡ 1)a2. ÷2 ˆ ÷S ‡ ÷ AX The next result gives the values of the characters ÷ S and ÷ A . we have ÷ S ( g) ˆ 1(÷ 2 ( g) ‡ ÷( g 2 )). By Proposition 19. . . and ÷ A to be the character of the CG-module A(V  V ).12. i € i. en of V such that e i g ˆ ë i e i (1 < i < n) for some complex numbers ë i . 19. Then (ei  ej À ej  ei ) g ˆ ë i ë j (ei  ej À ej  ei ). and the result follows. .13(2). j dim A(V  V ) > n(n À 1)a2X De®ne ÷ S to be the character of the CG-module S(V  V ). ÷ 2 ˆ ÷ S ‡ ÷ A .12. dim S(V  V ) ‡ dim A(V  V ) ˆ dim V  V ˆ n2 X Hence the above inequalities are equalities. and 2 ÷ A ( g) ˆ 1(÷ 2 ( g) À ÷( g 2 ))X 2 Proof By Proposition 9. and the vectors v i  v j À v j  v i (1 < i . Therefore i ˆ 2 ˆ ÷ 2 ( g) ˆ (÷( g))2 ˆ ëi ‡ 2 ë i ë j ˆ ÷( g 2 ) ‡ 2÷ A ( g)X 2 ë2 ei.14 Proposition For g P G. j Hence ÷ A ( g) ˆ 1(÷ 2 ( g) À ÷( g 2 ))X 2 Also. ˆ ÷ A ( g) ˆ ëi ë j X Now e i g ˆ € so ÷( g) ˆ i ë i and ÷( g2 ) ˆ i ë2 . which implies that ÷ S ( g) ˆ ÷ 2 ( g) À ÷ A ( g) ˆ 1(÷ 2 ( g) ‡ ÷( g 2 ))X 2 j .11 we can choose a basis e1 .198 Representations and characters of groups Proof Clearly the vectors v i  v j ‡ v j  v i (1 < i < j < n) are linearly independent elements of S(V  V ). j < n) are linearly independent elements of A(V  V ). and hence from Proposition 19. 19. gi |CG ( gi )| ÷ ÷S ÷A 1 24 3 6 3 (1 2) 4 1 2 À1 (1 2 3) 3 0 0 0 (1 2)(3 4) 8 À1 2 À1 (1 2 3 4) 4 À1 0 1 We ®nd that ÷ S ˆ ÷1 ‡ ÷3 ‡ ÷4 and ÷ A ˆ ÷5 .8. obtained by lifting the irreducible characters of GaG9. and use inner products to analyse ÷ S and ÷ A for new irreducible characters.14.16(4). The values of ÷. We illustrate this strategy with two examples. given one or two irreducible characters to start with. The strategy is simple: (1) Given a character ÷. G has conjugacy class representatives gi . then form ø S and ø A and repeat. the symmetric group of degree 5.16 Example The character table of S5 Let G ˆ S5 . and the values of ÷ S and ÷ A . Let ÷ ˆ ÷4 . form ÷ S and ÷ A .Tensor products 199 19. The techniques which we have developed so far give a useful method for ®nding new irreducible characters of a group. conjugacy class sizes and centralizer orders |CG ( gi )| as follows: gi Class size |CG ( gi )| 1 1 120 (1 2) 10 12 (1 2 3) (1 2)(3 4) (1 2 3 4) (1 2 3)(4 5) (1 2 3 4 5) 20 15 30 20 24 6 8 4 6 5 Thus G has exactly seven irreducible characters. G9 ˆ A5 and G has exactly two linear characters ÷1 . We have . The character table of G is given in Example 19. (2) If ø is a new character found in (1). appear below. (a) Linear characters By Example 17. ÷2 . By Example 12. given by Proposition 19.13.15 Example Let G ˆ S4 . À1. Proposition 17. By Proposition 19. if g is an odd permutationX Proposition 13. ÷3 i ˆ ÷3 ( g) ˆ jfix ( g)j À 1 ( g P G)X 42 22 12 (À1)2 (À1)2 ‡ ‡ ‡ ‡ ˆ 1X 120 12 6 6 5 Hence ÷3 is irreducible.200 Representations and characters of groups & ÷2 ( g) ˆ ÷1 ˆ 1 G . Next.24 gives us a character ÷3 (b) Permutation character of G with values Observe that h÷3 .14 the values of the characters ÷ S and ÷ A are gi |CG ( gi )| ÷S ÷A 1 120 10 6 (1 2) 12 4 0 (1 2 3) (1 2)(3 4) (1 2 3 4) (1 2 3)(4 5) (1 2 3 4 5) 6 8 4 6 5 1 0 2 À2 0 0 1 0 0 1 Thus. Write ÷ ˆ ÷3 . if g is an even permutation.20. At this point we have the following portion of the character table of G: gi |CG ( gi )| ÷1 ÷2 ÷3 ÷4 1 120 1 1 4 4 (1 2) 12 1 À1 2 À2 (1 2 3) (1 2)(3 4) (1 2 3 4) (1 2 3)(4 5) (1 2 3 4 5) 6 8 4 6 5 1 1 1 1 1 1 0 0 1 À1 0 0 1 À1 À1 1 1 1 À1 À1 (c) Tensor products We now use tensor products to construct the last three irreducible characters of G. .14 shows that ÷4 ˆ ÷3 ÷2 is also an irreducible character. and 1. by Theorem 14. The character table of S5 is as shown. of order 720.Tensor products h÷ A . Finally. 120 12 6 8 6 40 8 1 1 h÷ S . For ease of printing. 120 8 5 201 and so ÷ A is a new irreducible character. ÷1 i ˆ Therefore. 10 4 1 2 1 ‡ ‡ ‡ ‡ ˆ 1. it is convenient to label each conjugacy class by the cycle-shape of its elements. ÷7 ˆ ÷6 ÷2 is a different irreducible character of degree 5. Character table of S5 gi |CG ( gi )| ÷1 ÷2 ÷3 ÷4 ÷5 ÷6 ÷7 1 120 1 1 4 4 6 5 5 (1 2) 12 1 À1 2 À2 0 1 À1 (1 2 3) (1 2)(3 4) (1 2 3 4) (1 2 3)(4 5) (1 2 3 4 5) 6 8 4 6 5 1 1 1 1 0 À1 À1 1 1 0 0 À2 1 1 1 À1 0 0 0 À1 1 1 À1 À1 1 0 1 À1 1 1 À1 À1 1 0 0 19. 120 12 6 8 6 h÷ S . and 120 12 6 6 100 16 1 4 1 h÷ S . we use techniques similar to those of the previous example to ®nd 8 of the 11 irreducible characters of the symmetric group S6 .17 Example The character table of S6 In this example. Next. which we call ÷5 . Using . ÷ A i ˆ 36 4 1 ‡ ‡ ˆ 1. ÷ S ˆ ÷1 ‡ ÷3 ‡ ø. we then ®nd the last three irreducible characters by using the orthogonality relations. so that ÷6 ˆ ÷ S À ÷1 À ÷3 . We have now found all seven irreducible characters of S5 . ÷ S i ˆ ‡ ‡ ‡ ‡ ˆ 3. Let G ˆ S6 . where ø is an irreducible character of degree 5. Let ÷6 ˆ ø. ÷3 i ˆ ‡ ‡ À ˆ 1. 24. (a) Linear characters As with all symmetric groups Sn (n > 2).2) (6) 120 144 15 40 90 120 6 5 48 18 8 6 Since G has 11 conjugacy classes. À1. the conjugacy class sizes and centralizer orders are as follows (see Exercise 12.2) (5) (2. ÷3 i ˆ 1. by Proposition 13.2) 6 0 1 À1 (5) 5 0 0 0 (2.3) (4. ÷ S i ˆ 3X h÷ A . h÷ S .2) (3. . h÷ S . (b) Permutation character and tensor products is given by ÷3 ( g) ˆ jfix ( g)j À 1 The function ÷3 which ÷1 ˆ 1 G . 1.2) (3.3) (4. if g is odd ( g P G) is a character of G. where & ÷2 ( g) ˆ (see Example 17.3): Cycle-shape Class size |CG ( gi )| (1) (2) 1 15 720 48 (3) 40 18 (2. ÷1 i ˆ 1.202 Representations and characters of groups this notation. ÷ S and ÷ A are as follows: Class |CG ( gi )| ÷ ˆ ÷3 ÷S ÷A (1) (2) 720 48 5 15 10 3 7 2 (3) 18 2 3 1 (2.2. h÷ S . it has 11 irreducible characters.2) (6) 48 18 8 6 À1 3 À2 À1 0 1 À1 1 0 À1 0 1 We calculate that h÷3 .2) 45 16 (4) 90 8 (3. if g is even. ÷ A i ˆ 1. and we get exactly two linear characters ÷1 and ÷2 .2) 16 1 3 À2 (4) 8 1 1 0 (3. The values of ÷.13). ÷3 i ˆ 1.2. Let ÷ ˆ ÷3 . the derived subgroup is An . It will be shown later (Corollary 22. Class |CG ( gi )| ÷1 ÷2 ÷3 ÷4 ÷5 ÷6 ÷7 ÷8 (1) (2) 720 48 1 1 5 5 10 10 9 9 1 À1 3 À3 2 À2 3 À3 (3) 18 1 1 2 2 1 1 0 0 (2.2) 6 1 À1 0 0 À1 1 0 0 (5) 5 1 1 0 0 0 0 À1 À1 (2. . Ingeniously.3) (4.2. in the ordering which we have adopted. Further. The irreducible characters ÷1 . Thus the conjugacy classes of s and t correspond to the second and fourth columns of the character table. respectively. where ÷7 is another irreducible character. ÷8 are recorded in the following portion of the character table of G. .10 we know that ÷(s) and ÷(t) are integers for all characters ÷ of G. . Also.2) (6) 48 18 8 6 1 À1 À1 1 À2 2 3 À3 1 1 À1 À1 1 1 0 0 1 1 À1 À1 0 0 1 1 1 À1 À1 1 1 À1 0 0 (c) Orthogonality relations We now use the column orthogonality relations to complete the character table of G. as is ÷6 ˆ ÷5 ÷2 . From Corollary 13. Let s denote the permutation (1 2) and t denote the permutation (1 2)(3 4). It is therefore convenient ®rst to concentrate on elements of order 2. but for the moment we know for certain only that ÷( g) is an integer if g2 ˆ 1 (see Corollary 13.2) (3. ÷10 and ÷11 . ÷ S ˆ ÷1 ‡ ÷3 ‡ ÷7 . so is ÷4 ˆ ÷3 ÷2 . ÷8 ˆ ÷7 ÷2 is also irreducible.2) 16 1 1 1 1 À2 À2 1 1 (4) 8 1 À1 1 À1 0 0 À1 1 (3. of degree 9. ÷5 ˆ ÷ A is irreducible. The column orthogonality relations give 11 ˆ iˆ1 ÷ i (s)2 ˆ 48X .16) that all the entries in the character tables of all symmetric groups are integers. we call the three irreducible characters of G which have yet to be found ÷9 . . so that we can guarantee that the solutions to the equations which we deal with are integers. Finally.Tensor products 203 Therefore ÷3 is irreducible.10). . ad ‡ be ‡ cf ˆ 10X The only solution to these equations in integers with a . ÷9 ÷2 ˆ ÷10 X Once more. we lose no generality in assuming that ÷9 (s) ˆ 1. without loss of generality. . and is not equal to any of ÷1 . ÷11 (s)2 ˆ 0X Now ÷9 ÷2 is an irreducible character. ÷10 (s) ˆ À1X The plan is now to ®nd ÷ i (1) and ÷ i (t) for i ˆ 9. since ÷9 ÷2 (s) ˆ À÷9 (s). . whence a À b ˆ 0. d 2 ‡ e 2 ‡ f 2 ˆ 2. 11. ÷ i (t)÷ i (t) ˆ 16. f in the following portion of the character table: Element Class ÷9 ÷10 ÷11 1 (1) a b c s (2) 1 À1 0 t (2. . 10. ÷8 . 0 is d ˆ e ˆ 1.2) d e f The column orthogonality relations give 11 ˆ iˆ1 11 ˆ iˆ1 ÷ i (1)÷ i (s) ˆ 0. 0 and b . b. That is. 11 ˆ iˆ1 11 ˆ iˆ1 ÷ i (s)÷ i (t) ˆ 0. Moreover. Therefore. d. c. f ˆ 0. we aim to evaluate the integers a. we see that ÷9 ÷2 is not ÷9 or ÷11 . e. ÷ i (1)÷ i (t) ˆ 0. d À e ˆ 0. that ÷9 (s)2 ˆ ÷10 (s)2 ˆ 1. a ˆ b ˆ 5X .204 Hence Representations and characters of groups ÷9 (s)2 ‡ ÷10 (s)2 ‡ ÷11 (s)2 ˆ 2X We can assume. .2. we ®nd that c ˆ 16 by using the relation 11 ˆ iˆ1 205 ÷ i (1)2 ˆ 720X The above portion of the character table is therefore Element Class ÷9 ÷10 ÷11 1 (1) 5 5 16 s (2) 1 À1 0 t (2.2) 16 1 1 1 1 À2 À2 1 1 1 1 0 (4) 8 1 À1 1 À1 0 0 À1 1 À1 1 0 (3. v m .2) (6) 18 8 6 1 1 À1 À1 1 1 0 0 2 2 À2 1 1 À1 À1 0 0 1 1 À1 À1 0 1 À1 À1 1 1 À1 0 0 0 0 0 Direct products We conclude the chapter by showing that tensor products can be used to determine the character table of a direct product G 3 H. given the character tables of G and H.2) 6 1 À1 0 0 À1 1 0 0 1 À1 0 (5) 5 1 1 0 0 0 0 À1 À1 0 0 1 (2. Having done these calculations. . Character table of S6 Class |CG ( gi )| ÷1 ÷2 ÷3 ÷4 ÷5 ÷6 ÷7 ÷8 ÷9 ÷10 ÷11 (1) (2) 720 48 1 1 5 5 10 10 9 9 5 5 16 1 À1 3 À3 2 À2 3 À3 1 À1 0 (3) 18 1 1 2 2 1 1 0 0 À1 À1 À2 (2. and let W be a . with basis v1 .2) 1 1 0 We can now determine the three unknown entries in each further column. . Let V be a CG-module.3) (4. since the column orthogonality relations will give three independent equations in these unknowns (as the above 3 3 3 matrix is invertible).Tensor products Finally. we ®nd that the complete character table of S6 is as shown. .2) 48 1 À1 À1 1 À2 2 3 À3 À3 3 0 (3. . the character of V  W is ÷ 3 ø. j with 1 < i < m and 1 < j < n. we have (x. with basis w1. . ø l i H ˆ ä ik ä jl X (Here the subscripts G 3 H.4. h) ˆ v i g  wj h and extend this de®nition linearly to the whole of V  W. j i. w P W. where (÷ 3 ø)( g. Then G 3 H has precisely ab distinct irreducible characters. . k. G and H. and all g P G. y)À1 ( g. Let ÷ be the character of V and ø be the character of W. Then a proof similar to that of Proposition 19. h÷ i 3 ø j . respectively. By the proof of Proposition 19. h) ˆ ÷( g)ø(h) ( g P G. h P H. . y À1 hy)X . . l. for ë ij P C. j. y P H. ø b be the distinct irreducible characters of H. . For all i. ÷ k 3 ÷ l i G3 H ˆ ˆ 1 ÷ i ( g)ø j (h)÷ k ( g)ø l (h) jG 3 Hj gPG 2 ˆ 3 32 1 ˆ 1 ˆ ÷ i ( g)÷ k ( g) ø j (h)ø l (h) jGj gPG j Hj hP H hP H ˆ h÷ i .6. h) ˆ v g  wh. 2ˆ 3 ˆ ë ij (v i  wj ) ( g. . G and H indicate inner products of characters of G 3 H. 1 < j < b)X Proof For all i. h P H)X 19. . . Next.18 Theorem Let ÷1 . y) ˆ (x À1 gx. x P G and h.206 Representations and characters of groups C H-module. ÷ a be the distinct irreducible characters of G and let ø1 . . and these are ÷ i 3 ø j (1 < i < a. that is. j As in Proposition 19. h) ˆ ë ij (v i g  wj h)X i. note that for all g. we ®nd that (v  w)( g. . . h)(x. wn.5 shows that V  W is a C(G 3 H)-module. for all v P V. de®ne (v i  wj )( g.) Thus the ab characters ÷ i 3 ø j are distinct and irreducible. ÷ k i G hø j . . . À1). (1. . ga are representatives of the conjugacy classes of G and h1 . ((1 2 3). hj )| ÷1 ÷2 ÷3 ÷1 ÷2 ÷3 3 3 3 3 3 3 ø1 ø1 ø1 ø2 ø2 ø2 (1. 1). Character table of S3 3 C2 ( gi .18. In particular. alongside the character table of C2 . À1). if g1 . Character table of S3 gi |CG ( gi )| ÷1 ÷2 ÷3 1 6 1 1 2 (1 2) 2 1 À1 0 (1 2 3) 3 1 1 À1 Character table of C2 hi |CH (hi )| ø1 ø2 1 2 1 1 À1 2 1 À1 The conjugacy classes of S3 3 C2 are represented by (1. We reproduce it here. .3(1). G 3 H has exactly ab irreducible characters. . . À1) ((1 2 3). h9) of G 3 H are conjugate if and only if the elements g and g9 are conjugate in G and the elements h and h9 are conjugate in H. and by Theorem 19. G 3 H has precisely ab conjugacy classes. Consequently. h) and ( g9.Tensor products 207 Hence elements ( g. ((1 2).19 Example The character table of S3 3 C2 The character table of S3 ( D6 ) is given in Example 16. À1) ((1 2). 1) ((1 2 3). the character table of S3 3 C2 is as shown. hb are representatives of the conjugacy classes of H. 1). then the elements ( g i . À1). so the irreducible characters ÷ i 3 ø j which we have found must be all the irreducible characters of G 3 H. ((1 2). 1) ((1 2). . 1). À1) 12 4 6 12 4 6 1 1 2 1 1 2 1 À1 0 1 À1 0 1 1 À1 1 1 À1 1 1 2 À1 À1 À2 1 À1 0 À1 1 0 1 1 À1 À1 À1 1 . ((1 2 3). By Theorem 15. 1) (1. hj ) (1 < i < a. 1 < j < b) are representatives of the conjugacy classes of G 3 H.3. hj ) |CG3 H ( g i . . j 19. öi ˆ h÷. In Example 20. then so are ÷ S and ÷ A . if ÷ ˆ ø. if ÷ Tˆ øX 3. h÷ø. Prove that & 1. Let ÷. h) ˆ ÷( g)ø(h) for all g P G. 1 G i ˆ 0. (This shows that the hypothesis that ÷ is faithful cannot be dropped from Theorem 19. Suppose that ÷ and ø are irreducible characters of G. Show that h÷ø. Summary of Chapter 19 1.13 of the next chapter we shall show that there exist irreducible characters ÷ and ö of A5 which take the following values: . ÷öiX 2. Show that there is some irreducible character ø of G such that k÷ n . where we give the character table of D12 (Exercise 1.2. where ÷ S ( g) ˆ 1(÷ 2 ( g) ‡ ÷( g 2 )). If ÷ is a character of G. h P H. ø and ö be characters of the group G.5 shows that D12  S3 3 C2 ). The values of ÷ 3 ø are given by (÷ 3 ø)( g. Exercises for Chapter 19 1. øl ˆ 0 for all integers n with n > 0. 2 ÷ A ( g) ˆ 1(÷ 2 ( g) À ÷( g 2 )) 2 for all g P G.10. 3.208 Representations and characters of groups Compare the solution to Exercise 18. The product of any two characters of G is a character of G. Let ÷ be a character of G which is not faithful.) 4. The irreducible characters of G 3 H are those characters ÷ 3 ø. øöi ˆ hø. where ÷ is an irreducible character of G and ø is an irreducible character of H. 2. ö S and ö A . g7 . g1 . . g2 . ÷ A . g2 . g4 . A certain group G or order 24 has precisely seven conjugacy classes with representatives g1 . . Find ÷ S and ÷ A . g2 .Tensor products 1 ÷ ö 5 3 (1 2 3) À1 0 (1 2)(3 4) 1 À1 (1 2 3 4 5) (1 ‡ 0 p 5)a2 (1 3 4 5 2) (1 À 0 p 5)a2 209 Calculate the values of ÷ S . 5. Moreover. . g5 . further. G has a character ÷ with values as follows: gi |CG ( gi )| ÷ g1 24 2 g2 24 À2 g3 4 0 g4 6 Àù2 g5 6 Àù g6 6 ù g7 6 ù2 where ù ˆ e2ðia3 . ®nd the character table of G. . Express these characters as linear combinations of the irreducible characters ø1 . respectively. Write down the character table of D6 3 D6. g4 . g2 . . . . g5 . g2 . . g2 . 6.13. ø5 of A5 which are given in Example 20. g2 . and show that both are irreducible. . By forming products of the irreducible characters found so far. g2 are con1 2 3 4 5 6 7 jugate to g1 . 1 Example Let G ˆ D8 ˆ ka. We write this character of H as ÷ 5 H. if f: G 3 C is any function. As in Example 4. h P H if they hold for all g. More generally. then V is also a C H-module. and call it the restriction of V to H.5(1). let V be the CG-module with basis v1 . v2 a ˆ Àv1 . 20. If V is a CG-module then we write the corresponding C H-module as V 5 H. for example. The character of V 5 H is obtained from the character ÷ of V by evaluating ÷ on the elements of H only. and illustrate its use. v2 b ˆ Àv2 X 210 . and refer to it as the restriction of ÷ to H. which occurs.2 certainly hold for all g. then f 5 H denotes the restriction of f to H (so that ( f 5 H)(h) ˆ f (h) for all h P H). This simple way of converting a CG-module into a C H-module is known as restricting from G to H. bÀ1 ab ˆ aÀ1 l. Then C H is a subset of CG. h P G. we are going to look at ways of relating the representations of a group to the representations of its subgroups. when G ˆ S n and H ˆ A n . b: a4 ˆ b2 ˆ 1. since properties (1)±(5) of De®nition 4. We apply this result in the situation where H is of index 2 in G. Here. and Clifford's Theorem 20. Restriction Let H be a subgroup of the ®nite group G. v2 for which v1 a ˆ v2 . we introduce the elementary idea of restricting a CG-module to a subgroup H of G.8 gives important information in this case. If V is a CG-module.20 Restriction to a subgroup In this chapter and the next. v1 b ˆ v1 . The case where H is a normal subgroup of G is of particular interest. 2. Example 20. it might be the case that V is an irreducible CG-module while V 5 H is not an irreducible C H-module. then U 5 H is a C H-submodule of V 5 H. v2 for which v1 a2 ˆ Àv1 . 4} ®xing 5. The character ÷ of V is given by g ÷( g) 1 2 a 0 a2 À2 a3 0 b 0 ab 0 a2 b 0 a3 b 0 v1 b ˆ v1 . then V 5 H is the C Hmodule with basis v1 . 20. for if U is a CG-submodule of V.Restriction to a subgroup 211 If H is the subgroup {1. a2 . v2 a2 ˆ Àv2 . 3.1 illustrates this fact. By 18.2. the character table of H is gi jC H ( g i )j ø1 ø2 ø3 ø4 1 12 1 1 1 3 (1 2)(3 4) 4 1 1 1 À1 (1 2 3) 3 1 ù ù2 0 (1 3 2) 3 1 ù2 ù 0 . if V 5 H is an irreducible C H-module then V is an irreducible CG-module. b. v2 b ˆ Àv2 X and the character ÷ 5 H of V 5 H is given by h ÷(h) 1 2 a2 À2 b 0 a2 b 0 If V is a CG-module and H is a subgroup of G. then dim V ˆ dim (V 5 H). However. a2 b} of G. On the other hand.2 Example Let G ˆ S5 and let H be the subgroup A4 of G consisting of all even permutations of {1. and k . ÷ 6 5 H ˆ ø2 ‡ ø 3 ‡ ø 4 X Constituents of a restricted character To help us discuss the way in which a restricted character ÷ 5 H can be expressed in terms of the irreducible characters of H. W2 i G ˆ W1 ( g)W2 ( g). The character table of G is given in Example 19. we introduce the following notation. we calculate the character ÷ i 5 H as a sum of irreducible characters ø j . . . 20.212 Representations and characters of groups (where ù ˆ e2ðia3 ). For each i with 1 < i < 7.3 De®nitions The inner product k . ÷3 5 H. then 1 ˆ hW1 . . jGj gPG . with irreducible characters labelled ÷1 . ÷5 5 H and ÷6 5 H.16 we see that ÷1 5 H ˆ ÷2 5 H. . Thus. ÷5 5 H ˆ 2ø4 . l H is the inner product on the vector space of functions from H to C. From Example 19. if W1 and W2 are functions from G to C. l G is the inner product on the vector space of functions from G to C which we have de®ned earlier. ÷7 . ÷3 5 H ˆ ÷4 5 H and ÷6 5 H ˆ ÷7 5 HX Therefore we need only consider ÷1 5 H. ÷ 3 5 H ˆ ø1 ‡ ø 4 . These have values 1 ÷1 5 ÷3 5 ÷5 5 ÷6 5 H H H H 1 4 6 5 (1 2)(3 4) 1 0 À2 1 (1 2 3) (1 3 2) 1 1 0 À1 1 1 0 À1 It is easy to spot that ÷ 1 5 H ˆ ø1 .16. de®ned similarly. ö2 i H ˆ ö1 (h)ö2 (h)X j Hj hP H 213 If ÷ is a character of G and ø1 .17. ÷reg ( g) ˆ and ÷reg ˆ ÷ i (1)÷ i X 0 if g Tˆ 1. øi H X j Hj iˆ1 j Therefore h÷ i 5 H. d r which are given by d i ˆ h÷ 5 H. . . øi H Tˆ 0X Proof Let ÷1 . . it may be very dif®cult in practice to write down the restrictions ÷ 5 H in terms of irreducible characters of H. then 1 ˆ hö1 . Then there exists an irreducible character ÷ of G such that h÷ 5 H. Recall from Theorem 13. then by Theorem 14. Unfortunately. øi H Tˆ 0 for some i. The best .4. øi H ˆ ÷ i (1)h÷ i 5 H. . X X X .20 that the regular character ÷reg of G satis®es & k ˆ jGj if g ˆ 1. .Restriction to a subgroup and if ö1 and ö2 are functions from H to C. In the light of Proposition 20.4 Proposition Let H be a subgroup of G and let ø be a non-zero character of H. ÷ k be the irreducible characters of G. ø r are the irreducible characters of the subgroup H of G. Suppose that we know the character table of G. ø i i H X We say that ø i is a constituent of ÷ 5 H if the coef®cient d i in the above expression is non-zero. iˆ1 Now 0 Tˆ k ˆ jGj ø(1) ˆ h÷reg 5 H. . The next proposition shows that every irreducible character of H is a constituent of the restriction of some irreducible character of G. 20. . ÷ 5 H ˆ d 1 ø1 ‡ X X X ‡ d r ø r for some non-negative integers d 1 . we could hope to ®nd the character table of the subgroup H by restricting the irreducible characters ÷ of G to H. .19 and Proposition 13. we have equality in (20. 20. 1 ˆ h÷. jGj iˆ1 i € where K ˆ (1ajGj) ga H ÷( g)÷( g)X Now K > 0. as the following result shows. Proof By Theorem 14. The way to exploit the fact that H v G us revealed in the following proposition. we will see that all the constituents of ÷ 5 H have the same degree. .17. where the non-negative integers d 1 . since ÷ is irreducible. X X X . For example. we have r ˆ iˆ1 d 2 ˆ h÷ 5 H. ÷ 5 Hi H ˆ i 1 ˆ ÷(h)÷(h)X j Hj hP H Also. and let ø1 . ø r be the irreducible characters of H. and K ˆ 0 if and only P if ÷( g) ˆ 0 for all g with g P H.6) if and only if ÷( g) ˆ 0 for all elements g of G which lie outside H.214 Representations and characters of groups chance of doing this occurs when the index jG: Hj(ˆ jGjaj Hj) is small. d r satisfy (20X6) r ˆ iˆ1 d 2 < jG: HjX i Moreover.5 Proposition Let H be a subgroup of G. Then ÷ 5 H ˆ d 1 ø1 ‡ X X X ‡ d r ø r . ÷i G ˆ ˆ ˆ 1 ˆ ÷( g)÷( g) jGj gPG 1 ˆ ÷(h)÷(h) ‡ K jGj hP H r j Hj ˆ 2 d ‡ K. let ÷ be an irreducible character of G. since the restrictions ÷ 5 H then do not have many constituents. F F F . j We can say more about the constituents of ÷ 5 H in the case where H is a normal subgroup of G. The conclusions of the proposition a follow at once. then WgÀ1 is a C H-submodule of U. since V is irreducible. Further. Let V be an irreducible CG-module and U be an irreducible C H-submodule of V 5 H. proving that Ug is a C H-submodule of V 5 H. whence W ˆ {0} or Ug.12.Restriction to a subgroup 215 20. if W is a C Hsubmodule of Ug. u 3 ug (u P U) is an invertible linear transformation from U to Ug. (3) If g1 . we have ˆ Vˆ UgX gPG Then by Proposition 7. (2) As a C H-module. and the proof of the proposition is complete. as claimed. Ug is an irreducible C H-module. so dim U ˆ dim Ug. For every g P G let Ug ˆ fug: u P U g. Proof (1) Clearly. V is a direct sum of some of the C Hmodules Ug. Therefore. Then (1) The set Ug is an irreducible C H-submodule of V 5 H and dim Ug ˆ dim U . g P G and Ug1 and Ug2 are isomorphic C H-modules. De®ne è : Ug1 g 3 Ug2 g by è : wg 3 (wö) g (w P Ug 1 )X Then è is clearly an isomorphism of vector spaces. and (wgh)è ˆ (wh9 g)è ˆ (wh9ö) g ˆ (wö)h9 g ˆ (wö) gh ˆ (wgè)hX Therefore.7 Proposition Suppose that H v G. Therefore. since U is irreducible. g2 . WgÀ1 ˆ {0} or U. (3) Now let ö be a C H-isomorphism from Ug1 to Ug2 . è is a C H-isomorphism. Ug is a subspace of V. Then gh ˆ h9 g for some h9 P H. V is a direct sum of some of the C Hmodules Ug. then Ug1 g and Ug2 g are isomorphic C H-modules. so (ug)h ˆ u( ghg À1 ) g P Ug (u P U). Moreover. j . and since H v G we have ghgÀ1 P H for all h P H. (2) The sum of all the subspaces Ug with g P G is a CG-submodule of V. Suppose that h P H. Let e ˆ h÷ 5 H. Hence V has the form V ˆ X1 È X X X È X m where each Xi is a direct sum of e isomorphic C H-modules. if g P G then X1 g is a direct sum of isomorphic C H-modules. . Then it follows from Proposition 20. V is a sum of C Hmodules of the form X1 g. On the other hand. ø m are the constituents of ÷ 5 H. ÷ 5 H ˆ e(ø1 ‡ X X X ‡ ø m )X j Our main applications of Clifford's Theorem will concern the case where jG: Hj ˆ 2.7(3). Normal subgroups of index 2 We are shortly going to give more precise information about the constituents of ÷ 5 H when H is a normal subgroup of G of index 2 (that is. and (2) if ø1 . . and which is therefore a direct sum of e isomorphic C H-modules. Then V contains a C H-module X1 whose character is eø1 .14 in Chapter 22 to see an advanced use of the theorem. . ø1 i. jG: Hj ˆ 2).7(2).216 Representations and characters of groups We now come the fundamental theorem on the restriction of a character to a normal subgroup. Proof Let V be a CG-module with character ÷. parts (1) and (2). and Xi T X j if i Tˆ j. by Proposition 20. each having character ø1 . that all the constituents of ÷ 5 H have the same degree. Then (1) all the constituents of ÷ 5 H have the same degree. . Examples where this happens are G ˆ Sn . .7. but you might like to look at Corollary 22.8 Clifford's Theorem Suppose that H v G and that ÷ is an irreducible character of G. Therefore. 20. then ÷ 5 H ˆ e(ø1 ‡ X X X ‡ ø m ) for some positive integer e. say X 1 ˆ U1 È X X X È U e X By Proposition 20. since ë(h) ˆ 1 for all h P H. .10). and we then illustrate the results by ®nding the character table of A5 from that of S5 (which we have already obtained in Example 19. ë( g) ˆ À1 if g P HX a Note that for all irreducible characters ÷ of G. it is often desirable to have more details about the two cases in Proposition 20. Since GaH  C2 .16). or (2) ÷ 5 H is the sum of two distinct irreducible characters of H of the same degree. bÀ1 ab ˆ aÀ1 i.Restriction to a subgroup 217 H ˆ A n . Also. j with i Tˆ j. and let ÷ be an irreducible character of G. We describe this relationship in (20. Proof If ø1 .14). we may lift the non-trivial linear character of GaH to obtain a linear character ë of G which satis®es @ 1 if g P H. ø i and ø j have the same degree. the character tables of G and H are closely related. then H must be normal in G (see Exercise 1. or ÷ 5 H ˆ ø i ‡ ø j for some i. H ˆ hai.9. For practical purposes. 20. When H is a normal subgroup of index 2 in G. ÷ 5 H ˆ ÷ë 5 H. . In fact. d r are non-negative integers.5. 20.8 j €r ÷ 5 H ˆ d 1 ø1 ‡ X X X ‡ d r ø r . X X X . ø r are the irreducible characters of H. then by Proposition 20. we deduce that either ÷ 5 H ˆ ø i for some i. and that ÷ is . . .13) below. if H is a subgroup of index 2 in G.10 Proposition Suppose that H is a normal subgroup of index 2 in G. ÷ and ÷ë are irreducible characters of the same degree (see Proposition 17. Since d 1 . 2 where iˆ1 d i < 2. Then either (1) ÷ 5 H is irreducible. b: a n ˆ b2 ˆ 1. and we shall supply these next. by Clifford's Theorem 20. In the latter case.9 Proposition Suppose that H is a normal subgroup of index 2 in G. or G ˆ D2 n ˆ ha. Then the following three conditions are equivalent: (1) ÷ 5 H is irreducible. then ÷ 5 H is the sum of one or two irreducible characters of H. and ÷ is an irreducible character of G. ÷ë( g) ˆ À÷( g) if g P H. then either ö ˆ ÷ or ö ˆ ÷ë. Proof We have (÷ ‡ ÷ë)( g) ˆ Hence @ 2÷( g) 0 if g P H.5. a j According to Proposition 20. observe that @ ÷( g) if g P H.218 Representations and characters of groups an irreducible character of G. In the next proposition we consider the ®rst possibility. if g P HX a . If ö is an irreducible character of G which satis®es ö 5 H ˆ ÷ 5 H. a (3) the characters ÷ and ÷ë of G are not equal. To see that (2) is equivalent to (3). 20. since jG: Hj ˆ 2. and this happens if and only if ÷( g) Tˆ 0 for some g P G with g P H. ÷ 5 H is irreducible if and only if the inequality in (20.9.6) is strict. a so ÷( g) Tˆ 0 for some g with g P H if and only if ÷ë Tˆ ÷. if H is a normal subgroup of G of index 2.11 Proposition Suppose that H is a normal subgroup of index 2 in G. Thus (1) is equivalent to a (2). (2) ÷( g) Tˆ 0 for some g P G with g P H. and that ÷ is an irreducible character of G for which ÷ 5 H is irreducible. Proof We use Proposition 20. Therefore k÷ ‡ ÷ë. öi G ˆ ˆ 1 ˆ 2÷( g)ö( g) jGj gP H 1 ˆ ÷( g)ö( g) j Hj gP H 219 ˆ h÷ 5 H. since ÷ 5 H is irreducible and ö 5 H ˆ ÷ 5 H. by explaining how to list the irreducible characters of H on the assumption that we know the character table of G.11).Restriction to a subgroup h÷ ‡ ÷ë. say ÷ 5 H ˆ ø1 ‡ ø2. ÷ 5 Hi H X 2 If ö 5 H has ø1 or ø2 as a constituent. ÷l G Tˆ 0. Such characters of G occur in pairs (÷ and ÷ë) which have the same restriction to H (Propositions 20.12 Proposition Suppose that H is a normal subgroup of index 2 in an irreducible character of G for which ÷ 5 H is irreducible characters of H. 20. G. then hö 5 H. 20. so kö.10. we look at the case where ÷ 5 H is reducible. (1) Each irreducible character ÷ of G which is nonzero somewhere outside H restricts to be an irreducible character of H. ö 5 Hl H ˆ 1. (20. and hence ö ˆ ÷. a Therefore. j We summarize our results on subgroups of index 2. 1 ˆ 1 ˆ hö. . If ble character of G such that ö 5 H has ø1 or ø2 then ö ˆ ÷.10. ÷ 5 Hi H Tˆ 0. ÷i G ˆ ö( g)÷( g) ˆ ö( g)÷( g) jGj gPG jGj gP H ˆ 1hö 5 H.13) Let H be a normal subgroup of index 2 in the group G. Proof In view of Proposition 20. and that ÷ is the sum of two ö is an irreducias a constituent. öl G ˆ 1. and so either ö ˆ ÷ or j ö ˆ ÷ë. Finally. ÷( g) ˆ 0 for all g with g P H. ö 5 Hi H X Now k÷ 5 H. The conjugacy classes of H and their sizes are given in Example 12. extra work is needed to calculate the values taken by the two constituents of ÷ 5 H.4). ø5 are the distinct irreducible characters of H by (20. in practice case (1) occurs more frequently than case (2). . ø2 and ø3 . . They can also be veri®ed by calculating inner products. .12). 20. . Fortunately. 20. .18(2). The two characters of H which we get from ÷ in this way come from no other irreducible character of G (Propositions 20. . Also. ÷7 of S5 can be found in Example 19. and hence ø1 .13) of facts about characters of subgroups of index 2.13)(3). . Observe that ÷1 . as in parts (1) and (2) (Proposition 20.13)(1). respectively.13)(2). Call them ø1 . ÷5 5 H ˆ ø4 ‡ ø5 where ø4 and ø5 are distinct irreducible characters of H of degree 3. a so by (20.16. ÷1 5 H. ÷3 and ÷6 are non-zero somewhere outside H. ÷5 ( g) ˆ 0 for all g P H. (3) Every irreducible character of H appears among those obtained by restricting irreducible characters of G. ÷4 5 H ˆ ÷3 5 H and ÷7 5 H ˆ ÷6 5 H. and note that H is a normal subgroup of index 2 in the group S5 . 20.9. . In case (2) of (20. so by (20. The results we have obtained so far have been deduced from our summary (20. We have established that the character table of H is gi |CG ( g i )| ø1 ø2 ø3 ø4 ø5 1 60 1 4 5 3 3 (1 2 3) 3 1 1 À1 á2 â2 (1 2)(3 4) 4 1 0 1 á3 â3 (1 2 3 4 5) 5 1 À1 0 á4 â4 (1 3 4 5 2) 5 1 À1 0 á5 â5 . and the irreducible characters ÷1 .220 Representations and characters of groups (2) If ÷ is an irreducible character of G which is zero everywhere outside H. then ÷ restricts to be the sum of two distinct irreducible characters of H of the same degree.10. ÷3 5 H and ÷6 5 H are irreducible characters of H. Note that ÷2 5 H ˆ ÷1 5 H.13).14 Example The character table of A5 Write H ˆ A5 . á3 ˆ â3 ˆ À1. Hence by Proposition 13.Restriction to a subgroup 221 We use the column orthogonality relations to calculate the unknowns á i and â i . The values of á i ‡ â i for 2 < i < 5 are given by noting that ø4 ‡ ø5 ˆ ÷5 5 H (or by using the column orthogonality relations for column 1 and column i). all the numbers in the character table are real. á5 and â5 are 1(1 Æ 5). By the orthogonality relation for column i with itself (2 < i < 5). 3 3 5 ˆ 2 ‡ á2 ‡ â2 ˆ 2 ‡ á2 ‡ â2 X 4 4 5 5 Hence á2 ˆ â2 ˆ 0.13. ⠈ 1(1 À 2 p 5). we have 2 p p á5 ˆ 1(1 À 5). and we ®nd that á4 and â4 are the solutions of the quadratic x 2 À x À 1 ˆ 0. we obtain 3 ˆ 3 ‡ á2 ‡ â2 . We get á2 ‡ â2 ˆ 0. â5 ˆ 1(1 ‡ 5)X 2 2 Thus the character table of A5 is as shown. á3 ‡ â3 ˆ À2.9(4). 2 2 4 ˆ 2 ‡ á2 ‡ â2 . á4 ‡ â4 ˆ á5 ‡ â5 ˆ 1X Using Proposition 12. we see that each element of A5 is conjugate to its inverse. Since ø4 Tˆ ø5 . Since we have not yet distinguished between ø4 and ø5 . Character table of A5 gi |CG ( g i )| ø1 ø2 ø3 ø4 ø5 1 60 1 4 5 3 3 (1 2 3) 3 1 1 À1 0 0 (1 2)(3 4) 4 1 0 1 À1 À1 (1 2 3 4 5) 5 1 À1 0 á â (1 3 4 5 2) 5 1 À1 0 â á where á ˆ 1(1 ‡ 2 p 5). we may take p p á4 ˆ 1(1 ‡ 5). . â4 ˆ 1(1 À 5)X 2 2 p Similarly. Let G ˆ S4 and let H be the subgroup k(1 2 3 4). Use the restrictions of the irreducible characters of S6 . If ÷ is a character of G. express ÷ 5 H as a sum of irreducible characters of H. ø r are the irreducible characters of H. (1 3)l of G. Prove that ÷(1) < n for every irreducible character ÷ of G. Indeed.3 and 12. Suppose that G is a group with a subgroup H of index 3. if ø1 . Let G be a group with an abelian subgroup H of index n.17. then all the constituents of ÷ 5 H have the same degree. If H v G and ÷ is an irreducible character of G. . The values of ÷ 5 H are given by (÷ 5 H)(h) ˆ ÷(h) for all h P H. then r ˆ iˆ1 d 2 < jG : HjX i 3. then ÷ 5 H is a character of H. Summary of Chapter 20 Assume throughout that H is a subgroup of G.4. ÷ and ÷ 5 H have the same degree. (The seven conjugacy classes of A6 can be found by consulting the solutions to Exercises 12. (b) For each irreducible character ÷ of G (given in Section 18. to ®nd the character table of A6 .) 3. and ÷ 5 H ˆ d 1 ø1 ‡ X X X ‡ d r ø r. The number of irreducible constituents of ÷ 5 H is bounded above by jG : Hj. 2 or 3X Give examples to show that each possibility can occur.1). Prove that h÷ 5 H. given in Example 19. X X X . 4. Exercises for Chapter 20 1. 2.222 Representations and characters of groups Proposition 17.6 allows us to deduce from the character table that A5 is a simple group. In particular. 1. ÷ 5 Hi H ˆ 1. (a) Show that H  D8. and let ÷ be an irreducible character of G. 2. 14. It is known that the complete list of degrees of the irreducible characters of S7 is 1. 14. 1. 21.Restriction to a subgroup 223 5. 35. 6. A7 has nine conjugacy classes. 35X Also. 15. . 6. 21. 14. 14. Find the complete list of degrees of the irreducible characters of A7 . 15. 20. Before describing the process of induction. we require some results which connect C H-homomorphisms with CG-homomorphisms. As H is smaller than G. then there exists r P CG such that uW ˆ ru for all u P U X 224 . (us)W ˆ rus ˆ (uW)s. and induction is the main concern of this chapter. If W is a C H-homomorphism from U to CG.1 Proposition Assume that H < G. C H-homomorphisms and CG-homomorphisms Let U be a C H-submodule of the regular C H-module C H. We shall see many applications of this method in later chapters. it is usually the case that C H-modules are easier to understand and construct than CG-modules. We saw in the last chapter that restriction gives a simple way of converting a CG-module into a C H-module. so induction can often give us an important handle on the representations of a group if we know some representations of its subgroups. Much more subtle than this is the process of induction. which constructs a CG-module from a given C H-module. 21. and let U be a C H-submodule of C H.21 Induced modules and characters Throughout this chapter we assume that H is a subgroup of the ®nite group G. If r P CG. then W: u 3 ru (u P U ) de®nes a C H-homomorphism from U to CG. We now prove the striking fact that every C Hhomomorphism from U to CG has this simple form. since for all s P C H. v P V.3 Corollary Let U and V be CG-submodules of CG. Conversely. and rv ˆ 0 if v P V. so u‡v3 u (u P U . assume that for some r P CG we have ru ˆ u and rv ˆ 0 for all u P U. Let r ˆ 1ö. j . If x P U ’ V then rx ˆ x and rx ˆ 0. v P V ) is a function. it is a CG-homomorphism from U È V to CG (see Proposition 7. r(u ‡ v) ˆ uX Then ru ˆ u if u P U. and so W is of the required form. (2) there exists r P CG such that for all u P U. Then the sum U ‡ V is a direct sum. Consequently U ’ V ˆ {0}. 21. ru ˆ u and rv ˆ 0X Proof Assume that U ’ V ˆ {0}. the ®rst of which is just the case H ˆ G of the proposition. uW ˆ uö ˆ (1u)ö ˆ (1ö)u ˆ ru. De®ne ö: C H 3 CG by ö: u ‡ w 3 uW (u P U. w P W )X Then ö is easily seen to be a C H-homomorphism. Therefore by Corollary 21.2 Corollary Let U be a CG-submodule of CG. moreover. Then the following two statements are equivalent: (1) U ’ V ˆ {0}. v P V. Then every CG-homomorphism from U to CG has the form u 3 ru (u P U ) for some r P CG. j We give two corollaries of Proposition 21.2.1.Induced modules and characters 225 Proof By Maschke's Theorem 8.1. For u P U.11). there exists r P CG such that for all u P U. there is a C H-submodule W of C H such that C H ˆ U È W. 21. v P V. and so x ˆ 0. so C H is a subset of CG. a cyclic subgroup of G of order 3. W 2 4 G ˆ U4 X . a direct sum of irreducible CG-modules U i .4 De®nition Assume that H is a subgroup of G. Remember that H is a subgroup of G. W 2 ˆ sp (1 ‡ ùa ‡ ù2 a2 )X These are C H-submodules of C H (see Example 10. X (CG) ˆ sp fxg: x P X . b ‡ ùab ‡ ù2 a2 b)X Thus W 0 4 G ˆ U1 È U2 . U3 ˆ sp (1 ‡ ù2 a ‡ ùa2 . W 1 4 G ˆ U3 . W 0 4 G ˆ sp (1 ‡ a ‡ a2 . where U1 ˆ sp (1 ‡ a ‡ a2 ‡ b ‡ ab ‡ a2 b). b ‡ ab ‡ a2 b). W 1 ˆ sp (1 ‡ ù2 a ‡ ùa2 ). W 2 4 G ˆ sp (1 ‡ ùa ‡ ù2 a2 . b ‡ ùab ‡ ù2 a2 b)X Recall from Example 10. Let ù ˆ e2ðia3 . b ‡ ù2 ab ‡ ùa2 b). 21. b ‡ ù2 ab ‡ ùa2 b).5 Example Let G ˆ D6 ˆ ka. b: a3 ˆ b2 ˆ 1. bÀ1 ab ˆ aÀ1 l. and de®ne W 0 ˆ sp (1 ‡ a ‡ a2 ). and let H ˆ kal. we write X(CG) for the subspace of CG which is spanned by all the elements xg with x P X. 21. Let U be a C H-submodule of C H.8(1)). Then U 4 G is called the CG-module induced from U. U2 ˆ sp (1 ‡ a ‡ a2 À b À ab À a2 b). and let U 4 G denote the CG-module U(CG).8(2) that CG ˆ U1 È U 2 È U3 È U 4 . g P GgX Clearly.226 Representations and characters of groups Induction from H to G For any subset X of CG. W 1 4 G ˆ sp (1 ‡ ù2 a ‡ ùa2 . U4 ˆ sp (1 ‡ ùa ‡ ù2 a2 . X(CG) is then a CG-submodule of CG. g P G. That is. Clearly. Proof Let W: U 3 V be a C H-isomorphism. b P V 4 GX . Suppose that U and V are C H-submodules of C H and that U is C H-isomorphic to V. g P G). We now show that isomorphic C H-modules give isomorphic induced CG-modules. v P V.6 Proposition Assume that H < G. and also there exists s P CG such that vWÀ1 ˆ sv for all v P V. v P V X If a P U 4 G then a is a linear combination of elements ug (u P U.7 Proposition Assume that U and V are C H-submodules of C H with U ’ V ˆ {0}. 21. W 0 4 G is reducible. rsb ˆ b Hence the function b 3 sb (b P V 4 G) is the inverse of ö. and hence ra P V 4 G. Then U 4 G is CG-isomorphic to V 4 G. j The next proposition and its corollary enable us to de®ne the induced module U 4 G for an arbitrary C H-module U. g P G). so ra is a linear combination of elements rug. Therefore ö is a CG-isomorphism. for all a P U 4 G.Induced modules and characters 227 In particular. By Proposition 21. ö is a CG-homomorphism.1. while W 1 4 G and W 2 4 G are irreducible. proving that U 4 G is CG-isomorphic to V 4 G. 21. Therefore ö: a 3 ra (a P U 4 G) is a function from U 4 G to V 4 G. there exists r P CG such that uW ˆ ru for all u P U. Moreover. Then (U 4 G) ’ (V 4 G) ˆ f0g. we have sra ˆ a. Since sru ˆ u and rsv ˆ v for all u P U. as (aö) g ˆ rag ˆ (ag)ö (a P U 4 G. Consequently sru ˆ u and rsv ˆ v for all u P U . 7. j We can now de®ne the induced module U 4 G for an arbitrary C Hmodule U (where U is not necessarily a C H-submodule of C H).7 and 10. this implies that ru9 ˆ u9 and similarly. Now U ˆ U1 È V. g P G). V 4 G ˆ (U 2 4 G) È X X X È (U m 4 G). 21. using (2. a direct sum of C H-submodules U i .9 De®nition Let U be a C H-module. v P V. Therefore (U 4 G) ’ (V 4 G) ˆ f0g by Corollary 21. Then U 4 G ˆ (U1 4 G) È X X X È (U m 4 G)X Proof We prove this by induction on m.10).8 Corollary Let U be a C H-submodule of C H.3. as required. Then (by Theorems 8.3. and suppose that U ˆ U1 È X X X È Um . there exists r P CG such that ru ˆ u and rv ˆ 0 for all u P U. 21. we obtain U 4 G ˆ (U 1 4 G) È X X X È (Um 4 G). It is trivial for m ˆ 1. U 4 G ˆ (U1 4 G) È (V 4 G)X By induction. v P V and all g P G. rv9 ˆ 0 for all v9 P V 4 GX j for all u9 P U 4 G. and hence. The de®nition of U 4 G implies that U 4 G ˆ (U1 4 G) ‡ (V 4 G)X Therefore by Proposition 21.5). rug ˆ ug and rvg ˆ 0X Since U 4 G is spanned by elements of the form ug (u P U. U  U1 È X X X È Um . where V ˆ U2 È X X X È U m .228 Representations and characters of groups Proof By Corollary 21. Then for all u P U. then (U 4 K) 4 G  U 4 GX Proof Assume ®rst that U is a C H-submodule of C H. That is.10). We emphasize that the de®nition of the induced module U 4 G in the case where U is a C H-submodule of C H is a natural one: U 4 G ˆ U (CG)X We shall always prove results for general induced modules U 4 G by ®rst dealing with the special case where U is a C H-submodule of C H. and then applying the fact (which is immediate from De®nition 21. Now let U be an arbitrary C H-module. U 4 K  (U1 4 K) È X X X È (Um 4 K)X Since K < G.Induced modules and characters 229 for certain irreducible C H-submodules U i of C H. Then U(CK) is spanned by elements of the form uk ukg (21X12) (u P U . g P G)X (U 4 K) 4 G ˆ U 4 GX U  U1 È X X X È Um for certain irreducible C H-submodules U i of C H.8 ensure that this de®nition is consistent with De®nition 21.4. it follows that (U(CK))(CG) ˆ U(CG). (U(CK))(CG) is spanned by elements of the form (u P U . Then .11 Theorem Suppose that H and K are subgroups of G such that H < K < G. By (21. If U is a C H-module.9) that (21X10) (U1 È X X X È Um ) 4 G  (U1 4 G) È X X X È (U m 4 G)X Our ®rst major result on general induced modules is known as `the transitivity of induction'. k P K)X Therefore. k P K. 21.6 and Corollary 21. De®ne U 4 G to be the following (external) direct sum: U 4 G ˆ (U 1 4 G) È X X X È (Um 4 G)X Proposition 21. ÷7 are the irreducible characters of G (given in Example 19.2) then ÷ 1 5 H ˆ ø1 . . 21. The next example illustrates an important relationship between induced characters and restrictions of characters. by (21X12) j Induced characters 21. . ÷ 4 5 H ˆ ø1 ‡ ø4 . . the coef®cients which appear here are the values of h÷ i 5 H. ÷ 7 5 H ˆ ø 2 ‡ ø 3 ‡ ø4 X By Theorem 14. . as in Example 20. and is called the character induced from ø. ÷ 2 5 H ˆ ø1 .13 De®nition If ø is the character of a C H-module U. ø4 are the irreducible characters of H (given in 18. . . ø j i H for appropriate i. ÷ 6 5 H ˆ ø 2 ‡ ø 3 ‡ ø4 . then the character of the induced CG-module U 4 G is denoted by ø 4 G.2.230 Therefore Representations and characters of groups (U 4 K) 4 G  (U1 4 K) 4 G È X X X È (Um 4 K) 4 G by (21X10) ˆ (U1 4 G) È X X X È (U m 4 G) U4G by Definition 21. .17. . ÷5 5 H ˆ 2ø4 . and ø1 . We showed in that example that if ÷1 .16).14 Example Let G ˆ S5 and let H be the subgroup A4 of G. ÷ 3 5 H ˆ ø1 ‡ ø4 . j. We record these coef®cients in a .9. To be precise. The Frobenius Reciprocity Theorem Before proving the theorem. we need the following preliminary result. ø j 4 Gi G .Induced modules and characters matrix whose ij-entry is h÷ i 5 H. øi H for all characters ÷ of G and ø of H. Then the vector spaces HomCG (U 4 G. ø4 X Remarkably. .15 Proposition Assume that H < G. ø2 ‡ 0 . ø1 ‡ 0 . it turns out that the columns of the matrix tell us how the irreducible characters of H induce to G. which we already know to be equal to h÷ i 5 H. and let V be a CG-submodule of CG. the seven integers in column 1 give ø1 4 G ˆ 1 . ø 4 Gi G ˆ h÷ 5 H. is also equal to h÷ i . ø j i H : ÷1 ÷2 ÷3 ÷4 ÷5 ÷6 ÷7 H ø1 1 f 1 f f 1 f f 1 f f 0 f d 0 0 ø2 0 0 0 0 0 1 1 ø3 0 0 0 0 0 1 1 ø4 I 0 0 g g 1 g g 1 g g 2 g g 1 e 1 231 The rows of this matrix tell us how the irreducible characters of G restrict to H. V 5 H) have equal dimensions. For example. ÷3 ‡ 1 . ÷5 ‡ 0 . ÷2 ‡ 1 . ÷6 ‡ 0 . row 3 gives ÷3 5 H ˆ 1 . 21. V ) and HomC H (U . Let U be a C H-submodule of C H. ÷4 ‡ 0 . it is true that h÷. and ø4 4 G ˆ ÷3 ‡ ÷4 ‡ 2÷5 ‡ ÷6 ‡ ÷7 X Thus the ij-entry in our matrix. ø j i H . In fact. ÷7 X Similarly. ÷1 ‡ 1 . which is known as the Frobenius Reciprocity Theorem. and we devote the next section to a proof of this result. ø2 4 G ˆ ø3 4 G ˆ ÷6 ‡ ÷7 . ø3 ‡ 1 . Moreover. j 21.232 Representations and characters of groups Proof Suppose that W P HomCG (U 4 G.1. Let ö P HomC H (U . Finally. V ) to HomC H (U . De®ne W from U 4 G to CG by sW ˆ rs (s P U 4 G)X Then W P HomCG (U 4 G. V ) to HomC H (U . and there is a CG-submodule V of CG which has character ÷. Then by Proposition 21. ö ˆ W.2. then r1 s ˆ r2 s for all s P U 4 G. there is an element r P CG such that sW ˆ rs for all s P U 4 GX De®ne W: U 3 CG to be the restriction of W to U. since s is a linear combination of elements ug with u P U.16 The Frobenius Reciprocity Theorem Assume that H < G. Let ÷ be a character of G and let ø be a character of H. Thus the function W3W is a linear transformation from HomCG (U 4 G. uW ˆ ru for all u P U X Then W P HomC H (U . r2 P CG and r1 u ˆ r2 u for all u P U. ÷i G ˆ dim (HomCG (U 4 G. as required. Then hø 4 G. V 5 H). we have hø 4 G. ÷ 5 Hi H ˆ dim (HomC H (U . V )). Then there is a C H-submodule U of C H which has character ø. Then by Corollary 21. These two vector spaces therefore have the same dimension. that is. V 5 H).24. g P G. Therefore the function W 3 W is surjective. and hø. V ). V 5 H). ÷i G ˆ hø. We shall show that this linear transformation is invertible. By Theorem 14. there exists r P CG such that uö ˆ ru for all u P U. Hence the function W 3 W is injective. and so it is an invertible linear transformation from HomCG (U 4 G. note that if r1. V 5 H). V ). ÷ 5 Hi H X Proof First assume that the characters ÷ and ø are irreducible. V 5 H))X . ˆ hø. For the general case. Then for some integers di . de®ne the j . ø m be the irreducible characters of H. ÷ i 5 Hi H k ˆ iˆ1 by (21X17) B jˆ1 iˆ1 m ˆ jˆ1 C di÷i 5 H H ej ø j . f 5 Hi H X Proof This follows at once from the Frobenius Reciprocity Theorem. then hø 4 G. . ÷ k be the irreducible characters of G and let ø1 .18 Corollary If f is a class function on G. .4. . . and ø is a character of H. ÷ i i G ej d i hø j . we therefore deduce that (21X17) hø 4 G. k ˆ iˆ1 d i ÷i G m k ˆˆ jˆ1 iˆ1 m k ˆˆ ej d i hø j 4 G. let ÷1 .15. namely when ÷ and ø are irreducible. and for convenience of notation. f i G ˆ hø. ÷i G ˆ ˆ ˆ ˆ k ˆ iˆ1 d i ÷ i and ø ˆ m ˆ jˆ1 ej ø j X C B m ˆ jˆ1 ej ø j 4 G. ÷i G ˆ hø. ÷ 5 Hi H X This completes the proof of the Frobenius Reciprocity Theorem. . ej we have ÷ˆ Therefore hø 4 G. j The values of induced characters We now show how to evaluate induced characters. . Let ø be a character of the subgroup H of G. . .Induced modules and characters From Proposition 21. 21. ÷ 5 Hi H 233 in the special case we are considering. since by Corollary 15. the characters of G span the vector space of class functions on G. 234 Representations and characters of groups @ • function ø: G 3 C by • ø( g) ˆ ø( g) 0 if g P H. Therefore f is a class function. Then 1 ˆ h f . If w P G then 1 ˆ • f (w À1 gw) ˆ ø( y À1 w À1 gwy) ˆ f ( g) j Hj yPG since wy runs through G as y runs through G. and so by Corollary 15. ÷i G for all irreducible characters ÷ of G. ÷i G ˆ hø 4 G. if g P HX a 21. ÷i G ˆ f ( g)÷( g) jGj gPG ˆ Put x ˆ y À1 gy.4. Let ÷ be an irreducible character of G. Proof Let f: G 3 C be the function given by 1 ˆ • f ( g) ˆ ø( y À1 gy) ( g P G)X j Hj yPG We aim to prove that f ˆ ø 4 G. Then h f . ÷i G ˆ ˆ 1 1 ˆˆ • ø(x)÷( yxy À1 )X jGj j Hj xPG yPG 1 ˆ ø(x)÷(x) j Hj xP H 1 1 ˆˆ • ø( y À1 gy)÷( g)X jGj j Hj gPG yPG .19 Proposition The values of the induced character ø 4 G are given by 1 ˆ • (ø 4 G)( g) ˆ ø( y À1 gy) j Hj yPG for all g P G. it is suf®cient to show that h f . a formula for the values of induced characters different from that given in Proposition 21. de®ne the class function f G on G by x & 1 if y P x G f G ( y) ˆ ( y P G)X x 0 if y P x G a Thus f is the characteristic function of the conjugacy class x G . Alternatively. f G i G ˆ x ÷(x) X jCG (x)j . the stated degree of ø 4 G can be found using just the de®nition of induced modules (see Exercise 21.3). ÷i G ˆ hø. 21. and ÷( yxy À1 ) ˆ ÷(x) for all y P G. j For practical purposes. For x P G.19. ÷ 5 Hi H and so by the Frobenius Reciprocity Theorem. h f . so the proof is complete. Therefore a h f .19 is more useful. and we shall derive this next (it is given in Proposition 21. then h÷. ÷i G ˆ hø 4 G. then the degree of ø 4 G is given by (ø 4 G)(1) ˆ jGj ø(1)X j Hj Proof This follows immediately by evaluating (ø 4 G)(1) using Proposition 21.20 Corollary If ø is a character of the subgroup H of G. j 21.23 below).Induced modules and characters 235 • since ø(x) ˆ 0 if x P H.21 Proposition If ÷ is a character of G and x P G. ÷i G X This was the equation required to show that f ˆ ø 4 G. but if g P G then g G may contain 0. . then f G 5 H ˆ 0. 2 or more conjugacy classes of H. with representatives x1 . . x (2) If some element of x G lies in H.4. . xm P H such that H H f G 5 H ˆ f x1 ‡ X X X ‡ f x m X x Statement (2) just says that H ’ x G breaks up into m conjugacy classes of H. To put this another way. then there are elements x1 . then (ø 4 G)(x) ˆ 0. . we have (ø 4 G)(x) ˆ hø 4 G. f G 5 Hi H X x x jCG (x)j .22)).21 and Corollary 21.22) Suppose that x P G. (1) If no element of x G lies in H. and suppose that x P G. X X X . x Proof By Proposition 21. 1.23 Proposition Let ø be a character of the subgroup H of G. xm P H and f G 5 H ˆ f x1 ‡ F F F ‡ f x m (as in (21. (2) If some element of x G lies in H. we have: (21. f G i G ˆ x ˆ ˆ Note that a result similar to Proposition 21.236 Proof We have Representations and characters of groups 1 ˆ ÷( g) f G ( g) x jGj gPG 1 ˆ jx G j ÷(x) ÷( g) ˆ jGj gPx G jGj ÷(x) jCG (x)j by Theorem 12X8X j h÷.21 was used in the proof of Theorem 16. (1) If no element of x G lies in H. f G i G ˆ hø. .18. . 21. If H < G and h P H then h H  h G . . then   ø(x1 ) ø(xm ) (ø 4 G)(x) ˆ jCG (x)j ‡X X X‡ . jCH (x1 )j jCH (xm )j H H where x1 . . x m . and H H f G 5 H ˆ f x1 ‡ X X X ‡ f x m as in (21. a3 ˆ (1 4 3 2)g. f x1 i H ‡ X X X ‡ hø. . fb ˆ (1 3).Induced modules and characters 237 If no element of x G lies in H. G H H f (1 2)(3 4) 5 H ˆ f (1 3)(2 4) ‡ f (1 2)(3 4) . fa ˆ (1 2 3 4). b ˆ (1 3)X Then H  D8. the statement G H H f (1 2)(3 4) 5 H ˆ f (1 3)(2 4) ‡ f (1 2)(3 4) records the fact that the G-conjugacy class (1 2)(3 4) G contains exactly two H-conjugacy classes. where a ˆ (1 2 3 4).24 Example Let G ˆ S4 and let H ˆ ka. And if some element of x G lies in H. a2 b ˆ (2 4)g. then x (ø 4 G)(x) H H ˆ hø. G H f (1 3) 5 H ˆ f (1 3) . then f G 5 H ˆ 0. the conjugacy classes of H are f1g. and G H f (1 2 3 4) 5 H ˆ f (1 2 3 4) X For example. By (12. f x m i H ˆ ø(x1 ) ø(xm ) ‡X X X‡ jCH (x1 )j jCH (xm )j by Proposition 21X21X j The result follows. with representatives (1 3)(2 4) and (1 2)(3 4). fa2 ˆ (1 3)(2 4)g. f x1 ‡ X X X ‡ f x m i H jCG (x)j H H ˆ hø. since a4 ˆ b2 ˆ 1 and bÀ1 ab ˆ aÀ1 . 21.22)(2). We have G H f1 5 H ˆ f1 . G f (1 2 3) 5 H ˆ 0.12). and hence x (ø 4 G)(x) ˆ 0. bl. 3(3) for the irreducible characters ÷1 . Then according to Proposition 21.238 Representations and characters of groups The orders of the centralizers of the elements of H are as follows: h |CG (h)| |C H (h)| 1 24 8 (1 3)(2 4) 8 8 (1 2 3 4) 4 4 (1 3) 4 4 (1 2)(3 4) 8 4 Suppose that ø is a character of H. ÷5 of H  D8. 8 ø((1 3)) (ø 4 G)((1 3)) ˆ 4 . 8 4 (ø 4 G)((1 2 3 4)) ˆ 4 ø((1 2 3 4)) X 4 (ø 4 G)((1 2 3)) ˆ 0.25 Example (cf. . we therefore have 1 ÷1 4 G ÷2 4 G ÷3 4 G ÷4 4 G ÷5 4 G 3 3 3 3 6 (1 2) 1 À1 1 À1 0 (1 2 3) 0 0 0 0 0 (1 2)(3 4) 3 À1 À1 3 À2 (1 2 3 4) 1 1 À1 À1 0 In the next example. 4   ø((1 3)(2 4)) ø((1 2)(3 4)) (ø 4 G)((1 2)(3 4)) ˆ 8 ‡ . Exercise 17. . 21. we use induced characters to ®nd the character table of a group of order 21. we have (ø 4 G)(1) ˆ 24 ø(1) .2) De®ne permutations a. b in S7 by a ˆ (1 2 3 4 5 6 7). Referring to Example 16.23. b ˆ (2 3 5)(4 7 6) . . . Since kal v G and Gahai  C3. and since b P C G (a). ÷2 . bl of S7 . Using this. we obtain three linear characters ÷1 . 0 < j < 2. For 1 < k < 6. Their values are shown below: g |CG ( g)| ÷1 ÷2 ÷3 1 21 1 1 1 a 7 1 1 1 a3 7 1 1 1 b 3 1 ù ù2 b2 3 1 ù2 ù where ù ˆ e2ðia3 . 21. b and b2 to be representatives of the conjugacy classes. we see that the conjugacy classes of G are f1g. a3 . ÷3 of G as the lifts of the linear characters of Gahai. a. fai b: 0 < i < 6g. First we ®nd the conjugacy classes. 7 divides jC G (a)j. We shall obtain the last two irreducible characters of G by inducing linear characters of H.Induced modules and characters and let G be the subgroup ka. We aim to ®nd the character table of G. and similarly jC G (b)j ˆ 3. Hence jC G (a)j ˆ 7. there is a character ø k of H given by ø k (a j ) ˆ ç jk (0 < j < 6)X To use the formula in Proposition 21. Let ç ˆ e2ðia7 . a2 . fai b2 : 0 < i < 6gX We take 1. Also. Since hai < C G (a). a6 g. Check that a7 ˆ b3 ˆ 1. note that H H H f G 5 H ˆ f a ‡ f a2 ‡ f a4 a . fa3 . G has order 21. Notice that G has exactly ®ve irreducible characters. bÀ1 ab ˆ a2 X 239 It follows from these relations that the elements of G are all of the form ai bj with 0 < i < 6. a5 . a4 g.23 for calculating ø k 4 G. fa. Let H ˆ kal. a jC G (a)j . and (ø3 4 G)(b) ˆ (ø3 4 G)(b2 ) ˆ 0X Thus if we write ÷4 ˆ ø1 4 G and ÷5 ˆ ø3 4 G. b: a7 ˆ b3 ˆ 1. since ø1 . Thus ÷4 and ÷5 are our last two irreducible characters. . .240 Representations and characters of groups since no two of the elements a. . Hence by Proposition 21. (ø3 4 G)(a) ˆ ç3 ‡ ç5 ‡ ç6 . bÀ1 ab ˆ a2 i g |CG ( gi )| ÷1 ÷2 ÷3 ÷4 ÷5 1 21 1 1 1 3 3 a 7 1 1 1 ç ‡ ç2 ‡ ç4 ç3 ‡ ç5 ‡ ç6 a3 7 1 1 1 ç3 ‡ ç5 ‡ ç6 ç ‡ ç2 ‡ ç4 b 3 1 ù ù2 0 0 b2 3 1 ù2 ù 0 0 . . (ø3 4 G)(a3 ) ˆ ç ‡ ç2 ‡ ç4 . ÷5 i G ˆ 1. ø6 are linearly independent. a4 are conjugate in H. and the character table of G is as shown.23. (ø1 4 G)(a) ˆ ç ‡ ç2 ‡ ç4 and similarly (ø1 4 G)(a3 ) ˆ ç3 ‡ ç5 ‡ ç6 . Character table of ha. We now calculate that h÷4 . a2 . 21 7 7 3 3 and similarly h÷5 . Therefore ÷4 Tˆ ÷5 . ÷4 i G ˆ 9 2 2 0 0 ‡ ‡ ‡ ‡ ˆ 1. (ø1 4 G)(b) ˆ (ø1 4 G)(b2 ) ˆ 0X Similarly (ø3 4 G)(1) ˆ 3. then the values of ÷4 and ÷5 are 1 ÷4 ÷5 3 3 a ç ‡ ç2 ‡ ç4 ç3 ‡ ç5 ‡ ç6 a3 ç3 ‡ ç5 ‡ ç6 ç ‡ ç2 ‡ ç4 b 0 0 b2 0 0 Now ÷4 5 H ˆ ø1 ‡ ø2 ‡ ø4 and ÷5 5 H ˆ ø3 ‡ ø5 ‡ ø6 . (ø1 4 G)(1) ˆ 3X And since no G-conjugate of b or b2 lies in H. ÷ 5 Hi H . 3. where ø is a character of H and ÷ is a character of G. De®ne U to be the 1-dimensional subspace of C H spanned by 1 À a2 ‡ b À a2 bX (a) Check that U is a C H-submodule of C H. bl. (b) Find a basis of the induced CG-module U 4 G. as given in . ÷i G ˆ hø. 2. an induced CG-module U 4 G can be de®ned. If no element of g G lies in H. . For each C H-module U. b: a4 ˆ b2 ˆ 1. . Exercises for Chapter 21 1. (c) Write down the character of the C H-module U and the character of the CG-module U 4 G. then (ø 4 G)( g) ˆ 0X If some element of g G lies in H. then   ø(x1 ) ø(xm ) (ø 4 G)( g) ˆ jCG ( g)j ‡X X X‡ jCH (x1 )j jCH (xm )j H H where f G 5 H ˆ f x1 ‡ X X X ‡ f x m . If ø is a character of H then the induced character ø 4 G is given by 1 ˆ • (ø 4 G)( g) ˆ ø( y À1 gy)X j Hj yPG In particular. then U 4 G is simply U(CG). bÀ1 ab ˆ aÀ1 l and let H be the subgroup ka2 . Let G ˆ S4 and let H be the subgroup k(1 2 3)l  C3 . (a) If ÷1 . . The Frobenius Reciprocity Theorem states that hø 4 G. Is U 4 G irreducible? 2. the degree of ø:G is |G: H|ø(1). . g 4. Let G ˆ D8 ˆ ka. 241 1. If U is a C H-module of C H.Induced modules and characters Summary of Chapter 21 Assume that H is a subgroup of G. ÷5 are the irreducible characters of G. 6. ø2 . then jGj (ø 4 G)(1) ˆ ø(1)X j Hj 4. d k . Let ø be an irreducible character of H. . 3.) . b ˆ (2 3 5)(4 7 6). (b) Calculate the induced characters ø j 4 G (1 < j < 3) as sums of the irreducible characters ÷ i of G.25.1. bl. . work out the restrictions ÷ i 5 H (1 < i < 5) as sums of the irreducible characters ø1 . and use the Frobenius Reciprocity Theorem. as in Example 21. X X X . which are given by ø 4 G ˆ d 1 ÷1 ‡ X X X d k ÷k . ø3 of C3 . You are given that jC G (a)j ˆ 7 and jC G (b)j ˆ 18. Prove that (ø(÷ 5 H)) 4 G ˆ (ø 4 G)÷X (Hint: consider the inner product of each side with an arbitrary irreducible character of G. . Let H be a subgroup of G. Show that the integers d 1 . and let ÷1 . Let ö and ø be the irreducible characters of H which are given by gi |C H ( g i )| ö ø 1 21 1 3 a 7 a3 7 b 3 1 0 b2 3 1 0 1 1 ç ‡ ç2 ‡ ç4 ç3 ‡ ç5 ‡ ç6 where ç ˆ e2ðia7 (see Example 21. let ø be a character of H. Show direct from the de®nition that if H < G and ø is a character of H. and let ÷ be a character of G. . Let G ˆ S7 and let H ˆ ka. Suppose that H is a subgroup of G.242 Representations and characters of groups Section 18. ÷ k be the irreducible characters of G.) 5. where a ˆ (1 2 3 4 5 6 7).25).5. satisfy k ˆ iˆ1 d 2 < jG : HjX i (Compare Proposition 20. Calculate the values of the induced characters ö 4 G and ø 4 G. Discover and prove results for ø 4 G which are similar to those presented in Chapter 20 for the restriction of irreducible characters of G to H. .Induced modules and characters 243 7. and let ø be an irreducible character of H. Suppose that H is a normal subgroup of index 2 in G. we require that det (A À ëI) ˆ 0 for some square matrix A with integer entries. Algebraic integers 22. for ë to be an algebraic integer. all of whose entries are integers. Most of the results concern arithmetic properties of character values. We discuss properties of a group element g P G which ensure that ÷( g) is an integer for all characters ÷ of G.1 De®nition A complex number ë is an algebraic integer if and only if ë is an eigenvalue of some matrix. And we prove some useful congruence properties. using algebraic integers. if p is a prime number and g P G is an element of order pr for some r. perhaps the most opaque is that which states that the degree of an irreducible character of a ®nite group G must divide the order of G.22 Algebraic integers Among the properties of characters which may be regarded as fundamental. This is one of several results which we shall prove in this chapter. We remark that ë is an algebraic integer if and only if ë is a root of a polynomial of the form 244 . Thus. we have uA ˆ ëu for some non-zero row vector u. for the same matrix A. for example. Equivalently. then ÷( g)  ÷(1) mod p for any character ÷ of G for which ÷( g) is an integer. . and non-zero row vectors u and v.3 Theorem If ë and ì are algebraic integers. algebraic integers are usually de®ned in this way. where u is the row vector obtained from u by replacing each entry by its complex conjugate. . note that if A is an integer matrix and u is a row vector with uA ˆ ëu. ù nÀ1 . Then uA ˆ (ù.Algebraic integers x n ‡ a nÀ1 x nÀ1 ‡ X X X ‡ a1 x ‡ a0 245 where a0 . 1) ˆ ùuX This shows that every nth root of unity is an algebraic integer. since it is an eigenvalue of the 0 1 matrix X 2 0 (3) If ë is an algebraic integer. 22. ù. X X X . In fact. all of whose entries are integers. a nÀ1 are integers (see Exercise 22. 22. . then u(ÀA) ˆ (Àë)u and uA ˆ ëu. such that uA ˆ ëu. X X X .7). vB ˆ ìvX Suppose that A is an m 3 m matrix and B is an n 3 n matrix. (4) Let A be the n 3 n matrix H 0 0 f1 0 f f0 1 Aˆf f f d0 0 0 0 given by 0 XXX 0 XXX 0 XXX XXX 0 XXX 0 XXX 0 0 0 1 0 I 0 1 0 0g g 0 0g gX g g 0 0e 1 0 Suppose that ù is an nth root of unity. Proof There exist square matrices A and B. p (2) 2 is an algebraic integer. . . ù2 . To see this. ù nÀ1 ). then ëì and ë ‡ ì are also algebraic integers. then so are Àë and the complex conjugate ë of ë. since n is an eigenvalue of the 1 3 1 matrix (n).2 Examples (1) Every integer n is an algebraic integer. ù2 . and let u be the row vector (1. 9. by Example 22. Hence ÷( g) is an algebraic integer. . It is easy to check as in the proof of Proposition 19. ÷( g) is a sum of roots of unity. ( ë ij (ei  f j ))(A  B) ˆ ë ij (ei A  f j B)). respectively.3. and we deduce as above that ë ‡ ì is an algebraic integer. .246 Representations and characters of groups Let e1 . so any sum of roots of unity is an algebraic integer by Theorem 22. . 1 < j < n) has integer entries.3 shows that the set of all algebraic integers forms a subring of C. Each root of unity is an algebraic integer. Proof By Proposition 13. . 1 < j < n) form a basis of the tensor product space V ˆ C m  C n. j Theorem 22. Since the matrix of A  B relative to the basis ei  fj (1 < i < m. The next result provides a link between algebraic integers and characters. € € extending linearly (that is.4 that for all vectors x P C m . it follows that ëì is an algebraic integer. y P C n. 1 < j < n). De®ne an endomorphism A  B of V by (ei  f j )(A  B) ˆ ei A  f j B (1 < i < m. f n be a basis of C n . 22. Let Im and In denote the identity m 3 m and n 3 n matrices. X X X .4 Corollary If ÷ is a character of G and g P G.5 Proposition If ë is both a rational number and an algebraic integer. j 22. Then (u  v)(A  I n ‡ I m  B) ˆ uA  vI n ‡ uI m  vB ˆ ëu  v ‡ u  ìv ˆ (ë ‡ ì)(u  v). Then the vectors ei  fj (1 < i < m. then ÷( g) is an algebraic integer. we have (x  y)(A  B) ˆ xA  yBX Hence (u  v)(A  B) ˆ uA  vB ˆ ëu  ìv ˆ ëì(u  v)X Therefore ëì is an eigenvalue of A  B. . em be a basis of C m and f 1 .2(4). then ë is an integer. j The next result is an immediate consequence of Corollary 22. note that we have. as a special case of Proposition 22.4 and Proposition 22.5.6 Corollary Let ÷ be a character of G and let g P G. p the well known result that 2 is irrational. Therefore det (sA À rI) ˆ (Àr) n ‡ mp for some integer m. then ˆ Cˆ x P CGX xPC 22. (Example 22. then ÷( g) is an integer. the entries of sA À rI which are not on the diagonal are divisible by s. 22.7 Lemma Suppose that g P G and that C is the conjugacy class of G which contains g. we establish two preliminary lemmas. we deduce that det (sA À rI) Tˆ 0. which is enough to establish the proposition. Then . If ÷( g) is a rational number.5. Write ë ˆ ras. where r and s are coprime integers and s Tˆ Æ1. In passing.) The degree of every irreducible character divides |G| To prepare for the proof that |G| is divisible by the degree of each irreducible character of G. with character ÷. As p does not divide r (since r and s are coprime). Let p be a prime number which divides s. Let U be an irreducible CG-module. and hence also by p. Recall from De®nition 12. For every n 3 n matrix A of integers. We shall show that ë is not an algebraic integer.21 that if C is a conjugacy class of G.2(2) shows p that 2 is an algebraic integer. s and hence ë is not an algebraic integer.Algebraic integers 247 Proof Suppose that ë is a rational number which is not an integer. Thus  n 1 det (A À ëI) ˆ det (sA À rI) Tˆ 0. Then for 1 < i < n.248 Representations and characters of groups uC ˆ ëu for all u P U . gn be the elements of G. . we have gi r ˆ n ˆ jˆ1 aij g j . where each á g is an integer. As |C| ˆ |G:CG ( g)| by Theorem 12. we obtain ˆ ÷(x) ˆ ë÷(1). . then ˆ [x]B ˆ ëIX xPC Taking the traces of both sides of this equation.22). j 22. Then ë is an algebraic integer. . Proof Let g1 . where ëˆ jGj ÷( g) X jCG ( g)j ÷(1) Proof Since C lies in the centre of CG (see Proposition 12. this yields jCj÷( g) ˆ ë÷(1)X Thus ë ˆ jCj÷( g)a÷(1). where ë P C.8 Lemma € Let r ˆ gPG á g g P CG. . ˆ u x ˆ ëu for all u P U X xPC Consequently if B is a basis of U.14 that there exists ë P C such that uC ˆ ëu for all u P U. that is. Suppose that u is a non-zero element of CG such that ur ˆ ëu.8. the result follows. xPC and since ÷ is constant on the conjugacy class C. we know by Proposition 9. Proof Let U be an irreducible CG-submodule of CG with character ÷. gk be representatives of the conjugacy classes of G. where ù is an nth root of unity. then ëˆ is an algebraic integer.4. . then ÷(1) divides |G|.8. by Lemma 22. Therefore ë is an algebraic integer.) The i statement that ur ˆ ëu (with u Tˆ 0) says that ë is an eigenvalue of the integer matrix A ˆ (aij ). (In fact. Hence by Theorem 22.2(4). both jGj ÷( g i ) jCG ( g i )j ÷(1) and ÷( g i ) jGj ÷( g) jCG ( g)j ÷(1) are algebraic integers. j 22.Algebraic integers 249 for certain integers aij . .10 Corollary If ÷ is an irreducible character of G and g P G.3. . Proof Let g1 . and de®ne u ˆ 1 ‡ ùx À1 ‡ ù2 x À2 ‡ X X X ‡ ù nÀ1 x P CG. by Corollaries 22. and let C be the sum of the elements in the conjugacy class of G which contains g. 22. .7. Then for all i.8 con®rms that ù is an algebraic integer. Then uC ˆ ëu for all u P U. Then ux ˆ ùu and so Lemma 22.10 and 22. j 22.9 Example Let G ˆ Cn ˆ kx: x n ˆ 1l. aij ˆ á g where g ˆ gÀ1 gj . Therefore ë is an algebraic integer. Notice that this example is just a reworking of Example 22. k ˆ iˆ1 jGj ÷( g i )÷( g i ) jCG ( g i )j ÷(1) . by Lemma 22.11 Theorem If ÷ is an irreducible character of G. 11. That is. if jGj ˆ p2 then ÷(1) ˆ 1 for all irreducible characters ÷.11 also has the following interesting consequences concerning irreducible characters of simple groups. observe that G is non-abelian. then ÷(1) is a power of p for all irreducible characters ÷ of G.18. we recover the well known result that groups of order p2 are abelian. and so G9 ˆ G as G is simple.11.250 Representations and characters of groups is an algebraic integer. the number of linear characters of G divides |G|. (Recall that a group G is simple if it has no normal subgroups apart from {1} and G. using Proposition 9. By Theorem 17. Proof Suppose that G is a simple group which has an irreducible character ÷ of degree 2. Since Ker r v G and G is simple.5 now implies that jGja÷(1) is an integer. Theorem 16.12 Examples (1) If p is a prime number and G is a group of order pn for some n.7(a)). or they are 1.13 Corollary No simple group has an irreducible character of degree 2.11. (2) Let G be a group of order 2 p. But g 3 det ( gr) is a linear character of G (see Exercise 13. and this implies that det ( gr) ˆ 1 for all g P GX . by Proposition 9. Proposition 22. 2.5. p. 2 (with ( p À 1)a2 degrees 2)X (3) If G ˆ Sn then every prime p which divides the degree of an irreducible character of G also divides n!. we have Ker r ˆ f1g. as the sum of the squares of the degrees of the irreducible characters is equal to |G|. This algebraic integer is equal to jGja÷(1). Let r: G 3 GL(2. By Theorem 22. 1. and so r is injective. As jGja÷(1) is a rational number. First. (Note that ÷(1) . X X X . Hence G9 Tˆ 1. where p is prime. G has no non-trivial linear characters. and hence satis®es p < n. In particular.) 22. Therefore by Theorem 17.) Hence. j 22. Theorem 22.4(1). either the degrees of the irreducible characters of G are all 1. C) be a representation of G with character ÷. ÷(1) divides |G|. Hence. the degree of each irreducible character of G is 1 or 2 (it cannot be p for the reason noted in (1) above). by the row orthogonality relations. that is. by Theorem 22.14 Corollary Suppose that p is a prime and the degree of every irreducible character of G is a power of p. by our hypothesis. xr has order 2.8). we conclude that   À1 0 À1 T (xr)T ˆ X 0 À1 Thus xr ˆ T(ÀI)T À1 ˆ ÀIX Consequently (xr)( gr) ˆ ( gr)(xr) for all g P G. Since G is non-abelian.Algebraic integers 251 Now G has even order. the sum being over the irreducible characters ÷ of G for which ÷(1) . as G is simple. 1.6). Proof The result is correct if G is abelian (see Theorem 9. Therefore G contains an element x of order 2 (see Exercise 1. 1 for some irreducible character ÷ of G. This time. and jN j is coprime to p. In particular. Then ÷(1) is divisible by p. we assume that every irreducible character of G has degree a power of a prime p. Consider the 2 3 2 matrix xr. ÷(1) . and we deduce that G has an abelian normal pcomplement N. Theorems 11. 22. there is a 2 3 2 matrix T such that T À1 (xr)T is a diagonal matrix with diagonal entries Æ1. while jG: N j is a power of p. G is not simple unless G has prime order. As r is injective.11 give us the equation ˆ jGj ˆ jGaG9j ‡ ÷(1)2 .12 and 17. j Our next result again shows that information about character degrees can sometimes be used to learn about the structure of a group. so we assume that G is non-abelian. As r is injective. and by Proposition 9.11. Then G has an abelian normal p-complement. this means that xg ˆ gx for all g P G. so p divides |G| by . and hence hxi v GX This is a contradiction.11. N is an abelian normal subgroup of G. Since det (xr) ˆ 1. Theorem 22. Bearing in mind the dif®culties we encountered in constructing the character tables of Sn for small values of n (we reached n ˆ 6 in Example 19. øi Tˆ 0 for some irreducible character ø of H. Then g P H. for example. so it remains to prove that N v G. b). Also. assume that G is simple. it follows that GaG9 has a subgroup of index p. Before proving Theorem 22.8 shows that ø(1) divides ÷(1). We have that |N| is coprime to p and jG : N j is a power of p.17).16.1. Clifford's Theorem 20. since otherwise p divides the order of gH which in turn divides the order of g. Finally. Suppose that g P G and the order of g is coprime to p. so G is abelian. Next. if N ˆ G then G is again abelian.7).11. a similar argument shows that g P N. Since every ®nite abelian group is isomorphic to a direct product of cyclic groups. j A condition which ensures that ÷( g) is an integer In Theorem 22. Hence N consists of those elements of G whose order is coprime to p. But an abelian simple group has prime order. we require a preliminary lemma concerning roots of unity. Therefore. every entry in the character table of Sn is an integer (see Corollary 22. 22. then we denote their highest common factor by (a. that for all n. On the other hand. If a and b are positive integers.16 is evidently a useful result.4. so either N ˆ {1} or N ˆ G. Let ø be an irreducible character of H. This result implies. and we deduce from the equation above that p divides the order of the abelian group GaG9. and from this fact it follows easily that N v G. Hence G has a normal subgroup H of index p. we write d|n to denote the fact that d divides n. for integers d and n. If N ˆ f1g then G is a p-group so Z(G) Tˆ {1} (see Exercise 12.17).252 Representations and characters of groups Theorem 22. by Exercise 1. G has prime order. because G is simple.16 below we give a group-theoretic condition on an element g of G which implies that ÷( g) is an integer for every character ÷ of G.15 Lemma If ù is an nth root of unity. by Proposition 20. we have Z(G) ˆ G. then . Then h÷ 5 H. We may now apply induction on |G| to deduce that H has an abelian normal p-complement N. so ø(1) is a power of p. Then ÷( g) is an integer for all characters ÷ of G. It is trivial for n ˆ 1. €n i Now we partition the sum iˆ1 ù according to the highest common factor d of i and n: 0ˆ n ˆ iˆ1 ùi ˆ ˆ ˆ dj n 1<i< n (i. 1 ( j. (i. if ù ˆ 1 then the result is immediate.nad)ˆ1 If djn then ù d is an (nad)th root of unity. Proof We prove the result by induction on n. ( j. there is a basis B of V such that H I ù1 0 f g FF [ g]B ˆ d e F ùm 0 . Also. 1.n)ˆ1 ùi ˆ n ˆ iˆ1 ùi À ˆ ˆ ù dj P Z.n)ˆd ùi ˆ ˆ ˆ ù dj X dj n 1< j< nad.11. dj n. and if in addition d . ˆ ù dj P ZX 1< j< nad. Suppose that g is conjugate to gi for all i with 1 < i < n and (i.n)ˆ1 253 is an integer. Proof Let V be a CG-module with character ÷ of degree m. By Proposition 9. then by our induction hypothesis. 1< j< nad.nad)ˆ1 j as required.16 Theorem Let g be an element of order n in G. 22. So suppose that ù is an nth root of unity and ù Tˆ 1.nad)ˆ1 It follows that ˆ 1<i< n.Algebraic integers ˆ ùi 1<i< n. Then ù is a root of the polynomial €n (x n À 1)a(x À 1) ˆ x nÀ1 ‡ X X X ‡ x ‡ 1X Therefore iˆ1 ù i ˆ 0. d . n) ˆ 1. ( j. (i. and so i ÷( g i ) ˆ ù1 ‡ X X X ‡ ù im X Therefore by Lemma 22. .6. and hence are conjugate by Theorem 12. . n) ˆ 1. . then ÷( g)  ÷(1) mod p. .15. 22. ù m are nth roots of unity.16. namely that if ÷( g i ) P Z for all characters ÷ of G. Consequently ÷( g) is a rational number. one particularly useful consequence of our results is that if p is a prime number. then the permutations g and g i have the same cycle-shape. ù im . For example. .254 Representations and characters of groups where ù1 . and so ÷( g) is an integer by Corollary 22. we need to de®ne the p9-part of a group element. The de®nition will emerge from the following lemma. and hence s÷( g) P Z. g is an element of G of order pr for some r. . Before going into the character theory. then g is conjugate to g i whenever i is coprime to n.16.n)ˆ1 As g is conjugate to gi if 1 < i < n and (i. . . For 1 < i < n. the matrix i [ g i ]B has diagonal entries ù1 . j We remark that using Galois theory it is possible to prove the converse of Theorem 22. j The p9-part of a group element The rest of the chapter is devoted to some important congruence properties of character values.17 Corollary All the character values of symmetric groups are integers. (i. The result now follows from Theorem 22. . n) ˆ 1. where s is the number of integers i with 1 < i < n and (i. Proof If g P Sn and i is coprime to the order of g. and ÷ is a character of G such that ÷( g) P Z. we have ÷( g i ) ˆ ÷( g) for such i.15. ˆ ÷( g i ) P ZX 1<i< n. the order of x9 is a power of p and the order of y9 is coprime to p. and (3) the order of y is coprime to p. y P G such that (1) g ˆ xy ˆ yx. Therefore z ˆ 1. We must show that x ˆ x9 and y ˆ y9. y9 P G also satisfy (1)±(3). y9 commutes with y and y( y9)À1 has order coprime to p. xy ˆ g ˆ x9 y9. y u ˆ g bup ˆ 1X Hence the order of x is a power of p and the order of y divides u. where u. it follows that x À1 x9 has order a power of p. j v v v v v . the elements x and y of G which satisfy conditions (1)±(3) are unique.18 Lemma Let p be a prime number and let g P G. Then there exist x. so x9 commutes with g. x p ˆ g aup ˆ 1. hence also with gau ˆ x. (2) the order of x is a power of p. then we have shown that the order of z is both a power of p and coprime to p.Algebraic integers 255 22. Finally. Now suppose that x9. Therefore x and y satisfy conditions (1)±(3). Since both x and x9 have order a power of p. p) ˆ 1. We have x9 g ˆ x9 y9x9 ˆ gx9. Then there exist integers a. as required. so is coprime to p. so x À1 x9 ˆ y( y9)À1 X If z ˆ x À1 x9 ˆ y( y9)À1 . Moreover. v P Z and (u. g ˆ x9 y9 ˆ y9x9. and so x ˆ x9 and y ˆ y9. that is. Then xy ˆ yx ˆ g au‡bp ˆ g. b such that au ‡ bpv ˆ 1X Put x ˆ gau and y ˆ gbp . Proof Let the order of g be upv . Similarly. that is. a principal ideal of Z[æ]. We extract the following statement from the proof of Lemma 22.18 has x ˆ g3 . y ˆ gÀ2 . By de®nition. Every such coset contains an element of the form a0 ‡ a1 æ ‡ X X X ‡ a nÀ1 æ nÀ1 . . The ideals of Z[æ] which contain pZ[æ] are in . we conclude that Z[æ]a pZ[æ] is ®nite. . Z[æ] ˆ f f (æ): f (x) P Z[x]gX Clearly. De®ne Z[æ] to be the subring of C generated by Z and æ.20) Let the order of g be upv . (22. æ.19 De®nition We call the element y which appears in Lemma 22.21 Proposition There are only ®nitely many ideals I of Z[æ] which contain pZ[æ]. For example. . if p ˆ 2 and g has order 6. b with au ‡ bpv ˆ 1. 22.18. Proof Consider the factor ring Z[æ]a pZ[æ]. æ2 . . v Then the p9. 0 < ai < p À 1 for all iX As there are only ®nitely many such elements.256 Representations and characters of groups 22. so in fact Z[æ] ˆ f f (æ): f (x) P Z[x] of degree < n À 1gX Now let p be a prime number and let pZ[æ] ˆ f pr: r P Z[æ]g. the expression g ˆ xy in Lemma 22. then the p9-part of g is g . and choose integers a. v P Z and (u. Let n be a positive integer and let æ ˆ e2ðia n . æ nÀ1 . every element of Z[æ] is an integer combination of the powers 1. p) ˆ 1. we need a few basic facts about a certain subring of C in which all our character values will lie.part of g is gbp . where u. this has as its elements all the cosets pZ[æ] ‡ r with r P Z[æ]. À2 A little ring theory To prepare for our main result on congruence properties of character values. with ai P Z.18 the p9-part of g. that is. b with j am ‡ bp ˆ 1. (A proper ideal of Z[æ] is an ideal which is not equal to Z[æ]. For the last statement of the proposition. it follows that s P P. if r n P P for some positive integer n. we therefore have rZ[æ] ‡ P ˆ Z[æ]X Consequently. j . Therefore there are only ®nitely many such ideals. Thus pjm. b P P such that 1 ˆ ra ‡ bX Then s ˆ rsa ‡ sbX As rs P P and b P P. If p B m then there are integers a. then either r P P or s P P. which is false. We must show that s P P. As P is a maximal. Proof Assume that rs P P and r P P. a Since r P P. j We deduce from Proposition 22. the ideal rZ[æ] ‡ P of Z[æ] strictly contains P. P is a proper ideal which is contained in no larger proper ideal of Z[æ]. Proof Let m P P ’ Z. but this implies that 1 P P. Repeating this argument. In particular. Since n r ˆ rr nÀ1 . we conclude that r P P. this implies that either r P P or r nÀ1 P P. since P Tˆ Z[æ]. which establishes that P ’ Z  pZ.21 that there is a maximal ideal of P of Z[æ] which contains pZ[æ]. assume that r n P P.) We now prove two easy results about the maximal ideal P. then r P P. as required. 22. we also have pZ  P ’ Z.22 Proposition If r. and the proof is complete.Algebraic integers 257 bijective correspondence with the ideals of Z[æ]a pZ[æ] (the correspondence being I 3 Ia pZ[æ]). j 22.23 Proposition We have P ’ Z ˆ pZ. there exist a P Z[æ]. s P Z[æ] and rs P P. Since p P P.  v vÀ r v p bpv pv pv v pvÀ1 bpv (ù À ù ) ˆ ù À p ù ù ‡ X X X Æ ù p ù rbp r ‡ X X X ‡ (À1) p ù bp X For 0 . As in the previous section. so ÷( g) and ÷( y) are both sums of nth roots of unity. the binomial coef®cient ( rp ) is divisible by p. we have @ 0. The ring Z[æ] is of interest because all the character values of G lie in Z[æ] (see Proposition 9. Application of Proposition 22.22 now forces v v v v v 2v v v v v 2v v v 2v v v . Then y ˆ g bp (see (22. where u. Choose integers a. p) ˆ 1. 22.11). since ù p ˆ ù bp . b with au ‡ bpv ˆ 1.258 Representations and characters of groups Congruences At last we are ready to prove our results on congruences of character values. pv pv bp2v ù ‡ (À1) ù ˆ pv 2ù . v P Z and v (u. Moreover. where á P Z[æ]. Then v ù ˆ ù au‡bp . By the Binomial Theorem. Let G be a group of order n and let æ ˆ e2ðia n . if p ˆ 2.24 Theorem Let g P G and let y be the p9.part of g. Hence (ù À ù bp ) p ˆ ù p ‡ (À1) p ù bp ‡ pá. r . then ÷( g) À ÷( y) P PX Proof Suppose that g has order m ˆ upv . so it follows that (ù À ù bp ) p P pZ[æ]X Thus (ù À ù bp ) p lies in the maximal ideal P. pv . and hence lie in Z[æ]. and so v v 2v 2v ù p ˆ ù aup . Now let ù be an mth root of unity (so ù P Z[æ] as mjn).20)). if p Tˆ 2. ù bp ˆ ù bp X Consider the number (ù À ù bp ) p . let p be a prime number and let P be a maximal ideal of Z[æ] containing pZ[æ]. If ÷ is any character of G. The orders of g and of y divide n ˆ |G|. . If ÷ is a character of G such that ÷( g) and ÷( y) are both integers.27 extensively in our character calculations in Chapters 25±7. we just illustrate the results with reference to some character tables which we already know. Suppose that g P G and that y is the p9.27 Corollary Let p be a prime number. j Notice that Corollary 13. the p9-part of g is 1.Algebraic integers (22X25) (ù À ù bp ) P PX v 259 By Proposition 9.11.10 is the special case of Corollary 22. d v v which. j 22. For the moment. so the result is immediate from Corollary 22. j 22. then ÷( g)  ÷( y) mod pX Proof As ÷( g) and ÷( y) are both integers. If ÷ is a character of G such that ÷( g) P Z.24 and Proposition 22.part of g. . there are mth roots of unity ù1 . . Suppose that g P G and the order of g is a power of p. then ÷( g)  ÷(1) mod pX Proof As the order of g is a power of p.27 in which g has order 2.26 Corollary Let p be a prime number. We shall use the congruence results 22.23 give ÷( g) À ÷( y) P P ’ Z ˆ pZX Therefore ÷( g)  ÷( y) mod p. .26.25). lies in P. Theorem 22.24±22. by (22. ù d such that bp ÷( g) ˆ ù1 ‡ X X X ‡ ù d and ÷( y) ˆ ù1 ‡ X X X ‡ ù bp X d v v Then bp ÷( g) À ÷( y) ˆ (ù1 À ù1 ) ‡ X X X ‡ (ù d À ù bp ). . If we take p ˆ 5 and g ˆ (1 2 3 4 5). That is. Summary of Chapter 22 1.26 implies that ÷( g)  ÷(1) mod 3 whenever ÷( g) P Z. p p Then ( 5)2 P P. p p where á ˆ (1 ‡ 5)a2.22. 1(1 À 5) ˆ â 5X 2 Put æ ˆ e2ðia60 .28 Example Recall from Example 20. Since â P Z[æ] (see p Proposition 9. If g ˆ (1 2 3) then Corollary 22. ÷4 ((1 2 3 4 5)) ˆ á P Z. Similarly the entries in columns 1 and 3 are congruent modulo 2.11). . ⠈ (1 À 5)a2. We illustrate Theorem 22.260 Representations and characters of groups 22. 3X However. Character values are algebraic integers. as can be seen by inspecting the table. then the p9-part of g is 1. so 5 P P by Proposition 22. and p ÷4 ( g) À ÷4 (1) ˆ á À 3 ˆ 1(1 ‡ 5 À 6) 2 p p p ˆ 5 . 2. 2. ÷4 ( g) À ÷4 (1) P PX This illustrates Theorem 22.24. The degree of every irreducible character of G divides |G|.14 that the character table of A5 is as shown.24 for this a value. and let P be a maximal ideal of Z[æ] containing 5Z[æ]. we have â 5 P P. Also ÷ i ((1 2 3 4 5))  ÷ i (1) mod 5 for i ˆ 1. Thus the corresponding entries in columns 1 and 2 Character table of A5 1 ÷1 ÷2 ÷3 ÷4 ÷5 1 4 5 3 3 (1 2 3) 1 1 À1 0 0 (1 2)(3 4) 1 0 1 À1 À1 (1 2 3 4 5) 1 À1 0 á â (1 3 4 5 2) 1 À1 0 â á of the character table are congruent modulo 3. (a) Find the degrees of all the irreducible characters of G. Prove that the number of conjugacy classes in a group of order 16 is 7. If g P G and y is the p9-part of g. Prove that h÷. Let G be a group of order 15. 4. (d) Deduce that both a and b are integers. 17. (c) Deduce that ÷ ˆ 1 G . jGj where á is an algebraic integer. (c) Show that if ÷ is a non-trivial irreducible character of G. (a) Prove that if g P G and g ˆ gÀ1 . then g ˆ 1. (b) Show that |G9| ˆ p. for all characters ÷. Let G be a group and let ö be a character of G such that ö( g) ˆ ö(h) for all non-identity elements g and h of G. (a) Show that ö ˆ a1 G ‡ b÷reg for some a. (c) Show that q divides p À 1 and that G has q ‡ (( p À 1)aq) conjugacy classes. 5. Let p be a prime number. 1 G i ˆ 1 (÷(1) ‡ 2á). then b÷(1) is an integer. 3. 4. 10 or 16.11 and 22. Let p and q be prime numbers with p . and let G be a nonabelian group of order pq. Deduce that G is abelian. q. 2. Use Theorems 11. . (b) Show that a ‡ b and a ‡ b|G| are integers. Suppose that G is a group of odd order.Algebraic integers 261 3.12. (b) Now let ÷ be an irreducible character of G with ÷ ˆ ÷. then ÷( g) is an integer. then ÷( g)  ÷( y) mod p for all characters ÷ of G such that ÷( g) and ÷( y) are integers. This exercise shows that the only irreducible character ÷ of G such that ÷ ˆ ÷ is the trivial character. b P C. Exercises for Chapter 22 1.11 to show that every irreducible character of G has degree 1. If g is conjugate to g i for all integers i which are coprime to the order of g. 262 Representations and characters of groups 6. . 7. g7 with orders and centralizer orders as follows: gi Order of gi |CG ( gi )| g1 1 120 g2 2 12 g3 2 8 g4 3 6 g5 4 4 g6 6 6 g7 5 5 Using Corollary 22. (b) Use Corollary 22. Moreover. g. A certain group G of order 120 has exactly seven conjugacy classes. This exercise illustrates this point with the group G ˆ S5 . It is often possible to calculate the character table from limited arithmetic information about the group G. ®nd the character table of G. g2 . . . (a) Show that for every irreducible character ÷ of G. (c) Find ÷(1) and ÷( g) for all irreducible characters ÷ of G. Prove that a complex number ë is an algebraic integer if and only if ë is a root of a polynomial of the form x n ‡ a nÀ1 x nÀ1 ‡ X X X ‡ a1 x ‡ a0 . where each ar (0 < r < n À 1) is an integer.27 to deduce that G has two irreducible characters of degree 5. (d) You are given that all entries in the character table of G are integers. 1 or À1. . .26 and the column orthogonality relations. and contains an element g of order 5 such that jCG ( g)j ˆ 5. and that the conjugacy classes of G have representatives g1 . g3 and g4 are conjugate in G. ÷( g) is 0. but also gives delicate information about characters which often comes into play in more dif®cult calculations. Nevertheless. The material in this chapter is perhaps at a slightly more advanced level than that in the rest of the book. Often. This is used in the last section to prove a famous result of Brauer and Fowler concerning centralizers of involutions in ®nite simple groups. Various criteria for whether or not a character corresponds to a representation over R lead us to the remarkable Frobenius±Schur Count of Involutions. then the conjugacy class g G is also said to be 263 . we have always taken our representations to be over the ®eld C of complex numbers. Let r be a representation of G. the subject of real representations not only is elegant and interesting. There is a subtle interplay between representations over C and representations over R. which we shall explore in this chapter. However. and if g is real. If all the matrices gr ( g P G) have real entries. while there is no representation ó equivalent to r with all the matrices gó having real entries. and is not used in the ensuing chapters. the converse is not true: it can happen that the character of r is realvalued. which consist largely of the calculation of character tables and applications of character theory.23 Real representations Since Chapter 9. characters of CG-modules are real-valued. and the ®rst main result of the chapter describes the number of real-valued irreducible characters of G. then of course the character of r is real-valued. Real characters An element g of the ®nite group G is said to be real if g is conjugate to gÀ1 . However. some results in representation theory work equally well for the ®eld R of real numbers. . For every conjugacy class g G of G. Proof Let X denote the character table of G. so X can be obtained from X by permuting the rows. and the number of real conjugacy classes of G is tr (Q)X j As these numbers are equal. Hence there is a permutation matrix P such that PX ˆ X (see Exercise 4. by Proposition 13. Notice that if a conjugacy class is real. and let X denote the complex conjugate of the matrix X.15). Therefore X can be obtained from X be permuting the columns. since ( g À1 ) G ˆ fx À1 : x P g G g.5. Thus for example. then it contains the inverse of each of its elements. a character ÷ of G is real if ÷( g) is real for all g P G.2.264 Representations and characters of groups real. we have the number of real irreducible characters of G is tr (P).4). X is invertible. and so there is a permutation matrix Q such that XQ ˆ X By Proposition 16. the entries in the column of X which corresponds to g G are the complex conjugates of the entries in the column of X which corresponds to ( g À1 ) G . 23. Part of the following corollary was obtained by a different method in Exercise 22. the result is proved.2.1 Theorem The number of real irreducible characters of G is equal to the number of real conjugacy classes of G. For every irreducible character ÷ of G. Since the trace of a permutation matrix is equal to the number of points ®xed by the corresponding permutation. Therefore Q ˆ X À1 X ˆ X À1 PX X Consequently P and Q have the same trace. and the trivial character of G is real. the conjugacy class {1} of G is real. On the other hand. the complex conjugate ÷ is also an irreducible character (see Proposition 13. b: a4 ˆ b2 ˆ 1. .3 Examples (1) Let G ˆ D8 ˆ ka. the only real character of G is the trivial character. and there is a basis v1 .8. then no non-identity element of G is real (see the solution to Exercise 23. since     À1 0 0 1 . v n with real coef®cients. . v i g is a linear combination of v1 . br ˆ ar ˆ 0 1 À1 0 provides a representation r of G with character ÷ such that all the matrices gr ( g P G) have real entries.1). Proof If G has odd order. (2) Let G ˆ Q8 ˆ ka. . . . v n of V. . This is the same as saying that there is some CG-module V with character ÷. j Characters which can be realized over R Let ÷ be a character of the group G.3(3)). b: a4 ˆ 1. b2 ˆ a2 . Hence G has at least two real conjugacy classes. then by Exercise 1. If G has even order. Therefore by Theorem 23. We say that ÷ can be realized over R if there is a representation r: G 3 GL (n. and so G has at least two real irreducible characters by Theorem 23. C) with character ÷. 23. bÀ1 ab ˆ aÀ1 l.Real representations 265 23.1). {1} and g G. such that all the entries in each matrix gr ( g P G) are real. and let ÷ be the irreducible character of G of degree 2 (see Exercise 17. . bÀ1 ab ˆ aÀ1 l. and let ÷ be the irreducible character of G of degree 2 (see Example 16. such that for all g P G and 1 < i < n. Then ÷ can be realized over R. .2 Corollary The group G has a non-trivial real irreducible character if and only if the order of G is even. The values of ÷ are as follows: 1 ÷ 2 a2 À2 a 0 b 0 ab 0 .1.1. G has an element g of order 2. 2. Simply take a basis v1 . where F is R or C. Therefore we obtain a representation r: G 3 GL (n. R) is a representation then for each g P G. and consider the vector space over C with basis v1 . .4 Examples Let V be a 2-dimensional vector space over R. v2 a ˆ Àv1 . v n of the RG-module. . . with a multiplication by elements of G satisfying the conditions of De®nition 4. . v n . if we de®ne v1 x ˆ v2 . The construction is even easier to understand in terms of representations: if r: G 3 GL (n. the matrix gr has its entries in R.3(1)). where C3 ˆ kx: x 3 ˆ 1l. v1 b ˆ Àv1 . .3(2) tells us that the converse is false. Thus an RG-module is a vector space over R. (2) V becomes an RC3 -module. with basis v1 . v2 b ˆ v2 (compare Example 23. and hence also in C. In fact. This new vector space is clearly a CG-module (with v i g de®ned as before). but it is at the moment unclear how to prove this. RG-modules Recall that in Chapter 4 we de®ned an FG-module.) À1 À1 Every RG-module can easily be converted into a CG-module. . Example 23. . C). In this section we shall study the relationship between RG-modules and CGmodules.18(3) below.266 Representations and characters of groups Thus ÷ is real.) Although every character which can be realized over R is perforce a real character. (We shall eventually establish this in Example 23. v2 x ˆ Àv1 À v2 X   0 1 (This gives the representation x 3 of Exercise 3. 23. (1) V becomes an RD8 -module if we de®ne v1 a ˆ v2 .2. ÷ cannot be realized over R. v2 . . Notice that a character ÷ of G can be . X X X . then ÷( g) ˆ n ˆ kˆ1 zkk X The character of VR . and (À yjk v k ‡ xjk (iv k )) (1 < j < n). We de®ne a multiplication of VR by g by putting (23X5) vj g ˆ (iv j ) g ˆ n ˆ kˆ1 n ˆ kˆ1 (xjk v k ‡ yjk (iv k )). h P G. iv1 . 1 < j < nX It follows easily that. evaluated at g. is 2 n ˆ kˆ1 xkk ˆ ÷( g) ‡ ÷( g)X Hence the character of VR is ÷ ‡ ÷. . v n . Rather more subtle than this is the construction of an RG-module from a given CG-module. . iv n X Write z jk ˆ x jk ‡ iyjk with x jk . In this way we de®ne v g for all v P VR and all g P G.6. X X X . . v n . There exist complex numbers z jk such that vj g ˆ n ˆ kˆ1 zjk v k (1 < j < n)X Now let VR be the vector space over R with basis v1 .5) makes VR into an RGmodule. and extending linearly to de®ne v g for all v P VR . Let V be a CG-module with basis v1 . we have (v j g)h ˆ v j ( gh) and ((iv j ) g)h ˆ (iv j )( gh)X Hence using Proposition 4. we see that (23. and let g P G. Regarding v j as an element of the CG-module V.Real representations 267 realized over R if and only if there exists an RG-module with character ÷. y jk P R. we have (v j g)h ˆ v j ( gh) for all g. regarding v j and iv j as elements of VR . We summarize the basic properties of VR in the next proposition. If ÷ is the character of V. . . VR ˆ U È W where U is an RG-module with character ÷ and W is an RG-module with character ÷. v2 a ˆ Àiv2 . with character ÷. Proof We have already proved part (1). (2) If V is an irreducible CG-module and VR is a reducible RGmodule. Then VR has basis v1 . and let V be the 2dimensional CG-module with basis v1 . x is represented by the matrix   p À1a2 3a2 p X À 3a2 À1a2 (2) Let G ˆ D8 ˆ ka. v2 such that v1 a ˆ iv1 .6 Proposition Let V be a CG-module with character ÷.7 Examples (1) Let G ˆ C3 ˆ kx: x 3 ˆ 1l. iv1 . v1 b ˆ v2 . where v3 ˆ iv1 . v4 . and so ÷ can be realized over R.268 Representations and characters of groups 23. we obtain the representation r. bÀ1 ab ˆ aÀ1 l. in particular. dim VR ˆ 2 dim V. where H I H I 0 0 1 0 0 1 0 0 f 0 0 0 À1 g f g g. and with 2 respect to this basis. j 23. v2 b ˆ v1 X Then VR has basis v1 . v3 . v2 . b: a4 ˆ b2 ˆ 1. and let V be the 1-dimensional CGmodule with basis v1 such that p v1 x ˆ 1(À1 ‡ i 3)v1 2 p (note that 1(À1 ‡ i 3) ˆ e2ðia3 ). namely U. For part (2). Thus there is an RG-module. . br ˆ f 1 0 0 0 gX ar ˆ f d À1 0 0 e d0 0 0 1e 0 0 1 0 0 0 0 1 0 The subspace of VR which is spanned by v1 ‡ v4 and v2 ‡ v3 is an RG-submodule. v4 ˆ iv2 . Therefore the character of V can be realized over R. (1) The RG-module VR has character ÷ ‡ ÷. suppose that V is an irreducible CG-module and VR is a reducible RG-module. then ÷ can be realized over R. Then by part (1). With respect to this basis. A bilinear form â on V is a function which associates with each ordered pair (u. y) are both linear ± hence the term bilinear. A similar result for CG-modules was given in Exercise 8. 0 for all non-zero v P V X . v. we already know this from Example 23. Let V be a vector space over F. v) ‡ ë2 â(u2 .6.Real representations 269 by Proposition 23. then a bilinear form â on V is said to be Ginvariant if â(ug. u) for all u.) The bilinear form â is symmetric if â(u. v) ˆ Àâ(v. ë1 v1 ‡ ë2 v2 ) ˆ ë1 â(u. Bilinear forms The question of whether or not a given character can be realized over R turns out to be related to the existence of a certain bilinear form on the corresponding CG-module. vg) ˆ â(u.8 Theorem If V is an RG-module. the functions x 3 â(x. (Thus for ®xed u. In fact. â(u. v) of vectors in V an element â(u. then there exists a G-invariant symmetric bilinear form â on V such that â(v.6(2). v P V and g P GX Our next result shows that every RG-module has a G-invariant symmetric bilinear form with a strong positivity property. and which has the following properties: â(ë1 u1 ‡ ë2 u2 . u) for all u.3(1). v) ˆ â(v. v). v) . u1 . v2 ). v) for all u. v1 . u2 . v P V X And the bilinear form â is skew-symmetric if â(u. for all u. v) and y 3 â(u. v2 P V and ë1 . where F is R or C. v P V X If V is an FG-module. v1 ) ‡ ë2 â(u. v) of F. v) ˆ ë1 â(u1 . v. ë2 P F. 23. v) ˆ n ˆ jˆ1 ì2 . w) ˆ 0 for all u P U gX Proof It is easy to see that W is a subspace of V. If g P G. wg) ˆ â(ug À1 . j 23. W ˆ fw P V : â(u. and â(v. and that there exist u. de®ne ã(u. wgg À1 ) ˆ â(ug À1 . j 23. v n be a basis of V. 0 and â(v. 0. v) ˆ n ˆ jˆ1 vˆ ë j ì jX Then ã is a symmetric bilinear form on V. ì j P R. . . For u ˆ €n jˆ1 ì j v j P V with ë j . so W is an RG-submodule of V.10 Proposition Suppose that â is a G-invariant symmetric bilinear form on the RGmodule V. for non-zero v P V. v) . vx) (u. Moreover. v)X xPG Therefore â is G-invariant and the theorem is proved. w) ˆ 0X Thus wg P W. If U is an RG-submodule of V then so is . â is a symmetric bilinear form on V. . so â(u. vg) ˆ ã(ugx. Proof Let v1 . For all u P U. then gx runs through G as x runs through G. v gx) ˆ â(u. and hence ˆ â(ug. . Then V is a reducible RG-module. v) ˆ Now let â(u.270 Representations and characters of groups €n jˆ1 ë j v j . v P V )X Again. 0 for all non-zero v P V.9 Proposition Let V be an RG-module and let â be a G-invariant bilinear form on V. v) . . u) . Now let w P W and g P G. we have ugÀ1 P U. 0X j ˆ xPG ã(ux. ã(v. v P V with â(u. v2 ) . v) (u. The following two conditions are equivalent: (1) ÷ can be realized over R. v i ) ˆ 1 â(v1 . . €n But for all v ˆ iˆ1 ë i v i P V (ë i P R). . . (2) there exists a CG-module V with character ÷. v j ) ˆ 0 and â1 (v i . there is a basis v1 . â(v2 . v2 ) . v n of V such that â1 (v i . 23. We can now relate bilinear forms to the question of whether or not a given character of G can be realized over R. w) ˆ 0 for all v P V g. we have ã(v. . and a non-zero Ginvariant symmetric bilinear form on V . 0X Let â(v1 . v) ˆ â1 (u.11 Theorem Let ÷ be an irreducible character of G. so is ã. v P V )X x Since â and â1 are G-invariant symmetric bilinear forms on V. if we de®ne W ˆ fw P V : ã(v. 0 for all non-zero w P V X By a general result on bilinear forms (see Exercise 23. v2 ) ˆ 1 À â(v2 .9. and de®ne ã by 1 ã(u. v1 ) . v j ) ˆ â(v i .7). v1 ) ˆ ë1 ã(v1 . j if i Tˆ j.8 supplies us with a G-invariant symmetric bilinear form â1 on V such that â1 (w. v) À â(u.Real representations 271 Proof Theorem 23. Therefore V is a reducible RG-module. 0. 0. . v1 ) ˆ x. and is an RG-submodule of V by Proposition 23. v1 ) ˆ 0X Therefore. then W is non-zero. x so W Tˆ V. w) . Moreover. for all i. 1 ã(v2 . by Proposition 23.10. ” De®ne ã on V by 2 n 3 n n n ˆ ˆˆ ˆ ” ã ë j u j. v n of V. and let U be an RG-module with character ÷. . v1 ) ˆ 1X Extend v1 to a basis v1 . v P VR .8. v P V with â(u. There exist u. Let â(w. and for all w1 . Notice that ~ ~ â(v1 . Since â(u ‡ v. u k ) jˆ1 kˆ1 jˆ1 kˆ1 . This establishes that (2) implies (1) in the theorem. and let V be the vector space over C with basis u1 . all ë P R and all g P G. By Theorem 23. w) Tˆ 0. w2 . . v1 ) ˆ 1 and â(iv1 . Then â(v1 . and suppose that â is a non-zero G-invariant symmetric bilinear form on V. X X X . w) ˆ z and v1 ˆ z À1a2 w. . It now follows from Proposition 23. iv1 . that â is a Ginvariant symmetric bilinear form on VR . using the properties (23. v) ‡ 2â(u. we have (23X12) (w1 ‡ w2 )W ˆ w1 W ‡ w2 W. . u) Tˆ 0. . As explained earlier. . . u) ‡ â(v. . v). ì j P R)X jˆ1 jˆ1 jˆ1 Then W is a bijection. v) ˆ â(v. vW) (u.6(2) that ÷ can be realized over R. Conversely. X X X . . u n be a basis of U. there is a non-zero Ginvariant symmetric bilinear form ã on U. v P VR )X ~ You can readily check. v) ˆ the real part of â(uW. iv n is a basis of the RG-module VR . there exists w P V with â(w. .272 Representations and characters of groups Proof We ®rst show that (2) implies (1). u n . (ëv)W ˆ ë(vW). . V is a CG-module (with u i g de®ned as for U). iv1 ) ˆ À1X Therefore VR is a reducible RG-module. De®ne a function W from VR to V by n n n ˆ ˆ ˆ W: ë jv j ‡ ì j (iv j ) 3 (ë j ‡ iì j )v j (ë j . u ‡ v) ˆ â(u. Let u1 . suppose that ÷ can be realized over R. Let V be a CG-module with character ÷. (v g)W ˆ (vW) gX ~ Now de®ne a function â on ordered pairs of elements of VR by ~ â(u. .12). v n . ìk uk ˆ ë j ì k ã(u j . Then v1 . for irreducible characters ÷. if 1 G is a constituent of ÷ S . 1 or À1. and the proof of the theorem is complete. we have @ 0. then precisely one of ÷ S and ÷ A has 1 G as a constituent. b 0.Real representations 273 ” (where ë j . if ÷ is realX Let V be a CG-module with character ÷. 1 G i ˆ ÷( g)÷( g) ˆ h÷. Thus (1) implies (2). if 1G is a constituent of ÷ A X We call é the indicator function on the set of irreducible characters of G. j The indicator function We now associate with each irreducible character ÷ of G a certain number. which is always 0. and ÷ A is the character of the antisymmetric part of V  V. relating it to the internal structure of the group G. if ÷ is not real. ÷iX jGj gPG Therefore. Observe that 1 ˆ h÷ 2 . . b ` é÷ ˆ 1. called the indicator of ÷. and V has character ÷. The next result gives a signi®cant property of the indicator function. ì k P C). 1 G i ˆ 1. and ÷2 ˆ ÷S ‡ ÷ A . Hence if h÷ 2 . then we de®ne the indicator é÷ of ÷ by V if ÷ is not a constituent of ÷ S or ÷ A . where ÷ S is the character of the symmetric part of V  V. 2 h÷ . Then ã is a non-zero G-invariant symmetric bilinear form on the CG-module V. Recall from Chapter 19 that ÷ 2 is the character of the CG-module V  V. b b X À1. 1 G i ˆ 1. 23.13 De®nition If ÷ is an irreducible character of G. Note that é÷ Tˆ 0 if and only if ÷ is real. We shall see later that this number tells us whether or not ÷ can be realized over R. 23. and the result follows. ÷ where the sum is over all irreducible characters ÷ of G. The de®nition of é÷ gives é÷ ˆ h÷ S À ÷A . W is a linear combination of the irreducible characters of G. W ˆ (é÷)÷.14 Theorem For all x P G. The character table of G is 1 ÷1 ÷2 ÷3 1 1 2 (1 2) 1 À1 0 (1 2 3) 1 1 À1 j .274 23. since for g P G we have y 2 ˆ x D ( g À1 yg)2 ˆ g À1 xgX Therefore by Corollary 15.15 Example Let G ˆ S3 . Representations and characters of groups ˆ (é÷)÷(x) ˆ jf y P G: y 2 ˆ xgj. ÷iX € Therefore. Proof De®ne a function W: G 3 C by W(x) ˆ jf y P G: y 2 ˆ xgj (x P G)X Note that W is a class function on G. 1 G i 1 ˆ ˆ ÷( g 2 ) by Proposition 19X14 jGj gPG ˆ ˆ 1 ˆ ˆ ÷( g 2 ) jGj xPG gPG: g2 ˆx 1 ˆ W(x)÷(x) jGj xPG ˆ hW.4. we show that the indicator of an irreducible character determines whether or not it can be realized over R.14. and hence there is a non-zero CG-homomorphism W from V  V onto the trivial CG-module C. (1 3) and (2 3). Using this. By Proposition 8. g P G)X In this way. no elements square to be (1 2). namely 1. and hence the CG-module V  V has a trivial CG-submodule. and one element. (1 3 2). so (é÷)÷ ˆ ÷1 ‡ ÷2 ‡ ÷3 . (1 2). . Then 1 G is a constituent of ÷ 2 . (2) There exists a non-zero G-invariant symmetric bilinear form on V if and only if é÷ ˆ 1. (3) There exists a non-zero G-invariant skew-symmetric bilinear form on V if and only if é÷ ˆ À1.14 we calculate that é÷ ˆ 1 for each irreducible € character ÷ of G. 23. Back to reality We now relate the indicator function to the previous material on bilinear forms. squares to be (1 2 3). and de®ne a multiplication of C by elements of G by ëg ˆ ë (ë P C.8. C becomes a trivial CG-module. which takes the following values: 1 ÷ 1 ‡ ÷2 ‡ ÷3 4 (1 2) 0 (1 2 3) 1 Sure enough. in accordance with Theorem 23.16 Theorem Let V be an irreducible CG-module with character ÷. (1) Suppose that é÷ Tˆ 0. (1) There exists a non-zero G-invariant bilinear form on V if and only if é÷ Tˆ 0. there is a non-zero CG-homomorphism from V  V onto this trivial CG-submodule. Proof In this proof we regard C as a 1-dimensional vector space over C. and deduce the Frobenius±Schur Count of Involutions.Real representations 275 Using Proposition 19. four elements of G square to be 1. Thus. 1 < j < n) form a basis of V  V. it follows by Proposition 8. . Conversely. the symmetric part of V  V. Let v1 . W is well-de®ned. v j ) (1 < i. Conversely. v n be a basis of V. For g P G. As in (1). v n be a basis of V.8 that there is a non-zero CGhomomorphism W from S(V  V) onto the trivial CG-module C. v) ˆ (u  v ‡ v  u)W (u. De®ne W: V  V 3 C by putting (v i  v j )W ˆ â(v i . we have â(ug. and W is a non-zero CG-homomorphism from S(V  V) onto the trivial CG- . and therefore é÷ Tˆ 0. v P V and g P G. and de®ne W: S(V  V) 3 C by putting (v i  v j ‡ v j  v i )W ˆ â(v i . 1 < j < n) and extending linearly to the whole of V  V. we have ((v i  v j ) g)W ˆ (v i g  v j g)W ˆ â(v i g. Since â is symmetric. v j g) ˆ â(v i . v P V )X Then â is a non-zero bilinear form on V. v j ) (1 < i < n. X X X . De®ne â(u. v j ) as â is G-invariant ˆ (v i  v j )WX Hence W is a non-zero CG-homomorphism from V  V onto the trivial CG-module C. by Proposition 10. vg) ˆ (ug  v g)W ˆ ((u  v) g)W ˆ ((u  v)W) g ˆ (u  v)W ˆ â(u. v) ˆ (u  v)W (u. which is the character of the CG-module S(V  V). If follows that ÷ 2 has the trivial character 1 G as a constituent. . suppose that there is a non-zero G-invariant bilinear form â on V. Then 1 G is a constituent of ÷ S . v P V )X Then â is a non-zero G-invariant symmetric bilinear form on V. (2) Suppose that é÷ ˆ 1. suppose that there exists a non-zero G-invariant symmetric bilinear form â on V. Let v1 . so that v i  v j (1 < i < n. V  V has a trivial CGsubmodule. j < n) and extending linearly. .1. . v)X Thus â is G-invariant. and for u.276 Representations and characters of groups Now de®ne â by â(u. Real representations 277 module C. Hence ÷ S has the trivial character 1 G as a constituent, and so é÷ ˆ 1. (3) The proof of (3) is very similar to that of (2), and is omitted. j We can now relate real representations of G to involutions in G, where by an involution we mean an element of order 2. 23.17 Corollary (The Frobenius±Schur Count of Involutions) For each irreducible character ÷ of G, we have V if ÷ is not real, b 0, b ` é÷ ˆ 1, if ÷ can be realized over R, b b X À1, if ÷ is real, but ÷ cannot be realized over RX Moreover, for all x P G, ˆ (é÷)÷(x) ˆ jf y P G: y 2 ˆ xgj, ÷ where the sum is over all irreducible characters ÷ of G. In particular, ˆ (é÷)÷(1) ˆ 1 ‡ t, ÷ where t is equal to the number of involutions in G. Proof When we de®ned the indicator function, we showed that é÷ Tˆ 0 if and only if ÷ is real. And Theorems 23.11 and 23.16(2) show that ÷ can be realized over R if and only if é÷ ˆ 1. This proves that é÷ is determined as in the statement of the corollary. € The expression for ÷ (é÷)÷(x) was obtained in Theorem 23.14. € Putting x ˆ 1, we see that ÷ (é÷)÷(1) is equal to the number of elements y of G satisfying y 2 ˆ 1. These elements are just the involutions in G, together with the identity, so the number of them is precisely 1 ‡ t. j We conclude with some examples illustrating the use of Corollary 23.17. 23.18 Examples (1) Let ÷ be a linear character. Then é÷ ˆ 1 if ÷ is real, and é÷ ˆ 0 if ÷ is non-real. 278 Representations and characters of groups For an abelian group, the Frobenius±Schur Count of Involutions shows that the number of real irreducible (linear) characters is equal to the number of real conjugacy classes, since in this case g is conjugate to gÀ1 if and only if g2 ˆ 1. This special case of Theorem 23.1 can be proved directly without much dif®culty (see Exercise 23.2). (2) We know that all the irreducible characters of D8 ˆ ka, b: a4 ˆ b2 ˆ 1, bÀ1 ab ˆ aÀ1 l can be realized over R (see Example 23.3(1), and note that all four linear characters are real). Thus é÷ ˆ 1 for all irreducible characters ÷ of D8, and so ˆ (é÷)÷(1) ˆ 1 ‡ 1 ‡ 1 ‡ 1 ‡ 2 ˆ 6X ÷ The ®ve involutions in D8 which are predicted by the Frobenius± Schur Count of Involutions are a2 , b, ab, a2 b and a3 b. (3) In the group Q8 ˆ ka, b: a4 ˆ 1, a2 ˆ b2 , bÀ1 ab ˆ aÀ1 l, there is just one involution, namely a2 . Now é÷ ˆ 1 for each of the four linear characters, and ˆ (é÷)÷(1) ˆ 2 ÷ by the Frobenius±Schur Count of Involutions. Therefore, if ø is the irreducible character of degree 2, then éø ˆ À1. In particular ø cannot be realized over R. (4) The symmetric group S4 has ten elements whose square is 1, namely the identity, the six elements which are conjugate to (1 2), and the three elements which are conjugate to (1 2)(3 4). Since the degrees 1, 1, 2, 3, 3 of the irreducible characters of S4 sum to be 10 (see Section 18.1), we see that all the characters of S4 can be realized over R. The Brauer±Fowler Theorem We now apply Corollary 23.17 to give a proof of a famous theorem of Brauer and Fowler. 23.19 Brauer±Fowler Theorem Let n be a positive integer. Then there exist only ®nitely many nonisomorphic ®nite simple groups containing an involution with centralizer of order n. Real representations 279 Despite its fairly elementary proof, this result is of great historical importance in ®nite group theory. It led Brauer to propose the programme of determining, for each ®nite group C, all the simple groups G possessing an involution u with CG (u)  C. This programme was the start of the modern attempt to classify all ®nite simple groups, which was ®nally completed in the early 1980s. For further information about this, see the book by D. Gorenstein listed in the Bibliography. In Exercise 10 at the end of the chapter you are asked to carry out Brauer's programme in the case where C  C2 . This should not trouble you too much. In Chapter 30, Theorem 30.8, you will ®nd a much more sophisticated case, in which C  D8. For proof of Theorem 23.19 we require two preliminary lemmas. 23.20 Lemma € € If a1 , . . . , a n are real numbers, then a2 > … a i †2 an. i Proof This follows from the Cauchy±Schwarz inequality kvk kwk > jvXwj, taking v ˆ (a1 , . . . , a n ) and w ˆ (1, . . . , 1). j 23.21 Lemma Let G be a group of even order m, and let t be the number of involutions in G (so t . 0 by Exercise 8 of Chapter 1). Write a ˆ (m À 1)at. Then G contains a non-identity element x such that jG : C G (x)j < a2 . Proof By Corollary 23.17, we have ˆ t< ÷(1) ÷ where the sum is over all non-trivial irreducible characters ÷ of G. Writing k for the number of irreducible characters of G, we deduce using Lemma 23.20 and Theorem 11.12 that ˆ ˆ t2 < ( ÷(1))2 < (k À 1) ÷(1)2 ˆ (k À 1)(m À 1), ÷ and hence m À 1 < (k À 1)(m À 1)2 at 2 ˆ (k À 1)a2 . Now k À 1 is the number of non-identity conjugacy classes of G. If every nonidentity conjugacy class has size more then a2 , then (k À 1)a2 . 280 Representations and characters of groups jGj À 1 ˆ m À 1, a contradiction. Therefore some non-identity class x G has size at most a2 . Then jG : C G (x)j < a2 , giving the result. j Proof of Theorem 23.19 Suppose G is simple and contains an involution u such that jC G (u)j ˆ n. Let jGj ˆ m, and let t be the number of involutions in G. By Proposition 12.6, every element of the conjugacy class u G is an involution, and hence t > ju G j ˆ jG : C G (u)j ˆ manX Therefore (m À 1)at , n, and so by Lemma 23.21, there is a nonidentity element x P G such that jG : C G (x)j , n2 . Let H ˆ C G (x). If H ˆ G then x lies in Z(G), the centre of G, which is a normal subgroup of G. Since G is simple it follows that G ˆ Z(G), so G is abelian and therefore G  C2 . Now suppose that H Tˆ G. Write r ˆ jG : Hj, so r , n2 . By Exercise 9 at the end of the chapter, there is a non-trivial homomorphism è from G to the symmetric group S r . As G is simple, the normal subgroup Ker è ˆ f1g. Thus G is isomorphic to a subgroup of S r , hence of S n2 . In particular, given n, there are only ®nitely many possibilities for G. j Summary of Chapter 23 1. The number of real irreducible characters of G is equal to the number of real conjugacy classes of G. Let é be the indicator function, and let ÷ be an irreducible character of G. V if ÷ is non-real, b 0, b b b 1, ` if there exists an RG-module U with character ÷, 2X é÷ ˆ with character ÷ b b b b X À1, if ÷ is real, but there does not exist an RG-moduleX V if 1 G is not a constituent of ÷ S or ÷ A , b 0, ` 3X é÷ ˆ 1, if 1 G is a constituent of ÷ S , b X À1, if 1 G is a constituent of ÷ A X € 2 4. ÷ (é÷)÷(1) ˆ |{ g P G: g ˆ 1}|. Real representations Exercises for Chapter 23 281 1. Prove that if G is a group of odd order then no non-identity element of G is real. 2. Let G be a ®nite abelian group. Use the description of the irreducible characters of G, given in Theorem 9.8, to prove directly that the number of real irreducible characters of G is equal to the number of elements g in G for which g2 ˆ 1. 3. Let G ˆ D2 n and consult Section 18.3 for the character table of G. How many elements g of G satisfy g2 ˆ 1? Deduce that é÷ ˆ 1 for all irreducible characters ÷ of G. 4. Let r be an irreducible representation of degree 2 of a group G, and let ÷ be the character of r. Prove that ÷ A ( g) ˆ det ( gr) for all g P G. Deduce that é÷ ˆ À1 if and only if det ( gr) ˆ 1 for all g P G. 5. Let G ˆ T 4 n ˆ ha, b: a2 n ˆ 1, a n ˆ b2 , bÀ1 ab ˆ aÀ1 i, as in Exercise 17.6. Let V be a 2-dimensional vector space over C with basis v1 , v2 and let å be a (2n)th root of unity in C with å Tˆ Æ1. Exercise 17.6 shows that V becomes an irreducible CG-module if we de®ne v1 a ˆ åv1 , v2 a ˆ å À1 v2 , v1 b ˆ v2 , v2 b ˆ å n v1 X Let ÷ be the character of this CG-module V. (a) Note that å n ˆ Æ1. Use Exercise 4 to show that é÷ ˆ 1 if å n ˆ 1 and é÷ ˆ À1 if å n ˆ À1. (b) Let â be the bilinear form on V for which â(v1 , v1 ) ˆ â(v2 , v2 ) ˆ 0, â(v1 , v2 ) ˆ 1, â(v2 , v1 ) ˆ å n X Prove that the bilinear form â is G-invariant, and use Theorem 23.16 to provide a second proof that é÷ ˆ 1 if å n ˆ 1 and é÷ ˆ À1 if å n ˆ À1. (c) Prove that a n is the only element of order 2 in T4 n. (d) Use the character table of G, which appears in the solution to Exercise 18.3, to ®nd é÷ for each irreducible character ÷ of G. Check that 282 Representations and characters of groups ˆ (é÷)÷(1) ˆ 2, ÷ in agreement with the Frobenius±Schur Count of Involutions. 6. Prove that if ÷ is an irreducible character of a group G, and é÷ ˆ À1, then ÷(1) is even. (Hint: the solution uses a well known result about skew-symmetric bilinear forms.) 7. Suppose that V is a vector space over R and that â1 and â are symmetric bilinear forms on V. Assume that â1 (w, w) . 0 for all non-zero w in V. Prove that there is a basis e1 , X X X , e n of V such that â1 (ei , ei ) ˆ 1 for all i, and â1 (ei , ej ) ˆ â(ei , ej ) ˆ 0 for all i Tˆ jX 8. Schur's Lemma is crucial for the development of the theory of CG-modules. This exercise indicates the extent to which results like Schur's Lemma hold for RG-modules. Let V and W be irreducible RG-modules. (a) Prove that if W: V 3 W is an RG-homomorphism then either W is an RG-isomorphism or vW ˆ 0 for all v P V. (b) Prove that if W: V 3 V is an RG-isomorphism and V remains irreducible as a CG-module, then W ˆ ë1 V for some real number ë. . (c) Give an example of a group G, an irreducible RG-module V and an RG-homomorphism W: V 3 V which is not a multiple of 1 V . 9. Let G be a group with a subgroup H of index n. Let Ù be the set of n right cosets Hx of H in G. For g P G, de®ne a function r g : Ù 3 Ù by ( Hx)r g ˆ Hxg for all x P G. Prove that r g is a permutation of Ù, and that the function r : g 3 r g is a homomorphism from G to the symmetric group on Ù. „ Show that the kernel of r is xPG x À1 Hx. Deduce that if a group G has a subgroup H of index n, then there is a homomorphism G 3 S n with kernel contained in H. 10. Suppose that G is a ®nite group containing and involution t with CG (t)  C2 . Prove that |G : G9| ˆ 2. Deduce that if G is simple, then G  C2 . 24 Summary of properties of character tables In this short chapter we present no new results, but instead we gather together from previous chapters various properties which are helpful when we try to ®nd the character table of a particular group. In the next four chapters we shall calculate several character tables in detail. Usually we begin by working out the conjugacy classes and centralizer orders of our given ®nite group G. The size of the character table is determined by the number k of conjugacy classes of G; the character table is then a k 3 k matrix, with columns indexed by the conjugacy classes of G (the ®rst column corresponding to the conjugacy class {1}), and with rows indexed by the irreducible characters of G. When doing calculations, we commonly come across a new character ÷, which may or may not be irreducible. We can then calculate h÷, ÷i, which is given by h÷, ÷i ˆ 1 ˆ ÷( g)÷( g)X jGj gPG The character ÷ is irreducible if and only if h÷, ÷i ˆ 1 (see Theorem 14.20). If ÷ is reducible then we calculate h÷, ÷ i i for each of the irreducible characters ÷ i which we already know, and then ÷À ˆ h÷, ÷ i i÷ i i will also be a character. We can thus determine whether ÷ is a linear combination of the irreducible characters we already know; and if it is not, then we can obtain from ÷ a linear combination of irreducible characters, all of which are new. We have developed a number of methods for producing characters ÷ 283 284 Representations and characters of groups on which to perform such calculations. For example, every subgroup of Sn has a permutation character (see (13.22)); the product of two characters is again a character (Proposition 19.6); given a character ø we can form the symmetric and antisymmetric parts of its square, ø S and ø A (see Proposition 19.14); and if H is a subgroup of G then we can restrict characters of G to H, and induce characters of H to G. These and other properties of characters are summarized in the following list. Properties of characters Assume that ÷1 , . . . , ÷ k are the irreducible characters of G. (1) (Example 13.8(3)) There is a (trivial) character ÷ of G which is given by ÷( g) ˆ 1 for all g P GX (2) (Theorem 17.11) The group G has precisely jGaG9j linear characters. These are the characters ÷ given by ÷( g) ˆ ø( gG9) ( g P G) as ø varies over the irreducible (linear) characters of GaG9. (3) (Theorem 17.3) As a generalization of (2), if N v G and ø is an irreducible character of GaN , then we get an irreducible character ÷ of G which is given by ÷( g) ˆ ø( gN ) ( g P G) (÷ is the lift of ø). This method gives precisely those irreducible characters of G which have N contained in their kernel. (4) (Theorem 19.18) If G ˆ G1 3 G2 then all the irreducible characters ÷ of G are given by ÷( g1 , g 2 ) ˆ ö1 ( g 1 )ö2 ( g 2 ) ( g 1 P G1 , g 2 P G2 ), as ö i varies over the irreducible characters of Gi (i ˆ 1, 2). (5) (Proposition 13.24) If G is a subgroup of Sn , then the function í: G 3 C de®ned by í( g) ˆ jfix ( g)j À 1 is a character of G. ( g P G) Summary of properties of character tables 285 (6) (Theorems 11.12 and 22.11) The entries ÷ i (1) (1 < i < k) in the ®rst column of the character table of G are positive integers, and satisfy k ˆ iˆ1 ÷ i (1)2 ˆ jGjX Moreover, each integer ÷ i (1) divides |G|. (7) (Row orthogonality relations, Theorem 16.4(1)) For all i, j, we have h÷ i , ÷ j i ˆ ä ij . (8) (Column orthogonality relations, Theorem 16.4(2)) For all g, h P G, we have @ k ˆ jCG ( g)j, if g and h are conjugate, ÷ i ( g)÷ i (h) ˆ 0, otherwiseX iˆ1 (9) (Exercise 13.5) If ÷ is an irreducible character of G and z P Z(G), then there exists a root of unity å such that for all g P G, ÷(zg) ˆ å÷( g)X (10) (Proposition 13.9(2)) If g is an element of order n in G, and ÷ is a character of G, then ÷( g) is a sum of nth roots of unity. Moreover, |÷( g)| < ÷(1). (11) (Proposition 13.9(3, 4)) If g P G and ÷ is a character of G, then ÷( g À1 ) ˆ ÷( g)X In particular, if g is conjugate to gÀ1 then ÷( g) is real for all characters ÷ of G. (12) (Corollary 15.6) If g P G and g is not conjugate to gÀ1 , then ÷( g) is non-real for some character ÷ of G. (13) (Theorem 22.16) Let g P G. If g is conjugate to gi for all positive integers i which are coprime to the order of g, then ÷( g) is an integer for all characters ÷ of G. (14) (Corollary 22.26) Suppose that p is a prime number and that y is the p9-part of the element g of G. If ÷ is a character of G such that ÷( g) and ÷( y) are both integers, then ÷( g)  ÷( y) mod pX Two important normal subgroups are G9 and Z(G). if the order of g is a power of p. then so is ÷.1) The number of real irreducible characters of G is equal to the number of real conjugacy classes of G.13. The derived subgroup G9 consists of those elements g in G which satisfy ÷( g) ˆ 1 for all linear characters ÷ of G. where (ø 5 H)(h) ˆ ø(h) (h P H)X We have seen that the character table of a group G gives grouptheoretic information about G. with values given by Proposition 21.6) If ÷ and ø are characters of G. where for all g P G. then ÷( g)  ÷(1) mod pX (15) (Proposition 13. (21) (Chapter 20) If H is a subgroup of G and ø is a character of G. indeed. then so is the product ÷ø. then ÷ë is an irreducible character of G. these can be determined in the following ways.286 Representations and characters of groups In particular. where ÷ë( g) ˆ ÷( g)ë( g) ( g P G)X (18) (Proposition 19. For example. Proposition 21.6). 2 ÷ A ( g) ˆ 1(÷ 2 ( g) À ÷( g 2 ))X 2 (20) (De®nition 21.14) If ÷ is a character of G. We can see from the character table whether or not G is simple (Proposition 17. then ø 5 H is a character of H. we can ®nd all the normal subgroups of G (Proposition 17.23.14) If ÷ is an irreducible character of G and ë is a linear character of G. where ÷( g) ˆ ÷( g) ( g P G)X (16) (Theorem 23.15) If ÷ is an irreducible character of G. (17) (Proposition 17. the ®rst column determines |G| and jGaG9j (by (6) and (2)). where ÷ø( g) ˆ ÷( g)ø( g) ( g P G)X (19) (Proposition 19. The centre Z(G) can be found by noting which elements g of G satisfy .23) If H is a subgroup of G and ø is a character of H. then ø 4 G is a character of G.5). ÷ S ( g) ˆ 1(÷ 2 ( g) ‡ ÷( g 2 )). then so are ÷ S and ÷ A . the converse is false: in Exercise 17. In Chapter 30 we shall see some more impressive results about subgroups of G.1 we gave examples of non-isomorphic groups. the sum being over all irreducible characters ÷ of G. D8 and Q8 . . however.Summary of properties of character tables 287 € ÷( g)÷( g) ˆ |G|. which can be deduced from the character table. it is of course true that isomorphic groups have the same character table. As a ®nal remark. with the same character table. with addition and multiplication modulo p. is a ®eld.25 Characters of groups of order pq By the end of the next chapter. Z p is an abelian group under addition. but not at all obvious. In particular. and in this chapter we shall describe a class of Frobenius groups and ®nd the character tables of the groups in this class. generated by 1. that is. p will denote a prime number. We shall not provide a proof of Theorem 25. p À 1g.1. A number of these groups are so-called Frobenius groups. 1. and n r T 1 mod p for 0 . and Zà ˆ Z p À f0g is an abelian group p under multiplication. that is. Clearly Z p is a cyclic group under addition. we shall have determined the character tables of all groups of order less than 32. Primitive roots modulo p Recall that the set Z p ˆ f0. It is also true. there exists an integer n p such that n pÀ1  1 mod p. X X X . p À 1X An integer n of order p À 1 in Zà is called a primitive root modulo p p. that Zà is cyclic: p 25.1 Theorem The multiplicative group Zà is cyclic. this will give the character tables of all groups whose order is the product of two prime numbers. but for a good 288 . r . Throughout the chapter. 11 and 13. 5. Then   1 u À1 B AB ˆ ˆ Au . Fraleigh listed in the Bibliography.1). we refer you to Theorem 45. G is a group of order p( p À 1) (see Exercise 25. As an immediate consequence of Theorem 25.3 of the book by J.3 Proposition If q| p À 1 then there is an integer u such that u has order q modulo p ± that is.2 Example The number 2 is a primitive root modulo 3. Bˆ . B. Now let q| p À 1. 0 < j < q À 1. but not modulo 7. qX Frobenius groups of order pq. 25. where q| p 2 1 25. and 3 is a primitive root modulo 7. Bi. 0 1 and so we have the relations (25X5) Ap ˆ Bq ˆ I. such that u q  1 mod p.Characters of groups of order pq 289 account.4 Example De®ne Gˆ & 1 0  ' y Ã. and u r T 1 mod p for 0 .1 we have 25. r . These pq elements are dis- . BÀ1 AB ˆ Au X Using these relations. the subgroup of G generated by A and B. and let u be an element of order q in the multiplicative group Zà . De®ne p     1 1 1 0 Aˆ . y P Z X : xPZp p x Under matrix multiplication. we see that every element of F is of the form A i B j with 0 < i < p À 1. 0 1 0 u and let F ˆ hA. Suppose that H ˆ kal and Ga H ˆ h Hbi.q belong to a wider class of groups known as Frobenius groups.6 De®nition If p is a prime and q| p À 1. F p. Thus the order of u in the group Zà divides q. since they have prime order.q does not depend on which integer u of order q we choose (see Exercise 25.5) determine all products in F. Moreover the relations (25. bÀ1 ab ˆ au i. where u is an element of order q in Zà . Since bq P H but b does not have order pq (as G is non-abelian).7 Proposition Suppose that G is a group of order pq.3 that q divides p À 1 and G has a normal subgroup H of order p. so bÀ1 ab ˆ au for some integer u. it follows that b has order q. B: Ap ˆ Bq ˆ I. We shall not give the general de®nition of these here. b: a p ˆ b q ˆ 1. Further. S.q . B. then we write F p. The next result classi®es all groups whose order is the product of two distinct prime numbers. 25. or q divides p À 1 and G  F p. Fraleigh listed in the Bibliography). (Alternatively. up to isomorphism. these facts follow readily from Sylow's Theorems (see Chapter 18 of the book by J. the interested reader can ®nd more information in the book by D. p It is not hard to show that. The groups F p. Passman listed in the Bibliography. q.q for the group of order pq with presentation F p. so we have the presentation F ˆ hA. then G is generated by a and b. Proof Assume that G is non-abelian. Now H v G.q ˆ ha.3). Then either G is abelian.) Both H and Ga H are cyclic. BÀ1 AB ˆ Au iX 25. a ˆ bÀq abq ˆ a u q and so u q  1 mod p. where p and q are prime numbers with p . so jFj ˆ pq. p . It follows from Exercise 22. as we shall only be dealing with F p.q .290 Representations and characters of groups tinct. The character table of F p. Thus let G ˆ Fp. and has the form stated in the proposition. bÀ1 ab ˆ au . also the size of this conjugacy class is equal to |G: CG (av i )|. Hence (av i ) G has size q. j 25. and in Example 21. and choose coset representatives v1 .9 Proposition The conjugacy classes of G ˆ F p. and the groups of order 21 are C3 3 C7 and F7.8 Example By Proposition 25. isomorphic to C3 3 C5 ).3. already found the character tables of certain of the groups F p. p 25. (av i ) G ˆ fav i s : s P Sg (1 < i < r). Therefore the order of u is q. . Write r ˆ ( p À 1)aq.q . and G would be abelian.Characters of groups of order pq 291 If the order of u were 1 then we would have bÀ1 ab ˆ a. order of u in Zà is qX p Hence G  F p.7. Therefore the conjugacy class of av i has size at least q.q We have.3 . and u has order q modulo p. (bn ) G ˆ fam bn : 0 < m < p À 1g (1 < n < q À 1)X Proof The equation bÀ j av b j ˆ avu j shows that av is conjugate to avs for all s P S. bÀ1 ab ˆ au i where p is prime. Let S be the subgroup of Zà consisting of the powers of u. and since kal < CG (av i ). v r for S in Zà . every group of order 15 is abelian (indeed. in fact. this size is at most q. q| p À 1 (q not necessarily prime). b: ap ˆ bq ˆ 1.q in general.q : the dihedral group of order 2 p is the case where q ˆ 2.25 we dealt with F7. We now construct the character table of F p. Thus p jSj ˆ q.q ˆ ha. We have now established that a p ˆ b q ˆ 1.q are f1g. X X X . Hence (bn ) G ˆ fam bn : 0 < m < p À 1g and the proof is complete. By the Frobenius Reciprocity Theorem 21. ö j i G ˆ sPS øv j s ‡ ÷.23.292 Representations and characters of groups Since CG (bn ) contains kbl. . ö j i G X Hence ö j 5 hai ˆ hö j . so we seek q ‡ r irreducible characters. using Proposition 21. we have |CG (bn )| ˆ q. øv j s 4 Gi G ˆ hö j . øv j s ihai ˆ hö j . On the other hand. for all s P S. Let å ˆ e2ðia p . observe that the derived subgroup G9 ˆ kal.9. and ˆ (øv 4 G)(ax ) ˆ å vsx (0 < x < p À 1)X sPS Note that øv 4 G has degree q. and kbl has index p in G. and so the conjugacy class of bn has size p. every conjugate of bn has the form am bn for some m. G has precisely q linear characters.11. as Gahai is abelian. First. G has q ‡ r conjugacy classes. hö j 5 hai. denote by øv the character of kal which p is given by øv (ax ) ˆ å vx (0 < x < p À 1)X We calculate the values of the induced character øv 4 G. We obtain (øv 4 G)(ax by ) ˆ 0 if 1 < y < q À 1. so GaG9 has order q and therefore by Theorem 17. let p ö j ˆ øv j 4 GX We now prove that each ö j is irreducible. it follows that for n T 0 mod q. where ÷ n (a x b y ) ˆ e2ði nyaq (0 < x < p À 1. For v P Zà . and øv 4 G ˆ øvs 4 G if s P SX For each coset representative v j (1 < j < r) of S in Zà . j By Proposition 25. 0 < y < q À 1)X We shall show that G has r irreducible characters of degree q. These are given by ÷ n (0 < n < q À 1).16. 10 Theorem Let p be a prime number. We conclude by illustrating Theorem 25. 25. . and also that ˆ ö j 5 hai ˆ hö j . Taking degrees. 0 < y < q À 1g has q ‡ r irreducible characters. we deduce that kö j .Characters of groups of order pq 293 where ÷ is either 0 or a character of kal. . ö j of G (0 < n < q À 1. ö j i G X Since ö j (1) ˆ q ˆ jSj. . so we have the complete character table of G. Of these. ö r 5 kal are distinct. bÀ1 ab ˆ au i ˆ fax by : 0 < x < p À 1.11 Example Let G ˆ F p. ö j l G ˆ 1. X X X . .q ˆ ha. bÀ1 ab ˆ au i . b: a p ˆ b pÀ1 ˆ 1. q have degree 1 and are given by ÷ n (ax by ) ˆ e2ði nyaq (0 < n < q À 1) and r have degree q and are given by ö j (ax by ) ˆ 0 if 1 < y < q À 1. v r S are the cosets in Zà of the p subgroup S generated by u. where v1 S. 25. and hence ö1 5 kal. ö r are distinct. . ˆ ö j (ax ) ˆ e2ðiv j sxa p . This proves that ö j is irreducible. . 1 < j < r). q| p À 1 and r ˆ ( p À 1)aq. the characters øv (v P Zà ) are linearly indepenp dent. We have now found q ‡ r distinct irreducible characters ÷ n . We summarize in the following theorem. it follows that ö j (1) > jSjhö j . . .10 in some examples.23. b: a p ˆ bq ˆ 1. ö j i G øv j s X sPS By Theorem 14. Then the group F p. Consequently the irreducible characters ö1 . sPS for 1 < j < r. pÀ1 ˆ ha. and let á ˆ å ‡ å 5 ‡ å 8 ‡ å 12 .12 Example Let a. You may like . ⠈ å 2 ‡ å 3 ‡ å 10 ‡ å 11 . In Example 21. bÀ1 ab ˆ a5 iX Write å ˆ e2ðia13 . bl. Then G has p À 1 linear characters.4 gi |CG ( gi )| ÷0 ÷1 ÷2 ÷3 ö 1 20 1 1 1 1 4 a 5 1 1 1 1 À1 b 4 1 i À1 Ài 0 b2 4 1 À1 1 À1 0 b3 4 1 Ài À1 i 0 x if 1 < y < p À 2.13 Example We consider the case p ˆ 13. and one irreducible character ö of degree p À 1. b ˆ (2 3 5 4)X Check that a5 ˆ b4 ˆ 1. then G  F5.10. b: a13 ˆ b4 ˆ 1. the character table of F13.294 Representations and characters of groups where u is a primitive root modulo p. and so by the previous example the character table of G is as shown. if 1 < x < p À 1X 25.4 . 㠈 å4 ‡ å6 ‡ å7 ‡ å9 X By Theorem 25. with values given by ö(ax by ) ˆ 0 ö(a ) ˆ À1 25.3.4 ˆ ha. b P S5 be the permutations a ˆ (1 2 3 4 5). Here F13. q ˆ 4.25 we found the character table of F7.4 is as shown opposite. bÀ1 ab ˆ a2 X Hence if G ˆ ka. Character table of F5. Let u and v be two integers which are of order q modulo p. Let p and q be positive integers. is a group of order p( p À 1). p 0 x under matrix multiplication. 2.q are described in Theorem 25. Prove that  & ' 1 y : x P Zà . Character table of F13. 3.10. y P Z p .5 of order 55. and de®ne . If G has order pq.Characters of groups of order pq 295 to check that this agrees with the description of the character table provided by Theorem 25. 2. Exercises for Chapter 25 1. Let p be a prime number.4 gi |CG ( gi )| ÷0 ÷1 ÷2 ÷3 ö1 ö2 ö3 1 52 1 1 1 1 4 4 4 a 13 1 1 1 1 á â ã a2 13 1 1 1 1 â ã á a4 13 1 1 1 1 ã á â b 4 1 i À1 Ài 0 0 0 b2 4 1 À1 1 À1 0 0 0 b3 4 1 Ài À1 i 0 0 0 Summary of Chapter 25 1.q. Let u be an element of order q in Zà . b: ap ˆ bq ˆ 1. Then p Fp.q ˆ ha. Write down the character table of the non-abelian group F11. Let p and q be prime numbers with p . bÀ1 ab ˆ au iX The irreducible characters of Fp. q. Suppose that p is prime and q divides p À 1.10. with p prime and q| p À 1. then either G is abelian or G  F p. and ä ˆ À1 if p  À1 mod 4. Suppose that p is a prime number. By inducing linear characters of this subgroup. (c) Using the orthogonality relations.4.) 7. with p Tˆ 2. sPQ where Q is the set of quadratic residues modulo p (that is. G2 ˆ ha.296 Representations and characters of groups G1 ˆ ha. Let q ˆ ( p À 1)a2 and let G ˆ Fp. Show that the group E of Exercise 5 has the properties that Z(E) is cyclic. 6. Q ˆ f12 . where ä ˆ 1 if p  1 mod 4. bl is a normal subgroup of E which is isomorphic to C3 3 C3 . where u is an element of order q modulo p. cÀ1 bc ˆ bÀ1 i. Note that ka. bÀ1 ab ˆ au i. (a) Show that there exists an integer m such that u m  À1 mod p if and only if p  1 mod 4. bÀ1 ab ˆ av iX Prove that G1  G2 . (d) Deduce that if å ˆ e2ðia p then ˆ p å s ˆ (À1 Æ (ä p)). ab ˆ ba. b: ap ˆ bq ˆ 1. b: ap ˆ bq ˆ 1. cÀ1 ac ˆ aÀ1 .) 4. (Thus. (b) Deduce that a is conjugate to aÀ1 if and only if p  1 mod 4. E provides a counterexample to the converse of Proposition 9. (a) Find a group whose irreducible character degrees are . ö2 of G of degree q have values p 1 (ä p)) 2(À1 Æ on the element a. c: a3 ˆ b3 ˆ c2 ˆ 1. 5.6. X X X . (( p À 1)a2)2 g). as in Exercise 5. b.q in 25. but E has no faithful irreducible representation. Let E be the group of order 18 which is given by E ˆ ha. b: ap ˆ bq ˆ 1. show that the two irreducible characters ö1 . (This justi®es the comment which follows the de®nition of Fp. obtain the character table of E.q ˆ ha. bÀ1 ab ˆ au i.16. 22 . 6. 3. 3. 3. 2. 1. 3. 2. 297 . 3. 1.Characters of groups of order pq 1. 3. 2. 3. 6. 1. 1. 3. 1. 3. 1. 3. 1. 1. Let G be the group of order 54 which is given by G ˆ ha. 3. 3. 6. 3X (c) Find a group whose irreducible character degrees are 1. 3. 1. 3X (b) Find a group whose irreducible character degrees are 1. 3. b: a9 ˆ b6 ˆ 1. 1. bÀ1 ab ˆ a2 iX Find the character table of G. 3. 3. 1. 3. 6X 8. 1. 3. 15). Z(G) Tˆ {1}. We shall show how to obtain the character tables of all groups of order pn for n < 4. all of which have size a power of p. have an abelian subgroup of index p. we show that all groups of order pn with 1 < n < 4 do. H is a union of conjugacy classes of G. Proof (1) Since H v G. In the ®rst lemma we collect several well known properties of pgroups. and before explaining the method. with references. and H ’ Z(G) consists of those conjugacy classes in H which have size 1. that we have found the character tables of all groups of order less than 32. (3) If n < 2 then G is abelian.26 Characters of some p-groups Throughout this chapter. indeed. p will be a prime number. The method consists of examining the characters of those p-groups which contain an abelian subgroup of index p. Therefore 298 . At the end of the chapter we point out. 26. Recall that Z(G) denotes the centre of G (see De®nition 9. (1) If {1} Tˆ H v G then H ’ Z(G) Tˆ {1}. (2) If K < Z(G) and GaK is cyclic. We later give explicitly the irreducible characters of all groups of order p3 and of all groups of order 16. In particular.1 Lemma Let G be a group of order pn with n > 1. then G is abelian. Elementary properties of p-groups A p-group is a group whose order is a power of the prime number p. Moreover. Then G contains an abelian subgroup of index p. Let x1 . k2 P Z(G). we deduce that Z( H) ˆ H. G has an element x whose conjugacy class x G is of size p. (2) Suppose that GaK is cyclic.1(2). Then x1 ˆ g i k 1 . and so Z( H) > p2 . j and some k1 . Then by Theorem 12. Then there exists a normal subgroup K of G such that K < H ’ G9 ’ Z(G) and jKj ˆ pX . As K < Z( H) and. Ha Z( H) is not of order p. Then we can ®nd a subgroup H of G such that K < H and j Hj ˆ p nÀ1 . x2 P G.8. Proof The result is immediate if n ˆ 1. jGa Z(G)j < p nÀ1 . the only possibility is that |G| ˆ p4 and | Z(G)| ˆ p.7. it follows that x1 x2 ˆ x2 x1 . Therefore G is abelian. j For our ®nal result on the structure of p-groups. 26. Hence again Z( H) ˆ H by Lemma 26. k2 P K. Therefore H is an abelian subgroup of index p in G. Since Z(G) Tˆ f1g by Lemma 26. Let H ˆ CG (x).7). so suppose that 2 < n < 4. Z(G) and kxl are distinct nonidentity subgroups of Z( H).3 Lemma Let G be a non-abelian p-group which contains an abelian subgroup H of index p. recall that the derived subgroup of G is denoted by G9 (see De®nition 17. Since k1 . 26.1(1). j Hj ˆ jGjajx G j ˆ p3 .Characters of some p-groups j Hj ˆ j H ’ Z(G)j ‡ (a multiple of p)X 299 As | H| is a multiple of p and | H ’ Z(G)| Tˆ 0. Now assume that Z(G) has no subgroup of order p nÀ2 .1(2). x2 ˆ g j k 2 for some integers i. and H is an abelian subgroup of index p. (3) By (1). Then by Exercise 12. by Lemma 26.2 Lemma Let G be a group of order pn with 1 < n < 4. we deduce that H ’ Z(G) Tˆ {1}. Assume that Z(G) contains a subgroup K of order p nÀ2 . Hence if n < 2 then Ga Z(G) is j cyclic and so G is abelian by (2). generated by gK. 4 Theorem Assume that G is a non-abelian p-group which contains an abelian subgroup H of index p. The sum of the squares of the degrees of the irreducible characters obtained in this way is jGaKj ˆ p nÀ1 . and hence K < G9 < Ker ÷.3. or (2) ø 4 G. In the latter case. the degree of every irreducible character of G is a power of p). Since G is non-abelian and |G: H| ˆ p. Let K be a normal subgroup of G as in Lemma 26.8 that all the p nÀ1 irreducible characters of the abelian group H are linear. This establishes (26X5) if ÷(1) ˆ p and K T< Ker ÷. and hence G9 ’ Z(G) Tˆ {1} by Lemma 26. Let Ö denote the set of linear . then either ÷ is irreducible or ÷ is a sum of linear characters (since by Theorem 22.12. again by Theorem 11. Now K < Z(G) implies that K v G and that KH is an abelian subgroup of G (where KH ˆ {kh: k P K. we have {1} Tˆ G9 v G. for some linear character ø of H which satis®es K T< Ker ø. and therefore K < H. (à ) we shall then have obtained all the irreducible characters of G.300 Representations and characters of groups Proof Since G is non-abelian.11.1(1). each of degree p.3. By Theorem 17. as G9 is in the kernel of every linear character. we have G9 < Ker ÷. Since p nÀ1 ‡ ( p nÀ2 À p nÀ3 ) p2 ˆ pn ˆ jGj. the irreducible characters of GaK lift to give precisely those irreducible characters of G which have K in their kernel. we have KH ˆ H. h P H}).12. by Theorem 11. j Characters of p-groups with an abelian subgroup of index p In view of Lemma 26. Proof Let |G| ˆ pn . We shall construct p nÀ2 À p nÀ3 further irreducible characters of G. 26. the next theorem provides us with all the irreducible characters of non-abelian groups of order p3 or p4 . First note that if ÷ is a character of G of degree p. Then every irreducible character of G is given by either (1) the lift of an irreducible character of GaK.2. Let K be a subgroup of order p in G9 ’ Z(G). then ÷ is irreducibleX We know by Theorem 9. Now let G be a non-abelian group of order p3 . Suppose now that ø1 is a linear character of H such that ø 4 G ˆ ø1 4 G. Write Z ˆ Z(G). Then by the Frobenius Reciprocity Theorem 21.1. the abelian groups of order p3 are C p3 . Since the linear characters of H which do have K in their kernel are precisely the lifts of linear characters of HaK. and the proof is complete. Hence Ga Z  Cp 3 Cp and Z ˆ kzl  Cp . Choose aZ. 1 ˆ hø 4 G. Therefore fø 4 G: ø P Ög consists precisely of the p nÀ2 À p nÀ3 irreducible characters we seek. since K < Z(G). Groups of order p3 By Theorem 9.16. bZ such that Ga Z ˆ haZ. ø1 i H X Since (ø 4 G) 5 H has degree p. j We now use Theorem 26. By Proposition 21.Characters of some p-groups 301 characters of H which do not have K in their kernel.8.5).23. Therefore by (26. Z Tˆ {1} and Ga Z is non-cyclic. Then . C p2 3 Cp and Cp 3 Cp 3 Cp X The character tables of these groups are given by Theorem 9.4 to give an explicit construction of the irreducible characters of the non-abelian groups of order p3 . bZi. ø 4 G is an irreducible character of G.6. ø1 4 Gi G ˆ h(ø 4 G) 5 H. By Lemma 26. As we saw in (à ).4 further by constructing the character tables of all the non-abelian groups of order 16. G has at most p nÀ2 À p nÀ3 such characters. It follows that fø 4 G: ø P Ög gives at least jÖja p ˆ ( p nÀ1 À p nÀ2 )a p distinct irreducible characters of G of degree p which do not have K in their kernel. (ø 4 G)(k) ˆ pø(k) for all k P KX Thus ø 4 G has degree p and does not have K in its kernel. this implies that there are at most p elements ø1 of Ö such that ø1 4 G ˆ ø 4 G. We shall then illustrate Theorem 26. we have jÖj ˆ p nÀ1 À p nÀ2 X Let ø P Ö. v öu where for all r.v (ar bs z t ) ˆ å ru‡sv .8. 0 < v < p À 1). t < p À 1} be a non-abelian group of order p3 . ÷ u. zl. If ar is conjugate to an element g of G. @ ut if r ˆ s ˆ 0. Since ar P Z.v (ar bs Z) ˆ å ru‡sv X The lift to G of ø u.v (0 < u. and hence (ar ) G ˆ far z t : 0 < t < p À 1gX . ö u (ar bs z t ) ˆ 0.v which appears in the statement of the theorem. so ar Z ˆ gZ. s. 26. every element of G is of the form ar bs z t for some r. For 1 < u < p À 1. t. otherwiseX (0 < u < p À 1. choose a character ø u of H which satis®es ø u (z t ) ˆ å ut (0 < t < p À 1)X We shall calculate ø u 4 G. på . the irreducible characters of Ga Z are ø u. where ø u. t < p À 1. 0 < s < p À 1g and in particular. and (1 < u < p À 1). t with 0 < r. v < p À 1).v is the linear character ÷ u. s.302 Representations and characters of groups Ga Z ˆ far bs Z: 0 < r < p À 1. Then the irreducible characters of G are ÷ u. Proof By Theorem 9. Let H ˆ ka. so that H is an abelian subgroup of order p2 . and therefore g ˆ ar z t for some t. Let r be an integer with 1 < r < p À 1. a the conjugacy class (ar ) G does not have size 1. s. then ar Z is conjugate to gZ in the abelian group Ga Z. as above. Write å ˆ e2ðia p . s.6 Theorem Let G ˆ {ar bs z t : 0 < r. 12 ‡ ( p À 1) . and the sum of the squares of their degrees is p2 . v < p À 1) and ö u (1 < u < p À 1) are all distinct. p2 ˆ jGjX Hence we have found all the irreducible characters of G.23.6 is a special case of the proof of Theorem 26.Characters of some p-groups Then by Proposition 21. they are D8 and Q8. there are precisely two non-abelian groups of order p3 . j Notice that the calculation in the proof of Theorem 26. they are . (ø u 4 G)(ar z t ) ˆ ø u (ar ) ‡ ø u (ar z) ‡ X X X ‡ ø u (ar z pÀ1 ) ˆ ø u (ar ) ˆ ø u (ar ) ˆ 0X Also. up to isomorphism. (ø u 4 G)(z t ) ˆ pø u (z t ) ˆ på ut .v (0 < u. And if p is odd. ö u i G ˆ 3 ö u ( g)ö u ( g) p gPG ˆ ˆ 1 ˆ ö u ( g)ö u ( g) p3 gP Z 1 ˆ 2 p p3 gP Z ˆ 1X Therefore ö u is irreducible.4 (with K ˆ Z(G)). Clearly the irreducible characters ÷ u. If p ˆ 2. then ö u takes the values stated in the theorem. and (ø u 4 G)( g) ˆ 0 if g P HX a pÀ1 ˆ sˆ0 pÀ1 ˆ sˆ0 303 ø u (z s ) å us We have now established that if ö u ˆ ø u 4 G. We ®nd that 1 ˆ hö u . In fact. and H 2 ˆ ha. C8 3 C2 . . that all the nine groups G1 . b in H1 and H2 will serve for the elements a. Hence GaK  D8 . 26. These are G of order 16 with G1 ˆ ha.7). it is possible to see. using Exercise 26.8 The groups of order 16 It is known that. G3 ˆ ha.6. so the subgroup K described in Lemma 26.6. b chosen in the statement of Theorem 26. and their character tables are given by Theorem 9.1(2). bz ˆ zb. there are precisely fourteen groups of order 16 (see p. bÀ1 ab ˆ a p‡1 i. C4 3 C2 or C2 3 C2 3 C2 X We shall divide our descriptions into three parts. C4 3 C2 3 C2 and C2 3 C2 3 C2 3 C2 . G2 ˆ ha. Our descriptions will be in terms of presentations. 134 of the book by Coxeter and Moser listed in the Bibliography). By Theorem 9. b2 ˆ a4 . b: a8 ˆ b2 ˆ 1. The elements a.5. bÀ1 ab ˆ aziX 2 We have Z( H1 ) ˆ ka p l. We shall describe all these groups and their character tables. the abelian groups of order 16 are C16 . up to isomorphism. bÀ1 ab ˆ aÀ1 i. b: a p ˆ b p ˆ 1. b: a8 ˆ b2 ˆ 1.304 (26X7) Representations and characters of groups H 1 ˆ ha.8. b: a8 ˆ 1. . It is not C8 by Lemma 26.8. according to these three possibilities for GaK. bÀ1 ab ˆ aÀ1 i ˆ D16 . Z( H2 ) ˆ kzl. and it is not Q8 by Exercise 26. bÀ1 ab ˆ a3 iX . b. (A) There are three non-abelian groups GaK  D8 . az ˆ za. . z: a p ˆ b p ˆ z p ˆ 1. . G9 given below do indeed have order 16. For each of the nine non-abelian groups G of order 16 it is the case that |G9 ’ Z(G)| ˆ 2 (see Exercise 26. C4 3 C4 .3 is given by K ˆ G9 ’ Z(G)X Now GaK is a group of order 8. z 2 ˆ 1. they can be found by using the column orthogonality relations. of order 2). . a5 a5 . C7 . but not in G3 . a3 C5 a3 . b. az ˆ za.Characters of some p-groups 305 In each case K ˆ ka4 l. b2 ˆ z. z: a4 ˆ 1.4(2) as induced characters ø 4 G. G2 and G3 : Class Centralizer order C1 16 1 1 1 1 2 2 2 C2 16 1 1 1 1 2 À2 À2 C3 8 1 1 1 1 À2 0 0 C4 8 1 1 À1 À1 0 á â C5 8 1 1 À1 À1 0 â á C6 4 1 À1 1 À1 0 0 0 C7 4 1 À1 À1 1 0 0 0 where ሠ2 ˆ Àâ for G1 . G2 . a6 C4 a. bÀ1 ab ˆ azi. Note that a is conjugate to aÀ1 in G1 and G2 . . a7 a. The last two characters can be obtained as in Theorem 26. b. G2 G3 1 1 C2 a4 a4 C3 a2 . These are G4 ˆ ha. b. z: a4 ˆ 1. (B) There are three non-abelian groups G of order 16 with GaK  C4 3 C2 (where. . and these are given in the following table. as before. C1 G1 . p á ˆ i 2 ˆ Àâ for G3 X p The ®rst ®ve characters are the lifts of the irreducible characters of GaK  D8 .4. z: a4 ˆ z. K ˆ G9 ’ Z(G). we obtain the character tables of G1 . hence the values in the columns C4 and C5 are all real for G1 and G2 . bz ˆ zbiX . Each of the groups has seven conjugacy classes C1 . G5 ˆ ha. a6 a2 . where ø is a linear character of the abelian subgroup kal of index 2. a7 C6 ai b (i even) ai b (i even) C7 ai b (i odd) ai b (i odd) Using Theorem 26. b2 ˆ z 2 ˆ 1. alternatively. . b2 ˆ z 2 ˆ 1.6). but not for G3 (see Corollary 15. bÀ1 ab ˆ azi. bÀ1 ab ˆ az. G6 ˆ ha. . K ˆ kzl. a2 bz C9 ab. but they are in a form which allows us to describe the conjugacy classes C1 . bÀ1 ab ˆ aÀ1 . a3 bz a. G9 ˆ ha. C10 of all three groups G4 . b. The character tables of G4 . z is redundant). az ˆ za. there are three non-abelian groups G of order 16 with GaK  C2 3 C2 3 C2 (where K ˆ G9 ’ Z(G). These are G7 ˆ ha. z: a2 ˆ b2 ˆ z 4 ˆ 1. G5 . G8 ˆ ha.306 Representations and characters of groups The given presentations are somewhat cumbersome (for example. bz ˆ zbi  D8 3 C2 . abz C10 a3 b. az a3 . bz ˆ zbiX . G5 and G6 can again be found using Theorem 26. G6 X (C) Finally. z: a4 ˆ z 2 ˆ 1. . bz In each case. since a4 ˆ z in G4 . . b. bÀ1 ab ˆ aÀ1 . b.4: Class Centralizer order C1 16 1 1 1 1 1 1 1 1 2 2 C2 16 1 1 1 1 1 1 1 1 À2 À2 C3 16 1 À1 1 À1 1 À1 1 À1 á â C4 16 1 À1 1 À1 1 À1 1 À1 â á C5 8 1 i À1 Ài 1 i À1 Ài 0 0 C6 8 1 Ài À1 i 1 Ài À1 i 0 0 C7 8 1 1 1 1 À1 À1 À1 À1 0 0 C8 8 1 À1 1 À1 À1 1 À1 1 0 0 C9 8 1 i À1 Ài À1 Ài 1 i 0 0 C10 8 1 Ài À1 i À1 i 1 Ài 0 0 where á ˆ 2i ˆ Àâ á ˆ 2 ˆ Àâ for G4 . az ˆ za. a2 ˆ b2 . a3 z b. z: a4 ˆ b2 ˆ z 2 ˆ 1. bz ˆ zbi  Q8 3 C2 . of order 2). . bÀ1 ab ˆ az 2 . for G5 . az ˆ za. G6 simultaneously: C1 1 C2 z C3 a2 C4 a2 z C5 C6 C7 C8 a2 b. G8 G9 1 a2 1 z2 C5 C6 C7 C8 C9 C10 307 z a2 z a. az 2 az. a3 b abz. with references for their character tables. for G9 X 26. G8 . G8 .Characters of some p-groups Each of these groups has ten conjugacy classes. bz 3 ab. which are given by C1 C2 C3 C4 G7 . a3 bz b. a2 b bz. a2 bz ab. bz 2 bz. given by Theorem 26.9 The groups of order less than 32 At this point. and the character tables of G7 . abz3 We have Kˆ @ ha2 i hz 2 i for G7 . a3 az. Class Centralizer order C1 16 1 1 1 1 1 1 1 1 2 2 C2 16 1 1 1 1 1 1 1 1 À2 À2 C3 16 1 1 1 À1 1 À1 À1 À1 á â C4 16 1 1 1 À1 1 À1 À1 À1 â á C5 8 1 1 À1 1 À1 1 À1 À1 0 0 C6 8 1 1 À1 À1 À1 À1 1 1 0 0 C7 8 1 À1 1 1 À1 À1 1 À1 0 0 C8 8 1 À1 1 À1 À1 1 À1 1 0 0 C9 8 1 À1 À1 1 1 À1 À1 1 0 0 C10 8 1 À1 À1 À1 1 1 1 À1 0 0 where á ˆ 2 ˆ Àâ á ˆ 2i ˆ Àâ for G7 .3.4. abz 2 abz. for G9 . a3 z z z 3 a. G8 and G9 . are as follows: . the groups. whose character tables are given by Theorem 9. Apart from abelian groups and dihedral groups. az 3 b. are as follows.8 and Section 18. we have in fact found the character tables of all groups of order 31 or less. 3 Section 26.4 F7. . 3. H2 T28 D6 3 C5 . .2 Exercise 18.18 27 28 30 Summary of Chapter 26 In this chapter.1 Section 18.6 Exercise 18. 1. D6 3 C4 S4 SL (2.4: p-groups which contain an abelian subgroup of index p. . as follows.308 |G| 8 12 16 18 20 21 24 Representations and characters of groups G Q8 A4 T12 G1 . Theorem 26. G9 D6 3 C3 E T20 F5. Suppose that G is a group of order pn ( p prime.10 Theorem 25.6: groups of order p3 . A4 3 C2 . Theorem 26. G has pm linear characters and p nÀ2 À p mÀ2 irreducible characters of degree p.4 Exercise 18. we gave the irreducible characters of various nonabelian p-groups G. 2. with an abelian subgroup H of index p.8 Theorem 19.3 Theorem 25. Exercises for Chapter 26 1.1 Exercise 27. Show that for some integer m > 2.3 Exercise 18.3 D12 3 C2 . D10 3 C3 Reference for character table Exercise 17.18 Theorem 19. .3 Theorem 19.2 Exercise 18.5 Exercise 18.8: groups of order 16. 3) T24 U24 V24 H1 .18 Exercise 25.18 Section 18. Section 26.5 Theorem 26. Q8 3 C3 . . n > 2).10 Theorem 19. T12 3 C2 D8 3 C3 . 5.4. G3 . g 2 P h Zi. and ®nd all the irreducible representations of G. Find the conjugacy classes of H. b2 ˆ a8 . G6 . Let A. (b) Why do the remaining groups G5 . z: a3 ˆ b3 ˆ z 3 ˆ 1. G4 and G9 . . 4. 3. or otherwise. (b) Show that for all g in G. 0g e 0 I g 0g g g. . bz ˆ zb. Dl. G7 and G8 have no faithful irreducible representations? . az ˆ za. (Hint: use Corollary 9. fi 0 1 0g 0 0 e d 0 0 0 0 Ài À1 0 I 0 f f0 f Dˆf f0 d 1 H 0 Ài 0 0 0 1 1 0 0 0 0 1 0 I 0 0 g Ài g g g. and deduce that G is a 2group of order at most 32. B. G9 be the non-abelian groups of order 16 with presentations as given in the text. Let G1 . ®nd the character table of G. 0g e 0 and let G ˆ kA. Let H be the group of order 27 which is given by H ˆ ha. and deduce that G9 ˆ h Zi. b.) (d) Show that |G| ˆ 32. Let G be the group of order 32 which is given by G ˆ ha. bÀ1 ab ˆ azi 309 (see (26.3. B. (a) Prove that all pairs of generators commute modulo h Zi. G2 . b: a16 ˆ 1. C. . Ài g e 0 g 0g g g. and use Theorem 26. Write Z ˆ ÀI. (a) Find faithful irreducible representations of degree 2 for G1 . C. (c) Prove that the given representation of G of degree 4 is irreducible. . D H À1 f f 0 f Aˆf f 0 d 0 f fi f Cˆf f0 d 0 H i 0 0 0 be the following 4 3 4 matrices: I H 0 0 0 0 0 i g f f0 1 0 0g 0 0 g f Bˆf g.6 to write down the character table of H.Characters of some p-groups 2. bÀ1 ab ˆ aÀ1 iX Using Theorem 26.7)). b 3 d 1 0 0 e. . 0 0 1 H I H I i 0 0 0 1 0 G6 : a 3 d 0 Ài 0 e. then Ga(G9 ’ Z(G)) T Q8 . b 3 d 0 Ài 0 e. and hence that a2 P Z. (a) Prove that if G is any group.310 (c) Representations and characters of groups Check that the following give faithful representations of G5 and G6 : H I H I 0 1 0 i 0 0 G5 : a 3 d 1 0 0 e. a2  b2 mod Z. bZ: a4 P Z. . Let G be a non-abelian group of order p4 . (Hint: assume that Ga Z ˆ haZ.) 6. Prove that no two of the groups G1 . . Prove that a2 commutes with b. . . G9 are isomorphic. 0 0 i 0 0 1 H I À1 0 0 z 3 d 0 À1 0 e. (c) Deduce that |G9 ’ Z(G)| ˆ p. (b) Prove that |G9| ˆ p or p2 . and that if | Z(G)| ˆ p2 then G has p3 ‡ p2 À p conjugacy classes. G9 given in the text do indeed give groups of order 16. 8. . .) (b) Deduce from the result of Exercise 7 that if G is a group of order 16. and that if |G9| ˆ p2 then G has 2 p2 À 1 conjugacy classes. 0 0 i 0 0 1 H I À1 0 0 z 3 d 0 À1 0 eX 0 0 1 (d) Find faithful representations of degree 3 for G7  D8 3 C2 and G8  Q8 3 C2 . then Ga Z(G) T Q8 . (Note: This exercise can be used to con®rm that the presentations of G1 . . 7. bÀ1 ab  aÀ1 mod Zi. . (a) Prove that | Z(G)| ˆ p or p2 . Denote by SL (2. The next smallest is a certain group of order 168. and in this chapter we shall describe this group and ®nd its character table. Then SL (2. and recall that Z p is the ®eld which consists of the numbers 0. To calculate the order of the group SL (2.27 Character table of the simple group of order 168 Recall that a simple group is a non-trivial group G such that the only normal subgroups of G are f1g and G itself. b. p). p) is a group under matrix multiplication. We discussed brie¯y in Chapter 1 the signi®cance of simple groups in the theory of ®nite groups. And there are p2 ( p À 1) choices for a. and we begin with a description of this family. . c. Special linear groups Let p be a prime number. is the smallest non-abelian simple group. . A5 and A6 . . and d is determined by a). ad À bc ˆ 1)X c d If c ˆ 0. with addition and multiplication modulo p. . b. The group belongs to a whole family of simple groups. In fact the group A5 . p) the set of all 2 3 2 matrices M with entries in Z p such that det M ˆ 1. and is called the 2-dimensional special linear group over Z p. b. d which make ad À bc ˆ 1 (since a. we count the matrices   a b (a. d P Z p . b are arbitrary. c. Examples of simple groups which we have met so far are cyclic groups of prime order. then there are p( p À 1) choices for a. of order 60. except that a Tˆ 0. p À 1. d 311 . By Exercise 27. p).1 Lemma The group PSL (2. ÀIg (where I is the 2 3 2 identity matrix). d may be chosen arbitrarily. the group PSL (2. 7) has exactly six conjugacy classes. In the exercises. 7) 27. and it is easy to see that this group is isomorphic to S3 .19 of the book by J. . the order of CG ( gi ). Rotman listed in the Bibliography). such that ad À bc ˆ 1 (since a. p) has order 6. p)afÆIgX Since |SL (2. The following table records representatives gi (1 < i < 6) for the conjugacy classes. and the size of the conjugacy class containing gi . Thus PSL (2. The power of these techniques is therefore well illustrated. and then b is determined). 3)  A4 . Therefore jSL (2. together with the order of gi . p)j ˆ p( p2 À 1)a2X It is known that PSL (2. we indicate other ways of obtaining characters of G.1. The simple group G ˆ PSL (2. PSL (2. 5)  A5 . p) ˆ SL (2. using information about subgroups. p) is Z ˆ fI. J. so assume that p is an odd prime. notably the orthogonality relations (Theorem 16. and we shall construct the character table of this group. The factor group SL (2.312 Representations and characters of groups with c Tˆ 0. p) is simple (see Theorem 8. After ®nding the conjugacy classes of G. p)| ˆ p( p2 À 1). p)a Z is called the 2-dimensional projective special linear group. The conjugacy classes of PSL (2. the centre of SL (2. p)j ˆ p( p À 1) ‡ p2 ( p À 1) ˆ p( p2 À 1)X If p ˆ 2 then SL (2. and is written as PSL (2. we have jPSL(2. we shall ®nd the character table using only numerical calculations.4) and congruence properties (Corollary 22. 7) has order 168. and that for p > 5. c is any non-zero element of Z p .26). . . and then use direct calculation to ®nd all the elements of G which commute with g i . for example. Among g1 . . . g6 . we verify that gi has the stated order. . . MZ: M ˆ À2 4 2 4 1 0 0 1        ' 2 3 2 4 2 2 2 À2 . Consider. 6. CG ( gi ) ˆ k gi l for i ˆ 3. . the only elements with the same order are g5 . X 3 À2 4 À2 À2 2 2 2 Also.Character table of the simple group of order 168 Order of gi  g1 ˆ  g2 ˆ  g3 ˆ  g4 ˆ  g5 ˆ  g6 ˆ 1 0 0 1  Z  Z 2 4 3 7 7 8 4 3 7 7 21 42 56 24 24 1 168 |CG ( gi )| | gG | i 1 313 0 1 À1 0 2 2 2 0 1 0 1 0  À2 Z 2  0 Z 4  1 Z 1  À1 Z 1 Proof For each i.  a c  b . g4 . Consequently &   1 0 2 Z. d  ' Z X  4 0 0 2        3 À2 3 2 0 À1 1 0 . CG ( g 4 ) ˆ 0 1 0 Similarly CG ( g2 ) ˆ &  0 4 0 4  Z. 5. . . Suppose that   a b Z c d commutes with g4 . Then     a b 2 0 2 ˆÆ c d 0 4 0 and hence b ˆ c ˆ 0. so no two of these six elements are conjugate. . a Tˆ 0. for 1 < i < 4. Hence the conclusion follows from Theorem 22. Thus g5 is not conjugate to g6 . Suppose that gÀ1 g6 g ˆ g5 with   a b gˆ Z P GX c d Then gg5 ˆ g6 g. where ÷1 is the .2 Corollary (1) If 1 < i < 4 and ÷ is a character of G.8). . ÷6 be the irreducible characters of G. . and we have established that no two of the elements g1 . . gi is conjugate to ( gi ) k whenever gi and ( gi ) k have the same order. d ˆ aÀ1 and     a a‡b a b À aÀ1 ˆ X 0 aÀ1 0 aÀ1 Therefore a2 ˆ À1. g6 are conjugate. 27. since any normal subgroup is a union of conjugacy classes (see Proposition 12. and so    a a‡b aÀc ˆÆ c c‡d c bÀd d  with ad À bc ˆ 1X It follows that c ˆ 0. ÷( g5 ) is non-real.314 Representations and characters of groups and g6 . 5 Therefore (2) follows from Corollary 15. it also has six irreducible characters. it is easy to check that G is indeed simple. The size of the conjugacy class g G is obtained by dividing 168 by i |CG ( gi )| (Theorem 12. Proof (1) By Lemma 27. these exhaust the conjugacy i classes of G. Let ÷1 . . which is impossible for a P Z7 . so g5 is not conjugate to its inverse. j The character table of G PSL (2. . (2) For some character ÷ of G. . 7) Since G has six conjugacy classes. then ÷( gi ) is an integer.16. Since the sum of the sizes of the six conjugacy classes g G (1 < i < 6) is 168. j Notice that using Lemma 27.19).1. except possibly g5 and g6 . . (2) Notice that g6 ˆ gÀ1 .1.6. The entries must therefore be 1. Æ1. ÷6 . 0 in some order. Æ1. The entries in the column of g4 are integers. so di is the entry on row i of column 1. Æ2. . and so ÷( g 2 )  ÷( g 3 ) mod 2X Since we also know that 6 ˆ iˆ1 ÷ i ( g 3 )÷ i ( g 4 ) ˆ 0. ÷( g 2 )  ÷(1) mod 2. Æ1. by Corollary 27. we see that. Æ1.e. Æ1. g4 . .2. part of the character table of G is as follows: Class representative Centralizer order ÷1 ÷2 ÷3 ÷4 ÷5 ÷6 g2 8 1 Æ1 0 Æ1 Æ1 Æ2 g3 4 1 Æ1 0 Æ1 Æ1 0 g4 3 1 Æ1 Æ1 0 0 0 We shall determine the signs later. . 0. and the sum of the squares of these integers is equal to |CG ( g4 )| ˆ 3. We shall repeatedly exploit the column orthogonality relations. Æ1. we have by Corollary 22. Recall that the character table is the 6 3 6 matrix with ij-entry ÷ i ( gj ). 0.26 and 22. for which the character values are known to be integers.Character table of the simple group of order 168 315 trivial character (so that ÷1 ( g) ˆ 1 for all g P G). and the congruence properties given by Corollaries 22. Æ1.27. with a suitable ordering of ÷2 . . For the moment we concentrate on the entries in the ®rst column of the character table (i. 0 in some order. 0 in some order. Æ1. g3 . and ÷( g 3 )  ÷(1) mod 2. (We know that the entry in the ®rst row is ÷1 ( g4 ) ˆ 1.) Similarly the entries in the column of g3 are 1. Let di ˆ ÷ i (1). By . 0. and the entries in column g2 are 1.27 for the elements g2 . Theorem 16. Now for all characters ÷ of G. the degrees ÷ i (1)).4(2). and hence d6 ˆ 6. d 6 divides 168. Theorem 22. But 0ˆ 6 ˆ iˆ1 316 and and ÷ i ( g 2 )d i ˆ 1 Æ d 2 Æ 3 Æ 3 Æ 2d 6 . We have now found the ®rst column of the character table. so as d 2 < 168. d 4  1 mod 2. 8 in some order. we have d2 ˆ 7 and d3 ˆ 8.Representations and characters of groups €6 2 Corollary 22. Next. 2 3 so d 2 ‡ d 2 ˆ 113. d 2 < 168X 6 Therefore d6 is 6 or 12. we have d6 Tˆ 12. 2 Now 1 ‡ d 2 ‡ d 2 ‡ 32 ‡ 32 ‡ 62 ˆ 168. and have the following portion: Class representative Centralizer order ÷1 ÷2 ÷3 ÷4 ÷5 ÷6 g1 168 1 7 8 3 3 6 g2 8 1 Æ1 0 Æ1 Æ1 Æ2 g3 4 1 Æ1 0 Æ1 Æ1 0 g4 3 1 Æ1 Æ1 0 0 0 . d 6  0 mod 3. d5 ˆ 3. d 4 divides jGj ˆ 168. Since d 2  1 mod 2. d3 equal to 7. d3 2 3 positive integers have d2.11 and the fact that iˆ1 d i ˆ 168. we have d 4  0 mod 3. The only solutions to this equation with d2. d 6  0 mod 2. In the same way. d 2 < 168X 4 The only positive integer d4 which satis®es these conditions is d4 ˆ 3.27. ) Also. 0ˆ 6 ˆ iˆ1 ÷ i ( g 4 )÷ i ( g j ) ˆ 1 À ÷3 ( g j ) and so ÷3 ( g5 ) ˆ ÷3 ( g6 ) ˆ 1. there is an irreducible character ÷ of G such that ÷( g5 ) is non-real. g3 . the equation 1 ˆ h÷2 . ÷2 i ˆ ˆ 6 ˆ ÷2 ( g i )÷2 ( g i ) iˆ1 jCG ( g i )j 7X7 1 1 1 ÷2 ( g 5 )÷2 ( g 5 ) ÷2 ( g 6 )÷2 ( g 6 ) ‡ ‡ ‡ ‡ ‡ 168 8 4 3 7 7 gives ÷2 ( g5 ) ˆ ÷2 ( g6 ) ˆ 0. Thus the column for g5 is . Let ÷4 ( g5 ) ˆ ÷5 ( g 5 ) ˆ z. the complex conjugate ÷ will be a different character of the same degree.2. For this character ÷.Character table of the simple group of order 168 The equations 6 ˆ iˆ1 317 ÷ i ( g 1 )÷ i ( g j ) ˆ 0 for j ˆ 2. 6. but we could not use this fact as we were not sure that ÷2 ( g5 ) was an integer. and let ÷6 ( g5 ) ˆ t. for j ˆ 5. By Corollary 27. Hence ÷4 and ÷5 (being the only two irreducible characters with the same degree) must be complex conjugates of each other. We obtain Class representative Centralizer order ÷1 ÷2 ÷3 ÷4 ÷5 ÷6 g1 168 1 7 8 3 3 6 g2 8 1 À1 0 À1 À1 2 g3 4 1 À1 0 1 1 0 g4 3 1 1 À1 0 0 0 Next. g4 . 3. (Note that ÷2 ( g5 )  ÷2 (1) mod 7. 4 now enable us to determine the signs in the columns for g2 . 318 Representations and characters of groups Class representative Centralizer order ÷1 ÷2 ÷3 ÷4 ÷5 ÷6 g5 7 1 0 1 z z t Now 0ˆ 0ˆ 7ˆ 6 ˆ iˆ1 6 ˆ iˆ1 6 ˆ iˆ1 ÷ i ( g 2 )÷ i ( g 5 ) ˆ 1 À z À z ‡ 2t. . ÷ i ( g 3 )÷ i ( g 5 ) ˆ 1 ‡ z ‡ z. z ˆ (À1 Æ i 7)a2X Since g6 ˆ gÀ1 . g1 168 1 7 8 3 3 6 g2 8 1 À1 0 À1 À1 2 g3 4 1 À1 0 1 1 0 g4 3 1 1 À1 0 0 0 g5 7 1 0 1 á á À1 g6 7 1 0 1 á á À1 It is known that there are precisely ®ve non-abelian simple groups of order less than 1000. 7) Class representative Centralizer order ÷1 ÷2 ÷3 ÷4 ÷5 ÷6 p where á ˆ (À1 ‡ i 7)a2. ÷ i ( g 5 )÷ i ( g 5 ) ˆ 2 ‡ 2zz ‡ ttX Solving these equations. 7). we have ÷( g6 ) ˆ ÷( g 5 ) for all characters ÷ of G. as shown. We give you the character tables of all of these. we obtain p t ˆ À1. Character table of PSL (2. 5 We have now completely determined the character table of G ˆ PSL (2. as in Lemma 27. p) ˆ a c Summary of Chapter 27  ' b : a. Deduce directly from the character table of PSL (2. . Exercises for Chapter 27 1. 7) that this group is simple. 3).13 This chapter Exercise 20. b P Z7 7 0 aÀ1 (where Z ˆ {ÆI}). SL (2. p)) ˆ fÆIg.3 Exercise 27. where ÷ is an irreducible character of G.2 Exercise 28. 2. 8) PSL (2. In this exercise we present an alternative construction of the character table of G ˆ PSL (2. of order 21. Calculate the values of the induced character (1 T ) 4 G. 3. 7). 2. 11) Order of G 60 168 360 504 660 319 Reference for character table Example 20. 4. p)j ˆ p( p2 À 1). d jSL (2.6 & 1. ad À bc ˆ 1 . 7) A6 PSL (2.Character table of the simple group of order 168 as follows: G A5 PSL (2. We constructed the character table of PSL (2. the simple group of order 168. jPSL (2. Prove that Z(SL (2. (a) De®ne the subgroup T of G. p) ˆ SL (2. 7). p)j ˆ p( p2 À 1)a2 ( p odd). 3. and show that (1 T ) 4 G ˆ 1 G ‡ ÷. given the conjugacy classes of G. PSL (2. as follows: &  ' a b Tˆ Z : a P Zà . c. Find the character table of SL (2. p)afÆIg. d P Z p .1. b. we now have irreducible characters of G of degrees 1. obtain an irreducible character of G of degree 6. Calculate the values of ë 4 G and prove that this is an irreducible character of G. 7).14).320 Representations and characters of groups (b) Let ë be a non-trivial linear character of T. 7. (c). The character table of SL (2. (b). (c) By considering ÷ S (see Proposition 19. Let G ˆ SL (2. Use orthogonality relations to complete the character table of G. 8 and 6. with entries in the ®eld Z7 . the group of all 2 3 2 matrices of determinant 1. (a) Show that G has 11 conjugacy classes with representatives gi as follows: gi  g1 ˆ  g2 ˆ 1 0 0 1  1 2 4 8 8 3 6 7 14 7 14 336 336 8 8 8 6 6 14 14 14 14 1 1 42 42 42 56 56 24 24 24 24 Order of gi |CG ( gi )| | gG | i  À1 0 0 À1   0 1 g3 ˆ À1 0   2 À2 g4 ˆ 2 2   À2 2 g5 ˆ À2 À2   2 0 g6 ˆ 0 4   À2 0 g7 ˆ 0 À4   1 1 g8 ˆ 0 1   À1 À1 g9 ˆ 0 À1   1 À1 g 10 ˆ 0 1   À1 1 g 11 ˆ 0 À1 . 5. 7). (d) From (a). ÷10 . gÀ1 . we have ÷ j ( g) ˆ À÷ j (À g). show that the degrees of ÷7 . . . g3 and g6 (see Proposition 19. (f) Let ø be one of the irreducible characters of degree 4. By considering the values of ø A on g1 . (d) Prove that ÷ j ( g3 ) ˆ 0 for 7 < j < 11. . 4. Let G ˆ PSL (2. g8 . gÀ1 are conjugate to g5 . . ÷9 . ÷11 are 4. ÷8 . (c) Let ÷7 . The character table of PSL (2.7) to write down the six irreducible characters of G with kernel containing Z ˆ {ÆI}. respec5 6 7 8 tively. This group has eight conjugacy classes with representatives g1 . prove that ø A is equal to the irreducible character of G of degree 6 whose kernel contains Z. g6 .14). Deduce the values of the irreducible characters of degree 4 on all gi . and ®nd ÷ j ( g6 ) for 7 < j < 11. 6. 6. . gÀ1 . (g) Complete the character table of G. 8. 6. (e) By considering the column of g6 in the character table. Show that for any j with 7 < j < 11 and any g P G. . . . and congruences modulo 3. g8 having orders and centralizer orders as follows: gi Order of gi |CG ( gi )| g1 1 660 g2 2 12 g3 3 6 g4 6 6 g5 5 5 g6 5 5 g7 11 11 g8 11 11 Also. . gÀ1 . g2 . Find the character table of G.Character table of the simple group of order 168 321 (b) Use the character table of PSL (2. 11). ÷11 be the remaining irreducible characters of G. g7 . and deduce that ÷ j (1) is even. 11). we found the character tables of certain groups of 2 3 2 matrices with entries in Z7 and Z11 . b. q) We are now going to calculate the character tables of an important in®nite series of groups.28 Character table of GL(2. ‡) is an abelian group. In the last chapter and its exercises. the distributive law holds. First. if you are unfamiliar with ®nite ®elds then you might like to consult the book by J. 3) is a set F with two binary operations ‡ and 3 such that the following properties hold. B. we shall see that the conjugacy classes of our groups fall into four families. Fraleigh listed in the Bibliography. Recall that a ®eld (F. Consequently. Secondly. we can display the character values in a 4 3 4 matrix. R. this is a daunting task. with identity element 0. with the usual de®nitions of ‡ and 3. and we shall tell you which properties of these ®elds we will use. and one of the exercises will show you how to use the results of this chapter to determine the character tables of in®nitely many simple groups. We shall determine the character tables of some matrix groups with entries from an arbitrary ®nite ®eld. (F. At ®rst sight. 3) is an abelian group. The ®elds F q and F q 2 We consider ®nite ®elds. if we write F à ˆ Fnf0g. as do the irreducible characters. ‡. C and Z p ( p prime) are ®elds. The basic properties of ®nite ®elds which we will use without proof are these: 322 . Finally. then (F à . that is (a ‡ b)c ˆ ac ‡ bc for all a. For example. c P F. since the number of irreducible characters increases with the size of the ®eld. with identity element 1. However. 3) are abelian groups. Then there exists a ®eld F q of order q and every ®eld of order q is isomorphic to F q . ps ˆ 0. r 1‡q P S. and write q ˆ pn . We may write q r ˆ å m for some m and we let r ˆ ù m . (28. 28. q) (28. (1) The set S is a sub®eld of F of order q. .2 Proposition Let F ˆ F q 2 and S ˆ fs P F : s q ˆ sg. t P F q . Then (s ‡ t) q ˆ s q ‡ t q ˆ s ‡ t.Character table of GL(2. in short. You are now in a position to appreciate the statement of the main result of this chapter. We use this remark in the proof of the next proposition. it follows that (s ‡ t) p ˆ s p ‡ t p for all s. namely Theorem 28. Hereafter.3) Let å be a generator of the cyclic group FÃ2 and let q 2 ù ˆ e(2ðia(q À1)) .1) Let p be a prime and n be a positive integer. The group (Fà . This implies that (r ‡ r ) ˆ r ‡ r ˆ r ‡ r and (r 1‡q ) q ˆ j r 1‡q . and k k k hence (s ‡ t) p ˆ s p ‡ t p for all positive integers k. ‡) and (Snf0g.2 with the ®eld F q . Proof (1) Suppose that s. For every s P F q the sum of s with itself p times is zero. Moreq over. so s ‡ t P S. q 323 Notice that the binomial coef®cients ( ip ) with 1 < i < p À 1 are all divisible by p. We introduce the following useful notation. Then r 3 r is an irreducible character of FÃ2 . every irreducible character of FÃ2 has the form q r 3 r j for some integer j.5. 2 (2) Since FÃ2 is a group of order q 2 À 1. so S is a ®eld. so r ‡ r q . and hence S  F q . t P S. Suppose that r P FÃ2 . we see that r q ˆ r for all q q q q q2 q r P F. (2) If r P F then r ‡ r q . 3) is cyclic. we shall identify the sub®eld S of F q 2 in Proposition 28. r 1‡q P S. It is now easy to check that (S. cg ˆ fa9. c9g. q). since conjugate matrices have the same eigenvalues. and we talked about some special linear groups in the last chapter. b).324 Representations and characters of groups The conjugacy classes of GL(2. The number of such matrices is found by noting that (a. Let G ˆ GL(2. q). q) is de®ned to be the group of invertible 2 3 2 matrices with entries in F q . giving us q 2 À q choices. The matrices   s 0 sI ˆ (s P Fà ) q 0 s belong to the centre of G. jGj ˆ (q 2 À 1)(q 2 À q) ˆ q(q À 1)2 (q ‡ 1)X There are four families of conjugacy classes of G. Keep this in mind during the following discussion. b) can be any non-zero row vector. q). Therefore. Next. and once (a. The subgroup consisting of all matrices of determinant 1 is the special linear group SL(2.   a b 0 c can be conjugate to  a9 0 b9 c9  only if fa. consider the matrices  s us ˆ 0 Let give us q À 1 conjugacy classes of  1 s (s P Fà )X q . we are going to calculate the character table of GL(2. b) has been chosen. giving us q 2 À 1 choices. of which three are easy to describe. q) The general linear group GL(2. They size 1. First. and remember that the matrix   a b c d belongs to G if and only if its rows are linearly independent. Here. d) can be any row vector which is not a multiple of (a. (c. so. q)   a b P GX gˆ c d Then gus ˆ   and us g ˆ   325 as cs a ‡ bs c ‡ ds as cs d ‡ bs ds so g belongs to the centralizer of us if and only if c ˆ 0 and a ˆ d. v r P G. if s Tˆ t. consider   0 1 vr ˆ (r P F q 2 nF q )X Àr 1‡q r ‡ r q By Proposition 28. Thus. then we have that gd s. The characteristic polynomial of v r is det(xI À v r ) ˆ x(x À (r ‡ r q )) ‡ r 1‡q ˆ (x À r)(x À r q ). Thus. each conjugacy class contains q 2 À 1 elements. the q centralizer order is (q À 1)q. t ˆ P G (s. the centralizer order is (q À 1)2 .Character table of GL(2. 2 3 Àbr 1‡q a ‡ b(r ‡ r q ) gv r ˆ and Àdr 1‡q c ‡ d(r ‡ r q ) 2 3 c d vr g ˆ X Àar 1‡q ‡ c(r ‡ r q ) Àbr 1‡q ‡ d(r ‡ r q ) . so each conjugacy class contains q(q ‡ 1) elements. t g if and only if b ˆ c ˆ 0. Now.8. the matrices d s. Now. Since r P F q we see that v r lies in a none of the conjugacy classes we have constructed so far. t (s. let   s 0 d s. Finally.2. t P Fà . s Tˆ t) give us q (q À 1)(q À 2)a2 conjugacy classes. so v r has eigenvalues r and r q. t P Fà ) q 0 t and note that  0 1 1 0 À1  d s.s X On the other hand. by Theorem 12. t 0 1 1 0  ˆ d t. the matrices us (s P Fà ) give us q À 1 conjugacy classes. t ˆ d s. of classes qÀ1 us (q À 1)q qÀ1 d s. r q P F q. t is indexed by unordered pairs fs. t (q À 1)2 (q À 1)(q À 2)a2 vr q2 À 1 (q 2 À q)a2 The families of conjugacy class representatives sI and us are indexed by elements s of Fà . so it is not conjugate to v r unless t ˆ r or t ˆ r q. But this sum is equal to the order of GL(2. If these conditions hold. q). then ad À bc ˆ a2 ‡ ab(r ‡ r q ) ‡ b2 r 1‡q ˆ (a ‡ br)(a ‡ br q )X Since (a. Class rep. 28. q The family of conjugacy class representatives d s.326 Representations and characters of groups Hence gv r ˆ v r g only if c ˆ Àbr 1‡q and d ˆ a ‡ b(r ‡ r q ). described as follows. g sI |CG ( g)| (q 2 À 1)(q 2 À q) No. r q g of elements of FÃ2 nFà . We have now found all the conjugacy classes of G. j . each subset gives us a conjugacy class representative v r and different subsets give us representatives of different conjugacy classes. The matrix v t has eigenvalues t and t q. jCG (v r )j ˆ q 2 À 1. 0) and r. and the conjugacy class containing v r has size q 2 À q. Therefore. r q g. b) Tˆ (0. q). so we have found all the conjugacy classes. we see that a ‡ br and a ‡ br q a are non-zero. We therefore partition F q 2 nF q into subsets fr.4 Proposition There are q 2 À 1 conjugacy classes in GL(2. q The family of conjugacy class representatives v r is indexed by unordered pairs fr. q q Proof The conjugacy classes we have found account for (q À 1) ‡ (q À 1)(q 2 À 1) ‡ (q À 1)(q À 2)q(q ‡ 1)a2 ‡ (q 2 À q)(q 2 À q)a2 elements altogether. tg of distinct elements of Fà . g P CG (v r ) if and only if   a b gˆ X Àbr 1‡q a ‡ b(r ‡ r q ) Thus. Recall that å is our chosen generator for FÃ2 and q   0. we have the following restrictions on the subscripts. there are q À 1 characters ë i . Then jKj ˆ q 2 À 1. (c) For ø i.q) We are now in a position to describe the character table of G. each of degree q ‡ 1. sI ëi øi ø i. and of the remaining q 2 À q elements of K. we ®rst consider the set of integers j with 0 < j < q 2 À 1 and (q ‡ 1) T j j. . each of degree q À 1. and let r 3 r be the function from FÃ2 to C described in (28. Then the q irreducible characters of GL(2. Before we embark upon the task of calculating the irreducible characters of G. each of degree q. q) as in Proposition 28. each of degree 1. 1 X vå ˆ å ‡ åq Àå 1‡q 28. there are (q À 1)(q À 2)a2 characters ø i. there are q À 1 characters ø i .3). ÷ i as follows. j . 327 28. ø i . Thus. if j1 and j2 belong to this set and j1  j2 q mod (q 2 À 1) then we choose precisely one of j1 and j2 to belong to the indexing set for the characters ÷ i . j . we present a proposition which will be useful later. t (st) i (st) i si t j ‡ s j t i 0 vr r i(1‡q) Àr i(1‡q) 0 À(r i ‡ r iq ) Here. j < q À 2.5 Theorem Label the conjugacy classes of GL(2. The group K contains the q À 1 scalar matrices sI in G. Thus. (b) For ø i we have 0 < i < q À 2.Character table of GL(2. Thus. there are (q 2 À q)a2 characters ÷ i . q) are given by ë i . Hence. (d) For ÷ i. j ÷i s 2i qs 2i (q ‡ 1)s i‡ j (q À 1)s i us s 2i 0 s i‡ j Às i d s.4. ø i.6 Proposition Let K ˆ hvå i. (a) For ë i we have 0 < i < q À 2. j we have 0 < i . q) The characters of GL(2. in turn.8 Proposition For all integers i. j We will see later that the linear characters ë i (0 < i < q À 2) which appear in Proposition 28. j (q ‡ 1)s i‡ j us s i‡ j d s. j We shall construct.328 Representations and characters of groups precisely two belong to each of the conjugacy classes represented by v r with r P FÃ2 nFà .4. whose values appear in Theorem 28.5. Proof The map det : g 3 det g is a homomorphism from G onto Fà .5. and this case accounts for the q À 1 scalar matrices. sI ø i. so vå has order q 2 À 1. j and ÷ i which appear in Theorem 28. If å i Tˆ å iq then i i iq å P F q and vå and vå must be conjugate to vå i . i i i If å i ˆ å iq then vå ˆ å i I. 28. Hence two elements a of K belong to the conjugacy class of vå i . since vå has eigenvalues in F q and vå is diagonalizable. q q Proof The eigenvalues of vå are å and å q. t si t j ‡ s j t i vr 0 . the functions ë i : g 3 (det g) i ( g P G) give q À 1 distinct linear characters of G. q As i varies between 0 and q À 2 inclusive. ø i . the irreducible characters ë i .7 Proposition There are q À 1 linear characters ë i of G. j there is a character ø i. as described in Proposition 28.7 are all the linear characters of G. 28. j of G whose values on the conjugacy class representatives. are as follows.5. i iq The eigenvalues of vå and of vå are å i and å iq . and they are given in Theorem 28. ø i. j is a character of B.i which appears in Proposition 28.i ˆ ë i ‡ ø i . ë i i. j ( g) ˆ ø i. We let ø i. j ( g) for each conjugacy class representative g.i . t : g ˆ vr : ø i.9 Proposition For each integer i.i i and hø i. there is an irreducible character ø i of G whose values are given in Theorem 28. where g9 ˆ d t.i i ˆ (q ‡ 1)2 1 (q À 1) (q À 1) ‡ 2 À 1)(q 2 À q) (q À 1)q (q ‡ ˆ 2X Here. De®ne ë i. Remember that the complex conjugate of s i is s Ài .i . the ®rst term corresponds to the conjugacy classes of elements 4 (q À 1)(q À 2) (q À 1)2 2 .23 to calculate ø i. as follows. j ( g) ˆ jC G ( g)j ‡ . We use Proposition 21. j ( g9) ø i. ø i. j : 3 s i t jX 0 t Then ë i. Proof We shall demonstrate that the character ø i. j ( g) jC B ( g)j   ë i. j 4 G. j ( g) ë i. j : B 3 C by   s r ë i.5.s jC B ( g)j jC B ( g9)j ø i. j ( g) ˆ jC G ( g)j ë i. the values of ø i. ø i. g ˆ sI : g ˆ us : g ˆ d s.8 gives us ø i. We have hø i.i . j ( g) jC B ( g)j jC G ( g)j ë i. j are as stated in the proposition. The characters ø i for 0 < i < q À 2 are all different. q) Proof Let Bˆ &  ' PG X 329 a 0 b c Then B is a subgroup of G with jBj ˆ (q À 1)2 q. we calculate hø i. j ( g) ˆ 0X j Hence.Character table of GL(2. j ˆ ëi. 28. To this end. 1 3 s i .i .8.i (sI)ø i. Bˆ 2 À 1)(q 2 À q) (q À 1)q (q 1 1ˆ i j Cˆ (s t ‡ s j t i )(s Ài t À j ‡ s À j t Ài )X (q À 1)2 2 sTˆ t Aˆ and 2 (q À 1)(q À 2) 2 (q À 1) 2 The coef®cent 1 appears in C because we have just one conjugacy 2 class for each unordered pair fs. The remaining terms in hø i. ø i. tg of distinct elements of Fà .i .5. t : s. ø i. j . note that fd s.i ˆ ë i ‡ ø i for some irreducible character ø i. t P Fà g is an abelian group of q order (q À 1)2. Then the character ø i.i .8 is irreducible. j 28. ø i.i i ˆ 2 imply that ø i. j which are given in Proposition 28. Using the values of ø i. ë i i ˆ (q ‡ 1) 1 (q À 1) (q À 1) ‡ (q 2 À 1)(q 2 À q) (q À 1)q ‡ ˆ 1X The facts that hø i. q Hence the characters ø i for 0 < i < q À 2 are all different.i i are calculated in a similar fashion.330 Representations and characters of groups sI. j i ˆ A ‡ B ‡ C. j i ˆ 1. Next. ø i. j < q À 2. q To evaluate C. t 3 s i t j ‡ s j t i then ó is a sum of two . we obtain hø i.i . where (q ‡ 1)2 1 (q À 1) (q À 1). Let s be an element of Fà of order q À 1.i (sI) ˆ (q ‡ 1)2 . hø i.i to get the values of ø i as given in Theorem 28. (2) jC G (sI)j ˆ (q 2 À 1)(q 2 À q)X (3) There are q À 1 conjugacy classes with representatives of the forms sI. Then ø i : d s. (1) ø i. j which appears in Proposition 28. ë i i ˆ 1 and hø i. and the calculation of this ®rst term involves the following three observations. Proof We shall show that hø i. and if ó : d s. Subtract ë i from ø i. j .10 Proposition Suppose that 0 < i . j) Tˆ (i9. j9 on a conjugacy class j of G. We have   s b ë i. there exists complex (q À 1)th roots of unity s and t such that either s Tˆ t and s i t j ‡ s j t i Tˆ s i9 t j9 ‡ s j9 t i9 or s ˆ t and s i‡ j Tˆ s i9‡ j9 . Therefore. In either case. j . hó . we see that ø i. j9 . j < q À 2 are distinct irreducible Proof Suppose that 0 < i . sI öi q(q À 1)s i us 0 d s. Therefore. and (i. j9 ‡ ë j9.Character table of GL(2. ø i. 2 3 ˆ 1 i j j i Ài À j À j Ài 4(q À 1) ‡ (s t ‡ s t )(s t ‡ s t ) ˆ 2X (q À 1)2 sTˆ t Hence. t 0 vr r i ‡ r iq . j Tˆ ø i9.8. j is irreducible.i Tˆ ë i9.12 Proposition For each integer i. j ‡ ë j. ø i. Consider the group B and its linear characters ë i. j9 < q À 2. there exists a character ö i of G which takes the following values. 28. j Tˆ ø i9. Cˆ qÀ3 X qÀ1 And now we ®nd that A ‡ B ‡ C ˆ 1.11 Corollary The characters ø i.i9 . for 0 < i . hø i. j < q À 2 and 0 < i9 .i : 3 s i t j ‡ s j t iX 0 t Since ë i. j 28. j ‡ ë j. ó i ˆ 2X That is. j which were used in the proof of Proposition 28. j i ˆ 1. q) 331 inequivalent irreducible characters of this group. j differs from ø i9. and ø i. Thus. We must prove that ø i. j characters of G. j9). j9 . we shall the use the following lemma. as in Proposition 28. Then ˆ (r i ‡ r iq )(r Ài ‡ r Àiq ) ˆ 2(q À 1)2 X rPF q 2 nF q Proof Note that & r G1 ˆ 0 0 rq  : r P FÃ2 q ' and G2 ˆ & r 0 0 rq  : r P Fà q ' are abelian groups of orders q 2 À 1 and q À 1. Now. Hence á i ( g) ˆ r i or r iq and á i ( g) ‡ á i ( g q ) ˆ r i ‡ r iq X Let ö i ˆ á i 4 G. respectively.6.332 Representations and characters of groups Proof Let K ˆ hvå i. Also. by Proposition 28.6. ö i has the values stated in the proposition. In order to calculate ö i . Suppose that g P K and g is conjugate in G to v r . v r is conjugate to an element of g of K. j To be able to work out certain inner products involving our characters ö i . . ®rst recall that á i 4 G is zero on all elements which are not conjugate to an element of K. 28. by Proposition 28. If g ˆ sI with s P Fà then g P K and q ö i ( g) ˆ jC G ( g)j á i ( g) ˆ q(q À 1)s i X jC K ( g)j Suppose that r P F q 2 nF q . ö i is zero on the elements of the form u s and d s. Then. and consider the linear character á i of K which sends the generator vå of K to å i .13 Lemma Assume that i is an integer and (q ‡ 1) T j i.   á i ( g) ái( g q) ö i ( g) ˆ jC G ( g)j ‡ jC K ( g)j jC K ( g)j ˆ á i ( g) ‡ á i ( g q ) ˆ r i ‡ r iq X Thus. t (s Tˆ t). Then g has eigenvalues r and r q .6. Thus. let ÷ i be the class function on G with the following values. 28. sI ÷i (q À 1)s i us Às i d s. q)   r 0 3 r i ‡ r iq 0 rq 333 gives a character ÷ of degree 2 for each group.14 Proposition For each integer i. ø i and ö i given in Propositions 28. t 0 vr À(r i ‡ r iq ) If (q ‡ 1) T j i then ÷ i is an irreducible character of G. and for G2.9 and 28.i À ö i X . since r q ˆ r for r P FÃ. Recall the characters ø i.Character table of GL(2. the character ÷ is twice an irreducible character. ÷ i is the class function on G which is given by ÷ i ˆ ø0.Ài ø i À ø0. For G1.12. Proof We can justify the manoeuvre which we now perform only by saying that it gives the correct answer.8. Now. we get 1 ˆ i (r ‡ r iq )(r Ài ‡ r Àiq ) ˆ 2 2À1 q rPFà 2 q and doing the same for the character ÷ of G2 . j . we get 1 ˆ i (r ‡ r iq )(r Ài ‡ r Àiq ) ˆ 4X qÀ1 à rPF q Hence ˆ rPF q 2 nF q (r i ‡ r iq )(r Ài ‡ r Àiq ) ˆ 2(q 2 À 1) À 4(q À 1) ˆ 2(q À 1)2 X j 28. Taking the inner product of the q character ÷ of G1 with itself. since (q ‡ 1) T j i implies that å i Tˆ å iq . the character ÷ is a sum of two inequivalent irreducible characters. Since j T i. We work out h÷ i . so either s i Tˆ s j for some s P Fà or q r i ‡ r iq Tˆ r j ‡ r jq for some r P F q 2 nF q . as in Proposition 28. q) to ®nd the .13. since we have shown that the class functions given in the theorem are inequivalent irreducible characters. h÷ i . assume that (q ‡ 1) T j i. Therefore.15 Proposition Suppose that i and j are integers with (q ‡ 1) T j i and (q ‡ 1) T j j and j T i. t s Ài ‡ t Ài (st) i i s ‡ ti si ‡ t i 0 0 vr 0 Àr i(1‡q) 0 0 r i ‡ r iq À(r i ‡ r iq ) Next. sI ø0. ÷ i i using Lemma 28. j 28.Ài øi ø0. with integer coef®cients. ÷ i Tˆ ÷ j . Then the characters ÷ i and ÷ j of G are different. iq mod(q 2 À 1). It is possible to use the character table of GL(2. and consider the linear character á i of K which sends the generator vå of K to å i . ÷ i i ˆ 1 and ÷ i (1) . j We have now completed the proof of Theorem 28. Proof Let K ˆ hvå i. 0. the characters á i ‡ á iq and á j ‡ á jq of K are different. Suppose that g P K. iq mod(q 2 À 1). q If g is conjugate to v r where r P F q 2 nF q then (á i ‡ á iq )( g) ˆ r i ‡ r iq . it follows that ÷ i is an irreducible character of G. which is the same as the number of conjugacy classes of G.5. as we wished to show.334 Representations and characters of groups The table below allows us to verify this. If g ˆ sI where s P Fà then (á i ‡ á iq )( g) ˆ 2s i . and the number of them is q 2 À 1.i öi ÷i (q ‡ 1)s Ài qs 2i q(q ‡ 1)s i (q ‡ 1)s i q(q À 1)s i (q À 1)s i us s Ài 0 0 si 0 Às Ài d s. and h÷ i .Ài ø i ø0. ÷ i i ˆ (q À 1)2 1 (q À 1)2 (q À 1) ‡ 2 (q À 1) ‡ 2 ˆ 1X q Àq (q 2 À 1)(q 2 À q) q À1 Since ÷ i is a linear combination of irreducible characters of G.6. A. you are asked to consider the easiest case. and there are (q À 1)(q À 2)a2 0 t irreducible characters of degree q ‡ 1. this gives the character tables of an in®nite series of simple groups PSL(2. t ˆ (s Tˆ t). in 1955. q). q). q) was ®rst given in 1907. Since SL(2. q). Although the character table of GL(2. J. and there are q À 1 irreducible characters of 0 s degree 1. q) is rather easier to ®nd than that of SL(2. q) for all positive integers n. A challenging exercise is to determine the character table of PSL(2. since the answers are quite complicated. q) was found. though. q)  PSL(2.Character table of GL(2. q) ± compare Chapter 27 ± and so the character table of PSL(2. since most of the irreducible characters remain irreducible when restricted. q). In Exercise 28. those with kernel containing the centre of SL(2. q) when q  1 mod 4 or q  3 mod 4 from the character table of GL(2. it was not until the 1950's that the character table of GL(3.2. q) when q is a power of 2. q). and there are q À 1 irreducible characters of degree 0 s q. and they depend upon whether q is a power of 2 or q  1 mod 4 or q  3 mod 4. (d) There are (q 2 À q)a2 conjugacy classes with representatives of the . (b) There are q À 1 conjugacy classes with representatives of the form   s 1 us ˆ . Then. Among the characters of SL(2. q) has the following properties. (a) Thereare q  1 conjugacy classes with representatives of the form À s 0 sI ˆ . We do not go fully into this. q) 335 character table of SL(2. (c) There are (q À 1)(q À 2)a2 conjugacy classes with representatives   s 0 of the form d s. namely that where q is a power of 2. Summary of Chapter 28 The character table of GL(2. Green determined the character table of GL(n. q) provide the characters of the groups PSL(2. Let Z ˆ fsI : s P Fà g. and there are (q 2 À q)a2 irreducible form v r ˆ Àr 1‡q r ‡ r q characters of degree q À 1. 8). 2. Use Theorem 28. q)  Z 3 SL(2. q) is simple.2 to write down explicitly the character table of PSL(2.336 Representations and characters of groups   0 1 . Use your solution to Exercise 28. . Prove that q GL(2. q). q) from that of GL(2.5 to write down explicitly the character table of GL(2. Exercises for Chapter 28 1. Prove that if q Tˆ 2 then SL(2. q)X Deduce the character table of SL(2. 3. 3). Suppose that q is a power of 2. 29. An action of G on Ù is a homomorphism ö: G 3 Sym(Ù). F F F . (2) Let G ˆ Sn and let Ù be the set consisting of all pairs fi. If Ù is a set. De®nition Let G be a group and Ù a set. a subgroup of Sn for some n. In particular. 3g 3 f2. jg( gö) ˆ fig. F F F . i. (1 2)ö sends f1. denote by Sym(Ù) the group of all permutations of Ù. We also say that G acts on Ù (via ö). if Ù ˆ f1.12 below). and develop some useful results. De®ne ö: G 3 Sym(Ù) by setting fi. a fact which proved useful in many of our subsequent character table calculations. Group actions We begin with a more general notion than that of a permutation group. 3g. In this chapter we take the theory of permutation groups and characters somewhat further. particularly about irreducible characters of symmetric groups (see Theorem 29.29 Permutations and characters We have already seen in Chapter 13 that if G is a permutation group. it is called the action of Sn on pairs. ng.) Check that ö is an action of Sn . jg of elements of f1. jgg for all g P Sn and 1 < i . F F F . j < n.1 Examples (1) If G < Sn then the identity map is an action of G on f1. then G has a permutation character ð de®ned by ð( g) ˆ jfix( g)j for g P G. 2.e. ng. ng then Sym(Ù) ˆ Sn . (So for example. 337 . the fact that ö is a homomorphism simply says that ù( gh) ˆ (ù g)h for all ù P Ù and g. For example. if ö: G 3 Sym(Ù) is an action. b P F q . ö is an action of „ G. the group of invertible 2 3 2 matrices over the ®nite ®eld F q.338 Representations and characters of groups (3) Let G ˆ GL(2. The equivalence classes are called the orbits of G on Ù. b)i 3 h(a. Thus Ù is the disjoint union of the orbits of G. Write orb(G. there exists g P G such that á g ˆ â. for ù P Ù and g P G we usually just write ù g instead of ù( gö). Ù) ˆ 1. given any á. h P G. say. and let ö: G 3 S8 be the action . we have á $ â if and only if there exists g P G such that á g ˆ â. b) with a. de®ne a relation $ on Ù as follows: for á. q). â P Ù. generated by x. De®ne ö: G 3 Sym(Ù) by ( Hx)( gö) ˆ Hxg for all x. â P Ù. It is easy to see that $ is an equivalence relation on Ù. With this notation. Ù) for the number of orbits of G on Ù. Then by Exercise 9 of Chapter 23. Let V be the 2-dimensional vector space over F q consisting of all row vectors (a. (4) Let G be a group with a subgroup H of index n. 29. The group G is said to be transitive on Ù if orb(G. in other words. and Ker ö ˆ xPG x À1 Hx < H. To simplify notation.2 Examples (1) Let G ˆ C4 . as de®ned in Chapter 28. if   1 1 gˆ 0 1 then gö sends h(a. and let Ù be the set of all 1-dimensional subspaces hvi of V. and let Ù be the set of all right cosets Hx of H in G (so jÙj ˆ n). G is transitive if. a ‡ b)iX Then ö is an action of G on Ù. Adopting this notation. De®ne ö: G 3 Sym(Ù) by setting hvi( gö) ˆ hv gi for all hvi P Ù and g P G. g P G. Now let Ä be the set of right cosets Gù x of Gù in G. Hy P Ù. j Permutation characters Let G be a group acting on a ®nite set Ù. so ù G ˆ fù g : g P Gg. In other words. w P V there is an invertible 2 3 2 matrix A P GL(2. CÙ consists of all expressions of the form . Moreover. f5. 2. 8g. the element g ˆ x À1 y P G has the property that ( Hx) g ˆ Hy. 4g.1(2. the size of the orbit ù G is equal to the index of Gù in G. 29. 4). Gù x ˆ Gù y D xy À1 P Gù D ùxy À1 ˆ ù D ùx ˆ ù yX Hence we can de®ne an injective function ã : Ä 3 ù G by ã(Gù x) ˆ ùx for all x P G. and Gù contains the identity. y P G. and in Example (4). Observe that for x. Denote by CÙ the vector space over C for which Ù is a basis. as required. namely f1. jù G j ˆ jG : Gù jX Proof If g.Permutations and characters 339 de®ned by xö ˆ (1 2 3 4)(5 6)(7 8) (and of course x k ö ˆ ((1 2 3 4)(5 6)(7 8)) k for any k). Let G be a group acting on a set Ù. Also gÀ1 P Gù . 6g and f7. 8g. 3. so Gù is a subgroup. given two right cosets Hx. q) such that vA ˆ w. h P Gù then ù( gh) ˆ (ù g)h ˆ ùh ˆ ù. write ù G for the orbit of G which contains ù. Clearly ã is also surjective. (2) The group G is transitive on the set Ù in each of Examples 29. and hence jÄj ˆ jù G j. 3. and de®ne Gù ˆ f g P G : ù g ˆ ùgX We call Gù the stabilizer of ù in G. hence gh P Gù . F F F .3 Proposition The stabilizer Gù is a subgroup of G. This is clear in Example (2). that is. Then G has three orbits on Ù ˆ f1. simply observe that. to verify it for Example (3) you need to convince yourself that for any two non-zero row vectors v. For ù P Ù. where fixÙ ( g) ˆ fù P Ù : ù g ˆ ùg. Now de®ne Ö ˆ f(ù. ù g ˆ ùg. for 1 < i < t we have jÄ i j ˆ jù G j ˆ jG : Gù i jX i Hence jÄ i j jGù i j ˆ jGj. First. by de®ning 2 3 ˆ ˆ ëù ù g ˆ ëù (ù g) for all g P G. Ä t be the orbits of G on Ù. though elementary. Then 1 ˆ hð. We call ð the permutation character of G on Ù. but is in fact due to Cauchy and Frobenius. is rather famous. we can make CÙ into a CG-module. Ù)X jGj gPG Proof First note that hð. The next result. pick ù i P Ä i . 29.340 Representations and characters of groups ˆ ëù ù (ëù P C) ùPÙ with the obvious addition and scalar multiplication. ð( g) ˆ jfixÙ ( g)j.3. We calculate jÖj in two different ways. g) : ù P Ù. the number of ù P Ù such that ù g ˆ ù is equal to jfixÙ ( g)j. 1 G i ˆ 1 ˆ 1 ˆ ð( g) ˆ jfixÙ ( g)jX jGj gPG jGj gPG Let Ä1 .4 Proposition Let G be a group acting on a ®nite set Ù. It is often referred to as ``Burnside's Lemma''. F F F . g P G. and provides a basic link between the permutation character and the action of G. for each g. As in Chapter 13. 129) that if ð is the character of this permutation module. 1 G i ˆ jfixÙ ( g)j ˆ orb(G. hence . We see just as in Chapter 13 (p. and for each i. and let ð be the permutation character. then for g P G. By Proposition 29. called the permutation module. Permutations and characters ˆ jÖj ˆ jfixÙ ( g)jX gPG 341 Secondly. ù2 ) g ˆ (ù1 g. 1 G i ˆ 1. In the rest of the chapter we apply Proposition 29. ù2 g) for all ù i P Ù i . then ð( g) ˆ ð1 ( g)ð2 ( g) for all g P G.4. with corresponding permutation characters ð1 and ð2 respectively. ð2 i ˆ 1 ˆ 1 ˆ jfixÙ1 ( g)kfixÙ2 ( g)j ˆ jfixÙ1 3Ù2 ( g)j. jGj gPG jGj gPG j which is equal to orb(G. Ù1 3 Ù2 ) by Proposition 29. 29. . 29. and the conclusion follows. Ù1 3 Ù2 )X Proof We have hð1 . for each ù. Hence if ð is the permutation character of G on Ù1 3 Ù2 . Then hð1 . Now let G be a group.6 in a number of situations. and suppose that G acts on two sets Ù1 and Ù2 . It is clear that fixÙ1 3Ù2 ( g) ˆ fixÙ1 ( g) 3 fixÙ2 ( g) for any g P G. the ®rst being the case where Ù1 ˆ Ù2 . the number of g P G such that ù g ˆ ù is equal to jGù j. Then we can de®ne an action of G on the Cartesian product Ù1 3 Ù2 by setting (ù1 . with permutation characters ð1 and ð2 respectively.5 Corollary G is transitive on Ù if and only if hð. g P G. hence jÖj ˆ Therefore € gPG ˆ ùPÙ jGù j ˆ t ˆ iˆ1 jÄ i j jGù i j ˆ t ˆ 1 jGj ˆ tjGjX j jfixÙ ( g)j ˆ tjGj. ð2 i ˆ orb(G.6 Proposition Let G act on Ù1 and Ù2 . Then Ä ˆ f(ù. namely (ù1 . ù2 g) for all ù1 . j . The result follows. 29.8. 1 G i ˆ 1 by Corollary 29. there exists g P G such that á1 g ˆ â1 and á2 g ˆ â2 . Then G is said to be 2-transitive on Ù if r(G.5.342 Representations and characters of groups Suppose G acts on Ù. then ð ˆ 1 G ‡ ÷. Proof We have hð. for any ordered pairs (á1 . â1 Tˆ â2 .6. ði ˆ 2 by Proposition 29. written r(G. In other words. using Theorem 14. Ù) ˆ orb(G. ù2 P Ù. ù2 ) g ˆ (ù1 g.9 De®nition Let G be transitive on Ù. Thus r(G. with á1 Tˆ á2 . á2 ) and (â1 . Ù) ˆ hð. 29.10 Corollary If G is 2-transitive on Ù. The case where equality holds is of particular interest.8 Proposition Let G act on Ù. Ù 3 Ù)X The next result is immediate from Proposition 29. G is 2-transitive if. ù) : ù P Ùg is an orbit of G on Ù 3 Ù. and hence certainly r(G. ðiX Now suppose G is transitive on Ù and jÙj . Ù) > 2. â2 ) in Ù 3 Ù. 29. Then G also acts on Ù 3 Ù in the way de®ned above. 1.17. and hð. Then r(G. with permutation character ð. 29. where ÷ is an irreducible character of G. Ù) ˆ 2.7 De®nition The number of orbits of G on Ù 3 Ù is called the rank of G on Ù. g P G. with permutation character ð. Ù). jg ’ fk.1(2). jg. Each cycle-shape (including 1-cycles) is a sequence ë ˆ (ë1 . (3) Consider the action of Sn on pairs de®ned in Example 29. The linear transformation from V to V which sends v1 3 w1 . Since jÙj ˆ q ‡ 1 (see Exercise 1 at the end of the chapter).11 Examples (1) The symmetric group Sn is 2-transitive on f1. as claimed. v2 3 w2 is therefore invertible. In fact it is easy to see that the orbits of G ˆ Sn on Ù 3 Ù are Ä. lgj ˆ 1g.1. Then v1 . 3g).16. where Ä is as above. w2 are both bases of V. f2. lgj ˆ 0gX Thus hð. 4g) to (f1. ë s ) of positive integers ë i such that ë1 > ë2 > F F F > ë s and ë1 ‡ F F F ‡ ë s ˆ n. Some irreducible characters of S n By Theorem 12. lg) : jfi. with n > 4. for each partition ë. 19. Ù) ˆ 3. and so ð ˆ 1 G ‡ ÷ ‡ ø.10 is the character ø0 of degree q in Theorem 28. q) given in Example 29.Permutations and characters 343 29. let (hv1 i. there is no element of Sn which sends (f1. 19. since. hv2 i 3 hw2 i. A key aim is therefore to construct.17). 2g. for example. jg ’ fk.5. hw2 i) be two pairs of distinct 1-spaces in Ù. F F F . (2) Consider the action of G ˆ GL(2. We claim that G is 2-transitive on Ù.15 we know that the conjugacy classes of Sn are in bijective correspondence with the set of all possible cycle-shapes of permutations. By Theorem 15. F F F .1(3). f3. This action is not 2-transitive. the irreducible characters of Sn are also in bijective correspondence with the partitions ë of n.3. q) which sends hv1 i 3 hw1 i. the irreducible character ÷ of G given by Corollary 29. ng. 2g. where ÷ and ø are irreducible characters of Sn . Also An is 2-transitive. Ä1 and Ä2 . an irreducible character ÷ ë . Hence G is 2-transitive on Ù. hv2 i) and (hw1 i. fk. giving an element of GL(2. and we call such a sequence a partition of n. Hence these groups have an irreducible character ÷ given by ÷( g) ˆ jfix( g)j À 1X We have seen this irreducible character in a number of previous examples (see 18. provided n > 4. To see this. v2 and w1. and Ä1 ˆ f(fi. ði ˆ r(G. Ä2 ˆ f(fi. Here Ù is the set of all 1-dimensional subspaces of the 2-dimensional vector space V. fk. jg. lg) : jfi. Proof By Proposition 29. k) such that ð k ˆ ÷ ( n) ‡ ÷ ( nÀ1. 2. ÷ ( nÀ k. Observe that   n ð k (1) ˆ jI k j ˆ X k 29. k) X In particular. and much more. ÷ ( nÀ1.13 Theorem Let m ˆ na2 if n is even. de®ne I k to be the set consisting of all subsets of I of size k. F F F . ð l i ˆ orb(G. F F F . J s ˆ f(A. in a natural way. The ideas can be developed to carry out the aim in general. For an integer k < na2. but we do not do this. k) . Just as in Example 29. James listed in the Bibliography. The orbits of G ˆ Sn on I k 3 I l are easily seen to be J 0 . by induction on k. let Ag ˆ fi1 g. F F F . where for 0 < s < l. we refer you to the book by G. and m ˆ (n À 1)a2 if n is odd. k) ˆ ð k À ð kÀ1 .2) . ik ggX Let ð k be the permutation character of G in its action on I k . ÷ ( nÀ1. Let G ˆ Sn and I ˆ f1. B) P I k 3 I l : jA ’ Bj ˆ sgX Hence orb(G. hð k .344 Representations and characters of groups of Sn .1) ‡ F F F ‡ ÷ ( nÀ k.1(2) we can de®ne an action of G on I k as follows: for any subset A ˆ fi1 . k). j 29. ÷ ( nÀ k.1) ‡ F F F ‡ ÷ ( nÀ k. ÷ ( nÀ m.12 Proposition If l < k < na2. m) such that for all k < m. F F F .1) . Proof We prove the existence of irreducible characters ÷ ( n) .10. Then Sn has distinct irreducible characters ÷ ( n) ˆ 1 G . I k 3 I l ). ð k ˆ ÷ ( n) ‡ ÷ ( nÀ1. I k 3 I l ) ˆ l ‡ 1. We shall apply the material of this chapter to carry out this aim in the case where ë ˆ (n À k.13 below). giving the conclusion. ÷ ( nÀ2. then hð k . This holds for k ˆ 1 by Corollary 29. Now assume the statement holds for all values less than k. on the character theory of Sn . if you want to see this.6. F F F J l . Then . J 1 . ik g P I k and any g P G. ng. a 2-part partition (see Theorem 29. ð l i ˆ l ‡ 1. F F F .1) . hð k . we have ð k ˆ ÷ ( n) ‡ ÷ ( nÀ1. k.Permutations and characters 345 there exist irreducible characters ÷ ( n) . The irreducible characters ÷ ( nÀ k. ði.1) ‡ F F F ‡ ÷ ( nÀ k. 1 G i. ÷ (5. ÷ (3. ð1 i ˆ 2. k) ˆ ð k À ð kÀ1 makes it easy to calculate the values of the characters ÷ ( nÀ k. . 2. k) . as required. the irreducible characters ÷1 . k) . ÷ ( nÀ k‡1. the degree is     n n ( nÀ k. j 29. The orbits are the equivalence classes in Ù of the relation de®ned by á $ â D á g ˆ â for some g P G.1) ‡ F F F ‡ ÷ ( nÀi.2) . ÷7 .1) . F F F . k) corresponding to 2-part partitions have values given by Theorem 29. ð k i ˆ k ‡ 1X It follows that ð k ˆ ð kÀ1 ‡ ÷ for some irreducible character ÷.17. ÷ ( nÀ1.14 Examples (1) The formula ÷ ( nÀ k. ÷ (4. If G acts on Ù then CÙ is the permutation module. suppose n ˆ 7 and let us calculate the value of the irreducible character ÷ (5. Summary of Chapter 29 1.2) (123) ˆ ð2 (123) À ð1 (123) ˆ jfix I 2 (123)j À jfix I 1 (123)j ˆ 6 À 4 ˆ 2X (2) In the character table of S6 given in Example 19. k) . kÀ1) such that ð i ˆ ÷ ( n) ‡ ÷ ( nÀ1. hð k .i) for all i . hð k . Now by Proposition 29. F F F . k) ÷ (1) ˆ ð k (1) À ð kÀ1 (1) ˆ À X k kÀ1 As another example. hð k . where ð( g) ˆ jfixÙ ( g)j. The number of orbits is equal to hð. Ù) is the number of orbits of G on Ù 3 Ù. The size of the orbit ù G containing ù is jù G j ˆ jG : Gù j. ð kÀ1 i ˆ k. and r(G. ÷9 are equal to ÷ (6) . Writing ÷ ˆ ÷ ( nÀ k. An action of G on Ù is a homomorphism G 3 Sym(Ù). For example.2) on a 3-cycle: ÷ (5. 4. 3.12. The irreducible characters of Sn are in bijective correspondence with partitions of n.1) .3) . and the corresponding character of G is ð. If G is 2-transitive then r(G. ÷3 . 1 G i ˆ 1. Ù) ˆ 2 and ð ˆ 1 G ‡ ÷ with ÷ irreducible. Ù) ˆ hð. respectively.13. The rank r(G. 1(2). Let ð be the permutation character of G in this action.18. and take inner products with the irreducible characters of G given in 28. g P G. and de®ne an action ö : G 3 Sym(V à ) by v( gö) ˆ v g for v P V Ã. and ®nd the kernel of ö. 2 de®ne Ù i to be the set of right cosets of H i in G. q) and let V ˆ F2 as in Example 29.1(4).5. Let G be a ®nite group. and let H 1 . Prove that G contains an element g such that jfixÙ ( g)j ˆ 0. and ÷ 3 ÷ is the irreducible character of G 3 G given by Theorem 19. then H 1 ˆ H 2 X Give an example to show that this need not be the case in general. For i ˆ 1. then jÙj ˆ q ‡ 1. g. (c) Show that the rank of this action r(G 3 G.346 Representations and characters of groups Exercises for Chapter 29 1.5). Show that if Ù is the set of all 1-dimensional subspaces of a 2dimensional vector space over F q (as in Example 29.) 4. where the sum is over all irreducible characters ÷ of G. Let q V à ˆ V À f0g. (b) Find the stabilizer (G 3 G)1 of the identity 1 P G. so that G acts on Ù i as in Example 29. and de®ne a function ö : G 3 G 3 Sym(G) by x(( g. which is transitive.) . and the permutation character ð is ðˆ ˆ ÷ ÷ 3 ÷. 2. (Such an element is called a ®xed-point-free element of G. h P G. 5. Let G be a ®nite group. 3. G) is equal to the number of conjugacy classes of G. Let G be a ®nite group acting transitively on a set Ù of size greater than 1. H 2 be subgroups of G. q) (the latter are given by Theorem 28. Prove that if G is abelian. Let G ˆ GL(2. let ð i be the permutation character of G in the action on Ù i . (a) Show that ö is an action of G 3 G on G.1(3)). ( Hint: one way to do this is to write down the values of ð on the conjugacy classes of G. Suppose that ð1 ˆ ð2 . h)ö) ˆ g À1 xh for all x. Decompose ð as a sum of irreducible characters of GL(2. and calculate its value on the elements (12) and (123) of Sn . prove that ð( nÀ2. Let Sn act on Ù in the obvious way (namely.1) ? . and let the permutation character of Sn in this action be ð( nÀ2. j) with i. which irreducible character is equal to ÷ (4.Permutations and characters 347 6.1.2) ‡ ÷. jg) for g P Sn ). Writing ÷ ˆ ÷ ( nÀ2.1. j P f1. calculate the degree of ÷ ( nÀ2. (i.1) . By considering inner products as in the proof of Theorem 29. and let Ù be the set of all ordered pairs (i. F F F .1. In the character table of S6 given in Example 19.1) . j) g ˆ (ig.1.17. where ÷ is an irreducible character.1) ‡ ÷ ( nÀ2.1.13. ng and i Tˆ j.1) ˆ 1 ‡ 2÷ ( nÀ1. Let n be a positive integer.1) . . as we shall demonstrate. . The examples which we have come across so far ± ®nding the centre of the group. Using a little group theory and a lot of character theory we shall carry out such a study in the case where C ˆ D8 . and they can be used to investigate the subgroup structure of G.30 Applications to group theory There are several ways of using the character theory of a group to determine information about the structure of the group. . Recall from Proposition 12. The ®rst involves doing arithmetic with character values to determine certain numbers. In this chapter we present some rather deeper applications. known as the class algebra constants.19 motivates the study of simple groups containing an involution with centralizer isomorphic to a given group C. The second application takes this much further: the Brauer±Fowler Theorem 23. These constants carry information about the multiplication in G. 348 . C l form a basis for the centre of the group algebra CG € (where C i ˆ gPC i g). 30. . . . and so on ± require little calculation. . . the dihedral group of order 8. seeing whether or not the group is simple.1 Proposition There exist non-negative integers aijk such that for 1 < i < l and 1 < j < l.22 that the class sums C1 . Cl be the distinct conjugacy classes of G. Class algebra constants Let G be a ®nite group and let C1 . 4 Theorem Let gi P Ci for 1 < i < l. 30. b) with a P Ci .2 De®nition The integers aijk in the formula Ci C j ˆ l ˆ kˆ1 aijk C k are the class algebra constants of G. The result follows. the constants aijk determine the product of any two elements in the centre Z(CG) of the group algebra. you might suspect that the class algebra constants are determined by the character table of G. Then for all i. the numbers aijk carry information about the multiplication in G: (30X3) For all g P Ck and all i. j Another way of looking at Proposition 30. and is independent of the chosen element g of Ck . k. b P Cj and ab ˆ g. C l is a basis of Z(CG). since C1 . . j. . so it must be a linear combination of C1 . . . . 30. j we have aijk ˆ the number of pairs (a. . . we have aijk ˆ ˆ ÷( g i )÷( g j )÷( g k ) jGj jCG ( g i )j jCG ( g j )j ÷ ÷(1) where the sum is over all the irreducible characters ÷ of G. C l . This number is a non-negative integer. .Applications to group theory Ci C j ˆ l ˆ kˆ1 349 aijk C k X Proof For g P Ck the coef®cient of g in the product C i C j is equal to the number of pairs (a. From their very de®nition.1 is to note that C i C j belongs to Z(CG). Our next theorem shows that this is indeed the case. As the centre of the group algebra plays an important role in representation theory. b P Cj and ab ˆ gX Also. . b) with a P Ci . these results can readily be proved directly.6 Example In this example we shall use the class algebra constants to prove some facts about the elements and subgroups of the symmetric group S4 . Multiply both sides of equation (30. this yields aijk ˆ ˆ ÷( g i )÷( g j )÷( g k ) jGj X jCG ( g i )j jCG ( g j )j ÷ ÷(1) j Examples 30.350 Representations and characters of groups Proof Let ÷ be an irreducible character of G.1. and let U be a CGmodule with character ÷. to obtain l ˆ mˆ1 aijm ˆ ÷( g m )÷( g k ) ÷ jCG ( g m )j ˆ ˆ ÷( g i )÷( g j )÷( g k ) jGj X jCG ( g i )j jCG ( g j )j ÷ ÷(1) By the column orthogonality relations.5) by ÷( g k ) and sum over all irreducible characters ÷ of G. Let G ˆ S4 . but they serve as a useful illustration of the method. Theorem 16. Then by Lemma 22.4(2). By Section 18.7. we deduce that l ˆ mˆ1 aijm ÷( g i )÷( g j ) ÷( g m ) jGj ˆ X jCG ( g m )j jCG ( g i )j jCG ( g j )j ÷(1) Pick k with 1 < k < l. the character table of G is as shown: . for all u P U we have uC i ˆ Therefore uC i C j ˆ and l ˆ mˆ1 jGj÷( g i ) uX jCG ( g i )j÷(1) ÷( g i )÷( g j ) jGj2 u jCG ( g i )j jCG ( g j )j (÷(1))2 aijm uC m ˆ l ˆ mˆ1 aijm jGj÷( g m ) uX jCG ( g m )j÷(1) Since C i C j ˆ (30X5) € m aijm C m . 4.Applications to group theory Character table of S4 Class Ci gi |CG ( gi )| ÷1 ÷2 ÷3 ÷4 ÷5 C1 1 24 1 1 2 3 3 C2 (1 2) 4 1 À1 0 1 À1 C3 (1 2 3) 3 1 1 À1 0 0 C4 (1 2)(3 4) 8 1 1 2 À1 À1 351 C5 (1 2 3 4) 4 1 À1 0 À1 1 (1) We use Theorem 30.3). S4 is generated by a and b. we have x 4 ˆ 1. We deduce the fact (which we already know from Exercise 18. 43 so S4 has elements a of order 2 and b of order 3 with ab of order 4.1) that S4 has a subgroup which is isomorphic to D8. bl  D8. 24 a235 ˆ X (1 ‡ 1) ˆ 4. (2) By Theorem 30. We deduce from this that S4 does not have a subgroup which is isomorphic to the quaternion group Q8 : for Q8 does have two elements of order 4 with product of order 4. (3) Finally. We supply a . aÀ1 xa ˆ ba ˆ (ab)À1 ˆ x À1 . b: a2 ˆ b3 ˆ (ab)4 ˆ 1iX In other words. by (30.   24 1 1 a245 ˆ X 1‡1‡ ‡ ˆ 2X 48 3 3 Hence S4 has elements a. Writing x ˆ ab. S4 does not possess elements a. b of order 2 such that ab has order 4. so ka. In fact. and all products of elements of S4 are determined by the given relations.4 to calculate the class algebra constant a555 :   24 1 À1 0 À1 1 ‡ ‡ ‡ ‡ ˆ 0X a555 ˆ X 44 1 1 2 3 3 Hence. it can be shown that S4 has a presentation as follows: S4 ˆ ha. b of order 4 such that the product ab also has order 4. 7) with H isomorphic to S4 . 7) is as follows.4 to ®nd a subgroup H of the simple group PSL (2. Character table of PSL (2. 30.7 Example We use Theorem 30. We found in Chapter 27 that the character table of G ˆ PSL (2.352 Representations and characters of groups proof in the solution to Exercise 30. gi Order of gi |CG ( gi )| ÷1 ÷2 ÷3 ÷4 ÷5 ÷6 p where á ˆ (À1 ‡ i 7)a2. That such a subgroup exists is not obvious. and it is quite tricky to construct directly. g1 1 168 1 7 8 3 3 6 g2 2 8 1 À1 0 À1 À1 2 g3 4 4 1 À1 0 1 1 0 g4 3 3 1 1 À1 0 0 0 g5 7 7 1 0 1 á á À1 g6 7 7 1 0 1 á á À1 .6 ± in the meantime. you may wish to puzzle out the relevance of the ®gure above. 7) Class rep. 6. y has order 3 and xy has order 4. a dihedral group of order 8. we know that S4 ˆ ha. From Example 30. so H is isomorphic to S4 . V4 . We have chosen to present this result because it provides a wonderful illustration of the use of character theory in the service of group theory.Applications to group theory We calculate the class algebra constant a243 . A4 or S4 (see Example 12. Now Ker ö. The Brauer programme The Brauer±Fowler Theorem 23. b: a2 ˆ b3 ˆ (ab)4 ˆ 1iX Hence there is a homomorphism ö from S4 onto H (ö sends a to x and b to y). Then G has order 168 or 360. given a ®nite group C. Since H has an element of order 4. 30. G contains elements x and y such that x has order 2. is {1}. Gorenstein listed in the Bibliography). . an effort which was ®nally completed in the early 1980s (see the book by D. being a normal subgroup of S4 . all ®nite simple groups G possessing an involution t such that C G (t)  C.   168 1 a243 ˆ X 1 ‡ ‡ 0 ‡ 0 ‡ 0 ‡ 0 ˆ 8X 83 7 353 Hence. Let H be the subgroup kx. C2 or {1}.20).19 states that there are only ®nitely many non-isomorphic ®nite simple groups containing an involution with a given centralizer. This programme formed an important part of the effort of many mathematicians to classify all the ®nite simple groups.4. This fact led Brauer to initiate a programme to ®nd. The next result carries out part of Brauer's programme in the case where C ˆ D8 . we conclude that H  S4 X Thus we have shown that PSL (2. yl of G.10. by (30. It determines the possible orders of simple groups G having an involution t such that C G (t)  D8 .3). By Theorem 30. S4 aKer ö  H. S3 . namely xy. By Theorem 1. 7) has a subgroup which is isomorphic to S4 .8 Theorem Let G be a ®nite non-abelian simple group which has an involution t such that C G (t)  D8 . This means that in its action on Ù. hence is an odd permutation. we require a couple of preliminary results. such a subgroup is call a Sylow p-subgroup of G.2(4)). If u is an involution in G. Let Ù be the set of right cosets Qx of Q in G. Hence fixÙ (u) ˆ Æ. Now consider fixÙ (u) ˆ fù P Ù : ùu ˆ ùg If Qx P fixÙ (u). (3) if R is a subgroup of G with jRj ˆ p c for some c. Hence the subgroup f g P G : g acts as an even permutation on Ùg . We shall not prove this. 30.354 Representations and characters of groups Observe that PSL(2. where m is odd since P is a Sylow 2-subgroup. if P Q are Sylow p. then u is conjugate to an element of Q. then there exists g P G such that Q ˆ g Pg). and A6 is a simple group of order 360 with this property (see Exercise 7 at the end of the chapter). contrary to assumption. and let P be a Sylow 2subgroup of G. where a. a basic result in ®nite group theory. 7) and A6 are the only simple groups of order 168 or 360. Observe that jÙj ˆ 2jG : Pj ˆ 2m. Proof Suppose u is not conjugate to an element of Q. and de®ne an action of G on Ù by (Qx) g ˆ Qxg for x. Suppose Q is a subgroup of P with jP : Qj ˆ 2. the involution u is a product of m disjoint 2-cycles. but refer you to Theorems 18. Then (1) G contains a subgroup of order p a.3 and 18. one can show that PSL(2. b are positive integers and p T j b.8. then Qxu ˆ Qx and hence xux À1 P Q.e. À1 subgroups. 7) is a simple group of order 168 having an involution with centralizer D8 (see Lemma 27.10 Lemma Let G be a ®nite non-abelian simple group.1). The ®rst is Sylow's Theorem.9 Sylow's Theorem Let p be a prime number.4 of the book by J. then there is a Sylow p-subgroup of G containing R. g P G (see Example 29. and let G be a ®nite group of order p a b. Using some rather more sophisticated group theory than that covered in this book. Before embarking upon the proof of Theorem 30. 30. (2) all Sylow p-subgroups are conjugate in G (i. Fraleigh listed in the Bibliography. 1(1) we have Z(P) Tˆ 1. If n÷ > 0 for all ÷ then of course ø is a character. By Lemma 26. In particular.19 and Corollary 21. øi ˆ n2 ÷ for a generalized character ø as above. and each n÷ P Z. the degree ø(1) can be 0 or negative for a generalized character ø. This is impossible since G is non-abelian and simple.Applications to group theory 355 is a normal subgroup of index 2 in G. we de®ne the induced generalized character ø 4 G by ø 4 G ˆ (á 4 G) À ( â 4 G) where ø ˆ á À â as above. ⠈ À n÷ ÷X n÷ >0 n÷ . The generalized character ø can be expressed as a difference á À â.12)). so Z(P) < Z(D) ˆ hti. and hence Z(P) ˆ hti. if H is a subgroup of a group G. Certainly t commutes with itself. ÷i ˆ n÷ . so t P D.9(3). and hence Z(D) ˆ hti. there is a Sylow 2-subgroup P of G such that D < P.0 Finally.20 hold for generalized characters ø. where á and â are characters of H: take ˆ ˆ ሠn÷ ÷. This is simply a class function of the form ˆ øˆ n÷ ÷ x where the sum is over all the irreducible characters of H. but this need not be the case for a generalized character. Therefore P < . j We also need to introduce the idea of a generalized character of a group H. Then Z(P) < C G (t) ˆ D. The centre of D8 is a cyclic group of order 2 (see (12. we have t P Z(D).8 Let G be a ®nite non-abelian simple group with an involution t such that C G (t) ˆ D  D8 . It is clear from this de®nition that the formulae for the values of ø 4 G given in Proposition 21. the centre of D. By Theorem 30. Notice also that the orthogonality relations give the usual inner products ˆ hø. Proof of Theorem 30. hø. and as t commutes with all elements of D. This contradiction completes the proof. observe ®rst that by Frobenius reciprocity. Proposition 21. By Lemma 30. giving (30.12) that .20).11) D is a Sylow 2-subgroup of G. The rest is character theory. In other words. This is all the group theory we will need for the proof. è 4 Gi ˆ h(è 4 G) 5 D. we therefore have h(è 4 G) 5 D.356 Representations and characters of groups C G (t) ˆ D. and if C ’ g À1 Cg ˆ C then g P D. and so it follows from (30. 1 G i ˆ h1 C À ë. 1 C i ˆ 1. Also (è 4 G)(1) ˆ 0 (see Corollary 21.11). a generalized character of D. whence y À1 cy ˆ cÆ1 and è( y À1 cy) ˆ è(c). t G is the unique conjugacy class of involutions in G. Next. Referring to the character table of D8 in Example 16. èi ˆ hè. We summarise what we have proved so far: (30. (In particular. Then è takes the value 2 on a. we must have g À1 tg ˆ t. Since è vanishes on D À C. Write D ˆ ha.10. èi Now for 1 Tˆ c P C. if y À1 cy P C then y P D. è 4 Gi ˆ 3X To see this. è(1) ˆ 0. As t ˆ a2 is the only such involution. Let ë be the linear character of C such that ë(a) ˆ i. every involution of G is conjugate to an involution in C. And if y À1 cy P D À C then è( y À1 cy) ˆ 0. D is a Sylow 2-subgroup of G. and 0 elsewhere. hè 4 G. Let C ˆ hai be the cyclic subgroup of index 2 in D. we have è ˆ ÷1 ‡ ÷2 À ÷5 . Since t ˆ c or c2.12). and so P ˆ D. for any g P G we have C ’ gÀ1 Cg ˆ f1g or C. we conclude that t G is the unique conjugacy class of involutions in G. èi ˆ 3. èi ˆ 3. bi where a4 ˆ b2 ˆ 1 and bÀ1 ab ˆ aÀ1 . Now hè 4 G. We next establish (30X12) hè 4 G. hence g P C G (t) ˆ D and so g À1 Cg ˆ C. let g P G and suppose that g À1 cg P C for some non-identity element c P C.) Hence hè. and de®ne è ˆ (1 C 4 D) À (ë 4 D). aÀ1 . the value 4 on t. Then t ˆ a2 .19 gives 1 ˆ • À1 (è 4 G)(c) ˆ è( y cy)X 8 yPG By (30.3(3). It follows that (è 4 G)(c) ˆ è(c). Hence Theorem 30.4 yields the following. and hence x P D by (30.15) we deduce 2 3 á(t)2 â(t)2 jGj 1 ‡ (30X16) À ˆ 28 X á(1) â(1) This equation gives us enough number-theoretic information about jGj to ®nish the proof fairly quickly. Since we have shown that (è 4 G)(t) ˆ è(t) ˆ 4. (30. jDj2 ÷ ÷(1) where the sum is over all irreducible characters ÷ of G. We shall calculate the inner product of ã and è 4 G in two ways. from (30. ãi ˆ h1 C À ë. where á.3).13) we have . ã 5 Ci. ã 5 Ci ˆ 1 X4X((1 À i) ‡ 2 ‡ (1 ‡ i)) ˆ 4X jCj Hence from (30. Note that by Corollary 13. de®ne ã( g) to be number of ordered pairs (x. similarly y P D. Now calculation in D8 shows that ã(c) ˆ 4. If we write t G ˆ C i and g lies in the conjugacy class C k of G. By (30.13) and (30. then ã( g) ˆ a iik in the notation of (30. then x À1 cx ˆ yx ˆ cÀ1 . Therefore h1 C À ë. Write d ˆ á(1) and e ˆ á(t) P Z. we have now proved the following.13) We have è 4 G ˆ 1 G ‡ á À â. First. ãi ˆ 1‡ (30X15) À X 64 á(1) â(1) On the other hand. â are irreducible. by Frobenius Reciprocity.Applications to group theory è 4 G ˆ 1 G ‡ á À â. â are irreducible characters of G. Consider ã(c) for 1 Tˆ c P C.10. y P t G .14) we have 2 3 jGj á(t)2 â(t)2 hè 4 G. á(t) and â(t) are integers. If c ˆ xy with x. 357 where á.11).14) We have ㈠jGj ˆ ÷(t)2 ÷. We now introduce another class function of G into the picture. hè 4 G. For g P G. (30. 1 ‡ á(1) À â(1) ˆ 0 and 1 ‡ á(t) À â(t) ˆ 4. y) P t G 3 t G such that g ˆ xy. we deduce that d ‡ 2 ˆ 23 . suppose that e ˆ 2. giving d ˆ 6 and jGj ˆ 168.8. Suppose now that e ˆ 1. d d‡1 whence jGj ˆ 28 d(d ‡ 1) X (d À 1)2 Now the highest common factor hcf (d À 1.8. and hcf (d À 1.16) yields jGj ˆ 28 d(d ‡ 1) X (d ‡ 2)2 Reasoning as above. Then (30. one of the possibilities in the conclusion of Theorem 30. Moreover. Hence (d À 1)2 must divide 210 .4(2).358 Representations and characters of groups â(1) ˆ d ‡ 1. the class algebra constants of G can sometimes be used to determine whether or not G has a subgroup which isomorphic to H. d ‡ 1) is 1 or 2. Then (30. Finally. by using the formula aijk ˆ ˆ ÷( g i )÷( g j )÷( g k ) jGj X jCG ( g i )j jCG ( g j )j ÷ ÷(1) 2.16) gives   1 4 jGj 1 ‡ À ˆ 28 . This completes the proof of Theorem 30. â(t) ˆ e À 3X From the column orthogonality relations 16. we have 8 ˆ jC G (t)j > 1 ‡ á(t)2 ‡ â(t)2 ˆ 1 ‡ e 2 ‡ (e À 3)2 . and so d À 1 ˆ 2 r with r < 5. j Summary of Chapter 30 1. giving jGj ˆ 360. from which it follows that e ˆ 1 or 2. The class algebra constants aijk are given by ˆ Ci C j ˆ aijk C k X k They can be calculated from the character table. a Sylow 2-subgroup of G has order 8. Given groups G and H. It follows that r ˆ 3 and d ˆ 9. . d) ˆ 1. b has order 3 and ab has order 7. and that G has the character table shown. 11) is given in the solution to Exercise 27. Suppose that G is a group. it can be shown that any simple group possessing an involution with centralizer isomorphic to D8 must have order 168 or 360. to prove that PSL (2. (a) Show that G is a simple group of order 360. 7) contain a subgroup isomorphic to D14 ? (Hint: D14 ˆ ka. (b) Use the Frobenius±Schur Count of Involutions to obtain an .) For the next three exercises.Applications to group theory 359 3. The character table of PSL (2. b: a2 ˆ b3 ˆ (ab)5 ˆ 1iX 3. given at the end of Chapter 27. g1 ÷1 ÷2 ÷3 ÷4 ÷5 ÷6 ÷7 1 5 5 8 8 9 10 p g2 1 1 1 0 0 1 À2 g3 1 À1 À1 0 0 1 0 p 5)a2. b: a2 ˆ b2 ˆ 1. 11) contain a subgroup which is isomorphic to A5 ? 4. ⠈ (1 À 5. (ab)7 ˆ 1l. Does PSL (2. 7). 2. Prove that A5 is characterized by its character table ± that is. together with lots of ingenious character theory. then G  A5 .6.13). you may assume that A5 is a simple group. if G is a group with the same character table as A5 (see Example 20. Using Sylow's Theorem. and that A5 has the following presentation: A5 ˆ ha. Exercises for Chapter 30 1. g4 1 2 À1 À1 À1 0 1 g5 1 À1 2 À1 À1 0 1 g6 1 0 0 á â À1 0 g7 1 0 0 â á À1 0 where á ˆ (1 ‡ 5)a2. Use the character table of PSL (2. Does PSL (2. 7) contains elements a and b such that a has order 2. 9. p). (d) Using Exercise 23. 8. (Hint: look for a suitable simple group PSL(2.6(3) to show that every group G which is generated by two elements a and b which satisfy a2 ˆ b3 ˆ (ab)4 ˆ 1 has order at most 24. 7. 7) and A6 are simple groups of order 168.360 Representations and characters of groups upper bound for the number of involutions in G. and deduce that g2 has order 2 and g3 has order 4. (c) Prove that G has a subgroup H which is isomorphic to A5 . Find a simple group G having an involution C G (t)  D16 . both of which contain an involution with centralizer isomorphic to D8 . Use the ®gure which appears in Example 30. 6.) t such that . Prove that PSL(2. show that G  A6 . 360 respectively. In the ®rst edition of his book Theory of groups of ®nite order (1897). Let á be an algebraic number.2) which is concerned with character values. and let p(x) be a monic polynomial over Q of smallest possible degree having á as a root.31 Burnside's Theorem One of the most famous applications of representation theory is Burnside's Theorem. until H. Bender found one in 1972. Then p(x) is unique and irreducible. which states that if p and q are prime numbers and a and b are positive integers. Indeed. then no group of order pa q b is simple. many later attempts to ®nd a proof which does not use representation theory were unsuccessful. if ù is an nth root of unity then the minimal poly361 . but it was only after studying Frobenius's new theory of group representations that he was able to prove the theorem in general. The roots of p(x) are called the conjugates of á. An algebraic number is a complex number which is a root of some non-zero polynomial over Q. b. In order to establish this lemma we require some basic facts about algebraic integers and algebraic numbers. For example. which we now describe. A preliminary lemma We prepare for the proof of Burnside's Theorem with a lemma (31. it is called the minimal polynomial of á. see for instance the book by Pollard and Diamond listed in the Bibliography. We call a polynomial in x monic if the coef®cient of the highest power of x in it is 1. We omit proofs of these ± for a good account. Burnside presented group-theoretic arguments which proved the theorem for many special choices of the integers a. see Pollard and Diamond.1) can be proved easily using some Galois theory. j÷( g)a÷(1)j . where á9 is a conjugate of á. . and it turns out that the minimal polynomial of á also has integer coef®cients. where á9 is a conjugate of á and â9 is a conjugate of â.1). ù9 are roots of unity.362 Representations and characters of groups nomial of ù divides x n À 1.2 Lemma Let ÷ be a character of a ®nite group G. . Chapter V. 1 then ÷( g)a÷(1) is not an algebraic integer. and let g P G. ‡ ù d . . . Now suppose that ÷( g)a÷(1) is an algebraic integer and j÷( g)a÷(1)j . and let p(x) be the minimal polynomial of ã. If á is an algebraic integer. . Alternatively. For an elementary proof of this. it follows that j÷( g)a÷(1)j < 1. 1. Write 㠈 ÷( g)a÷(1). . so ÷( g)a÷(1) ˆ (ù1 ‡ X X X ‡ ù d )adX Since |÷( g)| ˆ |ù1 ‡ .1) Let á and â be algebraic numbers. Hence each conjugate of ã has 1 d . Moreover. if r P Q then every conjugate of rá is of the form rá9. We shall require the following fact about conjugates: (31. Section 3. ‡ ù d | < |ù1 | ‡ . We prove that ÷( g) ˆ 0. and so every conjugate of ù is also an nth root of unity. so that p(x) ˆ x n ‡ a nÀ1 x nÀ1 ‡ X X X ‡ a1 x ‡ a0 where ai P Z for all i. By (31. 31. then á is a root of a monic polynomial with integer coef®cients (see Chapter 22). Then j÷( g)a÷(1)j < 1. . (31. . Then every conjugate of á ‡ â is of the form á9 ‡ â9.9 we have ÷( g) ˆ ù1 ‡ . Proof Let ÷(1) ˆ d. . ‡ |ù d | ˆ d. and if 0 . . By Proposition 13. each conjugate of ã is of the form (ù9 ‡ X X X ‡ ù9 )ad 1 d where ù9 . where each ù i is a root of unity. Burnside's Theorem 363 modulus at most 1. It follows that if ë is the product of all the conjugates of ã (including ã), then jëj , 1X But the conjugates of ã are, by de®nition, the roots of the polynomial p(x), and the product of all these roots is equal to Æa0 . Thus ë ˆ Æa0 X Since a0 P Z and |ë| , 1, it follows that a0 ˆ 0. As p(x) is irreducible, this implies that p(x) ˆ x, which in turn forces 㠈 0. Thus ÷( g) ˆ 0, and the proof is complete. j Burnside's p a q b Theorem We deduce the main result, Theorem 31.4, from another interesting theorem of Burnside. 31.3 Theorem Let p be a prime number and let r be an integer with r > 1. Suppose that G is a ®nite group with a conjugacy class of size pr . Then G is not simple. Proof Let g P G with | gG | ˆ pr . Since pr . 1, G is not abelian and g Tˆ 1. As usual, denote the irreducible characters of G by ÷1 , . . . , ÷ k , and take ÷1 to be the trivial character. The column orthogonality relations, Theorem 16.4(2), applied to the columns corresponding to 1 and g in the character table of G, give 1‡ Therefore k ˆ iˆ2 k ˆ iˆ2 ÷ i ( g)÷ i (1) ˆ 0X ÷ i ( g) . ÷ i (1) 1 ˆÀ X p p Now À1a p is not an algebraic integer, by Proposition 22.5. Therefore, for some i > 2, ÷ i ( g)÷ i (1)a p is not an algebraic integer (see Theorem 22.3). Since ÷ i ( g) is an algebraic integer (Corollary 22.4), it follows 364 Representations and characters of groups that ÷ i (1)a p is not an algebraic integer; in other words, p does not divide ÷ i (1). Thus ÷ i ( g) Tˆ 0 and p T j ÷ i (1)X As | gG | ˆ pr , this means that ÷ i (1) and | gG | are coprime integers, and so there are integers a and b such that ajG:CG ( g)j ‡ b÷ i (1) ˆ 1X Hence a jGj÷ i ( g) ÷ i ( g) ‡ b÷ i ( g) ˆ X jCG ( g)j÷ i (1) ÷ i (1) By Corollaries 22.10 and 22.4, the left-hand side of this equation is an algebraic integer; and since ÷ i ( g) Tˆ 0, it is non-zero. Now Lemma 31.2 implies that j÷ i ( g)a÷ i (1)j ˆ 1X Let r be a representation of G with character ÷ i . By Theorem 13.11(1), there exists ë P C such that gr ˆ ëIX Let K ˆ Ker r, so that K is a normal subgroup of G. Since ÷ i is not the trivial character, K Tˆ G. If K Tˆ {1} then G is not simple, as required; so assume that K ˆ {1}, that is, r is a faithful representation. Since gr is a scalar multiple of the identity, gr commutes with hr for all h P G. As r is faithful, it follows that g commutes with all h P G; in other words, g P Z(G)X Therefore Z(G) Tˆ {1}. As Z(G) is a normal subgroup of G and Z(G) Tˆ G, we conclude that G is not simple. j We now come to the main result of the chapter, Burnside's Theorem. 31.4 Burnside's paqb Theorem Let p and q be prime numbers, and let a and b be non-negative integers with a ‡ b > 2. If G is a group of order pa q b , then G is not simple. Proof First suppose that either a ˆ 0 or b ˆ 0. Then the order of G is a power of a prime, so by Lemma 26.1(1) we have Z(G) Tˆ {1}. Burnside's Theorem 365 Choose g P Z(G) of prime order. Then k gl v G and k gl is not equal to {1} or G. Hence G is not simple. Now assume that a . 0 and b . 0. By Sylow's Theorem 30.9, G has a subgroup Q of order q b . We have Z(Q) Tˆ {1} by Lemma 26.1(1). Let g P Z(Q) with g Tˆ 1. Then Q < CG ( g), so j g G j ˆ jG:CG ( g)j ˆ pr for some r. If pr ˆ 1 then g P Z(G), so Z(G) Tˆ {1} and G is not simple as before. And if pr . 1 then G is not simple, by Theorem 31.3. j In fact Burnside's pa q b Theorem leads to a somewhat more informative result about groups of order pa q b : (31.5) Every group of order p a q b is soluble. Here, by a soluble group we mean a group G which has subgroups G0 , G1 , . . . , Gr with 1 ˆ G 0 , G1 , X X X , G r ˆ G such that for 1 < i < r, GiÀ1 v Gi and the factor group Gi aG iÀ1 is cyclic of prime order. We sketch a proof of (31.5), using induction on a ‡ b. The result is clear if a ‡ b < 1, so assume that a ‡ b > 2 and let G be a group of order pa q b . By Burnside's Theorem 31.4, G has a normal subgroup H such that H is not {1} or G. Both H and the factor group Ga H have order equal to a product of powers of p and q, and these orders are less than pa q b . Hence by induction, H and Ga H are both soluble. Therefore there are subgroups 1 ˆ G0 v G1 v X X X v Gs ˆ H, 1 ˆ Gs a H v G s‡1 a H v X X X v Gr a H ˆ Ga H with all factor groups Gi aG iÀ1 of prime order. Then the series 1 ˆ G0 v G 1 v X X X v G r ˆ G shows that G is soluble. Summary of Chapter 31 1. If G has a conjugacy class of size pr ( p prime, r > 1), then G is not simple. 366 Representations and characters of groups 2. If |G| ˆ pa q b ( p, q primes, a ‡ b > 2), then G is not simple. Exercises for Chapter 31 1. Show that a non-abelian simple group cannot have an abelian subgroup of prime power index. 2. Prove that if G is a non-abelian simple group of order less than 80, then |G| ˆ 60. (Hint: use Exercise 13.8.) 32 An application of representation theory to molecular vibration Representation theory is used extensively in many of the physical sciences. Such applications come about because every physical system has a symmetry group G, and certain vector spaces associated with the system turn out to be RG-modules. For example, the vibration of a molecule is governed by various differential equations, and the symmetry group of the molecule acts on the space of solutions of these equations. It is on this application ± the theory of molecular vibrations ± that we concentrate in this ®nal chapter. In order to keep our treatment elementary, we stay within the framework of classical mechanics throughout. (Quantum mechanical effects can be incorporated subsequently, but we shall not go into this ± for an account, consult the book by D. S. Schonland listed in the Bibliography.) Symmetry groups Let V be R2 or R3, and for v, w P V, let d(v, w) denote the distance between v and w ± in other words, if v ˆ (x1 , x2 , . . .) and w ˆ ( y1, y2, . . .), then 3 r2ˆ 2 d(v, w) ˆ (xi À yi ) X An isometry of V is an invertible endomorphism W of V such that d(vW, wW) ˆ d(v, w) for all v, w P V X The set of all isometries of V forms a group under composition, called the orthogonal group of V, and denoted by O(V). Any rotation of R3 about an axis through the origin is an example 367 368 Representations and characters of groups of an isometry; so is any re¯ection in a plane through the origin. The endomorphism À1R3 which sends every vector v to Àv is another example of an element in the orthogonal group O(R3 ). It turns out that the composition of two rotations is again a rotation, and that for every isometry g in O(R3 ), either g or À g is a rotation (see Exercise 32.1). The orthogonal group O(R3 ) therefore contains a subgroup of index 2 which consists of the rotations. The same is true of the group O(R2 ). If Ä is a subset of V , where V ˆ R2 or R3, then we de®ne G(Ä) to be the set of isometries which leave Ä invariant ± that is, G(Ä) ˆ f g P O(V ): Ä g ˆ Äg (where Ä g ˆ {v g: v P Ä}). Then G(Ä) is a subgroup of called the symmetry group of Ä. The subgroup of G(Ä) the rotations in G(Ä) is called the rotation group of Ä. the rotation group of Ä in the symmetry group G(Ä) is 1 O(V ), and is consisting of The index of or 2. 32.1 Example Let V ˆ R2, and let Ä be a regular n-sided polygon, with n > 3, centred at the origin. The symmetry group of Ä is easily seen to be the dihedral group D2n , which was de®ned as a group of n rotations and n re¯ections preserving Ä (see Example 1.1(3)). Now let V ˆ R3, and again let Ä be a regular n-sided polygon (n > 3) centred at the origin. This time, G(Ä) ˆ D2 n 3 C2 ; the extra elements arise because there is an isometry which ®xes all points of Ä, namely the re¯ection in the plane of Ä. 32.2 Example Let Ä be a regular tetrahedron in R3 centred at the origin: An application of representation theory to molecular vibration 369 Label the corners of the tetrahedron 1, 2, 3, 4. We claim that each permutation of the numbers 1, 2, 3, 4 corresponds to an isometry of Ä. For example, the 2-cycle (1 2) corresponds to a re¯ection in the plane which contains the origin and the edge 34; similarly each 2-cycle corresponds to a re¯ection. Since S4 is generated by the 2-cycles, each of the 24 permutations of 1, 2, 3, 4 corresponds to an isometry, as claimed. No non-identity endomorphism of R3 ®xes all the corners of Ä, since Ä contains three linearly independent vectors. Therefore we have found all the isometries, and G(Ä)  S4 . Notice that the rotation group of Ä is isomorphic to A4 ; for example, (1 2)(3 4) corresponds to a rotation through ð about the axis through the mid-points of the edges 12 and 34, and (1 2 3) corresponds to a rotation through 2ða3 about the axis through the origin and the corner 4. Finally, observe that the group G(Ä) is unchanged if we take Ä to consist of just the four corners of the tetrahedron. 32.3 Example In this example we describe the symmetry groups of the molecules H2 O (water), CH3 Cl (methyl chloride) and CH4 (methane). The symmetry group of a molecule is de®ned to be the group of isometries which not only preserve the position of the molecule in space, but also send each atom to an atom of the same kind. The shapes of the three molecules are as follows. 370 Representations and characters of groups We always assume that the centroid of our molecule lies at the origin in R3 . The CH4 molecule has four hydrogen atoms at the corners of a regular tetrahedron, and a carbon atom at the centre of the tetrahedron. So the symmetry group of the molecule CH4 is equal to the symmetry group of the tetrahedron, as given in Example 32.2. This group is isomorphic to S4 , permuting the four hydrogen atoms among themselves and ®xing the carbon atom. As for the CH3 Cl molecule, this possesses a rotation symmetry a of order 3 about the vertical axis, and three re¯ection symmetries in the planes containing the C, Cl and one of the H atoms. If b is one of these re¯ections, then the symmetry group is f1, a, a2 , b, ab, a2 bg and is isomorphic to S3 , permuting the three H atoms and ®xing the C and Cl atoms. Finally, the H2 O molecule possesses two re¯ection symmetries, one in the plane of the molecule, and one in a plane perpendicular to this one passing through the O atom; and it has a rotation symmetry of order 2. Hence the symmetry group is isomorphic to C2 3 C2 . Vibration of a physical system We prepare for a description of the general problem with an example. 32.4 Example Suppose we have a spring stretched between two points P and Q on a smooth horizontal table, with equal masses m attached at the points of trisection of the spring: The masses are displaced slightly, and released. What can we say about the subsequent motion of the system? To investigate this problem, we let x1 and x2 be the displacements of the two masses at time t. We measure x1 from left to right, and x2 from right to left, as indicated in the ®gure above. Let k be the An application of representation theory to molecular vibration 371 stiffness of the spring ± in other words, if the extension in the spring is x, then the restoring force is kx. The spring pulls the left-hand mass towards P with force kx1 and towards Q with force Àk(x1 ‡ x2 ). By dealing with the right-hand mass similarly, we obtain the following equations of motion of the system: m1 ˆ Àkx1 À k(x1 ‡ x2 ) ˆ À2kx1 À kx2 , x m2 ˆ Àkx2 À k(x1 ‡ x2 ) ˆ Àkx1 À 2kx2 , x  where xi denotes the second derivative of xi with respect to t. These are second order linear differential equations in two unknowns x1 and x2 , so the general solution involves four arbitrary constants. We shall ®nd the general solution, using a method which can be applied in a much wider context.  Write x ˆ (x1 , x2 ), x ˆ (1 , x2 ) and q ˆ kam. Then the equations of x  motion are equivalent to the matrix equation   À2q Àq  x ˆ xA, where A ˆ (32X5) X Àq À2q Notice that A is symmetric. Hence the eigenvalues of A are real, and A has two linearly independent eigenvectors. It is this property which we wish to emphasize and exploit in the present example. Before we explicitly ®nd the eigenvectors of A, let us explain why they allow us to solve the equation of motion (32.5). Suppose that u is an eigenvector of A, with eigenvalue Àù2 . For an arbitrary constant â, let x ˆ sin (ùt ‡ â) uX Then  x ˆ Àù2 sin (ùt ‡ â) u ˆ sin (ùt ‡ â) uA ˆ xAX Thus x is a solution of the equation of motion. If u1 and u2 are linearly independent eigenvectors of A, with eigenvalues Àù2 and 1 Àù2 , respectively, then 2 á1 sin (ù1 t ‡ â1 ) u1 ‡ á2 sin (ù2 t ‡ â2 ) u2 (since uA ˆ Àù2 u) so it is the general solution. We now adopt this line of attack in the problem to hand. At the equilibrium position of each atom. À1). we assign three coordinate axes. It follows that when we apply Newton's Second Law of Motion. x1 ˆ Àx2 ˆ sin ( q . which we use to measure the displacement of the atom. We assume that the internal forces are linear functions of the displacements. x1 ˆ x2 ˆ sin ( (3q) t ‡ â1 ) and the vibration is p Here. t ‡ â2 ) (1. for the moment. â2 . with corresponding eigenvectors (1. They are as follows.5). Thus. Therefore the general solution of the equation of motion (32. À1) The general molecular vibration problem Suppose we have a molecule consisting of n atoms which vibrate under internal forces. â1 .5) is p p á1 sin ( (3q) t ‡ â1 ) (1. t ‡ â2 ) (1. that  x ˆ xAX row vector in R3 n which measures the and A is a 3n 3 3n matrix with real the internal forces. 1) ‡ á2 sin ( q . 1) and (1. For the matrix given in (32. á2 . entries which are determined by Assume. t ‡ â2 ) and the vibration is Mode 2: p sin ( q .372 Representations and characters of groups is a solution of the equation of motion which involves four arbitrary constants á1 . at each atom the three coordinate axes . 1) Mode 1: p Here. À1)X The solutions which involve just one eigenvector of A are called the normal modes of vibration. p sin ( (3q) t ‡ â1 ) (1.5).) Here x is the displacements of all the atoms. the state of the molecule at a given time is described by a vector in the 3n-dimensional vector space R3 n . the eigenvalues are À3q and Àq. we obtain equations which may be expressed in the form (32X6) (Compare (32. 7 Proposition All the eigenvalues of A are real. from physical considerations. and for this solution all the atoms vibrate with the same frequency. we look for normal modes of the system. then  x ˆ Àù2 sin (ùt ‡ â) u ˆ sin (ùt ‡ â) uA ˆ xAX If uA ˆ 0 and x ˆ (t ‡ â)u. The general solution of the equation of motion is a linear combination of the normal modes of vibration. for the general case. and A has 3n linearly independent eigenvectors.9 Proposition Each normal mode of vibration is a solution of the equation of motion (32. 32. To solve the equation of motion (32. 32. Now. A has real eigenvalues. It can be shown.8 De®nition A normal mode of vibration for our molecule is a vector in R3 n of one of the following forms: (1) sin (ùt ‡ â) u ( â constant) where Àù2 is a non-zero eigenvalue of A and u is a corresponding eigenvector. (2) (t ‡ â) u ( â constant) where u is an eigenvector of A corresponding to the eigenvalue 0.6). 32.6). In particular.An application of representation theory to molecular vibration 373 which we have chosen are at right angles to each other. Therefore we have the following proposition. and A has 3n linearly independent eigenvectors. then  x ˆ 0 ˆ (t ‡ â)uA ˆ xAX This proves that the normal modes of vibration are solutions of the . that in this special case the matrix A is symmetric. where our chosen coordinate axes are not necessarily at right angles to each other. Proof If uA ˆ Àù2 A and x ˆ sin (ùt ‡ â) u. the effect of changing coordinate axes is merely to replace A by a matrix which is conjugate to A. which we de®ne next. each element of G acts as an endomorphism of the space R3 n of displacement vectors. this can be a huge and unwieldy task if it is attempted directly ± even writing down the matrix A for a given molecule can be a painful operation! The symmetry group of the molecule and its representation theory can often be used to simplify greatly the calculation of the eigenvectors of A.374 Representations and characters of groups equation of motion (32. the general linear combination of normal modes involves 6n arbitrary constants. ù or 0) in a normal mode. so it is the general solution to the equation of motion (32. Let G be the symmetry group of the molecule in question. Since each normal mode involves an arbitrary constant. Thus.9 reduces the problem of solving the equations of motion to that of ®nding all the eigenvalues and eigenvectors of the 3n 3 3n matrix A.6). Since G permutes the atoms among themselves. However. and . Note that A can have no strictly positive eigenvalue. Use of the symmetry group We continue the discussion of the previous section. with eigenvector u. then x ˆ e ë t u would be a solution to the equation of motion.6) consists of second order differential equations in 3n unknowns). let v i denote a unit vector along coordinate axis i.10 Example Let g be the rotation of order 2 of the H2 O molecule: Assign coordinate axes at the initial positions of each atom as shown. all the atoms vibrate with the same frequency (namely. for if ë were p such an eigenvalue. interchanges v4 and v7. Then g ®xes v1 . and we shall describe a method for doing this. negates v2 and v3.6) (as (32. Therefore there exist 3n linearly independent normal modes. By construction. 32.7. j Proposition 32. which is nonsense. R3 n is an RG-module. by Proposition 32. and for 1 < i < 9. vA ˆ ëv for some ë P R. In effect. fx P R3 n : xA ˆ ëxgX We can now present the crucial proposition which allows us to exploit the symmetry group G of our molecule. Then v speci®es the directions and relative magnitudes of the displacements of the atoms from the equilibrium position when the molecule is vibrating in a normal mode of frequency ù. x2 . Therefore g acts on R9 as follows: (x1 . x4 . x8 . This shows that the eigenspace for Àù2 is an RG-submodule of R3 n . and let g P G. and (v g)A ˆ ë(v g) ˆ (ëv) g ˆ (vA) gX Hence (xg)A ˆ (xA) g for all x P R3 n . v g is an eigenvector of A. Therefore. Àx8 . Proof Let Àù2 be a non-zero eigenvalue of A. x9 ) g ˆ (x1 . x3 . Àx9 . and the eigenspaces of A are RG-submodules of R3 n .11 Proposition For all g P G and x P R3 n . and hence . For all g in G. A similar argument applies to the eigenspace for the eigenvalue 0. it tells us that the function x 3 xA (x P R3 n ) is an RG-homomorphism from R3 n to itself. j The idea now is to use representation theory to express the RGmodule R3n as a direct sum of irreducible RG-submodules. x7 . The equations of motion are x ˆ xA. x4 . v g must also specify the directions and relative magnitudes of displacements in a normal mode of frequency ù. since the relative con®guration of the atoms is unaltered by applying g. Àx2 . x6 . 32. x7 .An application of representation theory to molecular vibration 375 interchanges v5 and v6 with the negatives of v8 and v9 . For all vectors v in the basis. (xg)A ˆ (xA) g. with eigenvalue Àù2 . The eigenspace for the eigenvalue ë of A is. Àx6 )X  We return to the general set-up. and let v be a corresponding eigenvector.7). Choose a basis of R3 n which consists of eigenvectors of A (see Proposition 32. Àx5 . Àx3 . and we are trying to ®nd the eigenvectors of A. by de®nition. x5 . and the normal modes of the molecule.8. problems like this are uncommon. By Proposition 11.) 32. 32. The problem of ®nding the eigenspaces of A is considerably simpli®ed as a consequence of our next proposition. xA P V÷ for all x P V÷ X Proof By Maschke's Theorem we may write R3n ˆ V÷ È W for some RG-module W. The function å: v ‡ w 3 w (v P V÷ . and no RG-submodule of W has character ÷. We can use character theory to see which irreducible RG-modules are contained in R3 n . Let V÷ denote the sum of those irreducible RG-submodules of R3 n which have character ÷. and if ÷ is the character of an irreducible RGmodule which occurs. We call V÷ a homogeneous component of R3 n .3 is stated in terms of CG-modules.3. so xA P V÷ for all x P V÷ . then the element ˆ ÷( g À1 ) g gPG sends R3 n onto the sum of those irreducible RG-submodules of R3 n which have character ÷ (see (14. this function is zero. the function x 3 (xA)å is an RG-homomorphism from V÷ into W. Therefore.376 Representations and characters of groups to determine the eigenspaces of A. w P W ) (x P V÷ ) is an RG-homomorphism. since the character of the RG-module R3 n might contain an irreducible character which cannot be realized over R ± but in practice. (Although Proposition 11.11. (This procedure sometimes needs to be modi®ed.12 De®nition Suppose that ÷ is the character of an irreducible RG-module.13 Proposition Each homogeneous component V÷ of R3 n is A-invariant ± that is.) j . by Proposition 32. its proof works equally well for RG-modules ± compare Exercise 23.27)). in turn. and ®nd the eigenvectors of A in V÷ i (see Proposition 32. so it must equal V÷ . 32. It is usually necessary to know the equations of motion in order to determine the frequency ù.14). This involves no extra work if V÷ i is irreducible (see Corollary 32.14 Corollary If V÷ is an irreducible RG-module.) Since V÷ is A-invariant.7. If V÷ i is reducible. we may choose v P V÷ such that v is an eigenvector of A. with eigenvalue Àù2 . (3) Calculate the character ÷ of the RG-module R3 n and express ÷ as a linear combination of the irreducible characters of G.13). to obtain R3n . as we shall illustrate in the examples which make up the rest of this chapter. or by some other method.19 below. (6) If v is an eigenvector of A. (4) Express R3 n as a direct sum of homogeneous components. each homogeneous component V÷ i of R3 n .15 Summary (1) Assign three coordinate axes at each of the n atoms of the molecule. . then sin (ùt ‡ â) v (or (t ‡ â)v if ù ˆ 0) is a normal mode. with eigenvalue ë. say. j We now summarize the steps in the procedure for ®nding the normal modes of vibration of a given molecule. or Exercise 32. to make further progress. This can € À1 3n be done by applying the element for each gPG ÷ i ( g ) g to R irreducible character ÷ i of G which appears in ÷. Proof (Compare the proof of Schur's Lemma. then all the non-zero vectors in V÷ are eigenvectors of A. This programme can often be successfully completed. Then the intersection of V÷ with the eigenspace for ë is a non-zero RG-submodule of V÷ . (2) Calculate the symmetry group G of the molecule. (5) Consider.An application of representation theory to molecular vibration 377 32. Then R3 n is an RG-module. where â is an arbitrary constant. then see Remark 32. 1). where u1 ˆ (1. . For simplicity. as this eases the calculations: Thus the position of the molecule is given by a vector (x1 . The displacement vectors (x1 . À1)X It follows that the normal modes of the system are given by sin (ù1 t ‡ â1 )(1. . where â1 . 1). The symmetry group (in two dimensions) of the molecule is the dihedral group D6. where the three identical atoms are at the corners of an equilateral triangle. sin (ù2 t ‡ â2 )(1. we consider only vibrations of the molecule in the plane. 32. the RG-submodules of R2 are sp (u1 ) and sp (u2 ).4. as shown. â2 are constants and ù1 . ù2 are the frequencies. x2 ) g ˆ (x2 . We choose to take our axes along the edges of the triangle. .16 Example We ®rst return to Example 32. so we assign two displacement coordinates to each atom. .378 Representations and characters of groups 32. Notice that we have determined the normal modes of vibration (but not their frequencies) using the symmetry group alone. x2 ) form an RG-module R2 . where xi is the displacement along axis i (1 < i < 6). u2 ˆ (1. where g is the re¯ection in the mid-point of PQ. À1). x1 ). Since (x1 . generated by a rotation a of order 3 and a re¯ection . with the spring and two vibrating masses: The symmetry group of this system is G ˆ h g: g 2 ˆ 1i. This agrees with the conclusion of Example 32. x6 ) in R6 .4.17 Example Consider a hypothetical triatomic molecule. x6 . And from the action of b given above. As a matter of notation. the character table of D6 is 1 ÷1 ÷2 ÷3 1 1 2 a 1 1 À1 b 1 À1 0 Hence ÷ ˆ ÷1 ‡ ÷2 ‡ 2÷3 . then we represent the displacement vector (v1 . we seek to express R6 as a direct sum of RD6 -submodules with characters ÷1 . These include the rotation and translation modes. ÷2 . v2 . ÷3 and ÷3 . x2 . v3 are 2-dimensional displacement vectors for the three atoms.3. x5 . then (x1 . we ®rst calculate the character ÷ of the module. Since the rotation a does not ®x any of the atoms. x3 .1). v2 . v3 ) P R6 pictorially by the diagram We ®rst calculate the normal modes of the form (t ‡ â)v. if v1 . It is easy to work out the action of each element of D6 on R6 . corresponding to the eigenspace of A for eigenvalue 0. if b is the re¯ection which ®xes the top atom. Thus the values of ÷ are 1 ÷ 6 a 0 b 0 By Section 18. Thus. x5 . x1 . For example. To do this. ÷(a) ˆ 0. x6 )b ˆ (x2 . x4 . we see that ÷(b) ˆ 0.An application of representation theory to molecular vibration 379 b (see Example 32. . x3 )X We want to express the RD6 -module R6 as a direct sum of irreducible RD6 -modules. x4 . which occur for every molecule. and is called the translation submodule. The modes are of the form (t ‡ â)v. 1. It is clearly an RD6 -submodule of R6 . the molecule rotates with constant angular velocity about the centre. À1. À1. where v is a vector in the span of v1 . 1. 0. À1. 1)). v2 . where ˆ å2 ˆ ÷2 ( g À1 ) g ˆ 1 ‡ a ‡ a2 À b À ab À a2 b gP D6 (compare (14. 0). sp (v) ˆ R6 å2 . Indeed. where v ˆ (1.380 Representations and characters of groups Rotation mode In this mode. 1. v3 ) has dimension 2. If ÷ R is the character of sp (v). so the character must be ÷3 . À1. and so ÷ R ˆ ÷2 . . Thus the character of the translation submodule is part of ÷ À ÷2 ˆ ÷1 ‡ 2÷3 . À1. pictorially.27)). the subspace sp (v1 . it does not contain the rotation submodule. v2 ˆ (1. these vectors being given pictorially by (thus v1 ˆ (À1. À1. ÷ R (a) ˆ 1. then ÷ R (1) ˆ 1. Since v1 ‡ v2 ‡ v3 ˆ 0. 1. 0. 0. We call sp (v) the rotation submodule of R6 . v2 and v3. À1). 0. 1. v3 ˆ (0. Translation modes These are modes in which all atoms move in the same constant direction with the same constant speed. 1. À1). ÷ R (b) ˆ À1. The mode is given by (t ‡ â)v. if w P R6 then the total component of w vib in each direction is zero. Since no mode in R6 can have vib vib any translation component. moreover. since D6 permutes the vectors u1 .11). and are called vibratory modes. with character ÷vib . The vibratory mode given by u1 ‡ u2 ‡ u3 is sometimes called the expansion±contraction mode (you will see the reason for this name in the picture in (32. Our calculation of the normal modes is now complete. and since R6 has vib vib dimension 3. Its character is ÷3 and it gives us the last eigenspace for the vib matrix A. The sum of these eigenspaces forms an RD6 -submodule R6 of R6 (by Proposition 32. u1 À u3 ) is an RD6 -submodule of R6 . and we summarize our ®ndings below. u2 . . so that total moment of each vector in R6 about the vib centre is zero. These constraints imply three independent linear equations in the coordinates of the vectors in R6 . every vector in R6 which satis®es these equations lies in R6 . Finally. u3 among themselves.18(3)) below). R6 has dimension 3. where vib ÷vib ˆ ÷ À (÷2 ‡ ÷3 ) ˆ ÷1 ‡ ÷3 X In particular. Therefore a basis of R6 is u1 . where vib vib Clearly sp (u1 ‡ u2 ‡ u3 ) is an RD6 -submodule of R6 with characvib ter ÷1 .An application of representation theory to molecular vibration 381 Vibratory modes The remaining normal modes correspond to eigenspaces of the matrix A with non-zero eigenvalues. u3 . u2 . R6 does not contain the rotation vib submodule. it is easy to see that sp (u1 À u2 . 382 (32.) We emphasize that we have found the normal modes of vibration without explicit knowledge of the equations of motion. In order to . u1 À u3 pictorially.18) Representations and characters of groups (1) Rotation mode: (2) Translation modes: linear combinations of (3) Vibratory mode: expansion±contraction mode (4) Vibratory modes: linear combinations of (We have chosen 2u2 À u1 À u3 . 2u1 À u2 À u3 as the basis for the vibratory modes in (4) merely because these modes look simpler than u1 À u2 . x6 ). x3 . PR À P9R9 ˆ (x1 ‡ x6 ) ‡ 1(x2 ‡ x5 ).An application of representation theory to molecular vibration 383 check our results. so that we may ignore second order terms. 2 k In the same way. Q9.) Similarly. x4 . denote the new positions of the atoms by P9. the difference in length between QR and Q9R9 is (x4 ‡ x5 ) ‡ 1(x3 ‡ x6 )X 2 (We always assume that x1 . m  x1 ˆ À(x1 ‡ x6 ) À 1(x2 ‡ x5 )X 2 k m  x2 ˆ À(x2 ‡ x3 ) À 1(x1 ‡ x4 ). For a general displacement (x1 . From the diagram. R9. x2 . x5 . . X X X . 2 PQ À P9Q9 ˆ (x2 ‡ x3 ) ‡ 1(x1 ‡ x4 )X 2 Hence the force on the molecule at P in the direction of the ®rst coordinate axis is Àk(PR À P9R9) ˆ Àk(x1 ‡ x6 ) À 1 k(x2 ‡ x5 )X 2 Therefore. x6 are small compared with the distance between the atoms. we now calculate the equations of motion. Let m be the mass of each atom. and assume that the magnitude of the force between two atoms is k times the decrease in distance between them. where we found that G is isomorphic to S4 .17. This illustrates a method which sometimes helps to deal with reducible homogeneous components. We determined the symmetry group G of this molecule in Example 32. since these homogenous components were irreducible (see Corollary 32. the character ÷ of the RG-module R6 was given by ÷ ˆ ÷1 ‡ ÷2 ‡ 2÷3 X All the non-zero vectors in the homogeneous components for ÷1 and ÷2 gave normal modes. but we were able to write it as a sum of two subspaces of eigenvectors (those appearing in (32. The matrix A for  which x ˆ xA is therefore given by H I 1 1a2 1a2 0 0 1 f 1a2 1 1 0 0 1a2 g f g Àk f 0 1 1 1a2 1a2 0 g f gX Aˆ 1a2 1a2 1 1 0 g m f 0 f g d 1a2 0 0 1 1 1a2 e 1 0 0 1a2 1a2 1 You should check that the vectors which we gave in (32. Label the corners of . 32. . .19 Remark In Example 32. .384 Representations and characters of groups   and we obtain similar equations for x3. In our next example.18)(2) and (4)) because V÷3 ’ R6 vib was an A-invariant RG-submodule of V÷3 different from {0} and V÷3 .14).18) are indeed eigenvectors of A. the situation is more complicated. .20 Example We analyse the normal modes for the methane molecule CH4 . 32. x6 . The homogeneous component V÷3 for ÷3 was reducible.2. The action of G on V is easy to describe. these four vectors span a 3-dimensional space. in all. w3 and w4 at the carbon atom. for example. 4. let v21 . We now introduce a new idea. giving twelve vectors v ij . v31 . v23 . v23 .An application of representation theory to molecular vibration 385 the tetrahedron 1. 23. v43 . w4. and let W be the vector space over R spanned by w1. as shown below. v41 . we have v ij g ˆ v ig. w3. v21 .jg for all i. 14. w2. v42 . Let V be the vector space over R with basis v12 . 3. (2 3 4). 24. w2. by taking four unit vectors w1. Let v12 . v14 . v24 . G permutes our twelve basis vectors of V. (2 4 3)X In order fully to exploit the symmetry of the methane molecule. jX Thus. v13 . v13 . Our main task is to ®nd RG-submodules of R15 ˆ V È W. respectively. v34 . v24 be unit vectors at corner 2 in the directions of the edges 21. and so on. similarly. v14 be unit vectors at corner 1 in the directions of the edges 12. for g in G. 13. Then V  R12. at each hydrogen atom we choose displacement axes along the edges of the tetrahedron. W  R3 and V and W are RG-modules. v32 . and we can quickly calculate the character ÷ of V: 1 ÷ 12 (1 2) 2 (1 2 3) 0 (1 2)(3 4) 0 (1 2 3 4) 0 . 2. and identify G with S4 . thus. with w i pointing towards corner i (1 < i < 4). Since w1 ‡ w2 ‡ w3 ‡ w4 ˆ 0. the rotations about the vertical axis through 1 are written as 1. ‡ w4 ˆ 0. . . 5. we have wi g ˆ wig (1 < i < 4)X After recalling that w1 ‡ . We ®nd that ÷ ˆ ÷1 ‡ ÷3 ‡ 2÷4 ‡ ÷5 . 387. we can ®nd RG-submodules with characters ÷1 . ö ˆ ÷4 X By applying the elements ˆ ÷ i ( g À1 ) g gPG (i ˆ 1. it is easy to calculate the character ö of W ± it is 1 ö 3 (1 2) 1 (1 2 3) 0 (1 2)(3 4) À1 (1 2 3 4) À1 By Section 18.386 Representations and characters of groups For example. The RG-submodule W1 with character ÷1 is spanned by ˆ v ij i. the character table of S4 is as shown at the top of p. all the basis vectors are moved by (1 2 3). for g in G. (1 2) ®xes the basis vectors v34 and v43 only. j and this gives the expansion±contraction normal mode: We next describe the RG-submodule W5 with character ÷5 . . ÷3 .27)).1. p2 ˆ (v31 À v13 ) ‡ (v14 À v41 ) ‡ (v43 À v34 ). Let p1 ˆ (v23 À v32 ) ‡ (v34 À v43 ) ‡ (v42 À v24 ). and so on. 4) to R15 . The group G acts on W as follows. 3. ÷5 and 3÷4 (see (14. p4 ˆ (v21 À v12 ) ‡ (v13 À v31 ) ‡ (v32 À v23 )X The vector pi gives a rotation about the axis through the corner i and the centroid of the tetrahedron.An application of representation theory to molecular vibration 387 Character table of S4 1 ÷1 ÷2 ÷3 ÷4 ÷5 1 1 2 3 3 (1 2) 1 À1 0 1 À1 (1 2 3) 1 1 À1 0 0 (1 2)(3 4) 1 1 2 À1 À1 (1 2 3 4) 1 À1 0 À1 1 p3 ˆ (v12 À v21 ) ‡ (v41 À v14 ) ‡ (v24 À v42 ). It should be clear from the pictures that for all i with 1 < i < 4 and . we have pi g ˆ Æ p j for some j. q2 ˆ (v13 ‡ v31 ) ‡ (v24 ‡ v42 ) À (v14 ‡ v41 ) À (v23 ‡ v32 ).) For all i with 1 < i < 4 and g in G. for example. so dim W5 ˆ 3. if we let W 5 ˆ sp ( p1 . q2 . Now p1 ‡ p2 ‡ p3 ‡ p4 ˆ 0. p4 ). (Compare. Therefore. Since q1 ‡ q2 ‡ q3 ˆ 0.) Now we construct the RG-submodule W3 of V with character ÷3 . we have q i g ˆ Æq j for some j. q3 ˆ (v14 ‡ v41 ) ‡ (v23 ‡ v32 ) À (v12 ‡ v21 ) À (v34 ‡ v43 )X (Each q i is associated with an `opposite pair of edges'. The RG-module W5 is the rotation submodule. Let q1 ˆ (v12 ‡ v21 ) ‡ (v34 ‡ v43 ) À (v13 ‡ v31 ) À (v24 ‡ v42 ). the picture for p4 with the picture for the rotation vector v in Example 32.388 Representations and characters of groups all g in G. Check that the character of W5 is ÷5 . Let W3 ˆ sp (q1 . .17. p2 . Then W3 is an RG-submodule of V. then W5 is an RG-submodule of V. the dimension of W3 is 2. p3 . q3 ). its character is ÷3 . r4 ˆ (v14 ‡ v41 ) ‡ (v24 ‡ v42 ) ‡ (v34 ‡ v43 ) À (v13 ‡ v31 ) À (v12 ‡ v21 ) À (v23 ‡ v32 )X (The vector ri is associated with corner i. by Corollary 32.) .An application of representation theory to molecular vibration 389 In the RG-submodules W 1 . all the non-zero vectors are eigenvectors of A. r3. De®ne the vectors r1. r3 ˆ (v13 ‡ v31 ) ‡ (v23 ‡ v32 ) ‡ (v34 ‡ v43 ) À (v12 ‡ v21 ) À (v14 ‡ v41 ) À (v24 ‡ v42 ). W 5 and W3 which we have found so far. We now come to the homogeneous component (V È W)÷4 of R15 . r2 ˆ (v12 ‡ v21 ) ‡ (v23 ‡ v32 ) ‡ (v24 ‡ v42 ) À (v13 ‡ v31 ) À (v14 ‡ v41 ) À (v34 ‡ v43 ).14. r2. r4 by r1 ˆ (v12 ‡ v21 ) ‡ (v13 ‡ v31 ) ‡ (v14 ‡ v41 ) À (v23 ‡ v32 ) À (v24 ‡ v42 ) À (v34 ‡ v43 ). r3. r4 among themselves. so r1. r4 span a 3-dimensional RGsubmodule W4 of V. de®ne the vectors s1 . r2. Next. s2 ˆ (v21 ‡ v23 ‡ v24 ) À (v12 ‡ v32 ‡ v42 ).390 Representations and characters of groups For each g in G and i with 1 < i < 4. s2 . s3 ˆ (v31 ‡ v32 ‡ v34 ) À (v13 ‡ v23 ‡ v43 ). s4 by s1 ˆ (v12 ‡ v13 ‡ v14 ) À (v21 ‡ v31 ‡ v41 ). The character of W4 is ÷4 (see Proposition 13. s3 . s1 ‡ s2 ‡ s3 ‡ s4 ˆ 0. we have ri g ˆ rig. r3. Thus G permutes the vectors r1. 1 < i < 4).24). Note that r1 ‡ r2 ‡ r3 ‡ r4 ˆ 0. . s4 ˆ (v41 ‡ v42 ‡ v43 ) À (v14 ‡ v24 ‡ v34 )X We have si g ˆ sig ( g P G. r2. look at the picture of the vector r1 ! We return now to the methane molecule. w1 ‡ w2 ‡ w3 ‡ w4 ˆ 0. in order to deal with the easier case of a molecule with identical atoms at the corners of the tetrahedron. r3. 1 < i < 4). so 9 (V È W )÷4 ˆ W 4 È W 4 È W X 9 We now break off temporarily from studying the methane molecule. and let U1 ˆ fv P (V È W )÷4 : vh ˆ v for all h P HgX Since (vh)A ˆ (vA)h for all v P V÷4 and all h P H.21) (1) The vectors r1 À 2s1 . w2 . so we cannot complete the work using only representation theory. we have wi g ˆ wig ( g P G. upon the constants which appear in the equations of motion. We shall therefore press on and explain how to reduce the dif®culty to that of calculating the eigenvectors of a 3 3 3 matrix. and the character of W is ÷4 . 9 (32. In this case. r4 span the subspace V÷4 ’ R12 of V÷4 . and so they give the ®nal 3-dimenvib sional space of eigenvectors (see Remark 32. (2) The vectors r1. r2. . r4 À 2s4 span the 3-dimensional space of translation modes. s3 . Let H be the subgroup of S4 consisting of those permutations which ®x the number 1. it is very dif®cult to calculate the eigenvectors of A in (V È W )÷4 directly. w3 . The normal mode (sin ùt)r1 is sometimes called an `umbrella mode'. W 4 and W is direct. and no central atom. r2 À 2s2 . To see why. The task in front of us is to ®nd the eigenvectors of A which lie in (V È W )÷4 ˆ W 4 È W 4 È W X 9 The solution of this problem depends. r3 À 2s3 . the space W does not enter our calculations. The sum of W 4 .19). s4 span a 3-dimensional RG-submodule W9 of V with 4 character ÷4 . s2 .An application of representation theory to molecular vibration 391 and s1 . w4 span W. in fact. and we can decompose V÷4 ˆ W 4 È W 4 in the following way. Now recall that w1 . Since dim (V È W )÷4 ˆ 9. it follows that U1 is A-invariant. s1 .392 We ®nd that Representations and characters of groups h3÷4 5 H. it is possible to calculate the 3 3 3 matrix B of the action of A on r1. The eigenvectors of B then give three eigenvectors of A. r3 À 2s3 ‡ 3 cos Ww3 ). and the matrix of A acting on r2. 1 H i H ˆ 3. w2 is again B. One eigenvector of A acting on r1. w2 ) is A-invariant. Therefore U 1 ˆ sp (r1 . s1 h ˆ s1 and w1 h ˆ w1 . A similar remark applies to U3. w1 is easy to ®nd. It is hard to imagine a more spectacular application of representation theory with which to conclude our text. s1 (1 2) ˆ s2 . and since A commutes with the action of G. namely the translation vector r1 À 2s1 ‡ 3 cos Ww1 . s2 . But for all h P H. w1 (see Exercise 32.5). Thus we obtain the translation submodule sp (r1 À 2s1 ‡ 3 cos Ww1 . the process of calculating the eigenvectors of the 3 3 3 matrix B gives nine eigenvectors of A which form a basis of (V È W )÷4 . s1 . r2 À 2s2 ‡ 3 cos Ww2 . and hence the matrix A. we have therefore reduced the initial problem of ®nding the eigenvectors of a 15 3 15 matrix A to that of calculating two of the eigenvectors of a 3 3 3 matrix. w1 )X Once the equations of motion. s3 . w1 (1 2) ˆ w2 . de®ned by U2 ˆ sp (r2 . s1 . r1 h ˆ r1. By means of representation theory. and so dim U1 ˆ 3. where U 3 ˆ sp (r3 . the space U 2 . where W is the angle between an edge of the tetrahedron and the line joining a corner on the edge to the centroid. . Better still. have been calculated. w3 )X Therefore. r1 (1 2) ˆ r2 . s2 . The space R3n of displacement vectors is an RG-module. 0. Prove successively that (i) C has a real eigenvalue. 0. 4. with eigenvalue Àù2 . To determine the eigenvectors of A (and hence all solutions of the equations of motion). 0). (b) Let C ˆ (det B)B. Exercises for Chapter 32 1. 2. e2 . If u is an eigenvector of A. 1). All solutions are linear combinations of normal modes. (ii) C has a positive eigenvalue. Deduce that det B ˆ Æ1. The equations of motion of the molecule have the form  x ˆ xA.An application of representation theory to molecular vibration 393 Summary of Chapter 32 1. e3 ˆ (0. and is called a normal mode. and Àb is a rotation otherwise. Suppose that b P O(R3 ). Suppose that G is the symmetry group of some molecule in R3 . then tr B ˆ 1 ‡ 2 cos ö. 2. If V÷ i is irreducible. (c) Deduce from (a) and (b) that b is a rotation if det B ˆ 1. 1. . (a) Show that the matrix B of b with respect to the basis e1 . 3. it is suf®cient to ®nd the eigenvectors of A restricted to each homogeneous component V÷ i of the RG-module R3 n . where x P R3 n and the 3n 3 3n matrix A has 3n linearly independent eigenvectors. The symmetry group G of a molecule with n atoms consists of those distance-preserving endomorphisms of R3 which send each atom to an atom of the same kind. 0). 5. and let e1 ˆ (1. e3 of R3 satis®es BBt ˆ I (where Bt denotes the transpose of B). then all non-zero vectors in V÷ i are eigenvectors of A. then x ˆ sin (ùt ‡ â)u (or x ˆ (t ‡ â)u if ù ˆ 0) is a solution of the equations of motion. e2 ˆ (0. The action of any g P G on R3 n commutes with A. (iii) 1 is an eigenvalue of C. (d) Show that if b is a rotation through an angle ö about some axis. X X X .7. w2 . Label the corners of the tetrahedron 1. r4 prompted us to use these vectors? 5. 2. and let the position vector of the molecule be ˆ iTˆ j xij v ij ‡ p 3 ˆ iˆ1 yi wi X (a) Prove that cos (/ 012) ˆ (2a3) and cos (/ 102) ˆ À1a3. Find a basis for this space which is simpler than the one which we have used. (See the paragraph before Proposition 32. b b X 0.17. r2.20. r3. Consider the space spanned by the vectors r1. r4 given in Example 32. if g is not a rotationX 3. What property of r1. w1 . r3. 3. The purpose of this exercise is to derive the equations of motion of the methane molecule.394 Representations and characters of groups Show that the sum of the character ÷ T of the translation submodule and the character ÷ R of the rotation submodule has value at g P G given by V b 2(1 ‡ 2 cos ö). and verify that A is symmetric.20. . v43 . 4 and let 0 denote the centroid of the tetrahedron.) 4. Consider the triatomic molecule studied in Example 32. w3 as described in Example 32. if g is a rotation b ` through angle ö (÷ T ‡ ÷ R )( g) ˆ about some axis. r2. v13 . Take the axes for the displacement coordinates as shown below:  Calculate the equations of motion x ˆ xA with respect to these axes. Work with 15 unit displacement vectors v12 . and so ®nd explicitly the 3 3 3 matrix B which appears at the end of Example 32.20. 24. 2 with similar expressions for the edges 13.An application of representation theory to molecular vibration 395 (b) Show that the decrease in the length of the edge 12 from its original length is x12 ‡ x21 ‡ 1(x13 ‡ x14 ‡ x23 ‡ x24 ). 23. x34 . j . x21 . x31 . Finally. Also. x14 . 14. i. q2 . p3 . and m2 the mass of a carbon atom. 03. show that the length of the edge 01 has decreased by p (2a3)(x12 ‡ x13 ‡ x14 ) ‡ y1 À 1( y2 ‡ y3 ). Finally. p2 . show that the length of the edge 04 has decreased by p (2a3)(x41 ‡ x42 ‡ x43 ) À 1( y1 ‡ y2 ‡ y3 )X 3 (c) Let m1 denote the mass of a hydrogen atom. 3   with similar expressions for y2 and y3 . 3 3         with similar expressions for x13. Prove that  m1 x12 ˆ À k 1 [x12 ‡ x21 ‡ 1(x13 ‡ x14 ‡ x23 ‡ x24 )] 2 p 1 À 3 k 2 [x12 ‡ x13 ‡ x14 ‡ (3a2)( y1 À 1 y2 À 1 y3 )]. x24 . (d) The equations in part (c) determine the 15 3 15 matrix A in the  equations of motion x ˆ xA. Assume that the magnitude of the force between hydrogen atoms is k1 times the decrease in distance between them. and the magnitude of the force between a hydrogen atom and a carbon atom is k2 times the decrease in distance between them. x23 . 3   with similar expressions for x42 and x43 . show  m1 x41 ˆ À k 1 [x14 ‡ x41 ‡ 1(x42 ‡ x43 ‡ x12 ‡ x13 )] 2 p 1 À 3 k 2 [x41 ‡ x42 ‡ x43 À 1 (3a2)( y1 ‡ y2 ‡ y3 )]. 3 with similar expressions for the edges 02. 34. x32 . q1 . p1 . Verify that the vectors ˆ v ij . show p  m2 y1 ˆ Àk 2 [ (2a3)(x12 ‡ x13 ‡ x14 À x41 À x42 À x43 ) ‡ 4 y1 ]. Also. 8. (Hint: compose the function w 3 wA with a projection. 6.  (b) Calculate the equations of motion. . Prove that if u and v are non-zero elements of U1. x ˆ xA. . where the vectors r1.20. È Um. Assume that the only internal forces are along the sides of the square.) We assume that ÷ i is the character of an irreducible RG-module which remains irreducible as a CGmodule. and use Exercise 23. s1 A ˆ b21 r1 ‡ b22 s1 ‡ b23 w1 . uW m . (f) Verify that p (1.15(5). a sum of m isomorphic irreducible RG-modules. Suppose that V÷ i ˆ U1 È . 6) is an eigenvector of B. (a) Find the normal modes of the molecule. . uW m ) with respect to the basis uW1 . . then Au ˆ Av . Show how to ®nd the eigenvectors of A in V÷ i . X X X . . (e) Find the entries b ij in the 3 3 3 matrix B which are given by r1 A ˆ b11 r1 ‡ b12 s1 ‡ b13 w1 . s1 . eigenvectors of A. . . In this exercise. (a) Prove that for all non-zero u in U1. w1 A ˆ b31 r1 ‡ b32 s1 ‡ b33 w1 . are eigenvectors of A. we derive a method for simplifying the problem of ®nding the eigenvectors of A when the homogeneous component V÷ i is reducible. Consider a hypothetical molecule in which there are four identical atoms at the corners of a square. indeed. w1 are as in Example 32. 7. For 1 < i < m. .20. .396 Representations and characters of groups which appear in Example 32. (See 32. (c) Assume that the eigenvectors of the m 3 m matrix A u are known.) (b) Let A u denote the matrix of the endomorphism w 3 wA of sp (uW1 . let W i be an RG-isomorphism from U1 to Ui . . sp (uW1 . and check that the vectors you found in part (a) are. uW m ) is an A-invariant vector space of dimension m. À2. . We reduce our problem to that of ®nding the eigenvectors of an m 3 m matrix. so kgl ˆ G. Therefore g p ˆ 1.4. Let D4 m ˆ ha. c2 }. If G were in®nite. and so G is cyclic of prime order. Let g be a non-identity element of G. where m is odd. and if Ker W ˆ G then H ˆ f1g. 2. Let x ˆ (c( m‡1)a2 . (b) Since b2 ë ˆ I but (bë)2 ˆ Y 2 ˆ ÀI. À1) and y ˆ (d. we may choose h P G with h P An . hence G is ®nite. b: a2 m ˆ b2 ˆ 1. Check using the method of Example 1. so G ’ An v G.4. Then kgl is a normal subgroup of G. (a) Using the method of Example 1. either Ker W ˆ f1g or Ker W ˆ G. Ker ö ˆ {1. d: cm ˆ d 2 ˆ 1. For all odd g in G. 5. Check that x 2 m ˆ y 2 ˆ 1. it is routine to verify that ö and ø are homomorphisms. and Ga(G ’ A n )  C2 . 3. Since G ’ An Tˆ G. Since G is simple and Ker W v G. First. the function W: D4 m 3 D2 m 3 C2 de®ned by 397 . Then h g p i is a normal subgroup of G which is not equal to G. The elements of D2 m 3 C2 are (c i d j . Note that all subgroups of G are normal. (À1) k ) for 0 < i < m À 1. Therefore G ’ An and (G ’ An )h are the only right cosets of G ’ An in G. 0 < j < 1. then h g 2 i would be a normal subgroup different from G and {1}. and G Tˆ {1} since G is simple. Let p be a prime number which divides |G|. since G is abelian. a2 } and Ker ø ˆ {1. G ’ An ˆ f g P G: g is even}. If Ker W ˆ f1g then W is an isomorphism. bÀ1 ab ˆ aÀ1 i. Also Ker ì ˆ {1} and Im ì ˆ L. 0 < k < 1. it follows that ë is not a homomorphism. y À1 xy ˆ x À1 X By Example 1. so ì is an isomorphism. 1).4 again that ì is a homomorphism. and D2 m ˆ hc. d À1 cd ˆ cÀ1 i. 4. we have g ˆ a (ghÀ1 )h P (G ’ An )h.Solutions to exercises Chapter 1 1. 398 Representations and characters of groups W: a i b j 3 x i y j (0 < i < 2m À 1. If H is a subgroup of G of order n. Choose k as small as possible such that k . by Lagrange's Theorem). so dj j. if g P G then g À1 ˆ 1a g P G. also H has order n. 0 and a k P H. we have r ˆ 0. by part (b). and so x is a power of y. h. B2 ˆ A4 and BÀ1 AB ˆ AÀ1 . k. B as follows:   eiða4 0 Aˆ . thus H is cyclic. Thus G is a group under multiplication. if jGj is even then there exists g P G such that g Tˆ 1 and the subset f g. where H ˆ { g P G: gn ˆ 1}. First observe that there exists i . 6. . It follows that f g P G: g n ˆ 1g ˆ had i. If g P G and g n ˆ 1. Let G be the set of non-zero complex numbers. De®ne matrices A. then g ˆ a j for some integer j and dnj jn. also 1 P G and 1g ˆ g1 ˆ g for all g P G. which is a cyclic group of order n. These relations show that every element of the group kA.     ei jða4 0 0 ei jða4 j j . Hence. We deduce that hxi ˆ H ˆ h yiX Thus x P h yi. Therefore a j ˆ a kq and so H ˆ kak l. h P G then gh Tˆ 0. g À1 g has size 1. and the identity element is in a subset of size 1. Bl has the form A j B k with 0 < j < 7. then x. Therefore H < f g P G: g n ˆ 1g ˆ he2ðia n iX Since j Hj ˆ n ˆ jhe2ðia n ij. kA. Each such subset has size 1 or 2. hence g P kad l. it contains x 2 ˆ (c. 1) and x m ˆ (1. Finally. (c) If x and y are elements of order n in the ®nite cyclic group G. it follows that H ˆ ke2ðia n l. we conclude that W is an isomorphism. If 1 Tˆ a j P H then j ˆ qk ‡ r for some integers q. Partition G into subsets f g. 7. 9. r with 0 < r . 0 < k < 1. with 0 < j < 7. Bl has order 16. so g ˆ gÀ1 and g has order 2. then h n ˆ 1 for all h P H (since the order of h divides n. 0 eÀiða4  Bˆ  1 X 0 0 À1 Check that A8 ˆ I. As |D4 m | ˆ |D2 m 3 C2 |. If g. If g. Since r . A Bˆ A ˆ X 0 eÀi jða4 ÀeÀi jða4 0 Since these matrices. yl. Since Im W ˆ kx. k P G then (gh)k ˆ g(hk). y P H. g À1 g ( g P G). (a) Let G ˆ kal and suppose that 1 Tˆ H < G. and gÀ1 g ˆ ggÀ1 ˆ 1. are all distinct. 8. Hence a r ˆ a j aÀqk ˆ a j (a k )Àq P H. k. Moreover. 0 < j < 1) is a homomorphism. Now kxl and k yl have order n. À1) and hence Im W ˆ D2 m 3 C2 . 0 such that ai P H. so gh P G. (b) Assume that G ˆ hai and jGj ˆ dn. w2 P W. . (ë(wWÀ1 ))W ˆ ë(wWÀ1 )W ˆ ëwX Hence (u ‡ w)WÀ1 ˆ uWÀ1 ‡ wWÀ1 and (ëw)WÀ1 ˆ ë(wWÀ1 ). 2. . the expression 0 ˆ 0 ‡ 0 is the unique expression for 0 as the sum of vectors in U and W. Conversely. Assume ®rst that V ˆ U È W. ur are linearly independent. This shows that V ˆ U È W. Since V ˆ U È W. . ur . . If u1 ‡ w1 ˆ u2 ‡ w2 with u1 . . w P W and ë P F.Chapter 2 399 10. Chapter 2 1. . and so u ˆ v. ‡ ë r ur ˆ ì1 w1 ‡ . . similarly ì i ˆ 0 for all i. 5. Therefore u1 . Let v P U ’ W. . And if g P H then H. so by Exercise 3. w1. Hence H v G. ws are linearly independent. v ˆ 0. Then u ˆ u1 ˆ u2 ‡ u3 for some ui P Ui (1 < i < 3). Then V ˆ U ‡ W. . since u is a linear combination of u1 . . w s . . it follows that v is a linear combination of u1 . . . this gives ë i ˆ ì j ˆ 0 for all i. ì j in F. . Hg are the two right cosets of H in G. so Im W ˆ V (by (2. Thus v ˆ 0 and so U ’ W ˆ {0}. Since W is a linear transformation. . Since such expressions are unique. . First suppose that V ˆ U È W. since u1 . ur . Since u1 ‡ 0 ‡ 0 ˆ 0 ‡ u2 ‡ u3 and the sum U1 ‡ U2 ‡ U3 is direct. Thus U ’ W ˆ {0}. Therefore u1 . . (a) Assume ®rst that V ˆ U1 È U2 È U3. Let u. u2 P U and w1. so u À v P Ker W ˆ {0}. If g P H then gÀ1 Hg ˆ H. ws are linearly independent. W is invertible. . we have u1 ˆ . If v P U ’ W then v ˆ ë1 u1 ‡ . suppose that u1 . F F F . . Now suppose that V ˆ U ‡ W and U ’ W ˆ {0}. . . (2) A (3): If Ker W ˆ {0} then dim (Im W) ˆ dim V (by (2. . v P V and uW ˆ vW then (u À v)W ˆ 0. . j. . ur . hence they form a basis of V. . Suppose that ë1 u1 ‡ X X X ‡ ë r ur ‡ ì1 w1 ‡ X X X ‡ ì s ws ˆ 0 with all ë i . and so ë1 u1 ‡ X X X ‡ ë r ur ˆ ì1 w1 ‡ X X X ‡ ì s ws ˆ 0X As u1 . F F F . F F F . . . V ˆ U È W. ur . . so Ker W ˆ {0}. Thus W is injective. Suppose jG: Hj ˆ 2 and let g P G. so W is surjective.7)). so WÀ1 is a linear transformation. . . ì j P F. Therefore Hg ˆ gH. . w1. . If v P V then v ˆ u ‡ w for some u P U and w P W. . . Let u P U1 ’ (U2 ‡ U3 ). w1 . As W is surjective and injective. . (3) A (1): Assume that Im W ˆ V.12). ws is a basis of V. . ur and w is a linear combination of w1 . while H. ‡ ì s ws for some ë i . . w1. Then v ˆ v ‡ 0 ˆ 0 ‡ v and this gives us two expressions for v as the sum of an element in U and an element in W. 4. 3. hence u1 ˆ u2 and w1 ˆ w2. . (1) A (2): If W is invertible then W is injective. this forces ë i ˆ 0 for all i. . . Ker W ˆ {0}. By (2. gH are a the two left cosets. u r . . ws span V. If u. and so gÀ1 Hg ˆ H again. we have (uWÀ1 ‡ wWÀ1 )W ˆ (uWÀ1 )W ‡ (wWÀ1 )W ˆ u ‡ w. . . .12)). w1. w s . then u1 À u2 ˆ w2 À w1 P U ’ W ˆ {0}. It is easy to see that V ˆ U ‡ W. 0)). If v P U ’ W then v ˆ vW ˆ Àv. by induction on r. Ker W ˆ sp ((0. Conversely. .400 Representations and characters of groups u2 ˆ u3 ˆ 0. 0)). suppose that r is a representation of G. . ws for Ker W. Let v P V. 1)). . . . . ‡ dim Ur. . Therefore V ˆ U1 È U2 È U3. . 0)X Then Im W ˆ sp ((1. 8. The representations r2 and r3 are faithful. y) 3 (x. 6. so dim V ˆ dim U1 ‡ . Then 1 2 3 u1 À u9 ˆ (u9 À u2 ) ‡ (u9 À u3 ) P U1 ’ (U2 ‡ U3 ) ˆ {0}. the diagonal entries being r 1's followed by s 0's. m À 1 and i . Check that A3 ˆ B3 ˆ C 3 ˆ I. 1 2(v Chapter 3 1. so v ˆ 0. and U1 ˆ sp ((1. 2 3 (b) Let V ˆ R2. . so 1(v ‡ vW) P U. 9. . 7. . . assume that Am ˆ I. (a i a j )r ˆ (a i‡ j )r ˆ A i‡ j ˆ A i A j ˆ (a i r)(a j r). . By Exercise 4. Thus V ˆ U ‡ W. by Exercise 4. .10)). . Then V ˆ Im W È Ker W by Proposition 2. ö: V 3 V by W: (x. U2 ’ (U1 ‡ U3 ) ˆ U 3 ’ (U1 ‡ U2 ) ˆ f0g. More generally. . 2 2 2 À vW) P W. . so V ˆ Im W È Ker W. the matrix [W]B is diagonal. ur for Im W and a basis w1. but r1 is not. 1 2 3 1 Similarly. . by Exercise 3. Therefore V ˆ U È W. 1)). . The construction of the basis B is similar to that in Solution 8. of V. ws is a basis. Therefore for all integers i. and so r is a representation. if V ˆ U È W then dim V ˆ dim U ‡ dim W. De®ne W. j. so V Tˆ Im ö È Ker ö. . so W is a projection. y) 3 ( y. ‡ dim Ur. ur . Suppose ®rst that W is a projection. Then u1 . u9 P Ui (1 < i < 3) and i u1 ‡ u2 ‡ u3 ˆ u9 ‡ u9 ‡ u9 . u2 ˆ u9 and u3 ˆ u9 . if V ˆ U1 È . . 0). Similarly. Now suppose that U1 ’ (U2 ‡ U3 ) ˆ U2 ’ (U1 ‡ U3 ) ˆ U3 ’ (U1 ‡ U2 ) ˆ {0}. 0) and ö: (x. È Ur ) (see (2. Similarly. Let V ˆ R2. if [W]B has the given form. dim (U2 È . . Hence by Exercise 1.32. . each r j is a representation. then clearly W2 ˆ W. so u1 ˆ u9 . . Then I ˆ 1r ˆ (am )r ˆ (ar) m ˆ Am X Conversely. First. Assume that ui . Take a basis u1 . U3 ˆ sp ((1. and Im ö ˆ Ker ö ˆ sp ((1. 0)). È Ur ) ˆ dim U2 ‡ . w1. . Therefore U1 ’ (U2 ‡ U3 ) ˆ {0}. È Ur then V ˆ U1 È (U2 È . Then v ˆ 1(v ‡ vW) ‡ 1(v À vW)X 2 2 Observe that 1(v ‡ vW)W ˆ 1(vW ‡ v). U2 ˆ sp ((0. 2. 1)). . Then (ai )r ˆ Ai for all integers i (including i . say B. . Since ui W ˆ ui for all i and wj W ˆ 0 for all j. then gô ˆ (ST)À1 (gr)(ST). The matrices A r Bs (0 < r < 5. .10. Similarly r4 is faithful. 6. (3) S ˆ ÀA. we have S 6 ˆ T 2 ˆ I. so r1 is faithful. hence r3 is not equivalent to any of the others. (2) S ˆ A3 . 0 < j < 1)X 401 It is easy to check that r is a representation of G. since a2 r2 ˆ I and a3 r3 ˆ I. so r is equivalent to ô. The representations r1 and r4 are equivalent: to see this. T ˆ B. bÀ1 ab ˆ aÀ1 l. (4) S ˆ C.2(1). De®ne r by (a i b j )r ˆ (À1) j (0 < i < n À 1. 7. 4. ®rst work out the eigenvectors of C. By Theorem 1. T À1 ST ˆ S À1 X It follows that each r k is a representation (see Example 1. (2) If r is equivalent to ó then there is an invertible matrix T such that gó ˆ T À1 ( gr)T for all g P G. T ˆ ÀB.) If j Tˆ 2. 0 < s < 1) is a faithful representation of D8 ˆ ka. then a3 r j Tˆ I. Check that in each of the cases (1) S ˆ A. But r2 and r3 are not faithful. GaKer r  Im r. let  Tˆ  1 1 X Ài i Then T À1 (gr4 )T ˆ gr1 for all g P G. (1) For all g P G. so ó is equivalent to r. then there are invertible matrices S and T such that gó ˆ SÀ1 ( gr)S and gô ˆ T À1 (gó)T for all g P G. b: a4 ˆ b2 ˆ 1. 0 < s < 1) are all different. And if j Tˆ 3. But Im r < GL (1. Compare Example 3. T ˆ B. hence r2 is not equivalent to any of the others. T ˆ D.4). 8. then a2 r j Tˆ I. F) and GL (1. hence r is equivalent to r. Therefore GaKer r is abelian. (To ®nd T. then gr ˆ (T À1 )À1 (gó)T À1. IÀ1 (gr)I ˆ gr. 0 1 1 B ˆ d0 0 H I 0 0 À1 0 eX 0 1 Then the function r: a r b s 3 A r Bs (0 < r < 3. No: let G be any non-abelian group and let r be the trivial representation.Chapter 3 3. 5. (3) If r is equivalent to ó and ó is equivalent to ô. F) is abelian. De®ne the matrices A and B by 0 A ˆ d À1 0 H I 1 0 0 0 e. h P Sn . h P An ) or vg ˆ Àv ˆ vh (if g. Let v P V and g. Assume ®rst that gh P An . Let g P Sn . v in V and ë in F. BÀ1 AB ˆ AÀ1 X Hence r: a i b j 3 A i B j (0 < i < 3.402 Representations and characters of groups Chapter 4 1.2.4(1). We have now checked all the conditions in De®nition 4. By Theorem 4. (u ‡ v) g ˆ ug ‡ v gX It remains to check (2) of De®nition 4. B2 ˆ A2 . h is in An and the other is not. Let A ˆ f d 0 0 0 À1 e and B ˆ d À1 0 0 0e 0 0 1 0 0 À1 0 0 Check that A4 ˆ I. Then v(gh) ˆ v. assume that gh P An . so V is an FG-module.2. 0 < j < 1) is a representation of Q8 over R. Let V ˆ R4. For all u. g H [ g]B 1 1 d0 0 H 1 d0 0 1 0 0 1 0e 0 1 I 0 0 1 0e 0 1 I 0 d1 0 H 1 d0 0 H (1 2) 1 0 0 0e 0 1 I 0 0 À1 0 e À1 1 I 0 d0 1 H 1 d0 0 H (1 3) I 0 1 1 0e 0 0 I 0 0 1 À1 e 0 À1 [ g]B 2 g H [ g]B 1 1 d0 0 H 1 d0 0 (2 3) 0 0 0 1e 1 0 I 0 0 0 1e 1 0 I H (1 2 3) 0 d0 1 H 1 d0 0 1 0 0 1e 0 0 I 0 0 À1 1 e À1 0 I H (1 3 2) 0 d1 0 H 1 d0 0 I 0 1 0 0e 1 0 I 0 0 0 À1 e 1 À1 [ g]B 2 All the matrices [g]B 2 have the form H I 1 0 0 d 0 j j eX 0 j j 2. Next. we have v g P V . since one of g. a a Then v(gh) ˆ Àv. v1 ˆ v. H I H I 0 0 1 0 0 1 0 0 f À1 0 0 f 0 0g 0 0 1g g f g. g P Q8. h P An ). If we put . and (vg)h ˆ Àv. since either vg ˆ v ˆ vh (if g. (ëv) g ˆ ë(v g). V becomes an RQ8 -module if we de®ne vg ˆ v( gr) for all v P V. 3. and (vg)h ˆ v. 0g g g g g e 1 I 0 then MA is obtained from A by swapping the ®rst two rows. â) À (á. 3. since ó is equivalent to r. where vg ˆ v(gr1 ) for . Therefore ó is reducible. If C is a matrix obtained from A by permuting the columns. 4. 0). . â) P U with (á. k . 0. À1) belongs to U. â) ‡ (á. 2. let g be the permutation in Sn which has the property that for all i. and AM is obtained from A by swapping the ®rst two columns. 1)). n. v3 ˆ (0. 0. 0). v i a and v i b are as required in the question. â)a ˆ (á À â. sp ((1. 0). r4 be the representations of G de®ned in Exercise 3. 1) or (1. and (á. 0. Suppose that r has degree n and r is reducible. ` 1. 1. Hence the FG-submodules of V are {0}. 1. the proof is similar to that for the rows. Let G ˆ D12 and let r1 . pij ˆ X 0. . First consider the FG-module V ˆ F 2. Then r is equivalent to a representation ô of the form H I Xg 0 e ( g P G) ô: g 3 d Yg Zg where X g is a k 3 k matrix and 0 . á ‡ â) P U. 0. v2 ˆ (0. 0. sp ((1. and the ij-entry of PA is n ˆ pik akj ˆ a ig. Let U be a non-zero FGsubmodule of V. Then ó is equivalent to ô. . then C ˆ AQ for some permutation matrix Q. À1)) and V. and let (á. It is easy to verify that V is an FG-module. â À á) P U.5. You may ®nd it helpful ®rst to check that if H 0 1 f1 0 f f 1 M ˆf FF f d F 403 v1 ˆ (1. j X kˆ1 Hence PA ˆ B. we deduce that (1. â) Tˆ (0. Since at least one of á ‡ â and á À â is non-zero. Then (á. v4 ˆ (0. if j Tˆ igX Then P is a permutation matrix. â)a ˆ (á ‡ â. 1). 0). row i of B ˆ row ig of AX Let P be the n 3 n matrix ( pij ) de®ned by V if j ˆ ig.Chapter 5 then for all i. 0. Chapter 5 1. . To solve the exercise. it follows that dim U > 2. since r1 and r4 are equivalent. 1) or (1.Cˆ çÀ1 1  1 X 0 Thus jGj < 18. (b) Let    å 0 ç Aˆ .) Therefore r1 is irreducible. and u and ub are linearly independent unless ç ˆ 1. Hence either (1. g P G. 1)) is an FG-submodule of V and r2 is reducible. C À1 AC ˆ AÀ1 and C À1 BC ˆ BÀ1 . there exists a cube root î of unity such that   î 0 gr ˆ À1 X 0 î But there are only three distinct cube roots of unity. b}. (c) For every element g of ka. Let V ˆ {0} and let 0 g ˆ 0 for all g P G. 1)b. then U is a CHsubmodule. if either å Tˆ 1 or ç Tˆ 1 then dim U ˆ 2 and so r is irreducible. Bˆ 0 å À1 0 Check that A3 ˆ B3 ˆ C 2 ˆ I. Then (1. where H is the subgroup {1. Consequently U ˆ V and so V is irreducible. c}. 1) and (1. bl. Then V is neither reducible nor irreducible. 4. (d) Let V ˆ C2 be the CG-module obtained by de®ning vg ˆ v(gr) for all v P C2 . is a multiple of 9 and jGj . bl Tˆ G. Now u and ua are linearly independent unless å ˆ 1. 1)) is a CG-submodule of V.5(2) for an alternative argument. (a) It is easy to check the given relations. By the solution to Exercise 1. if å ˆ ç ˆ 1 then sp ((1. g P G. À1)a are linearly independent. 1) ˆ (1. 1)a ˆ À(1. À1) (so that u P U). Hence. either (1. by Lagrange's Theorem. 9. Therefore r is never faithful. À1) and (1. 0 < j < 2. Hence r is a representation (compare Example 1. and also (1. it is clear Hence. let u be (1. so there exist distinct g1 . 1) or (1. by the solution to Exercise 1. Hence sp ((1. Since (1. Suppose that U is a non-zero FG-submodule of V. 0 < k < 1)X that |ka. À1) lies in U. Then U is an FH-module. bl with g1 r ˆ g2 r. Using the relations. accordingly.4).404 Representations and characters of groups v P V. where H is the subgroup {1. AB ˆ BA. yx ˆ À2X1 À a3 ‡ b ‡ 2a2 b ‡ 3a3 b. On the other hand. by an argument similar to that for r1. À1) lies in U.   0 0 . jGj Therefore jGj ˆ 18. also r4 is irreducible. However. 5. x 2 ˆ 4X1 ‡ a2 ‡ 4a3 X . Finally. so r is reducible. Now let V ˆ F 2 with vg ˆ v(gr2 ) for v P V. Chapter 6 1X (a) xy ˆ À2X1 À a3 ‡ ab ‡ 3a2 b ‡ 2a3 b. r3 is irreducible. If U is a non-zero CG-submodule of V. g P G. (See Example 5. 1)a are linearly independent. g2 P ka. 1) or (1. we may write every element of G in the form ai b j c k (0 < i < 2. bl| ˆ 9 and ka. so rz ˆ 2. . If r ˆ 1 À g. Let x P G. v1 b ˆ v2 and v2 b ˆ v1. and take r ˆ 1 ‡ a. hence W is an FG-homomorphism. . so (v i a)W ˆ v i‡1 W ˆ a i‡1 ˆ (v i W)a. Chapter 7 1. and u ‡ u ˆ u. and v0 ˆ v(0 ‡ 0) ˆ v0 ‡ v0.3 to prove that W is irreducible. Relative to the basis 1. v5 be the natural basis for the permutation module for G over F. Let V be the trivial FG-module and let 0 Tˆ v P V and 1 Tˆ g P G. For all u1 . Hence ch ˆ hc ˆ c. € € (b) c2 ˆ c hPG h ˆ hPG ch ˆ jGjc. (ab)r ˆ f d0 1 0 0e d1 0 0 0e 1 0 0 0 0 1 0 0 3. Let v1 ˆ 1 ‡ ù2 a ‡ ùa2 and v2 ˆ b ‡ ù2 ab ‡ ùa2 b. ab ˆ bal. (u1 g)Wö ˆ ((u1 W) g)ö ˆ ((u1 W)ö) g ˆ (u1 (Wö)) gX 2. Let C2 3 C2 ˆ ka. and bz ˆ 1 ‡ a2 ˆ zb. then u ˆ 0. s ˆ 1 À a. b: a2 ˆ b2 ˆ 1. Use the argument of either Example 5. 4. (ëu1 )Wö ˆ (ë(u1 W))ö ˆ ë(u1 (Wö)). g. b. Then W: ë1 v1 ‡ X X X ‡ ë5 v5 3 ë1 1 ‡ ë2 a ‡ ë3 a2 ‡ ë4 a3 ‡ ë5 a4 is the required FG-isomorphism.Chapter 7 405 (b) az ˆ ab ‡ a3 b ˆ a2 ba ‡ ba ˆ za. v2 a ˆ ù2 v2 . there exists a unique h in G such that gi h ˆ gj . g ë g g with ë g P C. ë P F and g P G. Note ®rst that if u is an element of a vector space. so do gh and hg. . Check that v1 a ˆ ùv1 . j. No: let G ˆ ka: a2 ˆ 1l. 6. then vr ˆ 0 and neither v nor r is 0. (a) As g runs through G. It is easy to show that V0 is an FG-submodule of V. u2 P U. the regular representation r is given by H I H I 0 1 0 0 1 0 0 0 f1 0 0 0g f0 1 0 0g f g f g 1r ˆ f g. a. Let a ˆ (1 2 3 4 5) and let v1 . (c) All the entries in [W]B are 1 (compare the solution to Exercise 2). If r P CG then ˆ € € rˆ ë g gz ˆ ë g zg ˆ zr. . ar ˆ f d0 0 0 1e d0 0 1 0e 0 0 1 0 0 0 0 1 H I H I 0 0 0 1 0 0 1 0 f0 0 1 0g f0 0 0 1g f g f g br ˆ f gX g. Hence a i b j z ˆ za i b j for all i. 5. (Note that v i W ˆ ai . Then . ab of F(C2 3 C2 ). Now 0r ˆ (0 ‡ 0)r ˆ 0r ‡ 0r. hence 0r ˆ v0 ˆ 0. we have (u1 ‡ u2 )Wö ˆ (u1 W ‡ u2 W)ö ˆ u1 (Wö) ‡ u2 (Wö). Hence W is a CG-submodule of CG.) 3. The reason is that for all i.5(2) or the solution to Exercise 5. j and so gz€ zg for all g P G. v3 ‡ v4 ) and (FG)0 ˆ sp g X gPG Since V0 and (FG)0 have different dimensions. V ˆ sp (Àùv1 ‡ v2 ) È sp (Àù2 v1 ‡ v2 ). restricted to V0 . (a) W is easily seen to be a linear transformation. ì g P C)X 0 ìg Then (gó)(hó) ˆ (hó)(gó) for all g. Chapter 8 1. ab ˆ ba). we see that W is an FGhomomorphism from V to V0 . g P G. Then RG ˆ sp (1 ‡ a ‡ b ‡ ab) È sp (1 ‡ a À b À ab) È sp (1 À a ‡ b À ab) È sp (1 À a À b ‡ ab)X 3. v2 . Let G ˆ {1. r is equivalent to a representation ó of the form   ëg 0 gó ˆ (ë g . . it follows from Exercise 4 that V and FG are not isomorphic FG-modules. 2ˆ 3 V0 ˆ sp (v1 ‡ v2 .) 2. (b) (á À â)(1 À x)W ˆ ((á À â) À (â À á))(1 À x) ˆ 2(á À â)(1 À x). and let V be a 2-dimensional vector space over C with basis v1 . De®ne vg ˆ v for all v P V. Then by Maschke's Theorem. noting that V W  V0 . hence W is surjective. ì P C) then W is a CG-homomorphism from V to V. No: let v1 . Hence W2 ˆ 2W. 5. since all diagonal matrices . . Let G be any group. . (vö)g ˆ (vg)ö ˆ vö. b. 1 ‡ x. (wöÀ1 )g ˆ (wg)öÀ1 ˆ wöÀ1 . If we let W: ëv1 ‡ ìv2 3 ëv2 (ë. v4 be the natural basis of the permutation module V for G over F. 4. 6. For all v P V0 . where ù ˆ e2ðia3 . In the notation of Exercise 3. If v P V0 then (vajGj)W ˆ v. and so V0 ö  W 0 . 4. Hence the function ö. Suppose that ö: V 3 W is an FG-isomorphism. Also (á1 ‡ âx)xW ˆ (â1 ‡ áx)W ˆ (â À á)(1 À x) ˆ (á À â)(1 À x)x ˆ (á1 ‡ âx)WxX Hence W is an FG-homomorphism. so W 0 öÀ1  V0 .406 Representations and characters of groups ˆ gPG vxg ˆ ˆ gPG vg ˆ ˆ gPG v gxX Hence (vx)W ˆ vW ˆ (vW)x. Suppose r is reducible. and Ker W ˆ Im W ˆ sp (v2 ). . a. this makes V into a CGmodule. ab}  C2 3 C2 (so a2 ˆ b2 ˆ 1. Let g P G. (c) Let B be the basis 1 À x. h P G. (Find eigenvectors for x. For all w P W 0 . is an FG-isomorphism from V0 to W 0 . if u Tˆ 0 then (ux. v g] ˆ (ugx. y)}. . Since G a is simple. r2 : 1r1 ˆ ar1 ˆ (1). 5. r3 : 1r1 ˆ br1 ˆ b2 r1 ˆ (1). 1). v]X xPG xPG (2) It is easy to prove that U c is a subspace of V.6).Chapter 9 407 commute with each other. h P G. 0. Also ˆ ˆ [ug. so [u. . 1r2 ˆ (1). v g] ˆ [ug À1 . È Ur. 1). hence also (gr)(hr) ˆ (hr)(gr) for all g. v gg À1 ] by part (1) ˆ [ug À1 . . This means that Ui is a faithful irreducible CG-module. Chapter 9 1. r2 . (1) It is straightforward to verify for [ . ux) . where U1. (3) Let W ˆ U c. [u. r2 . 7. Irreducible representations r1 . . vx) ˆ [u. . De®ne K ˆ fx P G: vx ˆ v for all v P Ui gX Check that K is a normal subgroup of G. U ˆ sp ((1. Then for all u P U. u] . 0 for all x P G. Let CG ˆ U1 È . Irreducible representations r1 . Let C2 ˆ ka: a2 ˆ 1l. r3 . Irreducible representations r1 . Therefore r is irreducible. (x. so there is no CG-submodule W of V with V ˆ U È W. For example. Then V ˆ U È W. also K Tˆ G since g P K. . . r} and g P G such that ug Tˆ u for some u P Ui (otherwise vg ˆ v for all v P CG). . ar2 ˆ (À1)X Let C3 ˆ kb: b ˆ 1l and let ù ˆ e2ðia3 . Then there exist i P {1. and so U c is a CG-submodule of V. Let v P U c and g P G. bi r2 ˆ (ù i ). This is a contradiction. ] the axioms of a complex inner product. v] ˆ 0 since ug À1 P U X Therefore vg P U c. v gx) ˆ (ux. 0)) is the only 1-dimensional CG-submodule of V. r4 : 3 . . we must therefore have K ˆ {1}. (1. 6. Ur are irreducible CG-submodules of CG. bi r3 ˆ (ù2i )X Let C2 3 C2 ˆ {(1. We know that the regular CG-module CG is faithful (Proposition 6. (x. y). and W is a CG-submodule of V by part (2). where x 2 ˆ y 2 ˆ 1. . 3). b3 (b) r: a 3 À1 0 0 À1 gives a faithful irreducible representation of D8 (see Example 5. and the result follows from Proposition 9. Let C4 3 C4 ˆ k(x. (x i . y j )r4 ˆ (À1) i‡ j X 2.408 Representations and characters of groups gr1 ˆ (1) for all g P C2 3 C2 . (1.     0 1 1 0 . (c) The centre of C2 3 D8 is isomorphic to C2 3 C2 . (b) If g1 ˆ (x 2 . Then H i I å11 f g i FF g r: ( g 11 . . y j ) 3 (À1) i . BÀ1 AB ˆ AÀ1 when A ˆ ar and B ˆ br. g2 and g1 g2 all have order 2. b) 3 Àù 0 0 Àù . so is not cyclic. 1) and g2 ˆ (1. a) 3 . Hence z P Z(CG). n2 ˆ 3. . (a) r: (x i . so b commutes with a ‡ aÀ1 . (x i . Then r: x j 3 (e2ði ja n ) is a faithful irreducible representation of Cn . (x. 4. X X X . y 2 ) then g1 . (a) Let Cn ˆ kx: x n ˆ 1l. n1 ˆ 2. let gj generate Cn j . Then xz ˆ z ˆ zx for all x P G. g irr ) 3 f d e F ir år 0 0 is a faithful representation of C n1 3 . r. y): x 4 ˆ y 4 ˆ 1l. If M(gr) ˆ (gr)M for g ˆ a and for g ˆ b.5(2)). Also bÀ1 (a ‡ aÀ1 )b ˆ aÀ1 ‡ a. Since (g1 g2 )ó ˆ (g1 ó)(g2 ó) for all representations ó. 3 C n r of degree r. similarly for ó. 6. Check that     0 ù ù 0 r: (x. 7.16 shows that C2 3 D8 has no faithful irreducible representation. (a) Clearly a commutes with a ‡ aÀ1 . Hence r is irreducible (Corollary 9. (d) Let C3 ˆ kx: x 3 ˆ 1l and let ù ˆ e2ðia3 .3). € 5. 3.14. then i i i i ó : ( g 11 . and let å j ˆ e2ðia n j . Hence r gives a representation. Notice that the matrix   5 À6 4 À5 commutes with gó for all g P G. Check that A4 ˆ B2 ˆ I. 1). Let z ˆ gPG g. For 1 < j < r. Therefore Proposition 9. g 22 ) 3 (å11 å22 ) is a faithful representation of degree 1 . we cannot have (g1 g2 )ó ˆ g1 ó ˆ g2 ó ˆ (À1). (x i . hence ó is reducible (Corollary 9. Yes: if r ˆ 2. (b) Check that w(a ‡ aÀ1 ) ˆ Àw for all w P W. then M ˆ ëI for some ë P C. y j )r3 ˆ (À1) i . y j )r2 ˆ (À1) j . w0 ). For 0 < j < 3. w3 3 v3.Chapter 10 gives a representation of C3 3 D8. so U ˆ sp (u) of for some u. namely V. w2 ) and sp (v3 . u2 .5(2) (or see Exercise 8.5 now shows that there are exactly ®ve non-isomorphic irreducible CG-modules.b3 X 0 Ài 1 0 . v2 ˆ 1 À a ‡ a2 À a3 . while u0 ˆ gPG g. We have sp (v0 . 2. u3 ˆ 1 À a ‡ a2 À a3 À b ‡ ab À a2 b ‡ a3 bX 4. v1 ˆ 1 ‡ ia À a2 À ia3 . Then V is a trivial CG-submodule of CG. b 3 (1) r1 : a 3 (1). the subspaces sp (v0 . Let U4 ˆ sp (v1 . Theorem 10. sp (v2 . Therefore every irreducible representation of D8 over C is equivalent to precisely one of the following: r0 : a 3 (1). w2 ) ˆ U2 È U3 . b 3 (À1) r2 : a 3 (À1). b 3 (À1) 2 3 2 3 i 0 0 1 r4 : a 3 . As in Example 5. U1 . w3 ). U3 and U4 . Let V ˆ sp ( gPG g). since there is a CG-isomorphism sending v1 3 w1. where Ui ˆ sp (ui ) (0 <€ 3) and u1 . Moreover U4  U5. u2 ˆ 1 À a ‡ a2 À a3 ‡ b À ab ‡ a2 b À a3 b. as in Example 10. It is irreducible (see for example Exercise 8. so |G|u ˆ u( gPG g) ˆ € ( gPG g)u P V. U5 ˆ sp (v3 . b 3 (1) r3 : a 3 (À1). U4 and U5 are irreducible CG-modules. Let G ˆ kx: x 4 ˆ 1l. Then CG ˆ sp (1 ‡ x ‡ x 2 ‡ x 3 ) È sp (1 ‡ ix À x 2 À ix 3 ) È sp (1 À x ‡ x 2 À x 3 ) È sp (1 À ix À x 2 ‡ ix 3 )X 3. w3 ). w0 ) ˆ U0 È U1 . 409 € 1. U2 . and so CG has exactly one trivial CGsubmodule. Then. sp (v1 . namely U0 . sp (v2 .4) and faithful. w1 ) are CG-submodules of CG. u3 are as in the solution to i< Exercise 3. Now suppose that U is an arbitrary trivial CG-submodule € CG. let wj ˆ bv j. We decompose CG as a direct sum of irreducible CG-submodules. v3 ˆ 1 À ia À a2 ‡ ia3 Chapter 10 (compare the solution to Exercise 2). Let v0 ˆ 1 ‡ a ‡ a2 ‡ a3 . Thus U ˆ V. Let u1 ˆ 1 ‡ a ‡ a2 ‡ a3 À b À ab À a2 b À a3 b.8(2). Then ug ˆ u for all g P G.4). w1 ). u2 ˆ b À iab À a2 b ‡ ia3 b. CG). 14 22 and 13 3 (where 112 means twelve 1s. Let V ˆ X1 È .6. by the proof of Proposition 11. For each g P G. Let v1 . Then W1 . ö2 is a basis of HomCG (U3. moreover. By Exercise 5. 18 2 means eight 1s and one 2. Chapter 11 1. . È Ys. U)) ˆ 1. w2 be the basis of U3 described in Example 10.18). Since G is non-abelian. Then ö1 . ö2 by uö1 ˆ u. Therefore. de®ne ö g : CG 3 CG by rö g ˆ gr (r P CG).12.1). . V is irreducible. For ë P C. dim (HomCG (V. . equals k ˆ iˆ1 jf(a. v n be the natural basis of V. by Corollary 11. . W2 is a basis of HomCG (CG. either by the method of Example 5. ‡ v n ) is the unique trivial CG-submodule of V (compare Exercise 10.3.5(2) or by Exercise 8. D12 has at least two inequivalent irreducible representations of degree 2. Therefore we have constructed in®nitely many CG-submodules Im öë of the required form. . Hence the answer for D12 is 14 22 . It will be shown later (Exercises 15. Let W: U1 3 U2 be a CG-isomorphism. 1. . u P Ker öë D u ‡ ëuW ˆ 0 D u ˆ 0. Let u1 ˆ 1 À ia À a2 ‡ ia3 . not all the dimensions are 1 (see Proposition 9.13 to see that the possible answers are 112 .3) that 18 2 cannot occur. V i )) ˆ d i . by Theorem 11. It is easy to check that if ë Tˆ ì then Im öë Tˆ Im ö ì . and similarly the number of integers b with Y b  V i is €k e i . 6. Then by (11. uö2 ˆ bu (u P U3 ). 4. Also. dim (HomCG (V. 2.4.410 Representations and characters of groups 5. the dimensions are 1. Then sp (u1 . 18 2. Thus U1  Im öë . .8(2). Compare Example 11. È Xr and W ˆ Y1 È . 5. Hence. v2 3 u2 . W)) ˆ iˆ1 d i e i . Hence by Corollary 11. u2 ) is a CG-submodule of CG which is isomorphic to V. .6. Let v1 . This. where each Xa and each Yb is an irreducible CG-module. A CG-isomorphism is given by v1 3 u1 . rW2 ˆ w2 r (r P CG). b) such that X a  Y b . 3. since the sum U1 ‡ U2 is direct. W)) is equal to the number of ordered pairs (a. de®ne the function öë : U1 3 V by öë : u 3 u ‡ ëuW (u P U 1 )X Then öë is easily seen to be a CG-homomorphism. dim (HomCG (V. de®ne ö1 . 6. 2. Then {ö g : g P G} gives a basis of HomCG (CG. .4. .2. in turn.8. De®ne W1 and W2 by rW1 ˆ v1 r.5)(3) and Proposition 11. b): X a  Yb  Vi gjX Now the number of integers a with X a  V i is dim (HomCG (V . . 17. Then sp (v1 ‡ . U3 ). CG) (compare the proof of Proposition 11. and so on).8). 15. where x is a permutation ®xing 1 and 2. b ‡ a2 b.17. For elements g of other cycle-shapes. n k. j. There are (4 ) choices for the numbers i. and a basis of Z(CQ8 ) is 1. j. We have Q8 ˆ ka. 12. 2 ˆ 40 elements in all.Chapter 12 Chapter 12 411 1. b: a4 ˆ 1. There are (3 ) choices for the numbers i. 12. The centralizer CG ((1 2)) consists of all elements x and (1 2)x. then we can make two different 3-cycles (k l m) and (k m l ) from the remaining numbers. Hence by Proposition 12. with i. and three permutations for each choice. ( j l )(k m) and ( j m)(k l ) of cycle shape (2. bÀ1 ab ˆ aÀ1 l. l (unordered). Now the required result follows from Theorem 12. a2 . There are ®ve choices for i. 2. The sizes of the conjugacy classes of S6 are given in the following table: Cycle-shape (1) Class size 1 (2) 15 (3) 40 (22 ) 45 (4) (3. the conjugacy classes of A5 have sizes 1. ab ‡ a3 bX 7. 2) (5) 90 120 144 (23 ) 15 (32 ) (4. fb. Assume that g. (1 2)(3 4) G consists of all permutations of the form (i j)(k l ) with i. The conjugacy classes of Q8 are f1g. l. namely (i j)(k l ). If H is a normal subgroup of A5 then j Hj divides 60.8). x A6 Tˆ x S6 . and H is a union of conjugacy classes of A5 . a2 bg. 4 . Thus |CG ((1 2))| ˆ 2´(n À 2)!. 20. then we can make three permutations ( j k)(l m). n 3. so 1 P CG (x). 2) from the remaining numbers. Then gx ˆ xg and hx ˆ xh. Also 1x ˆ x1. Note that x P CG (g) D x P CG ( gz). 2) (6) 40 90 120 4. Hence j Hj ˆ 1 or 60. namely (i j k) and (i k j). 3 ˆ 15. so hÀ1 x ˆ xhÀ1 and ghÀ1 x ˆ gxhÀ1 ˆ xghÀ1 . thus ghÀ1 P CG (x). in agreement with Theorem n 12. (i k)( j l ) and (i l )( j k). Hence |(1 2)(3 4)(5 6) G | ˆ 5 . (c) Every element of (1 2 3)(4 5 6) G has the form (1 i j)(k l m). 5. h P CG (x).18(2). By Example 12. n G (b) (1 2 3) consists of all 3-cycles (i j k). The elements of (1 2)(3 4)(5 6) G have the form (1 i)( j k)(l m). therefore A5 is simple. There are ®ve choices for i. fa2 g. g A6 ˆ g S6 .8. This gives 5 . (a) (1 2) G ˆ {(i j): 1 < i . k (unordered). so zx ˆ xz and z P CG (x). j. and 1 P H. j. The class equation gives . l distinct. then four choices for j. Therefore CG (x) is a subgroup of G. b2 ˆ a2 . If z P Z(G) then zg ˆ gz for all g P G. k. fa. 6. a3 g.8 (since (2 ) ˆ n3a(2X(n À 2)3)). a3 bg. j < n} and this set has size (2 ). fab. k. m distinct. a ‡ a3 . Each choice gives exactly two 3-cycles. An element x of cycle-shape (5) has CS6 (x) ˆ kxl (note that |x S6 | ˆ 144 and use Theorem 12. . If. (a) For all g. Conversely. we have ÷((1 2)) ˆ 5 and ÷((1 6)(2 3 5)) ˆ 2. i (b) If no conjugacy class of G has size p. and Im ä is a subgroup of the . then p2 divides |x G | for all i xi P Z(G). |G| > p3 . Let r be a representation with character ÷. I ˆ 1r ˆ z m r ˆ (zr) m ˆ ë m I. Let r be a representation with character ÷. The characters ÷ i of r i (i ˆ 1. a4 À1 2 b. 6. a4 b ab. Then zr ˆ ëI for some ë P C. a5 b 0 0 0 À2 Also Ker r1 ˆ {1. ÷4 of C4 are as follows: 1 ÷1 ÷2 ÷3 ÷4 1 1 1 1 x 1 i À1 Ài x2 1 À1 1 À1 x3 1 Ài À1 i We have ÷reg ˆ ÷1 ‡ ÷2 ‡ ÷3 ‡ ÷4 . Since ÷(g) ˆ |®x (g)|. for all g in G. 5. and hence ÷(zg) ˆ ë÷(g). a i i Therefore p divides |x G |. a5 À1 0 a2 . so ë m ˆ 1. . 2. by Proposition 9. and hence g P Z(G) since r is faithful. If ÷ is a non-zero character which is a homomorphism. in addition.9). 4. This is a contradiction. The required result now follows from Theorem 13. If g P Z(G) then gr ˆ ëI for some ë P C. so Z(G) Tˆ f1g.11(1). a4 }.8 and (12. then ÷(1) ˆ ÷(12 ) ˆ (÷(1))2 . a2 b. Chapter 13 1. 2) are as follows: 1 ÷1 ÷2 2 2 a3 2 0 a. 7. if gr ˆ ëI for some ë P C. (b) GaKer ä  Im ä by Theorem 1. . The irreducible characters ÷1 . . by Proposition 9. Hence g 3 (det (gr)) is a representation of G over C of degree 1. det ((gh)r) ˆ det ((gr)(hr)) ˆ det (gr) det (hr). Let C4 ˆ kx: x 4 ˆ 1l. p2 divides a |Z(G)|. h PG. We have now proved that gr ˆ ëI for some ë P C if and only if g P Z(G). a3 } and Ker r2 ˆ {1. 3. and so ä is a linear character of G. so ÷(1) ˆ 1. then by the class equation. Moreover.10.14. (zg)r ˆ (zr)(gr) ˆ ë(gr). Thus. then ( gr)(hr) ˆ (hr)(gr) for all h P G.412 Representations and characters of groups jGj ˆ j Z(G)j ‡ ˆ xiP Z(G) a jx G jX i (a) For xi P Z(G). |x G | divides pn and |x G | Tˆ 1 by Theorem 12. a2 . Hence p divides |Z(G)|.14. a3 b. a3 0 0 0 b. ÷i ˆ ‡ ‡0‡ ‡ ˆ 2. so ÷ i (x) Tˆ ÷ i (1) for some irreducible character ÷ i of G. Thus ä(x) ˆ À1. Say there are r entries 1 and s entries À1. (c) Im ä is a ®nite subgroup of Cà . 3).20 (but ÷ is not). which is abelian. 9. 3 (À1) . det ([x]B ) ˆ (À1) k ˆ À1. Therefore GaKer ä is abelian. (0 1 ). øi ˆ ‡ ‡0‡ ‡ ˆ 1X 24 4 8 4 Hence ø is irreducible by Theorem 14. g2 k of CG so as to obtain a basis B in which g and gx are adjacent for all g P G. 10. 24 4 8 4 3 .5(2). a3 b 0 0 À2 . Let r be the regular representation of G. we have ÷reg (x) Tˆ ÷reg (1) (see Proposition 13. and G has a normal subgroup of index 2 by Exercise 7. And if s is even then Às  s mod 4. As x Tˆ 1. (À1) (À1)(À1) h÷. so ÷( g) ˆ r À s  r ‡ s ˆ ÷(1) mod 4. 3 (À1)(À1) 3 . . øi ˆ ‡ ‡0‡ ‡ ˆ 0. 2. 3 (À1)(À1) h÷. De®ne ä as in Exercise 7. The values of these characters are as follows: Conjugacy class ÷1 ÷2 ÷3 1 2 2 2 a2 À2 À2 2 a. Then I H 0 1 g f1 0 g f g f 0 1 [x]B ˆ f gX g f 1 0 e d FF F 0 0 There are k blocks and since k is odd. so Im ä has even order. Also À1 P Im ä. Hence Im ä contains a subgroup H of index 2. 2.20). hence is cyclic. The required result now follows from Exercise 7.10)). 10 Chapter 14 1. Order the natural basis g1 . . by Theorem 13. . a2 b 0 0 0 ab.3 1.8. by Exercise 1.Chapter 14 413 multiplicative group Cà of non-zero complex numbers. G has an element x of order 2. Let ÷ i be the character of r i (i ˆ 1. .19. Using Proposition 14. and de®ne the character ä as in Exercise 7. 8. We may choose a basis B of V so that [g]B is diagonal with all diagonal entries Æ1 (see (9. By Exercise 1. 1 3 . we obtain 3 . Let V be a CG-module with character ÷. 24 4 8 4 3. If s is odd then ä( g) ˆ À1.1 (À1)(À1) (À1)(À1) hø.7. It is easy to check that {g P G: ä(g) P H} is a normal subgroup of G of index 2. ÷1 i ˆ 1(19 . 1 ‡ 3 . Every abelian group of order 12 has 12 conjugacy classes. then i exactly a of the integers di are 1 and the rest are 0.2.414 Representations and characters of groups By Theorem 14. hence r and ó are equivalent. 1 ‡ 3 . if G ˆ Z(G) ‘ Z(G)x then G ˆ Z(G). 6 6 3 ø 2 ˆ 1 ÷1 À 1 ÷2 . 4. and this gives the required matrix T.4 and Theorem 14. Hence k÷reg . ÷i ˆ 1 ˆ ÷reg ( g)÷( g)X jGj gPG But ÷reg (g) is |G| if g ˆ 1 and is 0 if g Tˆ 1. øi ˆ iˆ1 d 2 .21. if x P G then the subgroup generated by x and the elements of Z(G) is abelian (since the elements of Z(G) commute with powers of x). we have k÷. 6 h÷. ø is not a character of G. If |G| ˆ 12 .24. ÷2 i ˆ 1(19 . the rest are 0. Since the coef®cient of ÷2 is a negative integer. Therefore the centre of a group G never has index 2 in G. Recall that hø. Then 1 ˆ h÷. (a) For all groups G. the regular character of C2 . Hence if kø. (À2) . We have h÷reg . or exactly one of the di is 2. ÷3 i ˆ 1(19 . ÷l ˆ ÷(1). but r3 is not equivalent to r1 or r2. Since all the coef®cients here are non-negative integers. by Proposition 13. 1) ˆ 3. 1 ‡ 2 . By the method used in the solution to Exercise 1. 5. by Theorem 14. 8. ÷1 i ˆ ÷( g)X jGj gPG Since ÷(1) . Chapter 15 1X h÷. (À1)(À1) ‡ 2 . (À1) . €k 7. ÷1 l Tˆ 0. 0 and by hypothesis ÷(g) > 0 for all g P G. øl ˆ a where a ˆ 1. The representations r and ó have the same character. Hence. Theorem 14. Let ÷1 be the trivial character of G. (À2)(À1)) ˆ 7X 6 h÷. As ÷ Tˆ ÷1 . 6 Hence ÷ ˆ 2÷1 ‡ 3÷2 ‡ 7÷3 .17 shows that ÷ is reducible. 6. it follows that ÷ is a character of S3 . we obtain ø 1 ˆ 1 ÷1 ‡ 1 ÷2 ‡ 1 ÷3 . 2 or 3. If kø. 1) ˆ 2. 2 ‡ 0 ‡ 2 . 4. No: let G ˆ C2 and ÷ ˆ ÷reg . (À2) . 2 2 ø 3 ˆ 1 ÷1 ‡ 1 ÷2 À 1 ÷3 X 3 3 3 3. We ®nd that ø ˆ À÷2 ‡ ÷3 ‡ ÷5 ‡ 2÷6 . øl ˆ 4. This follows at once from Exercise 11. 2.21. 3. then either exactly four of the di are 1. r1 and r2 are equivalent. (b) Since the number of irreducible representations is equal to the number of conjugacy classes. together with the relation €4 €4 iˆ1 ÷ i (1)÷ i ( g 4 ) ˆ 0. If G is abelian (e. 1) 1 1 À1 À1 (1. y) 1 À1 À1 1 2. so there cannot be as many as 9 conjugacy classes in total. similarly ÷4 ( g3 ) ˆ (À1 ‡ 5)a2. .1) (1. the remaining conjugacy classes have size at least 2. Then iˆ1 ÷ i ( g 2 )÷ i ( g 4 ) ˆ 0 € gives ÷3 ( g2 ) ˆ 1. The character table of C2 3 C2 is (cf. y). at most 4 conjugacy classes of G have size 1 (see (12. (x. y) 1 À1 1 À1 (x. gives the values on g4 .Chapter 16 415 and G is non-abelian. if G ˆ D12 then G has 6 conjugacy classes (see (12. Corollary 13. ÷4 (1) are 1. and if G ˆ A4 then G has 4 conjugacy classes (see Example 12. (1. (x. similarly ÷3 (g3 ) ˆ 1. so |Z(G)| < 4. it follows from the solution to Exercise 11.2 and part (a) that G has 4. G ˆ C4 3 C3 ) then G has 12 conjugacy classes. iˆ1 Because g4 has order 2. Finally. Let C2 3 C2 ˆ {(1. 1) ÷1 ÷2 ÷3 ÷4 1 1 1 1 (x. Exercise 9.9)). 1). The complete character table of G is gi |CG ( gi )| ÷1 ÷2 ÷3 ÷4 g1 10 1 2 1 2 g2 5 1p (À1 ‡ 5)a2 1p (À1 À 5)a2 g3 5 1p (À1 À 5)a2 1p (À1 ‡ 5)a2 g4 2 1 0 À1 0 € The two unknown degrees ÷3 (1). 6 or 12 conjugacy classes.5(2)) g1 ÷5 2 g2 À2 g3 0 g4 0 g5 0 3.18(1)). 4 ÷ i (1)÷ i ( g 2 ) ˆ 0 gives iˆ1 p p ÷4 ( g 2 ) ˆ (À1 À 5)a2.g. 1). Example 16. then |Z(G)| divides 12 and |Z(G)| Tˆ 6 or 12. 2 since 4 ( ÷ i (1))2 ˆ 10.10.12)). y): x 2 ˆ y 2 ˆ 1}. Chapter 16 1. The last row of the character table is (cf. Therefore. fa.1. fb. p If G ˆ C3 then det C ˆ Æi3 3. By the column orthogonality relations. the last irreducible character of G is 1 ÷5 2 a2 À2 a 0 b 0 ab 0 The character table of Q8 is the same as that of D8. a2 bg and fab. G9b. Use Proposition 12. every element of G has the form am bn with . 5. The matrix C is obtained from C by rearranging the columns (see Proposition 13.9(3)).13 to see quickly that bÀ1 ab ˆ a2 . a3 bg. a2 g and GaG9 ˆ fG9. fa2 g. G9abg  C2 3 C2 . (b) G9 ˆ f1.9(3)). since æ is non-real. p Hence æ ˆ (À1 Æ i 7)a2. (b) The column which corresponds to the conjugacy class containing gÀ1 2 has values which are the complex conjugates of those in the column of g2 (see Proposition 13. this is a different column of the character table of G. if det C ˆ det C then det C is real.) Chapter 17 1. 2. G9a. Therefore det C ˆ Ædet C. (a) Using the relations. It is easy to see that a7 ˆ b3 ˆ 1. By the column orthogonality relations applied to the column €k corresponding to g. (a) The conjugacy classes of Q8 are f1g. 6. we have iˆ1 ÷ i ( g)÷ i ( g) ˆ jC G ( g)j. Let g P G.416 4. The character table of C2 3 C2 is given in the solution to Exercise 16. (a) €5 Representations and characters of groups iˆ1 ÷ i ( g 1 )÷ i ( g 2 ) € ˆ 0 gives 3 ‡ 3æ ‡ 3æ ˆ 0. This number is equal to |G| if and only if CG (g) ˆ G. C t C is the k 3 k diagonal matrix whose diagonal entries are |CG ( gi )| (1 < i < k). which occurs if and only if g P Z(G). Hence  jdet Cj2 ˆ jC G ( g i )j. Hence the linear characters of G are gi |CG ( gi )| ÷1 ÷2 ÷3 ÷4 1 8 1 1 1 1 a2 8 1 1 1 1 a 4 1 1 À1 À1 b 4 1 À1 1 À1 ab 4 1 À1 À1 1 (c) Using the column orthogonality relations. and 5 ÷ i ( g 2 )÷ i ( g2 ) ˆ 7 iˆ1 gives 3 ‡ 2ææ ˆ 7. a3 g. and if det C ˆ Àdet C then det C is purely imaginary. (The sign depends upon the ordering of rows and columns. Chapter 17 417 0 < m < 6. ÷2 ˆ ÷. (b) The conjugacy classes of G are f1g. i gi |CG ( gi )| | gG | i ÷1 ÷2 ÷3 ÷4 ÷5 ÷6 g1 12 1 1 1 1 1 2 2 g2 4 3 1 Ài À1 i 0 0 g3 4 3 1 i À1 Ài 0 0 g4 6 2 1 1 1 1 À1 À1 g5 6 2 1 À1 1 À1 À1 1 g6 12 1 1 À1 1 À1 2 À2 . If G has 3 or 4 linear characters then jGaG9j ˆ 3 or 4 and again G is not simple as G9 v G.11. all of these are irreducible characters by Proposition 17. fa. But a has order 7 and b has order 3.6). 4. a6 g. If there are 12. Therefore jGj ˆ 21. we have ÷(a) Tˆ ÷(a) (see Corollary 15. The number of linear characters of G divides jGj by Theorem 17. so we get three linear characters of G: gi |CG ( gi )| ÷1 ÷2 ÷3 1 21 1 1 1 a 7 1 1 1 a3 7 1 1 1 b 3 1 ù ù2 b2 3 1 ù2 ù where ù ˆ e2ðia3 . so 21 divides jGj by Lagrange's Theorem. fa3 .14. hence jGj < 21. Now consult the solution to Exercise 11. therefore for some irreducible character ÷. The centralizer orders are obtained by using the orthogonality relations. and the class sizes j g G j come from the equations i jGj ˆ jC G ( g i )jj g G j (Theorem 12. 3.18).2 to see that there are 3. then G is abelian (see Proposition 9. so G is certainly not simple.8). ÷6 ˆ ö÷. ÷4 ˆ ÷2 ÷3 . Applying the column orthogonality relations. ÷5 ˆ ö. a4 g. 4 or 12 linear characters. (c) First. we obtain 1 ÷4 ÷5 3 3 a á á a3 á á b 0 0 b2 0 0 p where á ˆ (À1 ‡ i 7)a2. Hence ÷4 and ÷5 must be complex conjugates of each other. G9 ˆ kal. 0 < n < 2. a2 . To ®nd the two remaining irreducible characters ÷4 and ÷5 . fa m b: 0 < m < 6g and fa m b2 : 0 < m < 6g. we have ÷1 ˆ 1 G . ÷3 ˆ ÷ 2 . we note that a is not conjugate to aÀ1 . In the character table below. a5 . they are not equivalent to any representation found earlier. . for å ˆ e2ði ka2 n with 0 < k < n À 1. 12 ˆ 4nX Hence we have found all the irreducible representations. n À 1. bi ˆ Ker ÷3 . these representations are irreducible and inequivalent. . . since b2 is in the kernel of each of these representations. (a) Check that the given matrices satisfy the relevant relations. The sum of the squares of the degrees of the irreducible representations we have found is n . (For further details on the representations of degree 1. hai ˆ Ker ÷2 . for å ˆ e2ði ja n with 0 < j < n À 1. ˆ . see the solution to Exercise 18. 0 å À1 0 1 0 å À1 ån 0 32 3À1 2 3À1 2 3 2 0 1 å 0 0 1 å 0 X ˆ ån 0 0 å À1 ån 0 0 å À1 Hence we have representations of T4 n (cf.418 Representations and characters of groups D8 ˆ Ker ÷1 .) 7.4). so jGaG9j ˆ 2n and there are 2n representations of degree 1. and inequivalent (consider the character values on a2 ). Moreover G9 ˆ ka2 l. ha2 . (a) Check that the given matrices satisfy the relevant relations. 22 ‡ 4 . Also G9 ˆ kbl. so we have obtained all the irreducible representations. . note that the structure of GaG9 depends upon whether n is even or odd. are irreducible (by Exercise 8. Now the sum of the squares of the degrees of the irreducible representations we have found so far is (n À 1) . (b) The representations in part (a) are irreducible unless å ˆ Æ1. since they have distinct characters. Note that b2 does not belong to the kernel of any of these representations. The normal subgroups of D8 are 6. Example 1. 0 çÀ1 1 0 where ç is any (2n)th root of unity in C.4. . ha2 .4) and inequivalent (their characters are distinct). We get further representations by     ç 0 0 1 a3 . 22 ‡ 2n . we get n À 1 irreducible representations. f1g ˆ Ker ÷5 X 5.4). Moreover.3. (b) The given representations. abi ˆ Ker ÷4 . by Exercise 8. ha2 i ˆ Ker ÷2 ’ Ker ÷3 . with r ˆ 1.b3 . 12 ˆ 6n.12. are irreducible (by Exercise 8.11). so jGaG9j ˆ 4 and there are four representations of degree 1 (see Theorem 17. (b) The given representations. For å ˆ e2ði ra2 n . no two of which are equivalent. by Theorem 11. 2. 8. (a) Check that the given matrices satisfy the relevant relations: 2 32 n 2 3n 2 3 2 32 å 0 1 0 å 0 0 1 ˆ . For ç ˆ e2ði ja2 n with 1 < j < n À 1. Chapter 18 419 Finally. Let ù ˆ e2ðia6 .3. 12 ˆ 8nX Chapter 18 1. since the sum of the squares of the irreducible representations given above is n .3(3) or Section 18. b2 l and GaG9  C2 3 C2. ù2 ‡ ùÀ2 ˆ ù4 ‡ ùÀ4 ˆ À1.) 2. Character table of D12 gi |CG ( gi )| ÷1 ÷2 ÷3 ÷4 ÷5 ÷6 1 12 1 1 1 1 2 2 a3 12 1 1 À1 À1 À2 2 a 6 1 1 À1 À1 1 À1 a2 6 1 1 1 1 À1 À1 b 4 1 À1 1 À1 0 0 ab 4 1 À1 À1 1 0 0 . Then ù ‡ ùÀ1 ˆ 1. The character table of D8 is as shown. using Section 18. Then ð takes the following values: 1 ð 4 a2 0 a 0 b 2 ab 0 Hence ð ˆ ÷1 ‡ ÷3 ‡ ÷5 . (Compare Example 14. Character table of D8 gi |CG ( gi )| ÷1 ÷2 ÷3 ÷4 ÷5 1 8 1 1 1 1 2 a2 8 1 1 1 1 À2 a 4 1 1 À1 À1 0 b 4 1 À1 1 À1 0 ab 4 1 À1 À1 1 0 (See Example 16.) Regarding D8 as the symmetry group of a square. We have now found all the irreducible representations. G9 ˆ ka2 . 22 ‡ 4 . so we get four representations of degree 1. Hence. take b to be a re¯ection in a diagonal of the square. 22 ‡ (n À 1) . the character table of D12 is as shown. where we took b to be a different re¯ection.28(2).3. then GaG9  C2 3 C2 and the linear characters are gi ÷1 ÷2 ÷3 ÷4 1 1 1 1 1 an 1 1 1 1 ar (1 < r < n À 1) 1 1 (À1) r (À1) r b 1 À1 1 À1 ab 1 À1 À1 1 Note that T4  C4 . a2 r b2 g. Exercise 17. for 0 < r < n À 1. The n ‡ 3 conjugacy classes of G are f1g. The remaining four irreducible characters of G are linear. abl ˆ Ker ÷4 . T8  Q8 and T12 is the Example in Section 18.4. aÀ r g(1 < r < n À 1). fan g. fa2 r‡1 . a2 r‡1 b. as shown. fa2 j b: 0 < j < n À 1g. then GaG9 ˆ hG9bi  C4 and the linear characters are gi ÷1 ÷2 ÷3 ÷4 1 1 1 1 1 an 1 À1 1 À1 ar (1 < r < n À 1) 1 (À1) r 1 (À1) r b 1 i À1 Ài ab 1 Ài À1 i If n is even. ka3 l ˆ Ker ÷6 and {1} ˆ Ker ÷5 . a2 r‡1 b2 gX We have G9 ˆ hbi and GaG9 ˆ hG9ai  C2 n . kal ˆ Ker ÷2 .7 gives us n irreducible characters ø k (0 < k < n À 1). 3. ka2 .6 as follows: gi |CG ( gi )| øj 1 4n 2 an 4n 2(À1) j ar (1 < r < n À 1) 2n ù rj ‡ ùÀ rj b 4 0 ab 4 0 where ù ˆ e2ðia2 n . ka2 . Hence we get 2n linear characters ÷ j (0 < j < 2n À 1). fa2 j‡1 b: 0 < j < n À 1gX We get n À 1 irreducible characters ø j (1 < j < n À 1) of G from Exercise 17. The 3n conjugacy classes of G are. 4. If n is odd. fa2 r b. ka2 l ˆ Ker ÷3 ’ Ker ÷4 .420 Representations and characters of groups Seven normal subgroups of D12 are G ˆ Ker ÷1 . . fa2 r g. far . bl ˆ Ker ÷3 . the character table of V24 is given at the top of p. 422. aÀ2s b2 g(1 < s < (n À 1)a2). Character table of V8 n gi |CG ( gi )| ÷1 ÷2 ÷3 ÷4 øj (0 < j < n À 1) öj (1 < j < n À 1) 1 8n 1 1 1 1 2 2 a2 r‡1 (0 < r < n À 1) 8n 4n 1 1 1 1 À2 2 1 1 À1 À1 b2 a2s a2s b2 (1 < s < (n À 1)a2) 4n 4n 1 1 1 1 ù4 js ‡ùÀ4 js ù2 js ‡ùÀ2 js 1 1 1 1 Àù4 js ÀùÀ4 js ù2 js ‡ùÀ2 js b 4 1 À1 1 À1 0 0 ab 4 1 À1 À1 1 0 0 ù2 j(2 r‡1) ÀùÀ2 j(2 r‡1) ù j(2 r‡1) ‡ùÀ j(2 r‡1) Note: ù ˆ e2ðia2 n . ÷4 . fa2s b2 . aÀ2 rÀ1 b2 g(0 < r < n À 1). and faj bk : j odd. 5.8. as shown below. fa2 r‡1 . and a further n À 1 characters ö j (1 < j < n À 1) of degree 2. k ˆ 1 or 3g. U12  T12 and U18  D6 3 C3 . For example. n characters ø j (0 < j < n À 1) of degree 2. The 2n ‡ 3 conjugacy classes of G are f1g. aÀ2s g. . . Observe that U6  D6. k ˆ 1 or 3gX a2 r 6n ù2 jr 2ù2 kr a2 r b 3n ù2 jr Àù2 kr a2 r‡1 2n ù j(2 r‡1) 0 421 Using Exercise 17. we get four linear characters ÷1 . fb2 g. faj bk : j even. .Chapter 18 Character table of U6 n gi |CG ( gi )| ÷j (0 < j < 2n À 1) øk (0 < k < n À 1) Note: ù ˆ e2ðia2 n . . fa2s . öl Tˆ 0. k÷ø. ö A ˆ ø4 X . Since ÷ is not faithful.15 and (14. Hence ö( g) ˆ ö(1) for all irreducible characters ö for which k÷ n . by Exercise 1. . ÷öl. Therefore k÷ n .422 Representations and characters of groups Character table of V24 gi |CG ( gi )| ÷1 ÷2 ÷3 ÷4 ø0 ø1 ø2 ö1 ö2 1 24 1 1 1 1 2 2 2 2 2 b2 24 1 1 1 1 À2 À2 À2 2 2 a 12 1 1 À1 À1 p0 ip 3 Ài 3 1 À1 a3 12 1 1 À1 À1 0 0 0 À2 2 a5 12 1 1 À1 À1 0 p Ài 3 p i 3 1 À1 a2 12 1 1 1 1 2 À1 À1 À1 À1 a2 b2 12 1 1 1 1 À2 1 1 À1 À1 b 4 1 À1 1 À1 0 0 0 0 0 ab 4 1 À1 À1 1 0 0 0 0 0 Chapter 19 1X h÷ø. By Proposition 15. k÷ø. øöiX jGj gPG jGj gPG Similarly.  V (n factors).14 we obtain 1 ÷S ÷A öS öA 15 10 6 3 (1 2 3) 0 1 0 0 (1 2)(3 4) 3 À2 2 À1 (1 2 3 4 5) 0 0 1 p (1 3 4 5 2) 0 0 1 p (1 ‡ 5)a2 (1 À 5)a2 Then ÷ S ˆ ø1 ‡ ø2 ‡ 2ø3 . øl ˆ 0. öl ˆ kø.13). ö S ˆ ø1 ‡ ø3 . Then wg ˆ w for all w P V  . there exists 1 Tˆ g P G with vg ˆ v for all v P V. Using Proposition 19. Let n be an integer with n > 0.5 there is an irreducible character ø of G such that ø(g) Tˆ ø(1). öi ˆ 1 ˆ 1 ˆ ÷( g)ø( g)ö( g) ˆ ÷( g)ø( g)ö( g) ˆ h÷. 1 G l ˆ k÷. øl. Let V be a CG-module with character ÷. 3. . Note that (1 2 3 4 5)2 is conjugate to (1 3 4 5 2) in A5 . The result now follows from Proposition 13. 4. 2. ÷ A ˆ ø2 ‡ ø4 ‡ ø5 . Take b to be the re¯ection in the axis shown: . Character table of D6 3 D6 ( gi . 5. below. b) (b.14. as in Example 1. these are irreducible by Propositions 13. b: a3 ˆ b2 ˆ 1.2) gi |CG ( gi )| ÷1 ÷2 ÷3 ÷4 ÷5 ÷6 ÷7 Note: ù ˆ e2ðia3 6. a) (b. Exercise 27. these characters are irreducible. The table also records the trivial character ÷1 . a) (1. 1) (a.Chapter 20 423 5. 4. 1) (1. bÀ1 ab ˆ aÀ1 l. ÷ i l ˆ 1 for i ˆ 2. 1) (b. Character table of G (cf.15 and 17. hj ) |CG ( gi . b) (a.1(3). Since k÷ i . ÷ S as ÷4 and ÷ A as ÷2 . We have recorded ÷ as ÷5 . a) (a. the character table of D6 3 D6 is as shown. hj )| ÷1 ÷2 ÷3 ÷1 ÷2 ÷3 ÷1 ÷2 ÷3 3 ÷1 3 ÷1 3 ÷1 3 ÷2 3 ÷2 3 ÷2 3 ÷3 3 ÷3 3 ÷3 (1. Since G has seven conjugacy classes. Taking D6 ˆ ka. b) 36 18 12 18 9 6 12 6 4 1 1 2 1 1 2 2 2 4 1 1 À1 1 1 À1 2 2 À2 1 À1 0 1 À1 0 2 À2 0 1 1 2 1 1 2 À1 À1 À2 1 1 À1 1 1 À1 À1 À1 1 1 À1 0 1 À1 0 À1 1 0 1 1 2 À1 À1 À2 0 0 0 1 1 À1 À1 À1 1 0 0 0 1 À1 0 À1 1 0 0 0 0 g1 24 1 1 1 3 2 2 2 g2 24 1 1 1 3 À2 À2 À2 g3 4 1 1 1 À1 0 0 0 g4 6 1 ù ù2 0 Àù2 Àù À1 g5 6 1 ù2 ù 0 Àù Àù2 À1 g6 6 1 ù2 ù 0 ù ù2 1 g7 6 1 ù ù2 0 ù2 ù 1 Chapter 20 1. the character table is complete. ÷3 ˆ ÷2 . ÷6 ˆ ÷5 and ÷7 ˆ ÷2 ÷5 . (a) Regard D8 as the subgroup of S4 which permutes the four corners of a square. ÷5 of S4 as in Section 18. or using (20. we ®nd that the characters ÷ i 5 A6 (i ˆ 1. 9) are distinct irreducible characters of A6 . We obtain ÷1 5 H ˆ ø1 . . ÷3 5 H ˆ ø1 ‡ ø4 . The inequality k÷ 5 H.13). ÷11 5 A6 l ˆ 2. .424 Representations and characters of groups Then a 3 (1 2 3 4). . . ÷11 be the irreducible characters of S6 . . ÷4 5 H ˆ ø3 ‡ ø5 . ‡ dr ø r for some non-negative integers di . ÷ 5 Hl H .2). 3.14. (b) Take the irreducible characters ÷1 . Either by direct calculation. For examples with d ˆ 1 or 2. Character table of A6 gi 1 (1 2 3) |C A6 ( g i )| 360 9 ø1 ø2 ø3 ø4 ø5 ø6 ø7 1 5 10 9 5 8 8 1 2 1 0 À1 À1 À1 p (1 2)(3 4) (1 2 3 4 5) (1 3 4 5 2) (1 2 3)(4 5 6) (1 2 3 4)(5 6) 8 5 5 9 4 1 1 À2 1 1 0 0 5)a2. and take the character table of H to be gi |CH ( gi )| ø1 ø2 ø3 ø4 ø5 1 8 1 1 1 1 2 (1 3)(2 4) 8 1 1 1 1 À2 (1 2 3 4) 4 1 1 À1 À1 0 (1 3) 4 1 À1 1 À1 0 (1 2)(3 4) 4 1 À1 À1 1 0 (see Example 16. . b 3 (1 3) gives the required isomorphism.5. Let ø1 . . . . k÷11 5 A6 . Let ÷1 . . Write d ˆ k÷ 5 H. ø5 in our character table below. Also. take G ˆ A4 . take G ˆ S3 and H a subgroup of order 2. ÷5 5 H ˆ ø2 ‡ ø5 X 2. . . ÷ 5 Hl H < 3 follows at once from Proposition 20. 7. and ÷ an irreducible character of G of degree d. these give the characters ø1 .17. .1.6) gives ÷(1) ˆ d 1 ‡ X X X ‡ d r < d 2 ‡ X X X ‡ d 2 < nX 1 r 4. ⠈ (1 À 1 0 0 À1 0 á â p 5)a2 1 0 0 À1 0 â á 1 À1 1 0 2 À1 À1 1 À1 0 1 À1 0 0 Note: á ˆ (1 ‡ 3.3(3) or Section 18. . we obtain from ÷11 5 A6 the two irreducible characters which we have called ø6 and ø7 . the inequality (20. . For an example with d ˆ 3. ø r be the irreducible characters of H. Since each ø i has degree 1. 5. Then ÷ 5 H ˆ d1 ø1 ‡ . . . ÷2 5 H ˆ ø4 . H ˆ V4 and ÷ an irreducible character of G of degree 3 (see Section 18. . Arguing as in Example 20. .3). as in Example 19. From the remaining fourteen irreducible characters of S7 .Chapter 21 425 5. the restriction of the irreducible character of degree 20 to A7 must be the sum of two different irreducible characters of degree 10. the elements u and ua form a basis of U 4 G. ÷4 5 H ˆ ÷5 5 H ˆ ø1 ‡ ø2 ‡ ø3 . ÷3 5 H ˆ ø2 ‡ ø3 . Then ua2 ˆ Àu and ub ˆ u.13). 14. the induced module U 4 G is irreducible. ø 4 Gi ˆ 1. Label the characters of H as follows: 1 ø1 ø2 ø3 1 1 1 (1 2 3) 1 ù ù2 (1 3 2) 1 ù2 ù where ù ˆ e2ðia3 . and we get precisely seven if and only if the restriction of each of the fourteen characters is irreducible. See (20. 21. 10. (a) ÷1 5 H ˆ ÷2 5 H ˆ ø1 . 10. (b) Since every element of G belongs to H or to Ha. 14. ø2 4 G ˆ ø3 4 G ˆ ÷3 ‡ ÷4 ‡ ÷5 X 3. we obtain ø1 4 G ˆ ÷1 ‡ ÷2 ‡ ÷4 ‡ ÷5 . 6. 35X Chapter 21 1. Hence sp (u) is a C H-submodule of CH. 15. We are told that A7 has exactly nine conjugacy classes. (b) Using the Frobenius Reciprocity Theorem. Let Hgj (1 < j < m) be the distinct right . (c) The character ø of U and the character ø 4 G of U 4 G are given by 1 ø 1 a2 À1 b 1 a2 b À1 1 ø4G 2 a2 À2 a 0 b 0 ab 0 Since hø 4 G. It is suf®cient to prove that if U is a CH-submodule of CH then dim (U 4 G) ˆ jG : Hjdim U . upon restriction to A7 we get at least seven irreducible characters of A7 . (a) Let u ˆ 1 À a2 ‡ b À a2 b. 2. Since 20 occurs only once in the list of degrees for S7 . Hence the irreducible characters of A7 have degrees 1. ) . By using the Frobenius Reciprocity Theorem twice. 7. Then ÷(1) ˆ 2ø(1) and k÷ 5 H. Cycle-shape ö4G ø4G (1) 240 720 (7) 2 À1 (3. we deduce. öi G ˆ hø(÷ 5 H).17 that (ø(÷ 5 H)) 4 G ˆ (ø 4 G)÷. ö 5 Hi H ˆ hø. together with the result of Exercise 19. hence ÷ 5 H is reducible. 3) 12 0 6. Also. Hence dim(U 4 G) ˆ dim(U (CG)) ˆ m dim U . By applying the result of Exercise 6. dim (Ugj ) ˆ dim U (since u 3 ugj (u P U) is a vector space isomorphism). Suppose ®rst that ø 4 G is irreducible. then for precisely one other irreducible character ö of H we have ø 4 G ˆ ö 4 G. The values of ö 4 G and ø 4 G are given by Proposition 21. Thus ÷ i (1) > di ø(1). . ÷ i 5 Hi H . ‡ Ugm . or (2) ø 4 G is the sum of two different irreducible characters of the same degree. that either (1) ø 4 G is irreducible. We have hø9 4 G. since ø is irreducible. 3). 4. øl H ˆ 1. the values are as follows. 5. we obtain h(ø(÷ 5 H)) 4 G. and therefore jG: Hjø(1) > (d 2 ‡ X X X ‡ d 2 )ø(1)X 1 k The required result follows. .23. as in the proof of Proposition 20. say ø 4 G ˆ ÷. by the Frobenius Reciprocity Theorem. ÷öi G ˆ h(ø 4 G)÷. . On elements of cycle-shapes (1). where Ugj ˆ {ugj : u P U}. ‡ dk ÷ k (1). The sum Ug1 ‡ . öi G X Since this holds for all irreducible characters ö of G. ÷ i i G ˆ hø. ÷ i 5 H ˆ di ø ‡ â where either â is a character of H or ⠈ 0. ‡ Ug m is direct (since the elements in Ugj are linear combinations of elements in the right coset Hgj ).1 (also twice). ÷ 5 Hi H Tˆ 0 D ø9 ˆ ø or öX Thus If ø 4 G is irreducible. . ÷i G Tˆ 0 D hø9. we deduce from Theorem 14. (Compare Proposition 20. (7) and (3. (÷ö) 5 Hi H ˆ hø 4 G. . where d i ˆ hø 4 G. Then U(CG) ˆ Ug1 ‡ .11.9. Let ö be an irreducible character of G. Hence. Now suppose that ø9 is an irreducible character of H.426 Representations and characters of groups cosets of H in G. . and on all other elements the values are zero. We have |G: H|ø(1) ˆ d1 ÷1 (1) ‡ . by the Frobenius Reciprocity Theorem. say ÷ 5 H ˆ ø ‡ ö. ÷l ˆ kb÷reg . ÷1 i G Tˆ 0 D hø9.Chapter 22 427 Suppose next that ø 4 G is reducible. ÷l P Z and k1 G .11). then r divides 16. 12 ‡ s . 3. (b) jG9j ˆ p by Theorem 17. see Chapter 25. 22 ˆ 16X Hence r ˆ 4 or 8 or 16.12). ÷l ˆ bjGj÷(1)ajGj ˆ b÷(1). since both sides of this equation take the same values on all elements of G. where r divides pq. øl H ˆ 1. But kö À a1 G . This time.12. then ø9 ˆ ø. öl and k÷reg . and jGj 1 jGj(a ‡ bjGj) ˆ a ‡ bjGjX jGj Since ö is a character. and ø9 is an irreducible character of H such that ø9 4 G has ÷1 or ÷2 as a constituent.11). and so G is abelian by Proposition 9.11 and 22. and if there are r characters of degree 1 and s of degree 2. say ø 4 G ˆ ÷1 ‡ ÷2. 2. . Then ö ˆ a1 G ‡ b÷reg . hence ÷1 5 H ˆ ø.) Chapter 22 1. and r . 17. (a) By hypothesis. (c) If ÷ is a non-trivial irreducible character of G. ÷1 5 Hi H Tˆ 0 D ø9 ˆ øX Thus If ø 4 G ˆ ÷1 ‡ ÷2 where ÷1 and ÷2 are irreducible characters of G. Now suppose that ø9 is an irreducible character of H.12. öi ˆ 1 (a ‡ bjGj ‡ (jGj À 1)a) ˆ a ‡ b. Hence kö À a1 G . and r ‡ s ˆ 7 or 10 or 16. Then ÷1 (1) ˆ ø(1) and k÷1 5 H. and the sum of the squares of the degrees of the irreducible characters is 15 (Theorem 11. (b) We have h1 G . b P C such that ö(g) ˆ a for all g Tˆ 1 and ö(1) ˆ a ‡ bjGj.11 again. each degree divides 15 (Theorem 22.11 show that there are r irreducible characters of degree 1 and s irreducible characters of degree q.11 and 22.12. Hence every irreducible character has degree 1. öi ˆ h÷reg . (c) The number of conjugacy classes of G is r ‡ s. öl are integers.11. there exist a. both k1 G .) 4. moreover. ÷l P Z. not every irreducible character has degree 1 (Proposition 9. The degree of each irreducible character is 1 or 2. (Compare Proposition 20. (For more information on groups of order pq. We have hø9 4 G. 1 < s and r ‡ sq 2 ˆ pqX Hence r ˆ q and s ˆ ( p À 1)/q. (a) Since G is non-abelian.18). ÷l ˆ 0. The number of linear characters divides 15 (Theorem 17. 17. then kö. Theorems 11.18. Use Theorems 11. À1). (c) By Corollary 22. and ÷ i (g) ˆ Æ1 for 4 < i < 7.27.) 6. ÷(g) is an integer for all characters ÷. Therefore. by Lagrange's Theorem. then k÷. and hence á ˆ À÷(1)a2. The only possibility which is consistent with equation (IV) is that the values of ÷ i (1) for 4 < i < 7 are 1. (b) For all g P G. and from equation (III) we have (IV) 7 ˆ (÷ i (1))2 ˆ 69X iˆ4 Hence the possibilities for the pairs of integers (÷ i (1). (6.2. By Corollary 22. 1. with ÷1 ˆ 1 G . we deduce that ÷2 (1) ˆ ÷3 (1) ˆ 5. say i ˆ 2. is an integer. 1).16. 4. or ÷ i (g) ˆ Æ2 for exactly one i and ÷ i (g) ˆ 0 for all other i . (b) We deduce from part (a) that ÷ i (g) ˆ 0 for two values of i. Partition Gnf1g into subsets by putting each element with its inverse. Therefore. This contradicts Proposition 22. (d) We now have the following part of the character table of G: . and (II) 1 ‡ 7 ˆ iˆ2 ÷ i (1)÷ i ( g) ˆ 0X From equation (I) we deduce that either ÷ i ( g) ˆ 0. the second possibility is ruled out by equation (II). 3. 4.1 and Corollary 23. The stated result follows. a. (a) If g P G then g has odd order. since 1 ˆ h÷. But À÷(1)a2 is a rational number which is not an integer (since ÷(1) divides jGj. By the column orthogonality relations. . Let ÷1 . 6 in some order. by part (a).27. 1). (Further information about the number of characters ÷ such that ÷ ˆ ÷ appears in Theorem 23. if g2 ˆ 1 then g ˆ 1. Each such subset has size 2. and ÷(g) is an algebraic integer. we have (I) 1‡ 7 ˆ iˆ2 (÷ i ( g))2 ˆ 5. (a) By Theorem 22. 1 G l ˆ 0. (4. ÷7 be the irreducible characters of G. Thus ÷ ˆ 1 G .428 Representations and characters of groups (d) Since ÷(1) divides jGj. part (c) implies that b|G| is an integer. . and hence also b. Hence ˆ ÷( g) ˆ ÷(1) ‡ 2á gPG for some algebraic integer á. Æ1 for all i. ÷ i ( g)) with 4 < i < 7 are (1. ÷2 (1)  ÷3 (1)  0 mod 5. . ÷ i (1)  ÷ i ( g) mod 5 for all i.5. we have ÷( g) ‡ ÷( g À1 ) ˆ ÷( g) ‡ ÷( g) ˆ 2÷( g). by part (b). 5. Also (III) 2 2 7 ˆ iˆ1 (÷ i (1))2 ˆ 120X Since 5 ‡ 10 . hence is odd). . 120. 1 G i ˆ ÷( g)X jGj gPG (c) If ÷ Tˆ 1 G in part (b). The character table of G is as shown. 7. we deduce that iˆ1 ÷ i (g3 ) ˆ Æ1 for 1 < i < 4 and€ the values of ÷ i ( g3 ) for 5 < i < 7 are that 7 0. 0. an (1) First. À1. € (4) We have ÷ i (g6 )  ÷ i (g4 ) mod 2 and 7 (÷ i (g6 ))2 ˆ 6. 1. from which €7 we see that ÷ i (g3 ) ˆ 1 for 1 < i < 4. gi Order of gi |CG ( gi )| ÷1 ÷2 ÷3 ÷4 ÷5 ÷6 ÷7 g1 1 120 1 5 5 1 4 4 6 g2 2 12 1 À1 1 À1 À2 2 0 g3 2 8 1 1 1 1 0 0 À2 g4 3 6 1 À1 À1 1 1 1 0 g5 4 4 1 1 À1 À1 0 0 0 g6 6 6 1 À1 1 À1 1 À1 0 g7 5 5 1 0 0 1 À1 À1 1 . We are told that ÷ i (gj ) is€ integer for all i. Therefore iˆ1 ÷ i (g6 ) ˆ Æ1 for 1 < i < 6 and ÷7 (g6 ) ˆ 0.Chapter 22 gi Order of gi |CG ( gi )| ÷1 ÷2 ÷3 ÷4 ÷5 ÷6 ÷7 g1 1 120 1 5 5 1 4 4 6 g2 2 12 1 g3 2 8 1 g4 3 6 1 g5 4 4 1 g6 6 6 1 429 g7 5 5 1 0 0 1 À1 À1 1 We successively calculate the ®ve remaining columns of the character table. Hence the values of iˆ1 ÷ i (g4 ) for 1 < i < 7 are 1. € (3) Since ÷ i (g3 )  ÷ i (1) mod 2 and 7 (÷ i (g3 ))2 ˆ 8. 1. 5 and 7 we obtain ÷2 (g6 ) ˆ À÷3 (g6 ) ˆ ÷4 (g6 ) ˆ À1 and (without loss of generality) ÷5 (g6 ) ˆ À÷6 (g6 ) ˆ 1. 1. Æ2 in some order. € 1. 4. Also iˆ1 ÷ i (g3 )÷ i ( gr ) ˆ 0 for r ˆ 4. € Hence ÷ i (g5 ) ˆ Æ1 iˆ1 for 1 < i < 4 and ÷ i (g5 ) ˆ 0 for 5 < i < 7. À1. (5) Only the entries in column 2 remain to be calculated. ÷ i (g5 )  ÷ i (1) mod 2 and 7 (÷ i (g5 ))2 ˆ 4. respectively. À2 in order from the top. These can be obtained from the column orthogonality relations. 0. 1. 0. iˆ1 we deduce that ÷4 ( g5 ) ˆ À1 and (without loss of generality) ÷2 (g5 ) ˆ À÷3 (g5 ) ˆ 1. ÷ i ( g4 )  ÷ i (1) mod 3 and 7 (÷ i ( g4 ))2 ˆ 6. By applying the column orthogonality relations involving column 6 and columns 3. j. Since 7 ÷ i (1)÷ i (g5 ) ˆ 0. From the relation iˆ1 ÷ i (1)÷ i ( g 3 ) ˆ 0 we now deduce that the entries in column 3 are 1. (2) Next. 0. 1. Let ë1 and ë2 be the eigenvalues of gr.13 of é÷ that é÷ ˆ À1 if and only if ÷ A ˆ 1 G . 3 C n r which is given by i i ÷( g 11 X X X g irr ) ˆ ë11 X X X ë irr is real if and only if ë i ˆ Æ1 for all i with 1 < i < r. 3.14). However. Since jGj is odd. . summing over all the irreducible characters ÷. . Since ÷(1) ˆ 2 we have ÷ A (1) ˆ 1. the elements g of G which satisfy g2 ˆ 1 are i precisely those elements g11 . . either i j ˆ 0 or n j is even and i j ˆ n j a2. . Now the ni th root of unity ë i can be À1 if and only if ni is even. Assume that x P G and x is real. Then det (A À ëIn ) ˆ 0. it follows that x ˆ 1. where m is the number of the integers n1 . so g2 P CG (x). . Adopt the notation of Theorem 9. € These numbers coincide with ÷(1). . m ˆ 2n ‡ 1 for some integer n. As p(ë) ˆ 0. ë is therefore an algebraic integer. Then gÀ1 xg ˆ x À1 for some g P G. all of whose entries are integers. Since x 2 ˆ 1 and x has odd order. . which is of the form x n ‡ anÀ1 x nÀ1 ‡ X X X ‡ a1 x ‡ a0 (ar P Z)X Conversely. Then g ˆ g2( n‡1) P CG (x). Hence gÀ2 xg2 ˆ x. ‡ a nÀ1 x nÀ1 ‡ x n (a r P Z). and hence ë is a root of the polynomial det (xIn À A). by Lagrange's Theorem. It now follows from the De®nition 23.430 Representations and characters of groups 7. nr which are even. 4. . 2. . a i b (0 < i < n À 1) (and also a na2 if n is even)X This gives n ‡ 1 elements if n is odd. Since é÷ < 1 for all ÷. . The result follows. Let I H 0 1 0 XXX 0 f 0 0 1 0 g g f f 0 0 0 0 g f F Aˆf F F gX F g F g f F d 0 0 0 1 e Àa0 Àa1 Àa2 XXX Àa nÀ1 Check that det (xIn À A) ˆ p(x). The character ÷ of G ˆ C n1 3 . The elements g of D2 n for which g2 ˆ 1 are 1. Chapter 23 1.8. it follows from the Frobenius±Schur Count of Involutions that we must have é÷ ˆ 1 for all ÷. . Hence the number of real irreducible characters is 2 m . and n ‡ 2 elements if n is even. Suppose ®rst that ë is an eigenvalue of an n 3 n matrix A. Since A has integer entries. Let m be the order of g. Therefore x À1 ˆ gÀ1 xg ˆ x. g irr where for each j. it follows that ë is an eigenvalue of A. assume that ë is a root of the polynomial p(x) ˆ a0 ‡ a1 x ‡ . The number of such elements is also 2 m . Then ÷ A (g) ˆ 1 2 2 2 2((ë1 ‡ ë2 ) À (ë1 ‡ ë2 )) ˆ ë1 ë2 ˆ det (gr) (see Proposition 19. By a well known property of symmetric matrices. (a) First. Let r be the representation obtained by using the basis v1 . Then det (ar) ˆ 1 and det (br) ˆ Àå n . Write Q ˆ (q ij ). Choose a basis f1 . 2} then â(v i g. . v1 bÀ1 ). so é÷ ˆ Æ1. as n ˆ ÷(1) the result is proved. . hence det (gr) ˆ 1 for all g P G if and only if å n ˆ À1. v j g À1 )X For example. the subspace {u P V: â(u. Therefore det (At ) ˆ (À1) n det A.16.3 for the characters ø j (1 < j < n À 1) and ÷ j (1 < j < 4) of T4 n. b ij ˆ â( f i . Also A is invertible by (à ). As â is G-invariant. . according to whether j is odd or even. fn of V and de®ne the symmetric n 3 n matrices A ˆ (aij ) and B ˆ (bij ) by aij ˆ â1 ( f i . respectively. v) ˆ 0 for all v P V} is a CGsubmodule of V. so det A Tˆ 0. it is easy to check that ÷( g) is real for all g P G. so det A ˆ (À1) n det A. . f j )X By applying the Gram±Schmidt orthogonalization process. according to whether n is odd or even. we get éø j ˆ À1 or 1. j. 7. f j ). v j ) ˆ â(v i . . en of V by ˆ ei ˆ qij f 9 X j j t t . QQt ˆ I) such that Q(PBPt )QÀ1 is diagonal. Hence ÷ (é÷)÷(1) ˆ 2. j P {1. v n of V and let A be the n 3 n matrix with ij-entry â(v i . since V is irreducible it follows that fu P V : â(u. we may construct a basis f 1 . there is an orthogonal matrix Q (i. (b) It is easy to check that if g ˆ a or b and i. Since â is skew-symmetric. we have At ˆ ÀA. and é÷2 ˆ é÷4 ˆ 0 or 1. (d) Refer the Exercise 18. v2 of V. . v1 ) ˆ å n ˆ â(v1 . . Hence â is G-invariant. Let V be a CG-module with character ÷. The de®nition of â shows that â is symmetric if å n ˆ 1 and â is skew-symmetric if å n ˆ À1. . X X X . a has order 2n and ai b has order 4. By part (a) (or part (b)) and the construction of the characters ø j in Exercise 17.Chapter 23 431 5. and de®ne the basis e1 . Hence an is the only element of order 2. f 9 of V such that â1 ( f 9. Clearly é÷1 ˆ é÷3 ˆ 1. The result now follows from Exercise 4. there exists a nonzero G-invariant skew-symmetric bilinear form â on V. respectively. according to whether n is odd or € even. € nÀ1 Therefore jˆ1 (éø j )ø j (1) ˆ 0 or À2. v1 ) ˆ â(v2 . The result now follows from Theorem 23. respectively. . f 9) ˆ ä ij for all i.e. v j ). . Since é÷ ˆ À1. v) ˆ 0 for all v P V g ˆ f0gX (à ) Pick a basis v1 . . å n v2 ) ˆ â(v1 . â(v1 b. 6. It follows that n is even. Let 9 n i j P ˆ ( pij ) be the n 3 n matrix which is given by ˆ f9ˆ pij f j X i j Then PAP ˆ I n and PBP is symmetric.6. (c) The elements of T4 n are ai and ai b (0 < i < 2n À 1). . . Hence G  C2 . ej ) ˆ 0 if i Tˆ j. . 0 x 0 x9 0 xx9 associativity is a property of matrix multiplication. . Further. we have G9 ˆ 1. r is a homomorphism G 3 Sym(Ù)  S n with kernel which is contained in H. Vx P G D g P ’ xPG xÀ1 HxX Finally. with ÷1 ˆ 1 G ). 0 1 0 x 0 x À1 Therefore G is a group. since a linear character must take the value Æ1 on t. (b) Let v1 . and write á ˆ å ‡ å 3 ‡ å 4 ‡ å 5 ‡ å 9 . identity is       1 0 1 y 1 À yx À1 . and let V be the RG-submodule of the regular RG-module which is spanned by 1 À a and 1 À a2 . and consider the vector space V9 over C with basis v1 . so we may take ÷1 (t) ˆ 1. Then V9 is a CG-module. if G is simple then since G9 v G. and since QPBPt Qt is diagonalX â(ei . ⠈ å 2 ‡ å 6 ‡ å 7 ‡ å 8 ‡ å 10 X . Chapter 25 1. Call it G. Hence |G : G9| ˆ 2 by Theorem 17. But v1 W ˆ ëv1 P V. (a) The proof is similar to that of part (1) of Schur's Lemma 9. xgx À1 Vx P G D „ xPG x À1 P H. 8. 2. so ë P R. .e. r g is a permutation. By Schur's Lemma there exists ë P C such that vW ˆ ëv for all v P V9. De®ne W: V 3 V by vW ˆ av (v P V). inverse of is . which we are assuming to be an irreducible CG-module.11. Let ç ˆ e2ðia5 and å ˆ e2ðia11 .432 Then Representations and characters of groups â1 (ei . It is clear that the given set of matrices has size p( p À 1). We have g P ker r D Hxg ˆ Hx. Let c1 . . . note that      1 y 1 y9 1 y9 ‡ yx ˆ . 9.1. c2 be the columns of the character table of G corresponding to the classes {1} and t G . as Hxg ˆ Hyg A Hx ˆ Hy. ÷2 (t) ˆ Æ1 and ÷ i (t) ˆ 0 for i > 3. . i. 10. . v n . ej ) ˆ ä ij . v n be a basis of the RG-module V. and r is a homomorphism as (Hx)(r gh ) ˆ Hxgh ˆ (Hx)(r g )(r h ). Then V is an irreducible RG-module. For the last part. G is abelian. Now the orthogonality of c1 and c2 gives ÷1 (1) ˆ ÷2 (1) ˆ 1 and ÷2 (t) ˆ À1. By the orthogonality relation for c2 we have € ÷ i (t)2 ˆ |CG (t)| ˆ 2 (the sum over all irreducible characters ÷ i . (c) Let G ˆ C3 ˆ ka: a3 ˆ 1l. ÷1 and ÷2 are the only linear characters. Hx. For closure. since QPAPt Qt ˆ I n . 2ö1 (a)ö1 (a) ˆ ( p ‡ 1)a2. Hence ˆ 0ˆ ÷(1)÷(a) ˆ q ‡ qö1 (a) ‡ qö2 (a)X ÷ irred m Therefore ö1 (a) ‡ ö2 (a) ˆ À1. If p  À1 mod 4. so ˆ pˆ ÷(a)÷(a) ˆ q ‡ ö1 (a)ö1 (a) ‡ ö2 (a)ö2 (a)X ÷ Hence ö1 (a)ö1 (a) ‡ ö2 (a)ö2 (a) ˆ ( p ‡ 1)a2. Hence ö1 (a) and ö2 (a) ÷. . Also. aG ˆ {au : m P Z}. and it follows from Corollary 15. b9À1 ab9 ˆ av iX Hence G1  G2 . bÀ m abm ˆ au ˆ av . m is coprime to q. b9: ap ˆ b9q ˆ 1. and we ®nd that p ö1 (a) and ö2 (a) are (À1 Æ i p)a2.6 that ö1 (a) and ö2 (a) are not both real.Chapter 25 Character table of F11. then a is conjugate to aÀ1 and so ÷(a) is real for all characters p so (ö1 (a))2 ‡ (ö2 (a))2 ˆ ( p ‡ 1)a2. (a) Note that À1 is the only element of order 2 in ZÃ. Recall that Zà is cyclic. |CG (a)| ˆ p. Therefore aÀ1 P aG D u m  À1 mod p for some m D p  1 mod 4X (c) ÷ i (a) ˆ 1 for all the q linear characters ÷ i of G. This time. Let b9 ˆ bm. are (À1 Æ p)a2. since both m and v have u order q modulo p. then a is not conjugate to aÀ1 . If p  1 mod 4.6(c). Hence ö2 (a) ˆ ö1 (a). 4. Then G1 ˆ ha. so by Exercise 1. Also. Also. 5 gi |CG ( gi )| ÷1 ÷2 ÷3 ÷4 ÷5 ÷6 ÷7 1 55 1 1 1 1 1 5 5 a 11 1 1 1 1 1 á â a2 11 1 1 1 1 1 â á b 5 1 ç ç2 ç3 ç4 0 0 b2 5 1 ç2 ç4 ç ç3 0 0 b3 5 1 ç3 ç ç4 ç2 0 0 433 b4 5 1 ç4 ç3 ç2 ç 0 0 3. there exists an integer m p such that u m  v mod p. Hence bm has order q.9. Hence p u m  À1 mod p for some m D the element u of Zà has even order p D q is even D p  1 mod 4X (b) By Proposition 25. say ÷1 and ÷2 . Hence we get six linear characters ÷1 . The conjugacy classes of G are f1g. fa r b5 : 0 < r < 8gX Let H1 ˆ kal. . . All the elements outside H form a single conjugacy class of E. ÷5 and ÷6 are obtained similarly. 6g. Then H2 v G and Ga H 2 ˆ h H 2 a. (a) F13. Lift the irreducible character of D6 of degree 2 to obtain ÷7 in the table below.18). ö1 (a) ˆ mˆ1 å u .3 (see Theorem 19. . fa r b2 : 3 B rg. fa3 . 8. 3. fa r b3 : 0 < r < 8g. as shown. Then for all h P H. E9 ˆ H. it follows that {u. a6 g. Let H2 ˆ ka3 . Then ÷ 4 E is the irreducible character ÷3 given in the table which follows. . fa r b4 : 3 B rg. u ( pÀ1)a2 } is precisely the set of quadratic residues modulo p. bl.3 (see Theorem 19. H 2 bi  D6 . Since Zà is cyclic of order p p À 1 and u has order ( p À 1)a2. b2 l. Character table of E gi |CG ( gi )| ÷1 ÷2 ÷3 ÷4 ÷5 ÷6 1 18 1 1 2 2 2 2 a 9 1 1 2 À1 À1 À1 b 9 1 1 À1 2 À1 À1 ab 9 1 1 À1 À1 2 À1 a2 b 9 1 1 À1 À1 À1 2 c 2 1 À1 0 0 0 0 6. Also. The characters ÷4 . A typical non-trivial linear character of H is 1 ÷ 1 a 1 a2 1 b ù b2 ù2 ab ù ab2 ù2 a2 b ù a2 b2 ù2 where ù ˆ e2ðia3 . 6g. . . Z(E) ˆ {1}. ÷6 of G. . fa r b: 0 < r < 8g. there exist gi P E such that gi Tˆ 1 but ÷ i (gi ) ˆ ÷ i (1) (so gi P Ker ÷ i ). and for all i with 1 < i < 6. . The ®nal irreducible character ÷10 can be found by using the column orthogonality relations.10).434 Representations and characters of groups €( pÀ1)a2 m (d) By Theorem 25. u 2 . so E has exactly two linear characters.10. (b) C2 3 F13. . 7. Then ÷8 ˆ ÷7 ÷2 and ÷9 ˆ ÷7 ÷3 are also irreducible. The result now follows from part (c). 3. fa r : 3 B rg. (c) D6 3 F13. 5. fa r b2 : r ˆ 0. Let H ˆ ka.18). the conjugacy class hE consists of h and hÀ1 . fa r b4 : r ˆ 0. Then H1 v G and Ga H 1  C6 .3 (see Theorem 25. hz. Assume that G has r linear characters and s irreducible characters of degree p. But ø(1) ˆ 1. {z} and {z 2 } are conjugacy classes of H. by Theorem 17X11. and r ‡ sp2 ˆ pn . Therefore h÷. m is at least 2. 2. by Corollary 21. hz 2 }. by Theorem 11X12X Since s ˆ p nÀ2 À p mÀ2 and s is an integer. bÀ1 ab ˆ a2 l gi |CG ( gi )| ÷1 ÷2 ÷3 ÷4 ÷5 ÷6 ÷7 ÷8 ÷9 ÷10 Note: ù ˆ e2ðia3 1 54 1 1 1 1 1 1 2 2 2 6 a3 27 1 1 1 1 1 1 2 2 2 À3 a 9 ab2 9 ab4 9 b 6 b2 18 b3 6 1 À1 1 À1 1 À1 0 0 0 0 b4 18 435 b5 6 1 1 1 1 1 1 ù2 ù Àù ù2 1 ù ù2 ù2 ù 1 1 1 À1 1 1 ù2 ù ù ù2 1 ù ù2 Àù2 ù À1 À1 À1 0 2 À1 Àù2 Àù 0 2ù2 À1 Àù Àù2 0 2ù 0 0 0 0 0 1 1 ù Àù2 ù2 ù 1 À1 ù ù2 ù2 Àù 2 0 2ù 0 2ù2 0 0 0 Chapter 26 1. øl H Tˆ 0 for some irreducible character ø of H. ø 4 Gi G Tˆ 0 by the Frobenius Reciprocity Theorem. and (ø 4 G)(1) ˆ p. and so ÷(1) ˆ 1 or p by Theorem 22. since H is abelian. b: a9 ˆ b6 ˆ 1. Hence ÷(1) < p. For all other elements h of H.Chapter 26 Character table of G ˆ ka. Then r ˆ pm for some m.11. Character table of H (a non-abelian group of order 27) gi |C H ( g i )| ÷00 ÷01 ÷02 ÷10 ÷11 ÷12 ÷20 ÷21 ÷22 ö1 ö2 1 27 1 1 1 1 1 1 1 1 1 3 3 z 27 1 1 1 1 1 1 1 1 1 3ù 3ù2 z2 27 1 1 1 1 1 1 1 1 1 3ù2 3ù a 9 1 1 1 ù ù ù ù2 ù2 ù2 0 0 a2 9 1 1 1 ù2 ù2 ù2 ù ù ù 0 0 b 9 1 ù ù2 1 ù ù2 1 ù ù2 0 0 ab 9 1 ù ù2 ù ù2 1 ù2 1 ù 0 0 a2 b 9 1 ù ù2 ù2 1 ù ù ù2 1 0 0 b2 9 1 ù2 ù 1 ù2 ù 1 ù2 ù 0 0 ab2 9 1 ù2 ù ù 1 ù2 ù2 ù 1 0 0 a2 b2 9 1 ù2 ù ù2 ù 1 ù 1 ù2 0 0 Note: ù ˆ e2ðia3 . Then k÷ 5 H. Let ÷ be an irreducible character of G. {1}. the conjugacy class hH ˆ {h.20. and GaK  D16 . 1}. we get a representation A i B j C k D l Z m 3 (À1) ir‡ js‡ kt‡ lu X Together with the irreducible representation of degree 4. and Gah Zi is abelian while G is non-abelian. bÀ1 ab ˆ aÀ1 l gi |CG ( g i )| ÷1 ÷2 ÷3 ÷4 ÷5 ÷6 ÷7 øj ( j ˆ 1. 5. Since Gah Zi is abelian. j. these are all the irreducible representations of G. Hence every element of G has the form A i B j C k D l Z m for some i. G has precisely 16 representations of degree 1. also G is a 2-group. CD ˆ ÀDC. s. (a) Check that AB ˆ ÀBA. These are as follows: for each (r. the group K which appears in Theorem 26. The character table of D16 is given in Section 26. 3. B. k. 1 32 a8 32 a 16 a2 16 a3 16 a4 16 a5 16 a6 16 a7 16 b 4 ab 4 1 À1 À1 1 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 À1 1 1 À1 1 À1 1 À1 1 À1 1 1 1 À1 1 À1 1 À1 1 À1 À1 2 2 p0 À2 0 2 0 0 p p À2 p0 2 2 2 0 Àp2 À2 Àp2 0 2 0 p p 2 2 À 2 0 2 À2 2 0 À 2 0 2 À2 cj c2 j c3 j c4 j c5 j c6 j c7 j 0 5. b2 ˆ a8 .10). ÷7 of G as shown in the table at the top of this page. 3. Combined with part (b). (c) A routine calculation shows that every matrix which commutes with each of A. BD ˆ ÀDB. Hence by Corollary 9. {a8 }. 7) come from inducing to G those linear characters ÷ of kal for which ÷(a8 ) ˆ À1. . t. we obtain the characters ÷1 . We obtain representations as follows: . Since G9 ˆ h Zi. it follows that g 2 P h Zi for all g P G. this shows that jGj ˆ 32.436 Representations and characters of groups Character table of G ˆ ka. (d) Since G has irreducible representations of degrees 1 and 4. (a) Let å ˆ e2ðia8 . {ar b: r odd}. a8 }. jGj > 12 ‡ 42 ˆ 17. . Then the four characters ø j ( j ˆ 1. u P {0. Therefore G9 ˆ h Zi (see Proposition 17. 4. AD ˆ DA. {ar b: r even}. s. . {ar . Here. AC ˆ ÀCA. the given representation is irreducible. 7) Note: cm ˆ e2ði ma16 ‡ eÀ2ði ma16 ˆ 2 cos (mð/8) 3. by Theorem 11. t. By lifting the irreducible characters of D16.3. Hence Z P G. aÀ r } (1 < r < 7). C and D has the form ëI for some ë P C. so jGj < 32. 1g. b: a16 ˆ 1. m P f0. 5. . (b) A2 ˆ ÀB2 ˆ ÀC2 ˆ D2 ˆ I. G has 11 conjugacy classes: {1}.3.4 is {1.8 (D16 ˆ G1 ) and in Section 18. l.12. u) with r. since g4 ˆ 1 for all g P G. BC ˆ CB. so the representations are faithful. i f G8 : a 3 d 0 0 H 0 0 0 À1 H I 0 0 0 1 0 1 0 0 g f g f g Ài 0 e. . Gj has no faithful irreducible representation. Assume that j Z(G)j ˆ p2 . while G8 aG8 9  C2 3 C2 3 C2 . (c) Check that the matrices satisfy the required relations. The following table records the numbers of elements of orders 1. But G5 aG5 9  C2 3 C4 .1(1) we have {1} Tˆ Z(G) Tˆ G. (d) The following give faithful representations: H I H I H I i 0 0 0 1 0 1 0 0 f g f g f g G7 : a 3 d 0 Ài 0 e.b3 3 . It is easy to see that the matrices generate groups with more than eight elements. G9 are isomorphic. Also jGa Z(G)j Tˆ p by Lemma 26. . À1 0 2 3 0 1 . z.1(2). and g P CG ( g). If g P Z(G) then Z(G) < CG (g) Tˆ G. Therefore j Z(G)j ˆ p or p2 . 6. .b3 . so G5 T G8 .b3 . b 3 d À1 0 0 e. 1 0 å3 3 2 3 0 0 1 . . 4 and 8 in G1 . . .b3 2 2 Chapter 26 0 1 1 0 0 1 3 .16.b3 . 7 or 8 then Z(Gj ) ˆ {1. . z 3 d 0 1 0 eX I H I 0 1 0 0 1 0 0 À1 0 1 0 0 1 6. (a) By Lemma 26. . a2 . z 3 d 0 1 0 e.z3 X À1 1 0 0 i (b) If j ˆ 5. b 3 d 1 0 0 e. 3 437 0 å 0 0 å À1 2 3 å 0 2 0 å 0 1 0 . Hence a . å5 1 0 3 2 3 2 3 0 0 1 i 0 . by Proposition 9. G9 : G1 Order Order Order Order 1 2 4 8 1 9 2 4 G2 1 1 10 4 G3 1 5 6 4 G4 1 3 4 8 G5 1 3 12 0 G6 1 7 8 0 G7 1 11 4 0 G8 1 3 12 0 G9 1 5 10 0 Therefore no two of G1 .2 G1 : a 3 2 G2 : a 3 G3 : a 3 G4 : a 3 2 G5 : a 3 å å 0 À1 3 . Since Z(Gj ) is not cyclic. so give representations. 7. 2. except possibly G5 and G8 . a2 z}  C2 3 C2 . either G9 ’ Z(G) ˆ G9. in which case Ga(G9 ’ Z(G)) is abelian. as r ‡ s is equal to the number of conjugacy classes of G. By parts (a) and (b). and if |G9| ˆ p2 then |Z(G)| ˆ p. 3) of order 3 which are not conjugate to each other. (a) Let Z ˆ Z(G). in which case Ga(G9 ’ Z(G)) T Q8 by part (a). and 1  1 z A c ˆ Àb. b and all elements in Z. (c) Note that G9 ’ Z(G) Tˆ {1} by Lemma 26. a2 Z ˆ b2 Z. and if r ˆ p2 then r ‡ s ˆ 2 p2 À 1. so a ˆ Æ1. 2. and . and since z P SL (2. we have a2 ˆ 1. a ˆ dX 0   1 1 1 ˆ 0 0 1    0 1 0 z ˆ À1 0 À1 z Therefore z ˆ aI. Therefore. G has p2 ‡ ( p4 À p2 )a p conjugacy classes. (b) Assume that G has r irreducible characters of degree 1 and s irreducible € characters of degree p. The element   À1 0 0 À1 lies in Z(G). p))X  1 z A c ˆ 0. p). bZi. bÀ1 abZ ˆ aÀ1 ZX Then a2 ˆ b2 z for some z P Z. or G9 ’ Z(G) ˆ Z(G). and assume that Ga Z ˆ haZ. Chapter 27 1. with a4 P Z. so r ‡ sp2 ˆ p4 . Therefore jGaG9j ˆ r ˆ p2 or p3 . and hence ba2 ˆ b3 z ˆ b2 zb ˆ a2 b. if |Z(G)| ˆ p2 then |G9| ˆ p.12).1(1). Assume that zˆ Then   a c b d  P Z(SL (2. there are no irreducible characters of degree greater than p.438 Representations and characters of groups jC G ( g)j ˆ p3 and j gG j ˆ p. Therefore Ga Z T Q8 . Since ÷(1)2 ˆ p4 (Theorem 11. Hence |G9 ’ Z(G)| ˆ p. Since a2 commutes with a. Check that   and   1 0 1 1 1 À1 0 1 are elements of G ˆ SL (2. Part (b) follows. (b) If G is a non-abelian group of order 16. we have a2 P Z. then by Exercise 7. 8. The characters ÷1 . they must be ÷6 and ÷7 in some order. a subgroup of S4 of order 12. Exercise 13. Since á is real.27. ÷(g5 ) ˆ ÷( g4 ) and ÷(g6 ) ˆ ÷( g7 ) for all ÷. 1). 0). g2 .1. Then á ˆ À1 since ÷5 ( g4 )  ÷5 (1) mod 3. by Corollary 22. Finally. 1). Since ÷5 ÷2 and ÷5 ÷3 are irreducible characters of G of degree 2. á ˆ Æ1. so we obtain a homomorphism ö: G 3 S4 . whose values on g4 are áù and áù2 . 6. . ÷7 on the elements g1 . ÷6 . Check that Ker ö ˆ {ÆI}. The € equation j ÷ j (g4 )÷ j ( g 4 ) ˆ 6 gives áá ˆ 1. Note that G has three real conjugacy classes. Now note that for j ˆ 5.5 implies that ÷ j ( g7 ) ˆ À÷ j ( g4 ). without loss of generality. which is given below. 1) and (1. ÷3 .Chapter 27  0 À1 1 0  439 has order 4. Hence GafÆIg  Im ö.2). g3 can be deduced from the column orthogonality relations. First observe that the vector space (Z3 )2 has exactly four 1-dimensional subspaces. where á is real. (1. Assume. Then ÷5 ( g4 ) ˆ á. Hence the following are conjugacy class representatives: g1 1 0 1 24 0 1 g2 À1 0 0 À1 2 24 g3 0 À1 4 4 1 0 g4         1 1 0 1 Order of gi |CG ( gi )| 3 6  g5 1 0 À1 1 3 6   g6 À1 1 0 À1 6 6   g7 À1 À1 0 À1 6 6  Order of gi |CG ( gi )| We now describe how to construct the character table of G. (2. The values of ÷5 . ÷2 . Also á Tˆ 0. ÷4 of G are obtained by lifting to G the irreducible characters of A4 (which are given in Section 18. that ÷5 is real. say ÷5 ÷2 ˆ ÷6 and ÷5 ÷3 ˆ ÷7 . therefore GafÆIg  A4 . The group G permutes these subspaces among themselves. one of ÷5 . ÷6 and ÷7 must be real. namely the spans of the vectors (0. so by Theorem 23. 7. h2 ˆ Z. so ë : G is irreducible. The values of 1 T : G and ë : G are as follows (see Proposition 21. h5 . h5 ˆ ZX 0 1 0 1 Two of the linear characters of T are hi |CT (h i )| 1T ë h1 21 1 1 h2 3 1 ù h3 3 1 ù2 h4 7 1 1 h5 7 1 1 g1 24 1 1 1 3 2 2 2 g2 24 1 1 1 3 À2 À2 À2 g3 4 1 1 1 À1 0 0 0 g4 6 1 ù ù2 0 À1 Àù Àù2 g5 6 1 ù2 ù 0 À1 Àù2 Àù g6 6 1 ù2 ù 0 1 ù2 ù g7 6 1 ù ù2 0 1 ù ù2 where ù ˆ e2ðia3 . Representatives of the conjugacy classes of T are h1 . Hence 1 T : G ˆ 1 G ‡ ÷. write ö ˆ ë : G. 4. 3) gi |CG ( gi )| ÷1 ÷2 ÷3 ÷4 ÷5 ÷6 ÷7 Note: ù ˆ e2ðia3 3. ë : Gl ˆ 1.2 and Example 21. (a) For the character table of T. kë : G. Apply Proposition 17. 1 G l ˆ 1. where 2 3 2 3 2 3 1 0 2 0 4 0 h1 ˆ Z. notice that T is isomorphic to the group of order 21 whose character table is found in Exercise 17.23): gi |CG ( g i )| 1T 4 G ë4G g1 168 8 8 g2 8 0 0 g3 4 0 0 g4 3 2 À1 g5 7 1 1 g6 7 1 1 We ®nd that k1 T : G. h3 ˆ Z. . Also. .440 Representations and characters of groups Character table of SL (2.6. where ÷ is an irreducible character of G. .25. . 1 T : Gl ˆ 2 and k1 T : G. 0 1 0 4 0 2 2 3 2 3 1 1 1 À1 h4 ˆ Z. . apply part (c). Note that because g2 lies in Z(G). in some order. By Corollary 22. ÷ j (1) is even. The congruences ÷(1)  ÷(g6 ) mod 3 now give the remaining values on g6 . ÷11 .) Also. are equal to 4.1.27. ÷(g6 ) P Z for all characters ÷. ø A has the following values on g1 . the (d) Since 11 ÷ j (g3 )÷ j ( g3 ) ˆ 8. gi and gi g2 have the same centralizer for all i.7). The values of ø are as shown above. ÷ and ø are the characters ÷1 . and ÷11 (1) ˆ 8. ÷l ˆ 1. ö. we deduce that ÷ j (g3 ) ˆ 0 for 7 < j < 11. (a) Compare the proof of Lemma 27. (c) Use Exercise 13. . (b) By lifting. 0.14. The remaining irreducible characters ÷4 . and none is a constituent of æ. Further. ÷6 in the character table shown below. in the character table of G given at the end of Chapter 27. ÷3 . 1 G l ˆ k÷ S . ÷5 can readily be calculated using the column orthogonality relations (noting that gi is real if and only if 1 < i < 4). the values of ÷ j (g6 ) for 7 < j < 11 must be Æ1. The only possibility is that two of ÷7 (1). . g2 . respectively. ö and ÷ are three of the six irreducible characters of G. we obtain the characters ÷1 .Chapter 27 (c) The values of ÷ and ÷ S are as shown below (see Proposition 19. 168. Since By 11 2 jˆ1 (÷ j (g6 )) ˆ 6. . Use part (c) to ®ll in the values on g2 and g7 . or æ is the sum of four distinct irreducible characters (cf. (f ) By Proposition 19. (d) The characters 1 G . by Corollary 22. two of ÷7 . . so either æ ˆ 2ø for some irreducible character ø. öl ˆ k÷ S . 5. have degrees 2divisible by 6. . . say ÷7 (1) and ÷8 (1).5 (noting that ÷ j (ÀI) Tˆ ÷ j (I) for 7 < j < 11. æ cannot therefore be the sum of four irreducible characters. 0. since ÀI is not in € kernel of these characters). 2 2 jˆ1 (÷ j (1)) ˆ 168. (e) € Theorem 22. Exercise 14. Since there are only six irreducible characters in all.14): gi |CG ( g i )| ÷ ÷S æ ø g1 168 7 28 12 6 g2 8 À1 4 4 2 g3 4 À1 0 0 0 g4 3 1 1 0 0 g5 7 0 0 À2 À1 g6 7 0 0 À2 À1 441 We ®nd that k÷ S . and so æ ˆ 2ø with ø irreducible. so ÷9 (1) ˆ ÷10 (1) ˆ 6. g3 and g6 : g1 øA 6 g2 6 g3 2 g6 0 .27 again. ÷11 (1).16. We calculate that kæ. ÷7 (1)2 ‡ ÷8 (1)2 ‡ ÷11 (1)2 ˆ 96. ÷2 and ÷6 . and 12 ‡ 6 . . Æ1. €11 say ÷9 and ÷10 . Hence there is a character æ of G such that ÷ S ˆ 1 G ‡ ö ‡ ÷ ‡ æX The values of æ are as shown above. Æ1. Now 1 G . æl ˆ 4. . Next. jˆ1 (Alternatively. ÷8 (1). Let Z ˆ {ÆI} and de®ne the subgroup T of G. Since g2 is conjugate to g8 . Let x ˆ ø(g8 ). we ®ll in the values of ÷7 and ÷8 . 6. Then the value of ø A on g6 shows that ÷1 is not a constituent of ø A . For all ÷. we have 11 ÷ j ( g i )÷ j ( g 6 ) ˆ 0.442 Representations and characters of groups Character table of SL (2. 0 1 0 6 and T has ®ve linear characters æ j (0 < j < 4). and therefore. Then 4 €11 € and 11 ÷ j ( g 4 )2 ˆ 8 imply that jˆ1 ÷ j ( g 1 )÷ j ( g 4 ) ˆ 0 p jˆ1 p ÷9 (g4 ) ˆ À÷10 (g4 ) ˆ Æ 2. € (g) For i Tˆ 6. ÷4 . Now g2 is conjugate to g3 . Say A ÷7 ( g 8 ) ˆ (1 À i 7)a2. Since g4 is conjugate to gÀ1 . the value on g3 forces ø A ˆ ÷6 . Therefore x ˆ (1 Æ i 7)a2. Then ÷8 ˆ ÷7 . Say ÷9 (g4 ) ˆ 2. using this fact and part (c). ÷( g 10 ) ˆ ÷(g8 ). ÷(g4 ) is real for all ÷. of order 55.bPZ Tˆ Z: a P Z 11 11 X 0 aÀ1 Then T is generated by     1 1 2 0 xˆ Z and y ˆ Z. The column orthogonality relations now let us ®nd the remaining values of ÷9 and ÷10 . 7) gi g1 Order of g i 1 |CG ( g i )| 336 ÷1 ÷2 ÷3 ÷4 ÷5 ÷6 ÷7 ÷8 ÷9 ÷10 ÷11 1 7 8 3 3 6 4 4 6 6 8 g2 2 336 1 7 8 3 3 6 À4 À4 À6 À6 À8 g3 4 8 g4 8 8 g5 8 8 g6 3 6 g7 6 6 1 1 À1 0 0 0 À1 À1 0 0 1 g8 7 14 1 0 1 á á À1 Àá Àá À1 À1 1 g9 14 14 1 0 1 á á À1 á á 1 1 À1 g10 7 14 1 0 1 á á À1 Àá Àá À1 À1 1 g11 14 14 1 0 1 á á À1 á á 1 1 À1 1 1 1 1 À1 À1 À1 1 0 0 0 À1 À1 1 1 0 À1 1 1 0 2 0 0 0 0 0 0 1 0 0 0 1 p p 0 0 p 2 Àp 2 0 À 2 2 0 0 0 0 À1 p Note: á ˆ (À1 ‡ i 7)a2 The values of ø A on g1 and g2 show that ø A is a linear combination of ÷1 . we get 8 p (x 2 À x)a2 ˆ øp( g8 ) ˆ À1. This allows us to ®ll in the jˆ1 values of ÷11 . ø(g4 ) ˆ 0. Similarly. ÷5 and ÷6 . ®nally. thereby completing the character table of G. where æ j : x u y v 3 e2ði jva5 X . by &  ' a b Ã. ø(g5 ) ˆ 0. Hence 4 (ø( g4 )2 À ø( g 3 ))a2 ˆ ø A ( g4 ) ˆ ÷6 ( g 4 ) ˆ 0. € 8 2 Now ÷(g4 )  ÷(g3 ) mod 2 for all ÷. and 8 ÷ j (g3 )2 ˆ 6. and 8 ÷ j ( g2 )2 ˆ 12. But 8 (÷ j (1))2 ˆ 250. € 6 and ÷ j (g2 ) ˆ Æ1 for j ˆ 7. ÷6 (1). € ÷ j (1)  0 mod 5 for 5 < j < 8. Next. 3 and 5 of the character table. By Corollary 22. we jˆ1 see that ÷7 (g2 ) ˆ ÷8 (g2 ) ˆ 1. respectively. and ÷5 (g2 ). (In calculating ÷3 (g5 ). ÷8 (1) are 10.16. The column orthogonality relations now enable us to ®nish the character table. We have now completed columns 1. ÷(g2 )  ÷(g4 ) mod 3 for all ÷. ÷4 . 3 for all irreducible ÷. ÷7 (1).27. and ˆ jˆ1 ÷ j (g4 ) € 6. . æ0 4 Gi ˆ 2. We may now conclude from the facts €8 2 that ÷(g2 )  ÷(g1 ) mod 2 and that ÷ j (g2 ) ˆ Æ2 for j ˆ 5. by Theorem 22. 5. Character table of PSL (2. without loss jˆ5 of generality. 8. ÷5 (1). ÷2 . ÷5 (g2 ) ˆ 2 ˆ À÷6 ( g2 ). jˆ1 ÷ j (g2 ) ˆ 12. By considering 8 ÷ j (1)÷ j (g2 ) ˆ 0. ⠈ (À1 À p 5)a2 and 㠈 (À1 ‡ i 11)a2 € Since 8 ÷ j (g5 )÷ j ( g 5 ) ˆ 5. ÷8 take the value 0 on g5 . € Since ÷(g4 )  ÷(g2 ) mod 3 for all characters ÷.Chapter 27 443 The characters æ1 4 G and æ2 4 G are irreducible. € Since ÷(1)  ÷(g3 ) mod 3 for all characters ÷. they are ÷3 and ÷4 in the p table. 5. 10. ÷1 i ˆ 1 and hæ0 4 G. namely ÷1 . note that e2ðia5 ‡ eÀ2ðia5 ˆ (À1 ‡ 5)a2X) Let ÷1 ˆ 1 G . ÷3 . we deduce that the remaining irreducible jˆ1 characters ÷5 . ÷6 (g2 ) have opposite signs. so ÷ j (g4 ) ˆ Æ1 for 5 < j < 8. without loss of generality. ÷7 . We have hæ0 4 G. Hence æ0 4 G ˆ ÷1 ‡ ÷2 for an irreducible character ÷2 of G. we jˆ1 can complete column 4. 2. ÷6 . Note that ÷(gj ) is an integer for 1 < j < 4 and all characters ÷. and 8 ÷ j (g4 )2 ˆ 6. We have now found four of the eight irreducible characters of G. hence. the jˆ1 values of ÷ j (g3 ) (5 < j < 8) are as shown. 11) gi Order of g i |CG ( g i )| ÷1 ÷2 ÷3 ÷4 ÷5 ÷6 ÷7 ÷8 Note: á ˆ (À1 ‡ p g1 1 660 1 11 12 12 10 10 5 5 g2 2 12 1 À1 0 0 2 À2 1 1 g3 3 6 1 À1 0 0 1 1 À1 À1 p g4 6 6 1 À1 0 0 À1 1 1 1 g5 5 5 1 1 á â 0 0 0 0 g6 5 5 1 1 â á 0 0 0 0 g7 11 11 1 0 1 1 À1 À1 ã ã g8 11 11 1 0 1 1 À1 À1 ã ã 5)a2. jˆ1 hence |÷( g2 )| . 3 2 3 2 3 2 0 2 0 1 1 2 g2 ˆ g3 ˆ g4 ˆ 1 0 2 0 1 0 3 2 3 2 3 2 0 0 1 0 1 0 g6 ˆ g7 ˆ g8 ˆ 2 2 0 1 2 1 where 3 1 2 1 1 3 X The character table of GL(2. g8 as representatives of the conjugacy classes.   a b Suppose that P GL(2. q You should have no dif®culty in proving that the conjugacy classes of SL(2. q) have representatives as follows. (a) The identity I has centralizer of order q 3 À q.444 Representations and characters of groups Chapter 28 1. q)  Z 3 SL(2. indexed by unordered pairs fs. Then s in F q      a b s 0 aas bas ˆ X 0 s cas das c d The ®rst matrix in the product is sI and the second belongs to SL(2. We take g 1 . . We may write ad À bc as s 2 for some c d à .1 ÷1 ÷2 ÷4 g1 48 1 1 3 3 4 2 2 2 g2 48 1 1 3 3 À4 À2 2 À2 g3 6 1 1 0 0 1 À1 À1 À1 g4 6 1 1 0 0 À1 1 À1 1 g5 4 1 À1 1 À1 0 0 0 0 g6 8 1 1 À1 À1 0 0 2 0 g7 8 1 À1 À1 1 p0  i 2 0 p Ài 2 g8 8 1 À1 À1 1 0 p Ài 2 p0  i 2 2. since r ˆ r q and q is even.   1 1 (b) The matrix u1 ˆ has centralizer of order q. Every element r of F q can be expressed as a square. Each such element has centralizer of order q À 1. X 2 1 g1 ˆ 0 2 1 g5 ˆ 0 X X . 3) is then as follows. 0 1 (c) There are (q À 2)a2 conjugacy classes with representatives   s 0 d s.s À1 ˆ . q). It now follows easily that GL(2. q) where Z ˆ fsI : s P Fà g. s À1 g of elements 0 s À1 from F q nF2 . gi |CG ( g i )| ë0 ë1 ø0 ø1 ø0. q). as follows.i ÷i 1 q q‡1 qÀ1 u1 1 0 1 À1 d s.Chapter 28 445 (d) There are qa2 conjugacy classes with representatives   0 1 vr ˆ . By restricting characters from GL(2. q) to SL(2. The polynomial x 3 ‡ x ‡ 1 is irreducible over F2. q) you will quickly be able to prove that the character table of SL(2. 1 ‡ ç ‡ ç2 g.i satisfy 1 < i < (q À 2)a2. b. g4 . f1 ‡ ç. Hence we may write F8 ˆ fa ‡ bç ‡ cç2 : a. The irreducible monic quadratics over F8 with constant term 1 are x 2 ‡ x ‡ 1. g 9 of the conjugacy classes of SL(2.s À1 1 1 s i ‡ s Ài 0 vr 1 À1 0 À(r i ‡ r Ài ) Here. and therefore SL(2. The subscripts for ø0. q) is as follows. X X X . x 2 ‡ ç2 x ‡ 1. Note ®rst that PSL(2. fç2 . and the subscripts for ÷ i satisfy 1 < i < qa2. g8 . 8). If q Tˆ 2 then the kernel of every non-trivial character is the identity subgroup. g 5 below. 3. 8). g 7 . Then 64 . 8)  SL(2. x 2 ‡ çx ‡ 1. c P F2 and ç3 ˆ 1 ‡ çgX The pairs fs.3). Each such element has centralizer of order q ‡ 1. x 2 ‡ (ç ‡ ç2 )x ‡ 1X The companion matrices for these quadratics give use the conjugacy class representatives g 6 . 2 3 2 3 1 0 1 1 g1 ˆ g2 ˆ 0 1 0 1 2 3 2 3 2 3 ç 0 1‡ç 0 0 ç2 g3 ˆ g4 ˆ g5 ˆ 0 1 ‡ ç2 0 ç ‡ ç2 0 1 ‡ ç ‡ ç2 2 3 2 3 2 3 2 3 0 1 0 1 0 1 0 1 g6 ˆ g9 ˆ X g7 ˆ g8 ˆ 1 1 1 ç 1 ç2 1 ç ‡ ç2 We may choose a generator å of Fà so that å 7 ‡ å À7 ˆ ç. 1 ‡ ç2 g. we have used the function r 3 r de®ned in (28. I ë0 ø0 ø0. q) is simple. r À1 g of elements 1 r ‡ r À1 from F q 2 nF q such that r 1‡q ˆ 1. g9 below We can now list representatives g 1 . ç ‡ ç2 gX These give us the conjugacy class representatives g3 . indexed by unordered pairs fr. s À1 g of elements from F8 nF2 are fç. 446 Representations and characters of groups å 14 ‡ å À14 ˆ ç2 , å 21 ‡ å À21 ˆ 1 and å 28 ‡ å À28 ˆ ç4 ˆ ç ‡ ç2 . The character table of SL(2, 8) is then as follows. gi |CG ( g i )| ë0 ø0 ø0,1 ø0,2 ø0,3 ÷3 ÷1 ÷2 ÷4 g1 504 1 8 9 9 9 7 7 7 7 g2 8 1 0 1 1 1 À1 À1 À1 À1 g3 7 1 1 g4 7 1 1 A 0 0 0 0 0 0 0 0 0 0 0 0 g5 7 1 1 g6 9 1 À1 0 0 0 À2 1 1 1 g7 9 1 À1 0 0 0 1 g8 9 1 À1 0 0 0 1 B g9 9 1 À1 0 0 0 1 Here, the 3 3 3 submatrices A and B are given by H I 2 cos(2ða7) 2 cos(4ða7) 2 cos(6ða7) f g A ˆ d 2 cos(4ða7) 2 cos(6ða7) 2 cos(2ða7) e 2 cos(6ða7) 2 cos(2ða7) 2 cos(4ða7) H À2 cos(2ða9) À2 cos(4ða9) À2 cos(8ða9) I f g B ˆ d À2 cos(4ða9) À2 cos(8ða9) À2 cos(2ða9) eX À2 cos(8ða9) À2 cos(2ða9) À2 cos(4ða9) Chapter 29 1. (a) It is straightforward to check that ö is a homomorphism. For x, y P G, the element (x, y) P G 3 G sends x to y, so that action is transitive. (b) (G 3 G)1 ˆ f( g, g) : g P Gg, and ker ö ˆ f(z, z) : z P Z(G)g. (c) Every orbit of G 3 G on G 3 G contains an ordered pair of the form (1, x), and if ( g, h) sends (1, x) to (1, y) then g ˆ h and y ˆ g À1 xg. Hence if C1 , X X X , C k are the conjugacy classes of G, and xi P C i, then (1, xi ) (1 < i < k) are orbit representatives for the action of G 3 G on G 3 G, and so the rank is equal to k. Since x(( g, h)ö) ˆ x if and only if xhx À1 ˆ g, we see that ð( g, h) ˆ jfix G ( g, h)j is equal to 0 if g is not conjugate to h, and is equal to jC G (x)j if g is conjugate to h (since in the latter case, if xhx À1 ˆ g then an arbitrary element y P G such that yhy À1 ˆ g is of the form y ˆ xc with c P € (x)). Hence using the column orthogonality relations we see CG that ð ˆ ÷ 3 ÷. 2. There are q 2 À 1 non-zero vectors in V, and each 1-dimensional subspace Chapter 29 447 contains q À 1 of them; also two 1-dimensional subspaces have no non-zero vectors in common. Hence jÙj ˆ (q 2 À 1)a(q À 1) ˆ q ‡ 1. 3. Use the notation for the conjugacy classes and irreducible characters of GL(2, q) given in Proposition 28.4 and Theorem 28.5. It is easy to check that ð takes the values q 2 À 1, q À 1, q À 1 on the classes with representatives I, u1 , d 1, t respectively, and takes the value 0 on all other classes. Taking inner products we ®nd hð, 1 G i ˆ hð, ø0 i ˆ hð, ø0, j i ˆ 1 (1 < j < q À 2)X € qÀ2 As 1 G ‡ ø0 ‡ 1 ø0, j has degree q 2 À 1 ˆ ð(1), we conclude that € qÀ2 ð ˆ 1 G ‡ ø0 ‡ 1 ø0, j . 4. Observe that the coset H 1 x is ®xed by g if and only if xgx À1 P H 1 . If G is abelian this amounts to g P H 1 , and hence we see that ð1 ( g) ˆ 0 if g P H 1 a and ð1 ( g) ˆ jG : H 1 j if g P H 1 . Thus H 1 ˆ f g P G : ð1 ( g) Tˆ 0g. Likewise for H 2 ; since ð1 ˆ ð2 we deduce that H 1 ˆ H 2 . As a counterexample for G non-abelian, take G ˆ D8 ˆ ha, b : a4 ˆ b2 ˆ 1, bÀ1 ab ˆ aÀ1 i with H 1 ˆ hbi, H 2 ˆ ha2 bi. Then ð1 ˆ ð2 but H 1 Tˆ H 2 . € € 1 5. By Proposition 29.4 we have 1 ˆ jGj gPG jfixÙ ( g)j, hence jfixÙ ( g)j ˆ jGj. Since jfixÙ ( g)j is a non-negative integer for each g, and jfixÙ (1)j ˆ jÙj . 1, we must have jfixÙ ( g)j ˆ 0 for some g. 6. Write ð ˆ ð( nÀ2,1,1) . Calculating inner products using Proposition 29.6, as in the proof of Theorem 29.13, we ®nd hð, ði ˆ 7, hð, 1i ˆ 1, hð, ð( nÀ1,1) i ˆ 3, hð, ð( nÀ2,2) i ˆ 4X Using Theorem 29.13 it follows that ð( nÀ2,1,1) ˆ 1 ‡ 2÷ ( nÀ1,1) ‡ ÷ ( nÀ2,2) ‡ ÷ with ÷ irreducible. Hence ÷ ˆ 1 ‡ ð À ð( nÀ1,1) À ð( nÀ2,2) , from which it is easy to calculate that ÷(1) ˆ 1(n À 1)(n À 2), 2 ÷(12) ˆ 1(n À 2)(n À 5), ÷(123) ˆ 1(n À 4)(n À 5). For n ˆ 6, ÷ (4,1,1) is the 2 2 character ÷5 in Example 19.17. Chapter 30 1. By Theorem 30.4, a245 ˆ 168/(8´3) ˆ 7. Hence, by (30.3), PSL (2, 7) contains elements a and b such that a has order 2, b has order 3 and ab has order 7. p p 2. No: a225 ˆ (1 ‡ (À1 ‡ i 7)a6 ‡ (À1 À i 7)a6 À 4a6)168a(8X8) ˆ 0, and similarly a226 ˆ 0. Hence PSL (2,7) does not contain two involutions whose product has order 7. 3. Yes. Number the conjugacy classes of PSL (2, 11) as in the solution to Exercise 27.6. Then   660 1 a235 ˆ X 1‡ ˆ 10X 12 6 11 Therefore PSL (2, 11) contains elements x and y such that x, y and xy have orders 2, 3 and 5, respectively. Let H be the subgroup kx, yl of PSL (2, 11). 448 Representations and characters of groups There is a homomorphism W from A5 onto H (W sends a 3 x, b 3 y). Since Ker W 3 A5 and A5 is simple, we deduce that H  A5 . 4. Suppose that G is a group whose p character table is p where á ˆ (1 ‡ 5)a2, ⠈ (1 À 5)a2. g1 ÷1 ÷2 ÷3 ÷4 ÷5 1 4 5 3 3 g2 1 1 À1 0 0 g3 1 0 1 À1 À1 g4 1 À1 0 á â g5 1 À1 0 â á € For 1 < i < 5 we have jC G ( g i )j ˆ 5 ÷ j ( g i )÷ j ( g i ). Therefore the jˆ1 centralizers of g1 , g2 , g3 , g4 , g5 have orders 60, 3, 4, 5, 5, respectively. Hence the orders of g2 , g4 and g5 are 3, 5 and 5; also the order of g3 must be 2, since for no other i (except i ˆ 1) is |CG (g i )| even. Now a324 ˆ 60a(4 . 3). Therefore G contains elements x and y such that x has order 2, y has order 3 and xy has order 5. As in the solution to Exercise 3, G has a subgroup H with H  A5 . Since jGj ˆ 60, we have G  A5 . € 5. (a) Using the fact that 7 ÷ j (g i )÷ j ( g i ) ˆ |CG (g i )|, we ®nd that the jˆ1 centralizer orders and class sizes are as follows: g1 |CG ( g i )| | gG | i 360 1 g2 8 45 g3 4 90 g4 9 40 g5 9 40 g6 5 72 g7 5 72 Hence jGj ˆ 360. Also G is simple, by Proposition 17.6. (b) By the Frobenius±Schur Count of Involutions (Corollary 23.17), the number of involutions t in G is bounded by 1‡ t < 7 ˆ jˆ1 ÷ j (1) ˆ 46X By considering jC G ( g i )j, we see that g i has even order only for i ˆ 2 and 3. Since t < 45, only g2 can be an involution. Hence g3 has order 4. (This information about the orders of g2 and g3 can also be deduced using Sylow's Theorem.) (c) Clearly g6 and g7 have order 5, and at least one of g4 and g5 has order 3. If j ˆ 4 or 5 and k ˆ 6 or 7 then Chapter 30 a2 jk ˆ ˆ ˆ ÷( g2 )÷( g j )÷ ( g k ) jGj jCG ( g 2 )j jCG ( g j )j ÷ ÷(1) 449 360 ˆ 5X 8.9 Therefore G contains elements x and y such that x has order 2, y has order 3 and xy has order 5. As in the solution to Exercise 3, the subgroup H ˆ kx, yl of G is isomorphic to A5 . (d) If g, h P G then ( gh)r: Hx 3 Hxgh, and ( gr)(hr): Hx 3 Hxg 3 HxghX Hence r is a homomorphism. (e) Since G is simple, Ker r ˆ {1}. Hence G is isomorphic to a subgroup K of S6 . Since jS6 : Kj ˆ 63a360 ˆ 2, K must be A6 . 6. Consider the ®gure in Example 30.6(3). We shall explain how to label the vertices by elements of G. Choose a vertex and label it 1. Label the vertices according to the following inductive rule. Assume that v is a vertex and an adjacent vertex u is labelled by g. Then label v by ga gb gb À1 if the edge uv has no arrow, if the edge uv has an arrow from u to v, if the edge uv has an arrow from v to uX For example, if you decided to label the bottom left-hand vertex by 1, then part of the labelling would be The relation a2 ˆ 1 ensures that the labelling is consistent along unmarked edges; since b3 ˆ 1, the labelling is consistent around triangles; and the relation abababab ˆ 1 deals with the octagons. Every element in G has the form given by the label of one of the 24 vertices, so jGj < 24. 7. The conjugacy classes of PSL(2, 7) are given in Lemma 27.1 The element g 2 is an involution with centralizer of order 8 given in the proof of     2 2 2 4 Lemma 27.1. Letting a ˆ , bˆ , we see that the À2 2 4 À2 centralizer is generated by a and b, and a4 ˆ b2 ˆ 1, bÀ1 ab ˆ aÀ1 À, hence the centralizer is isomorphic to D8 . As in Exercises 12.3 and 12.4, we see that C A6 ((12)(34)) has order 8 and is generated by (1324) and (13)(24), hence as above is isomorphic to D8 . 450 Representations and characters of groups 8. Let G be the simple group PSL(2, 17). In the ®eld Z17 the element 4 is a   4 0 fourth root of unity, so t ˆ Z is an involution. Calculate that 0 À4 C G (t) is generated by the group of diagonal matrices together with     0 1 3 0 bˆ Z, hence is generated by b and a ˆ Z. As À1 0 0 6 a 8 ˆ b2 ˆ 1 and bÀ1 ab ˆ aÀ1, we have C G (t)  D16 . Chapter 31 1. Assume that G has an abelian subgroup H of index p r ( p prime), and that |G| . p. If H ˆ {1} then |G| ˆ p r and so G is not simple (see Lemma 26.1(1)). So assume that H Tˆ {1}; pick 1 Tˆ h P H. Then H < CG (h) as H is abelian, so |G:CG (h)| is a power of p. If |G:CG (h)| ˆ 1 then khl v G and G is not simple. And if |G:CG (h)| . 1, then G is not simple by Theorem 31.3. 2. By Burnside's Theorem, jGj is divisible by at least three distinct primes. Since 3 . 5 . 7 . 80, jGj is even. Then by Exercise 13.8, |G| is divisible by 4. Since 4 . 3 . 7 . 80, the only possibility is that jGj ˆ 4 . 3 . 5 ˆ 60. Chapter 32 1. (a) The fact that BB ˆ I follows from the observation that for all i, j, d(ei b, ej b) ˆ d(ei , ej ) ˆ ä ij X Since 1 ˆ det I ˆ (det B)(det Bt ) ˆ (det B)2 , we have det B ˆ Æ1. (b) (i) The eigenvalues of C are the roots of det (C À xI), which is a cubic polynomial over R. Therefore, C has one or three real eigenvalues. (ii) Moreover, the product of the eigenvalues of C is det C ˆ 1. If C has three real eigenvalues, then they cannot all be negative; and if C has one real eigenvalue ë and a pair of conjugate non-real eigenvalues ì, ì, then ë ìì ˆ 1, and hence ë . 0. Therefore C has a real positive eigenvalue, say ë. (iii) Let v be an eigenvector for ë. Then d(v, v) ˆ d(vC, vC) ˆ d(ëv, ëv) ˆ ë2 d(v, v), and so ë ˆ 1. (c) Let c be the isometry v 3 vC. By (b), c ®xes a vector v; it is now easy to convince yourself that c must be a rotation about the axis through v. The required result for b follows from the de®nition of c. (d) Take three orthogonal axes, one of which is the axis of the rotation b. With respect to these axes, the matrix of b is H I 1 0 0 d 0 cos ö sin ö eX 0 Àsin ö cos ö Hence tr B ˆ 1 ‡ 2 cos ö. t Chapter 32 451 2. We regard G as a subgroup of O(R3 ). It is easy to see that the translation submodule T (which consists of all the translation modes) is isomorphic to the RG-module given by the natural action of G on R3 . Hence by part (d) of Exercise 1, V if g is a rotation through ö, about some b 1 ‡ 2 cos ö, b b b b axis, b ` ÷ T ( g) ˆ b À(1 ‡ 2 cos ö), if the element À g of O(R3 ) is a rotation b b b b b X through öX Now consider the rotation submodule R, which consists of all the rotation modes. A rotation mode is speci®ed by a 3-dimensional vector öv, where v is a unit vector along the axis of the rotation and ö denotes the angle of rotation, taken positive in the right-hand screw sense. Let g P G, and consider g acting on öv. It sends v to vg, and if g is a rotation, it preserves the sense of the rotation; however, if g is a re¯ection then it transforms a right-hand screw to a left-hand screw, and hence sends öv to (Àö)(vg). Therefore @ if g is a rotation, ÷ T ( g), ÷ R ( g) ˆ À÷T ( g), if g is not a rotation, and so ÷ T ‡ ÷ R has the required form. 3. The matrix A is I H p p À3a2 0 3a4 À 3a4 p3a4 3a4 p f 0 À1a2 À 3a4 p1a4 3a4 1a4 g g f p k f 3a4 À 3a4 À3a4 3a4 0 0 g gX f p p 3a4 À5a4 0 1 g m f À 3a4 p1a4 g f p d 3a4 3a4 0 0 À3a4 À 3a4 e p p 3a4 1a4 0 1 À 3a4 À5a4 4. A simpler basis is given by 1 2(r1 1 2(r1 1 2(r1 ‡ r2 ) ˆ (v12 ‡ v21 ) À (v34 ‡ v43 ), ‡ r3 ) ˆ (v13 ‡ v31 ) À (v24 ‡ v42 ), ‡ r4 ) ˆ (v14 ‡ v41 ) À (v23 ‡ v32 )X We chose r1, r2, r3, r4 to be the images of w1, w2, w3, w4 under an RGisomorphism. (Compare the construction of the matrix B, towards the end of Example 32.20.) 5. (a) This is a routine geometrical exercise. (b) Let the displaced positions of the atoms be 09, 19, 29, 39, 49. The distance of 19 from the plane through 1 perpendicular to 12 is x12 ‡ 1(x13 ‡ x14 ). Similarly, the distance of 29 from the plane through 2 2 perpendicular to 12 is x21 ‡ 1(x23 ‡ x24 ). Therefore 12 has decreased by 2 x12 ‡ x21 ‡ 1(x13 ‡ x14 ‡ x23 ‡ x24 ). The other calculations can be done in 2 the same way. (c) We express the force at each atom as a vector, and then write this vector 452 Representations and characters of groups as a linear combination of our three chosen unit vectors at the initial position of the atom. Let d ij denote the decrease in the length ij, as calculated in part (b). Then, for example, at vertex 1, the contributions to the component of the force vector in the direction 12 are as follows: Àk1 d12 from the force between atoms 1 and 2; zero from the force between atoms 1 and 3 and from the force between atoms 1 and 4; and Àk2 d10 sec (/012) from the force between atoms 1 and 0. Hence p m1 12 ˆ Àk 1 d 12 À 1 (3a2)k 2 d 10 X x 3 Upon substituting for d12 and d10 from part (b), we obtain the given expression for 12. x The other accelerations are calculated in the same way. (d) The entries in the 15 3 15 matrix A are the coef®cients which appear in the equations of motion  ˆ xA. If you write down the matrix A, then x you will easily verify that the given vectors are eigenvectors of A (with eigenvalues À(4k 1 ‡ k 2 )am1 , 0, 0, 0, Àk 1 am1 , Àk 1 am1 , p p I À2k 2 a3m1 À4k 2 p 2a(m2 p3) À2k 2 a3m1 p À4k 2 2a(m2 3) eX p À2k 2 3a(9m1 2) À4k 2 a3m2 p (f ) You will ®nd that (1, À2, 6) is an eigenvector of B, with eigenvalue 0. This agrees with the statement in Example 32.20 that the translation vector r1 À 2s1 ‡ 3 cos Ww1 is an eigenvector of A. 6. (a) Assign coordinate axes along the edges of the square, as shown below. respectively). (e) The matrix B is H À(6k 1 ‡ k 2 )a3m1 d À(3k 1 ‡ k 2 )a3m1 p p Àk 2 3a(9m1 2) The symmetry group is G ˆ D8. Let a denote the rotation sending P 3 Q 3 R 3 S 3 P, and let b denote the re¯ection in the axis PR. The character ÷ of the RG-module R8 is 1 ÷ 8 a2 0 a 0 b 0 ab 0 Chapter 32 453 We refer to the character table of D8 which is given in Example 16.3(3), and see that ÷ ˆ ÷1 ‡ ÷2 ‡ ÷3 ‡ ÷4 ‡ 2÷5 X The rotation mode is (t ‡ â)v, where v ˆ (1, À1, 1, À1, 1, À1, 1, À1) P V÷2 X The translation modes are (t ‡ â)v, where v is in the span of v1 , v2 and v1 ˆ (1, 0, 0, 1, À1, 0, 0, À1), v2 ˆ (0, 1, À1, 0, 0, À1, 1, 0)X The homogeneous components V÷1, V÷3 and V÷4 are spanned by the vectors (1, 1, 1, 1, 1, 1, 1, 1), (1, 1, À1, À1, 1, 1, À1, À1) and (1, À1, À1, 1, 1, À1, À1, 1), respectively. The ®nal set of eigen-vectors is given by V÷5 ’ R8 which is spanned by vib (1, 0, 0, À1, À1, 0, 0, 1) and (0, 1, 1, 0, 0, À1, À1, 0)X (b) The matrix A is 1 f0 f f0 f k f0 À f mf0 f f0 f d0 1 H 0 1 1 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 1 1 0 I 1 0g g 0g g 0g gX 0g g 0g g 0e 1 7. (a, b) Let å i (1 < i < m) be the projection which is given by å i : u1 ‡ X X X ‡ um 3 ui (where u k P Uk for all k). Then w 3 wAå j WÀ1 W i j (w P Ui ) gives an RG-homomorphism from Ui to Ui. By Exercise 23.8, there exist ë ij P R such that for all w P Ui, Since €m wAå j ˆ ë ij wWÀ1 W j X i is the identity endomorphism of U1 È . . . È Um, we have m ˆ ë ij wWÀ1 W j for all w P Ui X wA ˆ i jˆ1 jˆ1 å j Now take in turn w ˆ uW i and w ˆ vW i to obtain the results of parts (a) and (b) of the question. (c) Take a basis u1 , . . . , u n of U1. Assume that the eigenvectors of A u are known. For all k with 1 < k < n, the eigenvectors of A in sp (u k W1 , . . . , u k W m ) are given by the eigenvectors of A u (see part (b)). Hence we know all the eigenvectors of A in U1 È . . . È Um. Bibliography Books mentioned in the text H. S. M. Coxeter and W. J. O. Moser, Generators and Relations for Discrete Groups (Fourth Edition), Springer-Verlag, 1980. J. B. Fraleigh, A First Course in Abstract Algebra (Third Edition), AddisonWesley, 1982. D. Gorenstein, Finite Simple Groups: An Introduction to their Classi®cation, Plenum Press, New York, 1982. G. D. James, The Representation Theory of the Symmetric Groups, Lecture Notes in Mathematics No. 682, Spring-Verlag, 1978. D. S. Passman, Permutation Groups, Benjamin, 1968. H. Pollard and H. G. Diamond, The Theory of Algebraic Numbers (Second Edition), Carus Mathematical Monographs No. 9, Mathematical Association of America, 1975. J. J. Rotman, An Introduction to the Theory of Groups (Third Edition), Allyn and Bacon, 1984. D. S. Schonland, Molecular Symmetry ± an Introduction to Group Theory and its uses in Chemistry, Van Nostrand, 1965. Suggestions for further reading M. J. Collins, Representations and Characters of Finite Groups, Cambridge University Press, 1990. C. W. Curtis and I. Reiner, Methods of Representation Theory with Applications to Finite Groups and Orders, Volume I, Wiley-Interscience, 1981. W. Feit, Characters of Finite Groups, Benjamin, 1967. I. M. Isaacs, Character Theory of Finite Groups, Academic Press, 1976. W. Ledermann, Introduction to Group Characters (Second Edition), Cambridge University Press, 1987. J. P. Serre, Linear Representations of Finite Groups, Springer-Verlag, 1977. 454 415 D12  S3 3 C2 . 5. 2. 125. 269 symmetric. 222. 361 alternating group. 118 degree. 136. 355 induced. 236 integer-valued. 56 algebraic integer. 122.7). 362 algebraic number. 11.q . 220. 196. 112. 174 permutation. 269 skew-symmetric. 55. 443 Q8 . 205 SL(2. 223 An . 424 C2 .Index A4 . 2 basis. 116. 11. 308 A5 . 119 regular. 116. 160 C4 . 445 PSL(2. 291 GL(2. 359 A6 . 340 Burnside's Theorem. 15 natural. 129 product. 111. 247 faithful. 278 Burnside's Lemma. 127. 81. 423 E. 412 C2 3 C2 . 161 D10.5 . 112. 88 centralizer. 420 455 . 122 character table. 9. 269 Brauer±Fowler Theroem. 24 character. 360 A7 . 273 associative. 160 D8. 253 irreducible. 150 trivial. 153 change of basis. 106 centre of group. 130. 265 reducible. 3. 234. 107. 415 Cn . 186 T4 n. 180 S5 . 181 A5 . 318 PSL(2. 244. 111 antisymmetric part. 440 SL(2. 363. 54 bijection. 419 D2n ( n odd).q). 183 D6 3 D6. 442 SL(2. 433 Fp. 172. 434 F7. 417 F11. 5. 11. 160 C3 . 327 PSL(2. 262 S6 . 6 bilinear form.7). 416 S4 . 128. 181. 82 D6. 182 D2n ( n even).3). 83. 125 linear. 114. 2 Cn . 195 generalized. 119 kernel of.11). 45. 82. 201. 10. 263 realized over R. 122.3. 364 C. 176. 354.q).8). 337 algebra. 192 real. 9. 230. 207. 82 action. 116. 240. of order 18. 221. 312. 445 T12. 359. 85. 359 A6 . 159 A4 . 298 of group algebra. 343 abelian group. 82. 349 class equation. 56. 363. 2. 44. 85 faithful representation. 82. 2 general linear. 66 external. 3. 301 order pq. 6 surjective. 114 Clifford's Theorem. 421 V24 . 24 endomorphism. 277. 9 HomCG (V W). 91. 2. 134 involution. 4. 4. 104 conjugate. 3 GL(2. 49 FG-homomorphism. 343 general linear group. 107. 119 irreducible module. 6 invertible. 178. 206 order 16. 1 abelian. 53 factor group. 302. 63 U6n. 111 cyclic. 9. 46 even permutation. 88 D2n. 311. 259 conjugacy class. 24 eigenvector. 306. 6 bijective. 122. 353 irreducible character. 5 expansion±contraction mode. 30. 109. 368 group algebra. 32. 311 symmetric. 206 direct sum. 3 F n. 181 dimension. 291 p-group. 143. 17. see module . 257 proper. 178 dihedral. 3 order. 318. 254 symmetry. 307 order 27. 32. 101 completely reducible. 181 factor. 234. 9 ®nite. 290 FG. 353. 90 common. 249 derived subgroup. homogeneous component. 107. 104. 12. 305. 81.456 Representations and characters of groups FG-submodule. 96 . 308 order p3 . 5. 125. 10. 34 FG-module. 50. 88 dicyclic. 63 Frobenius group. 11. 10. 365 special linear. 230. 213 coset. 173 diagonalization.F). 312 quaternion. 5. 61 FG-isomorphism. 421 direct product. 187. 74. 61 ideal. 96 congruences. 12. 109 cyclic group. 152 class sum. 6. 91 irreducible representation. 12. 228 inner product. 2. 74 composition. 2 order p3 . 420 dihedral group. 107. 2. 376 homomorphism. 18 eigenvalue. 3. 257 index of subgroup. 300 class algebra constants. 175. 8 cycle-shape. 12. 79. 20 equivalent. 3 H v G. 216 complete set. 422 V8 n . 11. 116. 273 induced character. 15 Fp. 15 direct product. 290 Frobenius Reciprocity Theorem. 232 Frobenius±Schur Count of Involutions. 7. 6 GL(n. 236 induced module. 367 projective special linear. 195 faithful module. 226. 277 function. 368 simple. 361 constituent. 181 degree. 55 H < G. 6 injective. 107 class function.q . 324. 381 external direct sum. 364 soluble. 20. 12. 50. 304 orthogonal. 281. 79 isomorphism. 3 group. 2. 278. 2 composition factor. 250. 256 maximal. 95. 435 order . 18 F ˆ R or C. 82 alternating.q). 9 indicator function. 9 faithful character. 5 rotation. 8 cycle notation. 83 dicyclic group T4 n. 46 faithful. 122. 379 rotation submodule. 171. 318. 62 reducible. 9. 127. 34 representatives. 361 module. 3. 384 minimal polynomial. 50. 76 matrix. 177. 113. 67 projective special linear group. 116.q). 223 Sn . 116. 128. 30. 5 odd. 45 powers of characters. 373 p-complement. 129.7). 174 linear transformation. 45. 30 degree. 27. 34. 2 order of g. 50. 319. 173. 85 irreducible. 5. 416 quaternion group. 105 restriction.7). p). 74 Sylow's Theorem. 4 cyclic. 34 irreducible. 54 modes of vibration. 269 special linear group. 113. 278. 62. 150 regular module. 340 permutation matrix. 254. 56 trivial. 169 linear character. 161 PSL(2. 56. 251 subgroup. 372. 342 Rank±Nullity Theorem. 3 rank. 278. 85 permutation. 380. 15 linearly independent. 269 odd permutation. 215. 171. 43 natural normal normal normal 217 basis. 359. 298 p9-part. 56 regular representation. 50 reducible representation. 339 subgroup. 79. 320. 34. 262 S6 . 19. 39 completely reducible. 173 generated. 44. 4 derived. 15 Maschke's Theorem. 10. 70. 44. 216. 3. 250. 319. 18 linearly dependent. projection. 45 methane. 74 faithful. 21 invertible. p). 312. 311. 124 reducible. 368 rotation mode. 5. 111. 19 real character. 177. 365 symmetric bilinear form. 79 kernel of. 284 product of characters. 367 orthogonality relations. 263 real conjugacy class. 215. 311 stabilizer. 113. 4 normal. 340 permutation character. 210 rotation group. 26 identity. 3. 343. 125 Lagrange's Theorem. 23 permutation. 256. 217 submodule. 5 orbit. 21 change of basis. 394 S4 . 318.11). 50 regular. 32. 216. 360 PSL(2. 109. 354. 9.Index kernel. 3 primitive root. 110. 5 even. 78 simple group. 321. 311 SL(2. 205 S7 . 363. 56 trivial.3). 50 regular character. 312 p-group. 50 regular. 192 . 275 S5 . 10. 9 lift. 263 reducible character. 249 equivalent. 116. 124. 442 SL(2. 354. 312 457 Q8 . 119 reducible module. 416 R. 49 irreducible. 45. 258 permutation. 193 presentation. 344 SL(2. 56 representation. 175. 180. 201. 278. 45. 359 PSL(2. 336. 263 real element. 5 permutation module. 176. 338 order of G. 445 Schur's Lemma. 3. 364 skew-symmetric bilinear form. 440 SL(2. 24 diagonal. 4 orthogonal group. 116. 229 translation mode. 187. 190 tensor product space. 178. 194 vibratory modes. 188 trace. 273 symmetry group. 420 tensor product module. 187. 368 T4 n. 122 module. 421 V8n. 43 representation. 381 water. 379 translation submodule. 187. 178. 254 symmetric part. 369. 175. 374 Z. 109. 178. 196. 338.458 Representations and characters of groups trivial character. 5 . 281. 421 Vandermonde matrix. 394 transposition. 117 transitive. 34 U6n. 380. 3. 116. 341 transitivity of induction. 2 symmetric group.