Chapter IX Finite Matrix Groups

then K1‘,K2’, . . . ,K,’ is also a basis for W.Define R by R‘ = VRV-l = (r;,,) Then ( 5 ) takes the form K,‘A

I

=

q= 1

rLqK;,

1 < p Is

It follows that in our discussion the set ( R ) may be replaced by any equivalent set, by making the appropriate change of basis in W. We now appeal to Schur’s lemma. Sincethe matrices K,, K,, . . . ,K, are a basis, none of them is 0. Hence not all of the matrices PIcan be 0. It follows that either P, is a square nonsingular matrix for some i, in which case the sets {A} and { R } are equivalent, or the set (R}is reducible. Suppose first that P, is square and nonsingular for some i so that s = n and { A } and { R } are equivalent. Then after a suitable change of

168

IX FINITE MATRIX GROUPS

basis in W, we can in fact assume that they are the same. Equations (7) now become P,A = AP,, 1 5 i I n Corollary IX.1 now implies that P,is scalar:

P,= k,Z,

Thus

c i, J


kp, = kt6PI

Now from the fact that it follows that

i

1

kJ6piaiJ

='

9

n

C k p r I = 0,

I=

1

Hence

=o But A is nonsingular: Thus k,=O,

l l j s n

which in turn implies that all the matrices P, are zero, 1 < i n; a contradiction. Next, suppose that {R)is reducible. Then after a suitable change of basis has been performed we may assume that for some positive integer t, every R has the form

'1

R=[" T U

where the t x t matrices S are either all 0, or { S } is irreducible. If S = (spa),then the first t equations of (5) becomes

4. Burnside’s Theorem on Groups of Finite Exponent

169

If {S)is irreducible, we conclude as before that K,, K,, . . . ,K, are all 0. If all the S are 0, the same conclusion can be drawn directly from the equations KA=O, 1 < p < t since A is nonsingular. In either case we have arrived at a contradiction since no one of the matrices K,, K,, . . , K, can vanish. This concludes the proof.

.

An equivalent formulation is as follows:

Theorem IX.3. Let F be algebraically closed. Let G be an irreducible subgroup of GL(n, F). Then the dimension of the vector space spanned by the matrices of G is n2.

4. Burnside’s theorem on groups of finite exponent As an application of Burnside’s theorem, we prove Burnside’s celebrated result on groups of finite exponent. A group G is said to be of exponenr r if x‘ = 1 for every element x of G. Burnside’s theorem follows:

Theorem IX.4 (Burnside). Let G be a subgroup of GL(n, C) (C the complex field) of exponent r. Then G is finite. Proof. Suppose first that G is irreducible. If A E G, then tr(A) is the sum of n rth roots of unity, and so assumes only finitely many values (certainly not more than P ) . Therefore the set of traces {tr(A): A

E

G)

contains only finitely many distinct elements. Since G is irreducible, there are elements A,, A,, . . . ,A,, of G which are linearly independent. Let A be any element of G, and put A = (a,/), A, = (a:,), 1 k < n2. Then tr(AA

=

C a:,a,,, 1, I

l
Regarding these as n2 linear equations in the n2 unknowns Q,,, we see that the rows of the coefficient matrix are linearly independent, and hence a unique solution exists, depending only on the coefficient matrix

170

IX FINITE MATRIX GROUPS

and the values tr(A,A), 1 < k < n2. Since there are only finitely many possibilities for these, there are only finitely many possibilities for the u,,. Thus the group is finite in this case. We now proceed by induction on n. The theorem is certainly true for n = 1. Suppose the theorem proved for all dimensions < n. If G is irreducible, the discussion above shows that the theorem is true. If G is reducible, we may assume that G is already in reduced form; that is, that there are positive integers p, q such that every A E G has the form

'1

A = [ "C D

where B is p x p, Cis q x p, and D is q x q. Then the groups {B) and ED) are finite, by the induction hypothesis, since they are certainly of exponent r. Define G , = { A E G :B = I}, G, = { A E G : D = I ) Then G,, G, are normal subgroups of G of finite index. The same is therefore true of G, n G,. Let A E G, n G,. Then

Since A' = [tCt] = I, C = 0. Hence G, n G, = {I},and it follows that G is finite. This completes the proof. Almost the same proof shows that if G is any subgroup of GL(n, C ) with only finitely many conjugacy classes, then G is finite. The relevant point is that the set of traces contains only finitely many distinct elements. The theorem is false for arbitrary groups; for example, the direct product of countably many cyclic groups of order r, By a more complicated argument, Schur has shown that a finitely generated subgroup of GL(n, C ) , in which each element is of finite period, is finite. 5. Matrix representationsand characters

We now wish to derive some of the character relations for finite groups. For this purpose we change our notation slightly. Let G = ( x } be some fixed, finite group. Let a = {A(x):x E G} be a matrix repre-

171

6. Orthogonality Relationships of the First Kind

sentation of G, where the matrices A(x) belong to C a n , C),and satisfy

all x , y E G A(xy) = A(x)A(y), There is no need to assume that the representation is faithful: a may only be a homomorphic image of G. a is said to be of degree n. We set x E G

x,(x> = tr(A(x)),

The set

{xa(x): x E GI is known as the character of the representation a.

6. Orthogonality relationships of the first kind

Suppose now that dl =

{A(x):x E G},

/3 = {B(x):x

E

G)

are two irreducible matrix representations of G. Let a be of degree n, B of degree m, and put g = o(G). Let U be an arbitrary n x rn complex matrix, and form

Y=

c A(x)UB(x-') c A(x)UB(x)-' =

xEQ

xEQ

Let y be any element of G.Then A(y)VB(y)-'

=

c A(yx)UB(yx)-'

xCQ

=

v

since y x runs over G as x does. Comparing elements in the (i,j)position, we find that fJij

=

c a,p(-4%lqh,(x-1)

xCQ

Suppose first that a and /3 are not equivalent. Then Schur's lemma implies that V = 0; and since U was arbitrary, we find that (8)

C ~,~(x)b~,= ( x 0, -~)

XEQ

a, /3 not equivalent

In particular, setting p = i, q =j and summing over i , j we obtain

c x,(x)x&-')

a, /3 not equivalent Now consider the case when a = /3. Then Y commutes with every element of a, and the corollary to Schur's lemma implies that V is (9)

= 0,

172

IX FINITE MATRIX GROUPS

scalar: V = IZ. To determine I we compute the trace: tr(V)

= tr(

C A ( x ) U A ( x ) - ' ) = C tr(A(x) U A(x)-') = g tr(U) x€G

xEG

Hence

1 = (g/n)tr(U)

and so (g/n>(u,1

+

Ut2

+ - + 4,= xs5 %(x)upg~g,(x-l) * *

Comparing coefficients, we find that

c al,(x)ag,(x-l>

(10)

xSG

Setting p

=

= ( d n ) 4,Spg

i, q = j and summing as before over all i, j we find that

c Xa(x)Xa(x-')

(11)

xEG

=g

The two formulas (9) and (1 1) are known as the orthogonality relationships of the first kind. Formula (1 1) may be generalized slightly. Let y be any element of G. Multiply (10) by a,,(y) and sum over all p. The result, after some simplification, is

c ag,(x-' ) Q l l ( V )

(12)

= ( d n >S,,a,,(y)

xtG

Choose q

= j, I = i

and sum over all i,j. We obtain

C Xa(xy)Xa(x-'> = (g/n>Xa(Y)

(13)

xEG

An equivalent form of (13) is

7. A divisibility theorem We shall use (14) to derive a significant result for any irreducible finite subgroup of GL(n, C). Theorem IX.5. Let G be any irreducible finite subgroup of GL(n, C). Then n divides the order of G.

Proof. Denote the elements of G by x , , x 2 , . . . ,x,. Set zg

= Xa(xg),

Cpg

= Xa(xpx;'),

1I P, 4 I g

9. Lemmas on Traces

173

Then (14) becomes

The z, are not all 0 [since xa(Z) = n, for example] and so this system has a nontrivial solution. It follows that g/n is an eigenvalue of the Now the numbers cpqare algebraic integers of matrix C = (c,). since they are sums of gth roots of unity. Hence the eigenvalues of C are also algebraic integers, and so g/n is an algebraic integer. But g/n is rational. Hence it is a rational integer, and the proof is concluded.

Q(c,),

We are now very close to the proofs of the fundamental relations of representation theory, which depend on the so-called regular representation. These state that the number of pairwise inequivalentirreducible matrix representations of a finite group G is equal to the number of conjugacy classes of G, and that the sum of the squares of their degrees is equal to the order of G. Proofs of these statements along the lines outlined here may be found in [52]. We pursue the question no further, but go on now to finite integral matrix groups.

8. Integral matrix groups We assume then that G = {A} is a finite subgroup of GL(n,Z) of order g. Then if A E G, det(A) = f1. The matrices of G of determinant 1 are a normal subgroup of G of index 1or 2. It is sometimes more convenient to study this subgroup, which we may denote conveniently ' . by G 9. Lemmas on traces

We are going to derive bounds for the order of G in terms of n. For this purpose we prove Lemma IX.1. Let S be the set {Al, A,, . . . ,Ak}, where the elements of S are not necessarily distinct and belong to GL(n, 2). Suppose that S has the property A,S

= {A,Al, A,A,,

. . . ,A,A,)

=S

174

IX FINITE MATRIX GROUPS

for all i such that 1 < i

< k. Then k

C tr(A,) = 0 mod k M = C;"=,A,. We must show I= 1

Proof. Put

By the closure property of S, A,M =

3 A,A, k

I=

=

that tr(M)

k

CI A, = My

I=

15i

= 0 mod k.

5k

Summing over allj, we get that M 2= kM.Thus the eigenvalues of M are 0 and k, each with their proper multiplicity. Hence tr(M) is an integral multiple of k, as required.

Lemma IX.2. The sum t~,=

C {tr(A))&, k = 1,2,. ..

AEO

is always an integer divisible by g.

Proof. Let G(*)= {A(*)),where A(&' represents the kth Kronecker power of A. Then the distinct elements of G(k) form a subgroup of GL(nk,2) whose order divides g, since the mapping A -, A(&),A E G, is a homomorphism. Furthermore tr{Ack))= {tr(A))' The previous lemma now implies that ok is divisible by g, since the set {A(*)]obviously possesses the required closure property. 1 = g is also divisible by g. Lemma IX.2 at We note that CAEO once implies the following useful result:

Lemma IX.3. Let f ( A ) be any polynomial over 2. Then We also require

2f(tr(A)) = 0 mod g

Lemma IX.4. The only element of G with trace n is the identity. Proof. Let A E G, tr(A) = n. Since A. = I, the eigenvalues of A are roots of unity and therefore of absolute value 1. The sum of n numbers of absolute value 1 can be n if and only if each of them is 1. Thus all the eigenvalues of A are 1. This implies that all the eigenvalues of B =

11. CongruentialReults of Minkowski

I75

+ +- +

A . A g - 1 are g, so that B is nonsingular. But (A - I)B = 0, and so A = I. This concludes the proof.

Z

10. A bound for the order of a finite integral matrix group We now combine these lemmas to prove

Theorem IX.6. Let t , = n, t,, tr(A) as A runs over G. Then (15) (16)

(n - t,)(n - t , )

. . . , t , be the distinct values assumed by ..

a

(n - t,) = 0 mod g, (2n)! = 0 modg

Proof. The polynomial f(3.) = (3. - ?,)(A - t 3 ) *

*

- (3. - t,)

is an integral polynomial which has the property thatf(tr(A)) is different from 0 if and only if tr(A) = n ; hence if and only if A = Z, by Lemma IX.4.Thus Congruence (15) is now a consequence of Lemma IX.3.To prove (16), note that tr(A) is an integer such that Itr(A) 1 < n, since it is the sum of n gth roots of unity. Thus the only values possible for tr(A) are

0,fl, * . . ,f n It follows that (n - t,)(n - t o ) .-.(n - t,) is a divisor of l-J':;i (n - i) (n i ) = (2n)! This completes the proof.

n;=, +

11. Congruential resalts of Minkowski We now turn to some work of Minkowski, which was motivated by the problem of determining the automorphs of a positive definite quadratic form. We first prove

Lemma IX.5. Let p be a prime, and let m be an integer > 2. Let A E Z , satisfy A P = I, (17)

176

M

FINITE MATRIX GROUPS

and A

(18)

3

Zmodm

Then A = Z. Proof. Suppose that A # Z. Then because of (18) we may write A = Z mE, where E E Z,, E f 0. We may assume that the greatest common divisor 6 of the elements of E is 1, since otherwise m can be replaced by m6. From (17) we have that ( I mE)p = Z, which implies that

+

+

pE

+ ( $ ) r n E 2 = 0 mod m2

Thus pE = 0 mod m, and since 6 m = p, since m > 2. But now

=

1, p

= 0 mod m. This implies that

and since p is odd, pE = 0 mod p2,E = 0 mod p, a contradiction. This completes the proof. We use this lemma to prove

Theorem IX.7. Let k , m be positive integers. Let A (19)

Ak = Z,

E

Z , satisfy

(20) A = Imodm Then ifm > 2, A = I ; and if m = 2, A2 = I. Pvoof. Assume first that m > 2. The result is certainly true for k = 1, since A = Z then. Assume that it has been proved for all k with at most t - 1 prime factors, t 2 1; and let k have t prime factors. Write k = pk,, where p is prime and k, has t - 1 prime factors. Then (19) reads ( A k i ) p = I, and powering (20), we get Aka = Zmod m. By the lemma, Akl = I. By the induction hypothesis, A = I. This completes the proof for m > 2. Next, assume that m = 2. Write k = P k , , where a 2 0 and k, is odd. Put B = A Z a .Then (19) reads Bkf = Z, and powering (20), we get B = Zmod 2. Suppose that B f I. Put B = Z 2E, where E E Z,, E f 0. If the greatest common divisor 6 of the elements of E is > 1, the previous case applies (with 2 replaced by 26) and we deduce that

+

177

12. A Theorem on a Natural Homomorphism B = I. Suppose then that

2k,E

S = 1. Then as before we have

+ 2k1(k, - 1)E2

E 0 mod

8,

2k1E = 0 mod 4, k, =Ornod2

a contradiction. Hence B = I in this case as well. Thus we have shown A2" = I

I we are through. Suppose that a 2 2. We have the identity 0 = A'" - I = (A - I ) ( A I)(AZ I) * . . (A'"" I) Since A = I mod 2, A2'+I-21mod4, l
= 0 or

+

+

+

+

+

+

+

12. A theorem on a natural homomorphism The theorem we are leading up to is concerned with the natural homomorphism Q, of Z onto Z/(m), where m is any positive integer. This homomorphism induces a homomorphism of GL(n,Z )

into

GL(n,Z/(m))

and so a homomorphism of G onto a subgrup (pG of GL(n, Z/(m)).The important result that follows from Theorem IX.7 is the following:

Theorem IX.8. Suppose that m > 2. Then the kernel of Q, is the identity, so that G is isomorphic to some subgroup of GL(n,Z/(m)). Hence the order of G divides the order of GL(n, Z/(rn)). Similarly, the order of G+ divides the order of SL(n, Z/(m)). The kernel of Q, consists of all matrices A of G such that A E Z mod m. Since also As = I and m > 2, the result is a consequence of Theorem IX.7. Proof.

As was mentioned previously, this result of Minkowski's was motivated by his study of the automorphs of positive definite quadratic forms.

178

IX FINITE MATRIX GROUPS

13. Automorpbs Let M be a real n x n positive definite,symmetric matrix. Let x be a real n x 1 vector. Then if I is the largest eigenvalue of M and p the smallest eigenvalue of M , so that I 2 p > 0, pxTx IX T M X I I X T X (21) A matrix A E GL(n, Z ) is an automorph of M if

(22)

ATMA = M

The automorphs of M constitute a subgroup of GL(n, 2) which we denote by G ( M ) . This group is a finite group. For if A is written as the matrix of its column vectors, A

=

[C,,C,,

...

C,,l

then (22) becomes (CITMC,) = M

Comparing diagonal elements, we have CITMC, = mi, 1 Ii 5 n The diagonal elements mi, of M are positive, since they are values assumed by the quadratic form associated with M. Then (21) implies that pCTCi 5 CTMC, = mi,, (23) n CTC,I mirlp, 1 4 i I Inequality (23) has only finitely many solutions in integral vectors Ci. Hence G ( M ) is finite. Conversely, let G = { A } be any finite subgroup of GL(n, 2).Then G is a subgroup of some group G ( M ) , where M is a positive-definite symmetric matrix. In fact, we may choose M

=

C ATA

AEQ

so that M is clearly symmetric and positive definite. Furthermore if B is any element of G, then BTMB =

C BTATAB= C (AB)T(AB)= M

AEQ

since AB ruhs over G as A does.

AEO

14. The Finite Subgroups of GL(2,Z) and GL(3,Z)

179

Thus the maximal subgroups of GL(n, Z) appear as the automorph groups of M for some suitable symmetric positive-definite matrix M. This approach has been very fruitful, especially in the modern theory of quadratic forms (see [lo] and [%I). 14. The finite subgroups of GL(2,Z) and GL(3,Z)

We conclude this chapter with an applcation of the derived theorems to the determination of the finite subgroups of GL(2, Z) and GL(3,Z). We let C, stand for any cyclic group of order n, and D, for the dihedral group of order 2n. Thus D, may be presented as the group on two generators x, y with defining relations x2 = (xy)Z = y" = 1

In addition we let T stand for the tetrahedral group of order 12, and 0 for the octahedral group of order 24. Thus T may be presented as the group on two generators x, y with defining relations xz

z

(xy)3 = y3 == 1

and 0 may be presented as the group on two generators x, y with defining relations x2

= (xu)' = y3 = 1

Suppose now that A E SL(2, Z), A of finite period. Then the eigenvalues of A are reciprocals of one another (since det(A) = 1) and are algebraic integers of degree 5 2. Since A is of finite period, A must be similar to the diagonal matrix of its eigenvalues, say diag(l, l/A). It follows that r is the order of A if and only if it is the order of A. Thus l must be a primitive rth root of unity, and so is of degree &). Hence p(r) < 2, which implies that r = 1 , 2 , 3,4,6. This limits A to the values f l , fi, f p , f p 2 , where p is a primitive cube root of unity, and so the trace of A can only be 0, f l , k 2 . It follows by Theorem IX.6that the finite subgroups of SL(2,Z) have orders dividing (2 - 0)(2 - 1)(2

+ 1)(2 + 2) = 24

In fact the only possible orders are 1,2, 3,4, 6 and any such subgroup must be conjugate over GL(2, Z) to one of the cyclic groups

{-a{U),{VI, C-UI

I119 where U = [-: - f J , V = [-:

i], and V 2 = -I,

U3

= 1.

180

IX FINITE MATRIX GROUPS

Now consider any finite subgroup G of GL(2,Z).Then G' is of index 1 or 2 in G, and G+ = C,, C,, C3,C,, or C6. Since G contains C, as a normal subgroup of index 2 if and only if it is equal to C,,, C, x C,, or D,, a brief analysis shows that it is only necessary to add the groups D , , D,, D,, D,, D6 to the above list to obtain all finite subgroups of GL(2,Z). The number of new groups so obtained which are not conjugate over GL(2,Z) is 8, and in terms of the canonical generators x, y of D, [which satisfy x2 = (xy), = y" = 11 a complete list is the following:

D , : x - [ ~0 - 1

'1,

]

0 , : ~ - 0[ 1

'1,

1 0 ' D , : x + [ l 0 -1 0 1 D3 : x + [ "

'1,

1 0

-1,

Y

--+

Y

+

y

-

-1,

u,

0 -1

D,:x-r

'1, '1

Y-v,

1 0

D,:x+r 1 0 '

Y-[

'1

-1O 1

Thus there are in all 13 mutually nonconjugate finite subgroups of GL(2,Z). A similar discussion shows that the order of a finite subgroup of SL(3,Z) must also be a divisor of 24, and that an individual element of such a group must have order 1,2, 3 , 4 , 6 . A careful enumeration shows that the possible groups are cl,c2, c39

c4,

c6

D,, D6, D,, Di,

181

Exercises and Problems the tetrahedral group T of order 12 the octahedral group 0 of order 24

It can be shown that there are in all 70 mutually nonconjugate subgroups of GL(3,Z). The maximal finite subgroups of GL(4, Z) have been determined by Dade, up to conjugacy. There are nine of them, and they are described as the automorph groups of suitable symmetric positive definite matrices M. See Dade’s paper [7] for a complete discussion. EXERCISES AND PROBLEMS 1. Let G be a finite integral matrix group of order g and degree n. Let p be any prime dividing g. Then p 5 n 1. Hint: Since p I g, G must contain a matrix A of period p. Sincep is a prime, at least one eigenvalue A of A must be a primitive pth root of unity. Thus degA = p - 1. But also, degA < n. Thusp 5 n 1. 2. Show that the order of a finite subgroup of GL(4,Z) must divide 5760.

+

+