Representations and Characters of Groups, Second edition

Representations and Characters of Groups Now in its second edition, this text provides a modern introduction to the rep...

Author: Gordon James | Martin Liebeck

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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


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.

Preface to Second Edition

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.

1 Groups and homomorphisms

This book is devoted to the study of an aspect of group theory, so we begin with a reÂsumeÂ of facts about groups, most of which you should know already. In addition, we introduce several examples, such as dihedral groups and symmetric groups, which we shall use extensively to illustrate the later theory. An elementary course on abstract algebra would normally cover all the material in the chapter, and any book on basic group theory will supply you with further details. 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.

Groups A group consists of a set G, together with a rule for combining any two elements g, h of G to form another element of G, written gh; this rule must satisfy the following axioms: (1) for all g, h, k in G, ( gh)k g(hk); (2) there exists an element e in G such that for all g in G, eg ge g; (3) for all g in G, there exists an element gÿ1 in G such that gg ÿ1 g ÿ1 g e: We refer to the rule for combining elements of G as the product operation on G. 1

2


Axiom (1) states that the product operation is associative; the element e in axiom (2) is an identity element of G; and gÿ1 is an inverse of g in axiom (3). It is elementary to see that G has just one identity element, and that every g in G has just one inverse. Usually we write 1, rather than e, for the identity element of G. The product of an element g with itself, gg, is written g 2 ; similarly 3 g g 2 g, gÿ2 ( gÿ1 )2 , and so on. Also, g0 1. If the number of elements in G is ®nite, then we call G a ®nite group; the number of elements in G is called the order of G, and is written |G|. 1.1 Examples (1) Let n be a positive integer, and denote by C the set of all complex numbers. The set of nth roots of unity in C, with the usual multiplication of complex numbers, is a group of order n. It is written as Cn and is called the cyclic group of order n. If a e2ði= n , then Cn f1, a, a2 , : : : , a nÿ1 g, and an 1. (2) The set Z of all integers, under addition, is a group. (3) Let n be an integer with n > 3, and consider the rotation and re¯ection symmetries of a regular n-sided polygon.

There are n rotation symmetries: these are r0 , r1 , . . . , r nÿ1 where r k is the (clockwise) rotation about the centre O through an angle 2ðk=n. 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. These 2n rotations and re¯ections form a group under the product operation of composition (that is, for two symmetries f and g, the product fg means `®rst do f, then do g'). This group is called the dihedral group of order 2n, and is written D2 n. Let A be a corner of the polygon. Write b for the re¯ection in the

Groups and homomorphisms

3

line through O and A, and write a for the rotation r1 . Then the n rotations are 1, a, a2 , : : : , a nÿ1 (where 1 denotes the identity, which leaves the polygon ®xed); and the n re¯ections are b, ab, a2 b, : : : , a nÿ1 b: Thus all elements of D2 n are products of powers of a and b ± that is, D2 n is generated by a and b. Check that an 1, b2 1 and bÿ1 ab aÿ1 : These relations determine the product of any two elements of the group. For example, we have ba j aÿ j b (using the relation ba aÿ1 b), and hence (a i b)(a j b) a i ba j b a i aÿ j bb a iÿ j : We summarize all this in the presentation D2 n ha, b: an 1, b2 1, bÿ1 ab aÿ1 i: (4) For n a positive integer, the set of all permutations of {1, 2, . . . , n}, under the product operation of composition, is a group. It is called the symmetric group of degree n, and is written Sn . The order of Sn is n!. (5) Let F be either R (the set of real numbers) or C (the set of complex numbers). The set of all invertible n 3 n matrices with entries in F, under matrix multiplication, forms a group. This group is called the general linear group of degree n over F, and is denoted by GL(n, F). It is an in®nite group. The identity of GL(n, F) is of course the identity matrix, which we denote by In or just I. A group G is said to be abelian if gh hg for all g and h in G. While Cn and Z are abelian, most of the other examples given above are non-abelian groups. Subgroups Let G be a group. A subset H of G is said to be a subgroup if H is itself a group under the product operation inherited from G. We use the notation H < G to indicate that H is a subgroup of G.

4


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 2 H, and (2) if h, k 2 H then hkÿ1 2 H. 1.2 Examples (1) For every group G, both {1} and G are subgroups of G. (2) Let G be a group and g 2 G. The subset h gi f g n : n 2 Zg is a subgroup of G, called the cyclic subgroup generated by g. If gn 1 for some n > 1, then k gl is ®nite. In this case, let r be the least positive integer such that g r 1; then r is equal to the number of elements in k gl ± indeed, h gi f1, g, g 2 , : : : , g rÿ1 g: We call r the order of the element g. If G k gl for some g 2 G then we call G a cyclic group. The groups Cn and Z in Examples 1.1 are cyclic. (3) Let G be a group and let a, b 2 G. De®ne H to be the subset of G consisting of all elements which are products of powers of a and b ± that is, all elements of the form a i1 b j1 a i2 b j2 : : :a i n b j n for some n, where ik , jk 2 Z for 1 < k < n. Then H is a subgroup of G; we call H the subgroup generated by a and b, and write H ha, bi: Given any ®nite set S of elements of G, we can similarly de®ne hSi, the subgroup of G generated by S. This construction gives a powerful method of ®nding new groups as subgroups of given groups, such as general linear or symmetric groups. We illustrate the construction in the next example, and again in Example 1.5 below. (4) Let G GL(2, C), the group of invertible 2 3 2 matrices with entries in C, and let i 0 0 1 A , B : 0 ÿi ÿ1 0


5

Put H kA, Bl, the subgroup of G generated by A and B. Check that A4 I, A2 B2 , Bÿ1 AB Aÿ1 : Using the third relation, we see that every element of H has the form A i B j for some integers i, j; and using the ®rst two relations, we can take 0 < i < 3 and 0 < j < 1. Hence H has at most eight elements. Since the matrices Ai B j

(0 < i < 3, 0 < j < 1)

are all distinct, in fact j Hj 8. The group H is called the quaternion group of order 8, and is written Q8. The above three relations determine the product of any two elements of Q8 , so we have the presentation Q8 hA, B: A4 I, A2 B2 , Bÿ1 AB Aÿ1 i: (5) A transposition in the symmetric group Sn is a permutation which interchanges two of the numbers 1, 2, . . . , n and ®xes the other n ÿ 2 numbers. Every permutation g in Sn can be expressed as a product of transpositions. It can be shown that either all such expressions for g have an even number of transpositions, or they all have an odd number of transpositions; we call g an even or an odd permutation, accordingly. The subset An f g 2 Sn : g is an even permutationg is a subgroup of Sn , called the alternating group of degree n.

Direct products We describe a construction which produces a new group from given ones. Let G and H be groups, and consider G 3 H f( g, h): g 2 G and h 2 Hg: De®ne a product operation on G 3 H by ( g, h)( g9, h9) ( gg9, hh9) for all g, g9 2 G and all h, h9 2 H. With this product operation, G 3 H is a group, called the direct product of G and H.

6


More generally, if G1 , . . . , Gr are groups, then the direct product G1 3 : : : 3 Gr is f( g 1 , : : : , g r ): g i 2 Gi for 1 < i < rg, with product operation de®ned by ( g1 , : : : , g r )( g91 , : : : , g9r ) ( g 1 g91 , : : : , g r g9r ): If all the groups Gi are ®nite, then G1 3 . . . 3 Gr is also ®nite, of order |G1 | . . . |Gr |. 1.3 Example The group C2 3 . . . 3 C2 (r factors) has order 2 r and all its nonidentity elements have order 2. 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. In this book, we generally apply functions on the right ± that is, the image of g under a function W is written as gW, not as W g. We often indicate that W is a function from G to H by the notation W: G ! H. By an expression W: g ! h, where g 2 G and h 2 H, we mean that h gW. A function W: G ! H is invertible if there is a function ö: H ! G such that for all g 2 G, h 2 H, ( gW)ö g and (hö)W h: Then ö is called the inverse of W, and is written as Wÿ1 . A function W from G to H is invertible if and only if it is both injective (that is, g1 W g2 W for g1 , g2 2 G implies that g1 g2 ) and surjective (that is, for every h 2 H there exists g 2 G such that gW h). An invertible function is also called a bijection. Homomorphisms Given groups G and H, those functions from G to H which `preserve the group structure' ± the so-called homomorphisms ± are of particular importance. If G and H are groups, then a homomorphism from G to H is a function W: G ! H which satis®es ( g1 g 2 )W ( g 1 W)( g 2 W)

for all g 1 , g 2 2 G:


7

An invertible homomorphism is called an isomorphism. If there is an isomorphism W from G to H, then G and H are said to be isomorphic, and we write G H; also, Wÿ1 is an isomorphism from H to G, so H G. The following example displays a technique which can often be used to prove that certain functions are homomorphisms. 1.4 Example Let G D2 n ka, b: an b2 1, bÿ1 ab aÿ1 l, and write the 2n elements of G in the form ai b j with 0 < i < n ÿ 1, 0 < j < 1. Let H be any group, and suppose that H contains elements x and y which satisfy x n y 2 1, y ÿ1 xy x ÿ1 : We shall prove that the function W: G ! H de®ned by W: a i b j ! x i y j

(0 < i < n ÿ 1, 0 < j < 1)

is a homomorphism. Suppose that 0 < r < n ÿ 1, 0 < s < 1, 0 < t < n ÿ 1, 0 < u < 1. Then ar bs at bu a i b j for some i, j with 0 < i < n ÿ 1, 0 < j < 1. Moreover, i and j are determined by repeatedly using the relations an b2 1, bÿ1 ab aÿ1 : Since we have x n y 2 1, y ÿ1 xy x ÿ1 , we can also deduce that xr ys xt yu x i y j : Therefore, (ar bs at bu )W (a i b j )W x i y j x r y s x t y u (ar bs )W . (at bu )W, and so W is a homomorphism. We now demonstrate the technique of Example 1.4 in action. 1.5 Example Let G S5 and let x, y be the following permutations in G: x (1 2 3 4 5), y (2 5)(3 4):

8


(Here we adopt the usual cycle notation ± thus, (1 2 3 4 5) denotes the permutation 1 ! 2 ! 3 ! 4 ! 5 ! 1, and so on.) Check that x 5 y 2 1, y ÿ1 xy x ÿ1 : Let H be the subgroup kx, yl of G. Using the above relations, we see that H fx i y j : 0 < i < 4, 0 < j < 1g, a group of order 10. Now recall that D10 ha, b: a5 b2 1, bÿ1 ab aÿ1 i: By Example 1.4, the function W: D10 ! H de®ned by W: a i b j ! x i y j

(0 < i < 4, 0 < j < 1)

is a homomorphism. Since W is invertible, it is an isomorphism. Thus, H kx, yl D10.

Cosets Let G be a group and let H be a subgroup of G. For x in G, the subset Hx fhx: h 2 Hg of G is called a right coset of H in G. The distinct right cosets of H in G form a partition of G (that is, every element of G is in precisely one of the cosets). Suppose now that G is ®nite, and let Hx1 , . . . , Hxr be all the distinct right cosets of H in G. For all i, the function h ! hxi

(h 2 H)

is a bijection from H to Hxi , and so j Hxi j j Hj. Since G Hx1 [ : : : [ Hxr , and Hxi \ Hxj is empty if i 6 j, we deduce that jGj rj Hj: In particular, we have


9

1.6 Lagrange's Theorem If G is a ®nite group and H is a subgroup of G, then j Hj divides |G|. The number r of distinct right cosets of H in G is called the index of H in G, and is written as jG: Hj. Thus jG: Hj jGj=j Hj when G is ®nite. Normal subgroups A subgroup N of a group G is said to be a normal subgroup of G if gÿ1 Ng N for all g 2 G (where gÿ1 Ng f gÿ1 ng: n 2 Ng); we write N v G to indicate that N is a normal subgroup of G. Suppose that N v G and let G=N be the set of right cosets of N in G. The importance of the condition gÿ1 Ng N (for all g 2 G) is that it can be used to show that for all g, h 2 G, we have fxy: x 2 Ng and y 2 Nhg Ngh: Hence we can de®ne a product operation on G=N by (Ng)(Nh) Ngh

for all g, h 2 G:

This makes G=N into a group, called the factor group of G by N. 1.7 Examples (1) For every group G, the sub-groups {1} and G are normal subgroups of G. (2) For n > 1, we have An v Sn . If n > 2 then there are just two right cosets of An in Sn , namely An f g 2 Sn : g eveng, and An (1 2) f g 2 Sn : g oddg: Thus |Sn :An | 2, and so Sn =An C2 . (3) Let G D8 ka, b: a4 b2 1, ka2 l {1, a2 }. Then N v G and

bÿ1 ab aÿ1 l

and

let

N

G=N fN , Na, Nb, Nabg: Since (Na)2 (Nb)2 (Nab)2 N, we see that G=N C2 3 C2 . The subgroup kal is also normal in G, but the subgroup H kbl is not normal in G, since b 2 H while aÿ1 ba a2 b 2 = H.

10

Representations and characters of groups Simple groups

A group G is said to be simple if G 6 {1} and the only normal subgroups of G are {1} and G. For example, the cyclic group Cp , with p a prime number, is simple. We shall give examples of non-abelian simple groups in later chapters ± the smallest one is A5 . If G is a ®nite group which is not simple, then G has a normal subgroup N such that both N and G/N have smaller order than G; and in a sense, G is `built' out of these two smaller groups. Continuing this process with the smaller groups, we eventually see that G is `built' out of a collection of simple groups. (This is analogous to the fact that every positive integer is built out of its prime factors.) Thus, simple groups are fundamental to the study of ®nite groups.

Kernels and images To conclude the chapter, we relate normal subgroups and factor groups to homomorphisms. Let G and H be groups and suppose that W: G ! H is a homomorphism. We de®ne the kernel of W by (1:8)

Ker W f g 2 G: gW 1g:

Then Ker W is a normal subgroup of G. Also, the image of W is (1:9)

Im W f gW: g 2 Gg,

and Im W is a subgroup of H. The following result describes the way in which the kernel and image of W are related. 1.10 Theorem Suppose that G and H are groups and let W: G ! H be a homomorphism. Then G=Ker W Im W: An isomorphism is given by the function Kg ! gW where K Ker W.

( g 2 G)


11

1.11 Example The function W: Sn ! C2 given by 1, if g is an even permutation, W: g ! ÿ1, if g is an odd permutation, is a homomorphism. We have Ker W An , and for n > 2, Im W C2 . We know from Example 1.7(2) that Sn /An C2 , illustrating Theorem 1.10. Summary of Chapter 1 1. Examples of groups are Cn ka: an 1l, D2 n ka, b: an b2 1, bÿ1 ab aÿ1 l, Q8 ka, b: a4 1, a2 b2 , bÿ1 ab aÿ1 l, Sn the symmetric group of degree n, An the alternating group of degree n, GL(n, C) the group of invertible n 3 n matrices over C, G1 3 . . . 3 Gr , the direct product of the groups G1 , . . . , Gr . 2. A normal subgroup N of G is a subgroup such that gÿ1 Ng N for all g in G. The factor group G=N consists of the right cosets Ng ( g 2 G), with multiplication (Ng)(Nh) Ngh: 3. A homomorphism W: G ! H is a function such that ( g1 g 2 )W ( g 1 W)( g 2 W) for all g1 , g2 in G. The kernel, Ker W, is a normal subgroup of G, and the image, Im W, is a subgroup of H. The factor group G=Ker W is isomorphic to Im W. Exercises for Chapter 1 1. Show that if G is an abelian group which is simple, then G is cyclic of prime order. 2. Suppose that G and H are groups, with G simple, and that W: G ! H is a surjective homomorphism. Show that either W is an isomorphism or H {1}.

12


3. Suppose that G is a subgroup of Sn , and that G is not contained in An . Prove that G \ An is a normal subgroup of G, and G= (G \ An ) C2 : 4. Let G D8 ha, b: a4 b2 1, bÿ1 ab aÿ1 i, and H Q8 hc, d: c4 1, c2 d 2 , d ÿ1 cd cÿ1 i: (a) Let x, y be the permutations in S4 which are given by x (1 2), y (3 4), and let K be the subgroup kx, yl of S4 . Show that both the functions ö: G ! K and ø: H ! K, de®ned by ö: ar bs ! x r y s , ø: cr d s ! x r y s

(0 < r < 3, 0 < s < 1),

are homomorphisms. Find Ker ö and Ker ø. (b) Let X, Y be the 2 3 2 matrices which are given by 0 i 0 ÿ1 X ,Y , i 0 1 0 and let L be the subgroup hX , Y i of GL(2, C). Show that just one of the functions ë: G ! L and ì: H ! L, de®ned by ë: ar bs ! X r Y s , ì: cr d s ! X r Y s

(0 < r < 3, 0 < s < 1),

is a homomorphism. Prove that this homomorphism is an isomorphism. 5. Prove that D4 m D2 m 3 C2 if m is odd. 6. (a) Show that every subgroup of a cyclic group is cyclic. (b) Let G be a ®nite cyclic group, and let n be a positive integer which divides |G|. Prove that f g 2 G: g n 1g is a cylic subgroup of G of order n. (c) If G is a ®nite cyclic group and x, y are elements of G with the same order, show that x is a power of y.


13

7. Show that the set of non-zero complex numbers, under the usual multiplication, is a group. Prove that every ®nite subgroup of this group is cyclic. 8. Show that every group of even order contains an element of order 2. 9. Find elements A and B of GL(2, C) such that A has order 8, B has order 4, and B2 A4 and Bÿ1 AB Aÿ1 : Show that the group kA, Bl has order 16. (Hint: compare Q8 in Example 1.2(4).) 10. Suppose that H is a subgroup of G with |G: H| 2. Prove that H v G.

2 Vector spaces and linear transformations

An attractive feature of representation theory is that it combines two strands of mainstream mathematics, namely group theory and linear algebra. For reference purposes, we gather the results from linear algebra concerning vector spaces, linear transformations and matrices which we shall use later. Most of the material will be familiar to you if you have taken a ®rst course on linear algebra, so we omit the proofs. An exception occurs in the last section, where we deal with projections; here, we explain in detail how the results work, in case you have not come across projections before. Vector spaces Let F be either R (the set of real numbers) or C (the set of complex numbers). A vector space over F is a set V, together with a rule for adding any two elements u, v of V to form an element u v of V, 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.) Moreover, these rules must satisfy: (2.1)

(a) V is an abelian group under addition; (b) for all u, v in V and all ë, ì in F, (1) ë(u v) ëu ëv, (2) (ë ì)v ëv ìv, (3) (ëì)v ë(ìv), (4) 1v v.

The elements of V are called vectors, and those of F are called scalars. We write 0 for the identity element of the abelian group V under addition. 14

Vector spaces and linear transformations

15

2.2 Examples (1) Let R2 denote the set of all ordered pairs (x, y) where x and y are real numbers. De®ne addition and scalar multiplication on R2 by (x, y) (x9, y9) (x x9, y y9), ë(x, y) (ëx, ë y): Then R2 is a vector space over R. (2) More generally, for each positive integer n, we consider row vectors (x1 , x2 , : : : , xn ) where x1 , x2 , . . . , xn belong to F. We denote the set of all such row vectors by F n, and de®ne addition and scalar multiplication on F n by (x1 , : : : , xn ) (x91 , : : : , x9n ) (x1 x91 , : : : , xn x9n ), ë(x1 , : : : , xn ) (ëx1 , : : : , ëxn ): Then F n is a vector space over F.

Bases of vector spaces Let v1 , . . . , v n be vectors in a vector space V over F. A vector v in V is a linear combination of v1 , . . . , v n if v ë1 v1 : : : ë n v n for some ë1 , . . . , ë n in F. The vectors v1 , . . . , v n are said to span V if every vector in V is a linear combination of v1 , . . . , v n . We say that v1 , . . . , v n are linearly dependent if ë1 v1 : : : ë n v n 0 for some ë1 , . . . , ë n in F, not all of which are zero; otherwise, v1 , . . . , v n are linearly independent. The vectors v1 , . . . , v n form a basis of V if they span V and are linearly independent. Throughout this book, we shall consider only vector spaces V which are ®nite-dimensional ± this means that V has a basis consisting of ®nitely many vectors, as above. It turns out that any two bases of V have the same number of vectors. The number of vectors in a basis of V is called the dimension of V and is written as dim V. If V {0} then dim V 0. The vector space V is n-dimensional if dim V n.

16


2.3 Example Let V F n. Then (1, 0, 0, : : : , 0), (0, 1, 0, : : : , 0), : : : , (0, 0, 0, : : : , 1) is a basis of V, so dim V n. Another basis is (1, 0, 0, : : : , 0), (1, 1, 0, : : : , 0), : : : , (1, 1, 1, : : : , 1): Given a basis v1 , . . . , v n of a vector space V, each vector v in V can be written in a unique way as v ë1 v1 : : : ë n v n , with ë1 , . . . , ë n in F. The vector v therefore determines the scalars ë1 , . . . , ë n . Except in the case where V {0}, there are many bases of V. Indeed, the next result says that any linearly independent vectors can be extended to a basis. (2.4)

If v1 , . . . , v k are linearly independent vectors in V, then there exist v k1 , . . . , v n in V such that v1 , . . . , v n form a basis of V. 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. For a subset U of V to be a subspace, it is necessary and suf®cient that all the following conditions hold: (2.5)

(1) 0 2 U; (2) if u, v 2 U then u v 2 U; (3) if ë 2 F and u 2 U then ëu 2 U.

2.6 Examples (1) {0} and V are subspaces of V. (2) Let u1 , . . . , ur be vectors in V. We de®ne sp (u1 , . . . , ur ) to be the set of all linear combinations of u1 , . . . , ur ; that is, sp (u1 , : : : , ur ) fë1 u1 : : : ë r ur : ë1 , : : : , ë r 2 Fg: By (2.5), sp (u1 , . . . , ur ) is a subspace of V, and it is called the subspace spanned by u1 , . . . , ur .


17

Notice that the following fact is a consequence of (2.4). (2.7)

Suppose that U is a subspace of the vector space V. Then dim U < dim V. Also, dim U dim V if and only if U V. Direct sums of subspaces

If U1, . . . , Ur are subspaces of a vector space V, then the sum U1 . . . Ur is de®ned by U1 : : : Ur fu1 : : : ur : ui 2 Ui for 1 < i < rg: By (2.5), U1 . . . Ur is a subspace of V. We say that the sum U1 . . . Ur is a direct sum if every element of the sum can be written in a unique way as u1 . . . ur with ui 2 Ui for 1 < i < r. If the sum is direct, then we write it as U1 : : : Ur : 2.8 Examples (1) Suppose that v1 , . . . , v n is a basis of V, and for 1 < i < n, let Ui be the subspace spanned by v i. Then V U1 : : : Un : (2) Let U be a subspace of V and let v1 , . . . , v k be a basis of U. Extend v1, . . . , v k to a basis v1 , . . . , v n of V (see (2.4)), and let W sp (v k1 , . . . , v n ). Then V U W: From this construction it follows that there are in®nitely many subspaces W with V U W, unless U is {0} or V. The next result is frequently useful when dealing with the direct sum of two subspaces. You should consult the solutions to Exercises 2.3 and 2.4 if you have dif®culty with the proof. (2.9)

Suppose that V U W, that u1 , . . . , ur is a basis of U and that w1, . . . , ws is a basis of W. Then the following three conditions are equivalent: (1) V U W, (2) u1 , . . . , ur , w1, . . . , ws is a basis of V, (3) U \ W {0}.

18


Our next result, involving the direct sum of several subspaces, can be deduced immediately from the de®nition of a direct sum. (2.10)

Suppose that U, W, U1, . . . , Ua, W1, . . . , Wb are subspaces of a vector space V. If V U W and also U U1 : : : Ua , and W W1 : : : Wb , then

V U1 : : : Ua W 1 : : : W b :

We now introduce a construction for vector spaces which is analogous to the construction of direct products for groups. Let U1, . . . , Ur be vector spaces over F, and let V f(u1 , : : : , ur ): ui 2 U i for 1 < i < rg: De®ne addition and scalar multiplication on V as follows: for all ui , u9i in Ui (1 < i < r) and all ë in F, let (u1 , : : : , ur ) (u91 , : : : , u9r ) (u1 u91 , : : : , ur u9r ), ë(u1 , : : : , ur ) (ëu1 , : : : , ëur ): With these de®nitions, V is a vector space over F. If, for 1 < i < r, we put U 9i f(0, : : : , ui , : : : , 0): ui 2 U i g (where the ui is in the ith position), then it is immediate that V U 19 : : : U 9r : We call V the external direct sum of U1, . . . , Ur, and, abusing notation slightly, we write V U1 : : : Ur :

Linear transformations Let V and W be vector spaces over F. A linear transformation from V to W is a function W: V ! W which satis®es (u v)W uW vW

for all u, v 2 V , and

(ëv)W ë(vW)

for all ë 2 F and v 2 V :


19

Just as a group homomorphism preserves the group multiplication, so a linear transformation preserves addition and scalar multiplication. Notice that if W: V ! W is a linear transformation and v1, . . . , v n is a basis of V, then for ë1 , . . . , ë n in F we have (ë1 v1 : : : ë n v n )W ë1 (v1 W) : : : ë n (v n W): Thus, W is determined by its action on a basis. Furthermore, given any basis v1 , . . . , v n of V and any n vectors w1, . . . , wn in W, there is a unique linear transformation ö: V ! W such that v i ö wi for all i; the linear transformation ö is given by (ë1 v1 : : : ë n v n )ö ë1 w1 : : : ë n wn : We sometimes construct a linear transformation ö: V ! W in this way, by specifying the values of ö on a basis of V, and then saying èxtend the action of ö to be linear'.

Kernels and images Suppose that W: V ! W is a linear transformation. The kernel of W (written Ker W) and the image of W (written Im W) are de®ned as follows: (2:11)

Ker W fv 2 V : vW 0g, Im W fvW: v 2 V g:

Using (2.5), it is easy to check that Ker W is a subspace of V and Im W is a subspace of W. Their dimensions are connected by the following equation, which is known as the Rank±Nullity Theorem: (2:12)

dim V dim (Ker W) dim (Im W):

2.13 Examples (1) If W: V ! W is de®ned by vW 0 for all v 2 V, then W is a linear transformation, and Ker W V ,

Im W f0g:

(2) If W: V ! V is de®ned by vW 3v for all v 2 V, then W is a linear transformation, and Ker W f0g,

Im W V :

20


(3) If W: R3 ! R2 is given by (x, y, z)W (x 2 y z, ÿ y 3z) for all x, y, z 2 R, then W is a linear transformation; we have Ker W sp ((7, ÿ3, ÿ1)),

Im W R2 ,

so dim (Ker W) 1 and dim (Im W) 2. Invertible linear transformations Again, let V and W be vector spaces over F. A linear transformation W from V to W is injective if and only if Ker W {0}, and hence W is invertible precisely when W is surjective and Ker W {0}. It turns out that the inverse of an invertible linear transformation is also a linear transformation (see Exercise 2.1). If there exists an invertible linear transformation from V to W, then V and W are said to be isomorphic vector spaces. By applying (2.12), we see that isomorphic vector spaces have the same dimension. By also taking (2.7) into account, we obtain the next result (see Exercise 2.2). (2.14)

Let W be a linear transformation from V to itself. Then the following three conditions are equivalent: (1) W is invertible; (2) Ker W {0}; (3) Im W V. Endomorphisms

A linear transformation from a vector space V to itself is called an endomorphism of V. Suppose that W and ö are endomorphisms of V and ë 2 F. We de®ne the functions W ö, Wö and ëW from V to V by (2:15)

v(W ö) vW vö, v(Wö) (vW)ö, v(ëW) ë(vW),

for all v 2 V. Then W ö, Wö and ëW are endomorphisms of V. We write W2 for WW.


21

2.16 Examples (1) The identity function 1 V de®ned by 1 V : v ! v for all v 2 V is an endomorphism of V. If W is an endomorphism of V, then so is W ÿ ë1 V , for all ë 2 F. Note that Ker (W ÿ ë1 V ) fv 2 V : vW ëvg: (2) Let V R2, and let W, ö be the functions from V to V de®ned by (x, y)W (x y, x ÿ 2 y), (x, y)ö (x ÿ 2 y, ÿ2x 4 y): Then W and ö are endomorphisms of V, and W ö, Wö, 3W and W2 are given by (x, y)(W ö) (2x ÿ y, ÿx 2 y), (x, y)(Wö) (ÿx 5 y, 2x ÿ 10 y), (x, y)(3W) (3x 3 y, 3x ÿ 6 y), (x, y)W2 (2x ÿ y, ÿx 5 y):

Matrices Let V be a vector space over F, and let W be an endomorphism of V. Suppose that v1 , . . . , v n is a basis of V and call it B . Then there are scalars aij in F (1 < i < n, 1 < j < n) such that for all i, v i W a i1 v1 : : : ain v n : 2.17 De®nition The n 3 n matrix (aij ) is called the matrix of W relative to the basis B , and is denoted by [W]B . 2.18 Examples (1) If W 1 V (so that vW v for all v 2 V), then [W]B In for all bases B of V, where In denotes the n 3 n identity matrix. (2) Let V R2 and let W be the endomorphism (x, y) ! (x y, x ÿ 2 y) of V. If B is the basis (1, 0), (0, 1) of V and B 9 is the basis

22


(1, 0), (1, 1) of V, then

[W]B

1 , ÿ2

1 1

[W]B 9

0 3

1 : ÿ1

If we wish to indicate that the entries in a matrix A come from F, then we describe A as a matrix over F. Given two m 3 n matrices A (aij ) and B (bij ) over F, their sum A B is the m 3 n matrix over F whose ij-entry is aij bij for all i, j; and for ë 2 F, the matrix ëA is the m 3 n matrix over F obtained from A by multiplying all the entries by ë. As you know, the product of two matrices is de®ned in a less transparent way. Given an m 3 n matrix A (aij ) and an n 3 p matrix B (bij ), their product AB is the m 3 p matrix whose ij-entry is n X

aik bkj :

k1

2.19 Example Let

ÿ1 3

A Then A B BA

ÿ1

ÿ2

5

0 !

ÿ12

ÿ4

ÿ5

3

2 0 , B 1 2

! , 3A

ÿ3

6

9

3

ÿ4 : ÿ1

! , AB

4

2

2

ÿ13

! ,

:

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: (2.20)

Suppose that B is a basis of the vector space V, and that W and ö are endomorphisms of V. Then [W ö]B [W]B [ö]B , and [Wö]B [W]B [ö]B :


23

Also, for all scalars ë, [ëW]B ë[W]B : We showed you in (2.17) how to get a matrix from an endomorphism of a vector space V, given a basis of V. It is easy enough to reverse this process and use a matrix to de®ne an endomorphism. We concentrate on a particular way of doing this. Suppose that A is an n 3 n matrix over F, and let V F n, the vector space of row vectors (x1 , . . . , xn ) with each xi in F. Then for all v in V, the matrix product vA also lies in V. The following remark is easily justi®ed. (2.21)

If A is an n 3 n matrix over F, then the function v ! vA (v 2 F n ) is an endomorphism of F n.

2.22 Example Let

A

1 3

ÿ1 : 2

Then A gives us an endomorphism W of F 2, where 1 ÿ1 (x, y)W (x, y) (x 3 y, ÿx 2 y): 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 BA In . Such a matrix B, if it exists, is unique; it is called the inverse of A and is written as Aÿ1 . Write det A for the determinant of A. Then a necessary and suf®cient condition for A to be invertible is that det A 6 0. The connection between invertible endomorphisms and invertible matrices is straightforward, and follows from (2.20): given a basis B of V, an endomorphism W of V is invertible if and only if the matrix [W]B is invertible. Invertible matrices turn up when we relate two bases of a vector space. An invertible matrix converts one basis into another, and this same matrix is used to describe the way in which the matrix of an

24


endomorphism depends upon the basis. The precise meaning of these remarks is revealed in the de®nition (2.23) and the result (2.24) below. 2.23 De®nition Let v1 , . . . , v n be a basis B of the vector space V, and let v91 , . . . , v9n be a basis B 9 of V. Then for 1 < i < n, v9i t i1 v1 : : : tin v n for certain scalars tij . The n 3 n matrix T (tij ) is invertible, and is called the change of basis matrix from B to B 9. The inverse of T is the change of basis matrix from B 9 to B . (2.24)

If B and B 9 are bases of V and W is an endomorphism of V, then [W]B T ÿ1 [W]B 9 T , where T is the change of basis matrix from B to B 9.

2.25 Example Suppose that V R2. Let B be the basis (1, 0), (0, 1) and B 9 the basis (1, 0), (1, 1) of V. Then 1 0 1 0 ÿ1 T ,T : 1 1 ÿ1 1 If W is the endomorphism W: (x, y) ! (x y, x ÿ 2 y) of V, as in Example 2.18(2), then 1 0 0 1 1 0 1 1 [W]B T ÿ1 [W]B 9 T : ÿ1 1 3 ÿ1 1 1 1 ÿ2

Eigenvalues Let V be an n-dimensional vector space over F, and suppose that W is an endomorphism of V. The scalar ë is said to be an eigenvalue of W if vW ëv for some non-zero vector v in V. Such a vector v is called an eigenvector of W.


25

Now ë is an eigenvalue of W if and only if Ker (W ÿ ë1 V ) 6 {0}, which occurs if and only if W ÿ ë1 V is not invertible. Therefore, if B is a basis of V, then the eigenvalues of W are those scalars ë in F which satisfy the equation det ([W]B ÿ ëI n ) 0: Solving this equation involves ®nding the roots of a polynomial of degree n. Since every non-constant polynomial with coef®cients in C has a root in C, we deduce the following result. (2.26)

Let V be a non-zero vector space over C, and let W be an endomorphism of V. Then W has an eigenvalue.

2.27 Examples (1) Let V C2 and let W be the endomorphism of V which is given by (x, y)W (ÿ y, x): If B is the basis (1, 0), (0, 1) of V, then 0 1 [W]B : ÿ1 0 We have det ([W]B ÿ ëI2 ) ë2 1, so i and ÿi are the eigenvalues of W. Corresponding eigenvectors are (1, ÿi) and (1, i). Note that if B 9 is the basis (1, ÿi), (1, i) of V, then i 0 [W]B 9 : 0 ÿi (2) Let V R2 and let W again be the endomorphism which is given by (x, y)W (ÿ y, x): This time, V is a vector space over R, and W has no eigenvalues in R. Thus we depend upon F being C in result (2.26). For an n 3 n matrix A over F, the element ë of F is said to be an eigenvalue of A if vA ëv for some non-zero row vector v in F n. The eigenvalues of A are those elements ë of F which satisfy det (A ÿ ëI n ) 0:

26


2.28 Example We say that an n 3 n matrix A (aij ) is diagonal if aij 0 for all i and j with i 6 j. We often display such a matrix in the form 0 1 ë1 0 B C .. A@ A . ën 0 which indicates, in addition, that aii ë i for 1 < i < n. For this diagonal matrix A, the eigenvalues are ë1 , . . . , ë n . Projections If a vector space V is a direct sum of two subspaces U and W, then we can construct a special endomorphism of V which depends upon the expression V U W: 2.29 Proposition Suppose that V U W. De®ne ð: V ! V by (u w)ð u

for all u 2 U , w 2 W :

Then ð is an endomorphism of V. Further, Im ð U , Ker ð W and ð2 ð: Proof Since every vector in V has a unique expression in the form u w with u 2 U, w 2 W, it follows that ð is a function on V. Let v and v9 belong to V. Then v u w and v9 u9 w9 for some u, u9 in U and w, w9 in W. We have (v v9)ð (u u9 w w9)ð u u9 (u w)ð (u9 w9)ð vð v9ð: Also, for ë in F, (ëv)ð (ëu ëw)ð ëu ë(vð): Therefore, ð is an endomorphism of V. Clearly Im ð # U; and since uð u for all u in U, we have Im U. Also,


27

(u w)ð 0 , u 0 , u w 2 W , and so Ker ð W. Finally, (u w)ð2 uð u (u w)ð, and so ð2 ð.

j

2.30 De®nition An endomorphism ð of a vector space V which satis®es ð2 ð is called a projection of V. 2.31 Example The endomorphism (x, y) ! (2x 2 y, ÿx ÿ y) of R2 is a projection. We now show that every projection can be constructed using a direct sum, as in Proposition 2.29. 2.32 Proposition Suppose that ð is a projection of a vector space V. Then V Im ð Ker ð: Proof If v 2 V then v vð (v ÿ vð). Clearly the ®rst term vð belongs to Im ð; and the second term v ÿ vð lies in Ker ð, since (v ÿ vð)ð vð ÿ vð2 vð ÿ vð 0: This establishes that V Im ð Ker ð. Now suppose that v lies in Im ð \ Ker ð. As v 2 Im ð, we have v uð for some u 2 V. Therefore vð uð2 uð v: Since v 2 Ker ð, it follows that v vð 0. Thus Im ð \ Ker ð f0g, and (2.9) now shows that V Im ð Ker ð.

j

28


2.33 Example If ð: (x, y) ! (2x 2 y, ÿx ÿ y) is the projection of R2 which appears in Example 2.31, then Im ð f(2x, ÿx): x 2 Rg, Ker ð f(x, ÿx): x 2 Rg:

Summary of Chapter 2 1. All our vector spaces are ®nite-dimensional over F, where F C or R. For example, F n is the set of row vectors (x1 , . . . , x n ) with each xi in F, and dimF n n. 2. V U1 . . . Ur if each Ui is a subspace of V, and every element v of V has a unique expression of the form v u1 . . . ur (ui 2 Ui ). Also, V U W if and only if V U W and U \ W {0}. 3. A linear transformation W: V ! W satis®es (u v)W uW vW and (ëv)W ë(vW) for all u, v in V and all ë in F. Ker W is a subspace of V and Im W is a subspace of W, and dim V dim (Ker W) dim (Im W): 4. A linear transformation W: V ! W is invertible if and only if Ker W {0} and Im W W. 5. Given a basis B of the n-dimensional vector space V, there is a correspondence between the endomorphisms W of V and the n 3 n matrices [W]B over F. Given two bases B and B 9 of V, and an endomorphism W of V, there exists an invertible matrix T such that [W]B T ÿ1 [W]B 9 T: 6. Eigenvalues ë of an endomorphism W satisfy vW ëv for some nonzero v in V. 7. A projection is an endomorphism ð of V which satis®es ð2 ð. Exercises for Chapter 2 1. Show that if V and W are vector spaces and W: V ! W is an invertible linear transformation then Wÿ1 is a linear transformation.


29

2. Suppose that W is an endomorphism of the vector space V. Show that the following are equivalent: (1) W is invertible; (2) Ker W {0}; (3) Im W V. 3. Let U and W be subspaces of the vector space V. Prove that V U W if and only if V U W and U \ W = {0}. 4. Let U and W be subspaces of the vector space V. Suppose that u1 , . . . , ur is a basis of U and w1, . . . , ws is a basis of W. Show that V U W if and only if u1 , . . . , ur , w1, . . . , ws is a basis of V. 5. (a) Let U1, U2 and U3 be subspaces of a vector space V, with V U1 U2 U3. Show that V U1 U2 U3 , U 1 \ (U 2 U3 ) U 2 \ (U 1 U3 ) U 3 \ (U 1 U2 ) f0g: (b) Give an example of a vector space V with three subspaces U1, U2 and U3 such that V U1 U2 U3 and U1 \ U 2 U1 \ U3 U2 \ U 3 f0g, but V 6 U1 U2 U3. 6. Suppose that U1, . . . , Ur are subspaces of the vector space V, and that V U1 . . . Ur. Prove that dim V dim U1 : : : dim U r : 7. Give an example of a vector space V with endomorphisms W and ö such that V Im W Ker W, but V 6 Im ö Ker ö. 8. Let V be a vector space and let W be an endomorphism of V. Show that W is a projection if and only if there is a basis B of V such that [W]B is diagonal, with all diagonal entries equal to 1 or 0. 9. Suppose that W is an endomorphism of the vector space V and W2 1 V . Show that V U W, where U fv 2 V : vW vg, W fv 2 V : vW ÿvg: Deduce that V has a basis B such that [W]B is diagonal, with all diagonal entries equal to 1 or ÿ1.

3 Group representations

A representation of a group G gives us a way of visualizing G as a group of matrices. To be precise, a representation is a homomorphism from G into a group of invertible matrices. We set out this idea in more detail, and give some examples of representations. We also introduce the concept of equivalence of representations, and consider the kernel of a representation. Representations Let G be a group and let F be R or C. Recall from the ®rst chapter that GL (n, F) denotes the group of invertible n 3 n matrices with entries in F. 3.1 De®nition A representation of G over F is a homomorphism r from G to GL (n, F), for some n. The degree of r is the integer n. Thus if r is a function from G to GL (n, F), then r is a representation if and only if ( gh)r ( gr)(hr)

for all g, h 2 G:

Since a representation is a homomorphism, it follows that for every representation r: G ! GL (n, F), we have 1r I n , and g ÿ1 r ( gr)ÿ1

for all g 2 G,

where In denotes the n 3 n identity matrix. 30

Group representations

31

3.2 Examples (1) Let G be the dihedral group D8 ka, b: a4 b2 1, bÿ1 ab aÿ1 l. De®ne the matrices A and B by 0 1 1 0 A , B ÿ1 0 0 ÿ1 and check that A4 B2 I, Bÿ1 AB Aÿ1 : It follows (see Example 1.4) that the function r: G ! GL (2, F) which is given by r: a i b j ! A i B j

(0 < i < 3, 0 < j < 1)

is a representation of D8 over F. The degree of r is 2. The matrices gr for g in D8 are given in the following table: g gr

g gr

1 1 0 0 1

b

1 0 0 ÿ1

a 0 1 ÿ1 0

ab 0 ÿ1 ÿ1 0

a2

ÿ1 0 0 ÿ1

2 a b ÿ1 0 0 1

a3 0 ÿ1 1 0

3 a b 0 1 1 0

(2) Let G be any group. De®ne r: G ! GL (n, F) by gr I n

for all g 2 G,

where In is the n 3 n identity matrix, as usual. Then ( gh)r I n I n I n ( gr)(hr) for all g, h 2 G, so r is a representation of G. This shows that every group has representations of arbitrarily large degree. Equivalent representations We now look at a way of converting a given representation into another one.

32


Let r: G ! GL (n, F) be a representation, and let T be an invertible n 3 n matrix over F. Note that for all n 3 n matrices A and B, we have (T ÿ1 AT )(T ÿ1 BT ) T ÿ1 (AB)T: We can use this observation to produce a new representation ó from r; we simply de®ne gó T ÿ1 ( gr)T

for all g 2 G:

Then for all g, h 2 G, ( gh)ó T ÿ1 (( gh)r)T T ÿ1 (( gr)(hr))T T ÿ1 ( gr)T . T ÿ1 (hr)T ( gó )(hó ), and so ó is, indeed, a representation. 3.3 De®nition Let r: G ! GL (m, F) and ó : G ! GL (n, F) be representations of G over F. We say that r is equivalent to ó if n m and there exists an invertible n 3 n matrix T such that for all g 2 G, gó T ÿ1 ( gr)T: Note that for all representations r, ó and ô of G over F, we have (see Exercise 3.4): (1) r is equivalent to r; (2) if r is equivalent to ó then ó is equivalent to r; (3) if r is equivalent to ó and ó is equivalent to ô, then r is equivalent to ô. In other words, equivalence of representations is an equivalence relation. 3.4 Examples (1) Let G D8 ka, b: a4 b2 1, bÿ1 ab aÿ1 l, and consider the representation r of G which appears in Example 3.2(1). Thus ar A

Group representations and br B, where

A

1 1 , B 0 0

0 ÿ1

Assume that F C, and de®ne 1 T p 2

Then T

ÿ1

1 p 2

0 : ÿ1

1 : ÿi

1 i

33

1 ÿi : 1 i

In fact, T has been constructed so that T ÿ1 AT is diagonal; we have i 0 0 1 T ÿ1 AT , T ÿ1 BT , 0 ÿi 1 0 and so we obtain a representation ó of D8 for which 0 1 i 0 : , bó aó 1 0 0 ÿi The representations r and ó are equivalent. (2) Let G C2 ka: a2 1l and let ÿ5 A ÿ2

12 : 5

Check that A2 I. Hence r: 1 ! I, a ! A is a representation of G. If 2 ÿ3 , T 1 ÿ1 then T

ÿ1

AT

1 0

0 , ÿ1

and so we obtain a representation ó of G for which 1 0 1 0 , , aó 1ó 0 ÿ1 0 1 and ó is equivalent to r.

34


There are two easily recognized situations where the only representation which is equivalent to r is r itself; these are when the degree of r is 1, and when gr In for all g in G. However, there are usually lots of representations which are equivalent to r.

Kernels of representations We conclude the chapter with a discussion of the kernel of a representation r: G ! GL (n, F). In agreement with De®nition 1.8, this consists of the group elements g in G for which gr is the identity matrix. Thus Ker r f g 2 G: gr I n g: Note that Ker r is a normal subgroup of G. It can happen that the kernel of a representation is the whole of G, as is shown by the following de®nition. 3.5 De®nition The representation r: G ! GL (1, F) which is de®ned by gr (1) for all g 2 G, is called the trivial representation of G. 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. Of particular interest are those representations whose kernel is just the identity subgroup. 3.6 De®nition A representation r: G ! GL (n, F) is said to be faithful if Ker r {1}; that is, if the identity element of G is the only element g for which gr In . 3.7 Proposition A representation r of a ®nite group G is faithful if and only if Im r is isomorphic to G.


35

Proof We know that Ker r v G and by Theorem 1.10, the factor group G/ Ker r is isomorphic to Im r. Therefore, if Ker r {1} then G Im r. Conversely, if G Im r, then these two groups have the same (®nite) order, and so |Ker r| 1; that is, r is faithful. j 3.8 Examples (1) The representation r of D8 given by j i 1 0 0 1 i j (a b )r 0 ÿ1 ÿ1 0 as in Example 3.2(1) is faithful, since the identity is the only element g which satis®es gr I. The group generated by the matrices 0 1 1 0 and ÿ1 0 0 ÿ1 is therefore isomorphic to D8. (2) Since T ÿ1 AT In if and only if A In , it follows that all representations which are equivalent to a faithful representation are faithful. (3) The trivial representation of a group G if faithful if and only if G {1}. In Chapter 6 we shall show that every ®nite group has a faithful representation. The basic problem of representation theory is to discover and understand representations of ®nite groups.

Summary of Chapter 3 1. A representation of a group G is a homomorphism from G into GL(n, F), for some n. 2. Representations r and ó of G are equivalent if and only if there exists an invertible matrix T such that for all g 2 G, gó T ÿ1 ( gr)T : 3. A representation is faithful if it is injective.

36

Representations and characters of groups Exercises for Chapter 3

1. Let G be the cyclic group of order m, say G ka: am 1l. Suppose that A 2 GL (n, C), and de®ne r: G ! GL (n, C) by r: ar ! Ar

(0 < r < m ÿ 1):

Show that r is a representation of G over C if and only if Am I. 2. Let

A

1 0

1 0 , B 0 1

0

e2ði=3

,C

0 ÿ1

1 ÿ1

and let G ka: a3 1l C3 . Show that each of the functions r j : G ! GL (2, C) (1 < j < 3), de®ned by r1 : ar ! Ar , r2 : ar ! Br , r3 : a r ! C r

(0 < r < 2),

is a representation of G over C. Which of these representations are faithful? 3. Suppose that G D2 n ka, b: an b2 1, bÿ1 ab aÿ1 l, and F R or C. Show that there is a representation r: G ! GL (1, F) such that ar (1) and br (ÿ1). 4. Suppose that r, ó and ô are representations of G over F. Prove: (1) r is equivalent to r; (2) if r is equivalent to ó, then ó is equivalent to r; (3) if r is equivalent to ó, and ó is equivalent to ô, then r is equivalent to ô. 5. Let G D12 ka, b: a6 b2 1, bÿ1 ab aÿ1 l. De®ne the matrices A, B, C, D over C by 0 1 eið=3 0 A , B , 1 0 0 eÿið=3 p 1=2 3=2 1 0 p C , D : ÿ 3=2 1=2 0 ÿ1 Prove that each of the functions r k : G ! GL (2, C) (k 1, 2, 3, 4), given by


37

r1 : ar bs ! Ar Bs , r2 : ar bs ! A3 r (ÿB) s , r3 : ar bs ! (ÿA) r Bs , r 4 : ar bs ! C r D s

(0 < r < 5, 0 < s < 1),

is a representation of G. Which of these representations are faithful? Which are equivalent? 6. Give an example of a faithful representation of D8 of degree 3. 7. Suppose that r is a representation of G of degree 1. Prove that G= Ker r is abelian. 8. Let r be a representation of the group G. Suppose that g and h are elements of G such that ( gr)(hr) (hr)( gr). Does it follow that gh hg?

4 FG-modules

We now introduce the concept of an FG-module, and show that there is a close connection between FG-modules and representations of G over F. Much of the material in the remainder of the book will be presented in terms of FG-modules, as there are several advantages to this approach to representation theory.

FG-modules Let G be a group and let F be R or C. Suppose that r: G ! GL (n, F) is a representation of G. Write V F n , the vector space of all row vectors (ë1 , . . . , ë n ) with ë i 2 F. For all v 2 V and g 2 G, the matrix product v( gr), of the row vector v with the n 3 n matrix 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). We now list some basic properties of the multiplication v( gr). First, the fact that r is a homomorphism shows that v(( gh)r) v( gr)(hr) for all v 2 V and all g, h 2 G. Next, since 1r is the identity matrix, we have v(1r) v for all v 2 V. Finally, the properties of matrix multiplication give (ëv)( gr) ë(v( gr)), 38

FG-modules

39

(u v)( gr) u( gr) v( gr) for all u, v 2 V, ë 2 F and g 2 G. 4.1 Example Let G D8 ka, b: a4 b2 1, bÿ1 ab aÿ1 l, and let r: G ! GL (2, F) be the representation of G over F given in Example 3.2(1). Thus 1 0 0 1 : , br ar 0 ÿ1 ÿ1 0 If v (ë1 , ë2 ) 2 F 2 then, for example, v(ar) (ÿë2 , ë1 ), v(br) (ë1 , ÿë2 ), v(a3 r) (ë2 , ÿë1 ): Motivated by the above observations on the product v( gr), we now de®ne an FG-module. 4.2 De®nition Let V be a vector space over F and let G be a group. Then V is an FG-module if a multiplication v g (v 2 V, g 2 G) is de®ned, satisfying the following conditions for all u, v 2 V, ë 2 F and g, h 2 G: (1) (2) (3) (4) (5)

v g 2 V; v( gh) (v g)h; v1 v; (ëv) g ë(v g); (u v) g ug v g.

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 2 V). Note that conditions (1), (4) and (5) in the de®nition ensure that for all g 2 G, the function v ! vg is an endomorphism of V.

(v 2 V )

40


4.3 De®nition Let V be an FG-module, and let B be a basis of V. For each g 2 G, let [ g]B denote the matrix of the endomorphism v ! v g of V, relative to the basis B . The connection between FG-modules and representations of G over F is revealed in the following basic result. 4.4 Theorem (1) If r: G ! GL(n, F) is a representation of G over F, and V F n, then V becomes an FG-module if we de®ne the multiplication v g by v g v( gr)

(v 2 V , g 2 G):

Moreover, there is a basis B of V such that gr [ g]B

for all g 2 G:

(2) Assume that V is an FG-module and let B be a basis of V. Then the function g ! [ g]B

( g 2 G)

is a representation of G over F. Proof (1) We have already observed that for all u, v 2 F n, ë 2 F and g, h 2 G, we have v( gr) 2 F n , v(( gh)r) (v( gr))(hr), v(1r) v, (ëv)( gr) ë(v( gr)), (u v)( gr) u( gr) v( gr): n

Therefore, F becomes an FG-module if we de®ne v g v( gr)

for all v 2 F n , g 2 G:

Moreover, if we let B be the basis (1, 0, 0, : : : , 0), (0, 1, 0, : : : , 0), : : : , (0, 0, 0, : : : , 1) of F n, then gr [ g]B for all g 2 G.

FG-modules

41

(2) Let V be an FG-module with basis B . Since v( gh) (v g)h for all g, h 2 G and all v in the basis B of V, it follows that [ gh]B [ g]B [h]B : In particular, [1]B [ g]B [ g ÿ1 ]B for all g 2 G. Now v1 v for all v 2 V, so [1]B is the identity matrix. Therefore each matrix [ g]B is invertible (with inverse [ gÿ1 ]B ). We have proved that the function g ! [ g]B is a homomorphism from G to GL (n, F) (where n dim V ), and hence is a representation of G over F. j Our next example illustrates part (1) of Theorem 4.4. 4.5 Examples (1) Let G D8 ka, b: a4 b2 1, bÿ1 ab aÿ1 l and let r be the representation of G over F given in Example 3.2(1), so 0 1 1 0 ar , br : ÿ1 0 0 ÿ1 Write V F 2. By Theorem 4.4(1), V becomes an FG-module if we de®ne v g v( gr) (v 2 V , g 2 G): For instance,

0 (1, 0)a (1, 0) ÿ1

1 0

(0, 1):

If v1 , v2 is the basis (1, 0), (0, 1) of V, then we have v1 a v2 , v2 a ÿv1 ,

v1 b v1 , v2 b ÿv2 :

If B denotes the basis v1 , v2 , then the representation g ! [ g]B

( g 2 G)

is just the representation r (see Theorem 4.4(1) again). (2) Let G Q8 ka, b: a4 1, a2 b2 , bÿ1 ab aÿ1 l. In Example

42


1.2(4) we de®ned Q8 to be the subgroup of GL (2, C) generated by 0 1 i 0 , and B A ÿ1 0 0 ÿi so we already have a representation of G over C. To illustrate Theorem 4.4(1) we must this time take F C. We then obtain a CG-module with basis v1 , v2 such that v1 a iv1 , v2 a ÿiv2 ,

v1 b v2 , v2 b ÿv1 :

Notice that in the above examples, the vectors v1 a, v2 a, v1 b and v2 b determine v g for all v 2 V and g 2 G. For instance, in Example 4.5(1), (v1 2v2 )ab v1 ab 2v2 ab v2 b ÿ 2v1 b ÿv2 ÿ 2v1 : A similar remark holds for all FG-modules V: if v1 , . . . , v n is a basis of V and g1 , . . . , gr generate G, then the vectors v i g j (1 < i < n, 1 < j < r) determine v g for all v 2 V and g 2 G. Shortly, we shall show you various ways of constructing FG-modules directly, without using a representation. To do this, we turn a vector space V over F into an FG-module by specifying the action of group elements on a basis v1 , . . . , v n of V and then extending the action to be linear on the whole of V; that is, we ®rst de®ne v i g for each i and each g in G, and then de®ne (ë1 v1 : : : ë n v n ) g

(ë i 2 F)

to be ë1 (v1 g) : : : ë n (v n g): As you might expect, there are restrictions on how we may de®ne the vectors v i g. The next result will often be used to show that our chosen multiplication turns V into an FG-module. 4.6 Proposition Assume that v1 , . . . , v n is a basis of a vector space V over F. Suppose that we have a multiplication v g for all v in V and g in G which

FG-modules

43

satis®es the following conditions for all i with 1 < i < n, for all g, h 2 G, and for all ë1 , . . . , ë n 2 F: (1) (2) (3) (4)

v i g 2 V; v i ( gh) (v i g)h; vi 1 vi ; (ë1 v1 . . . ë n v n ) g ë1 (v1 g) . . . ë n (v n g).

Then V is an FG-module. Proof It is clear from (3) and (4) that v1 v for all v 2 V. Conditions (1) and (4) ensure that for all g in G, the function v ! v g (v 2 V) is an endomorphism of V. That is, vg 2 V, (ëv) g ë(v g), (u v) g ug v g, for all u, v 2 V, ë 2 F and g 2 G. Hence (4:7)

(ë1 u1 : : : ë n u n )h ë1 (u1 h) : : : ë n (un h)

for all ë1 , . . . , ë n 2 F, all u1 , . . . , un 2 V and all h 2 G. Now let v 2 V and g, h 2 G. Then v ë1 v1 . . . ë n v n for some ë1 , . . . , ë n 2 F, and v( gh) ë1 (v1 ( gh)) : : : ë n (v n ( gh))

by condition (4)

ë1 ((v1 g)h) : : : ë n ((v n g)h) by condition (2) (ë1 (v1 g) : : : ë n (v n g))h (v g)h

by (4:7)

by condition (4):

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 v

for all v 2 V , g 2 G:

44


(2) An FG-module V is faithful if the identity element of G is the only element g for which v g v for all v 2 V : 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 v ig : Then v i g 2 V and v i 1 v i . Also, for g, h in G, v i ( gh) v i( gh) v(ig) h (v i g)h: 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 : : : ë n v n ) g ë1 (v1 g) : : : ë n (v n g): Then V is an FG-module, by Proposition 4.6. 4.9 Example Let G S4 and let B denote the basis v1 , v2 , v3 , v4 of V. If g (1 2), then v1 g v2 , v2 g v1 , v3 g v3 , v4 g v4 : And if h (1 3 4), then v1 h v3 , v2 h v2 , v3 h v4 , v4 h v1 : We have

0

[ g]B

0 B1 B @0 0

1 0 0 0

0 0 1 0

0 1 0 0 B C 0C 0 , [h]B B @0 0A 1 1

0 1 0 0

1 0 0 0

1 0 0C C: 1A 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 v ig

for all i, and all g 2 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 C3 ka: a3 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 v1 , v2 1 v2 , v3 1 v3 , v1 a v2 , v2 a v3 , v3 a v1 , v1 a2 v3 , v2 a2 v1 , v3 a2 v2 : Of course, we de®ne v g, for v an arbitrary vector in V and g 1, a or a2 , by (ë1 v1 ë2 v2 ë3 v3 ) g ë1 (v1 g) ë2 (v2 g) ë3 (v3 g) for all ë1 , ë2 , ë3 2 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


module gives us many representations, all of the form g ! [ g]B

( g 2 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 ! [ g]B

( g 2 G):

(1) If B 9 is a basis of V, then the representation ö: g ! [ g]B 9

( g 2 G)

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 ! [ g]B 0

( g 2 G):

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 2 G, we have [ g]B T ÿ1 [ g]B 9 T: 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 T ÿ1 ( gó )T

for all g 2 G:

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 2 G, [ g]B T ÿ1 [ g]B 0 T , and so gó [ g]B 0 .

j

4.13 Example Again let G C3 ka: a3 1l. There is a representation r of G which is given by

1r

1 0

FG-modules 0 1 0 ÿ1 ÿ1 , a2 r , ar : ÿ1 ÿ1 1 1 0

47

(To see this, simply note that (ar)2 a2 r and (ar)3 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 a2 ÿv1 ÿ v2 ,

v1 1 v1 ,

v1 a v2 ,

v2 1 v2 ,

v2 a ÿv1 ÿ v2 ,

We then have 1 0 0 [1]B , [a]B 0 1 ÿ1

v2 a2 v1 :

1 ÿ1 2 , [a ]B ÿ1 1

ÿ1 : 0

Now let u1 v1 and u2 v1 v2 . Then u1 , u2 is another basis of V, which we call B 9. Since u1 1 u 1 ,

u1 a ÿu1 u2 ,

u2 1 u 2 ,

u2 a ÿu1 ,

u1 a2 ÿu2 ,

u2 a2 u1 ÿ u2 ,

we obtain the representation ö: g ! [ g]B 9 where 1 0 ÿ1 1 0 ÿ1 2 [1]B 9 , [a]B 9 , [a ]B 9 : 0 1 ÿ1 0 1 ÿ1 Note that if

T

1 0 1 1

then for all g in G, we have [ g]B T ÿ1 [ g]B 9 T , and so r and ö are equivalent, in agreement with Theorem 4.12(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


2. There is a correspondence between representations of G over F and FG-modules, as follows. (a) Suppose that r: G ! GL (n, F) is a representation of G. Then F n is an FG-module, if we de®ne v g v( gr) (v 2 F n , g 2 G): (b)

If V is an FG-module, with basis B , then r: g ! [ 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 , . . . , v n , and v i g v ig for all i with 1 < i < n, and all g in G. Exercises for Chapter 4 1. Suppose that G S3 , and that V 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 ÿ v2 , v1 ÿ 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 Sn and let V becomes an FG-module v, vg ÿv,

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 ka, b: a4 1, b2 a2 , bÿ1 ab aÿ1 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 v 2 ,

v2 a ÿv1 ,

v3 a ÿv4 , v4 a v3 , and

v1 b v 3 ,

v2 b v4 ,

v3 b ÿv1 , v4 b ÿv2 :

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 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 2 W for all w 2 W and all g 2 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 C3 ka: a3 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 v1 , v2 1 v2 , v3 1 v3 , v1 a v2 , v2 a v3 , v3 a v1 , v1 a2 v3 , v2 a2 v1 , v3 a2 v2 : Put w v1 v2 v3, and let W sp (w), the 1-dimensional subspace spanned by w. Since 49

50

Representations and characters of groups w1 wa wa2 w,

W is an FG-submodule of V. However, sp (v1 v2 ) is not an FGsubmodule, since (v1 v2 )a v2 v3 2 = sp (v1 v2 ): 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 ! GL (n, F) is irreducible if the corresponding FG-module F n given by v g v( gr) (v 2 F n , g 2 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 0 1 Xg 0 @ A (5:4) Yg Z g for some matrices Xg , Yg and Zg , where Xg is k 3 k (k 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 ! Xg and g ! Z g are representations of G: to see this, let g, h 2 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 C3 ka: a3 1l and let V be the 3-dimensional FGmodule with basis v1 , v2 , v3 such that

FG-submodules and reducibility

51

v1 a v2 , v2 a v3 , v3 a v1 , 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 sp (v1 v2 v3 ). Let B be the basis v1 v2 v3 , v1 , v2 of V. Then 0 0 1 1 1 0 0 1 0 0 B B C C B C [1]B B 0 1C @ 0 1 0 A, [a]B @ 0 A, 0 0 1 1 ÿ1 ÿ1 0 1 1 0 0 B C 2 C [a ]B B @ 1 ÿ1 ÿ1 A: 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 ÿ1 ÿ1 , a2 ! ,a! : 1! ÿ1 ÿ1 0 1 1 0 (2) Let G D8 and let V F 2 be the 2-dimensional FG-module described in Example 4.5(1). Thus G ka, bl, and for all (ë, ì) 2 V we have (ë, ì)a (ÿì, ë),

(ë, ì)b (ë, ÿì):

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 sp ((á, â)) for some á, â 2 F. As U is an FGmodule, (á, â)b is a scalar multiple of (á, â), and hence either á 0 or â 0. Since (á, â)a is also a scalar multiple of (á, â), this forces á â 0, so U {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.

52


1. Let G C2 ka: a2 1l, and let V F 2. For (á, â) 2 V, de®ne (á, â)1 (á, â) and (á, â)a (â, á). Verify that V is an FG-module and ®nd all the FG-submodules of V. 2. Let r and ó be equivalent representations of the group G over F. Prove that if r is reducible then ó is reducible. 3. Which of the four representations of D12 de®ned in Exercise 3.5 are irreducible? 4. De®ne the permutations a, b, c 2 S6 by a (1 2 3), b (4 5 6), c (2 3)(4 5), and let G ka, b, cl. (a) Check that a3 b3 c2 1, ab ba, cÿ1 ac aÿ1 and cÿ1 bc bÿ1 : Deduce that G has order 18. (b) Suppose that å and ç are complex cube roots of unity. Prove that there is a representation r of G over C such that å 0 ç 0 0 1 ar : , br , cr 0 å ÿ1 0 çÿ1 1 0 (c) For which values of å, ç is r faithful? (d) For which values of å, ç is r irreducible? 5. Let G C13 . Find a CG-module which is neither reducible nor irreducible.

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. In a sense, group algebras are the source of all you need to know about representation theory. In particular, 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. Group algebras are therefore of great interest. After de®ning the group algebra of G, we shall use it to construct an important faithful representation, known as the regular representation of G, which will be explored in greater detail later on.

The group algebra of G Let G be a ®nite group whose elements are g1 , . . . , g n , and let F be R or C. We de®ne a vector space over F with g1 , . . . , g n as a basis, and we call this vector space FG. Take as the elements of FG all expressions of the form ë1 g 1 : : : ë n g n

(all ë i 2 F):

The rules for addition and scalar multiplication in FG are the natural ones: namely, if u

n X

ë i g i and v

i1

n X i1

are elements of FG, and ë 2 F, then 53

ìi g i

54

Representations and characters of groups uv

n n X X (ë i ì i ) g i and ëu (ëë i ) g i : i1

i1

With these rules, FG is a vector space over F of dimension n, with basis g1 , . . . , g n . The basis g1 , . . . , g n is called the natural basis of FG. 6.1 Example Let G C3 ka: a3 el. (To avoid confusion with the element 1 of F, we write e for the identity element of G, in this example.) The vector space CG contains u e ÿ a 2a2 and v 12 e 5a: We have u v 32 e 4a 2a2 , 13 u 13 e ÿ 13 a 23 a2 : Sometimes we write elements of FG in the form X ë g g (ë g 2 F): g2G

Now, 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: ! ! X X X ëg g ìh h ë g ì h ( gh) g2G

h2G

g,h2G

XX (ë h ì hÿ1 g ) g g2G h2G

where all ë g , ì h 2 F. 6.2 Example If G C3 and u, v are the elements of CG which appear in Example 6.1, then uv (e ÿ a 2a2 )(12 e 5a) 5a ÿ 12 a ÿ 5a2 a2 10a3

1 2e

21 2e

92 a ÿ 4a2 :

Group algebras

55

6.3 De®nition The vector space FG, with multiplication de®ned by ! ! X X X ëg g ìh h ë g ì h ( gh) g2G

h2G

g,h2G

(ë g , ì h 2 F), is called the group algebra of G over F. The group algebra FG contains an identity for multiplication, namely the element 1e (where 1 is the identity of F and e is the identity of G). We write this element simply as 1. 6.4 Proposition Multiplication in FG satis®es the following properties, for all r, s, t 2 FG and ë 2 F: (1) (2) (3) (4) (5) (6) (7)

rs 2 FG; r(st) (rs)t; r1 1r r; (ër)s ë(rs) r(ës); (r s)t rt st; r(s t) rs rt; r0 0r 0.

Proof (1) It follows immediately from the de®nition of rs that rs 2 FG. (2) Let X X X r ë g g, s ì g g, t í g g, g2G

g2G

(ë g , ì g , í g 2 F). Then (rs)t

X

g2G

ë g ì h í k ( gh)k

g,h,k2G

X

ë g ì h í k g(hk)

g,h,k2G

r(st): We leave the proofs of the other equations as easy exercises.

j

56


In fact, any vector space equipped with a multiplication satisfying properties (1)±(7) of Proposition 6.4 is called an algebra. We shall be concerned only with group algebras, but it is worth pointing out that the axioms for an algebra mean that it is both a vector space and a ring. The regular FG-module We now use the group algebra to de®ne an important FG-module. Let V FG, so that V is a vector space of dimension n over F, where n |G|. For all u, v 2 V, ë 2 F and g, h 2 G, we have vg 2 V, v( gh) (v g)h, v1 v, (ëv) g ë(v g), (u v) g ug v g, by parts (1), (2), (3), (4) and (5) of Proposition 6.4, respectively. Therefore V is an FG-module. 6.5 De®nition Let G be a ®nite group and F be R or C. The vector space FG, with the natural multiplication v g (v 2 FG, g 2 G), is called the regular FG-module. The representation g ! [ g]B obtained by taking B to be the natural basis of FG is called the regular representation of G over F. Note that the regular FG-module has dimension equal to |G|. 6.6 Proposition The regular FG-module is faithful. Proof Suppose that g 2 G and v g v for all v 2 FG. Then 1 g 1, so g 1, and the result follows. j 6.7 Example Let G C3 ka: a3 el. The elements of FG have the form

Group algebras ë1 e ë2 a ë3 a2

57

(ë i 2 F):

We have (ë1 e ë2 a ë3 a2 )e ë1 e ë2 a ë3 a2 , (ë1 e ë2 a ë3 a2 )a ë3 e ë1 a ë2 a2 , (ë1 e ë2 a ë3 a2 )a2 ë2 e ë3 a ë1 a2 : By taking matrices relative to the basis e, a, a2 of FG, we obtain the regular representation of G: 0 1 0 1 0 1 1 0 0 0 1 0 0 0 1 e ! @ 0 1 0 A, a ! @ 0 0 1 A, a2 ! @ 1 0 0 A: 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, together with a multiplication v g for v 2 V and g 2 G (and the multiplication satis®es various axioms). Now, 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. This is done in the following natural way. 6.8 De®nition Suppose that V is an FG-module, and that v 2 V and r 2 FG; say P r g2G ì g g (ì g 2 F). De®ne vr by X vr ì g (v g): g2G

6.9 Examples (1) Let V be the permutation module for S4 , as described in Example 4.9. If r ë(1 2) ì(1 3 4)

(ë, ì 2 F)

then v1 r ëv1 (1 2) ìv1 (1 3 4) ëv2 ìv3 , v2 r ëv1 ìv2 , (2v1 v2 )r ëv1 (2ë ì)v2 2ìv3 :

58


(2) If V is the regular FG-module, then for all v 2 V and r 2 FG, the element vr is simply the product of v and r as elements of the group algebra, given by De®nition 6.3. Compare the next result with Proposition 6.4. 6.10 Proposition Suppose that V is an FG-module. Then the following properties hold for all u, v 2 V, all ë 2 F and all r, s 2 FG: (1) (2) (3) (4) (5) (6) (7)

vr 2 V; v(rs) (vr)s; v1 v; (ëv)r ë(vr) v(ër); (u v)r ur vr; v(r s) vr vs; v0 0r 0.

Proof All parts except (2) are straightforward, and we leave them to you. We shall give a proof of part (2), assuming the other parts. Let v 2 V, and let r, s 2 FG with X X r ë g g, s ì h h: g2G

Then X

v(rs) v

h2G

! ë g ì h ( gh)

g,h

X

ë g ì h (v( gh))

by (4) and (6)

g,h

X

ë g ì h ((v g)h)

g,h

X g

(vr)s:

! ! X ë g (v g) ìh h

by (4), (5), (6)

h j

Group algebras

59

Summary of Chapter 6 1. The group algebra FG of G over F consists of all linear combinations of elements of G, and has a natural multiplication de®ned on it. 2. The vector space FG, with the natural multiplication v g (v 2 FG, g 2 G) is the regular FG-module. 3. The regular FG-module is faithful.

Exercises for Chapter 6 1. Suppose that G D8 ka, b: a4 b2 1, bÿ1 ab aÿ1 l. (a) Let x and y be the following elements of CG: x a 2a2 , y b ab ÿ a2 : Calculate xy, yx and x 2 . (b) Let z b a2 b. Show that zg gz for all g in G. Deduce that zr rz for all r in CG. 2. Work out matrices for the regular representation of C2 3 C2 over F. 3. Let G C2 . For r and s in CG, does rs 0 imply that r 0 or s 0? 4. Assume that G is a ®nite group, say G { g1 , . . . , g n }, and write c Pn for the element i1 g i of CG. (a) Prove that ch hc c for all h in G. (b) Deduce that c2 |G|c. (c) Let W: CG ! CG be the linear transformation sending v to vc for all v in CG. What is the matrix [W]B , where B is the basis g1 , . . . , gn of CG? 5. If V is an FG-module, prove from the de®nition that 0r 0 for all r 2 FG, and v0 0 for all v 2 V , where the symbol 0 is used for the zero elements of V and FG. Show that for every ®nite group G, with |G| . 1, there exists an FG-module V and elements v 2 V, r 2 FG such that vr 0, but neither v nor r is 0.

60


6. Suppose that G D6 ka, b: a3 b2 1, bÿ1 ab aÿ1 l, and let ù e2ði=3 . Prove that the 2-dimensional subspace W of CG, de®ned by W sp (1 ù2 a ùa2 , b ù2 ab ùa2 b), is an irreducible CG-submodule of the regular CG-module.

7 FG-homomorphisms

For groups and vector spaces, the `structure-preserving' functions are, respectively, group homomorphisms and linear transformations. The analogous functions for FG-modules are called FG-homomorphisms, and we introduce these in this chapter. FG-homomorphisms 7.1 De®nition Let V and W be FG-modules. A function W: V ! W is said to be an FG-homomorphism if W is a linear transformation and (v g)W (vW) g

for all v 2 V , g 2 G:

In other words, if W sends v to w then it sends v g to wg. Note that if G is a ®nite group and W: V ! W is an FG-homomorphP ism, then for all v 2 V and r g2G ë g g 2 FG, we have (vr)W (vW)r since (vr)W

X

ë g (v g)W

g2G

X

ë g (vW) g (vW)r:

g2G

The next result shows that FG-homomorphisms give rise to FGsubmodules in a natural way. 7.2 Proposition Let V and W be FG-modules and let W: V ! W be an FG-homomorphism. Then Ker W is an FG-submodule of V, and Im W is an FGsubmodule of W. 61

62


Proof First note that Ker W is a subspace of V and Im W is a subspace of W, since W is a linear transformation. Let v 2 Ker W and g 2 G. Then (v g)W (vW) g 0 g 0, so vg 2 Ker W. Therefore Ker W is an FG-submodule of V. Now let w 2 Im W, so that w vW for some v 2 V. For all g 2 G, wg (vW) g (v g)W 2 Im W, and so Im W is an FG-submodule of W.

j

7.3 Examples (1) If W: V ! W is de®ned by vW 0 for all v 2 V, then W is an FGhomomorphism, and Ker W V, Im W {0}. (2) Let ë 2 F, and de®ne W: V ! V by vW ëv for all v 2 V. Then W is an FG-homomorphism. Provided ë 6 0, we have Ker W {0}, Im W V. (3) Suppose that G is a subgroup of S n . Let V sp (v1 , . . . , v n ) be the permutation module for G over F (see De®nition 4.10), and let W sp (w) be the trivial FG-module (see De®nition 4.8). We construct an FG-homomorphism W from V to W. De®ne ! n n X X W: ëivi ! ë i w (ë i 2 F): i1

i1

Thus v i W w for all i. Then W is a linear transformation, and for all P v ë i v i 2 V and all g 2 G, we have ! ! X X (v g)W ë i v ig W ë i w, and (vW) g

X

! ë i wg

X

! ë i w:

Therefore W is an FG-homomorphism. Here, ( n ) n X X Ker W ëi vi : ëi 0 , i1

Im W W :

i1

FG-homomorphisms

63

By Proposition 7.2, Ker W is an FG-submodule of the permutation module V.

Isomorphic FG-modules 7.4 De®nition Let V and W be FG-modules. We call a function W: V ! W an FGisomorphism if W is an FG-homomorphism and W is invertible. If there is such an FG-isomorphism, then we say that V and W are isomorphic FG-modules and write V W. In the next result, we check that if V W then W V. 7.5 Proposition If W: V ! W is an FG-isomorphism, then the inverse Wÿ1 : W ! V is also an FG-isomorphism. Proof Certainly Wÿ1 is an invertible linear transformation, so we need only show that Wÿ1 is an FG-homomorphism. For w 2 W and g 2 G, ((wWÿ1 ) g)W ((wWÿ1 )W) g

as W is an FG-homomorphism

wg ((wg)Wÿ1 )W: Hence (wWÿ1 ) g (wg)Wÿ1 , as required.

j

Suppose that W: V ! W is an FG-isomorphism. Then we may use W and Wÿ1 to translate back and forth between the isomorphic FGmodules V and W, and prove that V and W share the same structural properties. We list some examples below: (1) dim V dim W (since v1 , . . . , v n is a basis of V if and only if v1 W, . . . , v n W is a basis of W); (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); (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).

64


Just as we often regard isomorphic groups as being identical, we frequently disdain to distinguish between isomorphic FG-modules. For the moment, though, we continue simply to emphasize the similarity between isomorphic FG-modules. In the next result, we show that isomorphic FG-modules correspond to equivalent representations. 7.6 Theorem Suppose that V is an FG-module with basis B , and W is an FGmodule with basis B 9. Then V and W are isomorphic if and only if the representations r: g ! [ g]B and ó : g ! [ g]B 9 are equivalent.

Proof We ®rst establish the following fact: (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 [ g]B 2

for all g 2 G:

To see this, suppose ®rst that W is an FG-isomorphism from V to W, and let v1 , . . . , v n be a basis B 1 of V; then v1 W, . . . , v n W is a basis B 2 of W. Let g 2 G. Since (v i g)W (v i W) g for each i, it follows that [ g]B 1 [ g]B 2 . Conversely, suppose that v1 , . . . , v n is a basis B 1 of V and w1, . . . , w n is a basis B 2 of W such that [ g]B 1 [ g]B 2 for all g 2 G. Let W be the invertible linear transformation from V to W for which v i W w i for all i. Let g 2 G. Since [ g]B 1 [ g]B 2 , we deduce that (v i g)W (v i W) g for all i, and hence W is an FG-isomorphism. This completes the proof of (7.7). Now assume that V and W are isomorphic FG-modules. By (7.7), there are a basis B 1 of V and a basis B 2 of W such that [ g]B 1 [ g]B 2 for all g 2 G. De®ne a representation ö of G by ö: g ! [ g]B 1 . Then by Theorem 4.12(1), ö is equivalent to both r and ó. Hence r and ó are equivalent. Conversely, suppose that r and ó are equivalent. Then by Theorem 4.12(2), there is a basis B 0 of V such that gó [ g]B 0 for all g 2 G;

FG-homomorphisms

65

that is, [ g]B 9 [ g]B 0 for all g 2 G. Therefore V and W are isomorphic FG-modules, by (7.7). j 7.8 Example Let G ka: a3 1l, a cyclic group of order 3, and let W denote the regular FG-module. Then 1, a, a2 is a basis of W; call it B 9. We have 0 0 1 1 1 0 0 0 1 0 B B C C B C C [1]B 9 B @ 0 1 0 A, [a]B 9 @ 0 0 1 A, 0 0 1 1 0 0 0

0

B [a2 ]B 9 B @1 0

0

1

1

0

C 0C A:

1

0

Compare the FG-module V de®ned in Example 4.11, with basis v1 , v2 , v3 such that v1 a v2 , v2 a v3 , v3 a v1 : Writing B for the basis v1 , v2 , v3 of V, we have [ g]B [ g]B 9

for all g 2 G:

According to (7.7), the FG-modules V and W are therefore isomorphic. Indeed, the function W: ë1 v1 ë2 v2 ë3 v3 ! ë1 1 ë2 a ë3 a2

(ë i 2 F)

is an FG-isomorphism from V to W. 7.9 Example Let G D8 ka, b: a4 b2 1, bÿ1 ab aÿ1 l. In Example 3.4(1) we encountered two equivalent representations r and ó of G, where 0 1 1 0 ar , br ÿ1 0 0 ÿ1 and

aó

i 0

0 0 , bó ÿi 1

1 : 0

66


Let V be the CG-module with basis v1 , v2 for which v1 a v2 , v1 b v1 , v2 a ÿv1 , v2 b ÿv2 (see Example 4.5(1)), and, in a similar way, let W be the CG-module with basis w1, w2 for which w1 a iw1 , w2 a ÿiw2 ,

w1 b w2 , w2 b w1

Thus, if we write B for the basis v1 , v2 of V and B 9 for the basis w1, w2 of W, then for all g 2 G we have r: g ! [ g]B and ó : g ! [ g]B 9 : According to Theorem 7.6, the CG-modules V and W are isomorphic, since r and ó are equivalent. To verify this directly, let W: V ! W be the invertible linear transformation such that W: v1 ! w1 w2 , v2 ! iw1 ÿ iw2 : Then (v j a)W (v j W)a and (v j b)W (v j W)b for j 1, 2, and hence W is a CG-isomorphism from V to W. (Compare Example 3.4(1).)

Direct sums We conclude the chapter with a discussion of direct sums of FGmodules, and we show that these give rise to FG-homomorphisms. Let V be an FG-module, and suppose that V U W, where U and W are FG-submodules of V. Let u1 , . . . , u m be a basis B 1 of U, and w1, . . . , w n be a basis B 2 of W. Then by (2.9), u1 , . . . , u m , w1, . . . , w n is a basis B of V, and for g 2 G, 0 1 0 [ g]B 1 A: [ g]B @ 0 [ g]B 2 More generally, if V U1 . . . Ur, a direct sum of FG-submodules Ui, and B i is a basis of Ui, then we can amalgamate B 1 , . . . , B r to

FG-homomorphisms obtain a basis B of V, and for g 2 G, 0 [ g]B 1 B .. [ g]B @ (7:10) .

0

0

67 1 C A:

[ g]B r

The next result shows that direct sums give rise naturally to FGhomomorphisms. 7.11 Proposition Let V be an FG-module, and suppose that V U1 : : : U r where each Ui is an FG-submodule of V. For v 2 V, we have v u1 . . . ur for unique vectors ui 2 Ui, and we de®ne ð i : V ! V (1 < i < r) by setting vð i ui : Then each ð i is an FG-homomorphism, and is also a projection of V. Proof Clearly ð i is a linear transformation; and ð i is an FG-homomorphism, since for v 2 V with v u1 . . . u r (u j 2 U j for all j), and g 2 G, we have (v g)ð i (u1 g : : : ur g)ð i ui g (vð i ) g: Also, vð 2i ui ð i ui vð i , so ð2i ð i . Thus ð i is a projection (see De®nition 2.30).

j

We now present a technical result concerning sums of irreducible FG-modules which will be used at a later stage. 7.12 Proposition Let V be an FG-module, and suppose that V U1 : : : Ur , where each Ui is an irreducible FG-submodule of V. Then V is a direct sum of some of the FG-submodules Ui.

68


Proof The idea is to choose as many as we can of the FG-submodules U1, . . . , Ur so that the sum of our chosen FG-submodules is direct. To this end, choose a subset Y {W1, . . . , Ws } of {U1, . . . , Ur } which has the properties that W 1 : : : W s is direct (i:e: equal to W 1 : : : W s ), but W 1 : : : W s U i is not direct, if Ui 2 = Y: Let W W1 : : : Ws: We claim that Ui # W for all i. If Ui 2 Y this is clear, so assume that Ui 2 = Y. Then W Ui is not a direct sum, so W \ Ui 6 {0}. But W \ Ui is an FG-submodule of Ui, and Ui is irreducible; therefore W \ Ui Ui, and so Ui # W, as claimed. Since U i W for all i with 1 < i < r, we have V W W 1 . . . W s , as required. j Finally, we remark that if V1 , . . . , V r are FG-modules, then we can make the external direct sum V1 . . . Vr (see Chapter 2) into an FGmodule by de®ning (v1 , : : : , v r ) g (v1 g, : : : , v r g) for all v i 2 Vi (1 < i < r) and all g 2 G. Summary of Chapter 7 1. If V and W are FG-modules and W: V ! W is a linear transformation which satis®es (v g)W (vW) g for all v 2 V, g 2 G, then W is an FG-homomorphism. 2. Kernels and images of FG-homomorphisms are FG-modules. 3. Isomorphic FG-modules correspond to equivalent representations. Exercises for Chapter 7 1. Let U, V and W be FG-modules, and let W: U ! V and ö: V ! W be FG-homomorphisms. Prove that Wö: U ! W is an FG-homomorphism.

FG-homomorphisms

69

2. Let G be the subgroup of S5 which is generated by (1 2 3 4 5). Prove that the permutation module for G over F is isomorphic to the regular FG-module. 3. Assume that V is an FG-module. Prove that the subset V0 fv 2 V : v g v for all g 2 Gg is an FG-submodule of V. Show that the function X W: v ! v g (v 2 V ) g2G

is an FG-homomorphism from V to V0 . Is it necessarily surjective? 4. Suppose that V and W are isomorphic FG-modules. De®ne the FGsubmodules V0 and W0 of V and W as in Exercise 3. Prove that V0 and W0 are isomorphic FG-modules. 5. Let G be the subgroup of S4 which is generated by (1 2) and (3 4). Is the permutation module for G over F isomorphic to the regular FG-module? 6. Let G C2 kx: x 2 1l. (a) Show that the function W: á1 âx ! (á ÿ â)(1 ÿ x)

(á, â 2 F)

is an FG-homomorphism from the regular FG-module to itself. (b) Prove that W2 2W. (c) Find a basis B of FG such that 2 0 [W]B : 0 0

8 Maschke's Theorem

We now come to our ®rst major result in representation theory, namely Maschke's Theorem. A consequence of this theorem is that every FGmodule is a direct sum of irreducible FG-submodules, where as usual F R or C. (The assumption on F is important ± see Example 8.2(2) below.) This essentially reduces representation theory to the study of irreducible FG-modules. Maschke's Theorem 8.1 Maschke's Theorem Let G be a ®nite group, let F be R or C, and let V be an FG-module. If U is an FG-submodule of V, then there is an FG-submodule W of V such that V U W: Before proving Maschke's Theorem, we illustrate it with some examples. 8.2 Examples (1) Let G S3 and let V sp (v1 , v2 , v3 ) be the permutation module for G over F (see De®nition 4.10). Put u v1 v2 v3 and U sp (u): Then U is an FG-submodule of V, since ug u for all g 2 G. There are many subspaces W of V such that V U W, for instance sp (v2 , v3 ) and sp (v1 , v2 ÿ 2v3 ). But there is, in fact, only one FGsubmodule W of V with V U W. We shall ®nd this W in an 70

Maschke's Theorem

71

example after proving Maschke's Theorem (but you may like to look for it yourself now). (2) The conclusion of Maschke's Theorem can fail if F is not R or C. For example, let p be a prime number, let G C p ka: a p 1l, and take F to be the ®eld of integers modulo p. Check that the function 1 0 j a ! ( j 0, 1, : : : , p ÿ 1) j 1 is a representation from G to GL (2, F). The corresponding FG-module is V sp (v1 , v2 ), where, for 0 < j < p ÿ 1, v1 a j v1 , v2 a j jv1 v2 : Clearly, U sp (v1 ) is an FG-submodule of V. But there is no FGsubmodule W such that V U W, since U is the only 1-dimensional FG-submodule of V, as can easily be seen. Proof of Maschke's Theorem 8.1 We are given U, an FG-submodule of the FG-module V. Choose any subspace W0 of V such that V U W0: (There are many choices for W0 ± simply take a basis v1 , . . . , v m of U, extend it to a basis v1 , . . . , v n of V, and let W0 sp (v m1 , . . . , v n ).) For v 2 V, we have v u w for unique vectors u 2 U and w 2 W 0, and we de®ne ö: V ! V by setting vö u. By Proposition 2.29, ö is a projection of V with kernel W0 and image U. We aim to modify the projection ö to create an FG-homomorphism from V to V with image U. To this end, de®ne W: V ! V by (8:3)

vW

1 X v gö g ÿ1 jGj g2G

(v 2 V ):

It is clear that W is an endomorphism of V and Im W # U. We show ®rst that W is an FG-homomorphism. For v 2 V and x 2 G, (vx)W

1 X (vx) gö g ÿ1 : jGj g2G

72


As g runs over the elements of G, so does h xg. Hence 1 X (vx)W vhöhÿ1 x jGj h2G ! 1 X ÿ1 x vhöh jGj h2G (vW)x: Thus W is an FG-homomorphism. Next, we prove that W2 W. First note that for u 2 U, g 2 G, we have ug 2 U, and so (ug)ö ug. Using this, 1 X 1 X 1 X uW (8:4) ugö g ÿ1 (ug) g ÿ1 u u: jGj g2G jGj g2G jGj g2G Now let v 2 V. Then vW 2 U, so by (8.4) we have (vW)W vW. Consequently W2 W, as claimed. We have now established that W: V ! V is a projection and an FGhomomorphism. Moreover, (8.4) shows that Im W U. Let W Ker W. Then W is an FG-submodule of V by Proposition 7.2, and V U W by Proposition 2.32. This completes the proof of Maschke's Theorem. j 8.5 Example Let G S3 and let V sp (v1 , v2 , v3 ) be the permutation module, with submodule U sp (v1 v2 v3 ), as in Example 8.2(1). We use the proof of Maschke's Theorem to ®nd an FG-submodule W of V such that V U W. First, let W0 sp (v1 , v2 ). Then V U W0 (but of course W0 is not an FG-submodule). The projection ö onto U is given by ö: v1 ! 0, v2 ! 0, v3 ! v1 v2 v3 : Check now that the FG-homomorphism W given by (8.3) is W: v i ! 13(v1 v2 v3 ) (i 1, 2, 3): The required FG-submodule W is then Ker W, so W sp (v1 ÿ v2 , v2 ÿ v3 ): P P (In fact, W ë i v i : ë i 0 , the FG-submodule constructed in Example 7.3(3).)

Maschke's Theorem

73

Note that if B is the basis v1 v2 v3 , v1 , v2 of V, then for all g 2 G, the matrix [ g]B has the form 0 1 j 0 0 [ g]B @ j j j A: j

j

j

The zeros re¯ect the fact that U is an FG-submodule of V (see (5.4)). If instead we use v1 v2 v3 , v1 ÿ v2 , v2 ÿ v3 as a basis B 9, then we get 0 1 j 0 0 [ g]B 9 @ 0 j j A, 0 j j because sp (v1 ÿ v2 , v2 ÿ v3 ) is also an FG-submodule of V. This example illustrates the matrix version of Maschke's Theorem: for an arbitrary ®nite group G, if we can choose a basis B of an FGmodule V such that [ g]B has the form 0 1 0 @ A for all g 2 G (see (5.4)), then we can ®nd a basis B 9 such that [ g]B 9 has the form 0 1 0 @ A 0 for all g 2 G. To put this another way, suppose that r is a reducible representation of a ®nite group G over F of degree n. Then we know that r is equivalent to a representation of the form 0 1 Xg 0 A ( g 2 G), g !@ Yg Zg for some matrices X g , Yg, Z g , where X g is k 3 k with 0 , k , n.

74


Maschke's Theorem asserts further that r is equivalent to a representation of the form 0 1 Ag 0 A, g !@ 0 Bg where A g is also a k 3 k matrix. 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. (By an irreducible FG-submodule, we simply mean an FG-submodule which is an irreducible FG-module.) 8.6 De®nition An FG-module V is said to be completely reducible if V U1 . . . U r , where each Ui is an irreducible FG-submodule of V. 8.7 Theorem If G is a ®nite group and F R or C, then every non-zero FG-module is completely reducible. Proof Let V be a non-zero FG-module. The proof goes by induction on dim V. The result is true if dim V 1, since V is irreducible in this case. If V is irreducible then the result holds, so suppose that V is reducible. Then V has an FG-submodule U not equal to {0} or V. By Maschke's Theorem, there is an FG-submodule W such that V U W. Since dim U , dim V and dim W , dim V, we have, by induction, U U1 : : : Ur , W W 1 : : : W s , where each Ui and W j is an irreducible FG-module. Then by (2.10), V U1 : : : Ur W 1 : : : W s , a direct sum of irreducible FG-modules.

j

Another useful consequence of Maschke's Theorem is the next proposition.

Maschke's Theorem

75

8.8 Proposition Let V be an FG-module, where F R or C and G is a ®nite group. Suppose that U is an FG-submodule of V. Then there exists a surjective FG-homomorphism from V onto U. Proof By Maschke's Theorem, there is an FG-submodule W of V such that V U W. Then the function ð: V ! U which is de®ned by ð: u w ! u

(u 2 U , w 2 W )

is an FG-homomorphism onto U, by Proposition 7.11.

j

Theorem 8.7 tells us that every non-zero FG-module is a direct sum of irreducible FG-modules. Thus, in order to understand FG-modules, we may concentrate upon the irreducible FG-modules. We begin our study of these in the next chapter. Summary of Chapter 8 Assume that G is a ®nite group and F R or C. 1. Maschke's Theorem says that for every FG-submodule U of an FGmodule V, there is an FG-submodule W with V U W: 2. Every non-zero FG-module V is a direct sum of irreducible FGmodules: V U1 : : : Ur :

Exercises for Chapter 8 1. Let G kx: x 3 1l C3 , and let V be the 2-dimensional CGmodule with basis v1 , v2 , where v1 x v2 , v2 x ÿv1 ÿ v2 : (This is a CG-module, by Exercise 3.2.) Express V as a direct sum of irreducible CG-submodules. 2. If G C2 3 C2 , express the group algebra RG as a direct sum of 1-dimensional RG-submodules.

76


3. Find a group G, a CG-module V and a CG-homomorphism W: V ! V such that V 6 Ker W Im W. 4. Let G be a ®nite group and let r: G ! GL (2, C) be a representation of G. Suppose that there are elements g, h in G such that the matrices gr and hr do not commute. Prove that r is irreducible. (You may care to revisit Example 5.5(2) and Exercises 5.1, 5.3, 5.4, 6.6 in the light of this result.) 5. Suppose that G is the in®nite group 1 0 : n2Z n 1 and let V be the CG-module C2 , with the natural multiplication by elements of G (so that for v 2 V, g 2 G, the vector vg is just the product of the row vector v with the matrix g). Show that V is not completely reducible. (This shows that Maschke's Theorem fails for in®nite groups ± compare Example 8.2(2).) 6. An alternative proof of Maschke's Theorem for CG-modules. Let V be a CG-module with basis v1 , . . . , v n and suppose that U is a CG-submodule of V. De®ne a complex inner product ( , ) on V as follows (see (14.2) for the de®nition of a complex inner product): for ë i , ì j 2 C, ! n n n X X X ëi vi , ì jv j ëi ìi: i1

j1

i1

De®ne another complex inner product [ , ] on V by X [u, v] (ux, vx) (u, v 2 V ): x2G

(1) Verify that [ , ] is a complex inner product, which satis®es [ug, vg] [u, v] for all u, v 2 V and g 2 G: (2) Suppose that U is a CG-submodule of V, and de®ne U ? fv 2 V : [u, v] 0 for all u 2 U g: Show that U ? is a CG-submodule of V.

Maschke's Theorem

77

(3) Deduce Maschke's Theorem. (Hint: it is a standard property of complex inner products that V U U ? for all subspaces U of V.) 7. Prove that for every ®nite simple group G, there exists a faithful irreducible CG-module.

9 Schur's Lemma

Schur's Lemma is a basic result concerning irreducible modules. Though simple in both statement and proof, Schur's Lemma is fundamental to representation theory, and we give an immediate application by determining all the irreducible representations of ®nite abelian groups. Schur's Lemma concerns CG-modules rather than RG-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). Throughout, G denotes a ®nite group.

Schur's Lemma 9.1 Schur's Lemma Let V and W be irreducible CG-modules. (1) If W: V ! W is a CG-homomorphism, then either W is a CGisomorphism, or vW 0 for all v 2 V. (2) If W: V ! V is a CG-isomorphism, then W is a scalar multiple of the identity endomorphism 1 V . Proof (1) Suppose that vW 6 0 for some v 2 V. Then Im W 6 {0}. As Im W is a CG-submodule of W by Proposition 7.2, and W is irreducible, we have Im W W. Also by Proposition 7.2, Ker W is a CG-submodule of V; as Ker W 6 V and V is irreducible, Ker W {0}. Thus W is invertible, and hence is a CG-isomorphism. (2) By (2.26), the endomorphism W has an eigenvalue ë 2 C, and so Ker (W ÿ ë1 V ) 6 {0}. Thus Ker (W ÿ ë1 V ) is a non-zero CG-submodule 78

Schur's Lemma

79

of V. Since V is irreducible, Ker (W ÿ ë1 V ) V. Therefore v(W ÿ ë1 V ) 0

for all v 2 V :

That is, W ë1 V , as required.

j

Part (2) of Schur's Lemma has the following converse. 9.2 Proposition Let V be a non-zero CG-module, and suppose that every CG-homomorphism from V to V is a scalar multiple of 1 V . Then V is irreducible. Proof Suppose that V is reducible, so that V has a CG-submodule U not equal to {0} or V. By Maschke's Theorem, there is a CGsubmodule W of V such that V U W: Then the projection ð: V ! V de®ned by (u w)ð u for all u 2 U, w 2 W is a CG-homomorphism (see Proposition 7.11), and is not a scalar multiple of 1 V , which is a contradiction. Hence V is irreducible. j We next interpret Schur's Lemma and its converse in terms of representations. 9.3 Corollary Let r: G ! GL (n, C) be a representation of G. Then r is irreducible if and only if every n 3 n matrix A which satis®es ( gr)A A( gr)

for all g 2 G

has the form A ëI n with ë 2 C. Proof As in Theorem 4.4(1), regard C n as a CG-module by de®ning v g v( gr) for all v 2 C n , g 2 G. Let A be an n 3 n matrix over C. The endomorphism v ! vA of C n is a CG-homomorphism if and only if (v g)A (vA) g

for all v 2 C n , g 2 G;

80


that is, if and only if ( gr)A A( gr)

for all g 2 G:

The result now follows from Schur's Lemma 9.1 and Proposition 9.2. j

9.4 Examples (1) Let G C3 ka: a3 1l, and let r: G ! GL (2, C) be the representation for which ar

0 ÿ1

1 ÿ1

(see Exercise 3.2). Since the matrix

0 ÿ1

1 ÿ1

commutes with all gr ( g 2 G), Corollary 9.3 implies that r is reducible. (2) Let G D10 ka, b: a5 b2 1, bÿ1 ab aÿ1 l, and let ù e2ði=5 . Check that there is a representation r: G ! GL (2, C) for which ar

ù 0

0 , ùÿ1

br

0 1

1 : 0

Assume that the matrix A

á ã

â ä

commutes with both ar and br. The fact that (ar)A A(ar) forces â ã 0; and then (br)A A(br) gives á ä. Hence A

á 0

0 á

áI:

Consequently r is irreducible, by Corollary 9.3.

Schur's Lemma

81

Representation theory of ®nite abelian groups Let G be a ®nite abelian group, and let V be an irreducible CGmodule. Pick x 2 G. Since G is abelian, v gx vxg

for all g 2 G,

and hence the endomorphism v ! vx of V is a CG-homomorphism. By Schur's Lemma 9.1(2), this endomorphism is a scalar multiple of the identity 1 V , say ë x 1 V . Thus vx ë x v

for all v 2 V :

This implies that every subspace of V is a CG-submodule. As V is irreducible, we deduce that dim V 1. Thus we have proved 9.5 Proposition If G is a ®nite abelian group, then every irreducible CG-module has dimension 1. The next result is a major structure theorem for ®nite abelian groups. We shall not prove it here, but refer you to Chapter 9 of the book of J. B. Fraleigh listed in the Bibliography. 9.6 Theorem Every ®nite abelian group is isomorphic to a direct product of cyclic groups. We shall determine the irreducible representations of all direct products C n1 3 C n2 3 : : : 3 C n r where n1 , . . . , n r are positive integers. By Theorem 9.6, this covers the irreducible representations of all ®nite abelian groups. Let G C n1 3 . . . 3 C n r , and for 1 < i < r, let c i be a generator for C n i . Write g i (1, : : : , ci , : : : , 1)

(ci in ith position):

Then G h g1 , : : : , g r i, with g in i 1 and g i g j g j g i for all i, j: Now let r: G ! GL (n, C) be an irreducible representation of G

82


over C. Then n 1 by Proposition 9.5, so for 1 < i < r, there exists ë i 2 C such that g i r (ë i ) (where of course (ë i ) is a 1 3 1 matrix). As g i has order n i , we have ë in i 1; that is, ë i is an n i th root of unity. Also, the values ë1 , . . . , ë r determine r, since for g 2 G, we have g g1i1 . . . g irr for some integers i1 , . . . , i r , and then (9:7)

gr ( g 1i1 : : : g irr )r (ë1i1 : : :ë irr ):

For a representation r of G satisfying (9.7) for all i1 , . . . , i r , write r rë1 ,:::,ë r : Conversely, given any n i th roots of unity ë i (1 < i < r), the function g 1i1 : : : g irr ! (ë1i1 : : :ë irr ) is a representation of G. There are n1 n2 . . . n r such representations, and no two of them are equivalent. We have proved the following theorem. 9.8 Theorem Let G be the abelian group C n1 3 . . . 3 C n r . The representations rë1 ,:::,ë r of G constructed above are irreducible and have degree 1. There are |G| of these representations, and every irreducible representation of G over C is equivalent to precisely one of them. 9.9 Examples (1) Let G C n ka: a n 1l, and put ù e2ði= n . The n irreducible representations of G over C are rù j (0 < j < n ÿ 1), where ak rù j (ù jk ) (0 < k < n ÿ 1): (2) The four irreducible CG-modules for G C2 3 C2 k g1 , g2 l are V1 , V2 , V3 , V4 , where Vi is a 1-dimensional space with basis v i (i 1, 2, 3, 4) and v1 g 1 v1 ,

v1 g 2 v2 ;

v2 g 1 v2 ,

v2 g 2 ÿv2 ;

v3 g 1 ÿv3 ,

v3 g 2 v3 ;

v4 g 1 ÿv4 ,

v4 g 2 ÿv4 :

Schur's Lemma

83

Diagonalization Let H k gl be a cyclic group of order n, and let V be a non-zero C H-module. By Theorem 8.7, V U1 : : : Ur , a direct sum of irreducible C H-submodules Ui of V. Each Ui has dimension 1, by Proposition 9.5; let u i be a vector spanning Ui. Put ù e2ði= n . Then for each i, there exists an integer m i such that ui g ù m i u i : Thus if B is the basis u1 , . . . , u r of V, then 0 m1 1 ù 0 B C .. (9:10) [ g]B @ A: .

0

ù mr

The following useful result is an immediate consequence of this. 9.11 Proposition Let G be a ®nite group and V a CG-module. If g 2 G, then there is a basis B of V such that the matrix [ g]B is diagonal. If g has order n, then the entries on the diagonal of [ g]B are nth roots of unity. Proof Let H k gl. As V is also a C H-module, the result follows from (9.10). j Some further applications of Schur's Lemma Our next application concerns an important subspace of the group algebra CG. 9.12 De®nition Let G be a ®nite group. The centre of the group algebra CG, written Z(CG), is de®ned by Z(CG) fz 2 CG: zr rz for all r 2 CGg: Using (2.5), it is easy to check that Z(CG) is a subspace of CG. For abelian groups G, the centre Z(CG) is the whole group algebra. For arbitrary groups G, we shall see that Z(CG) plays a crucial role in

84


the study of representations of G (for example, its dimension is equal to the number of irreducible representations of G ± see Chapter 15). 9.13 Example P The elements 1 and g2G g lie in Z(CG). Indeed, if H is any normal subgroup of G, then X h 2 Z(CG): To see this, write z

P

h2 H

h. Then for all g 2 G, X X g ÿ1 zg g ÿ1 hg h z, h2 H

h2 H

h2 H

and so zg gz. Consequently zr rz for all r 2 CG. For example, if G D6 ka, b: a3 b2 1, bÿ1 ab aÿ1 l, then {1}, kal and G are normal subgroups of G, so the elements 1, 1 a a2 and 1 a a2 b ab a2 b lie in Z(CG). We shall see later that these elements in fact form a basis of Z(CG). We use Schur's Lemma to prove the following important property of the elements of Z(CG). 9.14 Proposition Let V be an irreducible CG-module, and let z 2 Z(CG). Then there exists ë 2 C such that vz ëv

for all v 2 V :

Proof For all r 2 CG and v 2 V, we have vrz vzr, and hence the function v ! vz is a CG-homomorphism from V to V. By Schur's Lemma 9.1(2), this CG-homomorphism is equal to ë1 V for some ë 2 C, and the result follows. j Some elements of the centre of CG are provided by the centre of G, which we now de®ne.

Schur's Lemma

85

9.15 De®nition The centre of G, written Z(G), is de®ned by Z(G) fz 2 G: zg gz for all g 2 Gg: Clearly Z(G) is a normal subgroup of G, and is a subset of Z(CG). Although we have seen in Proposition 6.6 that for every ®nite group G there is a faithful CG-module, it is not necessarily the case that there is a faithful irreducible CG-module. Indeed, the following result shows that the existence of a faithful irreducible CG-module imposes a strong restriction on the structure of G. 9.16 Proposition If there exists a faithful irreducible CG-module, then Z(G) is cyclic. Proof Let V be a faithful irreducible CG-module. If z 2 Z(G) then z lies in Z(CG), and hence by Proposition 9.14, there exists ë z 2 C such that vz ë z v for all v 2 V : Since V is faithful, the function z ! ëz

(z 2 Z(G))

is an injective homomorphism from Z(G) into the multiplicative group C of non-zero complex numbers. Therefore Z(G) {ë z : z 2 Z(G)}, which, being a ®nite subgroup of C , is cyclic (see Exercise 1.7). j We remark that the converse of Proposition 9.16 is false, since in Exercise 25.6, we give an example of a group G such that Z(G) is cyclic but there exists no faithful irreducible CG-module. 9.17 Example If G is an abelian group, then G Z(G), and so by Proposition 9.16, there is no faithful irreducible CG-module unless G is cyclic. For example, C2 3 C2 has no faithful irreducible representation (compare Example 9.9(2)). The irreducible representations of non-abelian groups are more dif®cult to construct than those of abelian groups. In particular, they

86


do not all have degree 1, as is shown by the following converse to Proposition 9.5. 9.18 Proposition Suppose that G is a ®nite group such that every irreducible CG-module has dimension 1. Then G is abelian. Proof By Theorem 8.7, we can write CG V1 : : : Vn , where each Vi is an irreducible CG-submodule of the regular CGmodule CG. Then dim Vi 1 for all i, since we are assuming that all irreducible CG-modules have dimension 1. For 1 < i < n, let v i be a vector spanning V i . Then v1 , . . . , v n is a basis of CG; call it B . For all x, y 2 G, the matrices [x]B and [ y]B are diagonal, and hence they commute. Since the representation g ! [ g]B

( g 2 G)

of G is faithful (see Proposition 6.6), we deduce that x and y commute. Hence G is abelian, as required. j Summary of Chapter 9 1. Schur's Lemma states that every CG-homomorphism between irreducible CG-modules is either zero or a CG-isomorphism. Also, the only CG-homomorphisms from an irreducible CG-module to itself are scalar multiples of the identity. 2. The centre Z(CG) of the group algebra CG consists of those elements which commute with all elements of CG. The elements of Z(CG) act as scalar multiples of the identity on all irreducible CGmodules. 3. All irreducible CG-modules for a ®nite abelian group G have dimension 1, and there are precisely |G| of them. Exercises for Chapter 9 1. Write down the irreducible representations over C of the groups C2 , C3 and C2 3 C2 .

Schur's Lemma

87

2. Let G C4 3 C4 . (a) Find a non-trivial irreducible representation r of G such that g2 r (1) for all g 2 G. (b) Prove that there is no irreducible representation ó of G such that gó (ÿ1) for all elements g of order 2 in G. 3. Let G be the ®nite abelian group C n1 3 . . . 3 C n r . Prove that G has a faithful representation of degree r. Can G have a faithful representation of degree less than r? 4. Suppose that G D8 ka, b: a4 b2 1, bÿ1 ab aÿ1 l. Check that there is a representation r of G over C such that ÿ7 10 ÿ5 6 ar , br : ÿ5 7 ÿ4 5 Find all 2 3 2 matrices M such that Hence determine whether or not r is Do the same for the representation 5 ÿ6 aó , bó 4 ÿ5

M( gr) ( gr)M for all g 2 G. irreducible. ó of G, where ÿ5 6 : ÿ4 5

5. Show that if V is an irreducible CG-module, then there exists ë 2 C such that X ! v g ëv for all v 2 V : g2G

6. Let G D6 ka, b: a3 b2 1, bÿ1 ab aÿ1 l. Write ù e2ði=3 , and let W be the irreducible CG-submodule of the regular CGmodule de®ned by W sp (1 ù2 a ùa2 , b ù2 ab ùa2 b) (see Exercise 6.6). (a) Show that a aÿ1 2 Z(CG). (b) Find ë 2 C such that w(a aÿ1 ) ëw for all w 2 W. (Compare Proposition 9.14.)

88


7. Which of the following groups have a faithful irreducible representation? (a) C n (n a positive integer); (b) D8 ; (c) C2 3 D8 ; (d) C3 3 D8.

10 Irreducible modules and the group algebra

Let G be a ®nite group and CG be the group algebra of G over C. Consider CG as the regular CG-module. By Theorem 8.7, we can write CG U1 : : : Ur where each Ui is an irreducible CG-module. We shall show in this chapter that every irreducible CG-module is isomorphic to one of the CG-modules U1, . . . , Ur. As a consequence, there are only ®nitely many non-isomorphic irreducible CG-modules (a result which has already been established for abelian groups in Theorem 9.8). Also, in theory, to ®nd all irreducible CG-modules, it is suf®cient to decompose CG as a direct sum of irreducible CG-submodules. However, this is not really a practical way of ®nding the irreducible CG-modules, unless G is a small group. Irreducible submodules of CG We begin with another consequence of Maschke's Theorem. 10.1 Proposition Let V and W be CG-modules and let W: V ! W be a CG-homomorphism. Then there is a CG-submodule U of V such that V Ker W U and U Im W. Proof Since Ker W is a CG-submodule of V by Proposition 7.2, there is by Maschke's Theorem a CG-submodule U of V such that V Ker W U . De®ne a function W: U ! Im W by uW uW

(u 2 U ): 89

90


We show that W is a CG-isomorphism from U to Im W. Clearly W is a CG-homomorphism, since W is a CG-homomorphism. If u 2 Ker W then u 2 Ker W \ U {0}; hence Ker W {0}. Now let w 2 Im W; so w vW for some v 2 V. Write v k u with k 2 Ker W, u 2 U. Then w vW kW uW uW uW: Therefore Im W Im W. We have now established that W: U ! Im W is an invertible CG-homomorphism. Thus U Im W, as required.

10.2 Proposition Let V be a CG-module, and write V U1 : : : Us , a direct sum of irreducible CG-submodules Ui. If U is any irreducible CG-submodule of V, then U Ui for some i. Proof For u 2 U, we have u u1 . . . us for unique vectors ui 2 Ui (1 < i < s). De®ne ð i : U ! Ui by setting uð i ui . Choosing i such that ui 6 0 for some u 2 U, we have ð i 6 0. Now ð i is a CG-homomorphism (see Proposition 7.11). As U and Ui are irreducible, and ð i 6 0, Schur's Lemma 9.1(1) implies that ð i is a CG-isomorphism. Therefore U Ui. j Of course it can happen that U is an irreducible CG-submodule of U1 . . . Us (each Ui irreducible) without U being equal to any Ui, as the following example shows. 10.3 Example Let G be any group and let V be a 2-dimensional CG-module, with basis v1 , v2 , such that v g v for all v 2 V and g 2 G. Then V U1 U2 , where U1 sp (v1 ) and U2 sp (v2 ) are irreducible CG-submodules. However, U sp (v1 v2 ) is an irreducible CG-submodule which is not equal to U1 or U2. 10.4 De®nitions (1) If V is a CG-module and U is an irreducible CG-module, then we say that U is a composition factor of V if V has a CG-submodule which is isomorphic to U.

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. We now come to the main result of the chapter, which shows that every irreducible CG-module is a composition factor of the regular CG-module. 10.5 Theorem Regard CG as the regular CG-module, and write CG U 1 : : : U r , a direct sum of irreducible CG-submodules. Then every irreducible CG-module is isomorphic to one of the CG-modules Ui. Proof Let W be an irreducible CG-module, and choose a non-zero vector w 2 W. Observe that {wr: r 2 CG} is a CG-submodule of W; since W is irreducible, it follows that (10:6)

W fwr: r 2 CGg:

Now de®ne W: CG ! W by rW wr

(r 2 CG):

Clearly W is a linear transformation, and Im W W by (10.6). Moreover, W is a CG-homomorphism, since for r, s 2 CG, (rs)W w(rs) (wr)s (rW)s: By Proposition 10.1, there is a CG-submodule U of CG such that CG U Ker W and U Im W W : As W is irreducible, so is U. By Proposition 10.2 we have U Ui for some i; then W Ui, and the result is proved. j Theorem 10.5 shows that there is a ®nite set of irreducible CGmodules such that every irreducible CG-module is isomorphic to one of them. We record this fact in the following corollary. 10.7 Corollary If G is a ®nite group, then there are only ®nitely many non-isomorphic irreducible CG-modules.

92


According to Theorem 10.5, to ®nd all the irreducible CG-modules we need only decompose the regular CG-module as a direct sum of irreducible CG-submodules. We now do this for a couple of examples; however, this is not a practical method for studying CG-modules in general. 10.8 Examples (1) Let G C3 ka: a3 1l, and write ù e2ði=3 . De®ne v0 , v1 , v2 2 CG by v0 1 a a2 , v1 1 ù2 a ùa2 , v2 1 ùa ù2 a2 , and let Ui sp (v i ) for i 0, 1, 2. Then v1 a a ù2 a2 ù1 ùv1 , and similarly vi a ùi vi

for i 0, 1, 2:

Hence Ui is a CG-submodule of CG for i 0, 1, 2. It is easy to check that v0 , v1 , v2 is a basis of CG, and hence CG U0 U 1 U2 , a direct sum of irreducible CG-submodules Ui. By Theorem 10.5, every irreducible CG-module is isomorphic to U0, U1 or U2. The irreducible representation of G corresponding to Ui is the representation rù i of Example 9.9(1). (2) Let G D6 ka, b: a3 b2 1, bÿ1 ab aÿ1 l. We decompose CG as a direct sum of irreducible CG-submodules. Let ù e2ði=3 and de®ne v0 1 a a2 ,

w0 bv0

v1 1 ù2 a ùa2 , 2 2

v2 1 ùa ù a ,

( b ba ba2 ),

w1 bv1 , w2 bv2 :

As in (1) above, v i a ù i v i for i 0, 1, 2, and so sp (v i ) and sp (wi ) are Ckal-modules. Next, note that v0 b w 0 ,

w0 b v0 ,

v1 b w2 ,

w1 b v2 ,

v2 b w1 ,

w2 b v1 :

Irreducible modules and the group algebra

93

Therefore, sp (v0 , w0 ), sp (v1 , w2 ) and sp (v2 , w1 ) are Ckbl-modules, and hence are CG-submodules of CG. By the argument in Example 5.5(2), the CG-submodules U3 sp (v1 , w2 ) and U4 sp (v2 , w1 ) are irreducible. However, sp(v0 , w0 ) is reducible, as U 1 sp(v0 w0 ) and U2 sp(v0 ÿ w0 ) are CG-submodules. Now v0, v1 , v2 , w0, w1, w2 is a basis of CG, and hence CG U1 U2 U3 U4 , a direct sum of irreducible CG-submodules. Note that U1 is the trivial CG-module, and U1 is not isomorphic to U2, the other 1-dimensional Ui. But U3 U4 (there is a CG-isomorphism sending v1 ! w1 , w2 ! v2 ). We conclude from Theorem 10.5 that there are exactly three nonisomorphic irreducible CG-modules, namely U1, U2 and U3. Correspondingly, every irreducible representation of D6 over C is equivalent to precisely one of the following: r1 : a ! (1), b ! (1); r2 : a ! (1), b ! (ÿ1); 0 1 ù 0 r3 : a ! ,b! : 1 0 0 ùÿ1

Summary of Chapter 10 1. Every irreducible CG-module occurs as a composition factor of the regular CG-module. 2. There are only ®nitely many non-isomorphic irreducible CGmodules.

Exercises for Chapter 10 1. Let G be a ®nite group. Find a CG-submodule of CG which is isomorphic to the trivial CG-module. Is there only one such CGsubmodule? 2. Let G C4 . Express CG as a direct sum of irreducible CGsubmodules. (Hint: copy the method of Example 10.8(1).)

94


3. Let G D8 ka, b: a4 b2 1, bÿ1 ab aÿ1 l. Find a 1-dimensional CG-submodule, sp (u1 ) say, of CG such that u1 a u1 , u1 b ÿu1 : Find also 1-dimensional CG-submodules, sp (u2 ) and sp (u3 ), such that u2 a ÿu2 ,

u2 b u2 , and

u3 a ÿu3 ,

u3 b ÿu3 :

4. Use the method of Example 10.8(2) to ®nd all the irreducible representations of D8 over C. 5. 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 U1 U2, isomorphic CG-modules. Show that there is a V which is not equal to U1 or U2, but is them.

6. Let G Q8 ka, b: a4 1, b2 a2 , bÿ1 ab aÿ1 l, and let V be the CG-module given in Example 4.5(2). Thus V has basis v1 , v2 and v1 a iv1 ,

v1 b v2 ,

v2 a ÿiv2 ,

v2 b ÿv1 :

Show that V is irreducible, and ®nd a CG-submodule of CG which is isomorphic to V.

11 More on the group algebra

We now go further into the structure of the group algebra CG of a ®nite group G. As in Chapter 10, we write CG U 1 : : : U r , a direct sum of irreducible CG-modules Ui. In Theorem 10.5 we proved that every irreducible CG-module U is isomorphic to one of the Ui. 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 11.9). Our proof of Theorem 11.9 is based on a study of the vector space of CG-homomorphisms from one CG-module to another. The space of CG-homomorphisms 11.1 De®nition Let V and W be CG-modules. We write HomCG (V , W ) for the set of all CG-homomorphisms from V to W. De®ne addition and scalar multiplication on HomCG (V , W ) as follows: for W, ö 2 HomCG (V , W ) and ë 2 C, de®ne W ö and ëW by v(W ö) vW vö, v(ëW) ë(vW) for all v 2 V . Then W ö, ëW 2 HomCG (V , W ). With these de®nitions, it is easily checked that HomCG (V, W ) is a vector space over C. We begin our study of the vector space HomCG (V, W) with an easy consequence of Schur's Lemma. 95

96


11.2 Proposition Suppose that V and W are irreducible CG-modules. Then 1, if V W , dim (HomCG (V , W )) 0, if V 6 W : Proof If V 6 W then this is immediate from Schur's Lemma 9.1(1). Now suppose that V W, and let W: V ! W be a CG-isomorphism. If ö 2 HomCG (V , W ), then öWÿ1 is a CG-isomorphism from V to V, so by Schur's Lemma 9.1(2), there exists ë 2 C such that öWÿ1 ë1 V : Then ö ëW, and so HomCG (V , W ) fëW: ë 2 Cg, a 1-dimensional space. j For the next result, recall the de®nition of a composition factor of a CG-module from 10.4. 11.3 Proposition Let V and W be CG-modules, and suppose that HomCG (V , W ) 6 f0g. Then V and W have a common composition factor. Proof Let W be a non-zero element of HomCG (V , W ). Then V Ker W U for some non-zero CG-module U, by Maschke's Theorem. Let X be an irreducible CG-submodule of U. Since XW 6 {0}, Schur's Lemma 9.1(1) implies that XW X. Therefore X is a common composition factor of V and W. j The next few results show how to calculate the dimension of HomCG (V , W ) in general. The key step is the following proposition. 11.4 Proposition Let V , V1 , V2 and W , W 1 , W 2 be CG-modules. Then (1) dim (HomCG (V, W1 W2 )) dim (HomCG (V, W1 )) dim (HomCG (V, W2 )), (2) dim (HomCG (V1 V2 , W )) dim (HomCG (V1 , W)) dim (HomCG (V2 , W )).

More on the group algebra Proof (1) De®ne W 1 W 2 ! W 2 by

the

functions

(w1 w2 )ð1 w1 ,

97

ð1 : W1 W2 ! W1

and

ð2 :

(w1 w2 )ð2 w2

for all w1 2 W 1 , w2 2 W 2 . By Proposition 7.11, ð1 and ð2 are CG-homomorphisms. If W 2 HomCG (V , W 1 W 2 ), then Wð1 2 HomCG (V , W 1 ) and Wð2 2 HomCG (V, W2 ) (see Exercise 7.1). We now de®ne a function f from HomCG (V, W1 W2 ) to the (external) direct sum of HomCG (V, W1 ) and HomCG (V, W2 ) by f : W ! (Wð1 , Wð2 )

(W 2 HomCG (V , W 1 W 2 )):

Clearly f is a linear transformation. We show that f is invertible. Given ö i 2 HomCG (V, Wi ) (i 1, 2), the function ö: v ! vö1 vö2

(v 2 V )

lies in HomCG (V, W1 W2 ), and the image of ö under f is (ö1 , ö2 ). Hence f is surjective. If W 2 Ker f, then vWð1 0 and vWð2 0 for all v 2 V, so vW vW(ð1 ð2 ) 0. Therefore W 0, so Ker f {0} and f is injective. We have established that f is an invertible linear transformation from HomCG (V, W1 W2 ) to HomCG (V, W1 ) HomCG (V, W2 ). Consequently these two vector spaces have equal dimensions, and (1) follows. (2) For W 2 HomCG (V1 V2 , W ), de®ne W Vi : Vi ! W (i 1, 2) to be the restriction of W to V i ; that is, W Vi is the function v i W Vi v i W

(v i 2 Vi ):

Then W Vi 2 HomCG (V i , W ) for i 1, 2. Now let h be the function from HomCG (V1 V2 , W ) HomCG (V1 , W ) HomCG (V2 , W ) which is given by h: W ! (W V1 , W V2 )

to

(W 2 HomCG (V1 V2 , W )):

Clearly h is an injective linear HomCG (Vi , W ) (i 1, 2), the function ö: v1 v2 ! v1 ö1 v2 ö2

transformation.

Given

öi 2

(v i 2 Vi for i 1, 2)

lies in HomCG (V1 V2 , W ) and has image (ö1 , ö2 ) under h. Hence h is surjective. We have shown that h is an invertible linear transformation, and (2) follows. j

98


Now suppose that we have CG-modules V, W, Vi, Wj (1 < i < r, 1 < j < s). By an obvious induction using Proposition 11.4, we have (11.5)

(1) dim (HomCG (V, W1 . . . Ws ))

s X

dim (HomCG (V, Wj )),

j1

(2) dim(HomCG (V1 . . . V r , W ))

r X

dim (HomCG (Vi, W)).

i1

These in turn imply (3) dim (HomCG (V1 . . . Vr , W1 . . . Ws ))

r X s X i1

dim (HomCG (Vi, Wj )).

j1

By applying (3) when all Vi and Wj are irreducible, and using Proposition 11.2, we can ®nd dim (HomCG (V, W )) in general. In the following corollary we single out the case where one of the CGmodules is irreducible. 11.6 Corollary Let V be a CG-module with V U1 : : : Us , where each Ui is an irreducible CG-module. Let W be any irreducible CG-module. Then the dimensions of HomCG (V , W ) and HomCG (W , V ) are both equal to the number of CG-modules Ui such that Ui W. Proof By (11.5), dim (HomCG (V , W ))

s X

dim (HomCG (Ui , W )), and

i1

dim (HomCG (W , V ))

s X i1

dim (HomCG (W , U i )):

More on the group algebra And by Proposition 11.2,

99

dim (HomCG (Ui , W )) dim (HomCG (W , Ui ))

1, 0,

if U i W , if U i 6 W:

The result follows.

j

11.7 Example For G D6, we saw in Example 10.8(2) that CG U1 U2 U3 U4 , a direct sum of irreducible CG-submodules, with U3 U4 but U3 not isomorphic to U1 or U2. Thus by Corollary 11.6, we have dim (HomCG (CG, U3 )) dim (HomCG (U3 , CG)) 2: You are asked in Exercise 11.5 to ®nd bases for these two vector spaces of CG-homomorphisms. The next proposition investigates the space of CG-homomorphisms from the regular CG-module to any other CG-module. When combined with Corollary 11.6, it will give the main result of this chapter. 11.8 Proposition If U is a CG-module, then dim (HomCG (CG, U )) dim U : Proof Let d dim U. Choose a basis u1 , . . . , ud of U. For 1 < i < d, de®ne ö i : CG ! U by rö i ui r

(r 2 CG):

Then ö i 2 HomCG (CG, U) since for all r, s 2 CG, (rs)ö i ui (rs) (ui r)s (rö i )s: We shall prove that ö1 , . . . , ö d is a basis of HomCG (CG, U ). Suppose that ö 2 HomCG (CG, U ). Then 1ö ë1 u1 : : : ë d u d for some ë i 2 C. Since ö is a CG-homomorphism, for all r 2 CG we have

100

Representations and characters of groups rö (1r)ö (1ö)r ë1 u1 r : : : ë d u d r r(ë1 ö1 : : : ë d ö d ):

Hence ö ë1 ö1 . . . ë d ö d . HomCG (CG, U ). Now assume that

Therefore

ö1 , . . . , ö d

span

ë1 ö1 : : : ë d ö d 0 (ë i 2 C): Evaluating both sides at the identity 1, we have 0 1(ë1 ö1 : : : ë d ö d ) ë1 u1 : : : ë d u d , which forces ë i 0 for all i. Hence ö1 , . . . , ö d is a basis of HomCG (CG, U ), which therefore has dimension d. j We now come to the main theorem of the chapter, which tells us how often each irreducible CG-module occurs in the regular CGmodule. 11.9 Theorem Suppose that CG U1 : : : Ur , a direct sum of irreducible CG-submodules. If U is any irreducible CG-module, then the number of CG-modules Ui with Ui U is equal to dim U. Proof By Proposition 11.8, dim U dim (HomCG (CG, U )), and by Corollary 11.6, this is equal to the number of Ui with Ui U. j

11.10 Example Recall again from Example 10.8(2) that if G D6 then CG U1 U 2 U3 U 4 , where U1, U2 are non-isomorphic 1-dimensional CG-modules, and

More on the group algebra

101

U3, U4 are isomorphic irreducible 2-dimensional CG-modules. This illustrates Theorem 11.9: U1 occurs once, dim U1 1; U2 occurs once, dim U2 1; U3 occurs twice, dim U3 2: We conclude the chapter with a signi®cant consequence of Theorem 11.9 concerning the dimensions of all irreducible CG-modules. 11.11 De®nition We say that the irreducible CG-modules V1 , : : : , V k form a complete set of non-isomorphic irreducible CG-modules if every irreducible CG-module is isomorphic to some V i , and no two of V1 , : : : , V k are isomorphic. (By Corollary 10.7, for any ®nite group G there exists a complete set of non-isomorphic irreducible CG-modules.) 11.12 Theorem Let V1 , : : : , V k form a complete set of non-isomorphic irreducible CG-modules. Then k X (dim Vi )2 jGj: i1

Proof Let CG U1 . . . Ur, a direct sum of irreducible CG-submodules. For 1 < i < k, write d i dim V i . By Theorem 11.9, for each i, the number of CG-modules Uj with Uj Vi is equal to di . Therefore dim CG dim U 1 : : : dim Ur

k X

d i (dim Vi )

i1

k X i1

As dim CG |G|, the result follows.

d 2i : j

11.13 Example Let G be a group of order 8, and let d1, . . . , dk be the dimensions of all the irreducible CG-modules. By Theorem 11.12, k X i1

d 2i 8:

102


Observe that the trivial CG-module is irreducible of dimension 1, and so di 1 for some i. Hence the possibilities for d1, . . . , dk are 1, 1, 1, 1, 1, 1, 1, 1

and

1, 1, 1, 1, 2: Both these possibilities do occur: the ®rst holds when G is an abelian group (see Proposition 9.5), and the second when G D8 (see Exercise 10.4). We shall see later that dim Vi divides |G| for all i, and this fact, combined with Theorem 11.12, is quite a powerful tool in ®nding the dimensions of irreducible CG-modules. Summary of Chapter 11 1. dim (HomCG (V1 : : : V r , W 1 : : : W s ))

r X s X

dim (HomCG (Vi , W j )):

i1 j1

2. dim (HomCG (CG, U )) dim U . 3. Let CG U1 . . . Ur, a direct sum of irreducible CG-modules, and let U be any irreducible CG-module. Then the number of Ui with Ui U is equal to dim U. 4. If V1 , : : : , V k is a complete set of non-isomorphic irreducible CGmodules, then k X (dim Vi )2 jGj: i1

Exercises for Chapter 11 1. If G is a non-abelian group of order 6, ®nd the dimensions of all the irreducible CG-modules. 2. If G is a group of order 12, what are the possible degrees of all the irreducible representations of G? Find the degrees of the irreducible representations of D12. (Hint: use Exercise 5.3.) 3. Let G be a ®nite group. Find a basis for HomCG (CG, CG).

More on the group algebra

103

4. Suppose that G Sn and V is the n-dimensional permutation module for G over C, as de®ned in 4.10. If U is the trivial CGmodule, show that HomCG (V, U) has dimension 1. 5. Let G D6 and let CG U1 U2 U3 U4, a direct sum of irreducible CG-modules, as in Example 10.8(2). Find a basis for HomCG (CG, U3 ) and a basis for HomCG (U3, CG). 6. Let V1 , : : : , V k be a complete set of non-isomorphic irreducible CG-modules, and let V, W be arbitrary CG-modules. Assume that for 1 < i < k, d i dim (HomCG (V , Vi )) and ei dim (HomCG (W , Vi )): Pk Show that dim (HomCG (V , W )) i1 d i ei .

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. After de®ning conjugacy classes, we develop enough theory to determine the conjugacy classes of dihedral, symmetric and alternating groups. At the end of the chapter we prove a result linking the conjugacy classes of a group to the structure of its group algebra. Throughout the chapter, G is a ®nite group. Conjugacy classes 12.1 De®nition Let x, y 2 G. We say that x is conjugate to y in G if y g ÿ1 xg

for some g 2 G:

The set of all elements conjugate to x in G is x G f g ÿ1 xg: g 2 Gg, and is called the conjugacy class of x in G. Our ®rst result shows that two distinct conjugacy classes have no elements in common. 12.2 Proposition If x, y 2 G, then either x G y G or x G \ y G is empty. Proof Suppose that x G \ y G is not empty, and pick z 2 x G \ y G. Then there exist g, h 2 G such that z gÿ1 xg hÿ1 yh: 104

Conjugacy classes

105

Hence x ghÿ1 yhgÿ1 kÿ1 yk, where k hgÿ1 . So a 2 x G ) a bÿ1 xb )ab

ÿ1 ÿ1

k

) a cÿ1 yc

for some b 2 G ykb where c kb

) a 2 yG : Therefore x G # y G. Similarly y G # x G (using y kxkÿ1 ), and so xG yG . j Since every element x of G lies in the conjugacy class x G (as x 1ÿ1 x1 with 1 2 G), G is the union of its conjugacy classes and so we deduce immediately 12.3 Corollary Every group is a union of conjugacy classes, and distinct conjugacy classes are disjoint. Another way of seeing this Corollary 12.3 is to observe that conjugacy is an equivalence relation, and that the conjugacy classes are the equivalence classes. 12.4 De®nition If G x1G [ . . . [ x Gl , where the conjugacy classes x1G , . . . , x Gl are distinct, then we call x1 , . . . , xl representatives of the conjugacy classes of G. 12.5 Examples (1) For every group G, 1 G {1} is a conjugacy class of G. (2) Let G D6 ka, b: a3 b2 1, bÿ1 ab aÿ1 l. The elements of G are 1, a, a2 , b, ab, a2 b. Since gÿ1 ag is a or a2 for every g 2 G, and bÿ1 ab a2 , we have aG fa, a2 g: Also, aÿi ba i aÿ2i b for all integers i, so bG fb, ab, a2 bg: Thus the conjugacy classes of G are f1g, fa, a2 g, fb, ab, a2 bg:

106


(3) If G is abelian then gÿ1 xg x for all x, g 2 G, and so x G {x}. Hence every conjugacy class of G consists of just one element. The next proposition is often useful when calculating conjugacy classes. 12.6 Proposition Let x, y 2 G. If x is conjugate to y in G, then x n is conjugate to y n in G for every integer n, and x and y have the same order. Proof Observe that for a, b 2 G, we have gÿ1 abg ( g ÿ1 ag)( g ÿ1 bg): Hence gÿ1 x n g ( gÿ1 xg) n . Suppose that x is conjugate to y in G, so that y gÿ1 xg for some g 2 G. Then y n gÿ1 x n g and therefore x n is conjugate to y n in G. Let x have order m. Then y m gÿ1 x m g 1, and for 0 , r , m, y r gÿ1 x r g 6 1, so y also has order m. 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. 12.7 De®nition Let x 2 G. The centralizer of x in G, written CG (x), is the set of elements of G which commute with x; that is, CG (x) f g 2 G: xg gxg: (So also CG (x) { g 2 G: gÿ1 xg x}.) It is easy to check that CG (x) is a subgroup of G (Exercise 12.1). Observe that x 2 CG (x) and indeed, kxl # CG (x) for all x 2 G. 12.8 Theorem Let x 2 G. Then the size of the conjugacy class x G is given by jx G j jG: CG (x)j jGj=jCG (x)j: In particular, |x G | divides |G|.

Conjugacy classes

107

Proof Observe ®rst that for g, h 2 G, we have g ÿ1 xg hÿ1 xh , hg ÿ1 x xhg ÿ1 , hg ÿ1 2 CG (x) , CG (x) g CG (x)h: By dint of this, 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 ÿ1 xg ! CG (x) g

( g 2 G):

Clearly f is surjective. Hence f is a bijection, proving that |x G | |G:CG (x)|. j Before summarizing our results on conjugacy classes, we make the observation that jx G j 1 , g ÿ1 xg x

(12:9)

for all g 2 G

, x 2 Z(G), where Z(G) is the centre of G, as de®ned in 9.15. We have now proved all parts of the following result. 12.10 The Class Equation Let x1 , . . . , xl be representatives of the conjugacy classes of G. Then X jx G jGj j Z(G)j i j, xi2 = Z(G)

where

|x G i |

|G:CG (xi )|, and both | Z(G)| and |x G i | divide |G|.

Conjugacy classes of dihedral groups We illustrate the use of Theorem 12.8 by ®nding the conjugacy classes of all dihedral groups. Let G D2 n, the dihedral group of order 2n. Thus G ha, b: an b2 1, bÿ1 ab aÿ1 i: In ®nding the conjugacy classes of G, it is convenient to consider separately the cases where n is odd and where n is even. (1) n odd First consider ai (1 < i < n ÿ 1). Since CG (ai ) contains kal, jG: CG (ai )j < jG: haij 2:

108


Also bÿ1 ai b aÿi , so {ai , aÿi } # (ai ) G . As n is odd, ai 6 aÿi , and so |(ai ) G | > 2. Using Theorem 12.8, we have 2 > jG: CG (ai )j j(ai ) G j > 2: Hence equality holds here, and CG (ai ) hai, (ai ) G fai , aÿi g: Next, CG (b) contains {1, b}; and as bÿ1 ai b aÿi , no element ai or ai b (with 1 < i < n ÿ 1) commutes with b. Thus CG (b) f1, bg: Therefore by Theorem 12.8, |bG | n. Since all the elements ai have been accounted for, bG must consist of the remaining n elements of G. That is, bG fb, ab, : : : , a nÿ1 bg: We have shown (12.11) The dihedral group D2 n (n odd) has precisely 12(n 3) conjugacy classes: {1}, {a, aÿ1 }, . . . , {a( nÿ1)=2 , aÿ( nÿ1)=2 }, {b, ab, . . . , a nÿ1 b}. (2) n even Write n 2m. As bÿ1 am b aÿ m am , the centralizer of am in G contains both a and b, and hence CG (am ) G. Therefore the conjugacy class of am in G is just fam g. As in case (1), (ai ) G {ai , aÿi } for 1 < i < m ÿ 1. For every integer j, a j baÿ j a2 j b, a j (ab)aÿ j a2 j1 b: It follows that bG fa2 j b: 0 < j < m ÿ 1g, (ab) G fa2 j1 b: 0 < j < m ÿ 1g: Hence (12.12) The dihedral group D2 n (n even, n 2m) has precisely m 3 conjugacy classes: {1}, {a m }, {a, aÿ1 }, . . . , {a mÿ1 , aÿ m1 }, {a2 j b: 0 < j < m ÿ 1}, {a2 j1 b: 0 < j < m ÿ 1}.

Conjugacy classes

109

Conjugacy classes of S n We shall later need to know the conjugacy classes of the symmetric group Sn . Our ®rst observation is simple but crucial. 12.13 Proposition Let x be a k-cycle (i1 i2 . . . ik ) in Sn , and let g 2 Sn . Then gÿ1 xg is the k-cycle (i1 g i2 g . . . ik g). Proof Write A {i1 , . . . , ik }. For ir 2 A, ir g( g ÿ1 xg) i r xg i r1 g (or i1 g if r k): Also, for 1 < i < n and i 2 = A, ig( gÿ1 xg) ixg ig: Hence gÿ1 (i1 i2 . . . ik ) g (i1 g i2 g . . . ik g), as required.

j

Now consider an arbitrary permutation x 2 Sn . Write x (a1 : : : ak 1 )(b1 : : : bk 2 ) : : : (c1 : : : ck s ), a product of disjoint cycles, with k1 > k2 > . . . > ks . By Proposition 12.13, for g 2 Sn we have (12.14) g ÿ1 xg g ÿ1 (a1 : : : a k 1 ) gg ÿ1 (b1 : : : b k 2 ) g : : : g ÿ1 (c1 : : : ck s ) g (a1 g : : : a k 1 g)(b1 g : : : b k 2 g) : : : (c1 g : : : c k s g): We call (k1 , . . . , ks ) the cycle-shape of x, and note that x and gÿ1 xg have the same cycle-shape. On the other hand, given any two permutations x, y of the same cycle-shape, say x (a1 : : : a k 1 ) : : : (c1 : : : ck s ), y (a91 : : : a9k 1 ) : : : (c91 : : : c9k s ), (products of disjoint cycles), there a1 ! a91 , . . . , ck s ! c9k s , and so by (12.14), g ÿ1 xg y: We have proved the following result.

exists

g 2 Sn

sending

110


12.15 Theorem For x 2 Sn , the conjugacy class x Sn of x in Sn consists of all permutations in Sn which have the same cycle-shape as x. 12.16 Examples (1) The conjugacy classes of S3 are Class

Cycle-shape

{1} {(1 2), (1 3), (2 3)} {(1 2 3), (1 3 2)}

(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 is f(1 2)(3 4), (1 3)(2 4), (1 4)(2 3)g: (3) There are precisely ®ve conjugacy classes of S4 , with representatives (see De®nition 12.4): 1, (1 2), (1 2 3), (1 2)(3 4), (1 2 3 4): To calculate the sizes of the conjugacy classes, we simply count the number of 2-cycles, 3-cycles, and so on. The number of 2-cycles is equal to the ÿ number of pairs that can be chosen from {1, 2, 3, 4}, which is 42 6. (The notation nr means the binomial coef®cient n!=(r!(n ÿ r)!).) 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). Similarly, there are three elements of cycle-shape (2, 2) and there are six 4-cycles. Thus for G S4 , the conjugacy class representatives g, the conjugacy class sizes | gG | and the centralizer orders |CG ( g)| (obtained using Theorem 12.8) 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

We check our arithmetic by noting that jS4 j 1 6 8 3 6:

(1 2 3 4) 6 4

Conjugacy classes

111

(4) Similarly, the corresponding table for G S5 is Rep. 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

(1 2 3)(4 5) (1 2 3 4 5) 20 24 6 5

Conjugacy classes of A n Given an even permutation x 2 A n , we have seen in Theorem 12.15 that the conjugacy class x S n consists of all permutations in S n which have the same cycle-shape as x. The conjugacy class x A n of x in A n , given by x A n f g ÿ1 xg: g 2 An g, is of course contained in x S n ; however, x A n might not be equal to x S n . For an easy example where equality does not hold, consider x (1 2 3) 2 A3 ; here x A3 fxg, while x S3 {x, x ÿ1 }. The next result determines precisely when x An and x Sn are equal, and what happens when equality fails. 12.17 Proposition Let x 2 An with n . 1. (1) If x commutes with some odd permutation in Sn , then x Sn x An . (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, with representatives x and (1 2)ÿ1 x(1 2). Proof (1) Assume that x commutes with an odd permutation g. Let y 2 x S n , so that y hÿ1 xh for some h 2 Sn . If h is even then y 2 x An ; and if h is odd then gh 2 An and y hÿ1 xh hÿ1 g ÿ1 xgh ( gh)ÿ1 x( gh), so again y 2 x An . Thus x Sn # x An , and so x Sn x An . (2) Assume that x does not commute with any odd permutation. Then CSn (x) CAn (x):

112


Hence by Theorem 12.8, jx An j jAn : CAn (x)j 12jSn : CAn (x)j

(as jAn j 12jSn j)

12jSn : CSn (x)j 12jx Sn j: Next, we observe that fhÿ1 xh: h is oddg ((1 2)ÿ1 x(1 2)) An since every odd permutation has the form (1 2)a for some a 2 An . Now x Sn fhÿ1 xh: h is eveng [ fhÿ1 xh: h is oddg x An [ ((1 2)ÿ1 x(1 2)) An : Since |x An | 12|x Sn |, the conjugacy classes x An and ((1 2)ÿ1 x(1 2)) An must be disjoint and of equal size, as we wished to show. j 12.18 Examples (1) We ®nd the conjugacy classes of A4 . The elements of A4 are the identity, together with the permutations of cycle-shapes (2, 2) and (3). Since (1 2)(3 4) commutes with the odd permutation (1 2), Proposition 12.17 implies that (1 2)(3 4) A4 (1 2)(3 4) S4 f(1 2)(3 4), (1 3)(2 4), (1 4)(2 3)g: However, the 3-cycle (1 2 3) commutes with no odd permutation: for if gÿ1 (1 2 3) g (1 2 3) then (1 2 3) (1 g 2 g 3 g) by Proposition 12.13, so g is 1, (1 2 3) or (1 3 2), an even permutation. Hence by Proposition 12.17, (1 2 3) S4 splits into two conjugacy classes in A4 of size 4, with representatives (1 2 3) and (1 2)ÿ1 (1 2 3)(1 2) (1 3 2). 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

(2) We ®nd the conjugacy classes of A5 . The non-identity even permutations in S5 are those of cycle-shapes (3), (2, 2) and (5). The elements (1 2 3) and (2 3)(4 5) commute with the odd permutation (4 5); but (1 2 3 4 5) commutes with no odd permutation. (Check this by using the argument in (1) above.) Hence by Proposition 12.17, the

Conjugacy classes

113

conjugacy classes of A5 are represented by 1, (1 2 3), (1 2)(3 4), (1 2 3 4 5) and (1 2)ÿ1 (1 2 3 4 5)(1 2) (1 3 4 5 2). Using Proposition 12.17(2), 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. 12.19 Proposition Let H be a subgroup of G. Then H v G if and only if H is a union of conjugacy classes of G. Proof If H is a union of conjugacy classes, then h 2 H, g 2 G ) g ÿ1 hg 2 H, so gÿ1 Hg # H. Thus H v G. Conversely, if H v G then for all h 2 H, g 2 G, we have gÿ1 hg 2 H, and so hG # H. Therefore [ H hG , h2 H

and so H is a union of conjugacy classes of G.

j

12.20 Example We ®nd all the normal subgroups of S4 . Let H v S4 . Then by Proposition 12.19, H is a union of conjugacy classes of S4 . As we saw in Example 12.16(3), these conjugacy classes have sizes 1, 6, 8, 3, 6. Since j Hj divides 24 by Lagrange's Theorem, and 1 2 H, there are just four possibilities: j Hj 1, 1 3, 1 8 3 or 1 6 8 3 6:

114


In the ®rst case H {1}, in the last case H S4 , and in the third case H A4 . In the case where j Hj 1 3, we have H 1 S4 [ (1 2)(3 4) S4 f1, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)g: This is easily checked to be a subgroup of S4 ; we write it as V4 (V stands for `Viergruppe', meaning `four-group'). We have now shown that S4 has exactly four normal subgroups: f1g, S4 , A4 and V4 f1, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)g: 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. Recall from De®nition 9.12 that the centre of CG is Z(CG) fz 2 CG: zr rz for all r 2 CGg: We know that Z(CG) is a subspace of the vector space CG. There is a convenient basis for this subspace which can be described in terms of the conjugacy classes of G. 12.21 De®nition Let C1 , . . . , Cl be the distinct conjugacy classes of G. For 1 < i < l, de®ne X Ci g 2 CG: g2Ci

The elements C1 , . . . , C l of CG are called class sums. 12.22 Proposition The class sums C1 , . . . , C l form a basis of Z(CG). Proof First we show that each C i belongs to Z(CG). Let Ci consist of ÿ1 the r distinct conjugates yÿ1 1 gy1, . . . , y r gyr of an element g, so Ci

r X j1

For all h 2 G, hÿ1 C i h

r X j1

yÿ1 j gyj :

hÿ1 yÿ1 j gyj h:

Conjugacy classes

115

As j runs from 1 to r, the elements hÿ1 yÿ1 j gyj h run through Ci , since ÿ1 ÿ1 ÿ1 hÿ1 yÿ1 y k gyk h , yÿ1 j gyj h h j gyj y k gyk :

Hence r X j1

hÿ1 yÿ1 j gyj h C i ,

and so hÿ1 C i h C i . That is, C i h hC i : Therefore each C i commutes with all h 2 G, hence with all P h2G ë h h 2 CG, and so C i 2 Z(CG). Next, observe that C1 , . . . , C l are linearly independent: for if Pl i1 ë i C i 0 (ë i 2 C), then all ë i 0 as the classes C1 , . . . , Cl are pairwise disjoint by Corollary 12.3. It remains to show that C1 , . . . , C l span Z(CG). Let r P ÿ1 g2G ë g g 2 Z(CG). For h 2 G, we have rh hr, so h rh r. That is, X X ë g hÿ1 gh ë g g: g2G

g2G

So for every h 2 G, the coef®cient ë g of g is equal to the coef®cient ë hÿ1 gh of hÿ1 gh. That is to say, the function g ! ë g is constant on Pl conjugacy classes of G. It follows that r i1 ë i C i where ë i is the j coef®cient ë gi for some gi 2 Ci . This completes the proof. 12.23 Examples (1) From Example 12.16(1), a basis for Z(CS3 ) is 1, (1 2) (1 3) (2 3), (1 2 3) (1 3 2): (2) From (12.12), a basis for Z(CD8 ) is 1, a2 , a a3 , b a2 b, ab a3 b:

Summary of Chapter 12 1. Every group is a union of conjugacy classes, and distinct conjugacy classes are disjoint. 2. For an element x of a group G, the centralizer CG (x) is the set of

116


elements of G which commute with x. It is a subgroup of G, and the number of elements in the conjugacy class x G is equal to |G:CG (x)|. 3. The conjugacy classes of Sn correspond to the cycle-shapes of permutations in Sn . 4. If x 2 An then x Sn x An if and only if x commutes with some odd permutation in Sn . 5. The class sums in CG form a basis for the centre of CG. Exercises for Chapter 12 1. If G is a group and x 2 G, show that CG (x) is a subgroup of G which contains Z(G). 2. Let G be a ®nite group and suppose that g 2 G and z 2 Z(G). Prove that the conjugacy classes gG and ( gz) G have the same size. 3. Let G Sn . (a) Prove that j(1 2) G j 2n and ®nd CG ((1 2)). Verify that your solution satis®es Theorem 12.8. (b) Show that j(1 2 3) G j 23n and j(1 2)(3 4) G j 34n . (c) Now let n 6. Show that j(1 2 3)(4 5 6) G j 40 and j(1 2)(3 4)(5 6) G j 15, and ®nd the sizes of the other conjugacy classes of S6 . (There are 11 conjugacy classes in all.) 4. What are the cycle-shapes of those permutations x 2 A6 for which x A6 6 x S6 ? 5. Show that A5 is a simple group. (Hint: use the method of Example 12.20.) 6. Find the conjugacy classes of the quaternion group Q8. Give a basis of the centre of the group algebra CQ8. 7. Let p be a prime number, and let n be a positive integer. Suppose that G is a group of order pn . (a) Use the Class Equation 12.10 to show that Z(G) 6 {1}. (b) Suppose that n > 3 and that | Z(G)| p. Prove that G has a conjugacy class of size p.

13 Characters

Suppose that r: G ! GL(n, C) is a representation of the ®nite group G. With each n 3 n matrix gr ( g 2 G) we associate the complex number given by adding all the diagonal entries of the matrix, and call this number ÷( g). The function ÷: G ! C is called the character of the representation r. Characters of representations have many remarkable properties, and they are the fundamental tools for performing calculations in representation theory. For example, we shall show later that two representations have the same character if and only if they are equivalent. Moreover, basic problems, such as deciding whether or not a given representation is irreducible, can be resolved by doing some easy arithmetic with the character of the representation. These facts are surprising, since from the de®nition of a representation r: G ! GL(n, C), it appears that we must keep track of all the n2 entries in each matrix gr, whereas the character records just one number for each matrix. The theory of characters will occupy a considerable portion of the rest of the book. In this chapter we present some basic properties and examples. The trace of a matrix 13.1 De®nition If A (aij ) is an n 3 n matrix, then the trace of A, written tr A, is given by tr A

n X

aii :

i1

That is, the trace of A is the sum of the diagonal entries of A. 117

118


13.2 Proposition Let A (aij ) and B (bij ) be n 3 n matrices. Then tr (A B) tr A tr B, and tr (AB) tr (BA): Moreover, if T is an invertible n 3 n matrix, then tr (T ÿ1 AT ) tr A:

Proof The ii-entry of A B is aii bii , and the ii-entry of AB is Pn j1 aij bji. Therefore tr (A B)

n X

(aii bii )

i1

n X

aii

n X

i1

bii tr A tr B,

i1

and tr (AB)

n X n X i1

aij bji

n X n X

j1

j1

bji aij tr (BA):

i1

For the last part, tr (T ÿ1 AT ) tr ((T ÿ1 A)T ) tr (T (T ÿ1 A))

(by the second part )

tr A:

j

Notice that, unlike the determinant function, the trace function is not multiplicative; that is, tr (AB) need not equal (tr A)(tr B).

Characters 13.3 De®nition Suppose that V is a CG-module with a basis B . Then the character of V is the function ÷: G ! C de®ned by ÷( g) tr [ g]B

( g 2 G):

The character of V does not depend on the basis B , since if B and B 9 are bases of V, then [ g]B 9 T ÿ1 [ g]B T

Characters

119

for some invertible matrix T (see (2.24)), and so by Proposition 13.2, tr [ g]B 9 tr [ g]B

for all g 2 G:

Naturally enough, we de®ne the character of a representation r: G ! GL(n, C) to be the character ÷ of the corresponding CGmodule C n , namely ÷( g) tr ( gr)

( g 2 G):

13.4 De®nition We say that ÷ is a character of G if ÷ is the character of some CGmodule. Further, ÷ is an irreducible character of G if ÷ is the character of an irreducible CG-module; and ÷ is reducible if it is the character of a reducible CG-module. You will have noticed that we are writing characters as functions on the left. That is, we write ÷( g) and not g÷. 13.5 Proposition (1) Isomorphic CG-modules have the same character. (2) If x and y are conjugate elements of the group G, then ÷(x) ÷( y) for all characters ÷ of G. Proof (1) Suppose that V and W are isomorphic CG-modules. Then by (7.7), there are a basis B 1 of V and a basis B 2 of W such that [ g]B 1 [ g]B 2

for all g 2 G:

Consequently tr [ g]B 1 tr [ g]B 2 for all g 2 G, and so V and W have the same character. (2) Assume that x and y are conjugate elements of G, so that x gÿ1 yg for some g 2 G. Let V be a CG-module, and let B be a basis of V. Then [x]B [ g ÿ1 yg]B [ g]ÿ1 B [ y]B [ g]B : Hence by Proposition 13.2, we have tr [x]B tr [ y]B . Therefore ÷(x) ÷( y), where ÷ is the character of V. j The result corresponding to Proposition 13.5(1) for representations is that equivalent representations have the same character.

120


Later, we shall prove an astonishing converse of Proposition 13.5(1): if two CG-modules have the same character, then they are isomorphic. 13.6 Examples (1) Let G D8 ka, b: a4 b2 1, bÿ1 ab aÿ1 l, and let r: G ! GL(2, C) be the representation for which 1 0 0 1 , br ar 0 ÿ1 ÿ1 0 (see Example 3.2(1)). Let ÷ be the character of this representation. The following table records g, gr and ÷( g) as g runs through G. (We obtain ÷( g) by adding the two entries on the diagonal of gr.) g

1

gr

1 0 0 1

÷( g)

2

g

a 0 1 ÿ1 0

a2 ÿ1 0 0 ÿ1

0

b

gr

1 0 0 ÿ1

÷( g)

0

a3 0 ÿ1 1 0

ÿ2

ab 0 ÿ1

ÿ1 0

0

a2 b ÿ1 0 0 1

0

a3 b 0 1

0

1 0

0

(2) Let G S3 , and take V to be the 3-dimensional permutation module for G over C (see De®nition 4.10). Let B be the natural basis of V; thus B is the basis v1 , v2 , v3 , where v i g v ig for 1 < i < 3 and all g 2 G. The matrices [ g]B ( g 2 G) are given by Exercise 4.1. We record these matrices, together with the character ÷ of V. g [ g]B ÷( g)

0

1 @0 0

1

1

0 0 1 0A 0 1 3

0

0 @1 0

(1 2)

1

1 0 0 0A 0 1 1

0

(1 3)

0 0 @0 1 1 0 1

1 1 0A 0

Characters g

(2 3)

0

1

[ g]B

1 0 0 @0 0 1A 0 1 0

÷( g)

1

0

121

(1 2 3)

0 1 @0 0 1 0

1

0 1A 0

0

(1 3 2)

0 @1 0

0

1 0 1 0 0A 1 0 0

(3) Let G C3 ha: a3 1 i. By Theorem 9.8, G has just three irreducible characters ÷1 , ÷2 , ÷3 , with values g

1

a

a2

÷1 ( g) ÷2 ( g) ÷3 ( g)

1 1 1

1 ù ù2

1 ù2 ù

where ù e2ði=3 . (4) Let G D6 ka, b: a3 b2 1, bÿ1 ab aÿ1 l (so G S3 ). In Example 10.8(2), we found a complete set of non-isomorphic irreducible CG-modules U1, U2, U3. Thus if ÷ i is the character of Ui for 1 < i < 3, then the irreducible characters of G are ÷1 , ÷2 and ÷3 . The values of these characters on the elements of G can be calculated from the corresponding representations r1 , r2 , r3 given in Example 10.8(2), and they are as follows: g

1

a

a2

b

ab

a2 b

÷1 ( g) ÷2 ( g) ÷3 ( g)

1 1 2

1 1 ÿ1

1 1 ÿ1

1 ÿ1 0

1 ÿ1 0

1 ÿ1 0

Notice that in all the above examples, the characters given take few distinct values. This re¯ects the fact that by Proposition 13.5(2), every character is constant on conjugacy classes of G. Moreover, 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. Nevertheless, the character encapsulates a great deal of information about the representation. This will become clear as the theory of characters develops.

122


13.7 De®nition If ÷ is the character of the CG-module V, then the dimension of V is called the degree of ÷. 13.8 Examples (1) In Example 13.6(1) we gave a character of D8 of degree 2; in 13.6(2) we gave a character of S3 of degree 3; and in 13.6(4) we saw that the irreducible characters of D6 have degrees 1, 1 and 2. (2) If V is any 1-dimensional CG-module, then for each g 2 G there is a complex number ë g such that v g ë gv

for all v 2 V :

The character ÷ of V is given by ÷( g) ë g

( g 2 G)

and ÷ has degree 1. Characters of degree 1 are called linear characters; they are, of course, irreducible characters. Observe that Theorem 9.8 gives all the irreducible characters of ®nite abelian groups; in particular, they are all linear characters. Every linear character of G is a homomorphism from G to the multiplicative group of non-zero complex numbers. In fact, these are the only non-zero characters of G which are homomorphisms (see Exercise 13.4). (3) The character of the trivial CG-module (see De®nition 4.8(1)) is a linear character, called the trivial character of G. We denote it by 1 G . Thus 1 G : g ! 1 for all g 2 G: Given any group G, we therefore know at least one of the irreducible characters of G, namely the trivial character. Finding all the irreducible characters is usually dif®cult. The values of a character The next result gives information about the complex numbers ÷( g), where ÷ is a character of G and g 2 G. 13.9 Proposition Let ÷ be the character of a CG-module V. Suppose that g 2 G and g has order m. Then

Characters (1) (2) (3) (4)

123

÷(1) dim V; ÷( g) is a sum of mth roots of unity; ÷( gÿ1 ) ÷( g); ÷( g) is a real number if g is conjugate to gÿ1 .

Proof (1) Let n dim V, and let B be a basis of V. Then the matrix [1]B of the identity element 1 relative to B is equal to In , the n 3 n identity matrix. Consequently ÷(1) tr [1]B tr I n n, and so ÷(1) dim V. (2) By Proposition 9.11 there is a basis B of V such that 0 1 ù1 0 B C .. [ g]B @ A . ùn 0 where each ù i is an mth root of unity. Therefore ÷( g) ù1 : : : ù n , a sum of mth roots of unity. (3) We have

0

B [ gÿ1 ]B @

ùÿ1 1

0

..

0 .

1 C A

ùÿ1 n

ÿ1 and so ÷( gÿ1 ) ùÿ1 1 . . . ù n . Every complex mth root of unity ù ÿ1 satis®es ù ù, since for all real W,

(eiW )ÿ1 eÿiW , which is the complex conjugate of eiW . Therefore ÷( g ÿ1 ) ù1 : : : ù n ÷( g): (4) If g is conjugate to gÿ1 then ÷( g) ÷( gÿ1 ) by Proposition 13.5(2). Also ÷( gÿ1 ) ÷( g) by (3), and so ÷( g) ÷( g); that is, ÷( g) is j real. When the element g of G has order 2, we can be much more speci®c about the possibilities for ÷( g):

124


13.10 Corollary Let ÷ be a character of G, and let g be an element of order 2 in G. Then ÷( g) is an integer, and ÷( g) ÷(1) mod 2: Proof By Proposition 13.9, we have ÷( g) ù1 : : : ù n , where n ÷(1) and each ù i is a square root of unity. Then each ù i is 1 or ÿ1. Suppose r of them are 1, and s are ÿ1, so that ÷( g) r ÿ s, and ÷(1) r s: Certainly then, ÷( g) 2 Z, and since r ÿ s r s ÿ 2s r s mod 2, we have ÷( g) ÷(1) mod 2. j Our next result gives the ®rst inkling of the importance of characters, showing that we can determine the kernel of a representation just from knowledge of its character. 13.11 Theorem Let r: G ! GL(n, C) be a representation of G, and let ÷ be the character of r. (1) For g 2 G, j÷( g)j ÷(1) , gr ëI n

for some ë 2 C:

(2) Ker r { g 2 G: ÷( g) ÷(1)}. Proof (1) Let g 2 G, and suppose that g has order m. If gr ëIn with ë 2 C, then ë is an mth root of unity, and ÷( g) në, so |÷( g)| n ÷(1). Conversely, suppose that |÷( g)| ÷(1). By Proposition 9.11, there is a basis B of C n such that 0 1 ù1 0 B C .. [ g]B @ A .

0

ùn

where each ù i is an mth root of unity. Then (13:12)

j÷( g)j jù1 : : : ù n j ÷(1) n:

Characters

125

Note now that for any complex numbers z1 , . . . , zn , we have jz1 : : : zn j < jz1 j : : : jzn j, with equality if and only if the arguments of z1 , . . . , zn are all equal. (To see this, consider the picture

in the Argand diagram.) Since |ù i | 1 for all i, we deduce from (13.12) that ù i ù j for all i, j. Thus 0 1 ù1 0 B C .. [ g]B @ A ù1 I n : .

0

ù1

Hence for all bases B 9 of C n we have [ g]B 9 ù1 In , and so gr ù1 In . This completes the proof of (1). (2) If g 2 Ker r then gr In , and so ÷( g) n ÷(1). Conversely, suppose that ÷( g) ÷(1). Then by (1), we have gr ëIn for some ë 2 C. This implies that ÷( g) ë÷(1), whence ë 1. Therefore gr In , and so g 2 Ker r. Part (2) follows. j Motivated by Theorem 13.11(2), we de®ne the kernel of a character as follows. 13.13 De®nition If ÷ is a character of G, then the kernel of ÷, written Ker ÷, is de®ned by Ker ÷ f g 2 G: ÷( g) ÷(1)g: By Theorem 13.11(2), if r is a representation of G with character ÷, then Ker r Ker ÷. In particular, Ker ÷ v G. We call ÷ a faithful character if Ker ÷ {1}. 13.14 Examples (1) According to Example 13.6(4), the irreducible characters of the group G D6 ka, b: a3 b2 1, bÿ1 ab aÿ1 l are ÷1 , ÷2 , ÷3 , with the following values:

126


g

1

a

a2

b

ab

a2 b

÷1 ( g) ÷2 ( g) ÷3 ( g)

1 1 2

1 1 ÿ1

1 1 ÿ1

1 ÿ1 0

1 ÿ1 0

1 ÿ1 0

Then Ker ÷1 G, Ker ÷2 kal and Ker ÷3 {1}. In particular, ÷3 is a faithful irreducible character of D6. (2) Let G D8 ka, b: a4 b2 1, bÿ1 ab aÿ1 l, and let ÷ be the character of G given in Example 13.6(1): g

1

a

a2

a3

b

ab

a2 b

a3 b

÷( g)

2

0

ÿ2

0

0

0

0

0

Then Ker ÷ {1}, so ÷ is a faithful character. And since |÷(a2 )| |ÿ2| ÷(1), Theorem 13.11(1) implies that if r: G ! GL(2, C) is a representation with character ÷, then a2 r ÿI. We next prove a result which is sometimes useful for constructing a new character from a given one. For a character ÷ of G, de®ne ÷: G ! C by ÷( g) ÷( g)

( g 2 G):

Thus the values of ÷ are the complex conjugates of the values of ÷. 13.15 Proposition Let ÷ be a character of G. Then ÷ is a character of G. If ÷ is irreducible, then so is ÷. Proof Suppose that ÷ is the character of a representation r: G ! GL(n, C). Thus ÷( g) tr ( gr) ( g 2 G): If A (aij ) is an n 3 n matrix over C, then we de®ne A to be the n 3 n matrix (a ij ). Observe that if A (aij ) and B (bij ) are n 3 n matrices over C, then (13:16)

(AB) A B,

Characters

127

since the ij-entry of A B is n X

a ik b kj ,

k1

Pn which is equal to the complex conjugate of k1 aik bkj , the ij-entry of AB. It follows from (13.16) that the function r: G ! GL(n, C) de®ned by gr ( gr)

( g 2 G)

is a representation of G. Since tr ( gr) tr ( gr) tr ( gr) ÷( g) ( g 2 G), the character of the representation r is ÷. It is clear that if r is reducible then r is reducible. Hence ÷ is irreducible if and only if ÷ is irreducible. j The regular character 13.17 De®nition The regular character of G is the character of the regular CG-module. We write the regular character as ÷reg . In Theorem 13.19, we shall express the regular character in terms of the irreducible characters of G. First we need a preliminary result. 13.18 Proposition Let V be a CG-module, and suppose that V U1 : : : Ur , a direct sum of irreducible CG-modules Ui. Then the character of V is equal to the sum of the characters of the CG-modules U1, . . . , Ur. Proof This is immediate from (7.10).

j

13.19 Theorem Let V1 , : : : , V k be a complete set of non-isomorphic irreducible CG-modules (see De®nition 11.11), and for i 1, . . . , k let ÷ i be the character of Vi and di ÷ i (1). Then ÷reg d 1 ÷1 : : : d k ÷ k :

128


Proof By Theorem 11.9, CG (V1 : : : V1 ) (V2 : : : V2 ) : : : (Vk : : : Vk ), where for each i there are di factors V i . Now the result follows from Proposition 13.18. j The values of ÷reg on the elements of G are easily described, and are given in the next result. 13.20 Proposition If ÷reg is the regular character of G, then ÷reg (1) jGj, and ÷reg ( g) 0 if g 6 1: Proof Let g1 , . . . , gn be the elements of G, and let B be the basis g1 , . . . , gn of CG. By Proposition 13.9(1), ÷reg (1) dim CG |G|. Now let g 2 G with g 6 1. Then for 1 < i < n, we have gi g gj for some j with j 6 i. Therefore the ith row of the matrix [ g]B has zeros in every place except column j; in particular, the ii-entry is zero for all i. It follows that ÷reg ( g) tr [ g]B 0:

j

13.21 Example We illustrate Theorem 13.19 and Proposition 13.20 for the group G D6 . By Example 13.6(4), the irreducible characters of G are ÷1 , ÷2 , ÷ 3 : g

1

a

a2

b

ab

a2 b

÷1 ( g) ÷2 ( g) ÷3 ( g)

1 1 2

1 1 ÿ1

1 1 ÿ1

1 ÿ1 0

1 ÿ1 0

1 ÿ1 0

We calculate ÷1 ÷2 2÷3 : (÷1 ÷2 2÷3 )( g)

6

0

0

0

0

0

Characters

129

This is the regular character of G, by Theorem 13.19; and it takes the value |G| on 1, and the value 0 on all non-identity elements of G, illustrating Proposition 13.20.

Permutation characters In the case where G is a subgroup of the symmetric group Sn , there is an easy construction using the permutation module which produces a character of degree n, and we now describe this. Suppose that G is a subgroup of Sn , so that G is a group of permutations of {1, . . . , n}. The permutation module V for G over C has basis v1 , . . . , v n , where for all g 2 G, v i g v ig

(1 < i < n)

(see De®nition 4.10). Let B denote the basis v1 , . . . , v n . Then the iientry in the matrix [ g]B is 0 if ig 6 i, and is 1 if ig i. Therefore the character ð of the permutation module V is given by ð( g) (the number of i such that ig i): For g 2 G, let fix ( g) fi: 1 < i < n and ig ig: Then (13:22)

ð( g) jfix( g)j

( g 2 G):

We call ð the permutation character of G. 13.23 Example Let G S4 . Then by Example 12.16(3), G has ®ve conjugacy classes, with representatives 1, (1 2), (1 2 3), (1 2)(3 4), (1 2 3 4): The permutation character ð takes the values gi

1

(1 2)

(1 2 3)

(1 2)(3 4)

(1 2 3 4)

ð( gi )

4

2

1

0

0

130


13.24 Proposition Let G be a subgroup of Sn . Then the function í: G ! C de®ned by í( g) jfix ( g)j ÿ 1

( g 2 G)

is a character of G. Proof Let v1 , . . . , v n be the basis of the permutation module V as above, and let u v1 : : : v n , and U sp (u): Observe that ug u for all g 2 G, so U is a CG-submodule of V. Indeed, U is isomorphic to the trivial CG-module, so the character of U is the trivial character 1 G (see Example 13.8(3)). By Maschke's Theorem 8.1, there is a CG-submodule W of V such that V U W: Let í be the character of W. Then ð 1 G í, so |®x( g)| 1 í( g) for all g 2 G, and therefore í( g) jfix( g)j ÿ 1 ( g 2 G):

j

13.25 Example Let G A4 , a subgroup of S4 . By Example 12.18(1), the conjugacy classes of G are represented by 1, (1 2)(3 4), (1 2 3), (1 3 2): The values of the character í of G are gi

1

(1 2)(3 4)

(1 2 3)

(1 3 2)

í( gi )

3

ÿ1

0

0

Summary of Chapter 13 1. A character is obtained from a representation by taking the trace of each matrix. 2. Characters are constant on conjugacy classes.

Characters

131

3. Isomorphic CG-modules have the same character. 4. For all characters ÷ of G, and all g 2 G, the complex number ÷( g) is a sum of roots of unity, and ÷( gÿ1 ) ÷( g). 5. The character of a representation determines the kernel of the representation. 6. The regular character ÷reg of G takes the value |G| on the identity and the value 0 on all other elements of G. 7. If G is a subgroup of Sn , then the function í which is given by í( g) jfix( g)j ÿ 1

( g 2 G)

is a character of G. Exercises for Chapter 13 1. Let G D12 ka, b: a6 b2 1, bÿ1 ab aÿ1 l, and let r1 , r2 be the representations of G for which ù 0 0 1 ar1 , br1 (where ù e2ði=3 ) 0 ùÿ1 1 0 and

ar2

ÿ1 0

0 1 , br2 1 0

0 : ÿ1

Find the characters of r1 and r2 . Find also Ker r1 and Ker r2 ; check that your answers are consistent with Theorem 13.11. 2. Find all the irreducible characters of C4 . Write the regular character of C4 as a linear combination of these. 3. Let ÷ be the character of the 7-dimensional permutation module for S7 . Find ÷(x) for x (1 2) and for x (1 6)(2 3 5). 4. Prove that the only non-zero characters of G which are homomorphisms are the linear characters. 5. Assume that ÷ is an irreducible character of G. Suppose that z 2 Z(G) and that z has order m. Prove that there exists an mth root of unity ë 2 C such that for all g 2 G, ÷(zg) ë÷( g): 6. Prove that if ÷ is a faithful irreducible character of the group G, then Z(G) { g 2 G: |÷( g)| ÷(1)}.

132


7. Let r be a representation of the group G over C. (a) Show that ä: g ! det ( gr) ( g 2 G) is a linear character of G. (b) Prove that G/Ker ä is abelian. (c) Assume that ä( g) ÿ1 for some g 2 G. Show that G has a normal subgroup of index 2. 8. Let g be a group of order 2k, where k is an odd integer. By considering the regular representation of G, show that G has a normal subgroup of index 2. 9. Let ÷ be a character of a group G, and let g be an element of order 2 in G. Show that either (1) ÷( g) ÷(1) mod 4, or (2) G has a normal subgroup of index 2. (Compare Corollary 13.10. Hint: use Exercise 7.) 10. Prove that if x is a non-identity element of the group G, then ÷(x) 6 ÷(1) for some irreducible character ÷ of G.

14 Inner products of characters

We establish some signi®cant properties of characters in this chapter, and in particular we prove the striking result (Theorem 14.21) that if two CG-modules have the same character then they are isomorphic. Also, we describe a method for decomposing a given CG-module as a direct sum of CG-submodules, using characters. The proofs rely on an inner product involving the characters of a group, and we describe this ®rst.

Inner products The characters of a ®nite group G are certain functions from G to C. The set of all functions from G to C forms a vector space over C, if we adopt the natural rules for adding functions and multiplying functions by complex numbers. That is, if W, ö are functions from G to C, and ë 2 C, then we de®ne W ö: G ! C by (W ö)( g) W( g) ö( g)

( g 2 G)

and we de®ne ëW: G ! C by ëW( g) ë(W( g))

( g 2 G):

(We write these functions on the left to agree with our notation for characters.) 14.1 Example Let G C3 ka: a3 1l, and suppose that W: G ! C and ö: G ! C are given by 133

134


W ö

1

a

a2

2 1

i 1

ÿ1 1

This means that W(1) 2, W(a) i, W(a2 ) ÿ1 and ö(1) ö(a) ö(a2 ) 1. Then W ö and 3W are given by

W ö 3W

1

a

a2

3 6

1i 3i

0 ÿ3

We shall often think of functions from G to C as row vectors, as in this example. 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. The de®nition of an inner product on a vector space over C runs as follows. With every ordered pair of vectors W, ö in the vector space, there is associated a complex number kW, öl which satis®es the following conditions: (14.2) (a) kW, öl hö, Wi for all W, ö; (b) kë1 W1 ë2 W2 , öl ë1 kW1 , öl ë2 kW2 , öl for all ë1 , ë2 2 C and all vectors W1 , W2 , ö; (c) kW, Wl . 0 if W 6 0. Notice that condition (a) implies that kW, Wl is always real, and that conditions (a) and (b) give hö, ë1 è1 ë2 W2 i ë1 hö, W1 i ë2 hö, W2 i for all ë1 , ë2 2 C and all vectors W1 , W2 , ö. We now introduce an inner product on the vector space of all functions from G to C. This will be of basic importance in our study of characters. 14.3 De®nition Suppose that W and ö are functions from G to C. De®ne 1 X hW, öi W( g)ö( g): jGj g2G

Inner products of characters

135

It is transparent that the conditions of (14.2) hold, so k , l is an inner product on the vector space of functions from G to C. 14.4 Example As in Example 14.1, suppose that G C3 ka: a3 1l and that W and ö are given by

W ö

1

a

a2

2 1

i 1

ÿ1 1

Then hè, öi 13(2 . 1 i . 1 ÿ 1 . 1) 13(1 i), hè, èi 13(2 . 2 i . i (ÿ1) . (ÿ1)) 2, hö, öi 13(1 . 1 1 . 1 1 . 1) 1:

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. 14.5 Proposition Assume that G has exactly l conjugacy classes, with representatives g1 , . . . , gl . Let ÷ and ø be characters of G. 1 X (1) h÷, øi hø, ÷i ÷( g)ø( g ÿ1 ), and this is a real number: jGj g2G (2) h÷, øi

l X ÷( g i )ø( g i ) i1

jCG ( g i )j

:

Proof (1) We have ø( g) ø( gÿ1 ) for all g 2 G, by Proposition 13.9(3). Therefore 1 X h÷, øi ÷( g)ø( g ÿ1 ): jGj g2G

136


Since { gÿ1 : g 2 G} G, we also have 1 X h÷, øi ÷( g ÿ1 )ø( g) hø, ÷i: jGj g2G Since kø, ÷l h÷, øi, it follows that h÷, øi is real. (We shall prove later that h÷, øi is, in fact, an integer.) (2) Recall that g G i denotes the conjugacy class of G which contains gi . Since characters are constant on conjugacy classes, X ÷( g)ø( g) j g G i j÷( g i )ø( g i ): g2 g G i

Now G

l [ i1

G gG i and j g i j jGj=jCG ( g i )j,

by Corollary 12.3 and Theorem 12.8. Hence h÷, øi

l X 1 X 1 X ÷( g)ø( g) ÷( g)ø( g) jGj g2G jGj i1 G g2 g i

l X i1

l X i1

j gG i j ÷( g i )ø( g i ) jGj 1 ÷( g i )ø( g i ): jCG ( g i )j

j

14.6 Example The alternating group A4 has four conjugacy classes, with representatives g 1 1,

g 2 (1 2)(3 4),

g 3 (1 2 3),

g 4 (1 3 2)

(see Example 12.18(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 )|

g1 12

g2 4

g3 3

g4 3

÷ ø

1 4

1 0

ù ù2

ù2 ù


137

(where ù e2ði=3 ). Using part (2) of Proposition 14.5, we have 1 . 4 1 . 0 ù . ù2 ù2 . ù 0, 12 4 3 3 4 . 4 0 . 0 ù 2 . ù2 ù . ù hø, øi 2: 12 4 3 3 h÷, øi

We advise you to check also that k÷, ÷l 1, and to ®nd the inner products of ÷ and ø with the trivial character (which takes the value 1 on all elements of A4 ). We are now going to pave the way to proving the key fact (Theorem 14.12) that the irreducible characters of G form an orthonormal set of vectors in the vector space of functions from G to C; that is, for distinct irreducible characters ÷ and ø of G, we have h÷, ÷i 1 and h÷, øi 0. Recall from Chapter 10 that the regular CG-module is a direct sum of irreducible CG-submodules, say CG U 1 : : : U r , and that every irreducible CG-module is isomorphic to one of the CGmodules U1, . . . , Ur. There are several ways of choosing CG-submodules W1 and W2 of CG such that CG W1 W2 and W1 and W2 have no common composition factor (see De®nition 10.4). For example, we may take W1 to be the sum of those irreducible CG-submodules Ui which are isomorphic to a given irreducible CG-module, and then let W2 be the sum of the remaining CG-modules U i . We shall investigate some consequences of writing CG like this; therefore, we temporarily adopt the following Hypothesis: 14.7 Hypothesis Let CG W 1 W 2 , where W1 and W2 are CG-submodules which have no common composition factor. Write 1 e1 e2 where e1 2 W1 and e2 2 W2. Among other results, we shall derive a formula for e1 in terms of the character of W 1 . We ®rst look at the effect of applying the elements e1 and e2 of CG to W1 and W 2 .

138


14.8 Proposition For all w1 2 W1 and w2 2 W2, we have w1 e1 w1 ,

w2 e1 0,

w1 e2 0,

w2 e 2 w2 :

Proof If w1 2 W1 then the function w2 ! w1 w2 (w2 2 W2 ) is clearly a CG-homomorphism from W2 to W 1 . But W2 and W1 have no common composition factor, so every CG-homomorphism from W2 to W1 is zero, by Proposition 11.3. Therefore w1 w2 0 for all w1 2 W 1 , w2 2 W 2 . Similarly w2 w1 0. In particular, w1 e2 w2 e1 0. Now w1 w1 1 w1 (e1 e2 ) w1 e1 , and w2 w2 1 w2 (e1 e2 ) w2 e2 , and this completes the proof.

j

14.9 Corollary For the elements e1 and e2 of CG which appear in Hypothesis 14.7, we have e21 e1 , e22 e2 and e1 e2 e2 e1 0: Proof In Proposition 14.8, take w1 e1 and w2 e2 .

j

Next, we evaluate e1 . 14.10. Proposition Let ÷ be the character of the CG-module W1 which appears in Hypothesis 14.7. Then 1 X e1 ÷( g ÿ1 ) g: jGj g2G Proof Let x 2 G. The function W: w ! we1 x ÿ1

(w 2 CG)

is an endomorphism of CG. We shall calculate the trace of W in two ways.


139

First, for w1 2 W1 and w2 2 W2 we have, in view of Proposition 14.8, w1 W w1 e1 x ÿ1 w1 x ÿ1 , w2 W w2 e1 x ÿ1 0: Thus W acts on W1 by w1 ! w1 x ÿ1 and on W2 by w2 ! 0. By the de®nition of the character ÷ of W 1 , the endomorphism w1 ! w1 x ÿ1 of W1 has trace equal to ÷(x ÿ1 ), and of course the endomorphism w2 ! 0 of W2 has trace 0. Therefore tr W ÷(x ÿ1 ): Secondly, e1 2 CG, so e1

X

ëg g

g2G

for some ë g 2 C. By Proposition 13.20, the endomorphism w ! wgx ÿ1 (w 2 CG) of CG has trace 0 if g 6 x and has trace |G| if g x. P Hence, as W: w ! w g2G ë g gx ÿ1 , we have tr W ë x jGj: Comparing our two expressions for tr W, we see that for all x 2 G, ë x ÷(x ÿ1 )=jGj: Therefore e1

1 X ÷( g ÿ1 ) g: jGj g2G

j

14.11 Corollary Let ÷ be the character of the CG-module W1 which appears in Hypothesis 14.7. Then h÷, ÷i ÷(1):

Proof Using the de®nition 6.3 of the multiplication in CG, we deduce from Proposition 14.10 that the coef®cient of 1 in e21 is 1 X 1 h÷, ÷i: ÷( g ÿ1 )÷( g) 2 jGj g2G jGj

140


On the other hand, we know from Corollary 14.9 that e21 e1 , and the coef®cient of 1 in e1 is ÷(1)/|G|. Hence k÷, ÷l ÷(1), as required. j

We can now prove the main theorem concerning the inner product k , l. 14.12 Theorem Let U and V be non-isomorphic irreducible CG-modules, with characters ÷ and ø, respectively. Then h÷, ÷i 1, and h÷, øi 0:

Proof Recall from Theorem 11.9 that CG is a direct sum of irreducible CG-submodules, say CG U1 : : : Ur , where the number of CG-submodules Ui which are isomorphic to U is dim U. Let m dim U, and de®ne W to be the sum of the m irreducible CG-submodules Ui which are isomorphic to U; let X be the sum of the remaining CG-submodules Ui. Then CG W X : Moreover, every composition factor of W is isomorphic to U, and no composition factor of X is isomorphic to U. In particular, W and X have no common composition factor. The character of W is m÷, since W is the direct sum of m CG-submodules, each of which has character ÷. We now apply Corollary 14.11 to the character of W, and obtain hm÷, m÷i m÷(1): As ÷(1) dim U m, this yields h÷, ÷i 1. Next, 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. Then CG Y Z,


141

and Y and Z have no common composition factor. The character of Y is m÷ nø, where n dim V. By Corollary 14.11, m÷(1) nø(1) hm÷ nø, m÷ nøi m2 h÷, ÷i n2 hø, øi mn(h÷, øi hø, ÷i): Now h÷, ÷i hø, øi 1, by the part of the theorem which we have already proved; and ÷(1) m, ø(1) n. Therefore h÷, øi hø, ÷i 0: By Proposition 14.5(1), k÷, øl kø, ÷l, and hence k÷, øl 0.

j

Applications of Theorem 14.12 Let G be a ®nite group, and let V1 , : : : , V k be a complete set of nonisomorphic irreducible CG-modules (see De®nition 11.11). If ÷ i is the character of Vi (1 < i < k), then by Theorem 14.12, we have (14:13)

h÷ i , ÷ j i ä ij

for all i, j,

where ä ij is the Kronecker delta function (that is, ä ij is 1 if i j and is 0 if i 6 j). In particular, this implies that the irreducible characters ÷1 , . . . , ÷ k are all distinct. Now let V be a CG-module. By Theorem 8.7, V is equal to a direct sum of irreducible CG-submodules. Each of these is isomorphic to some V i , so there are non-negative integers d1, . . . , dk such that (14:14)

V (V1 : : : V1 ) (V2 : : : V2 ) : : : (Vk : : : Vk ), where for each i, there are d i factors Vi :

Therefore the character ø of V is given by (14:15)

ø d 1 ÷1 : : : d k ÷ k :

Using (14.13), we obtain from this (14:16)

hø, ÷ i i h÷ i , øi d i for 1 < i < k, and hø, øi d 21 : : : d 2k :

Summarizing, we have

142


14.17 Theorem Let ÷1 , . . . , ÷ k be the irreducible characters of G. If ø is any character of G, then ø d 1 ÷1 : : : d k ÷ k for some non-negative integers d1, . . . , dk . Moreover, d i hø, ÷ i i hø, øi

k X i1

for 1 < i < k, and d 2i :

14.18 Example Recall from Example 13.6(4) that the irreducible characters of S3 D6 are ÷1 , ÷2 , ÷3 , taking the following values on the conjugacy class representatives 1, (1 2), (1 2 3): Now let ø be the character of the 3-dimensional permutation module gi |CS3 ( g i )|

1 6

(1 2) 2

(1 2 3) 3

÷1 ÷2 ÷3

1 1 2

1 ÿ1 0

1 1 ÿ1

for S3 . By Example 13.6(2), we know that ø(1) 3, ø(1 2) 1, ø(1 2 3) 0: Therefore, by Proposition 14.5(2), hø, ÷1 i

3.1 1.1 0 1: 6 2

Similarly, kø, ÷2 l 0 and kø, ÷3 l 1. Thus by Theorem 14.17, ø ÷1 ÷3 : (This can of course be checked immediately by comparing the values of ø and ÷1 ÷3 on each conjugacy class representative.) A more substantial calculation along these lines is given in Example 15.7.


143

We shall see many more applications of the important Theorem 14.17. 14.19 De®nition Suppose that ø is a character of G, and that ÷ is an irreducible character of G. We say that ÷ is a constituent of ø if kø, ÷l 6 0. Thus, the constituents of ø are the irreducible characters ÷ i of G for which the integer di in the expression ø d1 ÷1 . . . dk ÷ k is non-zero. The next result is another signi®cant consequence of Theorem 14.12. It gives us a quick and effective method of determining whether or not a given CG-module is irreducible. 14.20 Theorem Let V be a CG-module with character ø. Then V is irreducible if and only if kø, øl 1. Proof If V is irreducible then kø, øl 1 by Theorem 14.12. Conversely, assume that kø, øl 1. We have ø d 1 ÷1 : : : d k ÷ k for some non-negative integers di, and by (14.16), 1 hø, øi d 21 : : : d 2k : It follows that one of the integers di is 1 and the rest are zero. Then by (14.14), V Vi for some i, and so V is irreducible. j We are now in a position to prove the remarkable result that à CGmodule is determined by its character'. It is this fact which motivates our study of characters in much of the rest of the book, for it means that many questions about CG-modules can be answered using character theory. 14.21 Theorem Suppose that V and W are CG-modules, with characters ÷ and ø, respectively. Then V and W are isomorphic if and only if ÷ ø. Proof In Proposition 13.5 we proved the elementary fact that if V W then ÷ ø. It is the converse which is the substantial part of this theorem.

144


Thus, suppose that ÷ ø. Again let V1 , : : : , Vk be a complete set of non-isomorphic irreducible CG-modules with characters ÷1 , . . . , ÷ k . We know by (14.14) that there are non-negative integers ci , di (1 < i < k) such that V (V1 : : : V1 ) (V2 : : : V2 ) : : : (Vk : : : Vk ) with ci factors Vi for each i, and W (V1 : : : V1 ) (V2 : : : V2 ) : : : (Vk : : : Vk ) with di factors Vi for each i. By (14.16), ci h÷, ÷ i i, d i hø, ÷ i i (1 < i < k): Since ÷ ø, it follows that ci di for all i, and hence V W.

j

14.22 Example Let G C3 ka: a3 1l, and let r1 , r2 , r3 , r4 be the representations of G over C for which ! ! ù 0 ù 0 , ar1 , ar2 0 ù 0 ùÿ1 ! ! 0 1 1 ùÿ1 ar3 , ar4 ÿ1 ÿ1 0 ù (ù e2ði=3 ). The characters ø i of the representations r i (i 1, 2, 3, 4) are

ø1 ø2 ø3 ø4

1

a

a2

2 2 2 2

2ù ÿ1 ÿ1 1ù

2ù2 ÿ1 ÿ1 1 ù2

Hence by Theorem 14.21, the representations r2 and r3 are equivalent, but there are no other equivalences among r1 , r2 , r3 and r4 . The next theorem is another consequence of Theorem 14.12.


145

14.23 Theorem Let ÷1 , . . . , ÷ k be the irreducible characters of G. Then ÷1 , . . . , ÷ k are linearly independent vectors in the vector space of all functions from G to C. Proof Assume that ë1 ÷1 : : : ë k ÷ k 0 (ë i 2 C): Then for all i, using (14.13) we have 0 hë1 ÷1 : : : ë k ÷ k , ÷ i i ë i : Therefore ÷1 , . . . , ÷ k are linearly independent.

j

We now relate inner products of characters to the spaces of CGhomomorphisms which we constructed in Chapter 11. 14.24 Theorem Let V and W be CG-modules with characters ÷ and ø, respectively. Then dim (HomCG (V , W )) h÷, øi: Proof We know from (14.14) that there are non-negative integers ci , di (1 < i < k) such that V (V1 : : : V1 ) (V2 : : : V2 ) : : : (Vk : : : Vk ) with ci factors Vi for each i, and W (V1 : : : V1 ) (V2 : : : V2 ) : : : (Vk : : : Vk ) with di factors Vi for each i. By Proposition 11.2, for any i, j we have dim (HomCG (Vi , Vj )) ä ij : Hence, using (11.5)(3) we see that dim (HomCG (V , W ))

k X i1

ci d i :

146


On the other hand, ÷

k X

ci ÷ i and ø

i1

k X

di÷i

i1

and so (14.13) implies that h÷, øi

k X

ci d i :

i1

The result follows.

j

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, and we now describe a process for doing this. Once more we adopt Hypothesis 14.7: CG W 1 W 2 , where the CG-modules W1 and W2 have no common composition factor; and 1 e1 e2 with e1 2 W 1 , e2 2 W 2 . Let V be any CG-module. We can write V V1 V2 , where every composition factor of V1 is a composition factor of W1 and every composition factor of V2 is a composition factor of W2. 14.25 Proposition With the above notation, for all v1 2 V1 and v2 2 V2 we have v1 e1 v1 ,

v2 e1 0,

v1 e2 0,

v 2 e2 v 2 :

Proof If v1 2 V1 then the function w2 ! v1 w2 (w2 2 W2 ) is clearly a CG-homomorphism from W2 to V1. Since W2 and V1 have no common composition factor, we deduce the stated results just as in the proof of Proposition 14.8. j 14.26 Proposition If ÷ is an irreducible character of G, and V is any CG-module, then ! X ÿ1 V ÷( g ) g g2G


147

is equal to the sum of those CG-submodules of V which have character ÷ (where for r 2 CG, we de®ne Vr fvr: v 2 V g). Proof Write CG U 1 : : : U r , a direct sum of irreducible CG-submodules Ui. Let W1 be the sum of those CG-submodules Ui which have character ÷, and let W2 be the sum of the remaining CG-submodules Ui. Then the character of W1 is m÷ where m ÷(1), by Theorem 11.9. Also W1 and W2 satisfy Hypothesis 14.7, and by Proposition 14.10, the element e1 of W1 is given by m X e1 ÷( g ÿ1 ) g: jGj g2G Let V1 be the sum of those CG-submodules of V which have character ÷. Then Proposition 14.25 shows that Ve1 V1. Clearly we may omit the constant multiplier m/|G|, so ! X ÿ1 V1 V ÷( g ) g : g2G

j

Once the irreducible characters of our group G are known, Proposition 14.26 provides a useful practical tool for ®nding CG-submodules of a given CG-module V. The procedure is as follows: (14.27) (1) Choose a basis v1 , . . . , v n of V. (2) For each irreducible character ÷ of G, calculate the vectors P v i ( g2G ÷( gÿ1 ) g) for 1 < i < n, and let V÷ be the subspace of V spanned by these vectors. (3) Then V is the direct sum of the CG-modules V÷ as ÷ runs over the irreducible characters of G. The character of V÷ is a multiple of ÷. We illustrate this method with a couple of simple examples. Some more complicated uses of the method can be found in Chapter 32. 14.28 Examples (1) Let G be any ®nite group and let V be any non-zero CG-module. Taking ÷ to be the trivial character of G in Proposition 14.26, we see that

148

Representations and characters of groups X ! V g g2G

is the sum of all the trivial CG-submodules of V. For example, let G Sn and let V be the permutation module, with basis v1 , . . . , v n such that v i g v ig for all i and all g 2 G. Then X ! V g sp (v1 : : : v n ): g2G

Hence V has a unique trivial CG-submodule. (2) Let G be the subgroup of S4 which is generated by a (1 2 3 4) and b (1 2)(3 4): Then G D8 (compare Example 1.5). Here is a list of the irreducible characters ÷1 , . . . , ÷5 of D8 (see Example 16.3(3)):

÷1 ÷2 ÷3 ÷4 ÷5

1

a

a2

a3

b

ab

a2 b

a3 b

1 1 1 1 2

1 1 ÿ1 ÿ1 0

1 1 1 1 ÿ2

1 1 ÿ1 ÿ1 0

1 ÿ1 1 ÿ1 0

1 ÿ1 ÿ1 1 0

1 ÿ1 1 ÿ1 0

1 ÿ1 ÿ1 1 0

Let V be the permutation module for G, with basis v1 , v2 , v3 , v4 such that v i g v ig for all i and all g 2 G. For 1 < i < 5, let ÷ i (1) X ei ÷ i ( g ÿ1 ) g: 8 g2G For example, e5 12(1 ÿ a2 ). Then Ve1 sp (v1 v2 v3 v4 ), Ve2 0, Ve3 0, Ve4 sp (v1 ÿ v2 v3 ÿ v4 ), Ve5 sp (v1 ÿ v3 , v2 ÿ v4 ):


149

We have V Ve1 Ve4 Ve5 , and so we have expressed V as a direct sum of irreducible CGsubmodules whose characters are ÷1 , ÷4 and ÷5 , respectively. You might like to check that e1 : : : e5 1, e2i ei for 1 < i < 5, ei ej 0 for i 6 j: Compare these results with Corollary 14.9. Note that the procedure described in (14.27) 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). Summary of Chapter 14 1. The inner product of two functions W, ö from G to C is given by 1 X hW, öi W( g)ö( g): jGj g2G 2. The irreducible characters ÷1 , . . . , ÷ k of G form an orthonormal set; that is, h÷ i , ÷ j i ä ij for all i, j. 3. Every CG-module is determined by its character. 4. If ÷1 , . . . , ÷ k are the irreducible characters of G, and ø is any character, then ø d 1 ÷1 : : : d k ÷ k where d i hø, ÷ i i: Each di is a non-negative integer. Also, ø is irreducible if and only if kø, øl 1.

150


1. Let G S4 . 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 )|

1

(1 2)

(1 2 3)

(1 2)(3 4)

(1 2 3 4)

24

4

3

8

4

÷ ø

3 3

ÿ1 1

0 0

3 ÿ1

ÿ1 ÿ1

Calculate h÷, ÷i, h÷, øi and hø, øi. Which of ÷ and ø is irreducible? 2. Let G Q8 ka, b: a4 1, b2 a2 , bÿ1 ab aÿ1 l, and let r1 , r2 , r3 be the representations of G over C for which ! ! i 0 0 1 ar1 , br1 , 0 ÿi ÿ1 0 ! ! 0 i 0 ÿ1 ar2 , br2 , i 0 1 0 ! ! ÿ1 0 1 0 ar3 , br3 : 0 1 0 ÿ1 Show that r1 and r2 are equivalent, but r3 is not equivalent to r1 or r2. 3. Suppose that r and ó are representations of G, and that for each g in G there is an invertible matrix Tg such that gó T ÿ1 g ( gr)T g : Prove that there is an invertible matrix T such that for all g in G, gó T ÿ1 ( gr)T: 4. Suppose that ÷ is a non-zero, non-trivial character of G, and that ÷( g) is a non-negative real number for all g in G. Prove that ÷ is reducible. 5. If ÷ is a character of G, show that h÷reg , ÷i ÷(1):


151

6. If ð is the permutation character of Sn , prove that hð, 1 S n i 1: (Hint: you may ®nd Exercise 11.4 relevant.) 7. Let ÷1 , . . . , ÷ k be the irreducible characters of the group G, and suppose that ø d 1 ÷1 : : :d k ÷ k is a character of G. What can you say about the integers di in the cases kø, øl 1, 2, 3 or 4? 8. Suppose that ÷ is a character of G and that for every g 2 G, ÷( g) is an even integer. Does it follow that ÷ 2ö for some character ö?

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, and to some consequences of this theorem. Together with the material from Chapter 14, the theorem provides machinery for investigating characters which is used in the remainder of the book. Throughout, G is as usual a ®nite group. Class functions 15.1 De®nition A class function on G is a function ø: G ! C such that ø(x) ø( y) whenever x and y are conjugate elements of G (that is, ø is constant on conjugacy classes). By Proposition 13.5(2), the characters of G are class functions on G. The set C of all class functions on G is a subspace of the vector space of all functions from G to C. 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. Thus, if l is the number of conjugacy classes of G, then (15:2)

dim C l:

15.3 Theorem The number of irreducible characters of G is equal to the number of conjugacy classes of G. 152

The number of irreducible characters

153

Proof Let ÷1 , . . . , ÷ k be the irreducible characters of G, and let l be the number of conjugacy classes of G. By Theorem 14.23, ÷1 , . . . , ÷ k are linearly independent elements of C, so (15.2) implies that k < l. In order to prove the reverse inequality l < k, we consider the regular CG-module. If V1 , : : : , V k is a complete set of non-isomorphic irreducible CG-modules, we know from Theorem 8.7 that CG W 1 : : : W k , where for each i, W i is isomorphic to a direct sum of copies of V i . Since CG contains the identity element 1, we can write 1 f1 : : : fk with f i 2 W i for 1 < i < k. Now let z 2 Z(CG), the centre of CG. By Proposition 9.14, for each i there exists ë i 2 C such that for all v 2 Vi, vz ë i v: Hence wz ë i w for all w 2 W i , and in particular, f iz ëi f i

(1 < i < k):

It follows that z 1z ( f 1 : : : f k )z f 1 z : : : f k z ë1 f 1 : : : ë k f k : This shows that Z(CG) is contained in the subspace of CG spanned by f 1 , : : : , f k . Since Z(CG) has dimension l by Proposition 12.22, we deduce that l < k. This completes the proof that k l. j

15.4 Corollary The irreducible characters ÷1 , : : : , ÷ k of G form a basis of the vector space of all class functions on G. Indeed, if ø is a class function, then ø

k X

ëi÷i

i1

where ë i kø, ÷ i l for 1 < i < k. Proof Since ÷1 , . . . , ÷ k are linearly independent, they span a subspace of C of dimension k. By (15.2), dim C l, which is equal to k by

154


Theorem 15.3. Hence ÷1 , . . . , ÷ k span C, and so they form a basis of C. The last part follows, using (14.13). j Corollary 15.4 has the following useful consequence. 15.5 Proposition Suppose that g, h 2 G. Then g is conjugate to h if and only if ÷( g) ÷(h) for all characters ÷ of G. Proof If g is conjugate to h then ÷( g) ÷(h) for all characters ÷ of G, by Proposition 13.5(2). Conversely, suppose that ÷( g) ÷(h) for all characters ÷. Then by Corollary 15.4, ø( g) ø(h) for all class functions ø on G. In particular, 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. Then ø( g) ø(h) 1, and so g is conjugate to h. j 15.6 Corollary Suppose that g 2 G. Then g is conjugate to gÿ1 if and only if ÷( g) is real for all characters ÷ of G. Proof Since ÷( g) is real if and only if ÷( g) ÷( gÿ1 ) (see Proposition j 13.9(3)), the result follows immediately from Proposition 15.5. 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. As in previous examples, we regard a character ÷ of G as a row vector, whose k entries are the values of ÷ on the k conjugacy classes of G. 15.7 Example We shall see in Section 18.4 that there is a certain group G of order 12 which has exactly six conjugacy classes with representatives g1 , . . . , g6 (where g1 1), and six irreducible characters ÷1 , . . . , ÷6 given as follows:

The number of irreducible characters gi |CG ( g i )| ÷1 ÷2 ÷3 ÷4 ÷5 ÷6

155

g1 12

g2 12

g3 6

g4 6

g5 4

g6 4

1 1 1 1 2 2

1 ÿ1 1 ÿ1 2 ÿ2

1 ÿ1 1 ÿ1 ÿ1 1

1 1 1 1 ÿ1 ÿ1

1 i ÿ1 ÿi 0 0

1 ÿi ÿ1 i 0 0

Suppose we are given characters ÷ and ø of G as follows:

÷ ø

g1

g2

g3

g4

g5

g6

3 4

ÿ3 0

0 0

0 4

i 0

ÿi 0

Then it is easy to spot that ÷ ÷2 ÷6 ,

ø ÷1 ÷2 ÷3 ÷4 :

For example, the second entry in the row vector for ÷ is equal to minus the ®rst entry. Inspecting the values of the irreducible characters ÷ i , we see that ÷ must be a combination of ÷2 , ÷4 and ÷6 . The correct answer now comes quickly to mind. In fact, given any character ö of G whose degree is not large compared with the degrees of the ÷ i , it is not hard to use tactical guesswork to express ö as a combination of the irreducible characters. The reason for this is that the required coef®cients are known to be non-negative integers, and the entries in the column corresponding to g1 1 are positive integers (indeed, they are the degrees of the ÷ i ). We suggest that you use the `guesswork method' to express the following characters ë, ì of G as combinations of ÷1 , . . . , ÷6 :

ë ì

g1

g2

g3

g4

g5

g6

2 4

ÿ2 4

ÿ2 1

2 1

0 0

0 0

156


How do we cope with a class function or with a more dif®cult character, like the following one?

ö

g1

g2

g3

g4

11

3

ÿ3

5

g5 ÿ1 2i

g6 ÿ1 ÿ 2i

The answer is to use the inner product k , l. We know from Corollary 15.4 that the coef®cients ë i in the expression ö ë1 ÷1 : : : ë6 ÷6 are given by ë i hö, ÷ i i (1 < i < 6): Using Proposition 14.5(2), we calculate these inner products: 11 . 1 3 . 1 ÿ3 . 1 5 . 1 (ÿ1 2i) . 1 12 12 6 6 4 (ÿ1 ÿ 2i) . 1 1, 4 11 . 1 3 . (ÿ1) (ÿ3) . (ÿ1) 5 . 1 (ÿ1 2i) . (ÿi) hö, ÷2 i 12 12 6 6 4 (ÿ1 ÿ 2i) . i 3, 4 11 . 1 3 . 1 ÿ3 . 1 5 . 1 (ÿ1 2i) . (ÿ1) hö, ÷3 i 12 12 6 6 4 (ÿ1 ÿ 2i) . (ÿ1) 2, 4 hö, ÷1 i

and similarly kö, ÷4 l 1, kö, ÷5 l 2 and kö, ÷6 l 0. Therefore ö ÷1 3÷2 2÷3 ÷4 2÷5 :

Summary of Chapter 15 1. The number of irreducible characters of a group is equal to the number of conjugacy classes of the group.

The number of irreducible characters

157

2. The irreducible characters ÷1 , . . . , ÷ k of G form a basis of the vector space of all class functions on G. If ø is a class function, then ø

k X

ë i ÷ i where ë i hø, ÷ i i:

i1

Exercises for Chapter 15 1. The three irreducible characters of S3 are ÷1 , ÷2 , ÷3 :

÷1 ÷2 ÷3

1

(1 2)

(1 2 3)

1 1 2

1 ÿ1 0

1 1 ÿ1

Let ÷ be the class function on S3 with the following values:

÷

1

(1 2)

(1 2 3)

19

ÿ1

ÿ2

Express ÷ as a linear combination of ÷1 , ÷2 and ÷3 . Is ÷ a character of S3 ? 2. Let ø1 , ø2 and ø3 be the class functions on S3 taking the following values:

ø1 ø2 ø3

1

(1 2)

(1 2 3)

1 0 0

0 1 0

0 0 1

Express ø1 , ø2 and ø3 as linear combinations of the irreducible characters ÷1 , ÷2 and ÷3 of S3 .

158


3. Suppose that G is the group of order 12 in Example 15.7, with conjugacy class representatives g1 , . . . , g6 and irreducible characters ÷1 , . . . , ÷6 as in that example. Let ø be the class function on G taking the following values:

ø

g1

g2

g3

g4

6

0

3

ÿ3

g5 ÿ1 ÿ i

g6 ÿ1 i

Express ø as a linear combination of ÷1 , . . . , ÷6 . Is ø a character of G? 4. Let G be a group of order 12. (a) Show that G cannot have exactly 9 conjugacy classes. (Hint: show that Z(G) cannot have order 6.) (b) Using the solution to Exercise 11.2, prove that G has 4, 6 or 12 conjugacy classes. Find groups G in which each of these possibilities is realized.

16 Character tables and orthogonality relations

The irreducible characters of a ®nite group G are class functions, and the number of them is equal to the number of conjugacy classes of G. It is therefore convenient to record all the values of all the irreducible characters of G in a square matrix. This matrix is called the character table of G. The entries in a character table are related to each other in subtle ways, many of which are encapsulated in the orthogonality relations (Theorem 16.4). Much of the later material in the book will be devoted to understanding character tables. The motivation for this is Theorem 14.21, which tells us that every CG-module is determined by its character. Thus, many problems in representation theory can be solved by considering characters.

Character tables 16.1 De®nition Let ÷1 , : : : , ÷ k be the irreducible characters of G and let g 1 , : : : , g k be representatives of the conjugacy classes of G. The k 3 k matrix whose ij-entry is ÷ i ( g j ) (for all i, j with 1 < i < k, 1 < j < k) is called the character table of G. It is usual to number the irreducible characters and conjugacy classes of G so that ÷1 1 G , the trivial character, and g1 1, the identity element of G. Beyond this, the numbering is arbitrary. Note that in the character table, the rows are indexed by the irreducible characters of G and the columns are indexed by the conjugacy classes (or, in practice, by conjugacy class representatives). 159

160


16.2 Proposition The character table of G is an invertible matrix. Proof This follows immediately from the fact that the irreducible characters of G, and hence also the rows of the character table, are linearly independent (Theorem 14.23). j 16.3 Examples (1) Let G D6 ka, b: a3 b2 1, bÿ1 ab aÿ1 l. The irreducible characters of G are given in Example 13.6(4). We take 1, a, b as representatives of the conjugacy classes of G, and then the character table of G is

÷1 ÷2 ÷3

1

a

b

1 1 2

1 1 ÿ1

1 ÿ1 0

(2) We can write down the character table of any ®nite abelian group using Theorem 9.8. For example, the character table of C2 ha: a2 1i is

÷1 ÷2

1

a

1 1

1 ÿ1

and the character table of C3 ka: a3 1l is

÷1 ÷2 ÷3

1

a

a2

1 1 1

1 ù ù2

1 ù2 ù

(ù e2ði=3 )

(3) Let G D8 ka, b: a4 b2 1, bÿ1 ab aÿ1 l. You found all the irreducible representations of G in Exercise 10.4. The conjugacy classes

Character tables and orthogonality relations

161

of G are given by (12.12), and representatives are 1, a2 , a, b, ab. Hence the character table of G is

÷1 ÷2 ÷3 ÷4 ÷5

1

a2

a

b

ab

1 1 1 1 2

1 1 1 1 ÿ2

1 1 ÿ1 ÿ1 0

1 ÿ1 1 ÿ1 0

1 ÿ1 ÿ1 1 0

The character tables of all dihedral groups will be found in Chapter 18. The orthogonality relations We have already seen many uses for the relations (14.13), h÷ r , ÷ s i ä rs , among the irreducible characters ÷1 , . . . , ÷ k of G. These relations can be expressed in terms of the rows of the character table, by writing them as k X ÷ r ( g i )÷ s ( g i ) ä rs jCG ( g i )j i1 (see Proposition 14.5(2)). Similar relations exist between the columns of the character table, and these are given by part (2) of our next result. 16.4 Theorem Let ÷1 , . . . , ÷ k be the irreducible characters of G, and let g 1 , : : : , g k be representatives of the conjugacy classes of G. Then the following relations hold for any r, s 2 {1, . . . , k}. (1) The row orthogonality relations: k X ÷ r ( g i )÷ s ( g i ) i1

jCG ( g i )j

ä rs :

(2) The column orthogonality relations: k X i1

÷ i ( g r )÷ i ( g s ) ä rs jCG ( g r )j:

162


Proof The row orthogonality relations have already been proved. They are recorded here merely for comparison with the column relations. For 1 < s < k, let ø s be the class function which satis®es ø s ( g r ) ä rs

(1 < r < k):

By Corollary 15.4, ø s is a linear combination of ÷1 , . . . , ÷ k , say øs

k X

ëi ÷i

(ë i 2 C):

i1

We know that h÷ i , ÷ j i ä ij , so ë i hø s , ÷ i i

1 X ø s ( g)÷ i ( g): jGj g2G

Now ø s ( g) 1 if g is conjugate to g s , and ø s ( g) 0 otherwise; also there are jGj=jC G ( g s )j elements of G which are conjugate to g s , by Theorem 12.8. Hence ëi

1 X ÷ i ( gs ) ø s ( g)÷ i ( g) : jGj g2 g G jCG ( g s )j s

Therefore ä rs ø s ( g r )

k X

ëi ÷i ( gr )

i1

k X ÷ i ( g r )÷ i ( g s ) i1

jCG ( g s )j

and the column orthogonality relations follow.

, j

16.5 Examples We illustrate the column orthogonality relations. (1) Let G D6. We copy the character table of G from Example 16.3(1), 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 6

a 3

b 2

÷1 ÷2 ÷3

1 1 2

1 1 ÿ1

1 ÿ1 0

Character tables and orthogonality relations P Consider the sums 3i1 ÷ i ( g r )÷ i ( g s ) for various cases: r 1,

s 2:

1 . 1 1 . 1 2 . (ÿ1) 0;

r 2,

s 2:

1 . 1 1 . 1 (ÿ1) . (ÿ1) 3;

r 1,

s 3:

1 . 1 1 . (ÿ1) 2 . 0 0:

163

In each case, we read down columns r and s of the character table, taking the products of the numbers which appear. The sum of the products is 0 if r 6 s, and is the number at the top of the column (that is, the order of the centralizer of g r ) if r s. (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 )|

g1 12

g2 4

g3 3

g4 3

÷1 ÷2 ÷3 ÷4

1 1 1

1 1 1

1 ù ù2

1 ù2 ù

(where ù e2ði=3 ). We shall use the column orthogonality relations to determine the last row of the character table. The entries in the ®rst column of the character table are the degrees of the irreducible characters, so they are positive integers. By the column orthogonality relations with r s 1, the sum of the squares of these numbers is 12 (this also follows from Theorem 11.12). Hence the last entry in the ®rst column is 3. Let x denote the number at the foot of the second column. The column orthogonality relation 4 X

÷ i ( g 1 )÷ i ( g 2 ) 0

i1

gives

1 . 1 1 . 1 1 . 1 3x 0:

Therefore x ÿ1. By considering the orthogonality relations between the ®rst column and columns 3 and 4, we obtain the complete character table as

164


follows: gi |CG ( g i )|

g1 12

g2 4

g3 3

g4 3

÷1 ÷2 ÷3 ÷4

1 1 1 3

1 1 1 ÿ1

1 ù ù2 0

1 ù2 ù 0

Notice that the orthogonality relations hold between all pairs of columns, although our calculation has used only those relations which involve the ®rst column. For example, 4 X

÷ i ( g 2 )÷ i ( g 2 ) 1 . 1 1 . 1 1 . 1 (ÿ1) . (ÿ1) 4,

i1 4 X

÷ i ( g 3 )÷ i ( g 3 ) 1 . 1 ù . ù ù2 . ù2 0 . 0 3,

i1 4 X

÷ i ( g 3 )÷ i ( g 4 ) 1 . 1 ù . ù2 ù2 . ù 0 . 0 0:

i1

We shall see later that the character table which we have constructed here is that of A4 . Those column orthogonality relations which involve the ®rst column of the character table were proved in Chapter 13, since Theorem 13.19 and Proposition 13.20 give 8 k < jGj, if g 1, X d i ÷ i ( g) : i1 0, if g 6 1, where d i ÷ i (1). By taking the complex conjugate of each side of this equation, we get 8 < jGj, if g 1, k X ÷ i (1)÷ i ( g) : i1 0, if g 6 1,

Character tables and orthogonality relations

165

and these are just the column orthogonality relations which involve the ®rst column.

Rows versus columns Notice that in Example 16.5(2), where we were given three of the four irreducible characters of G, 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 , ÷4 i ä i4 to obtain four equations in the four unknown values ÷4 ( g j ) (1 < j < 4). Although the calculation with the column orthogonality relations was easier to perform, it is a fact that the column orthogonality relations contain precisely the same information as the row orthogonality relations, as we shall now show. The character table of G is a k 3 k matrix, and we adjust the entries ÷ i ( g j ) in this matrix to obtain another k 3 k matrix M, by letting the ij-entry of M be ÷ i ( gj ) : jCG ( g j )j1=2 Let M t denote the transpose of the complex conjugate of M. Now the rs-entry in M M t is k X ÷ r ( g i )÷ s ( g i ) i1

jCG ( g i )j

ä rs ,

by the row orthogonality relations, so M M t I. Indeed, the equation M M t I is just another way of expressing the row orthogonality relations. On the other hand, the rs-entry in M t M is k X 1 ÷ i ( g r )÷ i ( g s ) ä rs , jCG ( g r )j1=2 jCG ( g s )j1=2 i1

by the column orthogonality relations, so M t M I. Since the properties M t M I and M M t I of a square matrix M are equivalent to each other, we see that the row and column orthogonality relations are equivalent. We could have used the above argument to deduce the column orthogonality relations from the row ones. More importantly, the row and column orthogonality relations encapsulate the same information,

166


so when we are working with character tables, we can deduce exactly the same results using either set of relations. Summary of Chapter 16 Let G be a ®nite group with irreducible characters ÷1 , . . . , ÷ k and conjugacy class representatives g1 , . . . , gk . 1. The character table of G is the k 3 k matrix with ij-entry ÷ i ( g j ). 2. The row orthogonality relations state that for all r, s, k X ÷ r ( g i )÷ s ( g i ) i1

jCG ( g i )j

ä rs :

3. The column orthogonality relations state that for all r, s, k X

÷ i ( g r )÷ i ( g s ) ä rs jCG ( g r )j:

i1

Exercises for Chapter 16 1. Write down the character table of C2 3 C2 . 2. A certain group G of order 8 is known to have a total of ®ve conjugacy classes, with representatives g1 , . . . , g5 , and four linear characters ÷1 , . . . , ÷4 taking the following values: gi |CG ( g i )|

g1 8

g2 8

g3 4

g4 4

g5 4

÷1 ÷2 ÷3 ÷4

1 1 1 1

1 1 1 1

1 1 ÿ1 ÿ1

1 ÿ1 1 ÿ1

1 ÿ1 ÿ1 1

Find the complete character table of G. 3. There exists a group G of order 10 which has precisely four conjugacy classes, with representatives g1 , . . . , g4 , and has irreducible characters ÷1 , ÷2 as follows:

Character tables and orthogonality relations gi |CG ( g i )|

g1 10

g2 5

g3 5

g4 2

÷1 ÷2

1 2

1 á

1 â

1 0

167

p p where á (ÿ1 5)=2 and â (ÿ1 ÿ 5)=2. Find the complete character table of G. (Hint: ®rst ®nd the values of the remaining irreducible characters on g1 , then on g4 ± use Corollary 13.10.) 4. A certain group G has two columns of its character table as follows: gi |CG ( g i )|

g1 21

g2 7

÷1 ÷2 ÷3 ÷4 ÷5

1 1 1 3 3

1 1 1 æ æ

where g1 1 and æ 2 C. (a) Find æ. (b) Find another column of the character table. 5. Let ÷1 , . . . , ÷ k be the irreducible characters of G. Show that ( ) k X Z(G) g 2 G: ÷ i ( g)÷ i ( g) jGj : i1

6. Let G be a ®nite group with conjugacy class representatives g1 , : : : , g k and character table C. Show that det C is either real or purely imaginary, and that jdet Cj2

k Y i1

Find (det C) when G C3 .

jCG ( g i )j:

17 Normal subgroups and lifted characters

If N is a normal subgroup of the ®nite group G, and N 6 {1}, then the factor group G=N is smaller than G. The characters of G=N should therefore be easier to ®nd than the characters of G. In fact, we can use the characters of G=N to get some of the characters of G, by a process which is known as lifting. Thus, normal subgroups help us to ®nd characters of G. In the opposite direction, it is also true that the character table of G enables us to ®nd the normal subgroups of G; in particular, it is easy to tell from the character table whether or not G is simple. The linear characters of G (i.e. the characters of degree 1) are obtained by lifting the irreducible characters of G=N in the case where N is the derived subgroup of G. (The derived subgroup is de®ned below in De®nition 17.7.) The linear characters, in turn, can be used to get new irreducible characters from a given irreducible character, in a way which we shall describe. Lifted characters We begin by constructing a character of G from a character of G=N . 17.1 Proposition Assume that N v G, and let ~÷ be a character of G=N . De®ne ÷: G ! C by ÷( g) ~÷(Ng) ( g 2 G): Then ÷ is a character of G, and ÷ and ~÷ have the same degree. ~: G=N ! GL (n, C) be a representation of G=N with Proof Let r 168

Normal subgroups and lifted characters

169

character ~ ÷. The function r: G ! GL (n, C) which is given by the composition g ! Ng ! (Ng)~ r

( g 2 G)

is a homomorphism from G to GL (n, C). Thus r is a representation of G. The character ÷ of r satis®es ÷( g) tr ( gr) tr ((Ng)~ r) ~÷(Ng) for all g 2 G. Moreover, ÷(1) ~÷(N), so ÷ and ~÷ have the same degree. j 17.2 De®nition If N v G and ~ ÷ is a character of G=N , then the character ÷ of G which is given by ÷( g) ~÷(Ng)

( g 2 G)

is called the lift of ~ ÷ to G. 17.3 Theorem Assume that N v G. By associating each character of G=N with its lift to G, we obtain a bijective correspondence between the set of characters of G=N and the set of characters ÷ of G which satisfy N < Ker ÷. Irreducible characters of G=N correspond to irreducible characters of G which have N in their kernel. Proof If ~ ÷ is a character of G=N , and ÷ is the lift of ~÷ to G, then ~ ÷(N) ÷(1). Also, if k 2 N then ÷(k) ~÷(Nk) ~÷(N ) ÷(1), so N < Ker ÷. Now let ÷ be a character of G with N < Ker ÷. Suppose that r: G ! GL (n, C) is a representation of G with character ÷. If g 1 , ÿ1 g2 2 G and Ng1 Ng2 then g1 gÿ1 2 2 N, so ( g1 g2 )r I, and hence ~ : G=N ! GL (n, C) g1 r g2 r. We may therefore de®ne a function r by (Ng)~ r gr

( g 2 G):

Then for all g, h 2 G we have ((Ng)(Nh))~ r (Ngh)~ r ( gh)r ( gr)(hr) ((Ng)~ r)((Nh)~ r),

170


~ is a representation of G=N . If ~÷ is the character of r ~ then so r ~÷(Ng) ÷( g) ( g 2 G): ~. Thus ÷ is the lift of ÷ We have now established that the function which sends each character of G=N to its lift to G is a bijection between the set of characters of G=N and the set of characters of G which have N in their kernel. It remains to show that irreducible characters correspond to irreducible characters. To see this, let U be a subspace of C n , and note that u( gr) 2 U for all u 2 U , u(Ng)~ r 2 U for all u 2 U : Thus, U is a CG-submodule of C n if and only if U is a C(G=N )submodule of C n . The representation r is therefore irreducible if and ~ is irreducible. Hence ÷ is irreducible if and only if the representation r only if ~ ÷ is irreducible. j If we know the character table of G=N for some normal subgroup N of G, then Theorem 17.3 enables us to write down as many irreducible characters of G as there are irreducible characters of G=N . 17.4 Example Let G S4 and N V4 f1, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)g, so that N v G (see Example 12.20). If we put a N(1 2 3) and b N(1 2) then G=N ha, bi and a3 b2 N , bÿ1 ab aÿ1 , so G=N D6 . We know from Example 16.3(1) that the character table of G=N is

~÷1 ~÷2 ~÷3

N

N (1 2)

N(1 2 3)

1 1 2

1 ÿ1 0

1 1 ÿ1


171

To calculate the lift ÷ of a character ~÷ of G=N , we note that ÷((1 2)(3 4)) ~ ÷(N ) since (1 2)(3 4) 2 N , ÷((1 2 3 4)) ~ ÷(N (1 3)) since N (1 2 3 4) N (1 3): Hence the lifts of ~ ÷1 , ~ ÷2 , ~÷3 are ÷1 , ÷2 , ÷3 , which are given by

÷1 ÷2 ÷3

1

(1 2)

(1 2 3)

(1 2)(3 4)

(1 2 3 4)

1 1 2

1 ÿ1 0

1 1 ÿ1

1 1 2

1 ÿ1 0

Then ÷1 , ÷2 , ÷3 are irreducible characters of G, since ~÷1 , ~÷2 , ~÷3 are irreducible characters of G=N . Finding normal subgroups The character table contains accessible information about the structure of a group, as our next two propositions will demonstrate. First we shall show how to ®nd all the normal subgroups of G, once the character table of G is known. Recall that we can easily locate the kernel of an irreducible character ÷ from the character table, since Ker ÷ f g 2 G: ÷( g) ÷(1)g (see De®nition 13.13). Also Ker ÷ v G. Of course, any subgroup which is the intersection of the kernels of irreducible characters is a normal subgroup too. The following proposition shows that every normal subgroup arises in this way. 17.5 Proposition If N v G then there exist irreducible characters ÷1 , : : : , ÷ s of G such that N

s \

Ker ÷ i :

i1

Proof If g belongs to the kernel of each irreducible character of G, then ÷( g) ÷(1) for all characters ÷, so g 1 by Proposition 15.5. Hence the intersection of the kernels of all the irreducible characters of G is {1}.

172


~1 , . . . , ~ Now let ÷ ÷ s be the irreducible characters of G=N . By the above observation, s \

Ker ~÷ i fN g:

i1

For 1 < i < s, let ÷ i be the lift to G of ~÷ i . If g 2 Ker ÷ i then ~ ÷ i (N ) ÷ i (1) ÷ i ( g) ~÷ i (Ng), T T and so Ng 2 Ker ~ ÷ i. Therefore if g 2 Ker ÷ i then Ng 2 Ker ~÷ i {N}, and so g 2 N. Hence N

s \

Ker ÷ i :

i1

j

It is particularly easy to tell from the character table of G whether or not G is simple: 17.6 Proposition The group G is not simple if and only if ÷( g) ÷(1) for some non-trivial irreducible character ÷ of G, and some nonidentity element g of G. Proof Suppose there is a non-trivial irreducible character ÷ such that ÷( g) ÷(1) for some non-identity element g. Then g 2 Ker ÷, so Ker ÷ 6 {1}. If r is a representation of G with character ÷, then Ker ÷ Ker r by Theorem 13.11(2). Since ÷ is non-trivial and irreducible, Ker r 6 G; hence Ker ÷ 6 G. Thus Ker ÷ is a normal subgroup of G which is not equal to {1} of G, and so G is not simple. Conversely, suppose that G is not simple, so that there is a normal subgroup N of G with N 6 {1} and N 6 G. Then by Proposition 17.5, there is an irreducible character ÷ of G such that Ker ÷ is not {1} or G. As Ker ÷ 6 G, ÷ is non-trivial; and taking 1 6 g 2 Ker ÷, we have ÷( g) ÷(1). j Linear characters Recall that a linear character of a group is a character of degree 1. We shall show how to ®nd all linear characters of any group G, since the


173

®rst move in constructing the character table of G is often to write down the linear characters. As a preliminary step, it is necessary to determine the derived subgroup of G, which is de®ned in the following way. 17.7 De®nition For a group G, let G9 be the subgroup of G which is generated by all elements of the form g ÿ1 hÿ1 gh

( g, h 2 G):

Then G9 is called the derived subgroup of G. We abbreviate gÿ1 hÿ1 gh as [ g, h]. Thus G9 h[ g, h]: g, h 2 Gi: 17.8 Examples (1) If G is abelian then [ g, h] 1 for all g, h 2 G, so G9 {1}. (2) Let G S3 . Clearly [ g, h] is always an even permutation, so G9 < A3 . If g (1 2) and h (2 3) then [ g, h] (1 2 3). Hence G9 h(1 2 3)i A3 . 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 G=G9. One step is provided by the following proposition. 17.9 Proposition If ÷ is a linear character of G, then G9 < Ker ÷. Proof Let ÷ be a linear character of G. Then ÷ is a homomorphism from G to the multiplicative group of non-zero complex numbers. Therefore, for all g, h 2 G, ÷( g ÿ1 hÿ1 gh) ÷( g)ÿ1 ÷(h)ÿ1 ÷( g)÷(h) 1: Hence G9 < Ker ÷.

j

Next, we explore some group-theoretic properties of the derived subgroup.

174


17.10 Proposition Assume that N v G. (1) G9 v G. (2) G9 < N if and only if G=N is abelian. In particular, G=G9 is abelian. Proof (1) Note that for all a, b, x 2 G, we have x ÿ1 (ab)x (x ÿ1 ax)(x ÿ1 bx), and x ÿ1 aÿ1 x (x ÿ1 ax)ÿ1 : Now G9 consists of products of elements of the form [ g, h] and their inverses. Therefore, to prove that G9 v G it is suf®cient by the ®rst sentence to prove that x ÿ1 [ g, h]x 2 G9 for all g, h, x 2 G. But x ÿ1 [ g, h]x x ÿ1 g ÿ1 hÿ1 ghx (x ÿ1 gx)ÿ1 (x ÿ1 hx)ÿ1 (x ÿ1 gx)(x ÿ1 hx) [x ÿ1 gx, x ÿ1 hx]: Therefore G9 v G. (2) Let g, h 2 G. We have ghgÿ1 hÿ1 2 N , Ngh Nhg , (Ng)(Nh) (Nh)(Ng): Hence G9 < N if and only if G=N is abelian. Since we have proved that G9 v G, we deduce that G=G9 is abelian. j It follows from Proposition 17.10 that G9 is the smallest normal subgroup of G with abelian factor group. Given the derived subgroup G9, we can obtain the linear characters of G by applying the next theorem. 17.11 Theorem The linear characters of G are precisely the lifts to G of the irreducible characters of G=G9. In particular, the number of distinct linear characters of G is equal to jG=G9j, and so divides |G|. Proof Let m jG=G9j. Since G=G9 is abelian, Theorem 9.8 shows that G=G9 has exactly m irreducible characters ~÷1 , . . . , ~÷ m , all of degree 1. The lifts ÷1 , . . . , ÷ m of these characters to G also have degree 1, and by Theorem 17.3 they are precisely the irreducible characters ÷ of G


175

such that G9 < Ker ÷. In view of Proposition 17.9, the characters ÷1 , . . . , ÷ m are therefore all the linear characters of G. j 17.12 Example Let G S n . We shall show that G9 An . If n 1 or 2 then S n is abelian, so G9 {1} A n . We proved that S93 A3 in Example 17.8(2), so we assume that n > 4. As S n =A n C2 , we have G9 < A n by Proposition 17.10(2). If g (1 2), h (2 3) and k (1 2)(3 4), then [ g, h] (1 2 3), [h, k] (1 4)(2 3): Since G9 v G, all the elements in (1 2 3) G and (1 4)(2 3) G belong to G9. Therefore, by Theorem 12.15, G9 contains all 3-cycles and all elements of cycle-shape (2, 2). But every product of two transpositions is equal to the identity, a 3-cycle or an element of cycle-shape (2, 2); and A n consists of permutations, each of which is the product of an even number of transpositions. Therefore A n < G9. We have now proved that G9 A n . 17.13 Example We ®nd the linear characters of S n (n > 2). From the last example, we know that S9n An . Since Sn =S9n fAn , An (1 2)g C2 , the group S n =S9n has two linear characters ~÷1 and ~÷2, where ~÷1 (An (1 2)) 1, ~÷2 (An (1 2)) ÿ1: Therefore by Theorem 17.11, S n has exactly two linear characters ÷1 , ÷2 , which are given by ÷1 1 Sn , ( ÷2 ( g)

1,

if g 2 An ,

ÿ1,

if g 2 = An :

Not only are the linear characters of G important in being irreducible characters, but they can also be used to construct new irreducible characters from old, as the next result shows.

176


17.14 Proposition Suppose that ÷ is a character of G and ë is a linear character of G. Then the product ÷ë, de®ned by ÷ë( g) ÷( g)ë( g)

( g 2 G)

is a character of G. Moreover, if ÷ is irreducible, then so is ÷ë. Proof Let r: G ! GL (n, C) be a representation with character ÷. De®ne rë: G ! GL (n, C) by g(rë) ë( g)( gr)

( g 2 G):

Thus g(rë) is the matrix gr multiplied by the complex number ë( g). Since r and ë are homomorphisms it follows easily that rë is a homomorphism. The matrix g(rë) has trace ë( g) tr ( gr), which is ë( g)÷( g). Hence rë is a representation of G with character ÷ë. Now for all g 2 G, the complex number ë( g) is a root of unity, so ë( g)ë( g) 1. Therefore 1 X h÷ë, ÷ëi ÷( g)ë( g)÷( g)ë( g) jGj g2G

1 X ÷( g)÷( g) h÷, ÷i: jGj g2G

By Theorem 14.20, it follows that ÷ë is irreducible if and only if ÷ is irreducible. j The general case of a product of two characters will be discussed in Chapter 19.

Summary of Chapter 17 1. Characters of G=N correspond to characters ÷ of G for which N < Ker ÷. The character of G which corresponds to the character ~÷ of G=N is the lift of ~÷, and is given by ÷( g) ~÷(Ng) ( g 2 G). 2. The normal subgroups of G can be found from the character table of G. 3. The linear characters of G are precisely the lifts to G of the irreducible characters of G=G9.


177

Exercises for Chapter 17 1. Let G Q8 ka, b: a4 1, b2 a2 , bÿ1 ab aÿ1 l. (a) Find the ®ve conjugacy classes of G. (b) Find G9, and construct all the linear characters of G. (c) Complete the character table of G. Compare your table with the character table of D8 (Example 16.3(3)). 2. Let a and b be the following permutations in S7 : a (1 2 3 4 5 6 7), b (2 3 5)(4 7 6): Let G ka, bl. Check that a7 b3 1, bÿ1 ab a2 : (a) Show that G has order 21. (b) Find the conjugacy classes of G. (c) Find the character table of G. 3. Show that every group of order 12 has 3, 4 or 12 linear characters, and hence cannot be simple. 4. A certain group G of order 12 has precisely six conjugacy classes, with representatives g1 , . . . , g6 (where g1 1), and has irreducible characters ÷, ö with values as follows:

÷ ö

g1

g2

g3

g4

g5

g6

1 2

ÿi 0

i 0

1 ÿ1

ÿ1 ÿ1

ÿ1 2

Use Proposition 17.14 to complete the character table of G. What are the sizes of the conjugacy classes of G? 5. The character table of D8 is as shown (see Example 16.3(3)):

÷1 ÷2 ÷3 ÷4 ÷5

1

a2

1 1 1 1 2

1 1 1 1 ÿ2

a, a3 1 1 ÿ1 ÿ1 0

b, a2 b

ab, a3 b

1 ÿ1 1 ÿ1 0

1 ÿ1 ÿ1 1 0

178


Express each normal subgroup of D8 as an intersection of kernels of irreducible characters, as in Proposition 17.5. 6. You are given that the group T4 n ha, b: a2 n 1, an b2 , bÿ1 ab aÿ1 i has order 4n. (It is known as a dicyclic group.) (a) Show that if å is any (2n)th root of unity in C, then there is a representation of T4 n over C which sends å 0 0 1 a! , b ! : 0 å ÿ1 ån 0 (b) Find all the irreducible representations of T4 n. 7. For n > 1, the group U6 n ha, b: a2 n b3 1, aÿ1 ba bÿ1 i has order 6n. (a) Let ù e2ði=3 . Show that if å is any (2n)th root of unity in C, then there is a representation of U6n over C which sends 0 å ù 0 : a! ,b! å 0 0 ù2 (b) Find all the irreducible representations of U6 n. 8. Let n be an odd positive integer. The group V8 n ha, b: a2 n b4 1, ba aÿ1 bÿ1 , bÿ1 a aÿ1 bi has order 8n. (a) Show that if å is any nth root of unity in C, then there is a representation of V8 n over C which sends å 0 0 1 : a! ,b! 0 ÿå ÿ1 ÿ1 0 (b) Find all the irreducible representations of V8 n .

18 Some elementary character tables

We now illustrate the techniques we have presented so far by constructing the character tables of several groups, including the groups S4 and A4 , and all dihedral groups.

18.1 The group S4 In Example 17.4, we produced three irreducible characters ÷1 , ÷2 , ÷3 of S4 by lifting characters of the factor group S4 =V4 . We shall now use Proposition 17.14, which deals with the product of a character with a linear character, to complete the character table of S4 . Let ÷4 be the character ÷4 ( g) jfix ( g)j ÿ 1

( g 2 S4 )

which is given in Proposition 13.24. By Proposition 17.14, the product ÷4 ÷2 is also a character of S4 . The values of ÷2 , ÷4 and ÷4 ÷2 are as follows: gi |CG ( gi )| ÷2 ÷4 ÷4 ÷2

1 24

(1 2) 4

(1 2 3) 3

(1 2)(3 4) 8

(1 2 3 4) 4

1 3 3

ÿ1 1 ÿ1

1 0 0

1 ÿ1 ÿ1

ÿ1 ÿ1 1

Note that h÷4 , ÷4 i

9 1 1 1 1, 24 4 8 4 179

180


so ÷4 is irreducible. The character ÷4 ÷2 is also irreducible, either by using the same calculation or by quoting the result of Proposition 17.14. Let ÷5 ÷4 ÷2 . Since S4 has ®ve conjugacy classes, and we have produced ®ve irreducible characters, we have now found the complete character table of S4 , as shown. Character table of S4

gi |CG ( gi )|

1 24

(1 2) 4

(1 2 3) 3

(1 2)(3 4) 8

(1 2 3 4) 4

÷1 ÷2 ÷3 ÷4 ÷5

1 1 2 3 3

1 ÿ1 0 1 ÿ1

1 1 ÿ1 0 0

1 1 2 ÿ1 ÿ1

1 ÿ1 0 ÿ1 1

18.2 The group A4 Let G A4 , the alternating group of degree 4. Then |G| 12, and G has four conjugacy classes, with representatives 1, (1 2)(3 4), (1 2 3), (1 3 2) (see Example 12.18(1)). Let í be the character of A4 given by Proposition 13.24, so that í( g) |®x ( g)| ÿ 1 for all g 2 A4 . The values of í are as follows: gi |CG ( gi )|

1 12

(1 2)(3 4) 4

(1 2 3) 3

(1 3 2) 3

í

3

ÿ1

0

0

Note that hí, íi

9 1 1, 12 4

so í is an irreducible character of G of degree 3. Since G has four irreducible characters, and the sum of the squares of their degrees is 12, there must be exactly three linear characters of G. Thus jG=G9j 3 by Theorem 17.11. It is not dif®cult to con®rm this by showing that

Some elementary character tables

181

G9 V4 f1, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)g: Now G=G9 fG9, G9(1 2 3), G9(1 3 2)g C3, and the character table of G=G9 is

~÷1 ~÷2 ~÷3

G9

G9(1 2 3)

G9(1 3 2)

1 1 1

1 ù ù2

1 ù2 ù

(where ù e2ði=3 ). The lifts of ÷~1 , ~÷2 , ~÷3 to G, together with the character ÷4 í, give the complete character table of A4 : Character table of A4

gi |CG ( gi )|

1 12

(1 2)(3 4) 4

(1 2 3) 3

(1 3 2) 3

÷1 ÷2 ÷3 ÷4

1 1 1 3

1 1 1 ÿ1

1 ù ù2 0

1 ù2 ù 0

18.3 The dihedral groups Let G be the dihedral group D2 n of order 2n, with n > 3, so that G ha, b: an b2 1, bÿ1 ab aÿ1 i: We shall derive the character table of G. Write å e2ði= n . For each integer j with 1 < j , n=2, de®ne j 0 1 å 0 Aj , B : j 1 0 0 åÿ j Check that ÿ1 A nj B2j I, Bÿ1 j Aj Bj A j :

It follows that by de®ning r j : G ! GL(2, C) by (ar bs )r j (Aj ) r (Bj ) s

(r, s 2 Z),

we obtain a representation r j of G for each j with 1 < j , n=2.

182


Each r j is an irreducible representation, either by the proof of Example 5.5(2) or by applying the result of Exercise 8.4. If i and j are distinct integers with 1 < i , n=2 and 1 < j , n=2, then å i 6 å j and å i 6 å ÿ j , so ar i and ar j have different eigenvalues. Therefore there is no matrix T with ar i T ÿ1 (ar j )T, and so r i and r j are not equivalent. Let ø j be the character of r j . We have now constructed distinct irreducible characters ø j of G, one for each j which satis®es 1 < j , n=2. At this point it is convenient to consider separately the cases where n is odd and where n is even. Case 1: n odd By (12.11) the conjugacy classes of D2 n (n odd) are f1g, far , aÿ r g(1 < r < (n ÿ 1)=2), fas b: 0 < s < n ÿ 1g: Thus there are (n 3)=2 conjugacy classes. The (n ÿ 1)=2 irreducible characters ø1 , ø2 , : : : , ø( nÿ1)=2 each have degree 2. As G has (n 3)/2 irreducible characters in all, there are two more to be found. Since kal v G and G=hai C2, we obtain two linear characters ÷1 , ÷2 of G by lifting the irreducible characters of G=hai to G. These characters ÷1 and ÷2 are given by ÷1 1 G and 1 if g ar for some r, ÷2 ( g) ÿ1 if g ar b for some r: We have now found all the irreducible characters of D2 n (n odd). (Incidentally, we have proved that D92 n kal for n odd, in view of Theorem 17.11.) The character table of D2 n (n odd) is therefore as follows (where å e2ði= n ): gi |CG ( gi )|

1 2n

÷1 ÷2 øj (1 < j < (n ÿ 1)/2)

1 1 2

ar (1 < r < (n ÿ 1)/2) n

å jr

1 1 å ÿ jr

b 2 1 ÿ1 0


183

Case 2: n even If n is even, say n 2m, then the conjugacy classes of D2 n, as supplied by (12.12), are f1g, fam g, far , aÿ r g(1 < r < m ÿ 1), fas b: s eveng, fas b: s oddg: Hence G has m 3 irreducible characters, of which m ÿ 1 are given by ø1 , ø2 , : : : , ø mÿ1 : To ®nd the remaining four irreducible characters, we ®rst note that ha2 i fa j : j eveng is a normal subgroup of G and G=ha2 i fha2 i, ha2 ia, ha2 ib, ha2 iabg C2 3 C2 : Therefore G has four linear characters ÷1 , ÷2 , ÷3 , ÷4 (and G9 ka2 l). Since these linear characters are the lifts of the irreducible characters of G/ka2 l, they are easy to calculate, and their values appear in the following complete character table of D2 n (n even, n 2m, å e2ði= n ).

gi |CG ( gi )|

1 2n

am 2n

a r (1 < r < m ÿ 1) n

b 4

ab 4

÷1 ÷2 ÷3 ÷4 øj (1 < j < m ÿ 1)

1 1 1 1 2

1 1 (ÿ1) m (ÿ1) m 2(ÿ1) j

1 1 (ÿ1) r (ÿ1) r å jr å ÿ jr

1 ÿ1 1 ÿ1 0

1 ÿ1 ÿ1 1 0

18.4 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, and we shall construct the character table of G. It is in fact known that every non-abelian group of order

184


12 is isomorphic to A4 , D12 or G, but we shall not prove this result here. Let a and b be the following permutations in S12 : a (1 2 3 4 5 6)(7 8 9 10 11 12), b (1 7 4 10)(2 12 5 9)(3 11 6 8), and let G ka, bl, a subgroup of S12 . Since a has order 6 and b 2 = kal, the group G has at least 12 elements, namely ar , ar b (0 < r < 5): Check that a and b satisfy a6 1, a3 b2 , bÿ1 ab aÿ1 : It follows from these relations that every element of G has the form ar bs with 0 < r < 5, 0 < s < 1 as given above, and so |G| 12. The relations further imply that CG (a) hai, CG (a3 ) G, CG (b) f1, a3 , b, a3 bg: These, and similar facts, help us to ®nd the conjugacy classes of G, which are tabulated below: Conjugacy class

Representative gi

|CG ( gi )|

{1} {a3 } {a, aÿ1 } {a2 , aÿ2 } {b, a2 b, a4 b} {ab, a3 b, a5 b}

1 a3 a a2 b ab

12 12 6 6 4 4

Therefore G has six irreducible characters. Observe that ka2 l {1, a2 , a4 } v G, and G=ha2 i fha2 i, ha2 ia, ha2 ib, ha2 iabg: Since ka2 la ka2 lb2 , we have G=ha2 i C4 . By lifting the irreducible characters of C4 to G, we obtain the linear characters ÷1 , ÷2 , ÷3 , ÷4 of G given below:


185

gi |CG ( gi )|

1 12

a3 12

a 6

a2 6

b 4

ab 4

÷1 ÷2 ÷3 ÷4 ÷5 ÷6

1 1 1 1 á1 â1

1 ÿ1 1 ÿ1 á2 â2

1 ÿ1 1 ÿ1 á3 â3

1 1 1 1 á4 â4

1 i ÿ1 ÿi á5 â5

1 ÿi ÿ1 i á6 â6

It remains to ®nd the values á r , â r taken by the last two irreducible characters ÷5 , ÷6 . For this, we shall use the column orthogonality relations, Theorem 16.4(2). Observe that á1 , â1 are the degrees of ÷5 , ÷6 , so they are positive integers; also a3 is an element of order 2, so á2 and â2 are integers by Corollary 13.10. By the column orthogonality relations applied to columns 1 and 2, we have 4 á21 â21 12, 4 á22 â22 12, á1 á2 â1 â2 0: Since á1 , â1 are positive integers, the ®rst equation gives á1 â1 2. The other two equations then imply that á2 ÿâ2 2. Since we have not yet distinguished between ÷5 and ÷6 , we may take á2 2 and â2 ÿ2. For r . 2, the column orthogonality relations 6 X

÷ i ( g r )÷ i ( g 1 ) 0 and

i1

6 X

÷ i ( g r )÷ i ( g 2 ) 0

i1

now give us two equations involving 2á r 2â r and 2á r ÿ 2â r , respectively, so we can solve them for á r and â r . Explicitly: r 3:

2á3 2â3 0, 4 2á3 ÿ 2â3 0;

r 4:

4 2á4 2â4 0, 2á4 ÿ 2â4 0;

r 5:

2á5 2â5 0, 2á5 ÿ 2â5 0;

r 6:

2á6 2â6 0, 2á6 ÿ 2â6 0:

186


Hence á3 ÿ1,

â3 1,

á4 ÿ1,

â4 ÿ1,

á5 0,

â5 0,

á6 0,

â6 0:

The complete character table of G is therefore as follows: gi CG ( gi )|

1 12

a3 12

a 6

a2 6

b 4

ab 4

÷1 ÷2 ÷3 ÷4 ÷5 ÷6

1 1 1 1 2 2

1 ÿ1 1 ÿ1 2 ÿ2

1 ÿ1 1 ÿ1 ÿ1 1

1 1 1 1 ÿ1 ÿ1

1 i ÿ1 ÿi 0 0

1 ÿi ÿ1 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, without constructing the corresponding CG-modules. 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. (In fact, it is not hard to construct the representations of the above group G with characters ÷5 and ÷6 ± see Exercise 17.6.)

Summary of Chapter 18 In this chapter we gave the character tables of various groups, as follows. 1. Section 18.1: the group S4 . 2. Section 18.2: the group A4 . 3. Section 18.3: the dihedral groups.


187

Exercises for Chapter 18 1. Regard D8 as a subgroup of S4 permuting the four corners of a square, as in Example 1.1(3). Let ð be the corresponding permutation character of D8. Find the values of ð on the elements of D8, and express ð as a sum of irreducible characters. 2. Write down explicitly the character table of D12, and show that all its entries are integers. Use the character table to ®nd seven distinct normal subgroups of D12. (Hint: use Proposition 17.5.) 3. Let G T 4n ha, b: a2 n 1, an b2 , bÿ1 ab aÿ1 i, as in Exercise 17.6. Find the character table of G. (Hint: use the result of Exercise 17.6. It is a good idea to do the cases n odd and n even separately.) 4. Let G U6 n ka, b: a2n b3 1, aÿ1 ba bÿ1 l, as in Exercise 17.7. Find the character table of G. 5. Let G V8 n ha, b: a2n b4 1, ba aÿ1 bÿ1 , bÿ1 a aÿ1 bi, with n odd, as in Exercise 17.8. Find the character table of 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, and it can be extended to include the product of any pair of characters ÷ and ø. The value of the product ÷ø on an element g of G is simply ÷( g)ø( g). It is therefore straightforward to calculate the product, but a little ingenuity is required in order to justify the conclusion that the product ÷ø is a character of G. The plan is to take CG-modules V and W with characters ÷ and ø respectively, and to put them together to form a new CG-module, called the tensor product of V and W, which has character ÷ø. An important special case of the product ÷ø occurs when ÷ ø, so we consider the character ÷ 2 , and more generally ÷ 3 , ÷ 4 , and so on. If ÷ is not linear, then the degrees of ÷, ÷ 2 , . . . increase, and by taking successive powers of ÷ we obtain arbitrarily many new characters. Potentially, then, we have a chance of getting a large proportion of the character table of G from just one non-linear character ÷ of G; and indeed, products of characters provide a very good source of new characters from given ones. We shall illustrate this by constructing the character tables of S5 and S6 . At the end of the chapter, we apply tensor products in a different way, to ®nd all the irreducible characters of a direct product G 3 H, given those of G and H. Tensor product spaces Let V and W be vector spaces over C with bases v1 , . . . , v m and w1 , : : : , wn , respectively. For each i, j with 1 < i < m, 1 < j < n, we introduce a symbol v i wj. The tensor product space V W is de®ned to be the mn-dimensional vector space over C with a basis given by 188

Tensor products

189

fv i wj : 1 < i < m, 1 < j < ng: Thus V W consists of all expressions of the form X ë ij (v i wj ) (ë ij 2 C): i, j

Pm For v 2 V and w 2 W with v i1 ë i v i and (ë i , ì j 2 C), we de®ne v w 2 V W by X v w ë i ì j (v i wj ):

w

Pn

j1

ì j wj

i, j

For example, (2v1 ÿ v2 ) (w1 w2 ) 2v1 w1 2v1 w2 ÿ v2 w1 ÿ v2 w2 : Do not be misled by the notation into believing that every element of V W has the form v w, because this is not the case. For instance, it is impossible to express v1 w1 v2 w2 in the form v w. 19.1 Proposition (1) If v 2 V, w 2 W and ë 2 C, then v (ëw) (ëv) w ë(v w): (2) If x1 , . . . , xa 2 V and y1, . . . , yb 2 W, then ! 0 b 1 a X X X xi @ yj A xi yj : i1

Proof (1) Let v v (ëw)

Pm

P ë i v i and w nj1 ì j w j. Then ! ! X X X ëi vi

ëì j w j ëë i ì j (v i wj ), i1

i

(ëv) w

X i

ë(v w) ë

i, j

j1

X i, j

j

! ëë i v i

i, j

X

! ì jw j

j

ë i ì j (v i wj )

X

ëë i ì j (v i wj ),

i, j

X i, j

ëë i ì j (v i wj ):

190


Therefore v (ëw) (ëv) w ë(v w). The proof of part (2) is equally straightforward, and we leave it as an exercise. j Our construction of V W depended upon choosing a basis of V and a basis of W at the beginning; the next proposition shows that other bases of V and W work equally well. 19.2 Proposition If e1 , . . . , em is a basis of V and f1 , . . . , fn is a basis of W, then the elements in fei f j : 1 < i < m, 1 < j < ng give a basis of V W. Proof Write vi

m X

ë ik ek , wj

k1

n X

ì jl f l

(ë ik , ì jl 2 C):

l1

Then by Proposition 19.1, we have X v i wj ë ik ì jl (ek f l ): k, l

Now the elements V W , and hence span V W. Since elements ek fl are

v i wj (1 < i < m, 1 < j < n) are a basis of the mn elements ek fl (1 < k < m, 1 < l < n) V W has dimension mn, it follows that the also a basis of V W.

Tensor product modules We have introduced the tensor product of two vector spaces, so we are now in a position to de®ne the tensor product of two CG-modules. Let G be a ®nite group and let V and W be CG-modules with bases v1 , . . . , v m and w1 , : : : , w n, respectively. We know that the elements v i wj

(1 < i < m, 1 < j < n)

give a basis of V W. The multiplication of v i wj by an element of

Tensor products

191

G is de®ned in the following simple way, which is then extended linearly to a multiplication on the whole of V W. 19.3 De®nition Let g 2 G. For all i, j, de®ne (v i wj ) g v i g w j g and, more generally, let ! X X ë ij (v i wj ) g ë ij (v i g wj g) i, j

i, j

for arbitrary complex numbers ë ij . 19.4 Proposition For all v 2 V, w 2 W and all g 2 G, we have (v w) g v g wg: Proof Let v

Pm

(v w) g

P ë i v i and w nj1 ì j wj. Then ! X ë i ì j (v i wj ) g by Proposition 19:1

i1

i, j

X

ë i ì j (v i g wj g)

i, j

X

! ëivi g

i

v g wg:

X

! ì j wj g

by Proposition 19:1

j j

You should be warned that (v w)r 6 vr wr for most elements r in CG. For example, consider what happens when r is a scalar multiple of g. 19.5 Proposition The rule for multiplying an element of V W by an element of G, given in De®nition 19.3, makes the vector space V W into a CGmodule.

192


Proof Let 1 < i < m, 1 < j < n, and g, h 2 G. Then (v i wj ) g v i g wj g 2 V W , (v i wj )( gh) v i ( gh) wj ( gh) (v i g)h (wj g)h (v i g wj g)h

by Proposition 19:4

((v i wj ) g)h, (v i wj )1 v i wj , and X i, j

! X ë ij (v i wj ) g ë ij ((v i wj ) g): i, j

Therefore all the conditions of Proposition 4.6 are ful®lled, and V W is a CG-module. j We now calculate the character of V W. 19.6 Proposition Let V and W be CG-modules with characters ÷ and ø, respectively. Then the character of the CG-module V W is the product character ÷ø, where ÷ø( g) ÷( g)ø( g)

for all g 2 G:

Proof Let g 2 G. By Proposition 9.11 we can choose a basis e1 , . . . , em of V and a basis f1 , . . . , fn of W such that ei g ë i e i

(1 < i < m) and f j g ì j f j

(1 < j < n)

for some complex numbers ë i , ì j . Then ÷( g)

m X

ë i , ø( g)

i1

n X

ì j:

j1

Now for 1 < i < m and 1 < j < n, (ei f j ) g ei g f j g ë i ì j (ei f j ), and by Proposition 19.2, these vectors ei fj form a basis of V W. Hence, if ö is the character of V W then

Tensor products ! X X X ! ö( g) ëi ì j ëi ì j ÷( g)ø( g), i, j

i

193

j

as required.

j

19.7 Corollary The product of two characters of G is again a character of G. 19.8 Example The character table of S4 was given in Section 18.1. We reproduce it here, and calculate ÷3 ÷4 and ÷4 ÷4 . gi |CG ( gi )|

1 24

(1 2) 4

(1 2 3) 3

(1 2)(3 4) 8

(1 2 3 4) 4

÷1 ÷2 ÷3 ÷4 ÷5

1 1 2 3 3

1 ÿ1 0 1 ÿ1

1 1 ÿ1 0 0

1 1 2 ÿ1 ÿ1

1 ÿ1 0 ÿ1 1

÷3 ÷4 ÷4 ÷4

6 9

0 1

0 0

ÿ2 1

0 1

We see that ÷3 ÷4 ÷4 ÷5 , and ÷4 ÷4 ÷1 ÷3 ÷4 ÷5 :

Powers of characters Corollary 19.7 shows that if ÷ is a character of G then so is ÷ 2 , where ÷ 2 ÷÷, the product of ÷ with itself. More generally, for every nonnegative integer n, we de®ne ÷ n by ÷ n ( g) (÷( g)) n

for all g 2 G:

Thus ÷ 0 1 G . An inductive proof using Corollary 19.7 shows that ÷ n is a character of G. When ÷ is a faithful character (that is, Ker {1}), the powers of ÷ carry a lot of information about the whole character table of G, as can be seen from Theorem 19.10 below.

194


In the course of the proof of Theorem 19.10, we shall need the following result concerning the so-called `Vandermonde matrix'. (19.9) If á1 , . . . , á r are distinct 0 1 á1 B1 á 2 B AB @: : 1

ár

complex numbers, then the matrix 1 á21 : : : á1rÿ1 á22 : : : á2rÿ1 C C C : ::: : A á2r

:::

á rÿ1 r

is invertible. We ®rst sketch a proof of this result. Suppose that x1 , . . . , xr are indeterminates, and consider 0 1 1 x1 x21 : : : x1rÿ1 B 1 x x2 : : : x rÿ1 C 2 B C 2 2 Ä detB C: @: : : ::: : A 1 xr x2r : : : x rÿ1 r If i 6 j and xi xj then two rows of the given matrix are equal, so Ä 0. It follows that Ä is divisible by Y (xi ÿ x j ) (x1 ÿ x2 )(x1 ÿ x3 ) : : : (x1 ÿ xr ) i, j

3 (x2 ÿ x3 ) : : : (x2 ÿ xr ) .. . 3 (x rÿ1 ÿ xr ): Now the coef®cient of x1rÿ1 x2rÿ2 : : :x rÿ1 in this product is 1: for in the way we have displayed the product, the term x1rÿ1 must come from all the factors in the ®rst row, x2rÿ2 must come from all the factors in the second row, and so on. On the other hand, to obtain x1rÿ1 x2rÿ2 : : : x rÿ1 in the expansion of the determinant Ä, we must take x1rÿ1 from the ®rst row of the matrix, x2rÿ2 from the second row, and so on. Hence the coef®cient of x1rÿ1 x2rÿ2 . . . x rÿ1 in Ä is 1. It follows that Y Ä (xi ÿ xj ): i, j

Tensor products

195

To obtain (19.9), we substitute á i for xi (1 < i < r), and deduce that the matrix A is invertible since its determinant is non-zero. 19.10 Theorem Let ÷ be a faithful character of G, and suppose that ÷( g) takes precisely r different values as g varies over all the elements of G. Then every irreducible character of G is a constituent of one of the powers ÷ 0 , ÷ 1 , . . . , ÷ rÿ1 . Proof Let the r values taken by ÷ be á1 , . . . , á r , and for 1 < i < r, de®ne Gi f g 2 G: ÷( g) á i g: Take á1 ÷(1), so that G1 Ker ÷. As ÷ is faithful, G1 {1}. Now let ø be an irreducible character of G. We must show that h÷ j , øi 6 0 for some j with 0 < j < r ÿ 1. For 1 < i < r, let X âi ø( g), g2G i

and note that â1 ø(1) 6 0. Then for all j > 0, h÷ j , øi

r 1 X 1 X (÷( g)) j ø( g) (á i ) j â i : jGj g2G jGj i1

Let A be the r 3 r matrix with ij-entry (á i ) jÿ1 , and let b be the row vector which is given by b (â1 , : : : , â r ): Now A is invertible by (19.9), and b 6 0 since â1 6 0; hence bA 6 0. But the ( j 1)th entry in the row vector bA is equal to jGjh÷ j , øi, and thus h÷ j , øi 6 0 for some j with 0 < j < r ÿ 1, as we wished to prove. j 19.11 Examples (1) If G 6 {1} and ÷ is the regular character of G, then ÷( g) takes just two different values (see Proposition 13.20), so Theorem 19.10 says that every irreducible character of G is a constituent of 1 G or ÷; we know this already, by Theorem 10.5. (2) Let G S4 , and refer to Example 19.8. Let ÷ ÷4 . Then ÷( g) takes four different values. We have seen that ÷ 2 ÷1 ÷3 ÷4 ÷5

196


and we ®nd that

h÷ 3 , ÷2 i 1:

Thus ÷ 0 , ÷ 1 , ÷ 2 , ÷ 3 (indeed, in this case, just ÷ 2 , ÷ 3 ) have between them as constituents all the irreducible characters ÷1 , . . . , ÷5 of G, illustrating Theorem 19.10. Decomposing ÷ 2 In view of Theorem 19.10, it is of some importance to be able to decompose powers of a character ÷ into sums of irreducible characters. We are going to provide a method for decomposing ÷ 2 , the square of ÷. This special case is particularly useful in ®nding irreducible characters, as we shall see. Let V be a CG-module with character ÷. By Proposition 19.6, the module V V has character ÷ 2 . Let v1 , . . . , v n be a basis of V, and de®ne a linear transformation T: V V ! V V by (v i v j )T v j v i

for all i, j

and extending linearly ± that is, ! X X ë ij (v i v j ) T ë ij (v j v i ): i, j

i, j

Check that for all v, w 2 V, we have (v w)T w v: Hence T is independent of the choice of basis. Now de®ne subsets of V V as follows: S(V V ) fx 2 V V : xT xg, A(V V ) fx 2 V V : xT ÿxg, Since T is linear, it is easy to see that S(V V) and A(V V) are subspaces of V V (indeed, they are eigenspaces of T). The subspace S(V V) is called the symmetric part of V V, and the subspace A(V V ) is known as the antisymmetric part of V V. 19.12 Proposition The subspaces S(V V) and A(V V) are CG-submodules of V V. Also, V V S(V V ) A(V V ):

Tensor products

197

Proof For all ë ij 2 C and g 2 G, ! X X ë ij (v i v j ) Tg ë ij (v j g v i g) i, j

i, j

X i, j

ë ij (v i g v j g)T

X

! ë ij (v i v j ) gT :

i, j

Therefore T is a CG-homomorphism from V V to itself. Hence, for x 2 S(V V), y 2 A(V V) and g 2 G, we have (xg)T (xT ) g xg, and ( yg)T ( yT ) g ÿ yg, so xg 2 S(V V) and yg 2 A(V V). Thus S(V V) and A(V V) are CG-submodules of V V. If x 2 S(V V) \ A(V V) then x xT ÿx, so x 0. Further, for all x 2 V we have x 12(x xT ) 12(x ÿ xT ): Since T 2 is the identity, A(V V ). Therefore,

1 2(x

xT ) 2 S(V V ) and

V V S(V V ) A(V V ):

1 2(x

ÿ xT ) 2 j

Note that the symmetric part of V V contains all vectors which have the form v w w v with v, w 2 V, while the antisymmetric part of V V contains all vectors of the form v w ÿ w v. We now present bases of the symmetric and antisymmetric parts of V V which consist of elements like these. 19.13 Proposition Let v1 , . . . , v n be a basis of V. (1) The vectors v i v j v j v i (1 < i < j < n) form a basis of S(V V ). The dimension of S(V V ) is n(n 1)=2. (2) The vectors v i v j ÿ v j v i (1 < i , j < n) form a basis of A(V V). The dimension of A(V V ) is n(n ÿ 1)=2.

198


Proof Clearly the vectors v i v j v j v i (1 < i < j < n) are linearly independent elements of S(V V ), and the vectors v i v j ÿ v j v i (1 < i , j < n) are linearly independent elements of A(V V ). Hence dim S(V V ) > n(n 1)=2,

dim A(V V ) > n(n ÿ 1)=2:

By Proposition 19.12, dim S(V V ) dim A(V V ) dim V V n2 : Hence the above inequalities are equalities, and the result follows.

j

De®ne ÷ S to be the character of the CG-module S(V V ), and ÷ A to be the character of the CG-module A(V V ). By Proposition 19.12, ÷2 ÷S ÷ A: The next result gives the values of the characters ÷ S and ÷ A . 19.14 Proposition For g 2 G, we have ÷ S ( g) 12(÷ 2 ( g) ÷( g 2 )), and ÷ A ( g) 12(÷ 2 ( g) ÿ ÷( g 2 )): Proof By Proposition 9.11 we can choose a basis e1 , . . . , en of V such that e i g ë i e i (1 < i < n) for some complex numbers ë i . Then (ei ej ÿ ej ei ) g ë i ë j (ei ej ÿ ej ei ), and hence from Proposition 19.13(2), X ÷ A ( g) ëi ë j : P

i, j

P Now e i g 2 ë2i ei, so ÷( g) i ë i and ÷( g2 ) i ë2i . Therefore X 2 X ÷ 2 ( g) (÷( g))2 ëi 2 ë i ë j ÷( g 2 ) 2÷ A ( g): i

i, j

Hence ÷ A ( g) 12(÷ 2 ( g) ÿ ÷( g 2 )): Also, ÷ 2 ÷ S ÷ A , which implies that ÷ S ( g) ÷ 2 ( g) ÿ ÷ A ( g) 12(÷ 2 ( g) ÷( g 2 )):

j

Tensor products

199

19.15 Example Let G S4 . The character table of G is given in Example 19.8. Let ÷ ÷4 . The values of ÷, and the values of ÷ S and ÷ A , given by Proposition 19.14, appear below. gi |CG ( gi )|

1 24

(1 2) 4

(1 2 3) 3

(1 2)(3 4) 8

(1 2 3 4) 4

÷ ÷S ÷A

3 6 3

1 2 ÿ1

0 0 0

ÿ1 2 ÿ1

ÿ1 0 1

We ®nd that ÷ S ÷1 ÷3 ÷4 and ÷ A ÷5 . The techniques which we have developed so far give a useful method for ®nding new irreducible characters of a group, given one or two irreducible characters to start with. The strategy is simple: (1) Given a character ÷, form ÷ S and ÷ A , and use inner products to analyse ÷ S and ÷ A for new irreducible characters. (2) If ø is a new character found in (1), then form ø S and ø A and repeat. We illustrate this strategy with two examples. 19.16 Example The character table of S5 Let G S5 , the symmetric group of degree 5. By Example 12.16(4), G has conjugacy class representatives gi , 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. (a) Linear characters By Example 17.13, G9 A5 and G has exactly two linear characters ÷1 , ÷2 , obtained by lifting the irreducible characters of G=G9. We have

200

Representations and characters of groups ÷2 ( g)

÷1 1 G , and 1, ÿ1,

(b) Permutation character of G with values

if g is an even permutation, if g is an odd permutation: Proposition 13.24 gives us a character ÷3

÷3 ( g) jfix ( g)j ÿ 1 ( g 2 G): Observe that h÷3 , ÷3 i

42 22 12 (ÿ1)2 (ÿ1)2 1: 120 12 6 6 5

Hence ÷3 is irreducible, by Theorem 14.20. Next, Proposition 17.14 shows that ÷4 ÷3 ÷2 is also an irreducible character. At this point we have the following portion of the character table of G: gi |CG ( gi )| ÷1 ÷2 ÷3 ÷4

1 120

(1 2) 12

1 1 4 4

1 ÿ1 2 ÿ2

(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 ÿ1 0 0

1 ÿ1 ÿ1 1

1 1 ÿ1 ÿ1

(c) Tensor products We now use tensor products to construct the last three irreducible characters of G. Write ÷ ÷3 . By Proposition 19.14 the values of the characters ÷ S and ÷ A are gi |CG ( gi )|

1 120

(1 2) 12

÷S ÷A

10 6

4 0

Thus,

(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 ÿ2

0 0

1 0

0 1

Tensor products h÷ A , ÷ A i

201

36 4 1 1, 120 8 5

and so ÷ A is a new irreducible character, which we call ÷5 . Next, 10 4 1 2 1 1, 120 12 6 8 6 40 8 1 1 h÷ S , ÷3 i ÿ 1, and 120 12 6 6 100 16 1 4 1 h÷ S , ÷ S i 3, 120 12 6 8 6 h÷ S , ÷1 i

Therefore, ÷ S ÷1 ÷3 ø, where ø is an irreducible character of degree 5. Let ÷6 ø, so that ÷6 ÷ S ÿ ÷1 ÿ ÷3 . Finally, ÷7 ÷6 ÷2 is a different irreducible character of degree 5. We have now found all seven irreducible characters of S5 . The character table of S5 is as shown. Character table of S5 gi |CG ( gi )| ÷1 ÷2 ÷3 ÷4 ÷5 ÷6 ÷7

1 120

(1 2) 12

1 1 4 4 6 5 5

1 ÿ1 2 ÿ2 0 1 ÿ1

(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 ÿ1 ÿ1

1 1 0 0 ÿ2 1 1

1 ÿ1 0 0 0 ÿ1 1

1 ÿ1 ÿ1 1 0 1 ÿ1

1 1 ÿ1 ÿ1 1 0 0

19.17 Example The character table of S6 In this example, we use techniques similar to those of the previous example to ®nd 8 of the 11 irreducible characters of the symmetric group S6 ; we then ®nd the last three irreducible characters by using the orthogonality relations. Let G S6 , of order 720. For ease of printing, it is convenient to label each conjugacy class by the cycle-shape of its elements. Using

202


this notation, the conjugacy class sizes and centralizer orders are as follows (see Exercise 12.3): Cycle-shape Class size |CG ( gi )|

(1) (2) 1 15 720 48

(3) 40 18

(2,2) 45 16

(4) 90 8

(3,2) (5) (2,2,2) (3,3) (4,2) (6) 120 144 15 40 90 120 6 5 48 18 8 6

Since G has 11 conjugacy classes, it has 11 irreducible characters. (a) Linear characters As with all symmetric groups Sn (n > 2), the derived subgroup is An , and we get exactly two linear characters ÷1 and ÷2 , where ÷2 ( g)

÷1 1 G , 1, ÿ1,

if g is even, if g is odd

(see Example 17.13). (b) Permutation character and tensor products is given by ÷3 ( g) jfix ( g)j ÿ 1

The function ÷3 which

( g 2 G)

is a character of G, by Proposition 13.24. Let ÷ ÷3 . The values of ÷, ÷ S and ÷ A are as follows: Class |CG ( gi )|

(1) (2) 720 48

÷ ÷3 ÷S ÷A

5 15 10

3 7 2

(3) 18

(2,2) 16

(4) 8

(3,2) 6

(5) 5

2 3 1

1 3 ÿ2

1 1 0

0 1 ÿ1

0 0 0

We calculate that h÷3 , ÷3 i 1,

h÷ A , ÷ A i 1,

h÷ S , ÷1 i 1,

h÷ S , ÷3 i 1,

h÷ S , ÷ S i 3:

(2,2,2) (3,3) (4,2) (6) 48 18 8 6 ÿ1 3 ÿ2

ÿ1 0 1

ÿ1 1 0

ÿ1 0 1

Tensor products

203

Therefore ÷3 is irreducible; so is ÷4 ÷3 ÷2 . Also, ÷5 ÷ A is irreducible, as is ÷6 ÷5 ÷2 . Further, ÷ S ÷1 ÷3 ÷7 , where ÷7 is another irreducible character, of degree 9. Finally, ÷8 ÷7 ÷2 is also irreducible. The irreducible characters ÷1 , . . . , ÷8 are recorded in the following portion of the character table of G. Class |CG ( gi )|

(1) (2) 720 48

÷1 ÷2 ÷3 ÷4 ÷5 ÷6 ÷7 ÷8

1 1 5 5 10 10 9 9

1 ÿ1 3 ÿ3 2 ÿ2 3 ÿ3

(3) 18

(2,2) 16

(4) 8

(3,2) 6

(5) 5

1 1 2 2 1 1 0 0

1 1 1 1 ÿ2 ÿ2 1 1

1 ÿ1 1 ÿ1 0 0 ÿ1 1

1 ÿ1 0 0 ÿ1 1 0 0

1 1 0 0 0 0 ÿ1 ÿ1

(2,2,2) (3,3) (4,2) (6) 48 18 8 6 1 ÿ1 ÿ1 1 ÿ2 2 3 ÿ3

1 1 ÿ1 ÿ1 1 1 0 0

1 1 ÿ1 ÿ1 0 0 1 1

1 ÿ1 ÿ1 1 1 ÿ1 0 0

(c) Orthogonality relations We now use the column orthogonality relations to complete the character table of G. It will be shown later (Corollary 22.16) that all the entries in the character tables of all symmetric groups are integers, but for the moment we know for certain only that ÷( g) is an integer if g2 1 (see Corollary 13.10). It is therefore convenient ®rst to concentrate on elements of order 2, so that we can guarantee that the solutions to the equations which we deal with are integers. Let s denote the permutation (1 2) and t denote the permutation (1 2)(3 4). Thus the conjugacy classes of s and t correspond to the second and fourth columns of the character table, respectively, in the ordering which we have adopted. From Corollary 13.10 we know that ÷(s) and ÷(t) are integers for all characters ÷ of G. Ingeniously, we call the three irreducible characters of G which have yet to be found ÷9 , ÷10 and ÷11 . The column orthogonality relations give 11 X i1

÷ i (s)2 48:

204


Hence ÷9 (s)2 ÷10 (s)2 ÷11 (s)2 2: We can assume, without loss of generality, that ÷9 (s)2 ÷10 (s)2 1, ÷11 (s)2 0: Now ÷9 ÷2 is an irreducible character, and is not equal to any of ÷1 , . . . , ÷8 . Moreover, since ÷9 ÷2 (s) ÿ÷9 (s), we see that ÷9 ÷2 is not ÷9 or ÷11 . Therefore, ÷9 ÷2 ÷10 : Once more, we lose no generality in assuming that ÷9 (s) 1, ÷10 (s) ÿ1: The plan is now to ®nd ÷ i (1) and ÷ i (t) for i 9, 10, 11. That is, we aim to evaluate the integers a, b, c, d, e, f in the following portion of the character table: Element Class

1 (1)

s (2)

a b c

÷9 ÷10 ÷11

t (2,2)

1 ÿ1 0

d e f

The column orthogonality relations give 11 X

÷ i (1)÷ i (s) 0,

i1 11 X

11 X

÷ i (s)÷ i (t) 0,

i1

÷ i (t)÷ i (t) 16,

i1

11 X

÷ i (1)÷ i (t) 0,

i1

whence a ÿ b 0, d 2 e 2 f 2 2,

d ÿ e 0, ad be cf 10:

The only solution to these equations in integers with a . 0 and b . 0 is d e 1, f 0, a b 5:

Tensor products

205

Finally, we ®nd that c 16 by using the relation 11 X

÷ i (1)2 720:

i1

The above portion of the character table is therefore Element Class

1 (1)

÷9 ÷10 ÷11

5 5 16

s (2) 1 ÿ1 0

t (2,2) 1 1 0

We can now determine the three unknown entries in each further column, since the column orthogonality relations will give three independent equations in these unknowns (as the above 3 3 3 matrix is invertible). Having done these calculations, we ®nd that the complete character table of S6 is as shown. Character table of S6 Class |CG ( gi )|

(1) (2) 720 48

(3) 18

(2,2) 16

(4) 8

(3,2) 6

(5) 5

(2,2,2) 48

÷1 ÷2 ÷3 ÷4 ÷5 ÷6 ÷7 ÷8 ÷9 ÷10 ÷11

1 1 5 5 10 10 9 9 5 5 16

1 1 2 2 1 1 0 0 ÿ1 ÿ1 ÿ2

1 1 1 1 ÿ2 ÿ2 1 1 1 1 0

1 ÿ1 1 ÿ1 0 0 ÿ1 1 ÿ1 1 0

1 ÿ1 0 0 ÿ1 1 0 0 1 ÿ1 0

1 1 0 0 0 0 ÿ1 ÿ1 0 0 1

1 ÿ1 ÿ1 1 ÿ2 2 3 ÿ3 ÿ3 3 0

1 ÿ1 3 ÿ3 2 ÿ2 3 ÿ3 1 ÿ1 0

(3,3) (4,2) (6) 18 8 6 1 1 ÿ1 ÿ1 1 1 0 0 2 2 ÿ2

1 1 ÿ1 ÿ1 0 0 1 1 ÿ1 ÿ1 0

1 ÿ1 ÿ1 1 1 ÿ1 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. Let V be a CG-module, with basis v1 , . . . , v m , and let W be a

206


C H-module, with basis w1, . . . , wn. For all i, j with 1 < i < m and 1 < j < n, and all g 2 G, h 2 H, de®ne (v i wj )( g, h) v i g wj h and extend this de®nition linearly to the whole of V W, that is, for ë ij 2 C, ! X X ë ij (v i wj ) ( g, h) ë ij (v i g wj h): i, j

i, j

As in Proposition 19.4, we ®nd that (v w)( g, h) v g wh, for all v 2 V, w 2 W. Then a proof similar to that of Proposition 19.5 shows that V W is a C(G 3 H)-module. Let ÷ be the character of V and ø be the character of W. By the proof of Proposition 19.6, the character of V W is ÷ 3 ø, where (÷ 3 ø)( g, h) ÷( g)ø(h)

( g 2 G, h 2 H):

19.18 Theorem Let ÷1 , . . . , ÷ a be the distinct irreducible characters of G and let ø1 , . . . , ø b be the distinct irreducible characters of H. Then G 3 H has precisely ab distinct irreducible characters, and these are ÷ i 3 ø j (1 < i < a, 1 < j < b): Proof For all i, j, k, l, h÷ i 3 ø j , ÷ k 3 ÷ l i G3 H

X 1 ÷ i ( g)ø j (h)÷ k ( g)ø l (h) jG 3 Hj g2G h2 H

! ! 1 X 1 X ÷ i ( g)÷ k ( g) ø j (h)ø l (h) jGj g2G j Hj h2 H

h÷ i , ÷ k i G hø j , ø l i H ä ik ä jl : (Here the subscripts G 3 H, G and H indicate inner products of characters of G 3 H, G and H, respectively.) Thus the ab characters ÷ i 3 ø j are distinct and irreducible. Next, note that for all g, x 2 G and h, y 2 H, we have (x, y)ÿ1 ( g, h)(x, y) (x ÿ1 gx, y ÿ1 hy):

Tensor products

207

Hence elements ( g, h) and ( g9, 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. Consequently, if g1 , . . . , ga are representatives of the conjugacy classes of G and h1 , . . . , hb are representatives of the conjugacy classes of H, then the elements ( g i , hj )

(1 < i < a, 1 < j < b)

are representatives of the conjugacy classes of G 3 H. In particular, G 3 H has precisely ab conjugacy classes. By Theorem 15.3, G 3 H has exactly ab irreducible characters, so the irreducible characters ÷ i 3 ø j which we have found must be all the irreducible characters of G 3 H. j 19.19 Example The character table of S3 3 C2 The character table of S3 ( D6 ) is given in Example 16.3(1). We reproduce it here, alongside the character table of C2 . Character table of S3

Character table of C2

gi |CG ( gi )|

1 6

(1 2) 2

(1 2 3) 3

÷1 ÷2 ÷3

1 1 2

1 ÿ1 0

1 1 ÿ1

hi |CH (hi )|

1 2

ÿ1 2

ø1 ø2

1 1

1 ÿ1

The conjugacy classes of S3 3 C2 are represented by (1, 1), ((1 2), 1), ((1 2 3), 1), (1, ÿ1), ((1 2), ÿ1), ((1 2 3), ÿ1), and by Theorem 19.18, the character table of S3 3 C2 is as shown. Character table of S3 3 C2 ( gi , hj ) |CG3 H ( g i , hj )| ÷1 ÷2 ÷3 ÷1 ÷2 ÷3

3 3 3 3 3 3

ø1 ø1 ø1 ø2 ø2 ø2

(1, 1) ((1 2), 1) ((1 2 3), 1) (1, ÿ1) ((1 2), ÿ1) ((1 2 3), ÿ1) 12 4 6 12 4 6 1 1 2 1 1 2

1 ÿ1 0 1 ÿ1 0

1 1 ÿ1 1 1 ÿ1

1 1 2 ÿ1 ÿ1 ÿ2

1 ÿ1 0 ÿ1 1 0

1 1 ÿ1 ÿ1 ÿ1 1

208


Compare the solution to Exercise 18.2, where we give the character table of D12 (Exercise 1.5 shows that D12 S3 3 C2 ).

Summary of Chapter 19 1. The product of any two characters of G is a character of G. 2. If ÷ is a character of G, then so are ÷ S and ÷ A , where ÷ S ( g) 12(÷ 2 ( g) ÷( g 2 )), ÷ A ( g) 12(÷ 2 ( g) ÿ ÷( g 2 )) for all g 2 G. 3. The irreducible characters of G 3 H are those characters ÷ 3 ø, where ÷ is an irreducible character of G and ø is an irreducible character of H. The values of ÷ 3 ø are given by (÷ 3 ø)( g, h) ÷( g)ø(h) for all g 2 G, h 2 H.

Exercises for Chapter 19 1. Let ÷, ø and ö be characters of the group G. Show that h÷ø, öi h÷, øöi hø, ÷öi: 2. Suppose that ÷ and ø are irreducible characters of G. Prove that 1, if ÷ ø, h÷ø, 1 G i 0, if ÷ 6 ø: 3. Let ÷ be a character of G which is not faithful. Show that there is some irreducible character ø of G such that k÷ n , øl 0 for all integers n with n > 0. (This shows that the hypothesis that ÷ is faithful cannot be dropped from Theorem 19.10.) 4. In Example 20.13 of the next chapter we shall show that there exist irreducible characters ÷ and ö of A5 which take the following values:

Tensor products

÷ ö

1

(1 2 3)

(1 2)(3 4)

5 3

ÿ1 0

1 ÿ1

209 (1 2 3 4 5)

(1

0 p

(1 3 4 5 2)

5)=2

(1 ÿ

0 p

5)=2

Calculate the values of ÷ S , ÷ A , ö S and ö A . Express these characters as linear combinations of the irreducible characters ø1 , . . . , ø5 of A5 which are given in Example 20.13. 5. A certain group G or order 24 has precisely seven conjugacy classes with representatives g1 , . . . , g7 ; further, G has a character ÷ with values as follows: gi |CG ( gi )|

g1 24

g2 24

g3 4

g4 6

g5 6

g6 6

g7 6

÷

2

ÿ2

0

ÿù2

ÿù

ù

ù2

where ù e2ði=3 . Moreover, g21 , g22 , g23 , g24 , g25 , g26 , g27 are conjugate to g1 , g1 , g2 , g5 , g4 , g4 , g5 , respectively. Find ÷ S and ÷ A , and show that both are irreducible. By forming products of the irreducible characters found so far, ®nd the character table of G. 6. Write down the character table of D6 3 D6.

20 Restriction to a subgroup

In this chapter and the next, we are going to look at ways of relating the representations of a group to the representations of its subgroups. Here, we introduce the elementary idea of restricting a CG-module to a subgroup H of G, and illustrate its use. The case where H is a normal subgroup of G is of particular interest, and Clifford's Theorem 20.8 gives important information in this case. We apply this result in the situation where H is of index 2 in G, which occurs, for example, when G S n and H A n . Restriction Let H be a subgroup of the ®nite group G. Then C H is a subset of CG. If V is a CG-module, then V is also a C H-module, since properties (1)±(5) of De®nition 4.2 certainly hold for all g, h 2 H if they hold for all g, h 2 G. This simple way of converting a CG-module into a C H-module is known as restricting from G to H. If V is a CG-module then we write the corresponding C H-module as V # H, and call it the restriction of V to H. The character of V # H is obtained from the character ÷ of V by evaluating ÷ on the elements of H only. We write this character of H as ÷ # H, and refer to it as the restriction of ÷ to H. More generally, if f: G ! C is any function, then f # H denotes the restriction of f to H (so that ( f # H)(h) f (h) for all h 2 H). 20.1 Example Let G D8 ka, b: a4 b2 1, bÿ1 ab aÿ1 l. As in Example 4.5(1), let V be the CG-module with basis v1 , v2 for which v1 a v2 ,

v1 b v1 ,

v2 a ÿv1 ,

v2 b ÿv2 :

210

Restriction to a subgroup

211

If H is the subgroup {1, a2 , b, a2 b} of G, then V # H is the C Hmodule with basis v1 , v2 for which v1 a2 ÿv1 ,

v1 b v1 ,

v2 a2 ÿv2 ,

v2 b ÿv2 :

The character ÷ of V is given by g

1

a

a2

a3

b

ab

a2 b

a3 b

÷( g)

2

0

ÿ2

0

0

0

0

0

and the character ÷ # H of V # H is given by h

1

a2

b

a2 b

÷(h)

2

ÿ2

0

0

If V is a CG-module and H is a subgroup of G, then dim V dim (V # H). However, it might be the case that V is an irreducible CG-module while V # H is not an irreducible C H-module; Example 20.1 illustrates this fact. On the other hand, if V # H is an irreducible C H-module then V is an irreducible CG-module; for if U is a CG-submodule of V, then U # H is a C H-submodule of V # H.

20.2 Example Let G S5 and let H be the subgroup A4 of G consisting of all even permutations of {1, 2, 3, 4} ®xing 5. By 18.2, the character table of H is

gi jC H ( g i )j ø1 ø2 ø3 ø4

1 12

(1 2)(3 4) 4

(1 2 3) 3

(1 3 2) 3

1 1 1 3

1 1 1 ÿ1

1 ù ù2 0

1 ù2 ù 0

212


(where ù e2ði=3 ). The character table of G is given in Example 19.16, with irreducible characters labelled ÷1 , . . . , ÷7 . For each i with 1 < i < 7, we calculate the character ÷ i # H as a sum of irreducible characters ø j . From Example 19.16 we see that ÷1 # H ÷2 # H, ÷3 # H ÷4 # H and ÷6 # H ÷7 # H: Therefore we need only consider ÷1 # H, ÷3 # H, ÷5 # H and ÷6 # H. These have values

÷1 # ÷3 # ÷5 # ÷6 #

H H H H

1

(1 2)(3 4)

1 4 6 5

1 0 ÿ2 1

(1 2 3) (1 3 2) 1 1 0 ÿ1

1 1 0 ÿ1

It is easy to spot that ÷ 1 # H ø1 , ÷ 3 # H ø1 ø 4 , ÷5 # H 2ø4 , ÷ 6 # H ø2 ø 3 ø 4 :

Constituents of a restricted character To help us discuss the way in which a restricted character ÷ # H can be expressed in terms of the irreducible characters of H, we introduce the following notation. 20.3 De®nitions The inner product k , l G is the inner product on the vector space of functions from G to C which we have de®ned earlier, and k , l H is the inner product on the vector space of functions from H to C, de®ned similarly. Thus, if W1 and W2 are functions from G to C, then 1 X hW1 , W2 i G W1 ( g)W2 ( g), jGj g2G


213

and if ö1 and ö2 are functions from H to C, then 1 X hö1 , ö2 i H ö1 (h)ö2 (h): j Hj h2 H If ÷ is a character of G and ø1 , . . . , ø r are the irreducible characters of the subgroup H of G, then by Theorem 14.17, ÷ # H d 1 ø1 : : : d r ø r for some non-negative integers d 1 , : : : , d r which are given by d i h÷ # H, ø i i H : We say that ø i is a constituent of ÷ # H if the coef®cient d i in the above expression is non-zero. The next proposition shows that every irreducible character of H is a constituent of the restriction of some irreducible character of G. 20.4 Proposition Let H be a subgroup of G and let ø be a non-zero character of H. Then there exists an irreducible character ÷ of G such that h÷ # H, øi H 6 0: Proof Let ÷1 , . . . , ÷ k be the irreducible characters of G. Recall from Theorem 13.19 and Proposition 13.20 that the regular character ÷reg of G satis®es k X jGj if g 1, ÷reg ( g) and ÷reg ÷ i (1)÷ i : 0 if g 6 1, i1 Now 0 6

k X jGj ø(1) h÷reg # H, øi H ÷ i (1)h÷ i # H, øi H : j Hj i1

Therefore h÷ i # H, øi H 6 0 for some i.

j

Suppose that we know the character table of G. In the light of Proposition 20.4, we could hope to ®nd the character table of the subgroup H by restricting the irreducible characters ÷ of G to H. Unfortunately, it may be very dif®cult in practice to write down the restrictions ÷ # H in terms of irreducible characters of H. The best

214


chance of doing this occurs when the index jG: Hj( jGj=j Hj) is small, since the restrictions ÷ # H then do not have many constituents, as the following result shows. 20.5 Proposition Let H be a subgroup of G, let ÷ be an irreducible character of G, and let ø1 , : : : , ø r be the irreducible characters of H. Then ÷ # H d 1 ø1 : : : d r ø r , where the non-negative integers d 1 , . . . , d r satisfy r X

(20:6)

i1

d 2i < jG: Hj:

Moreover, we have equality in (20.6) if and only if ÷( g) 0 for all elements g of G which lie outside H. Proof By Theorem 14.17, we have r X i1

d 2i h÷ # H, ÷ # Hi H

1 X ÷(h)÷(h): j Hj h2 H

Also, since ÷ is irreducible, 1 h÷, ÷i G

1 X ÷( g)÷( g) jGj g2G

1 X ÷(h)÷(h) K jGj h2 H

r j Hj X d 2 K, jGj i1 i

P where K (1=jGj) g=2 H ÷( g)÷( g): Now K > 0, and K 0 if and only if ÷( g) 0 for all g with g 2 = H. The conclusions of the proposition follow at once. j We can say more about the constituents of ÷ # H in the case where H is a normal subgroup of G. For example, we will see that all the constituents of ÷ # H have the same degree. The way to exploit the fact that H v G us revealed in the following proposition.


215

20.7 Proposition Suppose that H v G. Let V be an irreducible CG-module and U be an irreducible C H-submodule of V # H. For every g 2 G let Ug fug: u 2 U g. Then (1) The set Ug is an irreducible C H-submodule of V # H and dim Ug dim U . (2) As a C H-module, V is a direct sum of some of the C Hmodules Ug. (3) If g1 , g2 , g 2 G and Ug1 and Ug2 are isomorphic C H-modules, then Ug1 g and Ug2 g are isomorphic C H-modules. Proof (1) Clearly, Ug is a subspace of V; and since H v G we have ghgÿ1 2 H for all h 2 H, so (ug)h u( ghg ÿ1 ) g 2 Ug

(u 2 U),

proving that Ug is a C H-submodule of V # H. Further, if W is a C Hsubmodule of Ug, then Wgÿ1 is a C H-submodule of U; since U is irreducible, Wgÿ1 {0} or U, whence W {0} or Ug. Therefore, Ug is an irreducible C H-module, as claimed. Moreover, u ! ug (u 2 U) is an invertible linear transformation from U to Ug, so dim U dim Ug. (2) The sum of all the subspaces Ug with g 2 G is a CG-submodule of V. Therefore, since V is irreducible, we have X V Ug: g2G

Then by Proposition 7.12, V is a direct sum of some of the C Hmodules Ug. (3) Now let ö be a C H-isomorphism from Ug1 to Ug2 . De®ne è : Ug1 g ! Ug2 g by è : wg ! (wö) g

(w 2 Ug 1 ):

Then è is clearly an isomorphism of vector spaces. Suppose that h 2 H. Then gh h9 g for some h9 2 H, and (wgh)è (wh9 g)è (wh9ö) g (wö)h9 g (wö) gh (wgè)h: Therefore, è is a C H-isomorphism, and the proof of the proposition is complete. j

216


We now come the fundamental theorem on the restriction of a character to a normal subgroup. 20.8 Clifford's Theorem Suppose that H v G and that ÷ is an irreducible character of G. Then (1) all the constituents of ÷ # H have the same degree; and (2) if ø1 , . . . , ø m are the constituents of ÷ # H, then ÷ # H e(ø1 : : : ø m ) for some positive integer e. Proof Let V be a CG-module with character ÷. Then it follows from Proposition 20.7, parts (1) and (2), that all the constituents of ÷ # H have the same degree. Let e h÷ # H, ø1 i. Then V contains a C H-module X1 whose character is eø1 , and which is therefore a direct sum of e isomorphic C H-modules, each having character ø1 ; say X 1 U1 : : : U e : By Proposition 20.7(3), if g 2 G then X1 g is a direct sum of isomorphic C H-modules. On the other hand, V is a sum of C Hmodules of the form X1 g, by Proposition 20.7(2). Hence V has the form V X1 : : : X m where each Xi is a direct sum of e isomorphic C H-modules, and Xi 6 X j if i 6 j. Therefore, ÷ # H e(ø1 : : : ø m ):

j

Our main applications of Clifford's Theorem will concern the case where jG: Hj 2, but you might like to look at Corollary 22.14 in Chapter 22 to see an advanced use of the theorem. Normal subgroups of index 2 We are shortly going to give more precise information about the constituents of ÷ # H when H is a normal subgroup of G of index 2 (that is, jG: Hj 2). Examples where this happens are G Sn ,


217

H A n , or G D2 n ha, b: a n b2 1, bÿ1 ab aÿ1 i, H hai. In fact, if H is a subgroup of index 2 in G, then H must be normal in G (see Exercise 1.10). When H is a normal subgroup of index 2 in G, the character tables of G and H are closely related. We describe this relationship in (20.13) below, and we then illustrate the results by ®nding the character table of A5 from that of S5 (which we have already obtained in Example 19.16). 20.9 Proposition Suppose that H is a normal subgroup of index 2 in G, and let ÷ be an irreducible character of G. Then either (1) ÷ # H is irreducible, or (2) ÷ # H is the sum of two distinct irreducible characters of H of the same degree. Proof If ø1 , . . . , ø r are the irreducible characters of H, then by Proposition 20.5, Pr

÷ # H d 1 ø1 : : : d r ø r ,

2 i1 d i

where < 2. Since d 1 , : : : , d r are non-negative integers, we deduce that either ÷ # H ø i for some i, or ÷ # H ø i ø j for some i, j with i 6 j. In the latter case, ø i and ø j have the same degree, by Clifford's Theorem 20.8 j For practical purposes, it is often desirable to have more details about the two cases in Proposition 20.9, and we shall supply these next. Since G=H C2 , we may lift the non-trivial linear character of G=H to obtain a linear character ë of G which satis®es ( 1 if g 2 H, ë( g) ÿ1 if g 2 = H: Note that for all irreducible characters ÷ of G, ÷ and ÷ë are irreducible characters of the same degree (see Proposition 17.14). Also, ÷ # H ÷ë # H, since ë(h) 1 for all h 2 H. 20.10 Proposition Suppose that H is a normal subgroup of index 2 in G, and that ÷ is

218


an irreducible character of G. Then the following three conditions are equivalent: (1) ÷ # H is irreducible; (2) ÷( g) 6 0 for some g 2 G with g 2 = H; (3) the characters ÷ and ÷ë of G are not equal. Proof We use Proposition 20.5; since jG: Hj 2, ÷ # H is irreducible if and only if the inequality in (20.6) is strict, and this happens if and only if ÷( g) 6 0 for some g 2 G with g 2 = H. Thus (1) is equivalent to (2). To see that (2) is equivalent to (3), observe that ( ÷( g) if g 2 H, ÷ë( g) ÿ÷( g) if g 2 = H, so ÷( g) 6 0 for some g with g 2 = H if and only if ÷ë 6 ÷.

j

According to Proposition 20.9, if H is a normal subgroup of G of index 2, and ÷ is an irreducible character of G, then ÷ # H is the sum of one or two irreducible characters of H. In the next proposition we consider the ®rst possibility. 20.11 Proposition Suppose that H is a normal subgroup of index 2 in G, and that ÷ is an irreducible character of G for which ÷ # H is irreducible. If ö is an irreducible character of G which satis®es ö # H ÷ # H, then either ö ÷ or ö ÷ë. Proof We have

( (÷ ÷ë)( g)

Hence

2÷( g)

if g 2 H,

0

if g 2 = H:

Restriction to a subgroup h÷ ÷ë, öi G

219

1 X 2÷( g)ö( g) jGj g2 H 1 X ÷( g)ö( g) j Hj g2 H

h÷ # H, ö # Hi H : Now k÷ # H, ö # Hl H 1, since ÷ # H is irreducible and ö # H ÷ # H. Therefore k÷ ÷ë, öl G 1, and so either ö ÷ or j ö ÷ë. Finally, we look at the case where ÷ # H is reducible. 20.12 Proposition Suppose that H is a normal subgroup of index 2 in an irreducible character of G for which ÷ # H is irreducible characters of H, say ÷ # H ø1 ø2. If ble character of G such that ö # H has ø1 or ø2 then ö ÷.

G, and that ÷ is the sum of two ö is an irreducias a constituent,

Proof In view of Proposition 20.10, ÷( g) 0 for all g with g 2 = H. Therefore, 1 X 1 X hö, ÷i G ö( g)÷( g) ö( g)÷( g) jGj g2G jGj g2 H 12hö # H, ÷ # Hi H : If ö # H has ø1 or ø2 as a constituent, then hö # H, ÷ # Hi H 6 0, so kö, ÷l G 6 0, and hence ö ÷. j We summarize our results on subgroups of index 2, by explaining how to list the irreducible characters of H on the assumption that we know the character table of G. (20.13) Let H be a normal subgroup of index 2 in the group G. (1) Each irreducible character ÷ of G which is nonzero somewhere outside H restricts to be an irreducible character of H. Such characters of G occur in pairs (÷ and ÷ë) which have the same restriction to H (Propositions 20.10, 20.11).

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. The two characters of H which we get from ÷ in this way come from no other irreducible character of G (Propositions 20.9, 20.10, 20.12). (3) Every irreducible character of H appears among those obtained by restricting irreducible characters of G, as in parts (1) and (2) (Proposition 20.4).

In case (2) of (20.13), extra work is needed to calculate the values taken by the two constituents of ÷ # H. Fortunately, in practice case (1) occurs more frequently than case (2). 20.14 Example The character table of A5 Write H A5 , and note that H is a normal subgroup of index 2 in the group S5 . The conjugacy classes of H and their sizes are given in Example 12.18(2), and the irreducible characters ÷1 , . . . , ÷7 of S5 can be found in Example 19.16. Observe that ÷1 , ÷3 and ÷6 are non-zero somewhere outside H, so by (20.13)(1), ÷1 # H, ÷3 # H and ÷6 # H are irreducible characters of H. Call them ø1 , ø2 and ø3 , respectively. Also, ÷5 ( g) 0 for all g 2 = H, so by (20.13)(2), ÷5 # H ø4 ø5 where ø4 and ø5 are distinct irreducible characters of H of degree 3. Note that ÷2 # H ÷1 # H, ÷4 # H ÷3 # H and ÷7 # H ÷6 # H, and hence ø1 , . . . , ø5 are the distinct irreducible characters of H by (20.13)(3). The results we have obtained so far have been deduced from our summary (20.13) of facts about characters of subgroups of index 2. They can also be veri®ed by calculating inner products. We have established that the character table of H is gi |CG ( g i )| ø1 ø2 ø3 ø4 ø5

1 60

(1 2 3) 3

(1 2)(3 4) 4

(1 2 3 4 5) 5

(1 3 4 5 2) 5

1 4 5 3 3

1 1 ÿ1 á2 â2

1 0 1 á3 â3

1 ÿ1 0 á4 â4

1 ÿ1 0 á5 â5


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 ÷5 # H (or by using the column orthogonality relations for column 1 and column i). We get á2 â2 0, á3 â3 ÿ2, á4 â4 á5 â5 1: Using Proposition 12.13, we see that each element of A5 is conjugate to its inverse. Hence by Proposition 13.9(4), all the numbers in the character table are real. By the orthogonality relation for column i with itself (2 < i < 5), we obtain 3 3 á22 â22 , 4 2 á23 â23 , 5 2 á24 â24 2 á25 â25 : Hence á2 â2 0, á3 â3 ÿ1, and we ®nd that á4 and â4 are the solutions of the quadratic x 2 ÿ x ÿ 1 0. Since we have not yet distinguished between ø4 and ø5 , we may take p p á4 12(1 5), â4 12(1 ÿ 5): p Similarly, á5 and â5 are 12(1 5). Since ø4 6 ø5 , we have p p á5 12(1 ÿ 5), â5 12(1 5): Thus the character table of A5 is as shown. Character table of A5 gi |CG ( g i )|

1 60

(1 2 3) 3

(1 2)(3 4) 4

(1 2 3 4 5) 5

(1 3 4 5 2) 5

1 4 5 3 3

1 1 ÿ1 0 0

1 0 1 ÿ1 ÿ1

1 ÿ1 0 á â

1 ÿ1 0 â á

ø1 ø2 ø3 ø4 ø5

where á 12(1

p

5), â 12(1 ÿ

p

5).

222


Proposition 17.6 allows us to deduce from the character table that A5 is a simple group. Summary of Chapter 20 Assume throughout that H is a subgroup of G. 1. If ÷ is a character of G, then ÷ # H is a character of H. The values of ÷ # H are given by (÷ # H)(h) ÷(h) for all h 2 H. In particular, ÷ and ÷ # H have the same degree. 2. The number of irreducible constituents of ÷ # H is bounded above by jG : Hj. Indeed, if ø1 , : : : , ø r are the irreducible characters of H, and ÷ # H d 1 ø1 : : : d r ø r, then r X i1

d 2i < jG : Hj:

3. If H v G and ÷ is an irreducible character of G, then all the constituents of ÷ # H have the same degree. Exercises for Chapter 20 1. Let G S4 and let H be the subgroup k(1 2 3 4), (1 3)l of G. (a) Show that H D8. (b) For each irreducible character ÷ of G (given in Section 18.1), express ÷ # H as a sum of irreducible characters of H. 2. Use the restrictions of the irreducible characters of S6 , given in Example 19.17, to ®nd the character table of A6 . (The seven conjugacy classes of A6 can be found by consulting the solutions to Exercises 12.3 and 12.4.) 3. Let G be a group with an abelian subgroup H of index n. Prove that ÷(1) < n for every irreducible character ÷ of G. 4. Suppose that G is a group with a subgroup H of index 3, and let ÷ be an irreducible character of G. Prove that h÷ # H, ÷ # Hi H 1, 2 or 3: Give examples to show that each possibility can occur.


223

5. It is known that the complete list of degrees of the irreducible characters of S7 is 1, 1, 6, 6, 14, 14, 14, 14, 15, 15, 20, 21, 21, 35, 35: Also, A7 has nine conjugacy classes. Find the complete list of degrees of the irreducible characters of A7 .

21 Induced modules and characters

Throughout this chapter we assume that H is a subgroup of the ®nite group G. We saw in the last chapter that restriction gives a simple way of converting a CG-module into a C H-module. Much more subtle than this is the process of induction, which constructs a CG-module from a given C H-module, and induction is the main concern of this chapter. As H is smaller than G, it is usually the case that C H-modules are easier to understand and construct than CG-modules, so induction can often give us an important handle on the representations of a group if we know some representations of its subgroups. We shall see many applications of this method in later chapters. Before describing the process of induction, we require some results which connect C H-homomorphisms with CG-homomorphisms. C H-homomorphisms and CG-homomorphisms Let U be a C H-submodule of the regular C H-module C H. If r 2 CG, then W: u ! ru (u 2 U ) de®nes a C H-homomorphism from U to CG, since for all s 2 C H, (us)W rus (uW)s. We now prove the striking fact that every C Hhomomorphism from U to CG has this simple form. 21.1 Proposition Assume that H < G, and let U be a C H-submodule of C H. If W is a C H-homomorphism from U to CG, then there exists r 2 CG such that uW ru for all u 2 U : 224

Induced modules and characters

225

Proof By Maschke's Theorem 8.1, there is a C H-submodule W of C H such that C H U W. De®ne ö: C H ! CG by ö: u w ! uW

(u 2 U, w 2 W ):

Then ö is easily seen to be a C H-homomorphism. Let r 1ö. For u 2 U, uW uö (1u)ö (1ö)u ru, and so W is of the required form.

j

We give two corollaries of Proposition 21.1, the ®rst of which is just the case H G of the proposition. 21.2 Corollary Let U be a CG-submodule of CG. Then every CG-homomorphism from U to CG has the form u ! ru (u 2 U ) for some r 2 CG. 21.3 Corollary Let U and V be CG-submodules of CG. Then the following two statements are equivalent: (1) U \ V {0}; (2) there exists r 2 CG such that for all u 2 U, v 2 V, ru u and rv 0: Proof Assume that U \ V {0}. Then the sum U V is a direct sum, so uv! u

(u 2 U , v 2 V )

is a function; moreover, it is a CG-homomorphism from U V to CG (see Proposition 7.11). Therefore by Corollary 21.2, there exists r 2 CG such that for all u 2 U, v 2 V, r(u v) u: Then ru u if u 2 U, and rv 0 if v 2 V. Conversely, assume that for some r 2 CG we have ru u and rv 0 for all u 2 U, v 2 V. If x 2 U \ V then rx x and rx 0, and so x 0. Consequently U \ V {0}. j

226

Representations and characters of groups Induction from H to G

For any subset X of CG, we write X(CG) for the subspace of CG which is spanned by all the elements xg with x 2 X, g 2 G. That is, X (CG) sp fxg: x 2 X , g 2 Gg: Clearly, X(CG) is then a CG-submodule of CG. Remember that H is a subgroup of G, so C H is a subset of CG. 21.4 De®nition Assume that H is a subgroup of G. Let U be a C H-submodule of C H, and let U " G denote the CG-module U(CG). Then U " G is called the CG-module induced from U. 21.5 Example Let G D6 ka, b: a3 b2 1, bÿ1 ab aÿ1 l, and let H kal, a cyclic subgroup of G of order 3. Let ù e2ði=3 , and de®ne W 0 sp (1 a a2 ), W 1 sp (1 ù2 a ùa2 ), W 2 sp (1 ùa ù2 a2 ): These are C H-submodules of C H (see Example 10.8(1)). Clearly, W 0 " G sp (1 a a2 , b ab a2 b), W 1 " G sp (1 ù2 a ùa2 , b ù2 ab ùa2 b), W 2 " G sp (1 ùa ù2 a2 , b ùab ù2 a2 b): Recall from Example 10.8(2) that CG U1 U 2 U3 U 4 , a direct sum of irreducible CG-modules U i , where U1 sp (1 a a2 b ab a2 b), U2 sp (1 a a2 ÿ b ÿ ab ÿ a2 b), U3 sp (1 ù2 a ùa2 , b ù2 ab ùa2 b), U4 sp (1 ùa ù2 a2 , b ùab ù2 a2 b): Thus W 0 " G U1 U2 , W 1 " G U3 , W 2 " G U4 :


227

In particular, W 0 " G is reducible, while W 1 " G and W 2 " G are irreducible. We now show that isomorphic C H-modules give isomorphic induced CG-modules. 21.6 Proposition Assume that H < G. Suppose that U and V are C H-submodules of C H and that U is C H-isomorphic to V. Then U " G is CG-isomorphic to V " G. Proof Let W: U ! V be a C H-isomorphism. By Proposition 21.1, there exists r 2 CG such that uW ru for all u 2 U, and also there exists s 2 CG such that vWÿ1 sv for all v 2 V. Consequently sru u and rsv v

for all u 2 U , v 2 V :

If a 2 U " G then a is a linear combination of elements ug (u 2 U, g 2 G), so ra is a linear combination of elements rug, and hence ra 2 V " G. Therefore ö: a ! ra

(a 2 U " G)

is a function from U " G to V " G. Moreover, ö is a CG-homomorphism, as (aö) g rag (ag)ö (a 2 U " G, g 2 G). Since sru u and rsv v for all u 2 U, v 2 V, we have sra a, rsb b

for all a 2 U " G, b 2 V " G:

Hence the function b ! sb

(b 2 V " G)

is the inverse of ö. Therefore ö is a CG-isomorphism, proving that U " G is CG-isomorphic to V " G. j The next proposition and its corollary enable us to de®ne the induced module U " G for an arbitrary C H-module U. 21.7 Proposition Assume that U and V are C H-submodules of C H with U \ V {0}. Then (U " G) \ (V " G) f0g.

228


Proof By Corollary 21.3, there exists r 2 CG such that ru u and rv 0 for all u 2 U, v 2 V. Then for all u 2 U, v 2 V and all g 2 G, rug ug and rvg 0: Since U " G is spanned by elements of the form ug (u 2 U, g 2 G), this implies that ru9 u9

for all u9 2 U " G,

rv9 0

for all v9 2 V " G:

and similarly, Therefore (U " G) \ (V " G) f0g by Corollary 21.3.

j

21.8 Corollary Let U be a C H-submodule of C H, and suppose that U U1 : : : Um , a direct sum of C H-submodules U i . Then U " G (U1 " G) : : : (U m " G): Proof We prove this by induction on m. It is trivial for m 1. Now U U1 V, where V U2 : : : U m . The de®nition of U " G implies that U " G (U1 " G) (V " G): Therefore by Proposition 21.7, U " G (U1 " G) (V " G): By induction, V " G (U 2 " G) : : : (U m " G), and hence, using (2.10), we obtain U " G (U 1 " G) : : : (Um " G), as required.

j

We can now de®ne the induced module U " G for an arbitrary C Hmodule U (where U is not necessarily a C H-submodule of C H). 21.9 De®nition Let U be a C H-module. Then (by Theorems 8.7 and 10.5), U U1 : : : Um


229

for certain irreducible C H-submodules U i of C H. De®ne U " G to be the following (external) direct sum: U " G (U 1 " G) : : : (Um " G): Proposition 21.6 and Corollary 21.8 ensure that this de®nition is consistent with De®nition 21.4. We emphasize that the de®nition of the induced module U " G in the case where U is a C H-submodule of C H is a natural one: U " G U (CG): We shall always prove results for general induced modules U " 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 21.9) that (21:10)

(U1 : : : Um ) " G (U1 " G) : : : (U m " G):

Our ®rst major result on general induced modules is known as `the transitivity of induction'. 21.11 Theorem Suppose that H and K are subgroups of G such that H < K < G. If U is a C H-module, then (U " K) " G U " G: Proof Assume ®rst that U is a C H-submodule of C H. Then U(CK) is spanned by elements of the form uk

(u 2 U , k 2 K):

Therefore, (U(CK))(CG) is spanned by elements of the form ukg

(u 2 U , k 2 K, g 2 G):

Since K < G, it follows that (U(CK))(CG) U(CG). That is, (21:12)

(U " K) " G U " G:

Now let U be an arbitrary C H-module. Then U U1 : : : Um for certain irreducible C H-submodules U i of C H. By (21.10), U " K (U1 " K) : : : (Um " K):

230


Therefore (U " K) " G (U1 " K) " G : : : (Um " K) " G by (21:10) (U1 " G) : : : (U m " G) U"G

by (21:12)

by Definition 21.9.

j

Induced characters 21.13 De®nition If ø is the character of a C H-module U, then the character of the induced CG-module U " G is denoted by ø " G, and is called the character induced from ø. The next example illustrates an important relationship between induced characters and restrictions of characters. 21.14 Example Let G S5 and let H be the subgroup A4 of G, as in Example 20.2. We showed in that example that if ÷1 , . . . , ÷7 are the irreducible characters of G (given in Example 19.16), and ø1 , . . . , ø4 are the irreducible characters of H (given in 18.2) then ÷ 1 # H ø1 , ÷ 2 # H ø1 , ÷ 3 # H ø1 ø4 , ÷ 4 # H ø1 ø4 , ÷5 # H 2ø4 , ÷ 6 # H ø 2 ø 3 ø4 , ÷ 7 # H ø 2 ø 3 ø4 : By Theorem 14.17, the coef®cients which appear here are the values of h÷ i # H, ø j i H for appropriate i, j. We record these coef®cients in a


231

matrix whose ij-entry is h÷ i # H, ø j i H : ÷1 ÷2 ÷3 ÷4 ÷5 ÷6 ÷7

0 ø1 1 B 1 B B 1 B B 1 B B 0 B @ 0 0

ø2 0 0 0 0 0 1 1

ø3 0 0 0 0 0 1 1

ø4 1 0 0 C C 1 C C 1 C C 2 C C 1 A 1

The rows of this matrix tell us how the irreducible characters of G restrict to H. For example, row 3 gives ÷3 # H 1 . ø1 0 . ø2 0 . ø3 1 . ø4 : Remarkably, it turns out that the columns of the matrix tell us how the irreducible characters of H induce to G. To be precise, the seven integers in column 1 give ø1 " G 1 . ÷1 1 . ÷2 1 . ÷3 1 . ÷4 0 . ÷5 0 . ÷6 0 . ÷7 : Similarly,

ø2 " G ø3 " G ÷6 ÷7 , and ø4 " G ÷3 ÷4 2÷5 ÷6 ÷7 :

Thus the ij-entry in our matrix, which we already know to be equal to h÷ i # H, ø j i H , is also equal to h÷ i , ø j " Gi G . In fact, it is true that h÷, ø " Gi G h÷ # H, øi H for all characters ÷ of G and ø of H, and we devote the next section to a proof of this result, which is known as the Frobenius Reciprocity Theorem. The Frobenius Reciprocity Theorem Before proving the theorem, we need the following preliminary result. 21.15 Proposition Assume that H < G. Let U be a C H-submodule of C H, and let V be a CG-submodule of CG. Then the vector spaces HomCG (U " G, V ) and HomC H (U , V # H) have equal dimensions.

232


Proof Suppose that W 2 HomCG (U " G, V ). Then by Corollary 21.2, there is an element r 2 CG such that sW rs

for all s 2 U " G:

De®ne W: U ! CG to be the restriction of W to U; that is, uW ru for all u 2 U : Then W 2 HomC H (U , V # H). Thus the function W!W is a linear transformation from HomCG (U " G, V ) to HomC H (U , V # H). We shall show that this linear transformation is invertible. Let ö 2 HomC H (U , V # H). Then by Proposition 21.1, there exists r 2 CG such that uö ru for all u 2 U. De®ne W from U " G to CG by sW rs (s 2 U " G): Then W 2 HomCG (U " G, V ). Moreover, ö W. Therefore the function W ! W is surjective. Finally, note that if r1, r2 2 CG and r1 u r2 u for all u 2 U, then r1 s r2 s for all s 2 U " G, since s is a linear combination of elements ug with u 2 U, g 2 G. Hence the function W ! W is injective, and so it is an invertible linear transformation from HomCG (U " G, V ) to HomC H (U , V # H). These two vector spaces therefore have the same dimension, as required. j 21.16 The Frobenius Reciprocity Theorem Assume that H < G. Let ÷ be a character of G and let ø be a character of H. Then hø " G, ÷i G hø, ÷ # Hi H : Proof First assume that the characters ÷ and ø are irreducible. Then there is a C H-submodule U of C H which has character ø, and there is a CG-submodule V of CG which has character ÷. By Theorem 14.24, we have hø " G, ÷i G dim (HomCG (U " G, V )), and hø, ÷ # Hi H dim (HomC H (U , V # H)):


233

From Proposition 21.15, we therefore deduce that (21:17)

hø " G, ÷i G hø, ÷ # Hi H

in the special case we are considering, namely when ÷ and ø are irreducible. For the general case, let ÷1 , . . . , ÷ k be the irreducible characters of G and let ø1 , . . . , ø m be the irreducible characters of H. Then for some integers di , ej we have ÷

k X

d i ÷ i and ø

m X

i1

Therefore

* hø " G, ÷i G

m X

ej ø j " G,

j1

ej ø j :

j1 k X

+ d i ÷i

i1

m X k X

G

ej d i hø j " G, ÷ i i G

j1 i1

m X k X

ej d i hø j , ÷ i # Hi H

j1 i1

*

m X j1

ej ø j ,

k X

by (21:17)

+ di÷i # H

i1

H

hø, ÷ # Hi H : This completes the proof of the Frobenius Reciprocity Theorem.

j

21.18 Corollary If f is a class function on G, and ø is a character of H, then hø " G, f i G hø, f # Hi H : Proof This follows at once from the Frobenius Reciprocity Theorem, since by Corollary 15.4, the characters of G span the vector space of class functions on 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, and for convenience of notation, de®ne the

234


_ G ! C by function ø:

(

_ g) ø(

ø( g)

if g 2 H,

0

if g 2 = H:

21.19 Proposition The values of the induced character ø " G are given by 1 X _ y ÿ1 gy) (ø " G)( g) ø( j Hj y2G for all g 2 G. Proof Let f: G ! C be the function given by 1 X _ y ÿ1 gy) ( g 2 G): f ( g) ø( j Hj y2G We aim to prove that f ø " G. If w 2 G then 1 X _ y ÿ1 w ÿ1 gwy) f ( g) f (w ÿ1 gw) ø( j Hj y2G since wy runs through G as y runs through G. Therefore f is a class function, and so by Corollary 15.4, it is suf®cient to show that h f , ÷i G hø " G, ÷i G for all irreducible characters ÷ of G. Let ÷ be an irreducible character of G. Then 1 X h f , ÷i G f ( g)÷( g) jGj g2G

1 1 XX _ y ÿ1 gy)÷( g): ø( jGj j Hj g2G y2G

h f , ÷i G

1 1 XX _ ø(x)÷( yxy ÿ1 ): jGj j Hj x2G y2G

Put x y ÿ1 gy. Then

1 X ø(x)÷(x) j Hj x2 H


235

_ since ø(x) 0 if x 2 = H, and ÷( yxy ÿ1 ) ÷(x) for all y 2 G. Therefore h f , ÷i G hø, ÷ # Hi H and so by the Frobenius Reciprocity Theorem, h f , ÷i G hø " G, ÷i G : This was the equation required to show that f ø " G, so the proof is complete. j 21.20 Corollary If ø is a character of the subgroup H of G, then the degree of ø " G is given by (ø " G)(1)

jGj ø(1): j Hj

Proof This follows immediately by evaluating (ø " G)(1) using Proposition 21.19. Alternatively, the stated degree of ø " G can be found using just the de®nition of induced modules (see Exercise 21.3). j For practical purposes, a formula for the values of induced characters different from that given in Proposition 21.19 is more useful, and we shall derive this next (it is given in Proposition 21.23 below). For x 2 G, de®ne the class function f G x on G by 1 if y 2 x G fG ( y) ( y 2 G): x 0 if y 2 = xG Thus f is the characteristic function of the conjugacy class x G . 21.21 Proposition If ÷ is a character of G and x 2 G, then h÷, f G x iG

÷(x) : jCG (x)j

236


Proof We have h÷, f G x iG

1 X ÷( g) f G x ( g) jGj g2G

1 X jx G j ÷(x) ÷( g) jGj g2x G jGj

÷(x) jCG (x)j

by Theorem 12:8:

j

Note that a result similar to Proposition 21.21 was used in the proof of Theorem 16.4. If H < G and h 2 H then h H h G ; but if g 2 G then g G may contain 0, 1, 2 or more conjugacy classes of H. To put this another way, we have: (21.22) Suppose that x 2 G. (1) If no element of x G lies in H, then f G x # H 0. (2) If some element of x G lies in H, then there are elements x1 , . . . , xm 2 H such that H H fG x # H f x1 : : : f x m :

Statement (2) just says that H \ x G breaks up into m conjugacy classes of H, with representatives x1 , : : : , x m . 21.23 Proposition Let ø be a character of the subgroup H of G, and suppose that x 2 G. (1) If no element of x G lies in H, then (ø " G)(x) 0. (2) If some element of x G lies in H, then ø(x1 ) ø(xm ) (ø " G)(x) jCG (x)j : : : , jCH (x1 )j jCH (xm )j H H where x1 , . . . , xm 2 H and f G x # H f x1 . . . f x m (as in (21.22)).

Proof By Proposition 21.21 and Corollary 21.18, we have (ø " G)(x) G hø " G, f G x i G hø, f x # Hi H : jCG (x)j


237

If no element of x G lies in H, then f G x # H 0, and hence (ø " G)(x) 0. And if some element of x G lies in H, and H H fG x # H f x1 : : : f x m as in (21.22)(2), then (ø " G)(x) hø, f xH1 : : : f xHm i H jCG (x)j hø, f xH1 i H : : : hø, f xHm i H

ø(x1 ) ø(xm ) : : : jCH (x1 )j jCH (xm )j

by Proposition 21:21:

The result follows.

j

21.24 Example Let G S4 and let H ka, bl, where a (1 2 3 4), b (1 3): Then H D8, since a4 b2 1 and bÿ1 ab aÿ1 . By (12.12), the conjugacy classes of H are f1g, fa2 (1 3)(2 4)g, fa (1 2 3 4), a3 (1 4 3 2)g, fb (1 3), a2 b (2 4)g, We have f 1G # H f 1H ,

G H f (1 3) # H f (1 3) ,

G f (1 2 3) # H 0,

G H H f (1 2)(3 4) # H f (1 3)(2 4) f (1 2)(3 4) , and G H f (1 2 3 4) # H f (1 2 3 4) :

For example, the statement G H H f (1 2)(3 4) # H 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, with representatives (1 3)(2 4) and (1 2)(3 4).

238


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. Then according to Proposition 21.23, we have ø(1) , 8 ø((1 3)) (ø " G)((1 3)) 4 , 4

(ø " G)(1) 24

(ø " G)((1 2 3)) 0,

ø((1 3)(2 4)) ø((1 2)(3 4)) (ø " G)((1 2)(3 4)) 8 , 8 4 (ø " G)((1 2 3 4)) 4

ø((1 2 3 4)) : 4

Referring to Example 16.3(3) for the irreducible characters ÷1 , . . . , ÷5 of H D8, we therefore have

÷1 " G ÷2 " G ÷3 " G ÷4 " G ÷5 " G

1

(1 2)

(1 2 3)

(1 2)(3 4)

(1 2 3 4)

3 3 3 3 6

1 ÿ1 1 ÿ1 0

0 0 0 0 0

3 ÿ1 ÿ1 3 ÿ2

1 1 ÿ1 ÿ1 0

In the next example, we use induced characters to ®nd the character table of a group of order 21. 21.25 Example (cf. Exercise 17.2) De®ne permutations a, b in S7 by a (1 2 3 4 5 6 7), b (2 3 5)(4 7 6)


239

and let G be the subgroup ka, bl of S7 . Check that a7 b3 1, bÿ1 ab a2 : It follows from these relations that the elements of G are all of the form ai bj with 0 < i < 6, 0 < j < 2. Also, G has order 21. We aim to ®nd the character table of G. First we ®nd the conjugacy classes. Since hai < C G (a), 7 divides jC G (a)j; and since b 2 = C G (a), jC G (a)j , 21. Hence jC G (a)j 7, and similarly jC G (b)j 3. Using this, we see that the conjugacy classes of G are f1g, fa, a2 , a4 g, fa3 , a5 , a6 g, fai b: 0 < i < 6g, fai b2 : 0 < i < 6g: We take 1, a, a3 , b and b2 to be representatives of the conjugacy classes. Notice that G has exactly ®ve irreducible characters. Since kal v G and G=hai C3, we obtain three linear characters ÷1 , ÷2 , ÷3 of G as the lifts of the linear characters of G=hai. Their values are shown below: g |CG ( g)| ÷1 ÷2 ÷3

1 21

a 7

a3 7

b 3

b2 3

1 1 1

1 1 1

1 1 1

1 ù ù2

1 ù2 ù

where ù e2ði=3 . Let H kal. We shall obtain the last two irreducible characters of G by inducing linear characters of H. Let ç e2ði=7 . For 1 < k < 6, there is a character ø k of H given by ø k (a j ) ç jk

(0 < j < 6):

To use the formula in Proposition 21.23 for calculating ø k " G, note that H H H fG a # H f a f a2 f a4

240


since no two of the elements a, a2 , a4 are conjugate in H. Hence by Proposition 21.23, (ø1 " G)(a) ç ç2 ç4 and similarly (ø1 " G)(a3 ) ç3 ç5 ç6 , (ø1 " G)(1) 3: And since no G-conjugate of b or b2 lies in H, (ø1 " G)(b) (ø1 " G)(b2 ) 0: Similarly (ø3 " G)(1) 3,

(ø3 " G)(a) ç3 ç5 ç6 ,

(ø3 " G)(a3 ) ç ç2 ç4 , and (ø3 " G)(b) (ø3 " G)(b2 ) 0: Thus if we write ÷4 ø1 " G and ÷5 ø3 " G, then the values of ÷4 and ÷5 are

÷4 ÷5

1

a

a3

b

b2

3 3

ç ç2 ç4 ç3 ç5 ç6

ç3 ç5 ç6 ç ç2 ç4

0 0

0 0

Now ÷4 # H ø1 ø2 ø4 and ÷5 # H ø3 ø5 ø6 . Therefore ÷4 6 ÷5 , since ø1 , . . . , ø6 are linearly independent. We now calculate that h÷4 , ÷4 i G

9 2 2 0 0 1, 21 7 7 3 3

and similarly h÷5 , ÷5 i G 1. Thus ÷4 and ÷5 are our last two irreducible characters, and the character table of G is as shown. Character table of ha, b: a7 b3 1, bÿ1 ab a2 i g |CG ( gi )| ÷1 ÷2 ÷3 ÷4 ÷5

1 21

a 7

a3 7

b 3

b2 3

1 1 1 3 3

1 1 1 ç ç2 ç4 ç3 ç5 ç6

1 1 1 ç3 ç5 ç6 ç ç2 ç4

1 ù ù2 0 0

1 ù2 ù 0 0


241

Summary of Chapter 21 Assume that H is a subgroup of G. 1. For each C H-module U, an induced CG-module U " G can be de®ned. If U is a C H-module of C H, then U " G is simply U(CG). 2. If ø is a character of H then the induced character ø " G is given by 1 X _ y ÿ1 gy): (ø " G)( g) ø( j Hj y2G In particular, the degree of ø:G is |G: H|ø(1). 3. If no element of g G lies in H, then (ø " G)( g) 0: If some element of g G lies in H, then ø(x1 ) ø(xm ) (ø " G)( g) jCG ( g)j : : : jCH (x1 )j jCH (xm )j where f Gg # H f xH1 : : : f xHm . 4. The Frobenius Reciprocity Theorem states that hø " G, ÷i G hø, ÷ # Hi H , where ø is a character of H and ÷ is a character of G.

Exercises for Chapter 21 1. Let G D8 ka, b: a4 b2 1, bÿ1 ab aÿ1 l and let H be the subgroup ka2 , bl. De®ne U to be the 1-dimensional subspace of C H spanned by 1 ÿ a2 b ÿ a2 b: (a) Check that U is a C H-submodule of C H. (b) Find a basis of the induced CG-module U " G. (c) Write down the character of the C H-module U and the character of the CG-module U " G. Is U " G irreducible? 2. Let G S4 and let H be the subgroup k(1 2 3)l C3 . (a) If ÷1 , . . . , ÷5 are the irreducible characters of G, as given in

242


Section 18.1, work out the restrictions ÷ i # H (1 < i < 5) as sums of the irreducible characters ø1 , ø2 , ø3 of C3 . (b) Calculate the induced characters ø j " G (1 < j < 3) as sums of the irreducible characters ÷ i of G. 3. Show direct from the de®nition that if H < G and ø is a character of H, then jGj (ø " G)(1) ø(1): j Hj 4. Let H be a subgroup of G, let ø be a character of H, and let ÷ be a character of G. Prove that (ø(÷ # H)) " G (ø " G)÷: (Hint: consider the inner product of each side with an arbitrary irreducible character of G, and use the Frobenius Reciprocity Theorem.) 5. Let G S7 and let H ka, bl, where a (1 2 3 4 5 6 7), b (2 3 5)(4 7 6), as in Example 21.25. 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

1 1 ç ç2 ç4 ç3 ç5 ç6

b 3

b2 3

1 0

1 0

where ç e2ði=7 (see Example 21.25). You are given that jC G (a)j 7 and jC G (b)j 18. Calculate the values of the induced characters ö " G and ø " G. 6. Suppose that H is a subgroup of G, and let ÷1 , . . . , ÷ k be the irreducible characters of G. Let ø be an irreducible character of H. Show that the integers d 1 , : : : , d k , which are given by ø " G d 1 ÷1 : : : d k ÷k , satisfy k X i1

(Compare Proposition 20.5.)

d 2i < jG : Hj:


243

7. Suppose that H is a normal subgroup of index 2 in G, and let ø be an irreducible character of H. Discover and prove results for ø " G which are similar to those presented in Chapter 20 for the restriction of irreducible characters of G to H.

22 Algebraic integers

Among the properties of characters which may be regarded as fundamental, 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. This is one of several results which we shall prove in this chapter, using algebraic integers. Most of the results concern arithmetic properties of character values. We discuss properties of a group element g 2 G which ensure that ÷( g) is an integer for all characters ÷ of G. And we prove some useful congruence properties; for example, if p is a prime number and g 2 G is an element of order pr for some r, then ÷( g) ÷(1) mod p for any character ÷ of G for which ÷( g) is an integer. Algebraic integers 22.1 De®nition A complex number ë is an algebraic integer if and only if ë is an eigenvalue of some matrix, all of whose entries are integers. Thus, for ë to be an algebraic integer, we require that det (A ÿ ëI) 0 for some square matrix A with integer entries. Equivalently, for the same matrix A, we have uA ëu for some non-zero row vector u. We remark that ë is an algebraic integer if and only if ë is a root of a polynomial of the form 244

Algebraic integers

245

x n a nÿ1 x nÿ1 : : : a1 x a0 where a0 , : : : , a nÿ1 are integers (see Exercise 22.7). In fact, algebraic integers are usually de®ned in this way. 22.2 Examples (1) Every integer n is an algebraic integer, since n is an eigenvalue of the 1 3 1 matrix (n). p (2) 2 is an algebraic integer, since it is an eigenvalue of the 0 1 matrix : 2 0 (3) If ë is an algebraic integer, then so are ÿë and the complex conjugate ë of ë. To see this, note that if A is an integer matrix and u is a row vector with uA ëu, then u(ÿA) (ÿë)u and uA ëu, where u is the row vector obtained from u by replacing each entry by its complex conjugate. (4) Let A be the n 3 n matrix 0 0 0 B1 0 B B0 1 AB B B @0 0 0 0

given by 0 ::: 0 ::: 0 ::: ::: 0 ::: 0 :::

0 0 0 1 0

1 0 1 0 0C C 0 0C C: C C 0 0A 1 0

Suppose that ù is an nth root of unity, and let u be the row vector (1, ù, ù2 , . . . , ù nÿ1 ). Then uA (ù, ù2 , : : : , ù nÿ1 , 1) ùu: This shows that every nth root of unity is an algebraic integer. 22.3 Theorem If ë and ì are algebraic integers, then ëì and ë ì are also algebraic integers. Proof There exist square matrices A and B, all of whose entries are integers, and non-zero row vectors u and v, such that uA ëu, vB ìv: Suppose that A is an m 3 m matrix and B is an n 3 n matrix.

246


Let e1 , . . . , em be a basis of C m and f 1 , : : : , f n be a basis of C n . Then the vectors ei fj (1 < i < m, 1 < j < n) form a basis of the tensor product space V C m C n. De®ne an endomorphism A B of V by (ei f j )(A B) ei A f j B (1 < i < m, 1 < j < n), P P extending linearly (that is, ( ë ij (ei f j ))(A B) ë ij (ei A f j B)). It is easy to check as in the proof of Proposition 19.4 that for all vectors x 2 C m , y 2 C n, we have (x y)(A B) xA yB: Hence

(u v)(A B) uA vB ëu ìv ëì(u v):

Therefore ëì is an eigenvalue of A B. Since the matrix of A B relative to the basis ei fj (1 < i < m, 1 < j < n) has integer entries, it follows that ëì is an algebraic integer. Let Im and In denote the identity m 3 m and n 3 n matrices, respectively. Then (u v)(A I n I m B) uA vI n uI m vB ëu v u ìv (ë ì)(u v), and we deduce as above that ë ì is an algebraic integer.

j

Theorem 22.3 shows that the set of all algebraic integers forms a subring of C. The next result provides a link between algebraic integers and characters. 22.4 Corollary If ÷ is a character of G and g 2 G, then ÷( g) is an algebraic integer. Proof By Proposition 13.9, ÷( g) is a sum of roots of unity. Each root of unity is an algebraic integer, by Example 22.2(4), so any sum of roots of unity is an algebraic integer by Theorem 22.3. Hence ÷( g) is an algebraic integer. j 22.5 Proposition If ë is both a rational number and an algebraic integer, then ë is an integer.

Algebraic integers

247

Proof Suppose that ë is a rational number which is not an integer. We shall show that ë is not an algebraic integer, which is enough to establish the proposition. Write ë r=s, where r and s are coprime integers and s 6 1. Let p be a prime number which divides s. For every n 3 n matrix A of integers, the entries of sA ÿ rI which are not on the diagonal are divisible by s, and hence also by p. Therefore det (sA ÿ rI) (ÿr) n mp for some integer m. As p does not divide r (since r and s are coprime), we deduce that det (sA ÿ rI) 6 0. Thus n 1 det (A ÿ ëI) det (sA ÿ rI) 6 0, s and hence ë is not an algebraic integer.

j

The next result is an immediate consequence of Corollary 22.4 and Proposition 22.5. 22.6 Corollary Let ÷ be a character of G and let g 2 G. If ÷( g) is a rational number, then ÷( g) is an integer. In passing, note that we have, as a special case of Proposition 22.5, p the well known result that 2 is irrational. (Example 22.2(2) shows p that 2 is an algebraic integer.) 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, we establish two preliminary lemmas. Recall from De®nition 12.21 that if C is a conjugacy class of G, then X C x 2 CG: x2C

22.7 Lemma Suppose that g 2 G and that C is the conjugacy class of G which contains g. Let U be an irreducible CG-module, with character ÷. Then

248

Representations and characters of groups uC ëu

for all u 2 U ,

where ë

jGj ÷( g) : jCG ( g)j ÷(1)

Proof Since C lies in the centre of CG (see Proposition 12.22), we know by Proposition 9.14 that there exists ë 2 C such that uC ëu for all u 2 U; that is, X u x ëu for all u 2 U : x2C

Consequently if B is a basis of U, then X [x]B ëI: x2C

Taking the traces of both sides of this equation, we obtain X ÷(x) ë÷(1), x2C

and since ÷ is constant on the conjugacy class C, this yields jCj÷( g) ë÷(1): Thus ë jCj÷( g)=÷(1). As |C| |G:CG ( g)| by Theorem 12.8, the result follows. j 22.8 Lemma P Let r g2G á g g 2 CG, where each á g is an integer. Suppose that u is a non-zero element of CG such that ur ëu, where ë 2 C. Then ë is an algebraic integer. Proof Let g1 , . . . , gn be the elements of G. Then for 1 < i < n, we have gi r

n X j1

aij g j

Algebraic integers

249

for certain integers aij . (In fact, aij á g where g gÿ1 i gj .) The statement that ur ëu (with u 6 0) says that ë is an eigenvalue of the integer matrix A (aij ). Therefore ë is an algebraic integer. j 22.9 Example Let G Cn kx: x n 1l, and de®ne u 1 ùx ÿ1 ù2 x ÿ2 : : : ù nÿ1 x 2 CG, where ù is an nth root of unity. Then ux ùu and so Lemma 22.8 con®rms that ù is an algebraic integer. Notice that this example is just a reworking of Example 22.2(4). 22.10 Corollary If ÷ is an irreducible character of G and g 2 G, then ë

jGj ÷( g) jCG ( g)j ÷(1)

is an algebraic integer. Proof Let U be an irreducible CG-submodule of CG with character ÷, and let C be the sum of the elements in the conjugacy class of G which contains g. Then uC ëu for all u 2 U, by Lemma 22.7. Therefore ë is an algebraic integer, by Lemma 22.8. j 22.11 Theorem If ÷ is an irreducible character of G, then ÷(1) divides |G|. Proof Let g1 , . . . , gk be representatives of the conjugacy classes of G. Then for all i, both jGj ÷( g i ) jCG ( g i )j ÷(1)

and

÷( g i )

are algebraic integers, by Corollaries 22.10 and 22.4. Hence by Theorem 22.3, k X i1

jGj ÷( g i )÷( g i ) jCG ( g i )j ÷(1)

250


is an algebraic integer. This algebraic integer is equal to jGj=÷(1), by the row orthogonality relations, Theorem 16.4(1). As jGj=÷(1) is a rational number. Proposition 22.5 now implies that jGj=÷(1) is an integer. That is, ÷(1) divides |G|. j 22.12 Examples (1) If p is a prime number and G is a group of order pn for some n, then ÷(1) is a power of p for all irreducible characters ÷ of G. In particular, if jGj p2 then ÷(1) 1 for all irreducible characters ÷. (Note that ÷(1) , p, as the sum of the squares of the degrees of the irreducible characters is equal to |G|.) Hence, using Proposition 9.18, we recover the well known result that groups of order p2 are abelian. (2) Let G be a group of order 2 p, where p is prime. By Theorem 22.11, the degree of each irreducible character of G is 1 or 2 (it cannot be p for the reason noted in (1) above). By Theorem 17.11, the number of linear characters of G divides |G|. Hence, either the degrees of the irreducible characters of G are all 1, or they are 1, 1, 2, : : : , 2 (with ( p ÿ 1)=2 degrees 2): (3) If G Sn then every prime p which divides the degree of an irreducible character of G also divides n!, and hence satis®es p < n. Theorem 22.11 also has the following interesting consequences concerning irreducible characters of simple groups. (Recall that a group G is simple if it has no normal subgroups apart from {1} and G.) 22.13 Corollary No simple group has an irreducible character of degree 2. Proof Suppose that G is a simple group which has an irreducible character ÷ of degree 2. Let r: G ! GL(2, C) be a representation of G with character ÷. Since Ker r v G and G is simple, we have Ker r f1g, and so r is injective. First, observe that G is non-abelian, by Proposition 9.5. Hence G9 6 1, and so G9 G as G is simple. Therefore by Theorem 17.11, G has no non-trivial linear characters. But g ! det ( gr) is a linear character of G (see Exercise 13.7(a)), and this implies that det ( gr) 1 for all g 2 G:

Algebraic integers

251

Now G has even order, by Theorem 22.11. Therefore G contains an element x of order 2 (see Exercise 1.8). Consider the 2 3 2 matrix xr. As r is injective, xr has order 2; and by Proposition 9.11, there is a 2 3 2 matrix T such that T ÿ1 (xr)T is a diagonal matrix with diagonal entries 1. Since det (xr) 1, we conclude that ÿ1 0 T ÿ1 (xr)T : 0 ÿ1 Thus xr T(ÿI)T ÿ1 ÿI: Consequently (xr)( gr) ( gr)(xr) for all g 2 G. As r is injective, this means that xg gx for all g 2 G, and hence hxi v G: This is a contradiction, as G is simple.

j

Our next result again shows that information about character degrees can sometimes be used to learn about the structure of a group. This time, we assume that every irreducible character of G has degree a power of a prime p, and we deduce that G has an abelian normal pcomplement N; that is, N is an abelian normal subgroup of G, and jN j is coprime to p, while jG: N j is a power of p. 22.14 Corollary Suppose that p is a prime and the degree of every irreducible character of G is a power of p. Then G has an abelian normal p-complement. In particular, G is not simple unless G has prime order. Proof The result is correct if G is abelian (see Theorem 9.6), so we assume that G is non-abelian. Theorems 11.12 and 17.11 give us the equation X jGj jG=G9j ÷(1)2 , the sum being over the irreducible characters ÷ of G for which ÷(1) . 1. Since G is non-abelian, ÷(1) . 1 for some irreducible character ÷ of G. Then ÷(1) is divisible by p, by our hypothesis, so p divides |G| by

252


Theorem 22.11; and we deduce from the equation above that p divides the order of the abelian group G=G9. Since every ®nite abelian group is isomorphic to a direct product of cyclic groups, it follows that G=G9 has a subgroup of index p. Hence G has a normal subgroup H of index p. Let ø be an irreducible character of H. Then h÷ # H, øi 6 0 for some irreducible character ø of H, by Proposition 20.4. Next, Clifford's Theorem 20.8 shows that ø(1) divides ÷(1), so ø(1) is a power of p. We may now apply induction on |G| to deduce that H has an abelian normal p-complement N. We have that |N| is coprime to p and jG : N j is a power of p, so it remains to prove that N v G. Suppose that g 2 G and the order of g is coprime to p. Then g 2 H, since otherwise p divides the order of gH which in turn divides the order of g; a similar argument shows that g 2 N. Hence N consists of those elements of G whose order is coprime to p, and from this fact it follows easily that N v G. Finally, assume that G is simple, so either N {1} or N G. If N f1g then G is a p-group so Z(G) 6 {1} (see Exercise 12.7); because G is simple, we have Z(G) G, so G is abelian. On the other hand, if N G then G is again abelian. But an abelian simple group has prime order, by Exercise 1.1. Therefore, G has prime order. j

A condition which ensures that ÷( g) is an integer In Theorem 22.16 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. This result implies, for example, that for all n, every entry in the character table of Sn is an integer (see Corollary 22.17). Bearing in mind the dif®culties we encountered in constructing the character tables of Sn for small values of n (we reached n 6 in Example 19.17), Theorem 22.16 is evidently a useful result. Before proving Theorem 22.16, we require a preliminary lemma concerning roots of unity. If a and b are positive integers, then we denote their highest common factor by (a, b). Also, for integers d and n, we write d|n to denote the fact that d divides n. 22.15 Lemma If ù is an nth root of unity, then

Algebraic integers X ùi

253

1< n, (i,n)1

is an integer. Proof We prove the result by induction on n. It is trivial for n 1. Also, if ù 1 then the result is immediate. So suppose that ù is an nth root of unity and ù 6 1. Then ù is a root of the polynomial Pn (x n ÿ 1)=(x ÿ 1) x nÿ1 : : : x 1: Therefore i1 ù i 0. Pn i Now we partition the sum i1 ù according to the highest common factor d of i and n: 0

n X

ùi

i1

X X

ùi

dj n 1< n (i,n)d

X

X

ù dj :

dj n 1< j< n=d, ( j,n=d)1

If djn then ù d is an (n=d)th root of unity, and if in addition d . 1, then by our induction hypothesis, X ù dj 2 Z: 1< j< n=d, ( j,n=d)1

It follows that X 1< n, (i,n)1

as required.

ùi

n X i1

ùi ÿ

X

X

ù dj 2 Z,

dj n, 1< j< n=d, d . 1 ( j,n=d)1 j

22.16 Theorem Let g be an element of order n in G. Suppose that g is conjugate to gi for all i with 1 < i < n and (i, n) 1. Then ÷( g) is an integer for all characters ÷ of G. Proof Let V be a CG-module with character ÷ of degree m. By Proposition 9.11, there is a basis B of V such that 0 1 ù1 0 B C .. [ g]B @ A . ù 0 m

254


where ù1 , . . . , ù m are nth roots of unity. For 1 < i < n, the matrix [ g i ]B has diagonal entries ù1i , . . . , ù im , and so ÷( g i ) ù1i : : : ù im : Therefore by Lemma 22.15,

X

÷( g i ) 2 Z:

1< n, (i,n)1

As g is conjugate to gi if 1 < i < n and (i, n) 1, we have ÷( g i ) ÷( g) for such i, and hence s÷( g) 2 Z, where s is the number of integers i with 1 < i < n and (i, n) 1. Consequently ÷( g) is a rational number, and so ÷( g) is an integer by Corollary 22.6. j We remark that using Galois theory it is possible to prove the converse of Theorem 22.16, namely that if ÷( g i ) 2 Z for all characters ÷ of G, then g is conjugate to g i whenever i is coprime to n. 22.17 Corollary All the character values of symmetric groups are integers. Proof If g 2 Sn and i is coprime to the order of g, then the permutations g and g i have the same cycle-shape, and hence are conjugate by Theorem 12.15. The result now follows from Theorem 22.16. j

The p9-part of a group element The rest of the chapter is devoted to some important congruence properties of character values. For example, one particularly useful consequence of our results is that if p is a prime number, g is an element of G of order pr for some r, and ÷ is a character of G such that ÷( g) 2 Z, then ÷( g) ÷(1) mod p. Before going into the character theory, we need to de®ne the p9-part of a group element. The de®nition will emerge from the following lemma.

Algebraic integers

255

22.18 Lemma Let p be a prime number and let g 2 G. Then there exist x, y 2 G such that (1) g xy yx, (2) the order of x is a power of p, and (3) the order of y is coprime to p. Moreover, the elements x and y of G which satisfy conditions (1)±(3) are unique. Proof Let the order of g be upv , where u, v 2 Z and (u, p) 1. Then there exist integers a, b such that au bpv 1: v

Put x gau and y gbp . Then v

xy yx g aubp g, v

v

x p g aup 1, v

y u g bup 1: Hence the order of x is a power of p and the order of y divides u, so is coprime to p. Therefore x and y satisfy conditions (1)±(3). Now suppose that x9, y9 2 G also satisfy (1)±(3); that is, g x9 y9 y9x9, the order of x9 is a power of p and the order of y9 is coprime to p. We must show that x x9 and y y9. We have x9 g x9 y9x9 gx9, so x9 commutes with g, hence also with gau x. Since both x and x9 have order a power of p, it follows that x ÿ1 x9 has order a power of p. Similarly, y9 commutes with y and y( y9)ÿ1 has order coprime to p. Finally, xy g x9 y9, so x ÿ1 x9 y( y9)ÿ1 : If z x ÿ1 x9 y( y9)ÿ1 , then we have shown that the order of z is both a power of p and coprime to p. Therefore z 1, and so x x9 and y y9, as required. j

256


22.19 De®nition We call the element y which appears in Lemma 22.18 the p9-part of g. We extract the following statement from the proof of Lemma 22.18. (22.20) Let the order of g be upv , where u, v 2 Z and (u, p) 1, and choose integers a, b with au bpv 1. v Then the p9- part of g is gbp . For example, if p 2 and g has order 6, then the p9-part of g is gÿ2 ; the expression g xy in Lemma 22.18 has x g3 , y gÿ2 . A little ring theory To prepare for our main result on congruence properties of character values, we need a few basic facts about a certain subring of C in which all our character values will lie. Let n be a positive integer and let æ e2ði= n . De®ne Z[æ] to be the subring of C generated by Z and æ; that is, Z[æ] f f (æ): f (x) 2 Z[x]g: Clearly, every element of Z[æ] is an integer combination of the powers 1, æ, æ2 , . . . , æ nÿ1 , so in fact Z[æ] f f (æ): f (x) 2 Z[x] of degree < n ÿ 1g: Now let p be a prime number and let pZ[æ] f pr: r 2 Z[æ]g, a principal ideal of Z[æ]. 22.21 Proposition There are only ®nitely many ideals I of Z[æ] which contain pZ[æ]. Proof Consider the factor ring Z[æ]= pZ[æ]. By de®nition, this has as its elements all the cosets pZ[æ] r with r 2 Z[æ]. Every such coset contains an element of the form a0 a1 æ : : : a nÿ1 æ nÿ1 , with ai 2 Z, 0 < ai < p ÿ 1 for all i: As there are only ®nitely many such elements, we conclude that Z[æ]= pZ[æ] is ®nite. The ideals of Z[æ] which contain pZ[æ] are in

Algebraic integers

257

bijective correspondence with the ideals of Z[æ]= pZ[æ] (the correspondence being I ! I= pZ[æ]). Therefore there are only ®nitely many such ideals, and the proof is complete. j We deduce from Proposition 22.21 that there is a maximal ideal of P of Z[æ] which contains pZ[æ]; that is, P is a proper ideal which is contained in no larger proper ideal of Z[æ]. (A proper ideal of Z[æ] is an ideal which is not equal to Z[æ].) We now prove two easy results about the maximal ideal P. 22.22 Proposition If r, s 2 Z[æ] and rs 2 P, then either r 2 P or s 2 P. In particular, if r n 2 P for some positive integer n, then r 2 P. Proof Assume that rs 2 P and r 2 = P. We must show that s 2 P. Since r 2 = P, the ideal rZ[æ] P of Z[æ] strictly contains P. As P is maximal, we therefore have rZ[æ] P Z[æ]: Consequently, there exist a 2 Z[æ], b 2 P such that 1 ra b: Then s rsa sb: As rs 2 P and b 2 P, it follows that s 2 P, as required. For the last statement of the proposition, assume that r n 2 P. Since r n rr nÿ1 , this implies that either r 2 P or r nÿ1 2 P. Repeating this argument, we conclude that r 2 P. j 22.23 Proposition We have P \ Z pZ. Proof Let m 2 P \ Z. If p Bj m then there are integers a, b with am bp 1; but this implies that 1 2 P, which is false, since P 6 Z[æ]. Thus pjm, which establishes that P \ Z pZ. Since p 2 P, we also have pZ P \ Z. j

258

Representations and characters of groups Congruences

At last we are ready to prove our results on congruences of character values. Let G be a group of order n and let æ e2ði= n . The ring Z[æ] is of interest because all the character values of G lie in Z[æ] (see Proposition 9.11). As in the previous section, let p be a prime number and let P be a maximal ideal of Z[æ] containing pZ[æ]. 22.24 Theorem Let g 2 G and let y be the p9- part of g. If ÷ is any character of G, then ÷( g) ÿ ÷( y) 2 P: Proof Suppose that g has order m upv , where u, v 2 Z and v (u, p) 1. Choose integers a, b with au bpv 1. Then y g bp (see (22.20)). The orders of g and of y divide n |G|, so ÷( g) and ÷( y) are both sums of nth roots of unity, and hence lie in Z[æ]. Now let ù be an mth root of unity (so ù 2 Z[æ] as mjn). Then v ù ù aubp , and so v v 2v 2v ù p ù aup . ù bp ù bp : v

v

Consider the number (ù ÿ ù bp ) p . By the Binomial Theorem, v v v v vÿ1 v vÿ r v p (ù ÿ ù bp ) p ù p ÿ pv ù p ù bp : : : ù p ù rbp r v

2v

: : : (ÿ1) p ù bp : v

For 0 , r , pv , the binomial coef®cient ( rp ) is divisible by p. Hence v

v

v

v

2v

(ù ÿ ù bp ) p ù p (ÿ1) p ù bp pá, v

2v

where á 2 Z[æ]. Moreover, since ù p ù bp , we have ( 0, if p 6 2, v v 2v ù p (ÿ1) p ù bp v 2ù p , if p 2, so it follows that v

v

(ù ÿ ù bp ) p 2 pZ[æ]: v

v

Thus (ù ÿ ù bp ) p lies in the maximal ideal P. Application of Proposition 22.22 now forces

Algebraic integers

259

v

(ù ÿ ù bp ) 2 P:

(22:25)

By Proposition 9.11, there are mth roots of unity ù1 , . . . , ù d such that v

v

÷( g) ù1 : : : ù d and ÷( y) ù1bp : : : ù bp d : Then v

v

÷( g) ÿ ÷( y) (ù1 ÿ ù1bp ) : : : (ù d ÿ ù bp d ), which, by (22.25), lies in P.

j

22.26 Corollary Let p be a prime number. Suppose that g 2 G and that y is the p9- part of g. If ÷ is a character of G such that ÷( g) and ÷( y) are both integers, then ÷( g) ÷( y) mod p: Proof As ÷( g) and ÷( y) are both integers, Theorem 22.24 and Proposition 22.23 give ÷( g) ÿ ÷( y) 2 P \ Z pZ: Therefore ÷( g) ÷( y) mod p.

j

22.27 Corollary Let p be a prime number. Suppose that g 2 G and the order of g is a power of p. If ÷ is a character of G such that ÷( g) 2 Z, then ÷( g) ÷(1) mod p: Proof As the order of g is a power of p, the p9-part of g is 1, so the result is immediate from Corollary 22.26. j Notice that Corollary 13.10 is the special case of Corollary 22.27 in which g has order 2. We shall use the congruence results 22.24±22.27 extensively in our character calculations in Chapters 25±7. For the moment, we just illustrate the results with reference to some character tables which we already know.

260


22.28 Example Recall from Example 20.14 that the character table of A5 is as shown. p p where á (1 5)=2, â (1 ÿ 5)=2. If g (1 2 3) then Corollary 22.26 implies that ÷( g) ÷(1) mod 3 whenever ÷( g) 2 Z. Thus the corresponding entries in columns 1 and 2

Character table of A5

÷1 ÷2 ÷3 ÷4 ÷5

1

(1 2 3)

(1 2)(3 4)

(1 2 3 4 5)

(1 3 4 5 2)

1 4 5 3 3

1 1 ÿ1 0 0

1 0 1 ÿ1 ÿ1

1 ÿ1 0 á â

1 ÿ1 0 â á

of the character table are congruent modulo 3, as can be seen by inspecting the table. Similarly the entries in columns 1 and 3 are congruent modulo 2. Also ÷ i ((1 2 3 4 5)) ÷ i (1) mod 5

for i 1, 2, 3:

However, ÷4 ((1 2 3 4 5)) á 2 = Z. We illustrate Theorem 22.24 for this value. If we take p 5 and g (1 2 3 4 5), then the p9-part of g is 1, and p ÷4 ( g) ÿ ÷4 (1) á ÿ 3 12(1 5 ÿ 6) p p p 5 . 12(1 ÿ 5) â 5: Put æ e2ði=60 , and let P be a maximal ideal of Z[æ] containing 5Z[æ]. p p Then ( 5)2 2 P, so 5 2 P by Proposition 22.22. Since â 2 Z[æ] (see p Proposition 9.11), we have â 5 2 P. That is, ÷4 ( g) ÿ ÷4 (1) 2 P: This illustrates Theorem 22.24. Summary of Chapter 22 1. Character values are algebraic integers. 2. The degree of every irreducible character of G divides |G|.

Algebraic integers

261

3. If g is conjugate to g i for all integers i which are coprime to the order of g, then ÷( g) is an integer, for all characters ÷. 4. Let p be a prime number. If g 2 G and y is the p9-part of g, then ÷( g) ÷( y) mod p for all characters ÷ of G such that ÷( g) and ÷( y) are integers.

Exercises for Chapter 22 1. Let G be a group of order 15. Use Theorems 11.12, 17.11 and 22.11 to show that every irreducible character of G has degree 1. Deduce that G is abelian. 2. Prove that the number of conjugacy classes in a group of order 16 is 7, 10 or 16. 3. Let p and q be prime numbers with p . q, and let G be a nonabelian group of order pq. (a) Find the degrees of all the irreducible characters of G. (b) Show that |G9| p. (c) Show that q divides p ÿ 1 and that G has q (( p ÿ 1)=q) conjugacy classes. 4. Let G be a group and let ö be a character of G such that ö( g) ö(h) for all non-identity elements g and h of G. (a) Show that ö a1 G b÷reg for some a, b 2 C. (b) Show that a b and a b|G| are integers. (c) Show that if ÷ is a non-trivial irreducible character of G, then b÷(1) is an integer. (d) Deduce that both a and b are integers. 5. Suppose that G is a group of odd order. This exercise shows that the only irreducible character ÷ of G such that ÷ ÷ is the trivial character. (a) Prove that if g 2 G and g gÿ1 , then g 1. (b) Now let ÷ be an irreducible character of G with ÷ ÷. Prove that h÷, 1 G i

1 (÷(1) 2á), jGj

where á is an algebraic integer. (c) Deduce that ÷ 1 G .

262


6. It is often possible to calculate the character table from limited arithmetic information about the group G. This exercise illustrates this point with the group G S5 . A certain group G of order 120 has exactly seven conjugacy classes, and contains an element g of order 5 such that jCG ( g)j 5. Moreover, g, g2 , g3 and g4 are conjugate in G. (a) Show that for every irreducible character ÷ of G, ÷( g) is 0, 1 or ÿ1. (b) Use Corollary 22.27 to deduce that G has two irreducible characters of degree 5. (c) Find ÷(1) and ÷( g) for all irreducible characters ÷ of G. (d) You are given that all entries in the character table of G are integers, and that the conjugacy classes of G have representatives g1 , . . . , 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 22.26 and the column orthogonality relations, ®nd the character table of G. 7. 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 nÿ1 x nÿ1 : : : a1 x a0 , where each ar (0 < r < n ÿ 1) is an integer.

23 Real representations

Since Chapter 9, we have always taken our representations to be over the ®eld C of complex numbers. However, some results in representation theory work equally well for the ®eld R of real numbers. There is a subtle interplay between representations over C and representations over R, which we shall explore in this chapter. Often, 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. Let r be a representation of G. If all the matrices gr ( g 2 G) have real entries, then of course the character of r is real-valued. However, the converse is not true: it can happen that the character of r is realvalued, while there is no representation ó equivalent to r with all the matrices gó having real entries. Various criteria for whether or not a character corresponds to a representation over R lead us to the remarkable Frobenius±Schur Count of Involutions. This is used in the last section to prove a famous result of Brauer and Fowler concerning centralizers of involutions in ®nite simple groups. The material in this chapter is perhaps at a slightly more advanced level than that in the rest of the book, and is not used in the ensuing chapters, which consist largely of the calculation of character tables and applications of character theory. Nevertheless, the subject of real representations not only is elegant and interesting, but also gives delicate information about characters which often comes into play in more dif®cult calculations. Real characters An element g of the ®nite group G is said to be real if g is conjugate to gÿ1 ; and if g is real, then the conjugacy class g G is also said to be 263

264


real. Notice that if a conjugacy class is real, then it contains the inverse of each of its elements, since ( g ÿ1 ) G fx ÿ1 : x 2 g G g. On the other hand, a character ÷ of G is real if ÷( g) is real for all g 2 G. Thus for example, the conjugacy class {1} of G is real, and the trivial character of G is real. 23.1 Theorem The number of real irreducible characters of G is equal to the number of real conjugacy classes of G. Proof Let X denote the character table of G, and let X denote the complex conjugate of the matrix X. For every irreducible character ÷ of G, the complex conjugate ÷ is also an irreducible character (see Proposition 13.15), so X can be obtained from X by permuting the rows. Hence there is a permutation matrix P such that PX X (see Exercise 4.4). For every conjugacy class g G of G, 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 ÿ1 ) G . Therefore X can be obtained from X be permuting the columns, and so there is a permutation matrix Q such that XQ X By Proposition 16.2, X is invertible. Therefore Q X ÿ1 X X ÿ1 PX : Consequently P and Q have the same trace, by Proposition 13.2. Since the trace of a permutation matrix is equal to the number of points ®xed by the corresponding permutation, we have the number of real irreducible characters of G is tr (P), and

the number of real conjugacy classes of G is tr (Q):

As these numbers are equal, the result is proved.

j

Part of the following corollary was obtained by a different method in Exercise 22.5.

Real representations

265

23.2 Corollary The group G has a non-trivial real irreducible character if and only if the order of G is even. Proof If G has odd order, then no non-identity element of G is real (see the solution to Exercise 23.1). Therefore by Theorem 23.1, the only real character of G is the trivial character. If G has even order, then by Exercise 1.8, G has an element g of order 2. Hence G has at least two real conjugacy classes, {1} and g G, and so G has at least two real irreducible characters by Theorem 23.1. j

Characters which can be realized over R Let ÷ be a character of the group G. We say that ÷ can be realized over R if there is a representation r: G ! GL (n, C) with character ÷, such that all the entries in each matrix gr ( g 2 G) are real. This is the same as saying that there is some CG-module V with character ÷, and there is a basis v1 , . . . , v n of V, such that for all g 2 G and 1 < i < n, v i g is a linear combination of v1 , . . . , v n with real coef®cients. 23.3 Examples (1) Let G D8 ka, b: a4 b2 1, bÿ1 ab aÿ1 l, and let ÷ be the irreducible character of G of degree 2 (see Example 16.3(3)). Then ÷ can be realized over R, since ÿ1 0 0 1 , br ar 0 1 ÿ1 0 provides a representation r of G with character ÷ such that all the matrices gr ( g 2 G) have real entries. (2) Let G Q8 ka, b: a4 1, b2 a2 , bÿ1 ab aÿ1 l, and let ÷ be the irreducible character of G of degree 2 (see Exercise 17.1). The values of ÷ are as follows:

÷

1

a2

a

b

ab

2

ÿ2

0

0

0

266


Thus ÷ is real. In fact, ÷ cannot be realized over R, but it is at the moment unclear how to prove this. (We shall eventually establish this in Example 23.18(3) below.) Although every character which can be realized over R is perforce a real character, Example 23.3(2) tells us that the converse is false.

RG-modules Recall that in Chapter 4 we de®ned an FG-module, where F is R or C. Thus an RG-module is a vector space over R, with a multiplication by elements of G satisfying the conditions of De®nition 4.2. In this section we shall study the relationship between RG-modules and CGmodules. 23.4 Examples Let V be a 2-dimensional vector space over R, with basis v1 , v2 . (1) V becomes an RD8 -module if we de®ne v1 a v2 ,

v1 b ÿv1 ,

v2 a ÿv1 ,

v2 b v2

(compare Example 23.3(1)). (2) V becomes an RC3 -module, where C3 kx: x 3 1l, if we de®ne v1 x v2 , v2 x ÿv1 ÿ v2 : 0 1 (This gives the representation x ! of Exercise 3.2.) ÿ1 ÿ1 Every RG-module can easily be converted into a CG-module. Simply take a basis v1 , . . . , v n of the RG-module, and consider the vector space over C with basis v1 , . . . , v n . This new vector space is clearly a CG-module (with v i g de®ned as before). The construction is even easier to understand in terms of representations: if r: G ! GL (n, R) is a representation then for each g 2 G, the matrix gr has its entries in R, and hence also in C. Therefore we obtain a representation r: G ! GL (n, C). Notice that a character ÷ of G can be


267

realized over R if and only if there exists an RG-module with character ÷. Rather more subtle than this is the construction of an RG-module from a given CG-module. Let V be a CG-module with basis v1 , . . . , v n , and let g 2 G. There exist complex numbers z jk such that vj g

n X

zjk v k

(1 < j < n):

k1

Now let VR be the vector space over R with basis v1 , : : : , v n , iv1 , : : : , iv n : Write z jk x jk iyjk with x jk , y jk 2 R. We de®ne a multiplication of VR by g by putting (23:5)

vj g

n X

(xjk v k yjk (iv k )), and

k1

(iv j ) g

n X

(ÿ yjk v k xjk (iv k ))

(1 < j < n),

k1

and extending linearly to de®ne v g for all v 2 VR . In this way we de®ne v g for all v 2 VR and all g 2 G. Regarding v j as an element of the CG-module V, we have (v j g)h v j ( gh) for all g, h 2 G, 1 < j < n: It follows easily that, regarding v j and iv j as elements of VR , we have (v j g)h v j ( gh) and ((iv j ) g)h (iv j )( gh): Hence using Proposition 4.6, we see that (23.5) makes VR into an RGmodule. If ÷ is the character of V, then ÷( g)

n X

zkk :

k1

The character of VR , evaluated at g, is 2

n X

xkk ÷( g) ÷( g):

k1

Hence the character of VR is ÷ ÷. We summarize the basic properties of VR in the next proposition.

268


23.6 Proposition Let V be a CG-module with character ÷. (1) The RG-module VR has character ÷ ÷; in particular, dim VR 2 dim V. (2) If V is an irreducible CG-module and VR is a reducible RGmodule, then ÷ can be realized over R. Proof We have already proved part (1). For part (2), suppose that V is an irreducible CG-module and VR is a reducible RG-module. Then by part (1), VR U W where U is an RG-module with character ÷ and W is an RG-module with character ÷. Thus there is an RG-module, namely U, with character ÷, and so ÷ can be realized over R. j 23.7 Examples (1) Let G C3 kx: x 3 1l, and let V be the 1-dimensional CGmodule with basis v1 such that p v1 x 12(ÿ1 i 3)v1 p (note that 12(ÿ1 i 3) e2ði=3 ). Then VR has basis v1 , iv1 , and with respect to this basis, x is represented by the matrix p ÿ1=2 3=2 p : ÿ 3=2 ÿ1=2 (2) Let G D8 ka, b: a4 b2 1, bÿ1 ab aÿ1 l, and let V be the 2dimensional CG-module with basis v1 , v2 such that v1 a iv1 ,

v1 b v2 ,

v2 a ÿiv2 ,

v2 b v1 :

Then VR has basis v1 , v2 , v3 , v4 , where v3 iv1 , v4 iv2 . With respect to this basis, we obtain the representation r, where 0 1 0 1 0 0 1 0 0 1 0 0 B 0 0 0 ÿ1 C B C C, br B 1 0 0 0 C: ar B @ ÿ1 0 0 @0 0 0 1A 0A 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. Therefore the character of V can be realized over R,


269

by Proposition 23.6(2). In fact, we already know this from Example 23.3(1).

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. Let V be a vector space over F, where F is R or C. A bilinear form â on V is a function which associates with each ordered pair (u, v) of vectors in V an element â(u, v) of F, and which has the following properties: â(ë1 u1 ë2 u2 , v) ë1 â(u1 , v) ë2 â(u2 , v), â(u, ë1 v1 ë2 v2 ) ë1 â(u, v1 ) ë2 â(u, v2 ), for all u, v, u1 , u2 , v1 , v2 2 V and ë1 , ë2 2 F. (Thus for ®xed u, v, the functions x ! â(x, v) and y ! â(u, y) are both linear ± hence the term bilinear.) The bilinear form â is symmetric if â(u, v) â(v, u) for all u, v 2 V : And the bilinear form â is skew-symmetric if â(u, v) ÿâ(v, u)

for all u, v 2 V :

If V is an FG-module, then a bilinear form â on V is said to be Ginvariant if â(ug, vg) â(u, v) for all u, v 2 V and g 2 G: Our next result shows that every RG-module has a G-invariant symmetric bilinear form with a strong positivity property. A similar result for CG-modules was given in Exercise 8.6. 23.8 Theorem If V is an RG-module, then there exists a G-invariant symmetric bilinear form â on V such that â(v, v) . 0

for all non-zero v 2 V :

270


Proof Let v1 , . . . , v n be a basis of V. For u Pn j1 ì j v j 2 V with ë j , ì j 2 R, de®ne ã(u, v)

n X

Pn

j1 ë j v j ,

v

ë j ì j:

j1

Then ã is a symmetric bilinear form on V. Moreover, for non-zero v 2 V, ã(v, v)

n X j1

Now let â(u, v)

X

ì2j . 0:

ã(ux, vx) (u, v 2 V ):

x2G

Again, â is a symmetric bilinear form on V, and â(v, v) . 0 for all non-zero v 2 V. If g 2 G, then gx runs through G as x runs through G, and hence X â(ug, vg) ã(ugx, v gx) â(u, v): x2G

Therefore â is G-invariant and the theorem is proved.

j

23.9 Proposition Let V be an RG-module and let â be a G-invariant bilinear form on V. If U is an RG-submodule of V, then so is W fw 2 V : â(u, w) 0 for all u 2 U g:

Proof It is easy to see that W is a subspace of V. Now let w 2 W and g 2 G. For all u 2 U, we have ugÿ1 2 U, so â(u, wg) â(ug ÿ1 , wgg ÿ1 ) â(ug ÿ1 , w) 0: Thus wg 2 W, so W is an RG-submodule of V.

j

23.10 Proposition Suppose that â is a G-invariant symmetric bilinear form on the RGmodule V, and that there exist u, v 2 V with â(u, u) . 0 and â(v, v) , 0. Then V is a reducible RG-module.


271

Proof Theorem 23.8 supplies us with a G-invariant symmetric bilinear form â1 on V such that â1 (w, w) . 0

for all non-zero w 2 V :

By a general result on bilinear forms (see Exercise 23.7), there is a basis v1 , . . . , v n of V such that â1 (v i , v j ) â(v i , v j ) 0 and

â1 (v i , v i ) 1

if i 6 j,

for all i,

â(v1 , v1 ) . 0, â(v2 , v2 ) , 0: Let â(v1 , v1 ) x, and de®ne ã by 1 ã(u, v) â1 (u, v) ÿ â(u, v) (u, v 2 V ): x Since â and â1 are G-invariant symmetric bilinear forms on V, so is ã. Pn But for all v i1 ë i v i 2 V (ë i 2 R), we have ã(v, v1 ) ë1 ã(v1 , v1 ) 0: Therefore, if we de®ne W fw 2 V : ã(v, w) 0 for all v 2 V g, then W is non-zero, and is an RG-submodule of V by Proposition 23.9. Moreover, 1 ã(v2 , v2 ) 1 ÿ â(v2 , v2 ) . 0, x so W 6 V. Therefore V is a reducible RG-module.

j

We can now relate bilinear forms to the question of whether or not a given character of G can be realized over R. 23.11 Theorem Let ÷ be an irreducible character of G. The following two conditions are equivalent: (1) ÷ can be realized over R; (2) there exists a CG-module V with character ÷, and a non-zero Ginvariant symmetric bilinear form on V.

272


Proof We ®rst show that (2) implies (1). Let V be a CG-module with character ÷, and suppose that â is a non-zero G-invariant symmetric bilinear form on V. There exist u, v 2 V with â(u, v) â(v, u) 6 0. Since â(u v, u v) â(u, u) â(v, v) 2â(u, v), there exists w 2 V with â(w, w) 6 0. Let â(w, w) z and v1 z ÿ1=2 w. Then â(v1 , v1 ) 1: Extend v1 to a basis v1 , . . . , v n of V. Then v1 , . . . , v n , iv1 , . . . , iv n is a basis of the RG-module VR . De®ne a function W from VR to V by n n n X X X W: ë jv j ì j (iv j ) ! (ë j iì j )v j (ë j , ì j 2 R): j1

j1

j1

Then W is a bijection, and for all w1 , w2 , v 2 VR , all ë 2 R and all g 2 G, we have (w1 w2 )W w1 W w2 W,

(23:12)

(ëv)W ë(vW), (v g)W (vW) g: ~ on ordered pairs of elements of VR by Now de®ne a function â ~ v) the real part of â(uW, vW) (u, v 2 VR ): â(u, ~ is a GYou can readily check, using the properties (23.12), that â invariant symmetric bilinear form on VR . Notice that ~ 1 , v1 ) 1 and â(iv ~ 1 , iv1 ) ÿ1: â(v Therefore VR is a reducible RG-module, by Proposition 23.10. It now follows from Proposition 23.6(2) that ÷ can be realized over R. This establishes that (2) implies (1) in the theorem. Conversely, suppose that ÷ can be realized over R, and let U be an RG-module with character ÷. By Theorem 23.8, there is a non-zero Ginvariant symmetric bilinear form ã on U. Let u1 , : : : , u n be a basis of U, and let V be the vector space over C with basis u1 , : : : , u n . As explained earlier, V is a CG-module (with u i g de®ned as for U). ^ on V by De®ne ã ! n n n X n X X X ^ ã ë j u j, ìk uk ë j ì k ã(u j , u k ) j1

k1

j1 k1


273

^ is a non-zero G-invariant symmetric (where ë j , ì k 2 C). Then ã bilinear form on the CG-module V, and V has character ÷. Thus (1) implies (2), and the proof of the theorem is complete. j The indicator function We now associate with each irreducible character ÷ of G a certain number, called the indicator of ÷, which is always 0, 1 or ÿ1. We shall see later that this number tells us whether or not ÷ can be realized over R. Observe that 1 X h÷ 2 , 1 G i ÷( g)÷( g) h÷, ÷i: jGj g2G Therefore, for irreducible characters ÷, we have ( 0, if ÷ is not real, h÷ 2 , 1 G i 1, if ÷ is real: Let V be a CG-module with character ÷. Recall from Chapter 19 that ÷ 2 is the character of the CG-module V V, and ÷2 ÷S ÷ A , where ÷ S is the character of the symmetric part of V V, and ÷ A is the character of the antisymmetric part of V V. Hence if h÷ 2 , 1 G i 1, then precisely one of ÷ S and ÷ A has 1 G as a constituent. 23.13 De®nition If ÷ is an irreducible character of G, then we de®ne the indicator é÷ of ÷ by 8 0, if ÷ is not a constituent of ÷ S or ÷ A , > > < é÷ 1, if 1 G is a constituent of ÷ S , > > : ÿ1, if 1G is a constituent of ÷ A : We call é the indicator function on the set of irreducible characters of G. Note that é÷ 6 0 if and only if ÷ is real. The next result gives a signi®cant property of the indicator function, relating it to the internal structure of the group G.

274 23.14 Theorem For all x 2 G,


X (é÷)÷(x) jf y 2 G: y 2 xgj, ÷

where the sum is over all irreducible characters ÷ of G. Proof De®ne a function W: G ! C by W(x) jf y 2 G: y 2 xgj

(x 2 G):

Note that W is a class function on G, since for g 2 G we have y 2 x , ( g ÿ1 yg)2 g ÿ1 xg: Therefore by Corollary 15.4, W is a linear combination of the irreducible characters of G. The de®nition of é÷ gives é÷ h÷ S ÿ ÷A , 1 G i 1 X ÷( g 2 ) by Proposition 19:14 jGj g2G

1 X X ÷( g 2 ) jGj x2G g2G: g2 x

1 X W(x)÷(x) jGj x2G

hW, ÷i:

Therefore, W

P (é÷)÷, and the result follows.

j

23.15 Example Let G S3 . The character table of G is

÷1 ÷2 ÷3

1

(1 2)

1 1 2

1 ÿ1 0

(1 2 3) 1 1 ÿ1


275

Using Proposition 19.14 we calculate that é÷ 1 for each irreducible P character ÷ of G, so (é÷)÷ ÷1 ÷2 ÷3 , which takes the following values:

÷ 1 ÷2 ÷3

1

(1 2)

(1 2 3)

4

0

1

Sure enough, in accordance with Theorem 23.14, four elements of G square to be 1, namely 1, (1 2), (1 3) and (2 3); no elements square to be (1 2); and one element, (1 3 2), squares to be (1 2 3).

Back to reality We now relate the indicator function to the previous material on bilinear forms. Using this, we show that the indicator of an irreducible character determines whether or not it can be realized over R, and deduce the Frobenius±Schur Count of Involutions. 23.16 Theorem Let V be an irreducible CG-module with character ÷. (1) There exists a non-zero G-invariant bilinear form on V if and only if é÷ 6 0. (2) There exists a non-zero G-invariant symmetric bilinear form on V if and only if é÷ 1. (3) There exists a non-zero G-invariant skew-symmetric bilinear form on V if and only if é÷ ÿ1. Proof In this proof we regard C as a 1-dimensional vector space over C, and de®ne a multiplication of C by elements of G by ëg ë

(ë 2 C, g 2 G):

In this way, C becomes a trivial CG-module. (1) Suppose that é÷ 6 0. Then 1 G is a constituent of ÷ 2 , and hence the CG-module V V has a trivial CG-submodule. By Proposition 8.8, there is a non-zero CG-homomorphism from V V onto this trivial CG-submodule, and hence there is a non-zero CG-homomorphism W from V V onto the trivial CG-module C.

276


Now de®ne â by â(u, v) (u v)W

(u, v 2 V ):

Then â is a non-zero bilinear form on V, and for u, v 2 V and g 2 G, we have â(ug, vg) (ug v g)W ((u v) g)W ((u v)W) g (u v)W â(u, v): Thus â is G-invariant. Conversely, suppose that there is a non-zero G-invariant bilinear form â on V. Let v1 , : : : , v n be a basis of V, so that v i v j (1 < i < n, 1 < j < n) form a basis of V V. De®ne W: V V ! C by putting (v i v j )W â(v i , v j ) (1 < i < n, 1 < j < n) and extending linearly to the whole of V V. For g 2 G, we have ((v i v j ) g)W (v i g v j g)W â(v i g, v j g) â(v i , v j ) as â is G-invariant (v i v j )W: Hence W is a non-zero CG-homomorphism from V V onto the trivial CG-module C. Thus, by Proposition 10.1, V V has a trivial CGsubmodule. If follows that ÷ 2 has the trivial character 1 G as a constituent, and therefore é÷ 6 0. (2) Suppose that é÷ 1. Then 1 G is a constituent of ÷ S , which is the character of the CG-module S(V V), the symmetric part of V V. As in (1), it follows by Proposition 8.8 that there is a non-zero CGhomomorphism W from S(V V) onto the trivial CG-module C. De®ne â(u, v) (u v v u)W

(u, v 2 V ):

Then â is a non-zero G-invariant symmetric bilinear form on V. Conversely, suppose that there exists a non-zero G-invariant symmetric bilinear form â on V. Let v1 , . . . , v n be a basis of V, and de®ne W: S(V V) ! C by putting (v i v j v j v i )W â(v i , v j ) (1 < i, j < n) and extending linearly. Since â is symmetric, W is well-de®ned; and W is a non-zero CG-homomorphism from S(V V) onto the trivial CG-


277

module C. Hence ÷ S has the trivial character 1 G as a constituent, and so é÷ 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 8 0, if ÷ is not real, > > < é÷ 1, if ÷ can be realized over R, > > : ÿ1, if ÷ is real, but ÷ cannot be realized over R: Moreover, for all x 2 G, X (é÷)÷(x) jf y 2 G: y 2 xgj, ÷

where the sum is over all irreducible characters ÷ of G. In particular, X (é÷)÷(1) 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÷ 6 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÷ 1. This proves that é÷ is determined as in the statement of the corollary. P The expression for ÷ (é÷)÷(x) was obtained in Theorem 23.14. P Putting x 1, we see that ÷ (é÷)÷(1) is equal to the number of elements y of G satisfying y 2 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÷ 1 if ÷ is real, and é÷ 0 if ÷ is non-real.

278


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 gÿ1 if and only if g2 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 ka, b: a4 b2 1, bÿ1 ab aÿ1 l can be realized over R (see Example 23.3(1), and note that all four linear characters are real). Thus é÷ 1 for all irreducible characters ÷ of D8, and so X (é÷)÷(1) 1 1 1 1 2 6: ÷

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 ka, b: a4 1, a2 b2 , bÿ1 ab aÿ1 l, there is just one involution, namely a2 . Now é÷ 1 for each of the four linear characters, and X (é÷)÷(1) 2 ÷

by the Frobenius±Schur Count of Involutions. Therefore, if ø is the irreducible character of degree 2, then éø ÿ1. 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.


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 P P If a1 , . . . , a n are real numbers, then a2i > a i 2 =n. Proof This follows from the Cauchy±Schwarz inequality kvk kwk > jv:wj, taking v (a1 , . . . , a n ) and w (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 (m ÿ 1)=t. Then G contains a non-identity element x such that jG : C G (x)j < a2 . Proof By Corollary 23.17, we have X 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 X X t2 < ( ÷(1))2 < (k ÿ 1) ÷(1)2 (k ÿ 1)(m ÿ 1), ÷

and hence m ÿ 1 < (k ÿ 1)(m ÿ 1)2 =t 2 (k ÿ 1)a2 . Now k ÿ 1 is the number of non-identity conjugacy classes of G. If every nonidentity conjugacy class has size more then a2 , then (k ÿ 1)a2 .

280


jGj ÿ 1 m ÿ 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 n. Let jGj 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 jG : C G (u)j m=n: Therefore (m ÿ 1)=t , n, and so by Lemma 23.21, there is a nonidentity element x 2 G such that jG : C G (x)j , n2 . Let H C G (x). If H 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 Z(G), so G is abelian and therefore G C2 . Now suppose that H 6 G. Write r 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 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. 8 0, if ÷ is non-real, > > > > < 1, if there exists an RG-module U with character ÷, 2: é÷ with character ÷ > > > > : ÿ1, if ÷ is real, but there does not exist an RG-module: 8 if 1 G is not a constituent of ÷ S or ÷ A , > < 0, 3: é÷ 1, if 1 G is a constituent of ÷ S , > : ÿ1, if 1 G is a constituent of ÷ A : P 2 4. ÷ (é÷)÷(1) |{ g 2 G: g 1}|.


281

Exercises for Chapter 23 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 1. 3. Let G D2 n and consult Section 18.3 for the character table of G. How many elements g of G satisfy g2 1? Deduce that é÷ 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) det ( gr) for all g 2 G. Deduce that é÷ ÿ1 if and only if det ( gr) 1 for all g 2 G. 5. Let G T 4 n ha, b: a2 n 1, a n b2 , bÿ1 ab aÿ1 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 6 1. Exercise 17.6 shows that V becomes an irreducible CG-module if we de®ne v1 a åv1 ,

v1 b v2 ,

v2 a å ÿ1 v2 ,

v2 b å n v1 :

Let ÷ be the character of this CG-module V. (a) Note that å n 1. Use Exercise 4 to show that é÷ 1 if å n 1 and é÷ ÿ1 if å n ÿ1. (b) Let â be the bilinear form on V for which â(v1 , v1 ) â(v2 , v2 ) 0, â(v1 , v2 ) 1, â(v2 , v1 ) å n : Prove that the bilinear form â is G-invariant, and use Theorem 23.16 to provide a second proof that é÷ 1 if å n 1 and é÷ ÿ1 if å n ÿ1. (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 X (é÷)÷(1) 2, ÷

in agreement with the Frobenius±Schur Count of Involutions. 6. Prove that if ÷ is an irreducible character of a group G, and é÷ ÿ1, 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 , : : : , e n of V such that â1 (ei , ei ) 1 for all i, and â1 (ei , ej ) â(ei , ej ) 0 for all i 6 j: 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 ! W is an RG-homomorphism then either W is an RG-isomorphism or vW 0 for all v 2 V. (b) Prove that if W: V ! V is an RG-isomorphism and V remains irreducible as a CG-module, then W 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 ! 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 2 G, de®ne a function r g : Ù ! Ù by ( Hx)r g Hxg for all x 2 G. Prove that r g is a permutation of Ù, and that the function r : g ! r g is a homomorphism from G to the symmetric group on Ù. T Show that the kernel of r is x2G x ÿ1 Hx. Deduce that if a group G has a subgroup H of index n, then there is a homomorphism G ! 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| 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

1 X ÷( g)÷( g): jGj g2G

The character ÷ is irreducible if and only if h÷, ÷i 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 ÷ÿ

X 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


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) 1

for all g 2 G:

(2) (Theorem 17.11) The group G has precisely jG=G9j linear characters. These are the characters ÷ given by ÷( g) ø( gG9)

( g 2 G)

as ø varies over the irreducible (linear) characters of G=G9. (3) (Theorem 17.3) As a generalization of (2), if N v G and ø is an irreducible character of G=N , then we get an irreducible character ÷ of G which is given by ÷( g) ø( gN ) ( g 2 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 G1 3 G2 then all the irreducible characters ÷ of G are given by ÷( g1 , g 2 ) ö1 ( g 1 )ö2 ( g 2 ) ( g 1 2 G1 , g 2 2 G2 ), as ö i varies over the irreducible characters of Gi (i 1, 2). (5) (Proposition 13.24) If G is a subgroup of Sn , then the function í: G ! C de®ned by í( g) jfix ( g)j ÿ 1 is a character of G.

( g 2 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 X

÷ i (1)2 jGj:

i1

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 ä ij . (8) (Column orthogonality relations, Theorem 16.4(2)) For all g, h 2 G, we have ( k X jCG ( g)j, if g and h are conjugate, ÷ i ( g)÷ i (h) 0, otherwise: i1 (9) (Exercise 13.5) If ÷ is an irreducible character of G and z 2 Z(G), then there exists a root of unity å such that for all g 2 G, ÷(zg) å÷( g): (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 2 G and ÷ is a character of G, then ÷( g ÿ1 ) ÷( g): In particular, if g is conjugate to gÿ1 then ÷( g) is real for all characters ÷ of G. (12) (Corollary 15.6) If g 2 G and g is not conjugate to gÿ1 , then ÷( g) is non-real for some character ÷ of G. (13) (Theorem 22.16) Let g 2 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 p:

286

Representations and characters of groups In particular, if the order of g is a power of p, then ÷( g) ÷(1) mod p:

(15) (Proposition 13.15) If ÷ is an irreducible character of G, then so is ÷, where ÷( g) ÷( g)

( g 2 G):

(16) (Theorem 23.1) The number of real irreducible characters of G is equal to the number of real conjugacy classes of G. (17) (Proposition 17.14) If ÷ is an irreducible character of G and ë is a linear character of G, then ÷ë is an irreducible character of G, where ÷ë( g) ÷( g)ë( g) ( g 2 G): (18) (Proposition 19.6) If ÷ and ø are characters of G, then so is the product ÷ø, where ÷ø( g) ÷( g)ø( g)

( g 2 G):

(19) (Proposition 19.14) If ÷ is a character of G, then so are ÷ S and ÷ A , where for all g 2 G, ÷ S ( g) 12(÷ 2 ( g) ÷( g 2 )), ÷ A ( g) 12(÷ 2 ( g) ÿ ÷( g 2 )): (20) (De®nition 21.13, Proposition 21.23) If H is a subgroup of G and ø is a character of H, then ø " G is a character of G, with values given by Proposition 21.23. (21) (Chapter 20) If H is a subgroup of G and ø is a character of G, then ø # H is a character of H, where (ø # H)(h) ø(h)

(h 2 H):

We have seen that the character table of a group G gives grouptheoretic information about G. For example, the ®rst column determines |G| and jG=G9j (by (6) and (2)). We can see from the character table whether or not G is simple (Proposition 17.6); indeed, we can ®nd all the normal subgroups of G (Proposition 17.5). Two important normal subgroups are G9 and Z(G); these can be determined in the following ways. The derived subgroup G9 consists of those elements g in G which satisfy ÷( g) 1 for all linear characters ÷ of G. The centre Z(G) can be found by noting which elements g of G satisfy

Summary of properties of character tables 287 P ÷( g)÷( g) |G|, the sum being over all irreducible characters ÷ of G. In Chapter 30 we shall see some more impressive results about subgroups of G, which can be deduced from the character table. As a ®nal remark, it is of course true that isomorphic groups have the same character table; however, the converse is false: in Exercise 17.1 we gave examples of non-isomorphic groups, D8 and Q8 , with the same character table.

25 Characters of groups of order pq

By the end of the next chapter, we shall have determined the character tables of all groups of order less than 32. A number of these groups are so-called Frobenius groups, and in this chapter we shall describe a class of Frobenius groups and ®nd the character tables of the groups in this class. In particular, this will give the character tables of all groups whose order is the product of two prime numbers. Throughout the chapter, p will denote a prime number.

Primitive roots modulo p Recall that the set Z p f0, 1, : : : , p ÿ 1g, with addition and multiplication modulo p, is a ®eld; that is, Z p is an abelian group under addition, and Zp Z p ÿ f0g is an abelian group under multiplication. Clearly Z p is a cyclic group under addition, generated by 1. It is also true, but not at all obvious, that Zp is cyclic: 25.1 Theorem The multiplicative group Zp is cyclic; that is, there exists an integer n such that n pÿ1 1 mod p, and n r 6 1 mod p for 0 , r , p ÿ 1: An integer n of order p ÿ 1 in Zp is called a primitive root modulo p. We shall not provide a proof of Theorem 25.1, but for a good 288

Characters of groups of order pq

289

account, we refer you to Theorem 45.3 of the book by J. B. Fraleigh listed in the Bibliography. 25.2 Example The number 2 is a primitive root modulo 3, 5, 11 and 13, but not modulo 7; and 3 is a primitive root modulo 7. As an immediate consequence of Theorem 25.1 we have 25.3 Proposition If q| p ÿ 1 then there is an integer u such that u has order q modulo p ± that is, such that u q 1 mod p, and u r 6 1 mod p for 0 , r , q:

Frobenius groups of order pq, where q| p 2 1 25.4 Example De®ne

G

1 0

y : x 2 Zp , y 2 Z p : x

Under matrix multiplication, G is a group of order p( p ÿ 1) (see Exercise 25.1). Now let q| p ÿ 1, and let u be an element of order q in the multiplicative group Zp . De®ne 1 1 1 0 A , B , 0 1 0 u and let F hA, Bi, the subgroup of G generated by A and B. Then 1 u Bÿ1 AB Au , 0 1 and so we have the relations (25:5)

Ap Bq I, Bÿ1 AB Au :

Using these relations, we see that every element of F is of the form A i B j with 0 < i < p ÿ 1, 0 < j < q ÿ 1. These pq elements are dis-

290


tinct, so jFj pq. Moreover the relations (25.5) determine all products in F, so we have the presentation F hA, B: Ap Bq I, Bÿ1 AB Au i: 25.6 De®nition If p is a prime and q| p ÿ 1, then we write F p,q for the group of order pq with presentation F p,q ha, b: a p b q 1, bÿ1 ab au i, where u is an element of order q in Zp . It is not hard to show that, up to isomorphism, F p,q does not depend on which integer u of order q we choose (see Exercise 25.3). The groups F p,q belong to a wider class of groups known as Frobenius groups. We shall not give the general de®nition of these here, as we shall only be dealing with F p,q ; the interested reader can ®nd more information in the book by D. S. Passman listed in the Bibliography. The next result classi®es all groups whose order is the product of two distinct prime numbers. 25.7 Proposition Suppose that G is a group of order pq, where p and q are prime numbers with p . q. Then either G is abelian, or q divides p ÿ 1 and G F p,q . Proof Assume that G is non-abelian. It follows from Exercise 22.3 that q divides p ÿ 1 and G has a normal subgroup H of order p. (Alternatively, these facts follow readily from Sylow's Theorems (see Chapter 18 of the book by J. B. Fraleigh listed in the Bibliography).) Both H and G= H are cyclic, since they have prime order. Suppose that H kal and G= H h Hbi; then G is generated by a and b. Since bq 2 H but b does not have order pq (as G is non-abelian), it follows that b has order q. Now H v G, so bÿ1 ab au for some integer u. Further, a bÿq abq a u

q

and so u q 1 mod p. Thus the order of u in the group Zp divides q.


291

If the order of u were 1 then we would have bÿ1 ab a, and G would be abelian. Therefore the order of u is q. We have now established that a p b q 1, bÿ1 ab au , order of u in Z is q: p

Hence G F p,q .

j

25.8 Example By Proposition 25.7, every group of order 15 is abelian (indeed, isomorphic to C3 3 C5 ); and the groups of order 21 are C3 3 C7 and F7,3. The character table of F p,q We have, in fact, already found the character tables of certain of the groups F p,q : the dihedral group of order 2 p is the case where q 2, and in Example 21.25 we dealt with F7,3 . We now construct the character table of F p,q in general. Thus let G Fp,q ha, b: ap bq 1, bÿ1 ab au i where p is prime, q| p ÿ 1 (q not necessarily prime), and u has order q modulo p. Let S be the subgroup of Zp consisting of the powers of u. Thus jSj q. Write r ( p ÿ 1)=q, and choose coset representatives v1 , : : : , v r for S in Zp . 25.9 Proposition The conjugacy classes of G F p,q are f1g, (av i ) G fav i s : s 2 Sg (1 < i < r), (bn ) G fam bn : 0 < m < p ÿ 1g (1 < n < q ÿ 1): Proof The equation bÿ j av b j avu

j

shows that av is conjugate to avs for all s 2 S. Therefore the conjugacy class of av i has size at least q; also the size of this conjugacy class is equal to |G: CG (av i )|, and since kal < CG (av i ), this size is at most q. Hence (av i ) G has size q, and has the form stated in the proposition.

292


Since CG (bn ) contains kbl, and kbl has index p in G, it follows that for n 6 0 mod q, we have |CG (bn )| q, and so the conjugacy class of bn has size p. On the other hand, as G=hai is abelian, every conjugate of bn has the form am bn for some m. Hence (bn ) G fam bn : 0 < m < p ÿ 1g and the proof is complete.

j

By Proposition 25.9, G has q r conjugacy classes, so we seek q r irreducible characters. First, observe that the derived subgroup G9 kal, so G=G9 has order q and therefore by Theorem 17.11, G has precisely q linear characters. These are given by ÷ n (0 < n < q ÿ 1), where ÷ n (a x b y ) e2ði ny=q

(0 < x < p ÿ 1, 0 < y < q ÿ 1):

We shall show that G has r irreducible characters of degree q. Let å e2ði= p . For v 2 Zp, denote by øv the character of kal which is given by øv (ax ) å vx

(0 < x < p ÿ 1):

We calculate the values of the induced character øv " G, using Proposition 21.23. We obtain (øv " G)(ax by ) 0 if 1 < y < q ÿ 1, and X (øv " G)(ax ) å vsx (0 < x < p ÿ 1): s2S

Note that øv " G has degree q, and øv " G øvs " G

if s 2 S:

For each coset representative v j (1 < j < r) of S in Zp , let ö j øv j " G: We now prove that each ö j is irreducible. By the Frobenius Reciprocity Theorem 21.16, for all s 2 S, hö j # hai, øv j s ihai hö j , øv j s " Gi G hö j , ö j i G : Hence ö j # hai hö j , ö j i G

X s2S

øv j s ÷,


293

where ÷ is either 0 or a character of kal. Taking degrees, it follows that ö j (1) > jSjhö j , ö j i G : Since ö j (1) q jSj, we deduce that kö j , ö j l G 1. This proves that ö j is irreducible, and also that X ö j # hai hö j , ö j i G øv j s : s2S

By Theorem 14.23, the characters øv (v 2 Zp ) are linearly independent, and hence ö1 # kal, . . . , ö r # kal are distinct. Consequently the irreducible characters ö1 , . . . , ö r are distinct. We have now found q r distinct irreducible characters ÷ n , ö j of G (0 < n < q ÿ 1, 1 < j < r), so we have the complete character table of G. We summarize in the following theorem. 25.10 Theorem Let p be a prime number, q| p ÿ 1 and r ( p ÿ 1)=q. Then the group F p,q ha, b: a p bq 1, bÿ1 ab au i fax by : 0 < x < p ÿ 1, 0 < y < q ÿ 1g has q r irreducible characters. Of these, q have degree 1 and are given by ÷ n (ax by ) e2ði ny=q

(0 < n < q ÿ 1)

and r have degree q and are given by ö j (ax by ) 0 if 1 < y < q ÿ 1, X ö j (ax ) e2ðiv j sx= p , s2S

for 1 < j < r, where v1 S, : : : , v r S are the cosets in Zp of the subgroup S generated by u. We conclude by illustrating Theorem 25.10 in some examples. 25.11 Example Let G F p, pÿ1 ha, b: a p b pÿ1 1, bÿ1 ab au i

294


where u is a primitive root modulo p. Then G has p ÿ 1 linear characters, and one irreducible character ö of degree p ÿ 1, with values given by ö(ax by ) 0

if 1 < y < p ÿ 2,

ö(ax ) ÿ1

if 1 < x < p ÿ 1:

25.12 Example Let a, b 2 S5 be the permutations a (1 2 3 4 5), b (2 3 5 4): Check that a5 b4 1, bÿ1 ab a2 : Hence if G ka, bl, then G F5,4 , and so by the previous example the character table of G is as shown. Character table of F5,4 gi |CG ( gi )| ÷0 ÷1 ÷2 ÷3 ö

1 20

a 5

b 4

b2 4

b3 4

1 1 1 1 4

1 1 1 1 ÿ1

1 i ÿ1 ÿi 0

1 ÿ1 1 ÿ1 0

1 ÿi ÿ1 i 0

25.13 Example We consider the case p 13, q 4. Here F13,4 ha, b: a13 b4 1, bÿ1 ab a5 i: Write å e2ði=13 , and let á å å 5 å 8 å 12 , â å 2 å 3 å 10 å 11 , ã å4 å6 å7 å9 : By Theorem 25.10, the character table of F13,4 is as shown opposite. In Example 21.25 we found the character table of F7,3. You may like


295

to check that this agrees with the description of the character table provided by Theorem 25.10. Character table of F13,4 gi |CG ( gi )| ÷0 ÷1 ÷2 ÷3 ö1 ö2 ö3

1 52

a 13

a2 13

a4 13

b 4

b2 4

b3 4

1 1 1 1 4 4 4

1 1 1 1 á â ã

1 1 1 1 â ã á

1 1 1 1 ã á â

1 i ÿ1 ÿi 0 0 0

1 ÿ1 1 ÿ1 0 0 0

1 ÿi ÿ1 i 0 0 0

Summary of Chapter 25 1. Suppose that p is prime and q divides p ÿ 1. Let u be an element of order q in Zp . Then Fp,q ha, b: ap bq 1, bÿ1 ab au i: The irreducible characters of Fp,q are described in Theorem 25.10. 2. Let p and q be prime numbers with p . q. If G has order pq, then either G is abelian or G F p,q.

Exercises for Chapter 25 1. Let p be a prime number. Prove that 1 y : x 2 Z p, y 2 Z p , 0 x under matrix multiplication, is a group of order p( p ÿ 1). 2. Write down the character table of the non-abelian group F11,5 of order 55. 3. Let p and q be positive integers, with p prime and q| p ÿ 1. Let u and v be two integers which are of order q modulo p, and de®ne

296

Representations and characters of groups G1 ha, b: ap bq 1, bÿ1 ab au i, G2 ha, b: ap bq 1, bÿ1 ab av i:

Prove that G1 G2 . (This justi®es the comment which follows the de®nition of Fp,q in 25.6.) 4. Suppose that p is a prime number, with p 6 2. Let q ( p ÿ 1)=2 and let G Fp,q ha, b: ap bq 1, bÿ1 ab au i, where u is an element of order q modulo p. (a) Show that there exists an integer m such that u m ÿ1 mod p if and only if p 1 mod 4. (b) Deduce that a is conjugate to aÿ1 if and only if p 1 mod 4. (c) Using the orthogonality relations, show that the two irreducible characters ö1 , ö2 of G of degree q have values p 1 (ä p)) 2(ÿ1 on the element a, where ä 1 if p 1 mod 4, and ä ÿ1 if p ÿ1 mod 4. (d) Deduce that if å e2ði= p then X p å s (ÿ1 (ä p)), s2Q

where Q is the set of quadratic residues modulo p (that is, Q f12 , 22 , : : : , (( p ÿ 1)=2)2 g). 5. Let E be the group of order 18 which is given by E ha, b, c: a3 b3 c2 1, ab ba, cÿ1 ac aÿ1 , cÿ1 bc bÿ1 i, as in Exercise 5.4. Note that ka, bl is a normal subgroup of E which is isomorphic to C3 3 C3 . By inducing linear characters of this subgroup, obtain the character table of E. 6. Show that the group E of Exercise 5 has the properties that Z(E) is cyclic, but E has no faithful irreducible representation. (Thus, E provides a counterexample to the converse of Proposition 9.16.) 7. (a) Find a group whose irreducible character degrees are

Characters of groups of order pq 1, 1, 1, 3, 3, 3, 3: (b) Find a group whose irreducible character degrees are 1, 1, 1, 1, 1, 1, 3, 3, 3, 3, 3, 3, 3, 3: (c) Find a group whose irreducible character degrees are 1, 1, 1, 1, 1, 1, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 6, 6, 6, 6: 8. Let G be the group of order 54 which is given by G ha, b: a9 b6 1, bÿ1 ab a2 i: Find the character table of G.

297

26 Characters of some p-groups

Throughout this chapter, p will be a prime number. We shall show how to obtain the character tables of all groups of order pn for n < 4. The method consists of examining the characters of those p-groups which contain an abelian subgroup of index p, and before explaining the method, we show that all groups of order pn with 1 < n < 4 do, indeed, have an abelian subgroup of index p. We later give explicitly the irreducible characters of all groups of order p3 and of all groups of order 16. At the end of the chapter we point out, with references, that we have found the character tables of all groups of order less than 32.

Elementary properties of p-groups A p-group is a group whose order is a power of the prime number p. In the ®rst lemma we collect several well known properties of pgroups. Recall that Z(G) denotes the centre of G (see De®nition 9.15). 26.1 Lemma Let G be a group of order pn with n > 1. (1) If {1} 6 H v G then H \ Z(G) 6 {1}. In particular, Z(G) 6 {1}. (2) If K < Z(G) and G=K is cyclic, then G is abelian. (3) If n < 2 then G is abelian. Proof (1) Since H v G, H is a union of conjugacy classes of G, all of which have size a power of p; and H \ Z(G) consists of those conjugacy classes in H which have size 1. Therefore 298

Characters of some p-groups

299

j Hj j H \ Z(G)j (a multiple of p): As | H| is a multiple of p and | H \ Z(G)| 6 0, we deduce that H \ Z(G) 6 {1}. (2) Suppose that G=K is cyclic, generated by gK. Let x1 , x2 2 G. Then x1 g i k 1 , x2 g j k 2 for some integers i, j and some k1 , k2 2 K. Since k1 , k2 2 Z(G), it follows that x1 x2 x2 x1 . Therefore G is abelian. (3) By (1), jG= Z(G)j < p nÿ1 . Hence if n < 2 then G= Z(G) is j cyclic and so G is abelian by (2). 26.2 Lemma Let G be a group of order pn with 1 < n < 4. Then G contains an abelian subgroup of index p. Proof The result is immediate if n 1, so suppose that 2 < n < 4. Assume that Z(G) contains a subgroup K of order p nÿ2 . Then we can ®nd a subgroup H of G such that K < H and j Hj p nÿ1 . As K < Z( H) and, by Lemma 26.1(2), H= Z( H) is not of order p, we deduce that Z( H) H. Therefore H is an abelian subgroup of index p in G. Now assume that Z(G) has no subgroup of order p nÿ2 . Since Z(G) 6 f1g by Lemma 26.1(1), the only possibility is that |G| p4 and | Z(G)| p. Then by Exercise 12.7, G has an element x whose conjugacy class x G is of size p. Let H CG (x). Then by Theorem 12.8, j Hj jGj=jx G j p3 . Moreover, Z(G) and kxl are distinct nonidentity subgroups of Z( H), and so Z( H) > p2 . Hence again Z( H) H by Lemma 26.1(2), and H is an abelian subgroup of index p. j For our ®nal result on the structure of p-groups, recall that the derived subgroup of G is denoted by G9 (see De®nition 17.7). 26.3 Lemma Let G be a non-abelian p-group which contains an abelian subgroup H of index p. Then there exists a normal subgroup K of G such that K < H \ G9 \ Z(G) and jKj p:

300


Proof Since G is non-abelian, we have {1} 6 G9 v G, and hence G9 \ Z(G) 6 {1} by Lemma 26.1(1). Let K be a subgroup of order p in G9 \ Z(G). Now K < Z(G) implies that K v G and that KH is an abelian subgroup of G (where KH {kh: k 2 K, h 2 H}). Since G is non-abelian and |G: H| p, we have KH H, and therefore K < H. j Characters of p-groups with an abelian subgroup of index p In view of Lemma 26.2, the next theorem provides us with all the irreducible characters of non-abelian groups of order p3 or p4 . 26.4 Theorem Assume that G is a non-abelian p-group which contains an abelian subgroup H of index p. Let K be a normal subgroup of G as in Lemma 26.3. Then every irreducible character of G is given by either (1) the lift of an irreducible character of G=K, or (2) ø " G, for some linear character ø of H which satis®es K 6< Ker ø. Proof Let |G| pn . By Theorem 17.3, the irreducible characters of G=K lift to give precisely those irreducible characters of G which have K in their kernel. The sum of the squares of the degrees of the irreducible characters obtained in this way is jG=Kj p nÿ1 , by Theorem 11.12. We shall construct p nÿ2 ÿ p nÿ3 further irreducible characters of G, each of degree p. Since p nÿ1 ( p nÿ2 ÿ p nÿ3 ) p2 pn jGj, ( ) we shall then have obtained all the irreducible characters of G, again by Theorem 11.12. First note that if ÷ is a character of G of degree p, then either ÷ is irreducible or ÷ is a sum of linear characters (since by Theorem 22.11, the degree of every irreducible character of G is a power of p). In the latter case, we have G9 < Ker ÷, as G9 is in the kernel of every linear character, and hence K < G9 < Ker ÷. This establishes (26:5)

if ÷(1) p and K 6< Ker ÷, then ÷ is irreducible:

We know by Theorem 9.8 that all the p nÿ1 irreducible characters of the abelian group H are linear. Let Ö denote the set of linear


301

characters of H which do not have K in their kernel. Since the linear characters of H which do have K in their kernel are precisely the lifts of linear characters of H=K, we have jÖj p nÿ1 ÿ p nÿ2 : Let ø 2 Ö. By Proposition 21.23, since K < Z(G), (ø " G)(k) pø(k) for all k 2 K: Thus ø " G has degree p and does not have K in its kernel. Therefore by (26.5), ø " G is an irreducible character of G. Suppose now that ø1 is a linear character of H such that ø " G ø1 " G. Then by the Frobenius Reciprocity Theorem 21.16, 1 hø " G, ø1 " Gi G h(ø " G) # H, ø1 i H : Since (ø " G) # H has degree p, this implies that there are at most p elements ø1 of Ö such that ø1 " G ø " G. It follows that fø " G: ø 2 Ög nÿ1

gives at least jÖj= p ( p ÿ p nÿ2 )= p distinct irreducible characters of G of degree p which do not have K in their kernel. As we saw in ( ), G has at most p nÿ2 ÿ p nÿ3 such characters. Therefore fø " G: ø 2 Ög consists precisely of the p nÿ2 ÿ p nÿ3 irreducible characters we seek, and the proof is complete. j We now use Theorem 26.4 to give an explicit construction of the irreducible characters of the non-abelian groups of order p3 . We shall then illustrate Theorem 26.4 further by constructing the character tables of all the non-abelian groups of order 16.

Groups of order p3 By Theorem 9.6, the abelian groups of order p3 are C p3 , C p2 3 Cp and Cp 3 Cp 3 Cp : The character tables of these groups are given by Theorem 9.8. Now let G be a non-abelian group of order p3 . Write Z Z(G). By Lemma 26.1, Z 6 {1} and G= Z is non-cyclic. Hence G= Z Cp 3 Cp and Z kzl Cp . Choose aZ, bZ such that G= Z haZ, bZi. Then

302

Representations and characters of groups G= Z far bs Z: 0 < r < p ÿ 1, 0 < s < p ÿ 1g

and in particular, every element of G is of the form ar bs z t for some r, s, t with 0 < r, s, t < p ÿ 1. 26.6 Theorem Let G {ar bs z t : 0 < r, s, t < p ÿ 1} be a non-abelian group of order p3 , as above. Write å e2ði= p . Then the irreducible characters of G are ÷ u,v

(0 < u < p ÿ 1, 0 < v < p ÿ 1), and

öu

(1 < u < p ÿ 1),

where for all r, s, t, ÷ u,v (ar bs z t ) å rusv , ( ut if r s 0, på , r s t ö u (a b z ) 0, otherwise:

Proof By Theorem 9.8, the irreducible characters of G= Z are ø u,v (0 < u, v < p ÿ 1), where ø u,v (ar bs Z) å rusv : The lift to G of ø u,v is the linear character ÷ u,v which appears in the statement of the theorem. Let H ka, zl, so that H is an abelian subgroup of order p2 . For 1 < u < p ÿ 1, choose a character ø u of H which satis®es ø u (z t ) å ut

(0 < t < p ÿ 1):

We shall calculate ø u " G. Let r be an integer with 1 < r < p ÿ 1. If ar is conjugate to an element g of G, then ar Z is conjugate to gZ in the abelian group G= Z, so ar Z gZ, and therefore g ar z t for some t. Since ar 2 = Z, the conjugacy class (ar ) G does not have size 1, and hence (ar ) G far z t : 0 < t < p ÿ 1g:


303

Then by Proposition 21.23, (ø u " G)(ar z t ) ø u (ar ) ø u (ar z) : : : ø u (ar z pÿ1 ) ø u (ar )

pÿ1 X

ø u (z s )

s0

ø u (ar )

pÿ1 X

å us

s0

0: Also,

(ø u " G)(z t ) pø u (z t ) på ut , and (ø u " G)( g) 0 if g 2 = H:

We have now established that if ö u ø u " G, then ö u takes the values stated in the theorem. We ®nd that 1 X hö u , ö u i G 3 ö u ( g)ö u ( g) p g2G

1 X ö u ( g)ö u ( g) p3 g2 Z

1 X 2 p p3 g2 Z

1: Therefore ö u is irreducible. Clearly the irreducible characters ÷ u,v (0 < u, v < p ÿ 1) and ö u (1 < u < p ÿ 1) are all distinct, and the sum of the squares of their degrees is p2 . 12 ( p ÿ 1) . p2 jGj: Hence we have found all the irreducible characters of G.

j

Notice that the calculation in the proof of Theorem 26.6 is a special case of the proof of Theorem 26.4 (with K Z(G)). In fact, up to isomorphism, there are precisely two non-abelian groups of order p3 . If p 2, they are D8 and Q8. And if p is odd, they are

304


(26:7)

2

H 1 ha, b: a p b p 1, bÿ1 ab a p1 i, and H 2 ha, b, z: a p b p z p 1, az za, bz zb, bÿ1 ab azi:

We have Z( H1 ) ka p l, Z( H2 ) kzl. The elements a, b in H1 and H2 will serve for the elements a, b chosen in the statement of Theorem 26.6.

26.8 The groups of order 16 It is known that, up to isomorphism, there are precisely fourteen groups of order 16 (see p. 134 of the book by Coxeter and Moser listed in the Bibliography). We shall describe all these groups and their character tables. By Theorem 9.6, the abelian groups of order 16 are C16 , C8 3 C2 , C4 3 C4 , C4 3 C2 3 C2 and C2 3 C2 3 C2 3 C2 , and their character tables are given by Theorem 9.8. For each of the nine non-abelian groups G of order 16 it is the case that |G9 \ Z(G)| 2 (see Exercise 26.7), so the subgroup K described in Lemma 26.3 is given by K G9 \ Z(G): Now G=K is a group of order 8. It is not C8 by Lemma 26.1(2), and it is not Q8 by Exercise 26.8. Hence G=K D8 , C4 3 C2 or C2 3 C2 3 C2 : We shall divide our descriptions into three parts, according to these three possibilities for G=K. Our descriptions will be in terms of presentations; it is possible to see, using Exercise 26.5, that all the nine groups G1 , . . . , G9 given below do indeed have order 16. (A) There are three non-abelian groups G=K D8 . These are

G of order 16 with

G1 ha, b: a8 b2 1, bÿ1 ab aÿ1 i D16 , G2 ha, b: a8 1, b2 a4 , bÿ1 ab aÿ1 i, G3 ha, b: a8 b2 1, bÿ1 ab a3 i:


305

In each case K ka4 l. Each of the groups has seven conjugacy classes C1 , . . . , C7 , and these are given in the following table.

G1 , G2 G3

C1

C2

C3

C4

C5

C6

C7

1 1

a4 a4

a2 , a6 a2 , a6

a, a7 a, a3

a3 , a5 a5 , a7

ai b (i even) ai b (i even)

ai b (i odd) ai b (i odd)

Using Theorem 26.4, we obtain the character tables of G1 , G2 and G3 : Class Centralizer order

where

C1

C2

C3

C4

C5

C6

C7

16

16

8

8

8

4

4

1 1 1 1 2 2 2

1 1 1 1 2 ÿ2 ÿ2

1 1 1 1 ÿ2 0 0

1 1 ÿ1 ÿ1 0 á â

1 1 ÿ1 ÿ1 0 â á

1 ÿ1 1 ÿ1 0 0 0

1 ÿ1 ÿ1 1 0 0 0

á

p

2 ÿâ for G1 , G2 , p á i 2 ÿâ for G3 :

The ®rst ®ve characters are the lifts of the irreducible characters of G=K D8 . The last two characters can be obtained as in Theorem 26.4(2) as induced characters ø " G, where ø is a linear character of the abelian subgroup kal of index 2; alternatively, they can be found by using the column orthogonality relations. Note that a is conjugate to aÿ1 in G1 and G2 , but not in G3 ; hence the values in the columns C4 and C5 are all real for G1 and G2 , but not for G3 (see Corollary 15.6). (B) There are three non-abelian groups G of order 16 with G=K C4 3 C2 (where, as before, K G9 \ Z(G), of order 2). These are G4 ha, b, z: a4 z, b2 z 2 1, bÿ1 ab azi, G5 ha, b, z: a4 1, b2 z, z 2 1, bÿ1 ab azi, G6 ha, b, z: a4 1, b2 z 2 1, bÿ1 ab az, az za, bz zbi:

306


The given presentations are somewhat cumbersome (for example, since a4 z in G4 , z is redundant), but they are in a form which allows us to describe the conjugacy classes C1 , . . . , C10 of all three groups G4 , G5 , G6 simultaneously: C1

C2

C3

C4

1

z

a2

a2 z

C5

C6

C7

a, az a3 , a3 z b, bz

C8

C9

C10

a2 b, a2 bz

ab, abz

a3 b, a3 bz

In each case, K kzl. The character tables of G4 , G5 and G6 can again be found using Theorem 26.4: Class Centralizer order

C1 16

C2 16

C3 16

C4 16

C5 8

C6 8

C7 8

C8 8

C9 8

C10 8

1 1 1 1 1 1 1 1 2 2

1 1 1 1 1 1 1 1 ÿ2 ÿ2

1 ÿ1 1 ÿ1 1 ÿ1 1 ÿ1 á â

1 ÿ1 1 ÿ1 1 ÿ1 1 ÿ1 â á

1 i ÿ1 ÿi 1 i ÿ1 ÿi 0 0

1 ÿi ÿ1 i 1 ÿi ÿ1 i 0 0

1 1 1 1 ÿ1 ÿ1 ÿ1 ÿ1 0 0

1 ÿ1 1 ÿ1 ÿ1 1 ÿ1 1 0 0

1 i ÿ1 ÿi ÿ1 ÿi 1 i 0 0

1 ÿi ÿ1 i ÿ1 i 1 ÿi 0 0

where á 2i ÿâ

for G4 ,

á 2 ÿâ

for G5 , G6 :

(C) Finally, there are three non-abelian groups G of order 16 with G=K C2 3 C2 3 C2 (where K G9 \ Z(G), of order 2). These are G7 ha, b, z: a4 b2 z 2 1, bÿ1 ab aÿ1 , az za, bz zbi D8 3 C2 , G8 ha, b, z: a4 z 2 1, a2 b2 , bÿ1 ab aÿ1 , az za, bz zbi Q8 3 C2 , G9 ha, b, z: a2 b2 z 4 1, bÿ1 ab az 2 , az za, bz zbi:


307

Each of these groups has ten conjugacy classes, which are given by C1 C2 C3 C4 G7 , G8 G9

1 a2 1 z2

C5

C6

C7

z a2 z a, a3 az, a3 z z z 3 a, az 2 az, az 3

We have

( K

C8

C9

C10

b, a2 b bz, a2 bz ab, a3 b abz, a3 bz b, bz 2 bz, bz 3 ab, abz 2 abz, abz3

ha2 i

for G7 , G8 ,

2

hz i

for G9 ,

and the character tables of G7 , G8 and G9 , given by Theorem 26.4, are as follows. Class Centralizer order

C1 16

C2 16

C3 16

C4 16

C5 8

C6 8

C7 8

C8 8

C9 8

C10 8

1 1 1 1 1 1 1 1 2 2

1 1 1 1 1 1 1 1 ÿ2 ÿ2

1 1 1 ÿ1 1 ÿ1 ÿ1 ÿ1 á â

1 1 1 ÿ1 1 ÿ1 ÿ1 ÿ1 â á

1 1 ÿ1 1 ÿ1 1 ÿ1 ÿ1 0 0

1 1 ÿ1 ÿ1 ÿ1 ÿ1 1 1 0 0

1 ÿ1 1 1 ÿ1 ÿ1 1 ÿ1 0 0

1 ÿ1 1 ÿ1 ÿ1 1 ÿ1 1 0 0

1 ÿ1 ÿ1 1 1 ÿ1 ÿ1 1 0 0

1 ÿ1 ÿ1 ÿ1 1 1 1 ÿ1 0 0

where á 2 ÿâ

for G7 , G8 ,

á 2i ÿâ

for G9 :

26.9 The groups of order less than 32 At this point, we have in fact found the character tables of all groups of order 31 or less. Apart from abelian groups and dihedral groups, whose character tables are given by Theorem 9.8 and Section 18.3, the groups, with references for their character tables, are as follows:

308


|G|

G

Reference for character table

8 12

Q8 A4 T12 G1 , . . . , G9 D6 3 C3 E T20 F5,4 F7,3 D12 3 C2 , A4 3 C2 , T12 3 C2 D8 3 C3 , Q8 3 C3 , D6 3 C4 S4 SL (2, 3) T24 U24 V24 H1 , H2 T28 D6 3 C5 , D10 3 C3

Exercise 17.1 Section 18.2 Exercise 18.3 Section 26.8 Theorem 19.18 Exercise 25.5 Exercise 18.3 Theorem 25.10 Theorem 25.10 Theorem 19.18 Theorem 19.18 Section 18.1 Exercise 27.2 Exercise 18.3 Exercise 18.4 Exercise 18.5 Theorem 26.6 Exercise 18.3 Theorem 19.18

16 18 20 21 24

27 28 30

Summary of Chapter 26 In this chapter, we gave the irreducible characters of various nonabelian p-groups G, as follows. 1. Theorem 26.4: p-groups which contain an abelian subgroup of index p. 2. Theorem 26.6: groups of order p3 . 3. Section 26.8: groups of order 16.

Exercises for Chapter 26 1. Suppose that G is a group of order pn ( p prime, n > 2), with an abelian subgroup H of index p. Show that for some integer m > 2, G has pm linear characters and p nÿ2 ÿ p mÿ2 irreducible characters of degree p.


309

2. Let H be the group of order 27 which is given by H ha, b, z: a3 b3 z 3 1, az za, bz zb, bÿ1 ab azi (see (26.7)). Find the conjugacy classes of H, and use Theorem 26.6 to write down the character table of H. 3. Let G be the group of order 32 which is given by G ha, b: a16 1, b2 a8 , bÿ1 ab aÿ1 i: Using Theorem 26.4, or otherwise, ®nd the character table of G. 4. Let A, B, C, D 0 ÿ1 B B 0 B AB B 0 @ 0

be the following 4 3 4 matrices: 1 0 0 0 0 0 0 i C B B0 1 0 0C 0 0 C B BB C, Bi 0 1 0C 0 0 A @

0 0

0

ÿ1 0

0 B Bi B CB B0 @

i

0

0

0

0

0

0

0

ÿi

1

C 0C C C, ÿi C A 0

0

0

ÿi

0 B B0 B DB B0 @

0 0

1

0 0

0 1 1 0

0

0 1

1

C ÿi C C C, 0C A 0 1

C 0C C C, 0C A 0

and let G kA, B, C, Dl. Write Z ÿI. (a) Prove that all pairs of generators commute modulo h Zi, and deduce that G9 h Zi. (b) Show that for all g in G, g 2 2 h Zi, and deduce that G is a 2group of order at most 32. (c) Prove that the given representation of G of degree 4 is irreducible. (Hint: use Corollary 9.3.) (d) Show that |G| 32, and ®nd all the irreducible representations of G. 5. Let G1 , . . . , G9 be the non-abelian groups of order 16 with presentations as given in the text. (a) Find faithful irreducible representations of degree 2 for G1 , G2 , G3 , G4 and G9 . (b) Why do the remaining groups G5 , G6 , G7 and G8 have no faithful irreducible representations?

310 (c)

Representations and characters of groups Check that the following give faithful representations of G5 and G6 : 0 1 0 1 0 1 0 i 0 0 G5 : a ! @ 1 0 0 A, b ! @ 0 ÿi 0 A, 0 0 i 0 0 1 0 1 ÿ1 0 0 z ! @ 0 ÿ1 0 A; 0 0 1 0 1 0 1 i 0 0 0 1 0 G6 : a ! @ 0 ÿi 0 A, b ! @ 1 0 0 A, 0 0 i 0 0 1 0 1 ÿ1 0 0 z ! @ 0 ÿ1 0 A: 0 0 1

(d) Find faithful representations of degree 3 for G7 D8 3 C2 and G8 Q8 3 C2 . (Note: This exercise can be used to con®rm that the presentations of G1 , . . . , G9 given in the text do indeed give groups of order 16.) 6. Prove that no two of the groups G1 , . . . , G9 are isomorphic. 7. Let G be a non-abelian group of order p4 . (a) Prove that | Z(G)| p or p2 , and that if | Z(G)| p2 then G has p3 p2 ÿ p conjugacy classes. (b) Prove that |G9| p or p2 , and that if |G9| p2 then G has 2 p2 ÿ 1 conjugacy classes. (c) Deduce that |G9 \ Z(G)| p. 8. (a) Prove that if G is any group, then G= Z(G) 6 Q8 . (Hint: assume that G= Z haZ, bZ: a4 2 Z, a2 b2 mod Z, bÿ1 ab aÿ1 mod Zi. Prove that a2 commutes with b, and hence that a2 2 Z.) (b) Deduce from the result of Exercise 7 that if G is a group of order 16, then G=(G9 \ Z(G)) 6 Q8 .

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. We discussed brie¯y in Chapter 1 the signi®cance of simple groups in the theory of ®nite groups. Examples of simple groups which we have met so far are cyclic groups of prime order, A5 and A6 . In fact the group A5 , of order 60, is the smallest non-abelian simple group. The next smallest is a certain group of order 168, and in this chapter we shall describe this group and ®nd its character table. The group belongs to a whole family of simple groups, and we begin with a description of this family.

Special linear groups Let p be a prime number, and recall that Z p is the ®eld which consists of the numbers 0, . . . , p ÿ 1, with addition and multiplication modulo p. Denote by SL (2, p) the set of all 2 3 2 matrices M with entries in Z p such that det M 1. Then SL (2, p) is a group under matrix multiplication, and is called the 2-dimensional special linear group over Z p. To calculate the order of the group SL (2, p), we count the matrices a b (a, b, c, d 2 Z p , ad ÿ bc 1): c d If c 0, then there are p( p ÿ 1) choices for a, b, d which make ad ÿ bc 1 (since a, b are arbitrary, except that a 6 0; and d is determined by a). And there are p2 ( p ÿ 1) choices for a, b, c, d 311

312


with c 6 0, such that ad ÿ bc 1 (since a, d may be chosen arbitrarily, c is any non-zero element of Z p ; and then b is determined). Therefore jSL (2, p)j p( p ÿ 1) p2 ( p ÿ 1) p( p2 ÿ 1): If p 2 then SL (2, p) has order 6, and it is easy to see that this group is isomorphic to S3 ; so assume that p is an odd prime. By Exercise 27.1, the centre of SL (2, p) is Z fI, ÿIg (where I is the 2 3 2 identity matrix). The factor group SL (2, p)= Z is called the 2-dimensional projective special linear group, and is written as PSL (2, p). Thus PSL (2, p) SL (2, p)=fIg: Since |SL (2, p)| p( p2 ÿ 1), we have jPSL(2, p)j p( p2 ÿ 1)=2: It is known that PSL (2, 3) A4 , PSL (2, 5) A5 , and that for p > 5, the group PSL (2, p) is simple (see Theorem 8.19 of the book by J. J. Rotman listed in the Bibliography). The simple group G PSL (2, 7) has order 168, and we shall construct the character table of this group. After ®nding the conjugacy classes of G, we shall ®nd the character table using only numerical calculations, notably the orthogonality relations (Theorem 16.4) and congruence properties (Corollary 22.26). The power of these techniques is therefore well illustrated. In the exercises, we indicate other ways of obtaining characters of G, using information about subgroups.

The conjugacy classes of PSL (2, 7) 27.1 Lemma The group PSL (2, 7) has exactly six conjugacy classes. The following table records representatives gi (1 < i < 6) for the conjugacy classes, together with the order of gi , the order of CG ( gi ), and the size of the conjugacy class containing gi .

Character table of the simple group of order 168

g1 g2 g3 g4 g5 g6

1 0

0 1

2 0 1 0 1 0

|CG ( gi )|

| gG i |

1

168

1

2

8

21

4

4

42

3

3

56

7

7

24

7

7

24

Z

0 1 ÿ1 0 2 2

Order of gi

313

Z

ÿ2 Z 2 0 Z 4 1 Z 1 ÿ1 Z 1

Proof For each i, we verify that gi has the stated order, and then use direct calculation to ®nd all the elements of G which commute with g i . Consider, for example, g4 . Suppose that a b Z c d commutes with g4 . Then a b 2 0 2 c d 0 4 0 and hence b c 0. Consequently 1 0 2 Z, CG ( g 4 ) 0 1 0 Similarly CG ( g2 )

0 4

0 4

a c

Z,

b , d

4 0 0 2

Z :

3 ÿ2 3 2 0 ÿ1 1 0 , , , , MZ: M ÿ2 4 2 4 1 0 0 1 2 3 2 4 2 2 2 ÿ2 , , , : 3 ÿ2 4 ÿ2 ÿ2 2 2 2

Also, CG ( gi ) k gi l for i 3, 5, 6. Among g1 , . . . , g6 , the only elements with the same order are g5

314


and g6 ; so no two of these six elements are conjugate, except possibly g5 and g6 . Suppose that gÿ1 g6 g g5 with a b g Z 2 G: c d Then gg5 g6 g, and so a ab aÿc c cd c

bÿd d

with ad ÿ bc 1:

It follows that c 0, a 6 0, d aÿ1 and a ab a b ÿ aÿ1 : 0 aÿ1 0 aÿ1 Therefore a2 ÿ1, which is impossible for a 2 Z7 . Thus g5 is not conjugate to g6 , and we have established that no two of the elements g1 , . . . , g6 are conjugate. The size of the conjugacy class g G i is obtained by dividing 168 by |CG ( gi )| (Theorem 12.8). Since the sum of the sizes of the six conjugacy classes g G i (1 < i < 6) is 168, these exhaust the conjugacy classes of G. j Notice that using Lemma 27.1, it is easy to check that G is indeed simple, since any normal subgroup is a union of conjugacy classes (see Proposition 12.19). 27.2 Corollary (1) If 1 < i < 4 and ÷ is a character of G, then ÷( gi ) is an integer. (2) For some character ÷ of G, ÷( g5 ) is non-real. Proof (1) By Lemma 27.1, for 1 < i < 4, gi is conjugate to ( gi ) k whenever gi and ( gi ) k have the same order. Hence the conclusion follows from Theorem 22.16. (2) Notice that g6 gÿ1 5 , so g5 is not conjugate to its inverse. Therefore (2) follows from Corollary 15.6. j The character table of G

PSL (2, 7)

Since G has six conjugacy classes, it also has six irreducible characters. Let ÷1 , . . . , ÷6 be the irreducible characters of G, where ÷1 is the


315

trivial character (so that ÷1 ( g) 1 for all g 2 G). Recall that the character table is the 6 3 6 matrix with ij-entry ÷ i ( gj ). We shall repeatedly exploit the column orthogonality relations, Theorem 16.4(2), and the congruence properties given by Corollaries 22.26 and 22.27 for the elements g2 , g3 , g4 , for which the character values are known to be integers, by Corollary 27.2. The entries in the column of g4 are integers, and the sum of the squares of these integers is equal to |CG ( g4 )| 3. The entries must therefore be 1, 1, 1, 0, 0, 0 in some order. (We know that the entry in the ®rst row is ÷1 ( g4 ) 1.) Similarly the entries in the column of g3 are 1, 1, 1, 1, 0, 0 in some order, and the entries in column g2 are 1, 1, 1, 1, 2, 0 in some order. Now for all characters ÷ of G, we have by Corollary 22.27, ÷( g 2 ) ÷(1) mod 2, and ÷( g 3 ) ÷(1) mod 2, and so ÷( g 2 ) ÷( g 3 ) mod 2: Since we also know that 6 X

÷ i ( g 3 )÷ i ( g 4 ) 0,

i1

we see that, with a suitable ordering of ÷2 , . . . , ÷6 , part of the character table of G is as follows: Class representative Centralizer order ÷1 ÷2 ÷3 ÷4 ÷5 ÷6

g2 8

g3 4

g4 3

1 1 0 1 1 2

1 1 0 1 1 0

1 1 1 0 0 0

We shall determine the signs later. For the moment we concentrate on the entries in the ®rst column of the character table (i.e. the degrees ÷ i (1)). Let di ÷ i (1), so di is the entry on row i of column 1. By

316

Representations and characters of groups P6 2 Corollary 22.27, Theorem 22.11 and the fact that i1 d i 168, we have d 4 0 mod 3, d 4 1 mod 2, d 4 divides jGj 168,

and

d 24 < 168: The only positive integer d4 which satis®es these conditions is d4 3. In the same way, d5 3. Next, d 6 0 mod 3, d 6 0 mod 2, d 6 divides 168,

and

d 26 < 168: Therefore d6 is 6 or 12. But 0

6 X

÷ i ( g 2 )d i 1 d 2 3 3 2d 6 ,

i1

so as d 22 < 168, we have d6 6 12, and hence d6 6. Now 1 d 22 d 23 32 32 62 168, so d 22 d 23 113. The only solutions to this equation with d2, d3 positive integers have d2, d3 equal to 7, 8 in some order. Since d 2 1 mod 2, we have d2 7 and d3 8. We have now found the ®rst column of the character table, and have the following portion: Class representative Centralizer order ÷1 ÷2 ÷3 ÷4 ÷5 ÷6

g1 168

g2 8

g3 4

g4 3

1 7 8 3 3 6

1 1 0 1 1 2

1 1 0 1 1 0

1 1 1 0 0 0


317

The equations 6 X

÷ i ( g 1 )÷ i ( g j ) 0

for j 2, 3, 4

i1

now enable us to determine the signs in the columns for g2 , g3 , g4 . We obtain Class representative Centralizer order

g1 168

g2 8

g3 4

g4 3

1 7 8 3 3 6

1 ÿ1 0 ÿ1 ÿ1 2

1 ÿ1 0 1 1 0

1 1 ÿ1 0 0 0

÷1 ÷2 ÷3 ÷4 ÷5 ÷6

Next, the equation 1 h÷2 , ÷2 i

6 X ÷2 ( g i )÷2 ( g i ) i1

jCG ( g i )j

7:7 1 1 1 ÷2 ( g 5 )÷2 ( g 5 ) ÷2 ( g 6 )÷2 ( g 6 ) 168 8 4 3 7 7

gives ÷2 ( g5 ) ÷2 ( g6 ) 0. (Note that ÷2 ( g5 ) ÷2 (1) mod 7, but we could not use this fact as we were not sure that ÷2 ( g5 ) was an integer.) Also, for j 5, 6, 0

6 X

÷ i ( g 4 )÷ i ( g j ) 1 ÿ ÷3 ( g j )

i1

and so ÷3 ( g5 ) ÷3 ( g6 ) 1. By Corollary 27.2, there is an irreducible character ÷ of G such that ÷( g5 ) is non-real. For this character ÷, the complex conjugate ÷ will be a different character of the same degree. Hence ÷4 and ÷5 (being the only two irreducible characters with the same degree) must be complex conjugates of each other. Let ÷4 ( g5 ) ÷5 ( g 5 ) z, and let ÷6 ( g5 ) t. Thus the column for g5 is

318

Representations and characters of groups Class representative Centralizer order

g5 7 1 0 1 z z t

÷1 ÷2 ÷3 ÷4 ÷5 ÷6

Now 0

6 X

÷ i ( g 2 )÷ i ( g 5 ) 1 ÿ z ÿ z 2t,

i1

0

6 X

÷ i ( g 3 )÷ i ( g 5 ) 1 z z,

i1

7

6 X

÷ i ( g 5 )÷ i ( g 5 ) 2 2zz tt:

i1

Solving these equations, we obtain

p t ÿ1, z (ÿ1 i 7)=2:

Since g6 gÿ1 5 , we have ÷( g6 ) ÷( g 5 ) for all characters ÷ of G. We have now completely determined the character table of G PSL (2, 7), as shown. Character table of PSL (2, 7) Class representative Centralizer order ÷1 ÷2 ÷3 ÷4 ÷5 ÷6

g1 168

g2 8

g3 4

g4 3

g5 7

g6 7

1 7 8 3 3 6

1 ÿ1 0 ÿ1 ÿ1 2

1 ÿ1 0 1 1 0

1 1 ÿ1 0 0 0

1 0 1 á á ÿ1

1 0 1 á á ÿ1

p where á (ÿ1 i 7)=2.

It is known that there are precisely ®ve non-abelian simple groups of order less than 1000. We give you the character tables of all of these,


319

as follows: G

Order of G

Reference for character table

A5 PSL (2, 7) A6 PSL (2, 8) PSL (2, 11)

60 168 360 504 660

Example 20.13 This chapter Exercise 20.2 Exercise 28.3 Exercise 27.6

1. SL (2, p)

a c

Summary of Chapter 27 b : a, b, c, d 2 Z p , ad ÿ bc 1 . d

jSL (2, p)j p( p2 ÿ 1). 2. PSL (2, p) SL (2, p)=fIg. jPSL (2, p)j p( p2 ÿ 1)=2

( p odd).

3. We constructed the character table of PSL (2, 7), the simple group of order 168. Exercises for Chapter 27 1. Prove that Z(SL (2, p)) fIg. 2. Find the character table of SL (2, 3). 3. Deduce directly from the character table of PSL (2, 7) that this group is simple. 4. In this exercise we present an alternative construction of the character table of G PSL (2, 7), given the conjugacy classes of G, as in Lemma 27.1. (a) De®ne the subgroup T of G, of order 21, as follows: a b T Z : a 2 Z 7 , b 2 Z7 0 aÿ1 (where Z {I}). Calculate the values of the induced character (1 T ) " G, and show that (1 T ) " G 1 G ÷, where ÷ is an irreducible character of G.

320


(b) Let ë be a non-trivial linear character of T. Calculate the values of ë " G and prove that this is an irreducible character of G. (c) By considering ÷ S (see Proposition 19.14), obtain an irreducible character of G of degree 6. (d) From (a), (b), (c), we now have irreducible characters of G of degrees 1, 7, 8 and 6. Use orthogonality relations to complete the character table of G. 5. The character table of SL (2, 7). Let G SL (2, 7), the group of all 2 3 2 matrices of determinant 1, with entries in the ®eld Z7 . (a) Show that G has 11 conjugacy classes with representatives gi as follows: gi g1

1 0

0 1

|CG ( gi )|

| gG i |

1

336

1

2

336

1

4

8

42

8

8

42

8

8

42

3

6

56

6

6

56

7

14

24

14

14

24

7

14

24

14

14

24

ÿ1 0 0 ÿ1 0 1 g3 ÿ1 0 2 ÿ2 g4 2 2 ÿ2 2 g5 ÿ2 ÿ2 2 0 g6 0 4 ÿ2 0 g7 0 ÿ4 1 1 g8 0 1 ÿ1 ÿ1 g9 0 ÿ1 1 ÿ1 g 10 0 1 ÿ1 1 g 11 0 ÿ1 g2

Order of gi


321

(b) Use the character table of PSL (2,7) to write down the six irreducible characters of G with kernel containing Z {I}. (c) Let ÷7 , ÷8 , ÷9 , ÷10 , ÷11 be the remaining irreducible characters of G. Show that for any j with 7 < j < 11 and any g 2 G, we have ÷ j ( g) ÿ÷ j (ÿ g). (d) Prove that ÷ j ( g3 ) 0 for 7 < j < 11, and deduce that ÷ j (1) is even. (e) By considering the column of g6 in the character table, and congruences modulo 3, show that the degrees of ÷7 , . . . , ÷11 are 4, 4, 6, 6, 8, and ®nd ÷ j ( g6 ) for 7 < j < 11. (f) Let ø be one of the irreducible characters of degree 4. By considering the values of ø A on g1 , g2 , g3 and g6 (see Proposition 19.14), prove that ø A is equal to the irreducible character of G of degree 6 whose kernel contains Z. Deduce the values of the irreducible characters of degree 4 on all gi . (g) Complete the character table of G. 6. The character table of PSL (2, 11). Let G PSL (2, 11). This group has eight conjugacy classes with representatives g1 , . . . , 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

ÿ1 ÿ1 ÿ1 Also, gÿ1 are conjugate to g5 , g6 , g8 , g7 , respec5 , g6 , g7 , g8 tively. Find the character table of G.

28 Character table of GL(2, q)

We are now going to calculate the character tables of an important in®nite series of groups, 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. In the last chapter and its exercises, we found the character tables of certain groups of 2 3 2 matrices with entries in Z7 and Z11 . We shall determine the character tables of some matrix groups with entries from an arbitrary ®nite ®eld. At ®rst sight, this is a daunting task, since the number of irreducible characters increases with the size of the ®eld. However, we shall see that the conjugacy classes of our groups fall into four families, as do the irreducible characters. Consequently, we can display the character values in a 4 3 4 matrix. The ®elds F q and F q 2 We consider ®nite ®elds, and we shall tell you which properties of these ®elds we will use; if you are unfamiliar with ®nite ®elds then you might like to consult the book by J. B. Fraleigh listed in the Bibliography. Recall that a ®eld (F, , 3) is a set F with two binary operations and 3 such that the following properties hold. First, (F, ) is an abelian group, with identity element 0. Secondly, if we write F Fnf0g, then (F , 3) is an abelian group, with identity element 1. Finally, the distributive law holds; that is (a b)c ac bc for all a, b, c 2 F. For example, R, C and Z p ( p prime) are ®elds, with the usual de®nitions of and 3. The basic properties of ®nite ®elds which we will use without proof are these: 322

Character table of GL(2, q)

323

(28.1) Let p be a prime and n be a positive integer, and write q pn . Then there exists a ®eld F q of order q and every ®eld of order q is isomorphic to F q . For every s 2 F q the sum of s with itself p times is zero; in short, ps 0. The group (Fq , 3) is cyclic. Notice that the binomial coef®cients ( ip ) with 1 < i < p ÿ 1 are all divisible by p; it follows that (s t) p s p t p for all s, t 2 F q , and k k k hence (s t) p s p t p for all positive integers k. We use this remark in the proof of the next proposition. 28.2 Proposition Let F F q 2 and S fs 2 F : s q sg. (1) The set S is a sub®eld of F of order q, and hence S F q . (2) If r 2 F then r r q , r 1q 2 S. Proof (1) Suppose that s, t 2 S. Then (s t) q s q t q s t, so s t 2 S. It is now easy to check that (S, ) and (Snf0g, 3) are abelian groups, so S is a ®eld. 2 (2) Since Fq 2 is a group of order q 2 ÿ 1, we see that r q r for all 2 r 2 F. This implies that (r r q ) q r q r q r r q and (r 1q ) q j r 1q , so r r q , r 1q 2 S. Hereafter, we shall identify the sub®eld S of F q 2 in Proposition 28.2 with the ®eld F q . We introduce the following useful notation. (28.3) Let å be a generator of the cyclic group Fq 2 and let 2 ù e(2ði=(q ÿ1)) . Suppose that r 2 Fq 2 . We may write r å m for some m and we let r ù m . Then r ! r is an irreducible character of Fq 2 . Moreover, every irreducible character of Fq 2 has the form r ! r j for some integer j. You are now in a position to appreciate the statement of the main result of this chapter, namely Theorem 28.5.

324

Representations and characters of groups The conjugacy classes of GL(2, q)

The general linear group GL(2, q) is de®ned to be the group of invertible 2 3 2 matrices with entries in F q . The subgroup consisting of all matrices of determinant 1 is the special linear group SL(2, q), and we talked about some special linear groups in the last chapter. Here, we are going to calculate the character table of GL(2, q). Let G GL(2, q), and remember that the matrix a b c d belongs to G if and only if its rows are linearly independent. The number of such matrices is found by noting that (a, b) can be any non-zero row vector, giving us q 2 ÿ 1 choices; and once (a, b) has been chosen, (c, d) can be any row vector which is not a multiple of (a, b), giving us q 2 ÿ q choices. Therefore, jGj (q 2 ÿ 1)(q 2 ÿ q) q(q ÿ 1)2 (q 1): There are four families of conjugacy classes of G, of which three are easy to describe. First, a b 0 c can be conjugate to

a9 0

b9 c9

only if fa, cg fa9, c9g, since conjugate matrices have the same eigenvalues. Keep this in mind during the following discussion. The matrices s 0 sI (s 2 Fq ) 0 s belong to the centre of G. They size 1. Next, consider the matrices s us 0 Let

give us q ÿ 1 conjugacy classes of

1 s

(s 2 Fq ):

Character table of GL(2, q) a b 2 G: g c d Then

gus

as cs

a bs c ds

and us g

as cs

325

d bs ds

so g belongs to the centralizer of us if and only if c 0 and a d. Thus, the matrices us (s 2 Fq ) give us q ÿ 1 conjugacy classes; the centralizer order is (q ÿ 1)q, so, by Theorem 12.8, each conjugacy class contains q 2 ÿ 1 elements. Now, let s 0 d s, t 2 G (s, t 2 Fq ) 0 t and note that

0 1 1 0

ÿ1

d s, t

0 1 1 0

d t,s :

On the other hand, if s 6 t, then we have that gd s, t d s, t g if and only if b c 0. Thus, the matrices d s, t (s, t 2 Fq , s 6 t) give us (q ÿ 1)(q ÿ 2)=2 conjugacy classes; the centralizer order is (q ÿ 1)2 , so each conjugacy class contains q(q 1) elements. Finally, consider 0 1 vr (r 2 F q 2 nF q ): ÿr 1q r r q By Proposition 28.2, v r 2 G. The characteristic polynomial of v r is det(xI ÿ v r ) x(x ÿ (r r q )) r 1q (x ÿ r)(x ÿ r q ), so v r has eigenvalues r and r q. Since r 2 = F q we see that v r lies in none of the conjugacy classes we have constructed so far. Now, ! ÿbr 1q a b(r r q ) gv r and ÿdr 1q c d(r r q ) ! c d vr g : ÿar 1q c(r r q ) ÿbr 1q d(r r q )

326


Hence gv r v r g only if c ÿbr 1q and d a b(r r q ). If these conditions hold, then ad ÿ bc a2 ab(r r q ) b2 r 1q (a br)(a br q ): Since (a, b) 6 (0, 0) and r, r q 2 = F q, we see that a br and a br q are non-zero. Therefore, g 2 CG (v r ) if and only if a b g : ÿbr 1q a b(r r q ) Thus, jCG (v r )j q 2 ÿ 1, and the conjugacy class containing v r has size q 2 ÿ q. The matrix v t has eigenvalues t and t q, so it is not conjugate to v r unless t r or t r q. We therefore partition F q 2 nF q into subsets fr, r q g; each subset gives us a conjugacy class representative v r and different subsets give us representatives of different conjugacy classes. We have now found all the conjugacy classes of G. 28.4 Proposition There are q 2 ÿ 1 conjugacy classes in GL(2, q), described as follows. Class rep. g sI |CG ( g)| (q 2 ÿ 1)(q 2 ÿ q) No. of classes qÿ1

us (q ÿ 1)q qÿ1

d s, t (q ÿ 1)2 (q ÿ 1)(q ÿ 2)=2

vr q2 ÿ 1 (q 2 ÿ q)=2

The families of conjugacy class representatives sI and us are indexed by elements s of Fq . The family of conjugacy class representatives d s, t is indexed by unordered pairs fs, tg of distinct elements of Fq . The family of conjugacy class representatives v r is indexed by unordered pairs fr, r q g of elements of Fq 2 nFq . Proof The conjugacy classes we have found account for (q ÿ 1) (q ÿ 1)(q 2 ÿ 1) (q ÿ 1)(q ÿ 2)q(q 1)=2 (q 2 ÿ q)(q 2 ÿ q)=2 elements altogether. But this sum is equal to the order of GL(2, q), so we have found all the conjugacy classes. j


327

The characters of GL(2,q) We are now in a position to describe the character table of G. 28.5 Theorem Label the conjugacy classes of GL(2, q) as in Proposition 28.4, and let r ! r be the function from Fq 2 to C described in (28.3). Then the irreducible characters of GL(2, q) are given by ë i , ø i , ø i, j , ÷ i as follows.

ëi øi ø i, j ÷i

sI

us

d s, t

vr

s 2i qs 2i (q 1)s i j (q ÿ 1)s i

s 2i 0 s i j ÿs i

(st) i (st) i si t j s j t i 0

r i(1q) ÿr i(1q) 0 ÿ(r i r iq )

Here, we have the following restrictions on the subscripts. (a) For ë i we have 0 < i < q ÿ 2. Thus, there are q ÿ 1 characters ë i , each of degree 1. (b) For ø i we have 0 < i < q ÿ 2. Thus, there are q ÿ 1 characters ø i , each of degree q. (c) For ø i, j we have 0 < i , j < q ÿ 2. Thus, there are (q ÿ 1)(q ÿ 2)=2 characters ø i, j , each of degree q 1. (d) For ÷ i, we ®rst consider the set of integers j with 0 < j < q 2 ÿ 1 and (q 1) 6 j j; if j1 and j2 belong to this set and j1 j2 q mod (q 2 ÿ 1) then we choose precisely one of j1 and j2 to belong to the indexing set for the characters ÷ i . Hence, there are (q 2 ÿ q)=2 characters ÷ i , each of degree q ÿ 1. Before we embark upon the task of calculating the irreducible characters of G, we present a proposition which will be useful later. Recall that å is our chosen generator for Fq 2 and 0, 1 : vå å åq ÿå 1q 28.6 Proposition Let K hvå i. Then jKj q 2 ÿ 1. The group K contains the q ÿ 1 scalar matrices sI in G, and of the remaining q 2 ÿ q elements of K,

328


precisely two belong to each of the conjugacy classes represented by v r with r 2 Fq 2 nFq . Proof The eigenvalues of vå are å and å q, so vå has order q 2 ÿ 1. The eigenvalues of våi and of våiq are å i and å iq . If å i 6 å iq then åi 2 = F q and våi and våiq must be conjugate to vå i . Hence two elements of K belong to the conjugacy class of vå i . If å i å iq then våi å i I, since våi has eigenvalues in F q and våi is diagonalizable, and this case accounts for the q ÿ 1 scalar matrices. j We shall construct, in turn, the irreducible characters ë i , ø i , ø i, j and ÷ i which appear in Theorem 28.5. 28.7 Proposition There are q ÿ 1 linear characters ë i of G, and they are given in Theorem 28.5. Proof The map det : g ! det g is a homomorphism from G onto Fq . As i varies between 0 and q ÿ 2 inclusive, the functions ë i : g ! (det g) i

( g 2 G)

give q ÿ 1 distinct linear characters of G, whose values appear in Theorem 28.5. j We will see later that the linear characters ë i (0 < i < q ÿ 2) which appear in Proposition 28.7 are all the linear characters of G. 28.8 Proposition For all integers i, j there is a character ø i, j of G whose values on the conjugacy class representatives, as described in Proposition 28.4, are as follows.

ø i, j

sI

us

d s, t

vr

(q 1)s i j

s i j

si t j s j t i

0

Character table of GL(2, q) Proof Let

B

a 0

b c

329

2G :

Then B is a subgroup of G with jBj (q ÿ 1)2 q. De®ne ë i, j : B ! C by s r ë i, j : ! s i t j: 0 t Then ë i, j is a character of B. We let ø i, j ëi, j " G. We use Proposition 21.23 to calculate ø i, j ( g) for each conjugacy class representative g, as follows. jC G ( g)j ë i, j ( g) jC B ( g)j

g sI :

ø i, j ( g)

g us :

ø i, j ( g)

g d s, t : g vr :

jC G ( g)j ë i, j ( g) jC B ( g)j ë i, j ( g) ë i, j ( g9) ø i, j ( g) jC G ( g)j , where g9 d t,s jC B ( g)j jC B ( g9)j

ø i, j ( g) 0:

Hence, the values of ø i, j are as stated in the proposition.

j

28.9 Proposition For each integer i, there is an irreducible character ø i of G whose values are given in Theorem 28.5. The characters ø i for 0 < i < q ÿ 2 are all different. Proof We shall demonstrate that the character ø i,i which appears in Proposition 28.8 gives us ø i,i ë i ø i . To this end, we calculate hø i,i , ø i,i i and hø i,i , ë i i. Remember that the complex conjugate of s i is s ÿi . We have hø i,i , ø i,i i

(q 1)2 1 (q ÿ 1) (q ÿ 1) (q ÿ 1)q (q 2 ÿ 1)(q 2 ÿ q)

4 (q ÿ 1)(q ÿ 2) 2 (q ÿ 1) 2

2: Here, the ®rst term corresponds to the conjugacy classes of elements

330


sI, and the calculation of this ®rst term involves the following three observations. (1) ø i,i (sI)ø i,i (sI) (q 1)2 . (2) jC G (sI)j (q 2 ÿ 1)(q 2 ÿ q): (3) There are q ÿ 1 conjugacy classes with representatives of the forms sI. The remaining terms in hø i,i , ø i,i i are calculated in a similar fashion. Next, hø i,i , ë i i

(q 2

(q 1) 1 (q ÿ 1) (q ÿ 1) 2 (q ÿ 1)q ÿ 1)(q ÿ q)

2 (q ÿ 1)(q ÿ 2) (q ÿ 1)2 2

1: The facts that hø i,i , ë i i 1 and hø i,i , ø i,i i 2 imply that ø i,i ë i ø i for some irreducible character ø i. Subtract ë i from ø i,i to get the values of ø i as given in Theorem 28.5. Let s be an element of Fq of order q ÿ 1. Then ø i : d s,1 ! s i . Hence the characters ø i for 0 < i < q ÿ 2 are all different. j 28.10 Proposition Suppose that 0 < i , j < q ÿ 2. Then the character ø i, j which appears in Proposition 28.8 is irreducible. Proof We shall show that hø i, j , ø i, j i 1. Using the values of ø i, j which are given in Proposition 28.8, we obtain hø i, j , ø i, j i A B C, where (q 1)2 1 (q ÿ 1) (q ÿ 1), B (q ÿ 1)q (q 2 ÿ 1)(q 2 ÿ q) 1 1X i j C (s t s j t i )(s ÿi t ÿ j s ÿ j t ÿi ): 2 (q ÿ 1) 2 s6 t A

and

The coef®cent 12 appears in C because we have just one conjugacy class for each unordered pair fs, tg of distinct elements of Fq . To evaluate C, note that fd s, t : s, t 2 Fq g is an abelian group of order (q ÿ 1)2, and if ó : d s, t ! s i t j s j t i then ó is a sum of two


331

inequivalent irreducible characters of this group. Thus, hó , ó i 2: That is, ! X 1 i j j i ÿi ÿ j ÿ j ÿi 4(q ÿ 1) (s t s t )(s t s t ) 2: (q ÿ 1)2 s6 t Hence, C

qÿ3 : qÿ1

And now we ®nd that A B C 1. Therefore, hø i, j , ø i, j i 1, and ø i, j is irreducible. j 28.11 Corollary The characters ø i, j characters of G.

for 0 < i , j < q ÿ 2 are distinct irreducible

Proof Suppose that 0 < i , j < q ÿ 2 and 0 < i9 , j9 < q ÿ 2, and (i, j) 6 (i9, j9). We must prove that ø i, j 6 ø i9, j9 . Consider the group B and its linear characters ë i, j which were used in the proof of Proposition 28.8. We have s b ë i, j ë j,i : ! s i t j s j t i: 0 t Since ë i, j ë j,i 6 ë i9, j9 ë j9,i9 , there exists complex (q ÿ 1)th roots of unity s and t such that either s 6 t and s i t j s j t i 6 s i9 t j9 s j9 t i9 or s t and s i j 6 s i9 j9 . In either case, we see that ø i, j differs from ø i9, j9 on a conjugacy class j of G. Therefore, ø i, j 6 ø i9, j9 . 28.12 Proposition For each integer i, there exists a character ö i of G which takes the following values.

öi

sI

us

d s, t

vr

q(q ÿ 1)s i

0

0

r i r iq

332


Proof Let K hvå i, as in Proposition 28.6, and consider the linear character á i of K which sends the generator vå of K to å i . Suppose that g 2 K and g is conjugate in G to v r . Then g has eigenvalues r and r q . Hence á i ( g) r i or r iq and á i ( g) á i ( g q ) r i r iq : Let ö i á i " G. In order to calculate ö i , ®rst recall that á i " G is zero on all elements which are not conjugate to an element of K. Thus, by Proposition 28.6, ö i is zero on the elements of the form u s and d s, t (s 6 t). If g sI with s 2 Fq then g 2 K and ö i ( g)

jC G ( g)j á i ( g) q(q ÿ 1)s i : jC K ( g)j

Suppose that r 2 F q 2 nF q . Then, by Proposition 28.6, v r is conjugate to an element of g of K. Also, á i ( g) ái( g q) ö i ( g) jC G ( g)j jC K ( g)j jC K ( g)j á i ( g) á i ( g q ) r i r iq : Thus, ö i has the values stated in the proposition.

j

To be able to work out certain inner products involving our characters ö i , we shall the use the following lemma. 28.13 Lemma Assume that i is an integer and (q 1) 6 j i. Then X (r i r iq )(r ÿi r ÿiq ) 2(q ÿ 1)2 : r2F q 2 nF q

Proof Note that r G1 0

0 rq

: r 2 Fq 2

and G2

r 0

0 rq

: r 2 Fq

are abelian groups of orders q 2 ÿ 1 and q ÿ 1, respectively. Now,

Character table of GL(2, q) r 0 ! r i r iq 0 rq

333

gives a character ÷ of degree 2 for each group. For G1, the character ÷ is a sum of two inequivalent irreducible characters, since (q 1) 6 j i implies that å i 6 å iq ; and for G2, the character ÷ is twice an irreducible character, since r q r for r 2 Fq . Taking the inner product of the character ÷ of G1 with itself, we get 1 X i (r r iq )(r ÿi r ÿiq ) 2 q2 ÿ 1 r2F 2 q

and doing the same for the character ÷ of G2 , we get 1 X i (r r iq )(r ÿi r ÿiq ) 4: qÿ1 r2F q Hence

X

(r i r iq )(r ÿi r ÿiq ) 2(q 2 ÿ 1) ÿ 4(q ÿ 1)

r2F q 2 nF q

2(q ÿ 1)2 :

j

28.14 Proposition For each integer i, let ÷ i be the class function on G with the following values.

÷i

sI

us

d s, t

vr

(q ÿ 1)s i

ÿs i

0

ÿ(r i r iq )

If (q 1) 6 j i then ÷ i is an irreducible character of G. Proof We can justify the manoeuvre which we now perform only by saying that it gives the correct answer. Recall the characters ø i, j , ø i and ö i given in Propositions 28.8, 28.9 and 28.12. Now, ÷ i is the class function on G which is given by ÷ i ø0,ÿi ø i ÿ ø0,i ÿ ö i :

334


The table below allows us to verify this.

ø0,ÿi øi ø0,ÿi ø i ø0,i öi ÷i

sI

us

d s, t

vr

(q 1)s ÿi qs 2i q(q 1)s i (q 1)s i q(q ÿ 1)s i (q ÿ 1)s i

s ÿi 0 0 si 0 ÿs ÿi

s ÿi t ÿi (st) i si t i si t i 0 0

0 ÿr i(1q) 0 0 r i r iq ÿ(r i r iq )

Next, assume that (q 1) 6 j i. We work out h÷ i , ÷ i i using Lemma 28.13. h÷ i , ÷ i i

(q ÿ 1)2 1 (q ÿ 1)2 (q ÿ 1) (q ÿ 1) 1: q2 ÿ q (q 2 ÿ 1)(q 2 ÿ q) q2 ÿ 1

Since ÷ i is a linear combination of irreducible characters of G, with integer coef®cients, and h÷ i , ÷ i i 1 and ÷ i (1) . 0, it follows that ÷ i is an irreducible character of G. j 28.15 Proposition Suppose that i and j are integers with (q 1) 6 j i and (q 1) 6 j j and j 6 i, iq mod(q 2 ÿ 1). Then the characters ÷ i and ÷ j of G are different. Proof Let K hvå i, as in Proposition 28.6, and consider the linear character á i of K which sends the generator vå of K to å i . Suppose that g 2 K. If g sI where s 2 Fq then (á i á iq )( g) 2s i . If g is conjugate to v r where r 2 F q 2 nF q then (á i á iq )( g) r i r iq . Since j 6 i, iq mod(q 2 ÿ 1), the characters á i á iq and á j á jq of K are different, so either s i 6 s j for some s 2 Fq or r i r iq 6 r j r jq for some r 2 F q 2 nF q . Therefore, ÷ i 6 ÷ j , as we wished to show. j We have now completed the proof of Theorem 28.5, since we have shown that the class functions given in the theorem are inequivalent irreducible characters; and the number of them is q 2 ÿ 1, which is the same as the number of conjugacy classes of G. It is possible to use the character table of GL(2, q) to ®nd the


335

character table of SL(2, q), since most of the irreducible characters remain irreducible when restricted. We do not go fully into this, since the answers are quite complicated, and they depend upon whether q is a power of 2 or q 1 mod 4 or q 3 mod 4. In Exercise 28.2, though, you are asked to consider the easiest case, namely that where q is a power of 2. Since SL(2, q) PSL(2, q) when q is a power of 2, this gives the character tables of an in®nite series of simple groups PSL(2, q). Among the characters of SL(2, q), those with kernel containing the centre of SL(2, q) provide the characters of the groups PSL(2, q) ± compare Chapter 27 ± and so the character table of PSL(2, q) is rather easier to ®nd than that of SL(2, q). A challenging exercise is to determine the character table of PSL(2, q) when q 1 mod 4 or q 3 mod 4 from the character table of GL(2, q). Although the character table of GL(2, q) was ®rst given in 1907, it was not until the 1950's that the character table of GL(3, q) was found. Then, in 1955, J. A. Green determined the character table of GL(n, q) for all positive integers n.

Summary of Chapter 28 The character table of GL(2, q) has the following properties. (a) Thereare q ÿ 1 conjugacy classes with representatives of the form s 0 sI , and there are q ÿ 1 irreducible characters of 0 s degree 1. (b) There are q ÿ 1 conjugacy classes with representatives of the form s 1 us , and there are q ÿ 1 irreducible characters of degree 0 s q. (c) There are (q ÿ 1)(q ÿ 2)=2 conjugacy classes with representatives s 0 of the form d s, t (s 6 t), and there are (q ÿ 1)(q ÿ 2)=2 0 t irreducible characters of degree q 1. (d) There are (q 2 ÿ q)=2 conjugacy classes with representatives of the

336

Representations and characters of groups 0 1 , and there are (q 2 ÿ q)=2 irreducible form v r ÿr 1q r r q characters of degree q ÿ 1. Exercises for Chapter 28

1. Use Theorem 28.5 to write down explicitly the character table of GL(2, 3). 2. Suppose that q is a power of 2. Let Z fsI : s 2 F g. Prove that q

GL(2, q) Z 3 SL(2, q): Deduce the character table of SL(2, q) from that of GL(2, q). Prove that if q 6 2 then SL(2, q) is simple. 3. Use your solution to Exercise 28.2 to write down explicitly the character table of PSL(2, 8).

29 Permutations and characters

We have already seen in Chapter 13 that if G is a permutation group, i.e. a subgroup of Sn for some n, then G has a permutation character ð de®ned by ð( g) jfix( g)j for g 2 G, a fact which proved useful in many of our subsequent character table calculations. In this chapter we take the theory of permutation groups and characters somewhat further, and develop some useful results, particularly about irreducible characters of symmetric groups (see Theorem 29.12 below). Group actions We begin with a more general notion than that of a permutation group. If Ù is a set, denote by Sym(Ù) the group of all permutations of Ù. In particular, if Ù f1, 2, . . . , ng then Sym(Ù) Sn . De®nition Let G be a group and Ù a set. An action of G on Ù is a homomorphism ö: G ! Sym(Ù). We also say that G acts on Ù (via ö). 29.1 Examples (1) If G < Sn then the identity map is an action of G on f1, . . . , ng. (2) Let G Sn and let Ù be the set consisting of all pairs fi, jg of elements of f1, . . . , ng. De®ne ö: G ! Sym(Ù) by setting fi, jg( gö) fig, jgg for all g 2 Sn and 1 < i , j < n. (So for example, (1 2)ö sends f1, 3g ! f2, 3g.) Check that ö is an action of Sn ; it is called the action of Sn on pairs. 337

338


(3) Let G GL(2, q), the group of invertible 2 3 2 matrices over the ®nite ®eld F q, as de®ned in Chapter 28. Let V be the 2-dimensional vector space over F q consisting of all row vectors (a, b) with a, b 2 F q , and let Ù be the set of all 1-dimensional subspaces hvi of V. De®ne ö: G ! Sym(Ù) by setting hvi( gö) hv gi for all hvi 2 Ù and g 2 G. For example, if 1 1 g 0 1 then gö sends h(a, b)i ! h(a, a b)i: Then ö is an action of G on Ù. (4) Let G be a group with a subgroup H of index n, and let Ù be the set of all right cosets Hx of H in G (so jÙj n). De®ne ö: G ! Sym(Ù) by ( Hx)( gö) Hxg for all x, g 2 G. Then by Exercise 9 of Chapter 23, ö is an action of T G, and Ker ö x2G x ÿ1 Hx < H. To simplify notation, if ö: G ! Sym(Ù) is an action, for ù 2 Ù and g 2 G we usually just write ù g instead of ù( gö). With this notation, the fact that ö is a homomorphism simply says that ù( gh) (ù g)h for all ù 2 Ù and g, h 2 G. Adopting this notation, de®ne a relation on Ù as follows: for á, â 2 Ù, we have á â if and only if there exists g 2 G such that á g â. It is easy to see that is an equivalence relation on Ù. The equivalence classes are called the orbits of G on Ù. Thus Ù is the disjoint union of the orbits of G. Write orb(G, Ù) for the number of orbits of G on Ù. The group G is said to be transitive on Ù if orb(G, Ù) 1; in other words, G is transitive if, given any á, â 2 Ù, there exists g 2 G such that á g â. 29.2 Examples (1) Let G C4 , generated by x, say, and let ö: G ! S8 be the action

Permutations and characters

339

de®ned by xö (1 2 3 4)(5 6)(7 8) (and of course x k ö ((1 2 3 4)(5 6)(7 8)) k for any k). Then G has three orbits on Ù f1, . . . , 8g, namely f1, 2, 3, 4g, f5, 6g and f7, 8g. (2) The group G is transitive on the set Ù in each of Examples 29.1(2, 3, 4). This is clear in Example (2); to verify it for Example (3) you need to convince yourself that for any two non-zero row vectors v, w 2 V there is an invertible 2 3 2 matrix A 2 GL(2, q) such that vA w; and in Example (4), simply observe that, given two right cosets Hx, Hy 2 Ù, the element g x ÿ1 y 2 G has the property that ( Hx) g Hy. Let G be a group acting on a set Ù. For ù 2 Ù, write ù G for the orbit of G which contains ù, so ù G fù g : g 2 Gg; and de®ne Gù f g 2 G : ù g ùg: We call Gù the stabilizer of ù in G. 29.3 Proposition The stabilizer Gù is a subgroup of G. Moreover, the size of the orbit ù G is equal to the index of Gù in G; that is, jù G j jG : Gù j: Proof If g, h 2 Gù then ù( gh) (ù g)h ùh ù, hence gh 2 Gù . Also gÿ1 2 Gù , and Gù contains the identity, so Gù is a subgroup. Now let Ä be the set of right cosets Gù x of Gù in G. Observe that for x, y 2 G, Gù x Gù y , xy ÿ1 2 Gù , ùxy ÿ1 ù , ùx ù y: Hence we can de®ne an injective function ã : Ä ! ù G by ã(Gù x) ùx for all x 2 G. Clearly ã is also surjective, and hence jÄj jù G j, as required. j Permutation characters Let G be a group acting on a ®nite set Ù. Denote by CÙ the vector space over C for which Ù is a basis. In other words, CÙ consists of all expressions of the form

340

Representations and characters of groups X ëù ù (ëù 2 C) ù2Ù

with the obvious addition and scalar multiplication. As in Chapter 13, we can make CÙ into a CG-module, called the permutation module, by de®ning ! X X ëù ù g ëù (ù g) for all g 2 G. We see just as in Chapter 13 (p. 129) that if ð is the character of this permutation module, then for g 2 G, ð( g) jfixÙ ( g)j, where fixÙ ( g) fù 2 Ù : ù g ùg. We call ð the permutation character of G on Ù. The next result, though elementary, is rather famous, and provides a basic link between the permutation character and the action of G. It is often referred to as ``Burnside's Lemma'', but is in fact due to Cauchy and Frobenius. 29.4 Proposition Let G be a group acting on a ®nite set Ù, and let ð be the permutation character. Then 1 X hð, 1 G i jfixÙ ( g)j orb(G, Ù): jGj g2G

Proof First note that hð, 1 G i

1 X 1 X ð( g) jfixÙ ( g)j: jGj g2G jGj g2G

Let Ä1 , . . . , Ä t be the orbits of G on Ù, and for each i, pick ù i 2 Ä i . By Proposition 29.3, for 1 < i < t we have jÄ i j jù G i j jG : Gù i j: Hence jÄ i j jGù i j jGj. Now de®ne Ö f(ù, g) : ù 2 Ù, g 2 G, ù g ùg. We calculate jÖj in two different ways. First, for each g, the number of ù 2 Ù such that ù g ù is equal to jfixÙ ( g)j, hence

Permutations and characters X jÖj jfixÙ ( g)j:

341

g2G

Secondly, for each ù, the number of g 2 G such that ù g ù is equal to jGù j, hence jÖj Therefore

P

X

jGù j

ù2Ù g2G

t X i1

jÄ i j jGù i j

t X

jGj tjGj:

1

jfixÙ ( g)j tjGj, and the conclusion follows.

j

29.5 Corollary G is transitive on Ù if and only if hð, 1 G i 1. Now let G be a group, and suppose that G acts on two sets Ù1 and Ù2 , with corresponding permutation characters ð1 and ð2 respectively. Then we can de®ne an action of G on the Cartesian product Ù1 3 Ù2 by setting (ù1 , ù2 ) g (ù1 g, ù2 g) for all ù i 2 Ù i , g 2 G. It is clear that fixÙ1 3Ù2 ( g) fixÙ1 ( g) 3 fixÙ2 ( g) for any g 2 G. Hence if ð is the permutation character of G on Ù1 3 Ù2 , then ð( g) ð1 ( g)ð2 ( g) for all g 2 G. 29.6 Proposition Let G act on Ù1 and Ù2 , with permutation characters ð1 and ð2 respectively. Then hð1 , ð2 i orb(G, Ù1 3 Ù2 ): Proof We have hð1 , ð2 i

1 X 1 X jfixÙ1 ( g)kfixÙ2 ( g)j jfixÙ1 3Ù2 ( g)j, jGj g2G jGj g2G

which is equal to orb(G, Ù1 3 Ù2 ) by Proposition 29.4.

j

In the rest of the chapter we apply Proposition 29.6 in a number of situations, the ®rst being the case where Ù1 Ù2 .

342


Suppose G acts on Ù. Then G also acts on Ù 3 Ù in the way de®ned above, namely (ù1 , ù2 ) g (ù1 g, ù2 g) for all ù1 , ù2 2 Ù, g 2 G. 29.7 De®nition The number of orbits of G on Ù 3 Ù is called the rank of G on Ù, written r(G, Ù). Thus r(G, Ù) orb(G, Ù 3 Ù): The next result is immediate from Proposition 29.6. 29.8 Proposition Let G act on Ù, with permutation character ð. Then r(G, Ù) hð, ði: Now suppose G is transitive on Ù and jÙj . 1. Then Ä f(ù, ù) : ù 2 Ùg is an orbit of G on Ù 3 Ù, and hence certainly r(G, Ù) > 2. The case where equality holds is of particular interest. 29.9 De®nition Let G be transitive on Ù. Then G is said to be 2-transitive on Ù if r(G, Ù) 2. In other words, G is 2-transitive if, for any ordered pairs (á1 , á2 ) and (â1 , â2 ) in Ù 3 Ù, with á1 6 á2 , â1 6 â2 , there exists g 2 G such that á1 g â1 and á2 g â2 . 29.10 Corollary If G is 2-transitive on Ù, with permutation character ð, then ð 1 G ÷, where ÷ is an irreducible character of G. Proof We have hð, 1 G i 1 by Corollary 29.5, and hð, ði 2 by Proposition 29.8. The result follows, using Theorem 14.17. j


343

29.11 Examples (1) The symmetric group Sn is 2-transitive on f1, . . . , ng. Also An is 2-transitive, provided n > 4. Hence these groups have an irreducible character ÷ given by ÷( g) jfix( g)j ÿ 1: We have seen this irreducible character in a number of previous examples (see 18.1, 19.16, 19.17). (2) Consider the action of G GL(2, q) given in Example 29.1(3). Here Ù is the set of all 1-dimensional subspaces of the 2-dimensional vector space V. We claim that G is 2-transitive on Ù. To see this, let (hv1 i, hv2 i) and (hw1 i, hw2 i) be two pairs of distinct 1-spaces in Ù. Then v1 , v2 and w1, w2 are both bases of V. The linear transformation from V to V which sends v1 ! w1 , v2 ! w2 is therefore invertible, giving an element of GL(2, q) which sends hv1 i ! hw1 i, hv2 i ! hw2 i. Hence G is 2-transitive on Ù, as claimed. Since jÙj q 1 (see Exercise 1 at the end of the chapter), the irreducible character ÷ of G given by Corollary 29.10 is the character ø0 of degree q in Theorem 28.5. (3) Consider the action of Sn on pairs de®ned in Example 29.1(2), with n > 4. This action is not 2-transitive, since, for example, there is no element of Sn which sends (f1, 2g, f3, 4g) to (f1, 2g, f2, 3g). In fact it is easy to see that the orbits of G Sn on Ù 3 Ù are Ä, Ä1 and Ä2 , where Ä is as above, and Ä1 f(fi, jg, fk, lg) : jfi, jg \ fk, lgj 1g, Ä2 f(fi, jg, fk, lg) : jfi, jg \ fk, lgj 0g: Thus hð, ði r(G, Ù) 3, and so ð 1 G ÷ ø, where ÷ and ø are irreducible characters of Sn . Some irreducible characters of S n By Theorem 12.15 we know that the conjugacy classes of Sn are in bijective correspondence with the set of all possible cycle-shapes of permutations. Each cycle-shape (including 1-cycles) is a sequence ë (ë1 , . . . , ë s ) of positive integers ë i such that ë1 > ë2 > . . . > ë s and ë1 . . . ë s n, and we call such a sequence a partition of n. By Theorem 15.3, the irreducible characters of Sn are also in bijective correspondence with the partitions ë of n. A key aim is therefore to construct, for each partition ë, an irreducible character ÷ ë

344


of Sn , in a natural way. We shall apply the material of this chapter to carry out this aim in the case where ë (n ÿ k, k), a 2-part partition (see Theorem 29.13 below). The ideas can be developed to carry out the aim in general, but we do not do this; if you want to see this, and much more, on the character theory of Sn , we refer you to the book by G. James listed in the Bibliography. Let G Sn and I f1, 2, . . . , ng. For an integer k < n=2, de®ne I k to be the set consisting of all subsets of I of size k. Just as in Example 29.1(2) we can de®ne an action of G on I k as follows: for any subset A fi1 , . . . , ik g 2 I k and any g 2 G, let Ag fi1 g, . . . , ik gg: Let ð k be the permutation character of G in its action on I k . Observe that n ð k (1) jI k j : k 29.12 Proposition If l < k < n=2, then hð k , ð l i l 1. Proof By Proposition 29.6, hð k , ð l i orb(G, I k 3 I l ). The orbits of G Sn on I k 3 I l are easily seen to be J 0 , J 1 , . . . J l , where for 0 < s < l, J s f(A, B) 2 I k 3 I l : jA \ Bj sg: Hence orb(G, I k 3 I l ) l 1, giving the conclusion.

j

29.13 Theorem Let m n=2 if n is even, and m (n ÿ 1)=2 if n is odd. Then Sn has distinct irreducible characters ÷ ( n) 1 G , ÷ ( nÿ1,1) , ÷ ( nÿ2,2) , . . . , ÷ ( nÿ m, m) such that for all k < m, ð k ÷ ( n) ÷ ( nÿ1,1) . . . ÷ ( nÿ k, k) : In particular, ÷ ( nÿ k, k) ð k ÿ ð kÿ1 . Proof We prove the existence of irreducible characters ÷ ( n) , ÷ ( nÿ1,1) , . . . , ÷ ( nÿ k, k) such that ð k ÷ ( n) ÷ ( nÿ1,1) . . . ÷ ( nÿ k, k) , by induction on k. This holds for k 1 by Corollary 29.10. Now assume the statement holds for all values less than k. Then


345

there exist irreducible characters ÷ ( n) , ÷ ( nÿ1,1) , . . . , ÷ ( nÿ k1, kÿ1) such that ð i ÷ ( n) ÷ ( nÿ1,1) . . . ÷ ( nÿi,i) for all i , k. Now by Proposition 29.12, hð k , 1 G i 1, hð k , ð1 i 2, . . . , hð k , ð kÿ1 i k, hð k , ð k i k 1: It follows that ð k ð kÿ1 ÷ for some irreducible character ÷. Writing ÷ ÷ ( nÿ k, k) , we have ð k ÷ ( n) ÷ ( nÿ1,1) . . . ÷ ( nÿ k, k) , as required. j 29.14 Examples (1) The formula ÷ ( nÿ k, k) ð k ÿ ð kÿ1 makes it easy to calculate the values of the characters ÷ ( nÿ k, k) . For example, the degree is n n ÷ ( nÿ k, k) (1) ð k (1) ÿ ð kÿ1 (1) ÿ : k kÿ1 As another example, suppose n 7 and let us calculate the value of the irreducible character ÷ (5,2) on a 3-cycle: ÷ (5,2) (123) ð2 (123) ÿ ð1 (123) jfix I 2 (123)j ÿ jfix I 1 (123)j 6 ÿ 4 2: (2) In the character table of S6 given in Example 19.17, the irreducible characters ÷1 , ÷3 , ÷7 , ÷9 are equal to ÷ (6) , ÷ (5,1) , ÷ (4,2) , ÷ (3,3) , respectively. Summary of Chapter 29 1. An action of G on Ù is a homomorphism G ! Sym(Ù). The orbits are the equivalence classes in Ù of the relation de®ned by á â , á g â for some g 2 G. The size of the orbit ù G containing ù is jù G j jG : Gù j. 2. If G acts on Ù then CÙ is the permutation module, and the corresponding character of G is ð, where ð( g) jfixÙ ( g)j. The number of orbits is equal to hð, 1 G i. 3. The rank r(G, Ù) is the number of orbits of G on Ù 3 Ù, and r(G, Ù) hð, ði. If G is 2-transitive then r(G, Ù) 2 and ð 1 G ÷ with ÷ irreducible. 4. The irreducible characters of Sn are in bijective correspondence with partitions of n. The irreducible characters ÷ ( nÿ k, k) corresponding to 2-part partitions have values given by Theorem 29.13.

346


1. Let G be a ®nite group, and de®ne a function ö : G 3 G ! Sym(G) by x(( g, h)ö) g ÿ1 xh for all x, g, h 2 G. (a) Show that ö is an action of G 3 G on G, which is transitive. (b) Find the stabilizer (G 3 G)1 of the identity 1 2 G, and ®nd the kernel of ö. (c) Show that the rank of this action r(G 3 G, G) is equal to the number of conjugacy classes of G, and the permutation character ð is ð

X

÷ 3 ÷,

÷

where the sum is over all irreducible characters ÷ of G, and ÷ 3 ÷ is the irreducible character of G 3 G given by Theorem 19.18. 2. Show that if Ù is the set of all 1-dimensional subspaces of a 2dimensional vector space over F q (as in Example 29.1(3)), then jÙj q 1. 3. Let G GL(2, q) and let V F2q as in Example 29.1(2). Let V V ÿ f0g, and de®ne an action ö : G ! Sym(V ) by v( gö) v g for v 2 V , g 2 G. Let ð be the permutation character of G in this action. Decompose ð as a sum of irreducible characters of GL(2, q) (the latter are given by Theorem 28.5). ( Hint: one way to do this is to write down the values of ð on the conjugacy classes of G, and take inner products with the irreducible characters of G given in 28.5.) 4. Let G be a ®nite group, and let H 1 , H 2 be subgroups of G. For i 1, 2 de®ne Ù i to be the set of right cosets of H i in G, so that G acts on Ù i as in Example 29.1(4); let ð i be the permutation character of G in the action on Ù i . Suppose that ð1 ð2 . Prove that if G is abelian, then H 1 H 2 : Give an example to show that this need not be the case in general. 5. Let G be a ®nite group acting transitively on a set Ù of size greater than 1. Prove that G contains an element g such that jfixÙ ( g)j 0. (Such an element is called a ®xed-point-free element of G.)


347

6. Let n be a positive integer, and let Ù be the set of all ordered pairs (i, j) with i, j 2 f1, . . . , ng and i 6 j. Let Sn act on Ù in the obvious way (namely, (i, j) g (ig, jg) for g 2 Sn ), and let the permutation character of Sn in this action be ð( nÿ2,1,1) . By considering inner products as in the proof of Theorem 29.13, prove that ð( nÿ2,1,1) 1 2÷ ( nÿ1,1) ÷ ( nÿ2,2) ÷, where ÷ is an irreducible character. Writing ÷ ÷ ( nÿ2,1,1) , calculate the degree of ÷ ( nÿ2,1,1) , and calculate its value on the elements (12) and (123) of Sn . In the character table of S6 given in Example 19.17, which irreducible character is equal to ÷ (4,1,1) ?

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. The examples which we have come across so far ± ®nding the centre of the group, seeing whether or not the group is simple, and so on ± require little calculation. In this chapter we present some rather deeper applications. The ®rst involves doing arithmetic with character values to determine certain numbers, known as the class algebra constants. These constants carry information about the multiplication in G, and they can be used to investigate the subgroup structure of G, as we shall demonstrate. The second application takes this much further: the Brauer±Fowler Theorem 23.19 motivates the study of simple groups containing an involution with centralizer isomorphic to a given group C. Using a little group theory and a lot of character theory we shall carry out such a study in the case where C D8 , the dihedral group of order 8.

Class algebra constants Let G be a ®nite group and let C1 , . . . , Cl be the distinct conjugacy classes of G. Recall from Proposition 12.22 that the class sums C1 , . . . , C l form a basis for the centre of the group algebra CG P (where C i g2C i g).

30.1 Proposition There exist non-negative integers aijk such that for 1 < i < l and 1 < j < l, 348

Applications to group theory Ci C j

l X

349

aijk C k :

k1

Proof For g 2 Ck the coef®cient of g in the product C i C j is equal to the number of pairs (a, b) with a 2 Ci , b 2 Cj and ab g. This number is a non-negative integer, and is independent of the chosen element g of Ck . The result follows. j Another way of looking at Proposition 30.1 is to note that C i C j belongs to Z(CG), so it must be a linear combination of C1 , . . . , C l . 30.2 De®nition The integers aijk in the formula Ci C j

l X

aijk C k

k1

are the class algebra constants of G. From their very de®nition, the numbers aijk carry information about the multiplication in G: (30:3)

For all g 2 Ck and all i, j we have

aijk the number of pairs (a, b) with a 2 Ci , b 2 Cj and ab g: Also, the constants aijk determine the product of any two elements in the centre Z(CG) of the group algebra, since C1 , . . . , C l is a basis of Z(CG). As the centre of the group algebra plays an important role in representation theory, you might suspect that the class algebra constants are determined by the character table of G. Our next theorem shows that this is indeed the case. 30.4 Theorem Let gi 2 Ci for 1 < i < l. Then for all i, j, k, we have aijk

X ÷( 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.

350


Proof Let ÷ be an irreducible character of G, and let U be a CGmodule with character ÷. Then by Lemma 22.7, for all u 2 U we have uC i

jGj÷( g i ) u: jCG ( g i )j÷(1)

Therefore uC i C j

÷( g i )÷( g j ) jGj2 u jCG ( g i )j jCG ( g j )j (÷(1))2

and l X

aijm uC m

m1

Since C i C j

P

l X

(30:5)

m1

m1

m aijm C m ,

aijm

l X

aijm

jGj÷( g m ) u: jCG ( g m )j÷(1)

we deduce that

÷( g i )÷( g j ) ÷( g m ) jGj : jCG ( g m )j jCG ( g i )j jCG ( g j )j ÷(1)

Pick k with 1 < k < l. Multiply both sides of equation (30.5) by ÷( g k ) and sum over all irreducible characters ÷ of G, to obtain l X m1

aijm

X ÷( g m )÷( g k ) ÷

jCG ( g m )j

X ÷( g i )÷( g j )÷( g k ) jGj : jCG ( g i )j jCG ( g j )j ÷ ÷(1)

By the column orthogonality relations, Theorem 16.4(2), this yields aijk


j

Examples 30.6 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 ; these results can readily be proved directly, but they serve as a useful illustration of the method. Let G S4 . By Section 18.1, the character table of G is as shown:

Applications to group theory

351

Character table of S4 Class Ci gi |CG ( gi )|

C1 1 24

C2 (1 2) 4

C3 (1 2 3) 3

C4 (1 2)(3 4) 8

C5 (1 2 3 4) 4

÷1 ÷2 ÷3 ÷4 ÷5

1 1 2 3 3

1 ÿ1 0 1 ÿ1

1 1 ÿ1 0 0

1 1 2 ÿ1 ÿ1

1 ÿ1 0 ÿ1 1

(1) We use Theorem 30.4 to calculate the class algebra constant a555 : 24 1 ÿ1 0 ÿ1 1 0: a555 : 44 1 1 2 3 3 Hence, by (30.3), S4 does not possess elements a, b of order 4 such that the product ab also has order 4. 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. (2) By Theorem 30.4, 24 1 1 a245 : 11 2: 48 3 3 Hence S4 has elements a, b of order 2 such that ab has order 4. Writing x ab, we have x 4 1, aÿ1 xa ba (ab)ÿ1 x ÿ1 , so ka, bl D8. We deduce the fact (which we already know from Exercise 18.1) that S4 has a subgroup which is isomorphic to D8. (3) Finally, 24 a235 : (1 1) 4, 43 so S4 has elements a of order 2 and b of order 3 with ab of order 4. In fact, it can be shown that S4 has a presentation as follows: S4 ha, b: a2 b3 (ab)4 1i: In other words, S4 is generated by a and b, and all products of elements of S4 are determined by the given relations. We supply a

352


proof in the solution to Exercise 30.6 ± in the meantime, you may wish to puzzle out the relevance of the ®gure above. 30.7 Example We use Theorem 30.4 to ®nd a subgroup H of the simple group PSL (2, 7) with H isomorphic to S4 . That such a subgroup exists is not obvious, and it is quite tricky to construct directly. We found in Chapter 27 that the character table of G PSL (2, 7) is as follows. Character table of PSL (2, 7) Class rep. gi Order of gi |CG ( gi )| ÷1 ÷2 ÷3 ÷4 ÷5 ÷6 p where á (ÿ1 i 7)=2.

g1 1 168

g2 2 8

g3 4 4

g4 3 3

g5 7 7

g6 7 7

1 7 8 3 3 6

1 ÿ1 0 ÿ1 ÿ1 2

1 ÿ1 0 1 1 0

1 1 ÿ1 0 0 0

1 0 1 á á ÿ1

1 0 1 á á ÿ1


353

We calculate the class algebra constant a243 . By Theorem 30.4, 168 1 a243 : 1 0 0 0 0 8: 83 7 Hence, by (30.3), G contains elements x and y such that x has order 2, y has order 3 and xy has order 4. Let H be the subgroup kx, yl of G. From Example 30.6, we know that S4 ha, b: a2 b3 (ab)4 1i: Hence there is a homomorphism ö from S4 onto H (ö sends a to x and b to y). By Theorem 1.10, S4 =Ker ö H. Now Ker ö, being a normal subgroup of S4 , is {1}, V4 , A4 or S4 (see Example 12.20), so H is isomorphic to S4 , S3 , C2 or {1}. Since H has an element of order 4, namely xy, we conclude that H S4 : Thus we have shown that PSL (2, 7) has a subgroup which is isomorphic to S4 .

The Brauer programme The Brauer±Fowler Theorem 23.19 states that there are only ®nitely many non-isomorphic ®nite simple groups containing an involution with a given centralizer. This fact led Brauer to initiate a programme to ®nd, given a ®nite group C, all ®nite simple groups G possessing an involution t such that C G (t) C. This programme formed an important part of the effort of many mathematicians to classify all the ®nite simple groups, an effort which was ®nally completed in the early 1980s (see the book by D. Gorenstein listed in the Bibliography). The next result carries out part of Brauer's programme in the case where C D8 , a dihedral group of order 8. It determines the possible orders of simple groups G having an involution t such that C G (t) D8 . 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. 30.8 Theorem Let G be a ®nite non-abelian simple group which has an involution t such that C G (t) D8 . Then G has order 168 or 360.

354


Observe that PSL(2, 7) is a simple group of order 168 having an involution with centralizer D8 (see Lemma 27.1); and A6 is a simple group of order 360 with this property (see Exercise 7 at the end of the chapter). Using some rather more sophisticated group theory than that covered in this book, one can show that PSL(2, 7) and A6 are the only simple groups of order 168 or 360. Before embarking upon the proof of Theorem 30.8, we require a couple of preliminary results. The ®rst is Sylow's Theorem, a basic result in ®nite group theory. We shall not prove this, but refer you to Theorems 18.3 and 18.4 of the book by J. Fraleigh listed in the Bibliography. 30.9 Sylow's Theorem Let p be a prime number, and let G be a ®nite group of order p a b, where a, b are positive integers and p 6 j b. Then (1) G contains a subgroup of order p a; such a subgroup is call a Sylow p-subgroup of G; (2) all Sylow p-subgroups are conjugate in G (i.e. if P, Q are Sylow psubgroups, then there exists g 2 G such that Q g ÿ1 Pg); (3) if R is a subgroup of G with jRj p c for some c, then there is a Sylow p-subgroup of G containing R. 30.10 Lemma Let G be a ®nite non-abelian simple group, and let P be a Sylow 2subgroup of G. Suppose Q is a subgroup of P with jP : Qj 2. If u is an involution in G, then u is conjugate to an element of Q. Proof Suppose u is not conjugate to an element of Q. Let Ù be the set of right cosets Qx of Q in G, and de®ne an action of G on Ù by (Qx) g Qxg for x, g 2 G (see Example 29.2(4)). Observe that jÙj 2jG : Pj 2m, where m is odd since P is a Sylow 2-subgroup. Now consider fixÙ (u) fù 2 Ù : ùu ùg If Qx 2 fixÙ (u), then Qxu Qx and hence xux ÿ1 2 Q, contrary to assumption. Hence fixÙ (u) Æ. This means that in its action on Ù, the involution u is a product of m disjoint 2-cycles, hence is an odd permutation. Hence the subgroup f g 2 G : g acts as an even permutation on Ùg


355

is a normal subgroup of index 2 in G. This is impossible since G is non-abelian and simple. This contradiction completes the proof. j We also need to introduce the idea of a generalized character of a group H. This is simply a class function of the form X ø n÷ ÷ x

where the sum is over all the irreducible characters of H, and each n÷ 2 Z. If n÷ > 0 for all ÷ then of course ø is a character, but this need not be the case for a generalized character. In particular, the degree ø(1) can be 0 or negative for a generalized character ø. Notice also that the orthogonality relations give the usual inner products X hø, ÷i n÷ , hø, øi n2÷ for a generalized character ø as above. The generalized character ø can be expressed as a difference á ÿ â, where á and â are characters of H: take X X á n÷ ÷, â ÿ n÷ ÷: n÷ >0

n÷ ,0

Finally, if H is a subgroup of a group G, we de®ne the induced generalized character ø " G by ø " G (á " G) ÿ ( â " G) where ø á ÿ â as above. It is clear from this de®nition that the formulae for the values of ø " G given in Proposition 21.19 and Corollary 21.20 hold for generalized characters ø. Proof of Theorem 30.8 Let G be a ®nite non-abelian simple group with an involution t such that C G (t) D D8 . Certainly t commutes with itself, so t 2 D; and as t commutes with all elements of D, we have t 2 Z(D), the centre of D. The centre of D8 is a cyclic group of order 2 (see (12.12)), and hence Z(D) hti. By Theorem 30.9(3), there is a Sylow 2-subgroup P of G such that D < P. Then Z(P) < C G (t) D, so Z(P) < Z(D) hti. By Lemma 26.1(1) we have Z(P) 6 1, and hence Z(P) hti. Therefore P
1 á(t)2 â(t)2 1 e 2 (e ÿ 3)2 , from which it follows that e 1 or 2. Suppose now that e 1. Then (30.16) gives 1 4 jGj 1 ÿ 28 , d d1 whence jGj 28

d(d 1) : (d ÿ 1)2

Now the highest common factor hcf (d ÿ 1, d 1) is 1 or 2, and hcf (d ÿ 1, d) 1. Hence (d ÿ 1)2 must divide 210 , and so d ÿ 1 2 r with r < 5. Moreover, a Sylow 2-subgroup of G has order 8. It follows that r 3 and d 9, giving jGj 360, one of the possibilities in the conclusion of Theorem 30.8. Finally, suppose that e 2. Then (30.16) yields jGj 28

d(d 1) : (d 2)2

Reasoning as above, we deduce that d 2 23 , giving d 6 and jGj 168. This completes the proof of Theorem 30.8. j

Summary of Chapter 30 1. The class algebra constants aijk are given by X Ci C j aijk C k : k

They can be calculated from the character table, by using the formula aijk


2. Given groups G and H, the class algebra constants of G can sometimes be used to determine whether or not G has a subgroup which isomorphic to H.


359

3. Using Sylow's Theorem, together with lots of ingenious character theory, it can be shown that any simple group possessing an involution with centralizer isomorphic to D8 must have order 168 or 360.

Exercises for Chapter 30 1. Use the character table of PSL (2, 7), given at the end of Chapter 27, to prove that PSL (2, 7) contains elements a and b such that a has order 2, b has order 3 and ab has order 7. 2. Does PSL (2, 7) contain a subgroup isomorphic to D14 ? (Hint: D14 ka, b: a2 b2 1, (ab)7 1l.) For the next three exercises, you may assume that A5 is a simple group, and that A5 has the following presentation: A5 ha, b: a2 b3 (ab)5 1i: 3. The character table of PSL (2, 11) is given in the solution to Exercise 27.6. Does PSL (2, 11) contain a subgroup which is isomorphic to A5 ? 4. Prove that A5 is characterized by its character table ± that is, if G is a group with the same character table as A5 (see Example 20.13), then G A5 .

÷1 ÷2 ÷3 ÷4 ÷5 ÷6 ÷7

g1

g2

g3

g4

g5

g6

g7

1 5 5 8 8 9 10

1 1 1 0 0 1 ÿ2

1 ÿ1 ÿ1 0 0 1 0

1 2 ÿ1 ÿ1 ÿ1 0 1

1 ÿ1 2 ÿ1 ÿ1 0 1

1 0 0 á â ÿ1 0

1 0 0 â á ÿ1 0

where á (1

p

5)=2, â (1 ÿ

p

5)=2.

5. Suppose that G is a group, and that G has the character table shown. (a) Show that G is a simple group of order 360. (b) Use the Frobenius±Schur Count of Involutions to obtain an

360


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 . (d) Using Exercise 23.9, show that G A6 . 6. Use the ®gure which appears in Example 30.6(3) to show that every group G which is generated by two elements a and b which satisfy a2 b3 (ab)4 1 has order at most 24. 7. Prove that PSL(2, 7) and A6 are simple groups of order 168, 360 respectively, both of which contain an involution with centralizer isomorphic to D8 . 8. Find a simple group G having an involution C G (t) D16 . (Hint: look for a suitable simple group PSL(2, p).)

t such that

31 Burnside's Theorem

One of the most famous applications of representation theory is Burnside's Theorem, which states that if p and q are prime numbers and a and b are positive integers, then no group of order pa q b is simple. In the ®rst edition of his book Theory of groups of ®nite order (1897), Burnside presented group-theoretic arguments which proved the theorem for many special choices of the integers a, b, but it was only after studying Frobenius's new theory of group representations that he was able to prove the theorem in general. Indeed, many later attempts to ®nd a proof which does not use representation theory were unsuccessful, until H. Bender found one in 1972.

A preliminary lemma We prepare for the proof of Burnside's Theorem with a lemma (31.2) which is concerned with character values. In order to establish this lemma we require some basic facts about algebraic integers and algebraic numbers, which we now describe. We omit proofs of these ± for a good account, see for instance the book by Pollard and Diamond listed in the Bibliography. An algebraic number is a complex number which is a root of some non-zero polynomial over Q. We call a polynomial in x monic if the coef®cient of the highest power of x in it is 1. Let á be an algebraic number; and let p(x) be a monic polynomial over Q of smallest possible degree having á as a root. Then p(x) is unique and irreducible; it is called the minimal polynomial of á. The roots of p(x) are called the conjugates of á. For example, if ù is an nth root of unity then the minimal poly361

362


nomial of ù divides x n ÿ 1, and so every conjugate of ù is also an nth root of unity. If á is an algebraic integer, then á is a root of a monic polynomial with integer coef®cients (see Chapter 22), and it turns out that the minimal polynomial of á also has integer coef®cients. We shall require the following fact about conjugates: (31.1) Let á and â be algebraic numbers. Then every conjugate of á â is of the form á9 â9, where á9 is a conjugate of á and â9 is a conjugate of â. Moreover, if r 2 Q then every conjugate of rá is of the form rá9, where á9 is a conjugate of á. For an elementary proof of this, see Pollard and Diamond, Chapter V, Section 3. Alternatively, (31.1) can be proved easily using some Galois theory. 31.2 Lemma Let ÷ be a character of a ®nite group G, and let g 2 G. Then j÷( g)=÷(1)j < 1, and if 0 , j÷( g)=÷(1)j , 1 then ÷( g)=÷(1) is not an algebraic integer. Proof Let ÷(1) d. By Proposition 13.9 we have ÷( g) ù1 . . . ù d , where each ù i is a root of unity, so ÷( g)=÷(1) (ù1 : : : ù d )=d: Since |÷( g)| |ù1 . . . ù d | < |ù1 | . . . |ù d | d, it follows that j÷( g)=÷(1)j < 1. Now suppose that ÷( g)=÷(1) is an algebraic integer and j÷( g)=÷(1)j , 1. We prove that ÷( g) 0. Write ã ÷( g)=÷(1), and let p(x) be the minimal polynomial of ã, so that p(x) x n a nÿ1 x nÿ1 : : : a1 x a0 where ai 2 Z for all i. By (31.1), each conjugate of ã is of the form (ù91 : : : ù9d )=d where ù91 , . . . , ù9d are roots of unity. Hence each conjugate of ã has

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 , 1: 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 a0 . Thus ë a0 : Since a0 2 Z and |ë| , 1, it follows that a0 0. As p(x) is irreducible, this implies that p(x) x, which in turn forces ã 0. Thus ÷( g) 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 2 G with | gG | pr . Since pr . 1, G is not abelian and g 6 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

k X

÷ i ( g)÷ i (1) 0:

i2

Therefore k X i2

÷ i ( g) .

÷ i (1) 1 ÿ : p p

Now ÿ1= p is not an algebraic integer, by Proposition 22.5. Therefore, for some i > 2, ÷ i ( g)÷ i (1)= p is not an algebraic integer (see Theorem 22.3). Since ÷ i ( g) is an algebraic integer (Corollary 22.4), it follows

364


that ÷ i (1)= p is not an algebraic integer; in other words, p does not divide ÷ i (1). Thus ÷ i ( g) 6 0 G

and

p 6 j ÷ i (1):

r

As | g | p , 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) 1: Hence a

jGj÷ i ( g) ÷ i ( g) b÷ i ( g) : 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) 6 0, it is non-zero. Now Lemma 31.2 implies that j÷ i ( g)=÷ i (1)j 1: Let r be a representation of G with character ÷ i . By Theorem 13.11(1), there exists ë 2 C such that gr ëI: Let K Ker r, so that K is a normal subgroup of G. Since ÷ i is not the trivial character, K 6 G. If K 6 {1} then G is not simple, as required; so assume that K {1}, that is, r is a faithful representation. Since gr is a scalar multiple of the identity, gr commutes with hr for all h 2 G. As r is faithful, it follows that g commutes with all h 2 G; in other words, g 2 Z(G): Therefore Z(G) 6 {1}. As Z(G) is a normal subgroup of G and Z(G) 6 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 0 or b 0. Then the order of G is a power of a prime, so by Lemma 26.1(1) we have Z(G) 6 {1}.

Burnside's Theorem

365

Choose g 2 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) 6 {1} by Lemma 26.1(1). Let g 2 Z(Q) with g 6 1. Then Q < CG ( g), so j g G j jG:CG ( g)j pr for some r. If pr 1 then g 2 Z(G), so Z(G) 6 {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 G 0 , G1 , : : : , G r G such that for 1 < i < r, Giÿ1 v Gi and the factor group Gi =G iÿ1 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 G= 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 G= H are both soluble. Therefore there are subgroups 1 G0 v G1 v : : : v Gs H, 1 Gs = H v G s1 = H v : : : v Gr = H G= H with all factor groups Gi =G iÿ1 of prime order. Then the series 1 G0 v G 1 v : : : v G r 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


2. If |G| 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| 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 2

3

Let V be R or R , and for v, w 2 V, let d(v, w) denote the distance between v and w ± in other words, if v (x1 , x2 , . . .) and w ( y1, y2, . . .), then ! r X 2 d(v, w) (xi ÿ yi ) : An isometry of V is an invertible endomorphism W of V such that d(vW, wW) d(v, w) for all v, w 2 V : 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


of an isometry; so is any re¯ection in a plane through the origin. The endomorphism ÿ1R3 which sends every vector v to ÿv 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 ÿ 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 R2 or R3, then we de®ne G(Ä) to be the set of isometries which leave Ä invariant ± that is, G(Ä) f g 2 O(V ): Ä g Äg (where Ä g {v g: v 2 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 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 R3, and again let Ä be a regular n-sided polygon (n > 3) centred at the origin. This time, G(Ä) 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 2ð=3 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


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 ÿk(x1 x2 ). By dealing with the right-hand mass similarly, we obtain the following equations of motion of the system: m x1 ÿkx1 ÿ k(x1 x2 ) ÿ2kx1 ÿ kx2 , m x2 ÿkx2 ÿ k(x1 x2 ) ÿkx1 ÿ 2kx2 , 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 (x1 , x2 ), x ( x1 , x2 ) and q k=m. Then the equations of motion are equivalent to the matrix equation ÿ2q ÿq x xA, where A (32:5) : ÿq ÿ2q 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 ÿù2 . For an arbitrary constant â, let x sin (ùt â) u: Then x ÿù2 sin (ùt â) u sin (ùt â) uA

(since uA ÿù2 u)

xA: Thus x is a solution of the equation of motion. If u1 and u2 are linearly independent eigenvectors of A, with eigenvalues ÿù21 and ÿù22 , respectively, then á1 sin (ù1 t â1 ) u1 á2 sin (ù2 t â2 ) u2

372


is a solution of the equation of motion which involves four arbitrary constants á1 , á2 , â1 , â2 , so it is the general solution. We now adopt this line of attack in the problem to hand. For the matrix given in (32.5), the eigenvalues are ÿ3q and ÿq, with corresponding eigenvectors (1, 1) and (1, ÿ1). Therefore the general solution of the equation of motion (32.5) is p p á1 sin ( (3q) t â1 ) (1, 1) á2 sin ( q . t â2 ) (1, ÿ1): The solutions which involve just one eigenvector of A are called the normal modes of vibration. They are as follows. p sin ( (3q) t â1 ) (1, 1) Mode 1: p Here, x1 x2 sin ( (3q) t â1 ) and the vibration is

Mode 2:

p sin ( q . t â2 ) (1, ÿ1)

p Here, x1 ÿx2 sin ( q . t â2 ) and the vibration is

The general molecular vibration problem Suppose we have a molecule consisting of n atoms which vibrate under internal forces. At the equilibrium position of each atom, we assign three coordinate axes, which we use to measure the displacement of the atom. Thus, the state of the molecule at a given time is described by a vector in the 3n-dimensional vector space R3 n . We assume that the internal forces are linear functions of the displacements. It follows that when we apply Newton's Second Law of Motion, we obtain equations which may be expressed in the form (32:6)

x xA:

(Compare (32.5).) Here x is the displacements of all the atoms, entries which are determined by Assume, for the moment, that

row vector in R3 n which measures the and A is a 3n 3 3n matrix with real the internal forces. at each atom the three coordinate axes

An application of representation theory to molecular vibration 373 which we have chosen are at right angles to each other. It can be shown, from physical considerations, that in this special case the matrix A is symmetric. In particular, A has real eigenvalues, and A has 3n linearly independent eigenvectors. Now, the effect of changing coordinate axes is merely to replace A by a matrix which is conjugate to A. Therefore we have the following proposition, for the general case, where our chosen coordinate axes are not necessarily at right angles to each other. 32.7 Proposition All the eigenvalues of A are real, and A has 3n linearly independent eigenvectors. To solve the equation of motion (32.6), we look for normal modes of the system, which we de®ne next. 32.8 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 ÿù2 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. 32.9 Proposition Each normal mode of vibration is a solution of the equation of motion (32.6), and for this solution all the atoms vibrate with the same frequency. The general solution of the equation of motion is a linear combination of the normal modes of vibration. Proof If uA ÿù2 A and x sin (ùt â) u, then x ÿù2 sin (ùt â) u sin (ùt â) uA xA: If uA 0 and x (t â)u, then x 0 (t â)uA xA: This proves that the normal modes of vibration are solutions of the

374


equation of motion (32.6). By construction, all the atoms vibrate with the same frequency (namely, ù or 0) in a normal mode. Note that A can have no strictly positive eigenvalue; for if ë were p such an eigenvalue, with eigenvector u, then x e ë t u would be a solution to the equation of motion, which is nonsense. Therefore there exist 3n linearly independent normal modes, by Proposition 32.7. Since each normal mode involves an arbitrary constant, the general linear combination of normal modes involves 6n arbitrary constants, so it is the general solution to the equation of motion (32.6) (as (32.6) consists of second order differential equations in 3n unknowns). j Proposition 32.9 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. However, 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, and we shall describe a method for doing this.

Use of the symmetry group We continue the discussion of the previous section. Let G be the symmetry group of the molecule in question. Since G permutes the atoms among themselves, each element of G acts as an endomorphism of the space R3 n of displacement vectors. Thus, R3 n is an RG-module. 32.10 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, and for 1 < i < 9, let v i denote a unit vector along coordinate axis i. Then g ®xes v1 , negates v2 and v3, interchanges v4 and v7, and

An application of representation theory to molecular vibration 375 interchanges v5 and v6 with the negatives of v8 and v9 . Therefore g acts on R9 as follows: (x1 , x2 , x3 , x4 , x5 , x6 , x7 , x8 , x9 ) g (x1 , ÿx2 , ÿx3 , x7 , ÿx8 , ÿx9 , x4 , ÿx5 , ÿx6 ): We return to the general set-up. The equations of motion are x xA, and we are trying to ®nd the eigenvectors of A. The eigenspace for the eigenvalue ë of A is, by de®nition, fx 2 R3 n : xA ëxg: We can now present the crucial proposition which allows us to exploit the symmetry group G of our molecule. In effect, it tells us that the function x ! xA (x 2 R3 n ) is an RG-homomorphism from R3 n to itself. 32.11 Proposition For all g 2 G and x 2 R3 n , (xg)A (xA) g, and the eigenspaces of A are RG-submodules of R3 n . Proof Let ÿù2 be a non-zero eigenvalue of A, and let v be a corresponding eigenvector. 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. For all g in G, 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. Therefore, v g is an eigenvector of A, with eigenvalue ÿù2 . This shows that the eigenspace for ÿù2 is an RG-submodule of R3 n . A similar argument applies to the eigenspace for the eigenvalue 0. Choose a basis of R3 n which consists of eigenvectors of A (see Proposition 32.7), and let g 2 G. For all vectors v in the basis, vA ëv for some ë 2 R, and (v g)A ë(v g) (ëv) g (vA) g: Hence (xg)A (xA) g for all x 2 R3 n .

j

The idea now is to use representation theory to express the RGmodule R3n as a direct sum of irreducible RG-submodules, and hence

376


to determine the eigenspaces of A, and the normal modes of the molecule. We can use character theory to see which irreducible RG-modules are contained in R3 n ; and if ÷ is the character of an irreducible RGmodule which occurs, then the element X ÷( g ÿ1 ) g g2G

sends R3 n onto the sum of those irreducible RG-submodules of R3 n which have character ÷ (see (14.27)). (This procedure sometimes needs to be modi®ed, since the character of the RG-module R3 n might contain an irreducible character which cannot be realized over R ± but in practice, problems like this are uncommon.) 32.12 De®nition Suppose that ÷ is the character of an irreducible RG-module. Let V÷ denote the sum of those irreducible RG-submodules of R3 n which have character ÷. We call V÷ a homogeneous component of R3 n . The problem of ®nding the eigenspaces of A is considerably simpli®ed as a consequence of our next proposition. 32.13 Proposition Each homogeneous component V÷ of R3 n is A-invariant ± that is, xA 2 V÷

for all x 2 V÷ :

Proof By Maschke's Theorem we may write R3n V÷ W for some RG-module W, and no RG-submodule of W has character ÷. The function å: v w ! w

(v 2 V÷ , w 2 W )

is an RG-homomorphism. Therefore, by Proposition 32.11, the function x ! (xA)å

(x 2 V÷ )

is an RG-homomorphism from V÷ into W. By Proposition 11.3, this function is zero, so xA 2 V÷ for all x 2 V÷ . (Although Proposition 11.3 is stated in terms of CG-modules, its proof works equally well for RG-modules ± compare Exercise 23.8.) j

An application of representation theory to molecular vibration 377 32.14 Corollary If V÷ is an irreducible RG-module, then all the non-zero vectors in V÷ are eigenvectors of A. Proof (Compare the proof of Schur's Lemma.) Since V÷ is A-invariant, we may choose v 2 V÷ such that v is an eigenvector of A, with eigenvalue ë, say. Then the intersection of V÷ with the eigenspace for ë is a non-zero RG-submodule of V÷ , so it must equal V÷ . j We now summarize the steps in the procedure for ®nding the normal modes of vibration of a given molecule. 32.15 Summary (1) Assign three coordinate axes at each of the n atoms of the molecule, to obtain R3n . (2) Calculate the symmetry group G of the molecule. Then R3 n is an RG-module. (3) Calculate the character ÷ of the RG-module R3 n and express ÷ as a linear combination of the irreducible characters of G. (4) Express R3 n as a direct sum of homogeneous components. This can P ÿ1 3n be done by applying the element for each g2G ÷ i ( g ) g to R irreducible character ÷ i of G which appears in ÷, or by some other method. (5) Consider, in turn, each homogeneous component V÷ i of R3 n , and ®nd the eigenvectors of A in V÷ i (see Proposition 32.13). This involves no extra work if V÷ i is irreducible (see Corollary 32.14). If V÷ i is reducible, then see Remark 32.19 below, or Exercise 32.7, to make further progress. (6) If v is an eigenvector of A, with eigenvalue ÿù2 , then sin (ùt â) v

(or (t â)v if ù 0)

is a normal mode, where â is an arbitrary constant. It is usually necessary to know the equations of motion in order to determine the frequency ù. This programme can often be successfully completed, as we shall illustrate in the examples which make up the rest of this chapter.

378


32.16 Example We ®rst return to Example 32.4, with the spring and two vibrating masses:

The symmetry group of this system is G h g: g 2 1i, where g is the re¯ection in the mid-point of PQ. The displacement vectors (x1 , x2 ) form an RG-module R2 . Since (x1 , x2 ) g (x2 , x1 ), the RG-submodules of R2 are sp (u1 ) and sp (u2 ), where u1 (1, 1), u2 (1, ÿ1): It follows that the normal modes of the system are given by sin (ù1 t â1 )(1, 1), sin (ù2 t â2 )(1, ÿ1), where â1 , â2 are constants and ù1 , ù2 are the frequencies. This agrees with the conclusion of Example 32.4. Notice that we have determined the normal modes of vibration (but not their frequencies) using the symmetry group alone. 32.17 Example Consider a hypothetical triatomic molecule, where the three identical atoms are at the corners of an equilateral triangle. For simplicity, we consider only vibrations of the molecule in the plane, so we assign two displacement coordinates to each atom. We choose to take our axes along the edges of the triangle, as shown, as this eases the calculations:

Thus the position of the molecule is given by a vector (x1 , . . . , x6 ) in R6 , where xi is the displacement along axis i (1 < i < 6). The symmetry group (in two dimensions) of the molecule is the dihedral group D6, generated by a rotation a of order 3 and a re¯ection

An application of representation theory to molecular vibration 379 b (see Example 32.1). It is easy to work out the action of each element of D6 on R6 . For example, if b is the re¯ection which ®xes the top atom, then (x1 , x2 , x3 , x4 , x5 , x6 )b (x2 , x1 , x6 , x5 , x4 , x3 ): We want to express the RD6 -module R6 as a direct sum of irreducible RD6 -modules. To do this, we ®rst calculate the character ÷ of the module. Since the rotation a does not ®x any of the atoms, ÷(a) 0. And from the action of b given above, we see that ÷(b) 0. Thus the values of ÷ are

÷

1

a

b

6

0

0

By Section 18.3, the character table of D6 is

÷1 ÷2 ÷3

1

a

b

1 1 2

1 1 ÿ1

1 ÿ1 0

Hence ÷ ÷1 ÷2 2÷3 . Thus, we seek to express R6 as a direct sum of RD6 -submodules with characters ÷1 , ÷2 , ÷3 and ÷3 . As a matter of notation, if v1 , v2 , v3 are 2-dimensional displacement vectors for the three atoms, then we represent the displacement vector (v1 , v2 , v3 ) 2 R6 pictorially by the diagram

We ®rst calculate the normal modes of the form (t â)v, corresponding to the eigenspace of A for eigenvalue 0. These include the rotation and translation modes, which occur for every molecule.

380


Rotation mode In this mode, the molecule rotates with constant angular velocity about the centre. The mode is given by (t â)v, where v (1, ÿ1, 1, ÿ1, 1, ÿ1); pictorially,

We call sp (v) the rotation submodule of R6 . If ÷ R is the character of sp (v), then ÷ R (1) 1, ÷ R (a) 1, ÷ R (b) ÿ1, and so ÷ R ÷2 . Indeed, sp (v) R6 å2 , where X å2 ÷2 ( g ÿ1 ) g 1 a a2 ÿ b ÿ ab ÿ a2 b g2 D6

(compare (14.27)). Translation modes These are modes in which all atoms move in the same constant direction with the same constant speed. The modes are of the form (t â)v, where v is a vector in the span of v1 , v2 and v3, these vectors being given pictorially by

(thus v1 (ÿ1, 1, 0, ÿ1, 1, 0), v2 (1, 0, ÿ1, 1, 0, ÿ1), v3 (0, ÿ1, 1, 0, ÿ1, 1)). Since v1 v2 v3 0, the subspace sp (v1 , v2 , v3 ) has dimension 2. It is clearly an RD6 -submodule of R6 , and is called the translation submodule; it does not contain the rotation submodule. Thus the character of the translation submodule is part of ÷ ÿ ÷2 ÷1 2÷3 , so the character must be ÷3 .

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, and are called vibratory modes. The sum of these eigenspaces forms an RD6 -submodule R6vib of R6 (by Proposition 32.11), with character ÷vib , where ÷vib ÷ ÿ (÷2 ÷3 ) ÷1 ÷3 : In particular, R6vib has dimension 3. Since no mode in R6vib can have any translation component, if w 2 R6vib then the total component of w in each direction is zero; moreover, R6vib does not contain the rotation submodule, so that total moment of each vector in R6vib about the centre is zero. These constraints imply three independent linear equations in the coordinates of the vectors in R6vib , and since R6vib has dimension 3, every vector in R6 which satis®es these equations lies in R6vib . Therefore a basis of R6vib is u1 , u2 , u3 , where

Clearly sp (u1 u2 u3 ) is an RD6 -submodule of R6vib with character ÷1 . 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 (32.18(3)) below). Finally, since D6 permutes the vectors u1 , u2 , u3 among themselves, it is easy to see that sp (u1 ÿ u2 , u1 ÿ u3 ) is an RD6 -submodule of R6vib . Its character is ÷3 and it gives us the last eigenspace for the matrix A. Our calculation of the normal modes is now complete, and we summarize our ®ndings below.

382 (32.18)

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 ÿ u1 ÿ u3 , 2u1 ÿ u2 ÿ u3 as the basis for the vibratory modes in (4) merely because these modes look simpler than u1 ÿ u2 , u1 ÿ u3 pictorially.) We emphasize that we have found the normal modes of vibration without explicit knowledge of the equations of motion. In order to

An application of representation theory to molecular vibration 383 check our results, 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.

For a general displacement (x1 , x2 , x3 , x4 , x5 , x6 ), denote the new positions of the atoms by P9, Q9, R9. From the diagram, the difference in length between QR and Q9R9 is (x4 x5 ) 12(x3 x6 ): (We always assume that x1 , : : : , x6 are small compared with the distance between the atoms, so that we may ignore second order terms.) Similarly, PR ÿ P9R9 (x1 x6 ) 12(x2 x5 ), PQ ÿ P9Q9 (x2 x3 ) 12(x1 x4 ): Hence the force on the molecule at P in the direction of the ®rst coordinate axis is ÿk(PR ÿ P9R9) ÿk(x1 x6 ) ÿ 12 k(x2 x5 ): Therefore,

m x1 ÿ(x1 x6 ) ÿ 12(x2 x5 ): k

In the same way, m x2 ÿ(x2 x3 ) ÿ 12(x1 x4 ), k

384


and we obtain similar equations for x3, . . . , x6 . The matrix A for which x xA is therefore given by 0 1 1 1=2 1=2 0 0 1 B 1=2 1 1 0 0 1=2 C B C B ÿk B 0 1 1 1=2 1=2 0 C C: A 1=2 1=2 1 1 0 C m B B 0 C @ 1=2 0 0 1 1 1=2 A 1 0 0 1=2 1=2 1 You should check that the vectors which we gave in (32.18) are indeed eigenvectors of A. 32.19 Remark In Example 32.17, the character ÷ of the RG-module R6 was given by ÷ ÷1 ÷2 2÷3 : All the non-zero vectors in the homogeneous components for ÷1 and ÷2 gave normal modes, since these homogenous components were irreducible (see Corollary 32.14). The homogeneous component V÷3 for ÷3 was reducible, but we were able to write it as a sum of two subspaces of eigenvectors (those appearing in (32.18)(2) and (4)) because V÷3 \ R6vib was an A-invariant RG-submodule of V÷3 different from {0} and V÷3 . This illustrates a method which sometimes helps to deal with reducible homogeneous components. In our next example, the situation is more complicated. 32.20 Example We analyse the normal modes for the methane molecule CH4 .

We determined the symmetry group G of this molecule in Example 32.2, where we found that G is isomorphic to S4 . Label the corners of

An application of representation theory to molecular vibration 385 the tetrahedron 1, 2, 3, 4, as shown below, and identify G with S4 ; thus, for example, the rotations about the vertical axis through 1 are written as 1, (2 3 4), (2 4 3):

In order fully to exploit the symmetry of the methane molecule, at each hydrogen atom we choose displacement axes along the edges of the tetrahedron. Let v12 , v13 , v14 be unit vectors at corner 1 in the directions of the edges 12, 13, 14, respectively; similarly, let v21 , v23 , v24 be unit vectors at corner 2 in the directions of the edges 21, 23, 24, and so on, giving twelve vectors v ij , in all. We now introduce a new idea, by taking four unit vectors w1, w2, w3 and w4 at the carbon atom, with w i pointing towards corner i (1 < i < 4). Since w1 w2 w3 w4 0, these four vectors span a 3-dimensional space. Let V be the vector space over R with basis v12 , v13 , v14 , v21 , v23 , v24 , v31 , v32 , v34 , v41 , v42 , v43 , and let W be the vector space over R spanned by w1, w2, w3, w4. Then V R12, W R3 and V and W are RG-modules. Our main task is to ®nd RG-submodules of R15 V W. The action of G on V is easy to describe; for g in G, we have v ij g v ig,jg

for all i, j:

Thus, G permutes our twelve basis vectors of V, and we can quickly calculate the character ÷ of V:

÷

1

(1 2)

(1 2 3)

(1 2)(3 4)

(1 2 3 4)

12

2

0

0

0

386


For example, (1 2) ®xes the basis vectors v34 and v43 only; all the basis vectors are moved by (1 2 3); and so on. The group G acts on W as follows; for g in G, we have wi g wig

(1 < i < 4):

After recalling that w1 . . . w4 0, it is easy to calculate the character ö of W ± it is

ö

1

(1 2)

(1 2 3)

(1 2)(3 4)

(1 2 3 4)

3

1

0

ÿ1

ÿ1

By Section 18.1, the character table of S4 is as shown at the top of p. 387. We ®nd that ÷ ÷1 ÷3 2÷4 ÷5 , ö ÷4 : By applying the elements X ÷ i ( g ÿ1 ) g

(i 1, 3, 5, 4)

g2G

to R15 , we can ®nd RG-submodules with characters ÷1 , ÷3 , ÷5 and 3÷4 (see (14.27)). The RG-submodule W1 with character ÷1 is spanned by X v ij i, j

and this gives the expansion±contraction normal mode:

We next describe the RG-submodule W5 with character ÷5 . Let p1 (v23 ÿ v32 ) (v34 ÿ v43 ) (v42 ÿ v24 ), p2 (v31 ÿ v13 ) (v14 ÿ v41 ) (v43 ÿ v34 ),

An application of representation theory to molecular vibration 387 Character table of S4

÷1 ÷2 ÷3 ÷4 ÷5

1

(1 2)

1 1 2 3 3

1 ÿ1 0 1 ÿ1

(1 2 3) 1 1 ÿ1 0 0

(1 2)(3 4) 1 1 2 ÿ1 ÿ1

(1 2 3 4) 1 ÿ1 0 ÿ1 1

p3 (v12 ÿ v21 ) (v41 ÿ v14 ) (v24 ÿ v42 ), p4 (v21 ÿ v12 ) (v13 ÿ v31 ) (v32 ÿ v23 ): The vector pi gives a rotation about the axis through the corner i and the centroid of the tetrahedron.

It should be clear from the pictures that for all i with 1 < i < 4 and

388


all g in G, we have pi g p j for some j. Therefore, if we let W 5 sp ( p1 , p2 , p3 , p4 ), then W5 is an RG-submodule of V. Now p1 p2 p3 p4 0, so dim W5 3. Check that the character of W5 is ÷5 . The RG-module W5 is the rotation submodule. (Compare, for example, the picture for p4 with the picture for the rotation vector v in Example 32.17.) Now we construct the RG-submodule W3 of V with character ÷3 . Let q1 (v12 v21 ) (v34 v43 ) ÿ (v13 v31 ) ÿ (v24 v42 ), q2 (v13 v31 ) (v24 v42 ) ÿ (v14 v41 ) ÿ (v23 v32 ), q3 (v14 v41 ) (v23 v32 ) ÿ (v12 v21 ) ÿ (v34 v43 ): (Each q i is associated with an òpposite pair of edges'.)

For all i with 1 < i < 4 and g in G, we have q i g q j for some j. Let W3 sp (q1 , q2 , q3 ). Then W3 is an RG-submodule of V. Since q1 q2 q3 0, the dimension of W3 is 2; its character is ÷3 .

An application of representation theory to molecular vibration 389 In the RG-submodules W 1 , W 5 and W3 which we have found so far, all the non-zero vectors are eigenvectors of A, by Corollary 32.14. We now come to the homogeneous component (V W)÷4 of R15 . De®ne the vectors r1, r2, r3, r4 by r1 (v12 v21 ) (v13 v31 ) (v14 v41 ) ÿ (v23 v32 ) ÿ (v24 v42 ) ÿ (v34 v43 ), r2 (v12 v21 ) (v23 v32 ) (v24 v42 ) ÿ (v13 v31 ) ÿ (v14 v41 ) ÿ (v34 v43 ), r3 (v13 v31 ) (v23 v32 ) (v34 v43 ) ÿ (v12 v21 ) ÿ (v14 v41 ) ÿ (v24 v42 ), r4 (v14 v41 ) (v24 v42 ) (v34 v43 ) ÿ (v13 v31 ) ÿ (v12 v21 ) ÿ (v23 v32 ): (The vector ri is associated with corner i.)

390


For each g in G and i with 1 < i < 4, we have ri g rig. Thus G permutes the vectors r1, r2, r3, r4 among themselves. Note that r1 r2 r3 r4 0, so r1, r2, r3, r4 span a 3-dimensional RGsubmodule W4 of V. The character of W4 is ÷4 (see Proposition 13.24). Next, de®ne the vectors s1 , s2 , s3 , s4 by s1 (v12 v13 v14 ) ÿ (v21 v31 v41 ), s2 (v21 v23 v24 ) ÿ (v12 v32 v42 ), s3 (v31 v32 v34 ) ÿ (v13 v23 v43 ), s4 (v41 v42 v43 ) ÿ (v14 v24 v34 ):

We have si g sig

( g 2 G, 1 < i < 4),

s1 s2 s3 s4 0,

An application of representation theory to molecular vibration 391 and s1 , s2 , s3 , s4 span a 3-dimensional RG-submodule W94 of V with character ÷4 . Now recall that w1 , w2 , w3 , w4 span W; we have wi g wig

( g 2 G, 1 < i < 4),

w1 w2 w3 w4 0, and the character of W is ÷4 . The sum of W 4 , W 49 and W is direct, so (V W )÷4 W 4 W 49 W : We now break off temporarily from studying the methane molecule, in order to deal with the easier case of a molecule with identical atoms at the corners of the tetrahedron, and no central atom. In this case, the space W does not enter our calculations, and we can decompose V÷4 W 4 W 49 in the following way. (32.21) (1) The vectors r1 ÿ 2s1 , r2 ÿ 2s2 , r3 ÿ 2s3 , r4 ÿ 2s4 span the 3-dimensional space of translation modes. (2) The vectors r1, r2, r3, r4 span the subspace V÷4 \ R12 vib of V ÷4 , and so they give the ®nal 3-dimensional space of eigenvectors (see Remark 32.19). The normal mode (sin ùt)r1 is sometimes called an ùmbrella mode'. To see why, look at the picture of the vector r1 ! We return now to the methane molecule. The task in front of us is to ®nd the eigenvectors of A which lie in (V W )÷4 W 4 W 49 W : The solution of this problem depends, in fact, upon the constants which appear in the equations of motion, so we cannot complete the work using only representation theory. Since dim (V W )÷4 9, it is very dif®cult to calculate the eigenvectors of A in (V W )÷4 directly. 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. Let H be the subgroup of S4 consisting of those permutations which ®x the number 1, and let U1 fv 2 (V W )÷4 : vh v for all h 2 Hg: Since (vh)A (vA)h for all v 2 V÷4 and all h 2 H, it follows that U1 is A-invariant.

392


We ®nd that h3÷4 # H, 1 H i H 3, and so dim U1 3. But for all h 2 H, r1 h r1, s1 h s1 and w1 h w1 . Therefore U 1 sp (r1 , s1 , w1 ): Once the equations of motion, and hence the matrix A, have been calculated, it is possible to calculate the 3 3 3 matrix B of the action of A on r1, s1 , w1 (see Exercise 32.5). The eigenvectors of B then give three eigenvectors of A. Better still, r1 (1 2) r2 , s1 (1 2) s2 , w1 (1 2) w2 , and since A commutes with the action of G, the space U 2 , de®ned by U2 sp (r2 , s2 , w2 ) is A-invariant, and the matrix of A acting on r2, s2 , w2 is again B. A similar remark applies to U3, where U 3 sp (r3 , s3 , w3 ): Therefore, the process of calculating the eigenvectors of the 3 3 3 matrix B gives nine eigenvectors of A which form a basis of (V W )÷4 . One eigenvector of A acting on r1, s1 , w1 is easy to ®nd, namely the translation vector r1 ÿ 2s1 3 cos Ww1 , where W is the angle between an edge of the tetrahedron and the line joining a corner on the edge to the centroid. Thus we obtain the translation submodule sp (r1 ÿ 2s1 3 cos Ww1 , r2 ÿ 2s2 3 cos Ww2 , r3 ÿ 2s3 3 cos Ww3 ). By means of representation theory, 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. It is hard to imagine a more spectacular application of representation theory with which to conclude our text.

An application of representation theory to molecular vibration 393 Summary of Chapter 32 1. 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. 2. The equations of motion of the molecule have the form x xA, 3n

where x 2 R and the 3n 3 3n matrix A has 3n linearly independent eigenvectors. 3. If u is an eigenvector of A, with eigenvalue ÿù2 , then x sin (ùt â)u (or x (t â)u if ù 0) is a solution of the equations of motion, and is called a normal mode. All solutions are linear combinations of normal modes. 4. The space R3n of displacement vectors is an RG-module. The action of any g 2 G on R3 n commutes with A. 5. To determine the eigenvectors of A (and hence all solutions of the equations of motion), it is suf®cient to ®nd the eigenvectors of A restricted to each homogeneous component V÷ i of the RG-module R3 n . If V÷ i is irreducible, then all non-zero vectors in V÷ i are eigenvectors of A. Exercises for Chapter 32 1. Suppose that b 2 O(R3 ), and let e1 (1, 0, 0), e2 (0, 1, 0), e3 (0, 0, 1). (a) Show that the matrix B of b with respect to the basis e1 , e2 , e3 of R3 satis®es BBt I (where Bt denotes the transpose of B). Deduce that det B 1. (b) Let C (det B)B. Prove successively that (i) C has a real eigenvalue, (ii) C has a positive eigenvalue, (iii) 1 is an eigenvalue of C. (c) Deduce from (a) and (b) that b is a rotation if det B 1, and ÿb is a rotation otherwise. (d) Show that if b is a rotation through an angle ö about some axis, then tr B 1 2 cos ö. 2. Suppose that G is the symmetry group of some molecule in R3 .

394


Show that the sum of the character ÷ T of the translation submodule and the character ÷ R of the rotation submodule has value at g 2 G given by 8 2(1 2 cos ö), if g is a rotation > > < through angle ö (÷ T ÷ R )( g) about some axis, > > : 0, if g is not a rotation: 3. Consider the triatomic molecule studied in Example 32.17. Take the axes for the displacement coordinates as shown below:

Calculate the equations of motion x xA with respect to these axes, and verify that A is symmetric. (See the paragraph before Proposition 32.7.) 4. Consider the space spanned by the vectors r1, r2, r3, r4 given in Example 32.20. Find a basis for this space which is simpler than the one which we have used. What property of r1, r2, r3, r4 prompted us to use these vectors? 5. The purpose of this exercise is to derive the equations of motion of the methane molecule, and so ®nd explicitly the 3 3 3 matrix B which appears at the end of Example 32.20. Label the corners of the tetrahedron 1, 2, 3, 4 and let 0 denote the centroid of the tetrahedron. Work with 15 unit displacement vectors v12 , v13 , : : : , v43 , w1 , w2 , w3 as described in Example 32.20, and let the position vector of the molecule be X

xij v ij

yi wi :

i1

i6 j

(a) Prove that cos (/ 012)

3 X

p

(2=3) and cos (/ 102) ÿ1=3.

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 12(x13 x14 x23 x24 ), with similar expressions for the edges 13, 14, 23, 24, 34. Also, show that the length of the edge 01 has decreased by p (2=3)(x12 x13 x14 ) y1 ÿ 13( y2 y3 ), with similar expressions for the edges 02, 03. Finally, show that the length of the edge 04 has decreased by p (2=3)(x41 x42 x43 ) ÿ 13( y1 y2 y3 ): (c) Let m1 denote the mass of a hydrogen atom, and m2 the mass of a carbon atom. 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. Prove that m1 x12 ÿ k 1 [x12 x21 12(x13 x14 x23 x24 )] p ÿ 13 k 2 [x12 x13 x14 (3=2)( y1 ÿ 13 y2 ÿ 13 y3 )], with similar expressions for x13, x14 , x21 , x23 , x24 , x31 , x32 , x34 . Also, show m1 x41 ÿ k 1 [x14 x41 12(x42 x43 x12 x13 )] p ÿ 13 k 2 [x41 x42 x43 ÿ 13 (3=2)( y1 y2 y3 )], with similar expressions for x42 and x43 . Finally, show p m2 y1 ÿk 2 [ (2=3)(x12 x13 x14 ÿ x41 ÿ x42 ÿ x43 ) 43 y1 ], with similar expressions for y2 and y3 . (d) The equations in part (c) determine the 15 3 15 matrix A in the equations of motion x xA. Verify that the vectors X v ij , p1 , p2 , p3 , q1 , q2 , i, j

396


which appear in Example 32.20, are eigenvectors of A. (e) Find the entries b ij in the 3 3 3 matrix B which are given by r1 A b11 r1 b12 s1 b13 w1 , s1 A b21 r1 b22 s1 b23 w1 , w1 A b31 r1 b32 s1 b33 w1 , where the vectors r1, s1 , w1 are as in Example 32.20. (f) Verify that p (1, ÿ2, 6) is an eigenvector of B. 6. Consider a hypothetical molecule in which there are four identical atoms at the corners of a square. Assume that the only internal forces are along the sides of the square. (a) Find the normal modes of the molecule. (b) Calculate the equations of motion, x xA, and check that the vectors you found in part (a) are, indeed, eigenvectors of A. 7. In this exercise, we derive a method for simplifying the problem of ®nding the eigenvectors of A when the homogeneous component V÷ i is reducible. (See 32.15(5).) We assume that ÷ i is the character of an irreducible RG-module which remains irreducible as a CGmodule. Suppose that V÷ i U1 . . . Um, a sum of m isomorphic irreducible RG-modules. We reduce our problem to that of ®nding the eigenvectors of an m 3 m matrix. For 1 < i < m, let W i be an RG-isomorphism from U1 to Ui . (a) Prove that for all non-zero u in U1, sp (uW1 , : : : , uW m ) is an A-invariant vector space of dimension m. (Hint: compose the function w ! wA with a projection, and use Exercise 23.8.) (b) Let A u denote the matrix of the endomorphism w ! wA of sp (uW1 , . . . , uW m ) with respect to the basis uW1 , . . . , uW m . Prove that if u and v are non-zero elements of U1, then Au Av . (c) Assume that the eigenvectors of the m 3 m matrix A u are known. Show how to ®nd the eigenvectors of A in V÷ i .

Solutions to exercises

Chapter 1 1. Note that all subgroups of G are normal, since G is abelian; and G 6 {1} since G is simple. Let g be a non-identity element of G. Then kgl is a normal subgroup of G, so kgl G. If G were in®nite, then h g 2 i would be a normal subgroup different from G and {1}; hence G is ®nite. Let p be a prime number which divides |G|. Then h g p i is a normal subgroup of G which is not equal to G. Therefore g p 1, and so G is cyclic of prime order. 2. Since G is simple and Ker W v G, either Ker W f1g or Ker W G. If Ker W f1g then W is an isomorphism; and if Ker W G then H f1g. 3. First, G \ An f g 2 G: g is even}, so G \ An v G. Since G \ An 6 G, we may choose h 2 G with h 2 = An . For all odd g in G, we have g (ghÿ1 )h 2 (G \ An )h. Therefore G \ An and (G \ An )h are the only right cosets of G \ An in G, and G=(G \ A n ) C2 . 4. (a) Using the method of Example 1.4, it is routine to verify that ö and ø are homomorphisms. Ker ö {1, a2 } and Ker ø {1, c2 }. (b) Since b2 ë I but (bë)2 Y 2 ÿI, it follows that ë is not a homomorphism. Check using the method of Example 1.4 again that ì is a homomorphism. Also Ker ì {1} and Im ì L, so ì is an isomorphism. 5. Let D4 m ha, b: a2 m b2 1, bÿ1 ab aÿ1 i, and D2 m hc, d: cm d 2 1, d ÿ1 cd cÿ1 i, where m is odd. The elements of D2 m 3 C2 are (c i d j , (ÿ1) k ) for 0 < i < m ÿ 1, 0 < j < 1, 0 < k < 1. Let x (c( m1)=2 , ÿ1) and y (d, 1). Check that x 2 m y 2 1, y ÿ1 xy x ÿ1 : By Example 1.4, the function W: D4 m ! D2 m 3 C2 de®ned by

397

398

Representations and characters of groups W: a i b j ! x i y j

(0 < i < 2m ÿ 1, 0 < j < 1)

is a homomorphism. Since Im W kx, yl, it contains x 2 (c, 1) and x m (1, ÿ1) and hence Im W D2 m 3 C2 . As |D4 m | |D2 m 3 C2 |, we conclude that W is an isomorphism. 6. (a) Let G kal and suppose that 1 6 H < G. First observe that there exists i . 0 such that ai 2 H. Choose k as small as possible such that k . 0 and a k 2 H. If 1 6 a j 2 H then j qk r for some integers q, r with 0 < r , k. Hence a r a j aÿqk a j (a k )ÿq 2 H. Since r , k, we have r 0. Therefore a j a kq and so H kak l; thus H is cyclic. (b) Assume that G hai and jGj dn. If g 2 G and g n 1, then g a j for some integer j and dnj jn, so dj j; hence g 2 kad l. It follows that f g 2 G: g n 1g had i, which is a cyclic group of order n. (c) If x and y are elements of order n in the ®nite cyclic group G, then x, y 2 H, where H { g 2 G: gn 1}. Now kxl and k yl have order n; also H has order n, by part (b). We deduce that hxi H h yi: Thus x 2 h yi, and so x is a power of y. 7. Let G be the set of non-zero complex numbers. If g, h 2 G then gh 6 0, so gh 2 G. If g, h, k 2 G then (gh)k g(hk); also 1 2 G and 1g g1 g for all g 2 G. Finally, if g 2 G then g ÿ1 1= g 2 G, and gÿ1 g ggÿ1 1. Thus G is a group under multiplication. If H is a subgroup of G of order n, then h n 1 for all h 2 H (since the order of h divides n, by Lagrange's Theorem). Therefore H < f g 2 G: g n 1g he2ði= n i: Since j Hj n jhe2ði= n ij, it follows that H ke2ði= n l. 8. Partition G into subsets f g, g ÿ1 g ( g 2 G). Each such subset has size 1 or 2, and the identity element is in a subset of size 1. Hence, if jGj is even then there exists g 2 G such that g 6 1 and the subset f g, g ÿ1 g has size 1; so g gÿ1 and g has order 2. 9. De®ne matrices A, B as follows: eið=4 0 A , 0 eÿið=4

B

0 ÿ1

1 : 0

Check that A8 I, B2 A4 and Bÿ1 AB Aÿ1 . These relations show that every element of the group kA, Bl has the form A j B k with 0 < j < 7, 0 < k < 1. Moreover, ei jð=4 0 0 ei jð=4 j , A Aj B : 0 eÿi jð=4 ÿeÿi jð=4 0 Since these matrices, with 0 < j < 7, are all distinct, kA, Bl has order 16.

Chapter 2

399

10. Suppose jG: Hj 2 and let g 2 G. If g 2 H then gÿ1 Hg H. And if g2 = H then H, Hg are the two right cosets of H in G, while H, gH are the two left cosets. Therefore Hg gH, and so gÿ1 Hg H again. Hence H v G.

Chapter 2 1. Let u, w 2 W and ë 2 F. Since W is a linear transformation, we have (uWÿ1 wWÿ1 )W (uWÿ1 )W (wWÿ1 )W u w, (ë(wWÿ1 ))W ë(wWÿ1 )W ëw: Hence (u w)Wÿ1 uWÿ1 wWÿ1 and (ëw)Wÿ1 ë(wWÿ1 ), so Wÿ1 is a linear transformation. 2. (1) ) (2): If W is invertible then W is injective, so Ker W {0}. (2) ) (3): If Ker W {0} then dim (Im W) dim V (by (2.12)), so Im W V (by (2.7)). (3) ) (1): Assume that Im W V, so W is surjective. By (2.12), Ker W {0}. If u, v 2 V and uW vW then (u ÿ v)W 0, so u ÿ v 2 Ker W {0}, and so u v. Thus W is injective. As W is surjective and injective, W is invertible. 3. First suppose that V U W. Then V U W. Let v 2 U \ W. Then v v 0 0 v and this gives us two expressions for v as the sum of an element in U and an element in W. Since such expressions are unique, v 0. Thus U \ W {0}. Now suppose that V U W and U \ W {0}. If u1 w1 u2 w2 with u1 , u2 2 U and w1, w2 2 W, then u1 ÿ u2 w2 ÿ w1 2 U \ W {0}; hence u1 u2 and w1 w2. This shows that V U W. 4. Assume ®rst that V U W. If v 2 V then v u w for some u 2 U and w 2 W; since u is a linear combination of u1 , . . . , ur and w is a linear combination of w1 , . . . , w s , it follows that v is a linear combination of u1 , . . . , u r , w1 , . . . , w s . Therefore u1 , . . . , ur , w1, . . . , ws span V. Suppose that ë1 u1 : : : ë r ur ì1 w1 : : : ì s ws 0 with all ë i , ì j in F. Since V U W, the expression 0 0 0 is the unique expression for 0 as the sum of vectors in U and W, and so ë1 u1 : : : ë r ur ì1 w1 : : : ì s ws 0: As u1 , . . . , ur are linearly independent, this forces ë i 0 for all i; similarly ì i 0 for all i. Therefore u1 , . . . , ur , w1, . . . , ws are linearly independent; hence they form a basis of V. Conversely, suppose that u1 , . . . , ur , w1, . . . , ws is a basis of V. If v 2 U \ W then v ë1 u1 . . . ë r ur ì1 w1 . . . ì s ws for some ë i , ì j 2 F; this gives ë i ì j 0 for all i, j, since u1 , . . . , ur , w1, . . . , ws are linearly independent. Thus v 0 and so U \ W {0}. It is easy to see that V U W, so by Exercise 3, V U W. 5. (a) Assume ®rst that V U1 U2 U3. Let u 2 U1 \ (U2 U3 ). Then u u1 u2 u3 for some ui 2 Ui (1 < i < 3). Since u1 0 0 0 u2 u3 and the sum U1 U2 U3 is direct, we have u1

400


u2 u3 0. Therefore U1 \ (U2 U3 ) {0}. Similarly, U2 \ (U1 U3 ) U 3 \ (U1 U2 ) f0g. Now suppose that U1 \ (U2 U3 ) U2 \ (U1 U3 ) U3 \ (U1 U2 ) {0}. Assume that ui , u9i 2 Ui (1 < i < 3) and u1 u2 u3 u91 u92 u93 . Then u1 ÿ u91 (u92 ÿ u2 ) (u93 ÿ u3 ) 2 U1 \ (U2 U3 ) {0}, so u1 u91 . Similarly, u2 u92 and u3 u93 . Therefore V U1 U2 U3. (b) Let V R2, and U1 sp ((1, 0)), U2 sp ((0, 1)), U3 sp ((1, 1)). 6. By Exercise 4, if V U W then dim V dim U dim W. More generally, if V U1 . . . Ur then V U1 (U2 . . . Ur ) (see (2.10)); by induction on r, dim (U2 . . . Ur ) dim U2 . . . dim Ur, so dim V dim U1 . . . dim Ur. 7. Let V R2. De®ne W, ö: V ! V by W: (x, y) ! (x, 0) and ö: (x, y) ! ( y, 0): Then Im W sp ((1, 0)), Ker W sp ((0, 1)), so V Im W Ker W; and Im ö Ker ö sp ((1, 0)), so V 6 Im ö Ker ö. 8. Suppose ®rst that W is a projection. Then V Im W Ker W by Proposition 2.32. Take a basis u1 , . . . , ur for Im W and a basis w1, . . . , ws for Ker W. Then u1 , . . . , ur , w1, . . . , ws is a basis, say B, of V, by Exercise 4. Since ui W ui for all i and wj W 0 for all j, the matrix [W]B is diagonal, the diagonal entries being r 1's followed by s 0's. Conversely, if [W]B has the given form, then clearly W2 W, so W is a projection. 9. Let v 2 V. Then v 12(v vW) 12(v ÿ vW): Observe that 12(v vW)W 12(vW v), so 12(v vW) 2 U. Similarly, ÿ vW) 2 W. Thus V U W. If v 2 U \ W then v vW ÿv, so v 0. Therefore V U W, by Exercise 3. The construction of the basis B is similar to that in Solution 8. 1 2(v

Chapter 3 1. First, suppose that r is a representation of G. Then I 1r (am )r (ar) m Am : Conversely, assume that Am I. Then (ai )r Ai for all integers i (including i . m ÿ 1 and i , 0). Therefore for all integers i, j, (a i a j )r (a i j )r A i j A i A j (a i r)(a j r), and so r is a representation. 2. Check that A3 B3 C 3 I. Hence by Exercise 1, each r j is a representation. The representations r2 and r3 are faithful, but r1 is not.

Chapter 3

401

3. De®ne r by (a i b j )r (ÿ1) j

(0 < i < n ÿ 1, 0 < j < 1):

It is easy to check that r is a representation of G. 4. (1) For all g 2 G, Iÿ1 (gr)I gr; hence r is equivalent to r. (2) If r is equivalent to ó then there is an invertible matrix T such that gó T ÿ1 ( gr)T for all g 2 G; then gr (T ÿ1 )ÿ1 (gó)T ÿ1, so ó is equivalent to r. (3) If r is equivalent to ó and ó is equivalent to ô, then there are invertible matrices S and T such that gó Sÿ1 ( gr)S and gô T ÿ1 (gó)T for all g 2 G; then gô (ST)ÿ1 (gr)(ST), so r is equivalent to ô. 5. Check that in each of the cases (1) S A, T B, (2) S A3 , T ÿB, (3) S ÿA, T B, (4) S C, T D, we have S 6 T 2 I, T ÿ1 ST S ÿ1 : It follows that each r k is a representation (see Example 1.4). The matrices A r Bs (0 < r < 5, 0 < s < 1) are all different, so r1 is faithful. Similarly r4 is faithful. But r2 and r3 are not faithful, since a2 r2 I and a3 r3 I. The representations r1 and r4 are equivalent: to see this, let T

1 1 : ÿi i

Then T ÿ1 (gr4 )T gr1 for all g 2 G. (To ®nd T, ®rst work out the eigenvectors of C.) If j 6 2, then a2 r j 6 I; hence r2 is not equivalent to any of the others. And if j 6 3, then a3 r j 6 I; hence r3 is not equivalent to any of the others. 6. De®ne the matrices A and B by 0

0 A @ ÿ1 0

1 1 0 0 0 A, 0 1

0

1 B @0 0

1 0 0 ÿ1 0 A: 0 1

Then the function r: a r b s ! A r Bs (0 < r < 3, 0 < s < 1) is a faithful representation of D8 ka, b: a4 b2 1, bÿ1 ab aÿ1 l. Compare Example 3.2(1). 7. By Theorem 1.10, G=Ker r Im r. But Im r < GL (1, F) and GL (1, F) is abelian. Therefore G=Ker r is abelian. 8. No: let G be any non-abelian group and let r be the trivial representation.

402

Representations and characters of groups Chapter 4

1. g

1 0

[ g]B 1

[ g]B 2

1 @0 0 0 1 @0 0

g

(1 2) 1

0 0 1 0A 0 1 1 0 0 1 0A 0 1

0

0 @1 0 0 1 @0 0

(2 3) 0

[ g]B 1

[ g]B 2

1 @0 0 0 1 @0 0

(1 3) 1

1 0 0 0A 0 1 1 0 0 ÿ1 0 A ÿ1 1

0

0 @0 1 0 1 @0 0

(1 2 3) 1

0 0 0 1A 1 0 1 0 0 0 1A 1 0

0

0 @0 1 0 1 @0 0

1 0 1 1 0A 0 0 1 0 0 1 ÿ1 A 0 ÿ1

(1 3 2) 1

1 0 0 1A 0 0 1 0 0 ÿ1 1 A ÿ1 0

0

0 @1 0 0 1 @0 0

1 0 1 0 0A 1 0 1 0 0 0 ÿ1 A 1 ÿ1

All the matrices [g]B 2 have the form 0 1 1 0 0 @ 0 j j A: 0 j j 2. Let g 2 Sn . For all u, v in V and ë in F, we have v g 2 V , v1 v, (ëv) g ë(v g), (u v) g ug v g: It remains to check (2) of De®nition 4.2. Let v 2 V and g, h 2 Sn . Assume ®rst that gh 2 An . Then v(gh) v; and (vg)h v, since either vg v vh (if g, h 2 An ) or vg ÿv vh (if g, h 2 = An ). Next, assume that gh 2 = An . Then v(gh) ÿv; and (vg)h ÿv, since one of g, h is in An and the other is not. We have now checked all the conditions in De®nition 4.2, so V is an FG-module. 0 1 0 1 0 1 0 0 0 0 1 0 B ÿ1 0 0 B 0 0C 0 0 1C C B C. 3. Let A B @ 0 0 0 ÿ1 A and B @ ÿ1 0 0 0A 0 0 1 0 0 ÿ1 0 0 Check that A4 I, B2 A2 , Bÿ1 AB Aÿ1 : Hence r: a i b j ! A i B j (0 < i < 3, 0 < j < 1) is a representation of Q8 over R. Let V R4. By Theorem 4.4(1), V becomes an RQ8 -module if we de®ne vg v( gr) for all v 2 V, g 2 Q8. If we put

Chapter 5

403

v1 (1, 0, 0, 0), v2 (0, 1, 0, 0), v3 (0, 0, 1, 0), v4 (0, 0, 0, 1), then for all i, v i a and v i b are as required in the question. 4. You may ®nd it helpful ®rst to check that if 0 0 1 B1 0 B B 1 M B .. B @ .

0

1

0C C C C C A

1

then MA is obtained from A by swapping the ®rst two rows, and AM is obtained from A by swapping the ®rst two columns. To solve the exercise, let g be the permutation in Sn which has the property that for all i, row i of B row ig of A: Let P be the n 3 n matrix ( pij ) de®ned by 8 if j ig, < 1, pij : 0, if j 6 ig: Then P is a permutation matrix, and the ij-entry of PA is n X pik akj a ig, j : k1

Hence PA B. If C is a matrix obtained from A by permuting the columns, then C AQ for some permutation matrix Q; the proof is similar to that for the rows.

Chapter 5 1. It is easy to verify that V is an FG-module. Let U be a non-zero FGsubmodule of V, and let (á, â) 2 U with (á, â) 6 (0, 0). Then (á, â) (á, â)a (á â, á â) 2 U, and (á, â) ÿ (á, â)a (á ÿ â, â ÿ á) 2 U. Since at least one of á â and á ÿ â is non-zero, we deduce that (1, 1) or (1, ÿ1) belongs to U. Hence the FG-submodules of V are {0}, sp ((1, 1)), sp ((1, ÿ1)) and V. 2. Suppose that r has degree n and r is reducible. Then r is equivalent to a representation ô of the form 0 1 Xg 0 A ( g 2 G) ô: g ! @ Yg Zg where X g is a k 3 k matrix and 0 , k , n. Then ó is equivalent to ô, since ó is equivalent to r. Therefore ó is reducible. 3. Let G D12 and let r1 , . . . , r4 be the representations of G de®ned in Exercise 3.5. First consider the FG-module V F 2, where vg v(gr1 ) for

404


v 2 V, g 2 G. Suppose that U is a non-zero FG-submodule of V. Then U is an FH-module, where H is the subgroup {1, b}. By the solution to Exercise 1, either (1, 1) or (1, ÿ1) lies in U. Since (1, 1) and (1, 1)a are linearly independent, and also (1, ÿ1) and (1, ÿ1)a are linearly independent, it follows that dim U > 2. Consequently U V and so V is irreducible. (See Example 5.5(2) for an alternative argument.) Therefore r1 is irreducible; also r4 is irreducible, since r1 and r4 are equivalent. Now let V F 2 with vg v(gr2 ) for v 2 V, g 2 G. Then (1, 1)a ÿ(1, 1) (1, 1)b. Hence sp ((1, 1)) is an FG-submodule of V and r2 is reducible. Finally, r3 is irreducible, by an argument similar to that for r1. 4. (a) It is easy to check the given relations. Using the relations, we may write every element of G in the form ai b j c k

(0 < i < 2, 0 < j < 2, 0 < k < 1):

Thus jGj < 18. However, it is clear Hence, by Lagrange's Theorem, jGj Therefore jGj 18. (b) Let å 0 ç A , B 0 å ÿ1 0

that |ka, bl| 9 and ka, bl 6 G. is a multiple of 9 and jGj . 9. 0 0 ,C çÿ1 1

1 : 0

Check that A3 B3 C 2 I, AB BA, C ÿ1 AC Aÿ1 and C ÿ1 BC Bÿ1 . Hence r is a representation (compare Example 1.4). (c) For every element g of ka, bl, there exists a cube root î of unity such that î 0 : gr 0 î ÿ1 But there are only three distinct cube roots of unity, so there exist distinct g1 , g2 2 ka, bl with g1 r g2 r. Therefore r is never faithful. (d) Let V C2 be the CG-module obtained by de®ning vg v(gr) for all v 2 C2 , g 2 G. If U is a non-zero CG-submodule of V, then U is a CHsubmodule, where H is the subgroup {1, c}. Hence either (1, 1) or (1, ÿ1) lies in U, by the solution to Exercise 1; accordingly, let u be (1, 1) or (1, ÿ1) (so that u 2 U). Now u and ua are linearly independent unless å 1, and u and ub are linearly independent unless ç 1. Hence, if either å 6 1 or ç 6 1 then dim U 2 and so r is irreducible. On the other hand, if å ç 1 then sp ((1, 1)) is a CG-submodule of V, so r is reducible. 5. Let V {0} and let 0 g 0 for all g 2 G. Then V is neither reducible nor irreducible.

Chapter 6 1: (a)

xy ÿ2:1 ÿ a3 ab 3a2 b 2a3 b, yx ÿ2:1 ÿ a3 b 2a2 b 3a3 b, x 2 4:1 a2 4a3 :

Chapter 7

405

(b) az ab a3 b a2 ba ba za, and bz 1 a2 zb. Hence a i b j zP za i b j for all i, j and so gzP zg for all P g 2 G. If r 2 CG then r ë g gz ë g zg zr. g ë g g with ë g 2 C, so rz 2. Let C2 3 C2 ka, b: a2 b2 1, ab bal. Relative to the basis 1, a, b, ab of F(C2 3 C2 ), the regular representation r is given by 0 1 0 1 0 1 0 0 1 0 0 0 B1 0 0 0C B0 1 0 0C B C B C 1r B C, C, ar B @0 0 0 1A @0 0 1 0A 0 0 1 0 0 0 0 1 0 1 0 1 0 0 0 1 0 0 1 0 B0 0 1 0C B0 0 0 1C B C B C br B C: C, (ab)r B @0 1 0 0A @1 0 0 0A 1 0 0 0 0 1 0 0 3. No: let G ka: a2 1l, and take r 1 a, s 1 ÿ a. 4. (a) As g runs P throughPG, so do gh and hg. Hence ch hc c. (b) c2 c h2G h h2G ch jGjc. (c) All the entries in [W]B are 1 (compare the solution to Exercise 2). The reason is that for all i, j, there exists a unique h in G such that gi h gj . 5. Note ®rst that if u is an element of a vector space, and u u u, then u 0. Now 0r (0 0)r 0r 0r, and v0 v(0 0) v0 v0; hence 0r v0 0. Let V be the trivial FG-module and let 0 6 v 2 V and 1 6 g 2 G. If r 1 ÿ g, then vr 0 and neither v nor r is 0. 6. Let v1 1 ù2 a ùa2 and v2 b ù2 ab ùa2 b. Check that v1 a ùv1 , v2 a ù2 v2 , v1 b v2 and v2 b v1. Hence W is a CG-submodule of CG. Use the argument of either Example 5.5(2) or the solution to Exercise 5.3 to prove that W is irreducible.

Chapter 7 1. For all u1 , u2 2 U, ë 2 F and g 2 G, we have (u1 u2 )Wö (u1 W u2 W)ö u1 (Wö) u2 (Wö), (ëu1 )Wö (ë(u1 W))ö ë(u1 (Wö)), (u1 g)Wö ((u1 W) g)ö ((u1 W)ö) g (u1 (Wö)) g: 2. Let a (1 2 3 4 5) and let v1 , . . . , v5 be the natural basis for the permutation module for G over F. Then W: ë1 v1 : : : ë5 v5 ! ë1 1 ë2 a ë3 a2 ë4 a3 ë5 a4 is the required FG-isomorphism. (Note that v i W ai , so (v i a)W v i1 W a i1 (v i W)a; hence W is an FG-homomorphism.) 3. It is easy to show that V0 is an FG-submodule of V. Let x 2 G. Then

406

Representations and characters of groups X

vxg

g2G

X

vg

g2G

X

v gx:

g2G

Hence (vx)W vW (vW)x; noting that V W V0 , we see that W is an FGhomomorphism from V to V0 . If v 2 V0 then (v=jGj)W v; hence W is surjective. 4. Suppose that ö: V ! W is an FG-isomorphism. Let g 2 G. For all v 2 V0 , (vö)g (vg)ö vö, and so V0 ö W 0 . For all w 2 W 0 , (wöÿ1 )g (wg)öÿ1 wöÿ1 , so W 0 öÿ1 V0 . Hence the function ö, restricted to V0 , is an FG-isomorphism from V0 to W 0 . 5. No: let v1 , . . . , v4 be the natural basis of the permutation module V for G over F. In the notation of Exercise 3, X ! V0 sp (v1 v2 , v3 v4 ) and (FG)0 sp g : g2G

Since V0 and (FG)0 have different dimensions, it follows from Exercise 4 that V and FG are not isomorphic FG-modules. 6. (a) W is easily seen to be a linear transformation. Also (á1 âx)xW (â1 áx)W (â ÿ á)(1 ÿ x) (á ÿ â)(1 ÿ x)x (á1 âx)Wx: Hence W is an FG-homomorphism. (b) (á ÿ â)(1 ÿ x)W ((á ÿ â) ÿ (â ÿ á))(1 ÿ x) 2(á ÿ â)(1 ÿ x). Hence W2 2W. (c) Let B be the basis 1 ÿ x, 1 x.

Chapter 8 2

1. V sp (ÿùv1 v2 ) sp (ÿù v1 v2 ), where ù e2ði=3 . (Find eigenvectors for x.) 2. Let G {1, a, b, ab} C2 3 C2 (so a2 b2 1, ab ba). Then RG sp (1 a b ab) sp (1 a ÿ b ÿ ab) sp (1 ÿ a b ÿ ab) sp (1 ÿ a ÿ b ab): 3. Let G be any group, and let V be a 2-dimensional vector space over C with basis v1 , v2 . De®ne vg v for all v 2 V, g 2 G; this makes V into a CGmodule. If we let W: ëv1 ìv2 ! ëv2

(ë, ì 2 C)

then W is a CG-homomorphism from V to V, and Ker W Im W sp (v2 ). 4. Suppose r is reducible. Then by Maschke's Theorem, r is equivalent to a representation ó of the form ëg 0 gó (ë g , ì g 2 C): 0 ìg Then (gó)(hó) (hó)(gó) for all g, h 2 G, since all diagonal matrices

Chapter 9

407

commute with each other; hence also (gr)(hr) (hr)(gr) for all g, h 2 G. This is a contradiction. Therefore r is irreducible. 5. U sp ((1, 0)) is the only 1-dimensional CG-submodule of V, so there is no CG-submodule W of V with V U W. 6. (1) It is straightforward to verify for [ , ] the axioms of a complex inner product. For example, if u 6 0 then (ux, ux) . 0 for all x 2 G, so [u, u] . 0. Also X X [ug, v g] (ugx, v gx) (ux, vx) [u, v]: x2G

(2) It is easy to prove that U Then for all u 2 U,

x2G ?

is a subspace of V. Let v 2 U ? and g 2 G.

[u, v g] [ug ÿ1 , v gg ÿ1 ]

by part (1)

ÿ1

[ug , v] 0 since ug ÿ1 2 U : Therefore vg 2 U ?, and so U ? is a CG-submodule of V. (3) Let W U ?. Then V U W, and W is a CG-submodule of V by part (2). 7. We know that the regular CG-module CG is faithful (Proposition 6.6). Let CG U1 . . . Ur, where U1, . . . , Ur are irreducible CG-submodules of CG. Then there exist i 2 {1, . . . , r} and g 2 G such that ug 6 u for some u 2 Ui (otherwise vg v for all v 2 CG). De®ne K fx 2 G: vx v for all v 2 Ui g: Check that K is a normal subgroup of G; also K 6 G since g 2 = K. Since G is simple, we must therefore have K {1}. This means that Ui is a faithful irreducible CG-module.

Chapter 9 2

1. Let C2 ka: a 1l. Irreducible representations r1 , r2 : 1r1 ar1 (1); 1r2 (1), ar2 (ÿ1): Let C3 kb: b3 1l and let ù e2ði=3 . Irreducible representations r1 , r2 , r3 : 1r1 br1 b2 r1 (1); bi r2 (ù i ); bi r3 (ù2i ): Let C2 3 C2 {(1, 1), (x, 1), (1, y), (x, y)}, where x 2 y 2 1. Irreducible representations r1 , r2 , r3 , r4 :

408

Representations and characters of groups gr1 (1) for all g 2 C2 3 C2 ; (x i , y j )r2 (ÿ1) j ; (x i , y j )r3 (ÿ1) i ; (x i , y j )r4 (ÿ1) i j :

2. Let C4 3 C4 k(x, 1), (1, y): x 4 y 4 1l. (a) r: (x i , y j ) ! (ÿ1) i . (b) If g1 (x 2 , 1) and g2 (1, y 2 ) then g1 , g2 and g1 g2 all have order 2. Since (g1 g2 )ó (g1 ó)(g2 ó) for all representations ó, we cannot have (g1 g2 )ó g1 ó g2 ó (ÿ1). 3. For 1 < j < r, let gj generate Cn j , and let å j e2ði= n j . Then 0 i 1 å11 B C .. C r: ( g 1i1 , : : : , g irr ) ! B @ A . å irr

0

0

is a faithful representation of C n1 3 . . . 3 C n r of degree r. Yes: if r 2, n1 2, n2 3, then ó : ( g 1i1 , g 2i2 ) ! (å1i1 å2i2 ) is a faithful representation of degree 1 , r. 4. Check that A4 B2 I, Bÿ1 AB Aÿ1 when A ar and B br. Hence r gives a representation; similarly for ó. If M(gr) (gr)M for g a and for g b, then M ëI for some ë 2 C. Hence r is irreducible (Corollary 9.3). Notice that the matrix 5 ÿ6 4 ÿ5 commutes with gó for all g 2 G; hence ó is reducible (Corollary 9.3). P 5. Let z g2G g. Then xz z zx for all x 2 G. Hence z 2 Z(CG), and the result follows from Proposition 9.14. 6. (a) Clearly a commutes with a aÿ1 . Also bÿ1 (a aÿ1 )b aÿ1 a, so b commutes with a aÿ1 . (b) Check that w(a aÿ1 ) ÿw for all w 2 W. 7. (a) Let Cn kx: x n 1l. Then r: x j ! (e2ði j= n ) is a faithful irreducible representation of C n . 0 1 1 0 , b! (b) r: a ! ÿ1 0 0 ÿ1 gives a faithful irreducible representation of D8 (see Example 5.5(2)). (c) The centre of C2 3 D8 is isomorphic to C2 3 C2 , so is not cyclic. Therefore Proposition 9.16 shows that C2 3 D8 has no faithful irreducible representation. (d) Let C3 kx: x 3 1l and let ù e2ði=3 . Check that 0 ù ù 0 r: (x, a) ! , (x, b) ! ÿù 0 0 ÿù

Chapter 10

409

gives a representation of C3 3 D8. It is irreducible (see for example Exercise 8.4) and faithful.

Chapter 10

P 1. Let V sp ( g2G g). Then V is a trivial CG-submodule of CG. Now suppose that U is an arbitrary trivial CG-submodule P of CG, so U sp (u) for some u. Then ug u for all g 2 G, so |G|u u( g2G g) P ( g2G g)u 2 V. Thus U V, and so CG has exactly one trivial CGsubmodule, namely V. 2. Let G kx: x 4 1l. Then CG sp (1 x x 2 x 3 ) sp (1 ix ÿ x 2 ÿ ix 3 ) sp (1 ÿ x x 2 ÿ x 3 ) sp (1 ÿ ix ÿ x 2 ix 3 ): 3. Let u1 1 a a2 a3 ÿ b ÿ ab ÿ a2 b ÿ a3 b, u2 1 ÿ a a2 ÿ a3 b ÿ ab a2 b ÿ a3 b, u3 1 ÿ a a2 ÿ a3 ÿ b ab ÿ a2 b a3 b: 4. We decompose CG as a direct sum of irreducible CG-submodules. Let v0 1 a a2 a3 ,

v1 1 ia ÿ a2 ÿ ia3 ,

v2 1 ÿ a a2 ÿ a3 ,

v3 1 ÿ ia ÿ a2 ia3

(compare the solution to Exercise 2). For 0 < j < 3, let wj bv j. Then, as in Example 10.8(2), the subspaces sp (v0 , w0 ), sp (v1 , w3 ), sp (v2 , w2 ) and sp (v3 , w1 ) are CG-submodules of CG. We have sp (v0 , w0 ) U0 U1 , sp (v2 , w2 ) U2 U3 , where Ui sp (ui ) (0
p3 , then by the class equation, p2 divides |Z(G)|. This is a contradiction.

Chapter 13 1. The characters ÷ i of r i (i 1, 2) are as follows:

÷1 ÷2

1

a3

a, a5

a2 , a4

2 2

2 0

ÿ1 0

ÿ1 2

b, a2 b, a4 b ab, a3 b, a5 b 0 0

0 ÿ2

Also Ker r1 {1, a3 } and Ker r2 {1, a2 , a4 }. 2. Let C4 kx: x 4 1l. The irreducible characters ÷1 , . . . , ÷4 of C4 are as follows:

÷1 ÷2 ÷3 ÷4

1

x

x2

x3

1 1 1 1

1 i ÿ1 ÿi

1 ÿ1 1 ÿ1

1 ÿi ÿ1 i

We have ÷reg ÷1 ÷2 ÷3 ÷4 . 3. Since ÷(g) |®x (g)|, we have ÷((1 2)) 5 and ÷((1 6)(2 3 5)) 2. 4. If ÷ is a non-zero character which is a homomorphism, then ÷(1) ÷(12 ) (÷(1))2 , so ÷(1) 1. 5. Let r be a representation with character ÷. Then zr ëI for some ë 2 C, by Proposition 9.14. Thus, for all g in G, (zg)r (zr)(gr) ë(gr), and hence ÷(zg) ë÷(g). Moreover, I 1r z m r (zr) m ë m I, so ë m 1. 6. Let r be a representation with character ÷. If g 2 Z(G) then gr ëI for some ë 2 C, by Proposition 9.14. Conversely, if gr ëI for some ë 2 C, then ( gr)(hr) (hr)(gr) for all h 2 G, and hence g 2 Z(G) since r is faithful. We have now proved that gr ëI for some ë 2 C if and only if g 2 Z(G). The required result now follows from Theorem 13.11(1). 7. (a) For all g, h 2G, det ((gh)r) det ((gr)(hr)) det (gr) det (hr). Hence g ! (det (gr)) is a representation of G over C of degree 1, and so ä is a linear character of G. (b) G=Ker ä Im ä by Theorem 1.10, and Im ä is a subgroup of the

Chapter 14

413

multiplicative group C of non-zero complex numbers, which is abelian. Therefore G=Ker ä is abelian. (c) Im ä is a ®nite subgroup of C , hence is cyclic, by Exercise 1.7. Also ÿ1 2 Im ä, so Im ä has even order. Hence Im ä contains a subgroup H of index 2. It is easy to check that {g 2 G: ä(g) 2 H} is a normal subgroup of G of index 2. 8. Let r be the regular representation of G, and de®ne the character ä as in Exercise 7. By Exercise 1.8, G has an element x of order 2. Order the natural basis g1 , . . . , g2 k of CG so as to obtain a basis B in which g and gx are adjacent for all g 2 G. Then 1 0 0 1 C B1 0 C B C B 0 1 [x]B B C: C B 1 0 A @ .. .

0

0

(01 10 ),

There are k blocks and since k is odd, det ([x]B ) (ÿ1) k ÿ1. Thus ä(x) ÿ1. The required result now follows from Exercise 7. 9. Let V be a CG-module with character ÷. We may choose a basis B of V so that [g]B is diagonal with all diagonal entries 1 (see (9.10)). Say there are r entries 1 and s entries ÿ1. De®ne ä as in Exercise 7. If s is odd then ä( g) ÿ1, and G has a normal subgroup of index 2 by Exercise 7. And if s is even then ÿs s mod 4, so ÷( g) r ÿ s r s ÷(1) mod 4. 10. As x 6 1, we have ÷reg (x) 6 ÷reg (1) (see Proposition 13.20), so ÷ i (x) 6 ÷ i (1) for some irreducible character ÷ i of G, by Theorem 13.19.

Chapter 14 1. Using Proposition 14.5(2), we obtain 3 . 3 (ÿ1)(ÿ1) 3 . 3 (ÿ1)(ÿ1) h÷, ÷i 0 2, 24 4 8 4 3 . 3 (ÿ1) . 1 3 . (ÿ1) (ÿ1)(ÿ1) h÷, øi 0 0, 24 4 8 4 3.3 1.1 (ÿ1)(ÿ1) (ÿ1)(ÿ1) hø, øi 0 1: 24 4 8 4 Hence ø is irreducible by Theorem 14.20 (but ÷ is not). 2. Let ÷ i be the character of r i (i 1, 2, 3). The values of these characters are as follows: Conjugacy class ÷1 ÷2 ÷3

1

a2

a, a3

b, a2 b

ab, a3 b

2 2 2

ÿ2 ÿ2 2

0 0 0

0 0 0

0 0 ÿ2

414


By Theorem 14.21, r1 and r2 are equivalent, but r3 is not equivalent to r1 or r2. 3. The representations r and ó have the same character, by Proposition 13.2; hence r and ó are equivalent, by Theorem 14.21, and this gives the required matrix T. 4. Let ÷1 be the trivial character of G. Then 1 X h÷, ÷1 i ÷( g): jGj g2G Since ÷(1) . 0 and by hypothesis ÷(g) > 0 for all g 2 G, we have k÷, ÷1 l 6 0. As ÷ 6 ÷1 , Theorem 14.17 shows that ÷ is reducible. 5. We have h÷reg , ÷i

1 X ÷reg ( g)÷( g): jGj g2G

But ÷reg (g) is |G| if g 1 and is 0 if g 6 1. Hence k÷reg , ÷l ÷(1). 6. This follows at once from Exercise 11.4 and Theorem 14.24. Pk 2 7. Recall that hø, øi i1 d i . Hence if kø, øl a where a 1, 2 or 3, then exactly a of the integers di are 1 and the rest are 0. If kø, øl 4, then either exactly four of the di are 1, or exactly one of the di is 2; the rest are 0. 8. No: let G C2 and ÷ ÷reg , the regular character of C2 .

Chapter 15 1:

h÷, ÷1 i h÷, ÷2 i h÷, ÷3 i

1 6(19 1 6(19 1 6(19

. 1 3 . (ÿ1) . 1 2 . (ÿ2) . 1) 2, . 1 3 . (ÿ1)(ÿ1) 2 . (ÿ2) . 1) 3, . 2 0 2 . (ÿ2)(ÿ1)) 7:

Hence ÷ 2÷1 3÷2 7÷3 . Since all the coef®cients here are non-negative integers, it follows that ÷ is a character of S3 . 2. By the method used in the solution to Exercise 1, we obtain ø1 16 ÷1 16 ÷2 13 ÷3 , ø2 12 ÷1 ÿ 12 ÷2 , ø3 13 ÷1 13 ÷2 ÿ 13 ÷3 : 3. We ®nd that ø ÿ÷2 ÷3 ÷5 2÷6 . Since the coef®cient of ÷2 is a negative integer, ø is not a character of G. 4. (a) For all groups G, if x 2 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). Hence, if G Z(G) [ Z(G)x then G Z(G). Therefore the centre of a group G never has index 2 in G. Every abelian group of order 12 has 12 conjugacy classes. If |G| 12

Chapter 16

415

and G is non-abelian, then |Z(G)| divides 12 and |Z(G)| 6 6 or 12, so |Z(G)| < 4. Therefore, at most 4 conjugacy classes of G have size 1 (see (12.9)); the remaining conjugacy classes have size at least 2, so there cannot be as many as 9 conjugacy classes in total. (b) Since the number of irreducible representations is equal to the number of conjugacy classes, it follows from the solution to Exercise 11.2 and part (a) that G has 4, 6 or 12 conjugacy classes. If G is abelian (e.g. G C4 3 C3 ) then G has 12 conjugacy classes; if G D12 then G has 6 conjugacy classes (see (12.12)); and if G A4 then G has 4 conjugacy classes (see Example 12.18(1)).

Chapter 16 1. Let C2 3 C2 {(1, 1), (x, 1), (1, y), (x, y): x 2 y 2 1}. The character table of C2 3 C2 is (cf. Exercise 9.1) (1, 1)

(x, 1)

(1, y)

(x, y)

1 1 1 1

1 1 ÿ1 ÿ1

1 ÿ1 1 ÿ1

1 ÿ1 ÿ1 1

÷1 ÷2 ÷3 ÷4

2. The last row of the character table is (cf. Example 16.5(2))

÷5

g1

g2

g3

g4

g5

2

ÿ2

0

0

0

3. The complete character table of G is gi |CG ( gi )| ÷1 ÷2 ÷3 ÷4

g1 10

g2 5

g3 5

g4 2

1 2 1 2

1p (ÿ1 5)=2 1p (ÿ1 ÿ 5)=2

1p (ÿ1 ÿ 5)=2 1p (ÿ1 5)=2

1 0 ÿ1 0

P The two unknown degrees ÷3 (1), ÷4 (1) are 1, 2 since 4i1 ( ÷ i (1))2 10. Because g4 has order 2, Corollary 13.10, together with the relation P4 P4 i1 ÷ i (1)÷ i ( g 4 ) 0, gives the values on g4 . Then i1 ÷ i ( g 2 )÷ i ( g 4 ) 0 P4 gives ÷3 ( g2 ) 1; similarly ÷3 (g3 ) 1. Finally, i1 ÷ i (1)÷ i ( g 2 ) 0 gives p p ÷4 ( g 2 ) (ÿ1 ÿ 5)=2; similarly ÷4 ( g3 ) (ÿ1 5)=2.

416


P 0 gives 3 3æ 3æ 0, and 5i1 ÷ i ( g 2 )÷ i ( g2 ) 7 gives 3 2ææ 7. p Hence æ (ÿ1 i 7)=2. (b) The column which corresponds to the conjugacy class containing gÿ1 2 has values which are the complex conjugates of those in the column of g2 (see Proposition 13.9(3)); since æ is non-real, this is a different column of the character table of G.

4. (a)

P5

i1 ÷ i ( g 1 )÷ i ( g 2 )

5. Let g 2 G. By the column orthogonality relations applied to the column Pk corresponding to g, we have i1 ÷ i ( g)÷ i ( g) jC G ( g)j. This number is equal to |G| if and only if CG (g) G, which occurs if and only if g 2 Z(G). 6. The matrix C is obtained from C by rearranging the columns (see Proposition 13.9(3)). Therefore det C det C; if det C det C then det C is real, and if det C ÿdet C then det C is purely imaginary. By the column orthogonality relations, C t C is the k 3 k diagonal matrix whose Q diagonal entries are |CG ( gi )| (1 < i < k). Hence jdet Cj2 jC G ( g i )j. p If G C3 then det C i3 3. (The sign depends upon the ordering of rows and columns.)

Chapter 17 1. (a) The conjugacy classes of Q8 are f1g, fa2 g, fa, a3 g, fb, a2 bg and fab, a3 bg. (b) G9 f1, a2 g and G=G9 fG9, G9a, G9b, G9abg C2 3 C2 . The character table of C2 3 C2 is given in the solution to Exercise 16.1. Hence the linear characters of G are gi |CG ( gi )|

1 8

a2 8

a 4

b 4

ab 4

÷1 ÷2 ÷3 ÷4

1 1 1 1

1 1 1 1

1 1 ÿ1 ÿ1

1 ÿ1 1 ÿ1

1 ÿ1 ÿ1 1

(c) Using the column orthogonality relations, the last irreducible character of G is

÷5

1

a2

a

b

ab

2

ÿ2

0

0

0

The character table of Q8 is the same as that of D8. 2. It is easy to see that a7 b3 1. Use Proposition 12.13 to see quickly that bÿ1 ab a2 . (a) Using the relations, every element of G has the form am bn with

Chapter 17

417

0 < m < 6, 0 < n < 2; hence jGj < 21. But a has order 7 and b has order 3, so 21 divides jGj by Lagrange's Theorem. Therefore jGj 21. (b) The conjugacy classes of G are f1g, fa, a2 , a4 g, fa3 , a5 , a6 g, fa m b: 0 < m < 6g and fa m b2 : 0 < m < 6g. (c) First, G9 kal, so we get three linear characters of G: gi |CG ( gi )| ÷1 ÷2 ÷3

1 21

a 7

a3 7

b 3

b2 3

1 1 1

1 1 1

1 1 1

1 ù ù2

1 ù2 ù

where ù e2ði=3 . To ®nd the two remaining irreducible characters ÷4 and ÷5 , we note that a is not conjugate to aÿ1 ; therefore for some irreducible character ÷, we have ÷(a) 6 ÷(a) (see Corollary 15.6). Hence ÷4 and ÷5 must be complex conjugates of each other. Applying the column orthogonality relations, we obtain

÷4 ÷5

1

a

a3

b

b2

3 3

á á

á á

0 0

0 0

p where á (ÿ1 i 7)=2. 3. The number of linear characters of G divides jGj by Theorem 17.11. Now consult the solution to Exercise 11.2 to see that there are 3, 4 or 12 linear characters. If there are 12, then G is abelian (see Proposition 9.18), so G is certainly not simple. If G has 3 or 4 linear characters then jG=G9j 3 or 4 and again G is not simple as G9 v G. 4. In the character table below, we have ÷1 1 G , ÷2 ÷, ÷3 ÷ 2 , ÷4 ÷2 ÷3 , ÷5 ö, ÷6 ö÷; all of these are irreducible characters by Proposition 17.14. The centralizer orders are obtained by using the orthogonality relations, and the class sizes j g G i j come from the equations jGj jC G ( g i )jj g G j (Theorem 12.8). i gi |CG ( gi )| | gG i | ÷1 ÷2 ÷3 ÷4 ÷5 ÷6

g1 12 1

g2 4 3

g3 4 3

g4 6 2

g5 6 2

g6 12 1

1 1 1 1 2 2

1 ÿi ÿ1 i 0 0

1 i ÿ1 ÿi 0 0

1 1 1 1 ÿ1 ÿ1

1 ÿ1 1 ÿ1 ÿ1 1

1 ÿ1 1 ÿ1 2 ÿ2

418


5. The normal subgroups of D8 are D8 Ker ÷1 , hai Ker ÷2 , ha2 , bi Ker ÷3 , ha2 , abi Ker ÷4 , ha2 i Ker ÷2 \ Ker ÷3 , f1g Ker ÷5 : 6. (a) Check that the given matrices satisfy the relevant !2 n !n ! å 0 1 0 å 0 , 0 å ÿ1 0 1 0 å ÿ1 ! !ÿ1 ! 0 1 å 0 0 1 å n ÿ1 n å 0 å å 0 0 0

relations: !2 0 1 ån 0

0 !ÿ1

å ÿ1

,

:

Hence we have representations of T4 n (cf. Example 1.4). (b) The representations in part (a) are irreducible unless å 1, by Exercise 8.4. For å e2ði r=2 n , with r 1, 2, . . . , n ÿ 1, we get n ÿ 1 irreducible representations, no two of which are equivalent, since they have distinct characters. Moreover G9 ka2 l, so jG=G9j 4 and there are four representations of degree 1 (see Theorem 17.11). Now the sum of the squares of the degrees of the irreducible representations we have found so far is (n ÿ 1) . 22 4 . 12 4n: Hence we have found all the irreducible representations, by Theorem 11.12. (For further details on the representations of degree 1, see the solution to Exercise 18.3; note that the structure of G=G9 depends upon whether n is even or odd.) 7. (a) Check that the given matrices satisfy the relevant relations. (b) The given representations, for å e2ði k=2 n with 0 < k < n ÿ 1, are irreducible (by Exercise 8.4), and inequivalent (consider the character values on a2 ). Also G9 kbl, so jG=G9j 2n and there are 2n representations of degree 1. The sum of the squares of the degrees of the irreducible representations we have found is n . 22 2n . 12 6n, so we have obtained all the irreducible representations. 8. (a) Check that the given matrices satisfy the relevant relations. (b) The given representations, for å e2ði j= n with 0 < j < n ÿ 1, are irreducible (by Exercise 8.4) and inequivalent (their characters are distinct). Note that b2 does not belong to the kernel of any of these representations. We get further representations by ç 0 0 1 a! ,b! , 0 çÿ1 1 0 where ç is any (2n)th root of unity in C. For ç e2ði j=2 n with 1 < j < n ÿ 1, these representations are irreducible and inequivalent. Moreover, they are not equivalent to any representation found earlier, since b2 is in the kernel of each of these representations.

Chapter 18

419

Finally, G9 ka2 , b2 l and G=G9 C2 3 C2, so we get four representations of degree 1. We have now found all the irreducible representations, since the sum of the squares of the irreducible representations given above is n . 22 (n ÿ 1) . 22 4 . 12 8n:

Chapter 18 1. The character table of D8 is as shown. Character table of D8 gi |CG ( gi )|

1 8

a2 8

a 4

b 4

ab 4

÷1 ÷2 ÷3 ÷4 ÷5

1 1 1 1 2

1 1 1 1 ÿ2

1 1 ÿ1 ÿ1 0

1 ÿ1 1 ÿ1 0

1 ÿ1 ÿ1 1 0

(See Example 16.3(3) or Section 18.3.) Regarding D8 as the symmetry group of a square, take b to be a re¯ection in a diagonal of the square. Then ð takes the following values:

ð

1

a2

a

b

ab

4

0

0

2

0

Hence ð ÷1 ÷3 ÷5 . (Compare Example 14.28(2), where we took b to be a different re¯ection.) 2. Let ù e2ði=6 . Then ù ùÿ1 1, ù2 ùÿ2 ù4 ùÿ4 ÿ1. Hence, using Section 18.3, the character table of D12 is as shown. Character table of D12 gi |CG ( gi )|

1 12

a3 12

a 6

a2 6

b 4

ab 4

÷1 ÷2 ÷3 ÷4 ÷5 ÷6

1 1 1 1 2 2

1 1 ÿ1 ÿ1 ÿ2 2

1 1 ÿ1 ÿ1 1 ÿ1

1 1 1 1 ÿ1 ÿ1

1 ÿ1 1 ÿ1 0 0

1 ÿ1 ÿ1 1 0 0

420


Seven normal subgroups of D12 are G Ker ÷1 , kal Ker ÷2 , ka2 , bl Ker ÷3 , ka2 , abl Ker ÷4 , ka2 l Ker ÷3 \ Ker ÷4 , ka3 l Ker ÷6 and {1} Ker ÷5 . 3. The n 3 conjugacy classes of G are f1g, fan g, far , aÿ r g(1 < r < n ÿ 1), fa2 j b: 0 < j < n ÿ 1g, fa2 j1 b: 0 < j < n ÿ 1g: We get n ÿ 1 irreducible characters ø j (1 < j < n ÿ 1) of G from Exercise 17.6 as follows: gi |CG ( gi )|

1 4n

an 4n

ar (1 < r < n ÿ 1) 2n

b 4

ab 4

øj

2

2(ÿ1) j

ù rj ùÿ rj

0

0

where ù e2ði=2 n . The remaining four irreducible characters of G are linear. If n is odd, then G=G9 hG9bi C4 and the linear characters are gi

1

an

ar (1 < r < n ÿ 1)

b

ab

÷1 ÷2 ÷3 ÷4

1 1 1 1

1 ÿ1 1 ÿ1

1 (ÿ1) r 1 (ÿ1) r

1 i ÿ1 ÿi

1 ÿi ÿ1 i

If n is even, then G=G9 C2 3 C2 and the linear characters are gi

1

an

÷1 ÷2 ÷3 ÷4

1 1 1 1

1 1 1 1

ar (1 < r < n ÿ 1)

b

ab

1 1 (ÿ1) r (ÿ1) r

1 ÿ1 1 ÿ1

1 ÿ1 ÿ1 1

Note that T4 C4 , T8 Q8 and T12 is the Example in Section 18.4. 4. The 3n conjugacy classes of G are, for 0 < r < n ÿ 1, fa2 r g, fa2 r b, a2 r b2 g, fa2 r1 , a2 r1 b, a2 r1 b2 g: We have G9 hbi and G=G9 hG9ai C2 n . Hence we get 2n linear characters ÷ j (0 < j < 2n ÿ 1), as shown. Exercise 17.7 gives us n irreducible characters ø k (0 < k < n ÿ 1).

Chapter 18

421

Character table of U6 n gi |CG ( gi )|

a2 r 6n

a2 r b 3n

a2 r1 2n

÷j (0 < j < 2n ÿ 1) øk (0 < k < n ÿ 1)

ù2 jr

ù2 jr

ù j(2 r1)

2ù2 kr

ÿù2 kr

0

Note: ù e2ði=2 n . Observe that U6 D6, U12 T12 and U18 D6 3 C3 . 5. The 2n 3 conjugacy classes of G are f1g, fb2 g, fa2 r1 , aÿ2 rÿ1 b2 g(0 < r < n ÿ 1), fa2s , aÿ2s g, fa2s b2 , aÿ2s b2 g(1 < s < (n ÿ 1)=2), faj bk : j even, k 1 or 3g, and faj bk : j odd, k 1 or 3g: Using Exercise 17.8, we get four linear characters ÷1 , . . . , ÷4 , n characters ø j (0 < j < n ÿ 1) of degree 2, and a further n ÿ 1 characters ö j (1 < j < n ÿ 1) of degree 2, as shown below. For example, the character table of V24 is given at the top of p. 422.

Character table of V8 n

8n

a2 r1 (0 < r < n ÿ 1) 8n 4n

a2s a2s b2 (1 < s < (n ÿ 1)=2) 4n 4n

1 1 1 1 2

1 1 1 1 ÿ2

2

2

1 1 1 1 ù4 js ùÿ4 js ù2 js ùÿ2 js

gi

1

|CG ( gi )| ÷1 ÷2 ÷3 ÷4 øj (0 < j < n ÿ 1) öj (1 < j < n ÿ 1)

Note: ù e2ði=2 n

b2

1 1 ÿ1 ÿ1

ù2 j(2 r1) ÿùÿ2 j(2 r1) ù j(2 r1) ùÿ j(2 r1)

1 1 1 1 ÿù4 js ÿùÿ4 js ù2 js ùÿ2 js

b

ab

4

4

1 ÿ1 1 ÿ1 0

1 ÿ1 ÿ1 1 0

0

0

422


Character table of V24 gi |CG ( gi )| ÷1 ÷2 ÷3 ÷4 ø0 ø1 ø2 ö1 ö2

1 24

b2 24

a 12

a3 12

a5 12

a2 12

a2 b2 12

b 4

ab 4

1 1 1 1 2 2 2 2 2

1 1 1 1 ÿ2 ÿ2 ÿ2 2 2

1 1 ÿ1 ÿ1 p0 i p3 ÿi 3 1 ÿ1

1 1 ÿ1 ÿ1 0 0 0 ÿ2 2

1 1 ÿ1 ÿ1 0 p ÿip 3 i 3 1 ÿ1

1 1 1 1 2 ÿ1 ÿ1 ÿ1 ÿ1

1 1 1 1 ÿ2 1 1 ÿ1 ÿ1

1 ÿ1 1 ÿ1 0 0 0 0 0

1 ÿ1 ÿ1 1 0 0 0 0 0

Chapter 19 1:

h÷ø, öi

1 X 1 X ÷( g)ø( g)ö( g) ÷( g)ø( g)ö( g) h÷, øöi: jGj g2G jGj g2G

Similarly, k÷ø, öl kø, ÷öl. 2. k÷ø, 1 G l k÷, øl, by Exercise 1. The result now follows from Proposition 13.15 and (14.13). 3. Let V be a CG-module with character ÷. Since ÷ is not faithful, there exists 1 6 g 2 G with vg v for all v 2 V. By Proposition 15.5 there is an irreducible character ø of G such that ø(g) 6 ø(1). Let n be an integer with n > 0. Then wg w for all w 2 V . . . V (n factors). Hence ö( g) ö(1) for all irreducible characters ö for which k÷ n , öl 6 0. Therefore k÷ n , øl 0. 4. Note that (1 2 3 4 5)2 is conjugate to (1 3 4 5 2) in A5 . Using Proposition 19.14 we obtain

÷S ÷A öS öA

1

(1 2 3)

(1 2)(3 4)

(1 2 3 4 5)

(1 3 4 5 2)

15 10 6 3

0 1 0 0

3 ÿ2 2 ÿ1

0 0 1 p

0 0 1 p

Then ÷ S ø1 ø2 2ø3 , ÷ A ø2 ø4 ø5 , ö S ø1 ø3 , ö A ø4 :

(1

5)=2

(1 ÿ

5)=2

Chapter 20

423

5. We have recorded ÷ as ÷5 , ÷ S as ÷4 and ÷ A as ÷2 , below. Since k÷ i , ÷ i l 1 for i 2, 4, 5, these characters are irreducible. The table also records the trivial character ÷1 , ÷3 ÷2 , ÷6 ÷5 and ÷7 ÷2 ÷5 ; these are irreducible by Propositions 13.15 and 17.14. Since G has seven conjugacy classes, the character table is complete. Character table of G (cf. Exercise 27.2) gi |CG ( gi )|

g1 24

g2 24

g3 4

g4 6

g5 6

g6 6

g7 6

÷1 ÷2 ÷3 ÷4 ÷5 ÷6 ÷7

1 1 1 3 2 2 2

1 1 1 3 ÿ2 ÿ2 ÿ2

1 1 1 ÿ1 0 0 0

1 ù ù2 0 ÿù2 ÿù ÿ1

1 ù2 ù 0 ÿù ÿù2 ÿ1

1 ù2 ù 0 ù ù2 1

1 ù ù2 0 ù2 ù 1

Note: ù e2ði=3 6. Taking D6 ka, b: a3 b2 1, bÿ1 ab aÿ1 l, the character table of D6 3 D6 is as shown. Character table of D6 3 D6 ( gi , hj ) |CG ( gi , 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, 1) (a, 1) (b, 1) (1, a) (a, a) (b, a) (1, b) (a, b) (b, b) 36 18 12 18 9 6 12 6 4 1 1 2 1 1 2 2 2 4

1 1 ÿ1 1 1 ÿ1 2 2 ÿ2

1 ÿ1 0 1 ÿ1 0 2 ÿ2 0

1 1 2 1 1 2 ÿ1 ÿ1 ÿ2

1 1 ÿ1 1 1 ÿ1 ÿ1 ÿ1 1

1 ÿ1 0 1 ÿ1 0 ÿ1 1 0

1 1 2 ÿ1 ÿ1 ÿ2 0 0 0

1 1 ÿ1 ÿ1 ÿ1 1 0 0 0

1 ÿ1 0 ÿ1 1 0 0 0 0

Chapter 20 1. (a) Regard D8 as the subgroup of S4 which permutes the four corners of a square, as in Example 1.1(3). Take b to be the re¯ection in the axis shown:

424


Then a ! (1 2 3 4), b ! (1 3) gives the required isomorphism. (b) Take the irreducible characters ÷1 , . . . , ÷5 of S4 as in Section 18.1, and take the character table of H to be gi |CH ( gi )|

1 8

(1 3)(2 4) 8

(1 2 3 4) 4

(1 3) 4

(1 2)(3 4) 4

ø1 ø2 ø3 ø4 ø5

1 1 1 1 2

1 1 1 1 ÿ2

1 1 ÿ1 ÿ1 0

1 ÿ1 1 ÿ1 0

1 ÿ1 ÿ1 1 0

(see Example 16.3(3) or Section 18.3). We obtain ÷1 # H ø1 , ÷2 # H ø4 , ÷3 # H ø1 ø4 , ÷4 # H ø3 ø5 , ÷5 # H ø2 ø5 : 2. Let ÷1 , . . . , ÷11 be the irreducible characters of S6 , as in Example 19.17. Either by direct calculation, or using (20.13), we ®nd that the characters ÷ i # A6 (i 1, 3, 5, 7, 9) are distinct irreducible characters of A6 ; these give the characters ø1 , . . . , ø5 in our character table below. Also, k÷11 # A6 , ÷11 # A6 l 2. Arguing as in Example 20.14, we obtain from ÷11 # A6 the two irreducible characters which we have called ø6 and ø7 . 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 ÿ1 ÿ1 ÿ1

Note: á (1

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 ÿ2 1 1 0 0 5)=2, â (1 ÿ

1 0 0 ÿ1 0 á â p

1 0 0 ÿ1 0 â á

1 ÿ1 1 0 2 ÿ1 ÿ1

1 ÿ1 0 1 ÿ1 0 0

5)=2

3. Let ø1 , . . . , ø r be the irreducible characters of H. Then ÷ # H d1 ø1 . . . dr ø r for some non-negative integers di . Since each ø i has degree 1, the inequality (20.6) gives ÷(1) d 1 : : : d r < d 21 : : : d 2r < n: 4. The inequality k÷ # H, ÷ # Hl H < 3 follows at once from Proposition 20.5. Write d k÷ # H, ÷ # Hl H . For examples with d 1 or 2, take G S3 and H a subgroup of order 2, and ÷ an irreducible character of G of degree d. For an example with d 3, take G A4 , H V4 and ÷ an irreducible character of G of degree 3 (see Section 18.2).

Chapter 21

425

5. See (20.13). Since 20 occurs only once in the list of degrees for S7 , the restriction of the irreducible character of degree 20 to A7 must be the sum of two different irreducible characters of degree 10. From the remaining fourteen irreducible characters of S7 , upon restriction to A7 we get at least seven irreducible characters of A7 ; and we get precisely seven if and only if the restriction of each of the fourteen characters is irreducible. We are told that A7 has exactly nine conjugacy classes. Hence the irreducible characters of A7 have degrees 1, 6, 14, 14, 15, 10, 10, 21, 35:

Chapter 21 2

2

1. (a) Let u 1 ÿ a b ÿ a b. Then ua2 ÿu and ub u. Hence sp (u) is a C H-submodule of CH. (b) Since every element of G belongs to H or to Ha, the elements u and ua form a basis of U " G. (c) The character ø of U and the character ø " G of U " G are given by

ø

ø"G

1

a2

b

a2 b

1

ÿ1

1

ÿ1

1

a2

a

b

ab

2

ÿ2

0

0

0

Since hø " G, ø " Gi 1, the induced module U " G is irreducible. 2. Label the characters of H as follows:

ø1 ø2 ø3

1

(1 2 3)

(1 3 2)

1 1 1

1 ù ù2

1 ù2 ù

where ù e2ði=3 . (a) ÷1 # H ÷2 # H ø1 , ÷3 # H ø2 ø3 , ÷4 # H ÷5 # H ø1 ø2 ø3 . (b) Using the Frobenius Reciprocity Theorem, we obtain ø1 " G ÷1 ÷2 ÷4 ÷5 , ø2 " G ø3 " G ÷3 ÷4 ÷5 : 3. It is suf®cient to prove that if U is a CH-submodule of CH then dim (U " G) jG : Hjdim U . Let Hgj (1 < j < m) be the distinct right

426


cosets of H in G. Then U(CG) Ug1 . . . Ugm , where Ugj {ugj : u 2 U}. The sum Ug1 . . . Ug m is direct (since the elements in Ugj are linear combinations of elements in the right coset Hgj ). Also, dim (Ugj ) dim U (since u ! ugj (u 2 U) is a vector space isomorphism). Hence dim(U " G) dim(U (CG)) m dim U . 4. Let ö be an irreducible character of G. By using the Frobenius Reciprocity Theorem twice, together with the result of Exercise 19.1 (also twice), we obtain h(ø(÷ # H)) " G, öi G hø(÷ # H), ö # Hi H hø, (÷ö) # Hi H hø " G, ÷öi G h(ø " G)÷, öi G : Since this holds for all irreducible characters ö of G, we deduce from Theorem 14.17 that (ø(÷ # H)) " G (ø " G)÷. 5. The values of ö " G and ø " G are given by Proposition 21.23. On elements of cycle-shapes (1), (7) and (3, 3), the values are as follows, and on all other elements the values are zero. Cycle-shape

(1)

(7)

(3, 3)

ö"G ø"G

240 720

2 ÿ1

12 0

6. We have |G: H|ø(1) d1 ÷1 (1) . . . dk ÷ k (1), where d i hø " G, ÷ i i G hø, ÷ i # Hi H , by the Frobenius Reciprocity Theorem. Hence, since ø is irreducible, ÷ i # H di ø â where either â is a character of H or â 0. Thus ÷ i (1) > di ø(1), and therefore jG: Hjø(1) > (d 21 : : : d 2k )ø(1): The required result follows. 7. By applying the result of Exercise 6, we deduce, as in the proof of Proposition 20.9, that either (1) ø " G is irreducible, or (2) ø " G is the sum of two different irreducible characters of the same degree. Suppose ®rst that ø " G is irreducible, say ø " G ÷. Then ÷(1) 2ø(1) and k÷ # H, øl H 1, by the Frobenius Reciprocity Theorem; hence ÷ # H is reducible, say ÷ # H ø ö. Now suppose that ø9 is an irreducible character of H. We have hø9 " G, ÷i G 6 0 , hø9, ÷ # Hi H 6 0 , ø9 ø or ö: Thus If ø " G is irreducible, then for precisely one other irreducible character ö of H we have ø " G ö " G. (Compare Proposition 20.11.)

Chapter 22

427

Suppose next that ø " G is reducible, say ø " G ÷1 ÷2. Then ÷1 (1) ø(1) and k÷1 # H, øl H 1; hence ÷1 # H ø. Now suppose that ø9 is an irreducible character of H. We have hø9 " G, ÷1 i G 6 0 , hø9, ÷1 # Hi H 6 0 , ø9 ø: Thus If ø " G ÷1 ÷2 where ÷1 and ÷2 are irreducible characters of G, and ø9 is an irreducible character of H such that ø9 " G has ÷1 or ÷2 as a constituent, then ø9 ø. (Compare Proposition 20.12.)

Chapter 22 1. The number of linear characters divides 15 (Theorem 17.11), and the sum of the squares of the degrees of the irreducible characters is 15 (Theorem 11.12); moreover, each degree divides 15 (Theorem 22.11). Hence every irreducible character has degree 1, and so G is abelian by Proposition 9.18. 2. Use Theorems 11.12, 17.11 and 22.11 again. The degree of each irreducible character is 1 or 2, and if there are r characters of degree 1 and s of degree 2, then r divides 16, and r . 12 s . 22 16: Hence r 4 or 8 or 16, and r s 7 or 10 or 16. 3. (a) Since G is non-abelian, not every irreducible character has degree 1 (Proposition 9.18). This time, Theorems 11.12, 17.11 and 22.11 show that there are r irreducible characters of degree 1 and s irreducible characters of degree q, where r divides pq, 1 < s and r sq 2 pq: Hence r q and s ( p ÿ 1)/q. (b) jG9j p by Theorem 17.11. (c) The number of conjugacy classes of G is r s. (For more information on groups of order pq, see Chapter 25.) 4. (a) By hypothesis, there exist a, b 2 C such that ö(g) a for all g 6 1 and ö(1) a bjGj. Then ö a1 G b÷reg , since both sides of this equation take the same values on all elements of G. (b) We have h1 G , öi h÷reg , öi

1 (a bjGj (jGj ÿ 1)a) a b, and jGj 1 jGj(a bjGj) a bjGj: jGj

Since ö is a character, both k1 G , öl and k÷reg , öl are integers. (c) If ÷ is a non-trivial irreducible character of G, then kö, ÷l 2 Z and k1 G , ÷l 0. Hence kö ÿ a1 G , ÷l 2 Z. But kö ÿ a1 G , ÷l kb÷reg , ÷l bjGj÷(1)=jGj b÷(1).

428


(d) Since ÷(1) divides jGj, part (c) implies that b|G| is an integer. Therefore, by part (b), a, and hence also b, is an integer. 5. (a) If g 2 G then g has odd order, by Lagrange's Theorem. Therefore, if g2 1 then g 1. (b) For all g 2 G, we have ÷( g) ÷( g ÿ1 ) ÷( g) ÷( g) 2÷( g), and ÷(g) is an algebraic integer. Partition Gnf1g into subsets by putting each element with its inverse. Each such subset has size 2, by part (a). Hence X ÷( g) ÷(1) 2á g2G

for some algebraic integer á. The stated result follows, since 1 X h÷, 1 G i ÷( g): jGj g2G (c) If ÷ 6 1 G in part (b), then k÷, 1 G l 0, and hence á ÿ÷(1)=2. But ÿ÷(1)=2 is a rational number which is not an integer (since ÷(1) divides jGj, hence is odd). This contradicts Proposition 22.5. Thus ÷ 1 G . (Further information about the number of characters ÷ such that ÷ ÷ appears in Theorem 23.1 and Corollary 23.2.) 6. (a) By Theorem 22.16, ÷(g) is an integer for all characters ÷. Let ÷1 , . . . , ÷7 be the irreducible characters of G, with ÷1 1 G . By the column orthogonality relations, we have (I)

1

7 X

(÷ i ( g))2 5, and

i2

(II) 1

7 X

÷ i (1)÷ i ( g) 0:

i2

From equation (I) we deduce that either ÷ i ( g) 0, 1 for all i, or ÷ i (g) 2 for exactly one i and ÷ i (g) 0 for all other i . 1; the second possibility is ruled out by equation (II). (b) We deduce from part (a) that ÷ i (g) 0 for two values of i, say i 2, 3, and ÷ i (g) 1 for 4 < i < 7. By Corollary 22.27, ÷2 (1) ÷3 (1) 0 mod 5. Also (III)

7 X

(÷ i (1))2 120:

i1

Since 52 102 . 120, we deduce that ÷2 (1) ÷3 (1) 5. (c) By Corollary 22.27, ÷ i (1) ÷ i ( g) mod 5 for all i, and from equation (III) we have (IV)

7 X (÷ i (1))2 69: i4

Hence the possibilities for the pairs of integers (÷ i (1), ÷ i ( g)) with 4 < i < 7 are (1, 1), (4, ÿ1), (6, 1). The only possibility which is consistent with equation (IV) is that the values of ÷ i (1) for 4 < i < 7 are 1, 4, 4, 6 in some order. (d) We now have the following part of the character table of G:

Chapter 22 gi Order of gi |CG ( gi )| ÷1 ÷2 ÷3 ÷4 ÷5 ÷6 ÷7

429

g1 1 120

g2 2 12

g3 2 8

g4 3 6

g5 4 4

g6 6 6

g7 5 5

1 5 5 1 4 4 6

1

1

1

1

1

1 0 0 1 ÿ1 ÿ1 1

We successively calculate the ®ve remaining columns of the character table. We are told that ÷ i (gj ) isP an integer for all i, j. (1) First, ÷ i ( g4 ) ÷ i (1) mod 3 and 7i1 (÷ i ( g4 ))2 6. Hence the values of ÷ i (g4 ) for 1 < i < 7 are 1, ÿ1, P ÿ1, 1, 1, 1, 0, respectively. (2) Next, ÷ i (g5 ) ÷ i (1) mod 2 and 7i1 (÷ i (g5 ))2 4. P Hence ÷ i (g5 ) 1 for 1 < i < 4 and ÷ i (g5 ) 0 for 5 < i < 7. Since 7i1 ÷ i (1)÷ i (g5 ) 0, we deduce that ÷4 ( g5 ) ÿ1 and (without loss of generality) ÷2 (g5 ) ÿ÷3 (g5 ) 1. P (3) Since ÷ i (g3 ) ÷ i (1) mod 2 and 7i1 (÷ i (g3 ))2 8, we deduce that ÷ i (g3 ) 1 for 1 < i < 4 andP that the values of ÷ i ( g3 ) for 5 < i < 7 are 0, 0, 2 in some order. Also 7i1 ÷ i (g3 )÷ i ( gr ) 0 for r 4, 7, from which we see that ÷ i (g3 ) 1 for 1 < i < 4. From the relation P 7 i1 ÷ i (1)÷ i ( g 3 ) 0 we now deduce that the entries in column 3 are 1, 1, 1, 1, 0, 0, ÿ2 in order from the top. P (4) We have ÷ i (g6 ) ÷ i (g4 ) mod 2 and 7i1 (÷ i (g6 ))2 6. Therefore ÷ i (g6 ) 1 for 1 < i < 6 and ÷7 (g6 ) 0. By applying the column orthogonality relations involving column 6 and columns 3, 4, 5 and 7 we obtain ÷2 (g6 ) ÿ÷3 (g6 ) ÷4 (g6 ) ÿ1 and (without loss of generality) ÷5 (g6 ) ÿ÷6 (g6 ) 1. (5) Only the entries in column 2 remain to be calculated. These can be obtained from the column orthogonality relations. The character table of G is as shown. gi Order of gi |CG ( gi )| ÷1 ÷2 ÷3 ÷4 ÷5 ÷6 ÷7

g1 1 120

g2 2 12

g3 2 8

g4 3 6

g5 4 4

g6 6 6

g7 5 5

1 5 5 1 4 4 6

1 ÿ1 1 ÿ1 ÿ2 2 0

1 1 1 1 0 0 ÿ2

1 ÿ1 ÿ1 1 1 1 0

1 1 ÿ1 ÿ1 0 0 0

1 ÿ1 1 ÿ1 1 ÿ1 0

1 0 0 1 ÿ1 ÿ1 1

430


7. Suppose ®rst that ë is an eigenvalue of an n 3 n matrix A, all of whose entries are integers. Then det (A ÿ ëIn ) 0, and hence ë is a root of the polynomial det (xIn ÿ A), which is of the form x n anÿ1 x nÿ1 : : : a1 x a0

(ar 2 Z):

Conversely, assume that ë is a root of the polynomial p(x) a0 a1 x . . . a nÿ1 x nÿ1 x n (a r 2 Z). Let 1 0 0 1 0 ::: 0 B 0 0 1 0 C C B B 0 0 0 0 C C AB . . B .. .. C: C B @ 0 0 0 1 A ÿa0

ÿa1

ÿa2

:::

ÿa nÿ1

Check that det (xIn ÿ A) p(x). As p(ë) 0, it follows that ë is an eigenvalue of A. Since A has integer entries, ë is therefore an algebraic integer.

Chapter 23 1. Assume that x 2 G and x is real. Then gÿ1 xg x ÿ1 for some g 2 G. Hence gÿ2 xg2 x, so g2 2 CG (x). Let m be the order of g. Since jGj is odd, m 2n 1 for some integer n, by Lagrange's Theorem. Then g g2( n1) 2 CG (x). Therefore x ÿ1 gÿ1 xg x. Since x 2 1 and x has odd order, it follows that x 1. 2. Adopt the notation of Theorem 9.8. The character ÷ of G C n1 3 . . . 3 C n r which is given by ÷( g 1i1 : : : g irr ) ë1i1 : : : ë irr is real if and only if ë i 1 for all i with 1 < i < r. Now the ni th root of unity ë i can be ÿ1 if and only if ni is even. Hence the number of real irreducible characters is 2 m , where m is the number of the integers n1 , . . . , nr which are even. However, the elements g of G which satisfy g2 1 are precisely those elements g1i1 . . . g irr where for each j, either i j 0 or n j is even and i j n j =2. The number of such elements is also 2 m . 3. The elements g of D2 n for which g2 1 are 1,

a i b (0 < i < n ÿ 1)

(and also a n=2 if n is even):

This gives n 1 elements if nPis odd, and n 2 elements if n is even. These numbers coincide with ÷(1), summing over all the irreducible characters ÷. Since é÷ < 1 for all ÷, it follows from the Frobenius±Schur Count of Involutions that we must have é÷ 1 for all ÷. 4. Let ë1 and ë2 be the eigenvalues of gr. Then ÷ A (g) 1 2 2 2 2((ë1 ë2 ) ÿ (ë1 ë2 )) ë1 ë2 det (gr) (see Proposition 19.14). Since ÷(1) 2 we have ÷ A (1) 1. It now follows from the De®nition 23.13 of é÷ that é÷ ÿ1 if and only if ÷ A 1 G . The result follows.

Chapter 23

431

5. (a) First, it is easy to check that ÷( g) is real for all g 2 G, so é÷ 1. Let r be the representation obtained by using the basis v1 , v2 of V. Then det (ar) 1 and det (br) ÿå n ; hence det (gr) 1 for all g 2 G if and only if å n ÿ1. The result now follows from Exercise 4. (b) It is easy to check that if g a or b and i, j 2 {1, 2} then â(v i g, v j ) â(v i , v j g ÿ1 ): For example, â(v1 b, v1 ) â(v2 , v1 ) å n â(v1 , å n v2 ) â(v1 , v1 bÿ1 ). Hence â is G-invariant. The de®nition of â shows that â is symmetric if å n 1 and â is skew-symmetric if å n ÿ1. The result now follows from Theorem 23.16. (c) The elements of T4 n are ai and ai b (0 < i < 2n ÿ 1); a has order 2n and ai b has order 4. Hence an is the only element of order 2. (d) Refer the Exercise 18.3 for the characters ø j (1 < j < n ÿ 1) and ÷ j (1 < j < 4) of T4 n. Clearly é÷1 é÷3 1, and é÷2 é÷4 0 or 1, according to whether n is odd or even, respectively. By part (a) (or part (b)) and the construction of the characters ø j in Exercise 17.6, we get éø j ÿ1 or 1, according to whether j is odd or even, respectively. P nÿ1 Therefore j1 (éø j )ø j (1) P 0 or ÿ2, according to whether n is odd or even, respectively. Hence ÷ (é÷)÷(1) 2. 6. Let V be a CG-module with character ÷. Since é÷ ÿ1, there exists a nonzero G-invariant skew-symmetric bilinear form â on V. As â is G-invariant, the subspace {u 2 V: â(u, v) 0 for all v 2 V} is a CGsubmodule of V; since V is irreducible it follows that fu 2 V : â(u, v) 0 for all v 2 V g f0g: ( ) Pick a basis v1 , . . . , v n of V and let A be the n 3 n matrix with ij-entry â(v i , v j ). Since â is skew-symmetric, we have At ÿA. Therefore det (At ) (ÿ1) n det A, so det A (ÿ1) n det A. Also A is invertible by ( ), so det A 6 0. It follows that n is even; as n ÷(1) the result is proved. 7. Choose a basis f1 , . . . , fn of V and de®ne the symmetric n 3 n matrices A (aij ) and B (bij ) by aij â1 ( f i , f j ), b ij â( f i , f j ): By applying the Gram±Schmidt orthogonalization process, we may construct a basis f 19 , : : : , f 9n of V such that â1 ( f 9i , f 9j ) ä ij for all i, j. Let P ( pij ) be the n 3 n matrix which is given by X f 9i pij f j : j

Then PAPt I n and PBPt is symmetric. By a well known property of symmetric matrices, there is an orthogonal matrix Q (i.e. QQt I) such that Q(PBPt )Qÿ1 is diagonal. Write Q (q ij ), and de®ne the basis e1 , . . . , en of V by X ei qij f 9j : j

432


Then â1 (ei , ej ) ä ij ,

since QPAPt Qt I n , and

â(ei , ej ) 0 if i 6 j,

since QPBPt Qt is diagonal:

8. (a) The proof is similar to that of part (1) of Schur's Lemma 9.1. (b) Let v1 , . . . , v n be a basis of the RG-module V, and consider the vector space V9 over C with basis v1 , . . . , v n . Then V9 is a CG-module, which we are assuming to be an irreducible CG-module. By Schur's Lemma there exists ë 2 C such that vW ëv for all v 2 V9. But v1 W ëv1 2 V, so ë 2 R. (c) Let G C3 ka: a3 1l, and let V be the RG-submodule of the regular RG-module which is spanned by 1 ÿ a and 1 ÿ a2 . Then V is an irreducible RG-module. De®ne W: V ! V by vW av (v 2 V). 9. r g is a permutation, as Hxg Hyg ) Hx Hy, and r is a homomorphism as (Hx)(r gh ) Hxgh (Hx)(r g )(r h ). We have g 2 ker r , Hxg Hx,

8x 2 G ,

xgx ÿ1 2 H, 8x 2 G , g 2 \ x2G xÿ1 Hx: Finally, r is a homomorphism G ! Sym(Ù) S n with kernel which is contained in H.

T

x2G x

ÿ1

Hx,

10. Let c1 , c2 be the columns of the character table of G corresponding to the classes {1} and t G . By the orthogonality relation for c2 we have P ÷ i (t)2 |CG (t)| 2 (the sum over all irreducible characters ÷ i , with ÷1 1 G ), so we may take ÷1 (t) 1, ÷2 (t) 1 and ÷ i (t) 0 for i > 3. Now the orthogonality of c1 and c2 gives ÷1 (1) ÷2 (1) 1 and ÷2 (t) ÿ1. Further, ÷1 and ÷2 are the only linear characters, since a linear character must take the value 1 on t. Hence |G : G9| 2 by Theorem 17.11. For the last part, if G is simple then since G9 v G, we have G9 1, i.e. G is abelian. Hence G C2 .

Chapter 25 1. It is clear that the given set of matrices has size p( p ÿ 1). Call it G. For closure, note that 1 y 1 y9 1 y9 yx ; 0 x 0 x9 0 xx9 associativity is a property of matrix multiplication, identity is 1 0 1 y 1 ÿ yx ÿ1 ; inverse of is . 0 1 0 x 0 x ÿ1

Therefore G is a group. 2. Let ç e2ði=5 and å e2ði=11 , and write á å å 3 å 4 å 5 å 9 , â å 2 å 6 å 7 å 8 å 10 :

Chapter 25

433

Character table of F11, 5 gi |CG ( gi )|

1 55

a 11

a2 11

b 5

b2 5

b3 5

b4 5

÷1 ÷2 ÷3 ÷4 ÷5 ÷6 ÷7

1 1 1 1 1 5 5

1 1 1 1 1 á â

1 1 1 1 1 â á

1 ç ç2 ç3 ç4 0 0

1 ç2 ç4 ç ç3 0 0

1 ç3 ç ç4 ç2 0 0

1 ç4 ç3 ç2 ç 0 0

3. Recall that Zp is cyclic, so by Exercise 1.6(c), there exists an integer m such that u m v mod p. Also, m is coprime to q, since both mu and v have order q modulo p. Hence bm has order q. Also, bÿ m abm au av . Let b9 bm. Then G1 ha, b9: ap b9q 1, b9ÿ1 ab9 av i: Hence G1 G2 . 4. (a) Note that ÿ1 is the only element of order 2 in Zp . Hence u m ÿ1 mod p for some m , the element u of Zp has even order , q is even , p 1 mod 4: m

(b) By Proposition 25.9, aG {au : m 2 Z}. Therefore aÿ1 2 aG , u m ÿ1 mod p for some m , p 1 mod 4: (c) ÷ i (a) 1 for all the q linear characters ÷ i of G. Hence X 0 ÷(1)÷(a) q qö1 (a) qö2 (a): ÷ irred

Therefore ö1 (a) ö2 (a) ÿ1. Also, |CG (a)| p, so X p ÷(a)÷(a) q ö1 (a)ö1 (a) ö2 (a)ö2 (a): ÷

Hence ö1 (a)ö1 (a) ö2 (a)ö2 (a) ( p 1)=2. If p 1 mod 4, then a is conjugate to aÿ1 and so ÷(a) is real for all characters p ÷, so (ö1 (a))2 (ö2 (a))2 ( p 1)=2. Hence ö1 (a) and ö2 (a) are (ÿ1 p)=2. If p ÿ1 mod 4, then a is not conjugate to aÿ1 , and it follows from Corollary 15.6 that ö1 (a) and ö2 (a) are not both real. Hence ö2 (a) ö1 (a). This time, 2öp1 (a)ö1 (a) ( p 1)=2, and we ®nd that ö1 (a) and ö2 (a) are (ÿ1 i p)=2.

434


P( pÿ1)=2 m (d) By Theorem 25.10, ö1 (a) m1 å u . Since Zp is cyclic of order p ÿ 1 and u has order ( p ÿ 1)=2, it follows that {u, u 2 , . . . , u ( pÿ1)=2 } is precisely the set of quadratic residues modulo p. The result now follows from part (c). 5. Let H ka, bl. Then for all h 2 H, the conjugacy class hE consists of h and hÿ1 . All the elements outside H form a single conjugacy class of E. Also, E9 H, so E has exactly two linear characters, say ÷1 and ÷2 . A typical non-trivial linear character of H is

÷

1

a

a2

b

b2

ab

ab2

a2 b

a2 b2

1

1

1

ù

ù2

ù

ù2

ù

ù2

where ù e2ði=3 . Then ÷ " E is the irreducible character ÷3 given in the table which follows. The characters ÷4 , ÷5 and ÷6 are obtained similarly. Character table of E gi |CG ( gi )|

1 18

a 9

b 9

ab 9

a2 b 9

c 2

÷1 ÷2 ÷3 ÷4 ÷5 ÷6

1 1 2 2 2 2

1 1 2 ÿ1 ÿ1 ÿ1

1 1 ÿ1 2 ÿ1 ÿ1

1 1 ÿ1 ÿ1 2 ÿ1

1 1 ÿ1 ÿ1 ÿ1 2

1 ÿ1 0 0 0 0

6. Z(E) {1}, and for all i with 1 < i < 6, there exist gi 2 E such that gi 6 1 but ÷ i (gi ) ÷ i (1) (so gi 2 Ker ÷ i ). 7. (a) F13,3 (see Theorem 25.10). (b) C2 3 F13,3 (see Theorem 19.18). (c) D6 3 F13,3 (see Theorem 19.18). 8. The conjugacy classes of G are f1g, fa3 , a6 g, fa r : 3 B rg, fa r b2 : 3 B rg, fa r b4 : 3 B rg, fa r b2 : r 0, 3, 6g, fa r b4 : r 0, 3, 6g, fa r b: 0 < r < 8g, fa r b3 : 0 < r < 8g, fa r b5 : 0 < r < 8g: Let H1 kal. Then H1 v G and G= H 1 C6 . Hence we get six linear characters ÷1 , . . . , ÷6 of G, as shown. Let H2 ka3 , b2 l. Then H2 v G and G= H 2 h H 2 a, H 2 bi D6 . Lift the irreducible character of D6 of degree 2 to obtain ÷7 in the table below. Then ÷8 ÷7 ÷2 and ÷9 ÷7 ÷3 are also irreducible. The ®nal irreducible character ÷10 can be found by using the column orthogonality relations.

Chapter 26

435

Character table of G ka, b: a9 b6 1, bÿ1 ab a2 l gi |CG ( gi )|

1 54

a3 27

÷1 ÷2 ÷3 ÷4 ÷5 ÷6 ÷7 ÷8 ÷9 ÷10

1 1 1 1 1 1 2 2 2 6

1 1 1 1 1 1 2 2 2 ÿ3

a 9

ab2 9

ab4 9

b 6

b2 18

1 1 1 1 1 1 ù2 ù ÿù ù2 1 ù ù2 ù2 ù 1 1 1 ÿ1 1 1 ù2 ù ù ù2 1 ù ù2 ÿù2 ù ÿ1 ÿ1 ÿ1 0 2 ÿ1 ÿù2 ÿù 0 2ù2 ÿ1 ÿù ÿù2 0 2ù 0 0 0 0 0

b3 6 1 ÿ1 1 ÿ1 1 ÿ1 0 0 0 0

b4 18

b5 6

1 1 ù ÿù2 ù2 ù 1 ÿ1 ù ù2 ù2 ÿù 2 0 2ù 0 2ù2 0 0 0

Note: ù e2ði=3

Chapter 26 1. Let ÷ be an irreducible character of G. Then k÷ # H, øl H 6 0 for some irreducible character ø of H. Therefore h÷, ø " Gi G 6 0 by the Frobenius Reciprocity Theorem. But ø(1) 1, since H is abelian; and (ø " G)(1) p, by Corollary 21.20. Hence ÷(1) < p, and so ÷(1) 1 or p by Theorem 22.11. Assume that G has r linear characters and s irreducible characters of degree p. Then r pm for some m, by Theorem 17:11, and r sp2 pn , by Theorem 11:12: Since s p nÿ2 ÿ p mÿ2 and s is an integer, m is at least 2. 2. {1}, {z} and {z 2 } are conjugacy classes of H. For all other elements h of H, the conjugacy class hH {h, hz, hz 2 }. Character table of H (a non-abelian group of order 27) gi |C H ( g i )|

1 27

z 27

z2 27

a 9

a2 9

b 9

ab 9

a2 b 9

b2 9

ab2 9

a2 b2 9

÷00 ÷01 ÷02 ÷10 ÷11 ÷12 ÷20 ÷21 ÷22 ö1 ö2

1 1 1 1 1 1 1 1 1 3 3

1 1 1 1 1 1 1 1 1 3ù 3ù2

1 1 1 1 1 1 1 1 1 3ù2 3ù

1 1 1 ù ù ù ù2 ù2 ù2 0 0

1 1 1 ù2 ù2 ù2 ù ù ù 0 0

1 ù ù2 1 ù ù2 1 ù ù2 0 0

1 ù ù2 ù ù2 1 ù2 1 ù 0 0

1 ù ù2 ù2 1 ù ù ù2 1 0 0

1 ù2 ù 1 ù2 ù 1 ù2 ù 0 0

1 ù2 ù ù 1 ù2 ù2 ù 1 0 0

1 ù2 ù ù2 ù 1 ù 1 ù2 0 0

Note: ù e2ði=3

436


Character table of G ka, b: a16 1, b2 a8 , bÿ1 ab aÿ1 l gi |CG ( g i )| ÷1 ÷2 ÷3 ÷4 ÷5 ÷6 ÷7 øj ( j 1, 3,

1 32

a8 32

a 16

a2 16

a3 16

a4 16

a5 16

a6 16

a7 16

b 4

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ÿ1 1 1 ÿ1 1 ÿ1 1 ÿ1 1 ÿ1 1 1 1 ÿ1 1 ÿ1 1 ÿ1 1 ÿ1 ÿ1 2 2 p0 ÿ2 2 0 p0 p0 ÿ2 p0 2 2 2 0 ÿ 2 ÿ2 ÿ 2 0 2 0 p p p p 2 2 ÿ 2 0 2 ÿ2 2 0 ÿ 2 0 2 ÿ2 cj c2 j c3 j c4 j c5 j c6 j c7 j 0 5, 7)

ab 4 1 ÿ1 ÿ1 1 0 0 0 0

Note: cm e2ði m=16 eÿ2ði m=16 2 cos (mð/8)

3. G has 11 conjugacy classes: {1}, {a8 }, {ar , aÿ r } (1 < r < 7), {ar b: r even}, {ar b: r odd}. Here, the group K which appears in Theorem 26.4 is {1, a8 }, and G=K D16 . The character table of D16 is given in Section 26.8 (D16 G1 ) and in Section 18.3. By lifting the irreducible characters of D16, we obtain the characters ÷1 , . . . , ÷7 of G as shown in the table at the top of this page. Then the four characters ø j ( j 1, 3, 5, 7) come from inducing to G those linear characters ÷ of kal for which ÷(a8 ) ÿ1. 4. (a) Check that AB ÿBA, AC ÿCA, AD DA, BC CB, BD ÿDB, CD ÿDC. Hence Z 2 G, and G=h Zi is abelian while G is non-abelian. Therefore G9 h Zi (see Proposition 17.10). (b) A2 ÿB2 ÿC2 D2 I. Since G=h Zi is abelian, it follows that g 2 2 h Zi for all g 2 G. Hence every element of G has the form A i B j C k D l Z m for some i, j, k, l, m 2 f0, 1g, so jGj < 32; also G is a 2-group, since g4 1 for all g 2 G. (c) A routine calculation shows that every matrix which commutes with each of A, B, C and D has the form ëI for some ë 2 C. Hence by Corollary 9.3, the given representation is irreducible. (d) Since G has irreducible representations of degrees 1 and 4, jGj > 12 42 17. Combined with part (b), this shows that jGj 32. Since G9 h Zi, G has precisely 16 representations of degree 1. These are as follows: for each (r, s, t, u) with r, s, t, u 2 {0, 1}, we get a representation A i B j C k D l Z m ! (ÿ1) ir js kt lu : Together with the irreducible representation of degree 4, these are all the irreducible representations of G, by Theorem 11.12. 5. (a) Let å e2ði=8 . We obtain representations as follows:

å

G1 : a !

0

0 å ÿ1 å

G2 : a !

0

,b! !

0 å ÿ1 ! å 0

G3 : a !

0 å3 å

G4 : a !

0

G5 : a !

,b! ,b!

!

0 å5 1

Chapter 26

!

0

0 ÿ1

,b! ! ,b!

0 1

;

1 0 0 1 ÿ1 0 ! 0 1 ; 1 0 ! 0 1 ; 1 0 ! 0 1 1 0

437

!

! ;

,z!

i

0

0

i

! :

(b) If j 5, 6, 7 or 8 then Z(Gj ) {1, a2 , z, a2 z} C2 3 C2 . Since Z(Gj ) is not cyclic, Gj has no faithful irreducible representation, by Proposition 9.16. (c) Check that the matrices satisfy the required relations, so give representations. It is easy to see that the matrices generate groups with more than eight elements, so the representations are faithful. (d) The following give faithful representations: 0 1 0 1 0 1 i 0 0 0 1 0 1 0 0 B C B C B C G7 : a ! @ 0 ÿi 0 A, b ! @ 1 0 0 A, z ! @ 0 1 0 A; 0

0

i B G8 : a ! @ 0 0

0 1

0 0 ÿ1 0 1 0 0 0 1 0 1 0 0 C B C B C ÿi 0 A, b ! @ ÿ1 0 0 A, z ! @ 0 1 0 A: 1

0

0 1

0 0 1

1

0 0 1

0

0 ÿ1

6. The following table records the numbers of elements of orders 1, 2, 4 and 8 in G1 , . . . , G9 :

Order Order Order Order

1 2 4 8

G1

G2

G3

G4

G5

G6

G7

G8

G9

1 9 2 4

1 1 10 4

1 5 6 4

1 3 4 8

1 3 12 0

1 7 8 0

1 11 4 0

1 3 12 0

1 5 10 0

Therefore no two of G1 , . . . , G9 are isomorphic, except possibly G5 and G8 . But G5 =G5 9 C2 3 C4 , while G8 =G8 9 C2 3 C2 3 C2 , so G5 6 G8 . 7. (a) By Lemma 26.1(1) we have {1} 6 Z(G) 6 G. Also jG= Z(G)j 6 p by Lemma 26.1(2). Therefore j Z(G)j p or p2 . Assume that j Z(G)j p2 . If g 2 = Z(G) then Z(G) < CG (g) 6 G, and g 2 CG ( g). Hence

438


jC G ( g)j p3 and j gG j p. Therefore, G has p2 ( p4 ÿ p2 )= p conjugacy classes. (b) Assume that G has r irreducible of degree 1 and s irreducible P characters characters of degree p. Since ÷(1)2 p4 (Theorem 11.12), there are no irreducible characters of degree greater than p, so r sp2 p4 . Therefore jG=G9j r p2 or p3 , and if r p2 then r s 2 p2 ÿ 1. Part (b) follows, as r s is equal to the number of conjugacy classes of G. (c) Note that G9 \ Z(G) 6 {1} by Lemma 26.1(1). By parts (a) and (b), if |Z(G)| p2 then |G9| p; and if |G9| p2 then |Z(G)| p. Hence |G9 \ Z(G)| p. 8. (a) Let Z Z(G), and assume that G= Z haZ, bZi, with a4 2 Z, a2 Z b2 Z, bÿ1 abZ aÿ1 Z: Then a2 b2 z for some z 2 Z, and hence ba2 b3 z b2 zb a2 b. Since a2 commutes with a, b and all elements in Z, we have a2 2 Z. Therefore G= Z 6 Q8 . (b) If G is a non-abelian group of order 16, then by Exercise 7, either G9 \ Z(G) G9, in which case G=(G9 \ Z(G)) is abelian, or G9 \ Z(G) Z(G), in which case G=(G9 \ Z(G)) 6 Q8 by part (a).

Chapter 27 1. Assume that

z

Then

a c

b d

2 Z(SL (2, p)):

1 1 1 z 0 0 1 0 1 0 z ÿ1 0 ÿ1

1 z ) c 0, and 1 1 z ) c ÿb, a d: 0

Therefore z aI; and since z 2 SL (2, p), we have a2 1, so a 1. 2. Check that

1 0

1 1

and

1 ÿ1 0 1

are elements of G SL (2, 3) of order 3 which are not conjugate to each other. The element ÿ1 0 0 ÿ1 lies in Z(G), and

Chapter 27

0 ÿ1

1 0

439

has order 4. Hence the following are conjugacy class representatives:

g1 1 0

1 24

Order of gi |CG ( gi )|

Order of gi |CG ( gi )|

0 1

g5 1 0

ÿ1 1 3 6

g2 ÿ1 0 0 ÿ1

2 24

g6 ÿ1 1 0 ÿ1 6 6

g3 0 ÿ1

1 0

4 4

g7 ÿ1 ÿ1 0 ÿ1

1 1 0 1

g4

3 6

6 6

We now describe how to construct the character table of G, which is given below. First observe that the vector space (Z3 )2 has exactly four 1-dimensional subspaces, namely the spans of the vectors (0, 1), (1, 1), (2, 1) and (1, 0). The group G permutes these subspaces among themselves, so we obtain a homomorphism ö: G ! S4 . Check that Ker ö {I}. Hence G=fIg Im ö, a subgroup of S4 of order 12; therefore G=fIg A4 . The characters ÷1 , ÷2 , ÷3 , ÷4 of G are obtained by lifting to G the irreducible characters of A4 (which are given in Section 18.2). The values of ÷5 , ÷6 , ÷7 on the elements g1 , g2 , g3 can be deduced from the column orthogonality relations. Note that G has three real conjugacy classes, so by Theorem 23.1, one of ÷5 , ÷6 and ÷7 must be real. Assume, without loss of generality, that ÷5 is real. Then ÷5 ( g4 ) á, where á is real. Also á 6 0, by Corollary 22.27. Since ÷5 ÷2 and ÷5 ÷3 are irreducible characters of G of degree 2, whose values on g4 are áù and áù2 , they must be ÷P 6 and ÷7 in some order, say ÷5 ÷2 ÷6 and ÷5 ÷3 ÷7 . The equation j ÷ j (g4 )÷ j ( g 4 ) 6 gives áá 1. Since á is real, á 1. Then á ÿ1 since ÷5 ( g4 ) ÷5 (1) mod 3. Now note that for j 5, 6, 7, Exercise 13.5 implies that ÷ j ( g7 ) ÿ÷ j ( g4 ). Finally, ÷(g5 ) ÷( g4 ) and ÷(g6 ) ÷( g7 ) for all ÷.

440


Character table of SL (2, 3) gi |CG ( gi )|

g1 24

g2 24

g3 4

g4 6

g5 6

g6 6

g7 6

÷1 ÷2 ÷3 ÷4 ÷5 ÷6 ÷7

1 1 1 3 2 2 2

1 1 1 3 ÿ2 ÿ2 ÿ2

1 1 1 ÿ1 0 0 0

1 ù ù2 0 ÿ1 ÿù ÿù2

1 ù2 ù 0 ÿ1 ÿù2 ÿù

1 ù2 ù 0 1 ù2 ù

1 ù ù2 0 1 ù ù2

Note: ù e2ði=3 3. Apply Proposition 17.6. 4. (a) For the character table of T, notice that T is isomorphic to the group of order 21 whose character table is found in Exercise 17.2 and Example 21.25. Representatives of the conjugacy classes of T are h1 , . . . , h5 , where ! ! ! 1 0 2 0 4 0 h1 Z, h2 Z, h3 Z, 0 1 0 4 0 2 ! ! 1 1 1 ÿ1 h4 Z, h5 Z: 0 1 0 1 Two of the linear characters of T are hi |CT (h i )| 1T ë

h1 21

h2 3

h3 3

h4 7

h5 7

1 1

1 ù

1 ù2

1 1

1 1

where ù e2ði=3 . The values of 1 T : G and ë : G are as follows (see Proposition 21.23): gi |CG ( g i )|

g1 168

g2 8

g3 4

g4 3

g5 7

g6 7

1T " G ë"G

8 8

0 0

0 0

2 ÿ1

1 1

1 1

We ®nd that k1 T : G, 1 T : Gl 2 and k1 T : G, 1 G l 1. Hence 1 T : G 1 G ÷, where ÷ is an irreducible character of G. Also, kë : G, ë : Gl 1, so ë : G is irreducible; write ö ë : G.

Chapter 27

441

(c) The values of ÷ and ÷ S are as shown below (see Proposition 19.14): gi |CG ( g i )|

g1 168

g2 8

g3 4

g4 3

g5 7

g6 7

÷ ÷S æ ø

7 28 12 6

ÿ1 4 4 2

ÿ1 0 0 0

1 1 0 0

0 0 ÿ2 ÿ1

0 0 ÿ2 ÿ1

We ®nd that k÷ S , 1 G l k÷ S , öl k÷ S , ÷l 1. Hence there is a character æ of G such that ÷ S 1 G ö ÷ æ: The values of æ are as shown above. We calculate that kæ, æl 4, so either æ 2ø for some irreducible character ø, or æ is the sum of four distinct irreducible characters (cf. Exercise 14.7). Now 1 G , ö and ÷ are three of the six irreducible characters of G, and none is a constituent of æ. Since there are only six irreducible characters in all, æ cannot therefore be the sum of four irreducible characters, and so æ 2ø with ø irreducible. The values of ø are as shown above. (d) The characters 1 G , ö, ÷ and ø are the characters ÷1 , ÷3 , ÷2 and ÷6 , respectively, in the character table of G given at the end of Chapter 27. The remaining irreducible characters ÷4 , ÷5 can readily be calculated using the column orthogonality relations (noting that gi is real if and only if 1 < i < 4). 5. (a) Compare the proof of Lemma 27.1. Note that because g2 lies in Z(G), gi and gi g2 have the same centralizer for all i. (b) By lifting, we obtain the characters ÷1 , . . . , ÷6 in the character table shown below. (c) Use Exercise 13.5 (noting that ÷ j (ÿI) 6 ÷ j (I) for 7 < j < 11, since ÿI is not in P the kernel of these characters). (d) Since 11 j1 ÷ j (g3 )÷ j ( g 3 ) 8, we deduce that ÷ j (g3 ) 0 for 7 < j < 11. (Alternatively, apply part (c).) Also, by Corollary 22.27, ÷ j (1) is even. (e) P By Theorem 22.16, ÷(g6 ) 2 Z for all characters ÷. Since 11 2 j1 (÷ j (g6 )) 6, the values of ÷ j (g6 ) for 7 < j < 11 must be 1, 1, 1, 0, 0, in some order. By Corollary 22.27 again, two of ÷7 , . . . , ÷11 , say ÷9 and ÷10 , have degrees divisible by 6. Further, P 11 2 2 2 j1 (÷ j (1)) 168, and 12 6 . 168, so ÷9 (1) ÷10 (1) 6. 2 2 2 Next, ÷7 (1) ÷8 (1) ÷11 (1) 96. The only possibility is that two of ÷7 (1), ÷8 (1), ÷11 (1), say ÷7 (1) and ÷8 (1), are equal to 4, and ÷11 (1) 8. The congruences ÷(1) ÷(g6 ) mod 3 now give the remaining values on g6 . Use part (c) to ®ll in the values on g2 and g7 . (f ) By Proposition 19.14, ø A has the following values on g1 , g2 , g3 and g6 :

øA

g1

g2

g3

g6

6

6

2

0

442


Character table of SL (2, 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 ÿ4 ÿ4 ÿ6 ÿ6 ÿ8

g3 4 8

g4 8 8

g5 8 8

g6 3 6

g7 6 6

g8 7 14

g9 14 14

g10 7 14

g11 14 14

1 1 1 1 ÿ1 ÿ1 ÿ1 1 0 0 0 ÿ1 ÿ1 1 1 0 ÿ1 1 1 0 2 0 0 0 0 0 0 1 0 1 p0 p0 0 2 ÿ 2 0 p p 0 ÿ 2 2 0 0 0 0 ÿ1

1 1 ÿ1 0 0 0 ÿ1 ÿ1 0 0 1

1 0 1 á á ÿ1 ÿá ÿá ÿ1 ÿ1 1

1 0 1 á á ÿ1 á á 1 1 ÿ1

1 0 1 á á ÿ1 ÿá ÿá ÿ1 ÿ1 1

1 0 1 á á ÿ1 á á 1 1 ÿ1

p Note: á (ÿ1 i 7)=2

The values of ø A on g1 and g2 show that ø A is a linear combination of ÷1 , ÷4 , ÷5 and ÷6 . Then the value of ø A on g6 shows that ÷1 is not a constituent of ø A ; ®nally, the value on g3 forces ø A ÷6 . Now g24 is conjugate to g3 . Hence (ø( g4 )2 ÿ ø( g 3 ))=2 ø A ( g4 ) ÷6 ( g 4 ) 0, and therefore, ø(g4 ) 0. Similarly, ø(g5 ) 0. Let x ø(g8 ). Since g28 is conjugate to g8 , we get p (x 2 ÿ x)=2 øp A ( g 8 ) ÿ1. Therefore x (1 i 7)=2. Say ÷7 ( g 8 ) (1 ÿ i 7)=2. Then ÷8 ÷7 . For all ÷, ÷( g 10 ) ÷(g8 ); using this fact and part (c), weP®ll in the values of ÷7 and ÷8 . (g) For i 6 6, we have 11 j1 ÷ j ( g i )÷ j ( g 6 ) 0. This allows us to ®ll in the values to gÿ1 4 , ÷(g4 ) is real for all ÷. Then P11 of ÷11 . Since g4 is conjugate P11 2 ÷ ( g )÷ ( g ) 0 and ÷ ( g ) j 1 j 4 j 4 j1 j1 p p8 imply that ÷9 (g4 ) ÿ÷10 (g4 ) 2. Say ÷9 (g4 ) 2. The column orthogonality relations now let us ®nd the remaining values of ÷9 and ÷10 , thereby completing the character table of G. 6. Let Z {I} and de®ne the subgroup T of G, of order 55, by a b ,b2Z T Z: a 2 Z : 11 11 0 aÿ1 Then T is generated by 1 1 2 0 x Z and y Z, 0 1 0 6 and T has ®ve linear characters æ j (0 < j < 4), where æ j : x u y v ! e2ði jv=5 :

Chapter 27

443

The characters æ1 " G and æ2 " G are irreducible; they are ÷3 and p ÷4 in the table. (In calculating ÷3 (g5 ), note that e2ði=5 eÿ2ði=5 (ÿ1 5)=2:) Let ÷1 1 G . We have hæ0 " G, ÷1 i 1 and hæ0 " G, æ0 " Gi 2. Hence æ0 " G ÷1 ÷2 for an irreducible character ÷2 of G. We have now found four of the eight irreducible characters of G, namely ÷1 , ÷2 , ÷3 , ÷4 .

Character table of PSL (2, 11) gi Order of g i |CG ( g i )| ÷1 ÷2 ÷3 ÷4 ÷5 ÷6 ÷7 ÷8 Note: á (ÿ1

p

g1 1 660

g2 2 12

g3 3 6

g4 6 6

g5 5 5

g6 5 5

g7 11 11

g8 11 11

1 11 12 12 10 10 5 5

1 ÿ1 0 0 2 ÿ2 1 1

1 ÿ1 0 0 1 1 ÿ1 ÿ1

1 ÿ1 0 0 ÿ1 1 1 1

1 1 á â 0 0 0 0

1 1 â á 0 0 0 0

1 0 1 1 ÿ1 ÿ1 ã ã

1 0 1 1 ÿ1 ÿ1 ã ã

5)=2, â (ÿ1 ÿ

p

p 5)=2 and ã (ÿ1 i 11)=2

P Since 8j1 ÷ j (g5 )÷ j ( g 5 ) 5, we deduce that the remaining irreducible characters ÷5 , ÷6 , ÷7 , ÷8 take the value P 0 on g5 . By Corollary 22.27, ÷ j (1) 0 mod 5 for 5 < j < 8. But 8j5 (÷ j (1))2 250; hence, without loss of generality, ÷5 (1), ÷6 (1), ÷7 (1), ÷8 (1) are 10, 10, 5, 5, respectively. Note that ÷(gj ) is an integer for 1 < j < 4 and all characters ÷, by Theorem 22.16. P Since ÷(1) ÷(g3 ) mod 3 for all characters ÷, and 8j1 ÷ j (g3 )2 6, the values of ÷ j (g3 ) (5 < j < 8) are as shown. P Now ÷(g4 ) ÷(g3 ) mod 2 for all ÷, and 8j1 ÷ j (g4 )2 P 6, so ÷ j (g4 ) 1 for 5 < j < 8. Next, ÷(g2 ) ÷(g4 ) mod 3 for all ÷, and 8j1 ÷ j ( g2 )2 12; hence |÷( g2 )| , 3 for all irreducible ÷. We may now conclude from the facts P that ÷(g2 ) ÷(g1 ) mod 2 and 8j1 ÷ j (g2 )2 12, P that ÷ j (g2 ) 2 for j 5, 6 and ÷ j (g2 ) 1 for j 7, 8. By considering 8j1 ÷ j (1)÷ j (g2 ) 0, we see that ÷7 (g2 ) ÷8 (g2 ) 1, and ÷5 (g2 ), ÷6 (g2 ) have opposite signs; without loss of generality, ÷5 (g2 ) 2 ÿ÷6 ( g2 ). We have now completed columns 1, 2, 3 and 5 of the P character table. Since ÷(g4 ) ÷(g2 ) mod 3 for all characters ÷, and 8j1 ÷ j (g4 )2 6, we can complete column 4. The column orthogonality relations now enable us to ®nish the character table.

444

Representations and characters of groups Chapter 28

1. We take g 1 , : : : , g8 as representatives of the conjugacy classes, ! ! ! 1 0 2 0 1 1 2 g1 g2 g3 g4 0 1 0 2 0 1 0 ! ! ! 1 0 0 1 0 1 0 g5 g6 g7 g8 0 2 2 0 1 2 1

where ! 1 2 1 1

! :

The character table of GL(2, 3) is then as follows. gi |CG ( g i )| ë0 ë1 ø0 ø1 ø0,1 ÷1 ÷2 ÷4

g1 48

g2 48

g3 6

g4 6

g5 4

g6 8

g7 8

g8 8

1 1 3 3 4 2 2 2

1 1 3 3 ÿ4 ÿ2 2 ÿ2

1 1 0 0 1 ÿ1 ÿ1 ÿ1

1 1 0 0 ÿ1 1 ÿ1 1

1 ÿ1 1 ÿ1 0 0 0 0

1 1 ÿ1 ÿ1 0 0 2 0

1 ÿ1 ÿ1 1 p0 i 2 0 p ÿi 2

1 ÿ1 ÿ1 1 0 p ÿi 2 p0 i 2

2. Every element r of F q can be expressed as a square, since r r q and q is even. a b Suppose that 2 GL(2, q). We may write ad ÿ bc as s 2 for some c d s in Fq . Then a b s 0 a=s b=s : c d 0 s c=s d=s The ®rst matrix in the product is sI and the second belongs to SL(2, q). It now follows easily that GL(2, q) Z 3 SL(2, q) where Z fsI : s 2 Fq g. 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 ÿ q. 1 1 (b) The matrix u1 has centralizer of order q. 0 1 (c) There are (q ÿ 2)=2 conjugacy classes with representatives s 0 ÿ1 d s,s ÿ1 ÿ1 , indexed by unordered pairs fs, s g of elements 0 s from F q nF2 . Each such element has centralizer of order q ÿ 1.

Chapter 28

445

(d) There are q=2 conjugacy classes with representatives 0 1 vr , indexed by unordered pairs fr, r ÿ1 g of elements 1 r r ÿ1 from F q 2 nF q such that r 1q 1. Each such element has centralizer of order q 1. By restricting characters from GL(2, q) to SL(2, q) you will quickly be able to prove that the character table of SL(2, q) is as follows.

ë0 ø0 ø0,i ÷i

I

u1

d s,s ÿ1

vr

1 q q1 qÿ1

1 0 1 ÿ1

1 1 s i s ÿi 0

1 ÿ1 0 ÿ(r i r ÿi )

Here, we have used the function r ! r de®ned in (28.3). The subscripts for ø0,i satisfy 1 < i < (q ÿ 2)=2, and the subscripts for ÷ i satisfy 1 < i < q=2. If q 6 2 then the kernel of every non-trivial character is the identity subgroup, and therefore SL(2, q) is simple. 3. Note ®rst that PSL(2, 8) SL(2, 8). The polynomial x 3 x 1 is irreducible over F2. Hence we may write F8 fa bç cç2 : a, b, c 2 F2 and ç3 1 çg: The pairs fs, s ÿ1 g of elements from F8 nF2 are fç, 1 ç2 g, fç2 , 1 ç ç2 g, f1 ç, ç ç2 g: These give us the conjugacy class representatives g3 , g4 , g 5 below. The irreducible monic quadratics over F8 with constant term 1 are x 2 x 1, x 2 çx 1, x 2 ç2 x 1, x 2 (ç ç2 )x 1: The companion matrices for these quadratics give use the conjugacy class representatives g 6 , g 7 , g8 , g9 below We can now list representatives g 1 , : : : , g 9 of the conjugacy classes of SL(2, 8), as follows. ! ! 1 0 1 1 g1 g2 0 1 0 1 ! ! ! ç 0 1ç 0 0 ç2 g3 g4 g5 0 1 ç2 0 ç ç2 0 1 ç ç2 ! ! ! ! 0 1 0 1 0 1 0 1 g6 g9 : g7 g8 1 1 1 ç 1 ç2 1 ç ç2 We may choose a generator å of F so that å 7 å ÿ7 ç. Then 64

446


å 14 å ÿ14 ç2 , å 21 å ÿ21 1 and å 28 å ÿ28 ç4 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

g2 8

g3 7

g4 7

g5 7

g6 9

g7 9

g8 9

g9 9

1 8 9 9 9 7 7 7 7

1 0 1 1 1 ÿ1 ÿ1 ÿ1 ÿ1

1 1

1 1

1 1

1 ÿ1 0 0 0 ÿ2 1 1 1

1 ÿ1 0 0 0 1

1 ÿ1 0 0 0 1

1 ÿ1 0 0 0 1

A 0 0 0 0

0 0 0 0

0 0 0 0

B

Here, the 3 3 3 submatrices A and B are given by 0 1 2 cos(2ð=7) 2 cos(4ð=7) 2 cos(6ð=7) B C A @ 2 cos(4ð=7) 2 cos(6ð=7) 2 cos(2ð=7) A 2 cos(6ð=7) 2 cos(2ð=7) 2 cos(4ð=7) 0

ÿ2 cos(2ð=9) ÿ2 cos(4ð=9) ÿ2 cos(8ð=9)

1

B C B @ ÿ2 cos(4ð=9) ÿ2 cos(8ð=9) ÿ2 cos(2ð=9) A: ÿ2 cos(8ð=9) ÿ2 cos(2ð=9) ÿ2 cos(4ð=9)

Chapter 29 1. (a) It is straightforward to check that ö is a homomorphism. For x, y 2 G, the element (x, y) 2 G 3 G sends x to y, so that action is transitive. (b) (G 3 G)1 f( g, g) : g 2 Gg, and ker ö f(z, z) : z 2 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 h and y g ÿ1 xg. Hence if C1 , : : : , C k are the conjugacy classes of G, and xi 2 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) x if and only if xhx ÿ1 g, we see that ð( g, h) 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 ÿ1 g then an arbitrary element y 2 G such that yhy ÿ1 g is of the form y xc with c 2 P C G (x)). Hence using the column orthogonality relations we see that ð ÷ 3 ÷. 2. There are q 2 ÿ 1 non-zero vectors in V, and each 1-dimensional subspace

Chapter 29

447

contains q ÿ 1 of them; also two 1-dimensional subspaces have no non-zero vectors in common. Hence jÙj (q 2 ÿ 1)=(q ÿ 1) 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 ÿ 1, q ÿ 1, q ÿ 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 hð, ø0 i hð, ø0, j i 1 (1 < j < q ÿ 2): P As 1 G ø0 1qÿ2 ø0, j has degree q 2 ÿ 1 ð(1), we conclude that P ð 1 G ø0 1qÿ2 ø0, j . 4. Observe that the coset H 1 x is ®xed by g if and only if xgx ÿ1 2 H 1 . If G is abelian this amounts to g 2 H 1 , and hence we see that ð1 ( g) 0 if g 2 = H1 and ð1 ( g) jG : H 1 j if g 2 H 1 . Thus H 1 f g 2 G : ð1 ( g) 6 0g. Likewise for H 2 ; since ð1 ð2 we deduce that H 1 H 2 . As a counterexample for G non-abelian, take G D8 ha, b : a4 b2 1, bÿ1 ab aÿ1 i with H 1 hbi, H 2 ha2 bi. Then ð1 ð2 but H 1 6 H 2 . P P 1 5. By Proposition 29.4 we have 1 jGj jfixÙ ( g)j g2G jfixÙ ( g)j, hence jGj. Since jfixÙ ( g)j is a non-negative integer for each g, and jfixÙ (1)j jÙj . 1, we must have jfixÙ ( g)j 0 for some g. 6. Write ð ð( nÿ2,1,1) . Calculating inner products using Proposition 29.6, as in the proof of Theorem 29.13, we ®nd hð, ði 7, hð, 1i 1, hð, ð( nÿ1,1) i 3, hð, ð( nÿ2,2) i 4: Using Theorem 29.13 it follows that ð( nÿ2,1,1) 1 2÷ ( nÿ1,1) ÷ ( nÿ2,2) ÷ with ÷ irreducible. Hence ÷ 1 ð ÿ ð( nÿ1,1) ÿ ð( nÿ2,2) , from which it is easy to calculate that ÷(1) 12(n ÿ 1)(n ÿ 2), ÷(12) 12(n ÿ 2)(n ÿ 5), ÷(123) 12(n ÿ 4)(n ÿ 5). For n 6, ÷ (4,1,1) is the character ÷5 in Example 19.17.

Chapter 30 1. By Theorem 30.4, a245 168/(8´3) 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 (1 (ÿ1 i 7)=6 (ÿ1 ÿ i 7)=6 ÿ 4=6)168=(8:8) 0, and similarly a226 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 : 1 10: 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


There is a homomorphism W from A5 onto H (W sends a ! x, b ! y). Since Ker W 3 A5 and A5 is simple, we deduce that H A5 . 4. Suppose that G is character table is p a group whose p where á (1 5)=2, â (1 ÿ 5)=2.

÷1 ÷2 ÷3 ÷4 ÷5

g1

g2

g3

g4

g5

1 4 5 3 3

1 1 ÿ1 0 0

1 0 1 ÿ1 ÿ1

1 ÿ1 0 á â

1 ÿ1 0 â á

P For 1 < i < 5 we have jC G ( g i )j 5j1 ÷ j ( g i )÷ j ( g i ). Therefore the 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 1) is |CG (g i )| even. Now a324 60=(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 60, we have G A5 . P 5. (a) Using the fact that 7j1 ÷ j (g i )÷ j ( g i ) |CG (g i )|, we ®nd that the centralizer orders and class sizes are as follows:

|CG ( g i )| | gG i |

g1

g2

g3

g4

g5

g6

g7

360 1

8 45

4 90

9 40

9 40

5 72

5 72

Hence jGj 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
> > > > axis, > < ÷ T ( g) > ÿ(1 2 cos ö), if the element ÿ g of O(R3 ) is a rotation > > > > > : through ö: 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 2 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 (ÿö)(vg). Therefore ( if g is a rotation, ÷ T ( g), ÷ R ( g) ÿ÷T ( g), if g is not a rotation, and so ÷ T ÷ R has the required form. 3. The matrix A is 1 0 p p ÿ3=2 0 3=4 ÿ 3=4 p3=4 3=4 p B 0 ÿ1=2 ÿ 3=4 p1=4 3=4 1=4 C C B p kB 3=4 ÿ 3=4 ÿ3=4 3=4 0 0 C C: B p p 3=4 ÿ5=4 0 1 C mB C B ÿ 3=4 p1=4 p @ 3=4 3=4 0 0 ÿ3=4 ÿ 3=4 A p p 3=4 1=4 0 1 ÿ 3=4 ÿ5=4 4. A simpler basis is given by 1 2(r1 1 2(r1 1 2(r1

r2 ) (v12 v21 ) ÿ (v34 v43 ), r3 ) (v13 v31 ) ÿ (v24 v42 ), r4 ) (v14 v41 ) ÿ (v23 v32 ):

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 12(x13 x14 ). Similarly, the distance of 29 from the plane through 2 perpendicular to 12 is x21 12(x23 x24 ). Therefore 12 has decreased by x12 x21 12(x13 x14 x23 x24 ). The other calculations can be done in 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: ÿk1 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 ÿk2 d10 sec (/012) from the force between atoms 1 and 0. Hence p m1 x12 ÿk 1 d 12 ÿ 13 (3=2)k 2 d 10 :

Upon substituting for d12 and d10 from part (b), we obtain the given expression for x12. 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 x xA. If you write down the matrix A, then you will easily verify that the given vectors are eigenvectors of A (with eigenvalues ÿ(4k 1 k 2 )=m1 , 0, 0, 0, ÿk 1 =m1 , ÿk 1 =m1 , respectively). (e) The matrix B is 0 ÿ(6k 1 k 2 )=3m1 @ ÿ(3k 1 k 2 )=3m1 p p ÿk 2 3=(9m1 2)

p p 1 ÿ2k 2 =3m1 ÿ4k 2 p 2=(m2 p3) ÿ2k ÿ4k 2 2=(m2 3) A: p 2 =3m1 p ÿ2k 2 3=(9m1 2) ÿ4k 2 =3m2 p (f ) You will ®nd that (1, ÿ2, 6) is an eigenvector of B, with eigenvalue 0. This agrees with the statement in Example 32.20 that the translation vector r1 ÿ 2s1 3 cos Ww1 is an eigenvector of A. 6. (a) Assign coordinate axes along the edges of the square, as shown below.

The symmetry group is G D8. Let a denote the rotation sending P ! Q ! R ! S ! P, and let b denote the re¯ection in the axis PR. The character ÷ of the RG-module R8 is

÷

1

a2

a

b

ab

8

0

0

0

0

Chapter 32

453

We refer to the character table of D8 which is given in Example 16.3(3), and see that ÷ ÷1 ÷2 ÷3 ÷4 2÷5 : The rotation mode is (t â)v, where v (1, ÿ1, 1, ÿ1, 1, ÿ1, 1, ÿ1) 2 V÷2 : The translation modes are (t â)v, where v is in the span of v1 , v2 and v1 (1, 0, 0, 1, ÿ1, 0, 0, ÿ1), v2 (0, 1, ÿ1, 0, 0, ÿ1, 1, 0): 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, ÿ1, ÿ1, 1, 1, ÿ1, ÿ1) and (1, ÿ1, ÿ1, 1, 1, ÿ1, ÿ1, 1), respectively. The ®nal set of eigen-vectors is given by V÷5 \ R8vib which is spanned by (1, 0, 0, ÿ1, ÿ1, 0, 0, 1) and (0, 1, 1, 0, 0, ÿ1, ÿ1, 0): (b) The matrix A is

0

1 B0 B B0 B kB 0 ÿ B mB B0 B0 B @0 1

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

1 1 0C C 0C C 0C C: 0C C 0C C 0A 1

7. (a, b) Let å i (1 < i < m) be the projection which is given by å i : u1 : : : um ! ui (where u k 2 Uk for all k). Then w ! wAå j Wÿ1 j Wi

(w 2 Ui )

gives an RG-homomorphism from Ui to Ui. By Exercise 23.8, there exist ë ij 2 R such that for all w 2 Ui, Since

Pm

j1 å j

wAå j ë ij wWÿ1 i W j: is the identity endomorphism of U1 . . . Um, we have m X ë ij wWÿ1 for all w 2 Ui : wA i Wj j1

Now take in turn w uW i and w 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 . . . 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

Index

A4 , 112, 130, 136, 181, 308 A5 , 10, 112, 116, 220, 312, 359 A6 , 116, 222, 354, 360 A7 , 223 An , 5, 9, 11, 111, 343 abelian group, 3, 11, 81, 82 action, 337 algebra, 55, 56 algebraic integer, 244, 362 algebraic number, 361 alternating group, 5, 9, 11, 111 antisymmetric part, 196, 273 associative, 2

permutation, 129 product, 176, 192 real, 263 realized over R, 265 reducible, 119 regular, 127, 128, 150 trivial, 122 character table, 159 A4 , 181 A5 , 221, 359 A6 , 359, 424 C2 , 160 C3 , 160 C4 , 412 C2 3 C2 , 415 Cn , 82 D6, 160 D8, 161 D10, 415 D12 S3 3 C2 , 207, 419 D2n ( n odd), 182 D2n ( n even), 183 D6 3 D6, 423 E, of order 18, 434 F7,3, 240, 417 F11,5 , 433 Fp,q , 291 GL(2,q), 327 PSL(2,7), 318 PSL(2,8), 445 PSL(2,11), 443 Q8 , 416 S4 , 180 S5 , 201, 262 S6 , 205 SL(2,3), 440 SL(2,7), 442 SL(2,q), 445 T12, 186 T4 n, 420

basis, 15 natural, 45, 54 bijection, 6 bilinear form, 269 symmetric, 269 skew-symmetric, 269 Brauer±Fowler Theroem, 278 Burnside's Lemma, 340 Burnside's Theorem, 363, 364 C, 2 Cn , 2, 82, 88 centralizer, 106 centre of group, 85, 107, 116, 298 of group algebra, 83, 114, 153 change of basis, 24 character, 118 degree, 122, 247 faithful, 125, 195 generalized, 355 induced, 230, 234, 236 integer-valued, 253 irreducible, 119 kernel of, 125 linear, 122, 172, 174

455

456


U6n, 421 V24 , 422 V8 n , 421 direct product, 206 order 16, 305, 306, 307 order 27, 435 order , 32, 308 order p3 , 301 order pq, 291 p-group, 300 class algebra constants, 349 class equation, 107 class function, 152 class sum, 114 Clifford's Theorem, 216 complete set, 101 completely reducible, 74 composition, 2 composition factor, 90 common, 91, 96 congruences, 259 conjugacy class, 104 conjugate, 104, 361 constituent, 143, 213 coset, 8 cycle notation, 8 cycle-shape, 109 cyclic group, 2, 4, 12, 82, 88 D2n, 2, 12, 107, 181 degree, 30, 122, 249 derived subgroup, 173 diagonalization, 83 dicyclic group T4 n, 178, 187, 281, 420 dihedral group, 2, 12, 107, 181 dimension, 15 direct product, 5, 206 direct sum, 17, 66 external, 18 eigenvalue, 24 eigenvector, 24 endomorphism, 20 equivalent, 32, 46 even permutation, 5 expansion±contraction mode, 381 external direct sum, 18 F R or C, 3 F n, 15 Fp,q , 290 FG, 53 factor group, 9 faithful character, 125, 195 faithful module, 44, 56, 85 faithful representation, 34 FG-module, see module

FG-submodule, 49 FG-homomorphism, 61 FG-isomorphism, 63 Frobenius group, 290 Frobenius Reciprocity Theorem, 232 Frobenius±Schur Count of Involutions, 277 function, 6 bijective, 6 injective, 6 invertible, 6 surjective, 6 GL(n,F), 3 GL(2,q), 324, 343 general linear group, 3 group, 1 abelian, 3, 11, 81, 82 alternating, 5, 9, 11, 111 cyclic, 2, 4, 12, 82, 88 dicyclic, 178 dihedral, 2, 12, 107, 181 factor, 9 ®nite, 2 general linear, 3 order, 2 order p3 , 302, 304 orthogonal, 367 projective special linear, 312 quaternion, 5 rotation, 368 simple, 10, 250, 278, 311, 318, 353, 363, 364 soluble, 365 special linear, 311 symmetric, 3, 109, 116, 175, 254 symmetry, 368 group algebra, 55 H < G, 3 H v G, 9 HomCG (V, W), 95, 96 homogeneous component, 376 homomorphism, 6, 10, 61 ideal, 256 maximal, 257 proper, 257 index of subgroup, 9 indicator function, 273 induced character, 230, 234, 236 induced module, 226, 228 inner product, 134 involution, 277, 353 irreducible character, 119 irreducible module, 50, 74, 79, 91 irreducible representation, 50, 79 isomorphism, 7, 20, 63

Index kernel, 10, 19, 34, 124, 125 Lagrange's Theorem, 9 lift, 169 linear character, 122, 173, 174 linear transformation, 18 linearly dependent, 15 linearly independent, 15 Maschke's Theorem, 70, 76 matrix, 21 change of basis, 24 diagonal, 26 identity, 3, 21 invertible, 23 permutation, 45 methane, 384 minimal polynomial, 361 module, 39 completely reducible, 74 faithful, 44, 56, 85 irreducible, 50, 79, 85 permutation, 45, 62 reducible, 50 regular, 56 trivial, 43 natural normal normal normal 217

basis, 45, 54 modes of vibration, 372, 373 p-complement, 251 subgroup, 9, 113, 171, 215, 216,

odd permutation, 5 orbit, 338 order of G, 2 order of g, 4 orthogonal group, 367 orthogonality relations, 161 PSL(2,7), 312, 318, 319, 354, 359, 360 PSL(2,11), 321, 359 PSL(2, p), 312 p-group, 298 p9-part, 256, 258 permutation, 3, 5 even, 5 odd, 5 permutation module, 45, 62, 340 permutation character, 129, 340 permutation matrix, 45 powers of characters, 193 presentation, 3 primitive root, 284 product of characters, 176, 192

457 projection, 27, 67 projective special linear group, 312 Q8 , 5, 116, 177, 278, 416 quaternion group, 5, 116, 177, 278, 416 R, 3 rank, 342 Rank±Nullity Theorem, 19 real character, 263 real conjugacy class, 263 real element, 263 reducible character, 119 reducible module, 50 reducible representation, 50 regular character, 127, 128, 150 regular module, 56 regular representation, 56 representation, 30 degree, 30, 249 equivalent, 32, 46 faithful, 34 irreducible, 50, 79 kernel of, 34, 124 reducible, 50 regular, 56 trivial, 34 representatives, 105 restriction, 210 rotation group, 368 rotation mode, 379 rotation submodule, 380, 394 S4 , 44, 110, 113, 180, 275 S5 , 111, 201, 262 S6 , 116, 205 S7 , 223 Sn , 3, 109, 116, 175, 254, 343, 344 SL(2,3), 319, 440 SL(2,7), 320, 442 SL(2, p), 311 SL(2,q), 336, 445 Schur's Lemma, 78 simple group, 10, 250, 278, 311, 318, 363, 364 skew-symmetric bilinear form, 269 special linear group, 311 stabilizer, 339 subgroup, 3, 4 cyclic, 4 derived, 173 generated, 4 normal, 9, 113, 171, 215, 216, 217 submodule, 49 irreducible, 74 Sylow's Theorem, 354, 365 symmetric bilinear form, 269

458


symmetric group, 3, 109, 116, 175, 254 symmetric part, 196, 273 symmetry group, 368

trivial character, 122 module, 43 representation, 34

T4 n, 178, 187, 281, 420 tensor product module, 190 tensor product space, 188 trace, 117 transitive, 338, 341 transitivity of induction, 229 translation mode, 379 translation submodule, 380, 394 transposition, 5

U6n, 178, 187, 421 V8n, 178, 187, 421 Vandermonde matrix, 194 vibratory modes, 381 water, 369, 374 Z, 2

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