Chapter 38 Applications of Characters

Chapter 38

Applications of Characters This chapter is devoted exclusively to applications of characters. Our applications are confined to two areas : group theory and integral group rings. Among other results, we prove the famous solvability criterion of Burnside and some classical theorems of Wielandt pertaining to existence and uniqueness of relative normal complements. We then establish a theorem due to Brauer and Suzuki which shows that if G is a group with a generalized ) z is the inquaternion Sylow 2-subgroup S , then G = 0 2 t ( G ) C ~ ( zwhere volution of s. In particular, any such group G cannot be simple. The proof of the most difficult case where IS1 = 8 is based on a work of Glauberman (1974). As a final group-theoretic application, we provide a characterization of PSL2(q), q odd, due to Brauer, Suzuki and Wall (1958). The second part of the chapter is devoted exclusively to the study of U ( Z G ) ,where G is a finite abelian group. Special attention is drawn to the problem of effective construction of units of !&'which is of fundamental importance in algebraic topology (see Milnor (1966, Theorem 12.8 and Corollary 12.9)) and unitary K-theory (see Novikov (1970, p.502))

1

Burnside's paqb theorem

In this section, we prove the famous solvability criterion of Burnside. All groups below are assumed to be finite and all characters are C-characters. The essence of the argument is contained in the following result.

Theorem 1.1. (Burnside (1904)). Let p be a prime and let a group G have a conjugacy class C with ICl = p" f o r some n 2 1. Then G is not 1179

Applications of Characters

1180

simple.

Proof. Assume by way of contradiction that G is simple. Fix g E C and let x be a nonprincipal irreducible character of G. Since G is simple, K e r x = 1. Since ICI > 1, G is nonabelian and so Z ( x ) = Z(G) = 1. Let I r r ( G ) be the set of all irreducible characters of G, let

x = {x f W G ) I P 1 x(%,

y = {A E WG)lPIX(1)}, and let p be the regular character of G. If x E X ,then x ( g ) = 0 by virtue

of Theorem 21.2.3 (v). Hence

Denoting by R the ring of algebraic integers of C , it follows that -1 = p a , where This contradiction completes the proof of the theorem.

Theorem 1.2. (Burnside (1904)). Let G be a group of order paqb where p and q Q W primes. Then G is solvable.

Proof. We argue by induction on (GI. We may assume that G is nonabelian and that G has a nonidentity Sylow q-subgroup Q. Then there exists 1 # g E Z ( Q ) and if C is the conjugacy class of G containing g , then (C(= (G : C&)) divides pa. If (C( = 1, then 1 # g E Z(G). Hence Z(G) is a nontrivial normal subgroup of G. If ( C (# 1 , then by Theorem 1.1 G has a nontrivial normal subgroup. Thus, in all cases, G has a nontrivial normal subgroup N . Since both N and GIN satisfy the hypothesis of the theorem, it follows by induction that both N and GIN are solvable. Thus G is solvable and the result follows. W

2

Wielandt's theorems

This section provides some classical results of Wielandt (1958). Our first aim is to prove a generalized version of the Frobenius theorem which guarantees the existence of the F'robenius kernel in Frobenius groups (see Theorem 36.1.8). All groups considered below are finite and all characters are (c -characters.

2 Wielandt's theorems

1181

Lemma 2.1. Let H o d H be subgroups of a group G and let S be a subset of G . Assume that every irreducible character of H with Ho C K e r x can be extended to a character X of G with S K e r A. Then there exists N d G such that S C N and H n N = H o .

x

Proof. As usual, we identify the characters of H / H o with characters of H having HO in their kernels. Let p be the regular character of H / H o . Then, by hypothesis, p can be extended to a character p of G with S C I i e r p . Setting N = Ii'erp, it follows that N a G, S C N and H n N = Ho, as required. I Let H be a subgroup of a group G and let H o d H . Recall that a subgroup Go of G is said to be a normal complement of H over Ho if Go d G, G = GoH and Ho = Go n H .

Theorem 2.2. G such that

(Wielandt (1958)). Let Ho<3 H be subgroups of a group

H n z-'Hz

Then

Ho for 011 z E G - H

(1)

Go = G - (U,EGX-~(H- Ho).)

is a unique normal complement of H over Ho. Proof. If Ho = H , then Go = G and there is nothing to prove. We may therefore assume that Ho # H . We divide the rest of the proof into three steps. Step 1. Uniqueness of the normal complement. Assume that N is a normal complement of H over Ho. Then, since H n N = Ho, the set ( H - H o ) n N is empty and so

z-'(H - H o ) z n N = 0 for all z E G Hence N

Go. On the other hand, by (l),the subsets 2

-1

( H - H 0 ) q Y-'(H - Ho)Y

(X,Y

E G)

are either disjoint or coincide depending on whether or not y is contained in Hx. Hence the number of elements in Go is equal to

Applications of Characters

1182

i.e. lGol = (G : H)IHol. But, since GIN % H / H o , we have

Thus lGol = IN1 and so N = Go, as required. Step 2. Our aim here is to demonstrate that every irreducible character $ of H with Ho E K e r $ can be extended to a character x of G with Go Kerx. We may clearly assume that 11, # 1 ~Let . x be a class function of G defined by

Then x extends t,b and we show that x is in fact a character of G, from which will follow that x is a desired character. Put Q = x - n l ~where , n = x(1) = $(1). Then (Y is a class function of G such that

-

...

t a r x r , a; E Write a = alxl iirreducible characters of G. Then

(c,

where x1 = 1 G , X 2 , . . . , x r are all

Let T be a right transversal for H in G. Since a vanishes on Go and class function, we have :

Thus

LY

is a

2 Wielandt's theorems

In particular, since T)

#

1183

l ~ ,

ul = IHI-' c(T)(g - n)) = -n SEH

Moreover, since CYH is a generalized character, we see that each a; E 2Z Next we note that

=

p1-l

c

.

la(h)12 = n2 t 1

hEH

since a vanishes off H and a~ = $ - n - 1 ~But . a1 = -n and Elzlus = n2 t 1. Hence, for i > 1, all a; except for one, say a j , are equaI to zero and aj = f l . Thus ct = fX - n l ~ where , X # lc is an irreducible character of G. Hence x = fX and, since x(1) = n > 0, we conclude that x = X is a character of G. Step 3. Completion of the proof. By Step 2 and Lemma 2.1, there exists a normal subgroup N of G such that Go C N and H n N = Ho.Hence we are left to verify that G = N H . Since

the result follows. We next observe that condition (1) of Theorem 2.2 forces H = Ho or # N G ( H ) , then we may choose z E N G ( H ) - H in which case H n Z - ~ H Z = H H~

H = N a ( H ) . Indeed, if H

which forces H = Ho. Thus Theorem 2.2 is of practical use only in the case where H = N c ( H ) . The following result has a wider scope of applications.

Theorem 2.3. (Wielandt (1958)). Let H o d H be subgroups of a group G such that the following two conditions hold : (a) ((NG(H) : H ) , ( H : Ho))= 1 (b) H n s - l H z E HO for all 2 E G - N G ( H )

Applications of Characters

1184

Then

(i) G has at most one normal complement of H over Ho. (ii) G has a normal complement of H over HO if and only if there exists a subgroup M of G such that Ho

cM

4 N G ( H ) and

( N G ( H ) : M ) = ( H : Ho)

(2)

(iii) There exists at most one M satisfying (2) and, for any such M ,

GO= G - U , ~ Q X - ' ( N G ( H-) M ) z is a unique normal complement of H over H o ,

Proof. It is clear that (i) is a consequence of (ii) and (iii). Assume that there exists a subgroup M of G satisfying (2) and put N = N G ( H ) . Then ( N : M H ) divides both ( N : M ) and ( N : H ) . Hence, by (a), N = M H and so

( H : ( H n M))= (N : M ) = (H: Ho)

HO C H n M , we have H n M = HO and thus HO a N . Hence M / H o a N / H o . Since IM/Ho) = ( N : H ) and ( N / H o : M / H o ) = ( H : H o ) are coprime, we see that M / H o is a unique subgroup of N / H o of order ( N : H ) . Hence there exists at most one such M . Assume that M satisfies (2). We now show that Since

N n z-'Nz

C_

M

for all z E G - N

(3)

Indeed, assume that g E N n z-'Nz for some 2 E G - N . Put n = ( N : H ) , rn = ( H : Ho) = ( N : M ) Since g E N , we have g n E H and similarly g n E z-'Hz. Hence, by (b), gn E

H n z - ' H x C Ho

M

On the other hand, g n E M . Since (m,n) = 1, we deduce that g E M , proving (3). Applying (3) and Theorem 2.2, we see that the group Go given in (iii) is a unique normal complement of N over M . It is clear that GOis a normal complement of H over Ho. Finally, assume that K is a normal complement of H over Ho and put

2 Wielandt's theorems

M = N n K. Then Ho C H N , Ho clear that M a N . Moreover,

1185

Ii and so Ho C N n K = M . It is

so that G = K N and ( N : M ) = (G : I<) = ( H : Ho). Hence M satisfies (2) and, by the preceding paragraph, Ii = Go. This proves (ii) and (iii), as desired. W

Theorem 2.4. (Wielandt (1958)). Let H o d H be subgroups of a group G such that the following two conditions hold : (u) H n x - l H z C Ho f o r all x E G - N G ( H ) (6) ( ( H : Ho), (G : H ) I H o l ) = 1 If the equation xm = 1, where m = (G : H)IHol, has precisely m solutions in G , then these solutions form u characteristic subgroup of G which is a unique normal complement of H over Ho.

Proof. Condition (b) ensures that HO is a normal Hall subgroup of H . In particular, Ho is a characteristic subgroup of H and therefore Ho d N , where N = N G ( H ) . Note also that, by (b),

and so H / H o is a normal Hall subgroup of N / H o . Therefore, by the SchurZassenhaus theorem (Theorem 26.1.4), there exists a subgroup M / H o of N / H o with IM/Ho( = ( N : H ) . Observe also that, by (4) and (a), H o , H and G satisfy the hypotheses of Theorem 2.3. We now claim that M d N , from which will follow that M satisfies (2). Put .Mo = { g E Nlg" = 1). To prove that M a N , it suffices to show that M = Mo. Since (MI = (Hol(N : H ) divides m, we have M c Mo. Hence we need only verify that = IMol. To this end, put n = ( H : H o ) , t = ( N : H ) and s = (Hal. Then IN( = t n s and so ( N - Mol 5 t n s - ts. Put Go = ( 9 E GIs" = 1) and let y E G - GO. Then the order of y is not divisible by m and so there exists a prime p with pln and plo(y). Hence there is a power t of y with o ( z ) = p . Let Q be a Sylow p-subgroup of G with z E Q. Since p 4 (G : H ) , Q = u - l P u for some u E G and some Sylow psubgroup P of H . Then z = y-'ty E y-'u-'Puy and therefore

Applications of Characters

1186

If uyu"

E G - N, then by (a),

P n u~-~u-'PuYu-~ c H nU

c H~

Y - ~ U - ~ H ~ ~ U - ~

which is impossible since (p, IHol) = 1 and o ( z ) = p. Thus uyu-' E N and therefore G - Go = U Z E ~ ~ - ' ( NMO)Z (5) Because the number of subgroups conjugate to N is at most (G : N ) and IN - Mol 5 tns - ts, we deduce that

(G- Go1 5 (G : N)(tns - ts) = !GI-

m

By hypothesis, the set G - Go contains precisely IGl - m elements. Thus

IN - Mol = t n s - t s and lMol = t s Hence M = MO and so M Q N. Therefore, applying ( 5 ) and Theorem 2.3 (iii), we conclude that Go is a unique normal complement of H over Ho.

3

Generalized quaternion Sylow subgroups

All groups considered below are assumed to be finite and all characters are (r: -characters.

Lemma 3.1. Let X I , . . . ,xr be all irreducible characters of G, let S be a subset of G and let C be a conjugacy class of G such that SnCC = 0 . Let a = clxl t * t c,xr, c; E (1: , be a class function of G which vanishes on G - S . Then, for any x E C ,

Proof. Let s E S. Then, in the notation of Lemma 36.3.2, nc(s) = 0.

Therefore, if

then

3 Generalized quaternion Sylow subgroups

-

IGl ICI

--pc(s)

=0

1187

(by Lemma 36.3.2)

Hence /3(s-') = 0 for all s E S. Since, by hypothesis, a(g) = 0 for all g E G - S, we deduce that

as desired. H

Lemma 3.2. Let G be a nonabelian simple group, let x be a character of G and let g be an involution in G. Then x ( g ) E Z and x ( g ) I x ( l ) ( m o d 4 ) . Proof. Let p be a representation of G which affords x. We may assume that p(g) = diag(e1, ...,En) where n = x ( 1 ) and E i E { l , - l ) , 1 5 i 5 n. Since x ( g ) = ~1 , , + E ~ we , see that x ( g ) E Z . Since G is a simple nonabelian group, detp(g) = 1 and therefore ~ 1 . . . E = ~ 1. Hence, in the sequence ~ 1 , ... , E ~ -, 1 occurs 2m times for some m 2 0. Thus

+ .

x ( g ) = ~1

+ - . - + E ~ = ( n - 2m)- 2m = n - 4m

as desired. H

Theorem 3.3. (Brauer and Suzuki (1959)). Let G be a group with a generalized quaternion Sylow 2-subgroup S. Then

where x is the involution of S . In particular, (i) The centre of G/Ozt(G) is of order 2. (ii) G is not simple.

Applications of Characters

1188

Proof. For the sake of clarity, we divide the proof into three steps. Step 1. Here we demonstrate that it suffices to show that any such group G is not simple. Indeed, assume that every group with a generalized quaternion Sylow 2-subgroup is not simple. Then the group G/Opn(G) is not simple. Hence G has a proper normal subgroup N of even order. By Lemma 37.3.6, the Sylow 2-subgroups of N are either cyclic or generalized quternion. In the first case N has a normal 2-complement 0 2 1 ( N )by Corollary 26.1.17 and so N = 0 2 1 ( N ) C ~ ( zfor c ) some involution z E N . In the second case, the same is true by induction. Since all involutions in N are conjugate, we deduce that

proving the first assertion. Since 021(G/O,n(G)) = 1, it follows that the centre of G/O2n(G) contains an involution. On the other hand, the centre of G/Ozt(G) is a 2-group. Since the Sylow 2-subgroups of G/Oz#(G)are generalized quaternion, we deduce that the centre of G/Oy(G)is of order 2. Step 2. Here we treat the most difficult case where IS1 = 8, i.e.

S =< a,bla4 = 1, b2 = a2, b-'ab = a3 > We follow the arguments given by Glauberman (1974). If b is not G-conjugate to an element of < a >, then the image of b under the transfer homomorphism G + S/ < a > is nonidentity. Then G is not simple and therefore we may assume that all elements of G of order 4 are conjugate. Put C = C G ( ~and ) N = N G ( < a >). Then C = B x < a > where B is a subgroup of odd order and N = CS = BS. It is clear that

is a trivial intersection set in G and one easily verifies that N = N G ( D ) . Let p be a nonprincipal linear character of C with kernel B x < u2 >, i.e. if y E B x < a2 > P(Y) = if ED

{ '1

Setting X = lc - p , we have

if if

ED

yEC-D

3 Generalized quaternion Sylow subgroups

1189

Hence, for any given y E D ,

where A’ = X on N and A’ vanishes on G - N . Similarly, we find that for any y E N - D,X*(y) = 0. Thus

which implies that < A N , A N >= 4 and < A N , 1~ >= 1. By Frobenius reciprocity and Theorem 36.1.7, we have

< l G , X G >=< l N , X N >= 1 and

=
N,AN >=4

Since AG = ( 1 ~- )pG~is a generalized character of G, we see that

where l ~xI,x2, , x 3 are distinct irreducible characters of G and ~i E (1, -l}, i = 1,2,3. Owing to Theorem 36.1.7, we have XG(y)=O if y € G - D G and

~ ~ ( =y~ ) ~ ( yif ) y

Since AN(y) = 4 for any y E

ED

D ,it follows that

Now none of the elements of D G is a product of two involutions, since G cannot have subgroups of the form < z > < y >, o ( z ) = 2m, m 2 1, o(y) = 2, with < y > normalizing < z >. Hence DG n I I = 8, where I is the set of all involutions in G. (Observe that, in view of the structure of S, all involutions

Applications of Characters

1190

in G are conjugate). Hence, by Lemma 3.1, for any involution t of G, we have

Now put n; = xi(l), x; = x;(a2) = X ; ( t ) , y; = ni - xi. By Lemma 3.2, we may assume that xi E 1z and 2; ni(mod4), i = 1 , 2 , 3 (since otherwise G is not simple). If ls;l = n;, then t E Z ( x ; ) and G is not simple. It is also clear that G is not simple if G has a nonprincipal linear character. Thus we may assume that Iz;l < ni,n; > 1, i = 1,2,3. Hence each y;, i = 1 , 2 , 3 , is a positive integer divisible by 4. Applying (1) for y = 1 and y = t , we have

1 t €121 t E Z Z Z t '53x3 = 0 Equality (2) in the new notation can be written as E1Xi

Ezxg

&,Xi

12.1

n2

723

lt-t-t-=O

It foUows from (3), (4) and ( 5 ) that ElYl

t E2Y2 t E3Y3 = 0

-I E1Y12

E2Y2

E3Y32

nl

n2

n3

=0

It will now be shown that for any generalized character x of G,

c

< X,XG >= p\-l x ( t a ) zEB

Indeed, by F'robenius reciprocity,

< xc,X > = =

IcI-' 1CI-l

< x,XG >=< xc,X >. Moreover,

cX(Y>Xo cX(Y>Xo c

YEC

YED

= 2lcl-l

Y ED

X(Y) = 21cl-f

c

zEBx

x(4

3 Generalized quaternion Sylow subgroups

Since

b-l(Ba)b

= Ba-' and

1191

x is a class function, we have

which proves (8). Let E be a primitive complex /GI-th root of unity. Then the values of characters of G lie in Z ( E )Let . u E Gal( Q (&)/a)). Then each "xi, i = 1,2,3 is an irreducible character of G. Since the values of XG are integers and E; E B , we see that

Thus the irreducible character G

= 1G

Oxi

+

has the coefficient E ; in

ElXl

+ E2X2 t E3X3

Assume by way of contradiction that O X ; # xi. For example, let O x 1 = x 2 . Then, since x1(a2) and x l ( 1 ) are integers, we have 2 2 = q ,n2 = nl and y2 = y1. Moreover, 61 = ~ 2 Hence, . by (3) and (6) we find that E3713

= 1 - 2E1711,

E3Y3

= -2E1Y1

Substituting these values in (7), we obtain 2ElY,2 711

4yf =0 1 - 2&1711

whence y1 = 0, a contradiction. Thus O x ; = xj for all a E Gal(Q ( E ) / Q ), i = 1,2,3, which implies that x;(g) E Q n W E )= Z , i = 1 , 2 , 3 . It is easy to find an integer which is congruent to 1 modulo 4 and congruent to -1 modulo IGl/S. Hence there exists a E Gal(Q ( E ) / Q ) which fixes a

Applications of Characters

1192

primitive 4-th root of 1 and sends each m-th root of 1, m odd, to its inverse. Now let z E B . Then, by Schur's lemma applied to < z > x < a >, there exists a representation p; with character x; such that

We conclude therefore that for ad z E B,i E { 1,2,3},

(9)

x;(za)= x;(z-la) Further, from (8) we have E;

=< & A G >= p1-l

c

x:(za)

ZEB

Denote by B1 the subset of B - { 1) which contains precisely one element of the pair { z , z - ' } with z E B - (1). Then

C Xi(za) = xi(a> -t 2 C xi(za> zEBi

zEB

and therefore

GIBI = xi@) t 2

Since IBI and E ; are odd and Xi(za) E from (8),we have 2

G

IBI
> =

c

x(za>

zEBi

Z ,we see that x;(a)is odd. Similarly,

CX;(Z~)

ZEB

3 Generalized quaternion Sylow subgroups

1193

which shows that < x?,XG> is odd. Assume that there exists i E {1,2,3} with x ; ( a ) # ~ i .Then, since x i ( a ) is an odd integer, Ix;(a)l > 1 or x;(a) = - E ; . In any case, we have x?(a) > E;x;(u). Since the character x;(i = 1,2,3) assumes only integral values, we have x:((y) 2 E i x j ( y ) for all y E G. Thus, applying (8),we obtain

=

Ei(IBI-*

C xi(za>)

zEB

< Xi,XG >

=

E;

=

EiE;

= 1,

i.e. < x?,XG >> 1. Since < xf,XG > is odd, we have < x?,XG > 2 3. Because < x i , lc >= 0, we obtain CyEGxi(y) = 0 and therefore

Bearing in mind that if if and that x?(1)

Thus

> -&;x;(l), we deduce that

ED^

yEG-DG

Applications of Characters

1194

a contradiction. Thus xi(a) = E; for all i E {1,2,3). Note further that for any i E {1,2,3}, YES

= n; t x; t 6xi(a) = n; +xi t 6 ~ ;

+

and therefore nj zj 4- 6 ~ =i O(mod8). Since = n; - y; and 6 ~ ; -2ej(mod8), we have 2( n; - E ; )

- y;

O( mod 8)

As we noted above, 41~;. Therefore nj - E ; is even and n; is odd. Moreover, if y; is divisible by 8, then n; - E* E O(mod4), i = 1,2,3. Let 2k be the highest power of 2 which divides each of the numbers y1, y2, y3, k 2 2. To reach the final contradiction, we put m; = ~ ; / 2 i~=, 1,2,3. Then, by (7), we have

o = ~ 1 n 2 n 32mt~~2n1n3mit ~3n1n2m:

(10)

Therefore, since E ; , ni(i = 1,2,3) are odd, we may harmlessly assume that ml is even (and hence 81yl) and m2,m3 are odd. Invoking (lo), we deduce the following congruences modulo 4 :

+

Thus ~ 2 n 2 &3n2= O(rnod4), But, by (3),

3 Generalized quaternion Sylow subgroups

1195

+

and so 41(nl ~ 1 ) .It follows from the fact that 81yl that 4)(nl - ~ 1 ) .But then 4 divides (n1 -t ~ 1 -) (n1 - ~ 1 =) 2 ~ 1 a, contradiction. Step 3. We now complete the proof by treating the easier case where IS[ 2 16. In this case

S =< a,bla2n = 1, b2 =

b-'ab = a-1

>

(n 1 3 )

Setting Q =< a2"-* >, we see that Q is a subgroup of S of order 4. It is clear that CG(Q)= B < a > where B is a normal subgroup of CG(Q)of odd order. Setting N = N G ( Q ) ,it follows from

that N = BS with B a N . It is clear that ( B x where x is the involution of 5'. Now put

D =B



< 2 >)aN and B < a > Q N ,

-Bx < x >

Then D is an invariant subset of N. Let g E G and D n g-'Dg # 8. Then g-' < dl > g =< d2 > for some dl,d2 E D. Because the Sylow 2-subgroups of < dl > and < d2 > contain Q ,we have g E N . This shows that D is a trivial intersection set in G and N = N c ( D ) . Moreover, by Example 36.3.1, D is an exceptional subset in G . Now any element of N can be written in the form ta%' with t E B and b , s integers. Let A1

:N

+ (I: *

and

X2

: N + GL2((I:)

be given by

Xl(takba) = (-1)'

where E is a primitive 2n-1-th root of unity. It is clear that A 1 and X 2 are (I: -representations of N . Of course, X1 has degree 1 and so A1 is irreducible. We claim that is also irreducible. Indeed, if not, then X 2 is a sum of one-dimensional representations and so & ( N ) is abelian. But, since 6 # E - ~ and K e r A2 = B < z >, Az(a)Az(b) # A2(b)X2(a),a contradiction. Thus A 2 is an irreducible representation.

Applications of Characters

1196

Consider the generalized character a = 1~ -t x 1 - x 2 of N , where x; is the character of Xi, i = 1,2. By the definitions of D and N, each element from N - D is of the form t , t x or ta'b, where t E B. This easily implies that cr vanishes on N - D. Note further that, in view of the structure of S , no involution inverts an element of D. Hence, by Lemma 36.3.4, all involutions in G generate a proper normal subgroup of G. Thus G is not simple and the result is established.

4

The Brauer-Suzuki-Wall theorem

We begin by introducing the following terminology. Given a prime power q, let GL,(q) be the group of all 2 x 2 nonsingular matrices with entries in the finite field of q elements. This group with certain closely related groups plays an important role in group theory. The group GL2(q) is known as the general linear group (in dimension 2); its subgroup SL2(q) consisting of matrices of determinant 1 is c d e d the special linear g r o u p . The centre of GL2(q) consists of scalar matrices and the corresponding factor group PGL2(q) is called the projective linear g r o u p . Finally, the image PSL,(q) of SL2(q) in PGL2(q) is called the projective special linear group Our aim is to provide a characterization of PSLz(q) for q odd. This characterization constitutes an important part of the classical paper of Brauer, Suzuki and Wall (1958). By combining this characterization with some other results pertaining to the case where q is a power of 2, Brauer, Suzuki and Wall proved the following assertion. Let G be a group of even order having no subgroups of index 2. Assume that if A , B are cyclic subgroups of the same even order and A n B # 1, then A = B. Then either G E P S L 2 ( q ) for some q 1 3 or the Sylow 2-subgroup of G is normal and elementary abelian. A simplified version of the characterization of PSL2(q) for q odd was given by Bender (1974). Our presentation is based on Bender's approach.

.

T h e o r e m 4.1. (Brauer, Suzuki and Wall (1958)). Let t be an involution of a finite group G , let H = &(t) and let A be an abelian subgroup of H containing t and satisfying the following properties : (i) H = A < s > with an involution s # A . (ii) CA(S)=< t > and s-'as = a-* for aEI a E A . (iii) A n g-'Ag = 1 for all g E G - H .

4 The Brauer-Suzuki-Wall theorem

1197

(iv) All involutions in G are conjugate. Then G S P S L 2 ( q ) for some odd prime power q.

Proof. We wish to show that G has a subgroup Q of odd order q such that the following properties hold : ( 4 IGI = d!l4- l ) ( q (b) CG(X) = Q for all x E Q - (1). (c) NG(Q)= QB with an abelian subgroup B of order (l/2)(q - 1). (d) If B # 1, then for any subgroup B1 # 1 of B , Nc(B1) = N c ( B ) = B < u > with an involution u $ B inverting B . At the end of the proof, by appealing to a result of Zassenhaus (1936), we will show that this will complete the proof. In what follows, Steps 1-10 treat exclusively the most difficult case where IAl > 4. Step 1. Since A is abelian, it has exactly IAI irreducible characters which are all linear. Furthermore, each linear character of A takes values 1 and -1 on the involution t . Since the group A / < t > has precisely IAl/2 linear characters, we see that A has ]A1/2 linear characters x with X(t) = 1 and the remaining IA)/2linear characters of A satisfy ~ ( t=) -1. Given a character x of A , define xs by ~ " ( a=) x ( s - ' a s ) for all a E A . Then, if x E I r r ( A ) , then x" = x if and only if x ( a 2 )= 1 for all a E A since by hypothesis s-las = a-l for all a E A. Setting A0 =< a E Ala2 = 1 >, we see that x" = x if and only if A0 K e r x . By (i) and (ii), t is a unique involution in A . Hence the Sylow 2-subgroup of A is cyclic and so ( A : Ao) = 2. In particular, A has precisely two irreducible characters containing A0 in their kernel. Therefore A has precisely two irreducible characters x with x = x". Because, by hypothesis, IA1/2 2 3 we deduce that there exist linear characters X and p of A with X # A", p # p", X ( t ) = 1 and p ( t ) = -1. Step 2. Here we examine the induced characters A H and p H . Firts of all, by definition, we have

w.

X(h)+ X(h-')

if if

hE A hEH-A

p(h)t p(h-')

if if

hEA hEH-A

In particular, XH(t )= 2, p H ( t )= -2 and so AH # p H . Furthermore, since X # As and p # p", the induced characters AH and p H are irreducible.

Applications of Characters

1198

Step 3. Put D = A - { l}, a1 = ( 1 ~ -) AH~ and a 2 = AH - p H . Then it is easy to see that D is an exceptional subset of G, H = N G ( D ) and a l ,a2 vanish on H - D . Our aim is to show that

where xi = E B ~i, = 1,2,3, E E (1, -1) and 81,82,83 are distinct nonprincipal irreducible characters of G . First of all, we note that

(lA>H(x) =

[

2 0

if if

xEA X E H - A

and therefore

~ >= , 1. Hence ( 1 ~ =) 1~ ~ and < ( l ~ )1~ character of H and

+ 7,where 7 is an irreeducible

It now follows that q ( 1 ) = 0, q ( t ) = 0, 4 1 ) = 0, a2(t) = 4, y ( t ) = 1 7 ( s )= 7 ( s t ) = -1, P ( t ) = 2, P ( s ) = A H ( s t ) = 0 pH(t) = -2, pH(s) = pH(st) = 0 ctl

= lHt Y - x ~ ,a2 = X~ - p H

where l ~7, ,AH and p H are distinct irreducible characters of H . We also have

< a 2 , a 2 >= 2, < a1,CYz >= -1

4 The Brauer-Suzuki-Wall theorem

1199

Applying Frobenius reciprocity together with Theorem 36.1.7, we have a1G (1) = a f ( 1 ) = 0,

G "1 ( t ) =

a&) = 0

( t ) = a2(t)= 4, < l& >= 1, < lc,af >= 0 < ay,a:: >= 3, < "2G ,a2G >-- 2, < aF,af >= -1, G

"2

which proves (1). Step 4. Here we demonstrate that

X3(q2

-Xl(q2

=0

(3)

To this end, we first apply Suzuki's formula (Theorem 36.3.3) with a = al, to obtain

Note that O;(t)E = x;(t), i = 1,2; p ( I + ) = B P ( t ) , since I = tG and 111 = (G : H ) ( p is any character of G). For any character T of H , we have T ( ( I ~H ) + ) =

~ ( tt) +s) IAl

t TI4~ ( ~ t )

since I r l H consists of three H-conjugacy classes of involutions in H , with representatives t , s and st. The sizes of these classes are 1, respectively. Hence, substituting the values of characters, we obtain

y,y,

Applications of Characters

1200

proving (2). Similarly, applying Suzuki's formula for Q = a2 and taking into account that x3(1) = xl(1) (since af(1) = 0), we obtain (3). Step 5. Our aim is to show that there exists 6 E {-1, l} such that

(G : H ) = (21-41

+ S)(lAl t 6)

To this end, we first note that from a f ( t ) = xs(t) have xs(t) = 2 and xl(t) = -2. Because

- xl(t) = 4 and

(4) (3), we

= 1 t x d t ) - X2(t) = 0,

we have xz(t) = -1. Hence (2) can be rewritten in the form 21A12Xl(l)X2(1)= (G : H)m where

(5)

m = X1(1)Xz(1) t 4X2(1) - Xl(1)

(6) Because ay(1) = 1 t xl(1) - xz(1) = 0, we see that xl(1) and xz(1) are coprime. Thus (m,xz(l)) = 1 and so, by ( 5 ) , xz(1) divides (G : H). Consequently, ~ 4 1 is) odd and so 81rn by ( 5 ) . Since 81m and x2(1) is odd, we see that x1(1)(~2(1)- 1) is not divisible by 8. Hence xl(1) is not divisible by 4. The equality 1 t xl(1) - xz(1) = 0 implies that xl(1) is even. Therefore, applying (6) and the fact that (xl(l),Xz(l)) = 1, we see that (xl(l),rn) = 2. By (iv), H contains a Sylow 2-subgroup of G, so (G : H ) is odd. Furthermore, by (iii) and (iv), H is a Hall subgroup of G. Since A H , we therefore deduce that (21AI2, (G : H)) = 1. Applying ( 5 ) , we see that

(G : H ) = (1/2)Xl(l)X2(1)

(7)

and, since xl(1) = x ~ ( 1-) 1, 4[AI2 = m = xl(l)(x2(1) - 1) t 4 x ~ ( 1 ) = (Xz(1) - 1)' 4Xz(1)

+

= (XZ(1) t

Therefore xz(1) t 1 = -21A16, xl(1) = ~ 4 1 -) 1 = -21A16 - 2 for some 6 E {-1,1), which implies (4)by virtue of (7). Step 6. Setting q = 2lAl 6, it follows from ]HI = 21AI and (4) that

+

IGl = 4(4 t l)(q - 1)/2

4 The Brauer-Suzuki-Wall theorem

1201

The rest of the proof is based on group-theoretic arguments. Because H A tG, we have H G = AG and AG contains the centralizer of any of its nonidentity element. Let p ( z ) be defined by p ( 5 ) = I{(u,?I)Iuv= 5 , o ( u ) = .(?I)

Then

p ( z ) = 1{u E Glu #

5, O(U)

= 2,

U-'ZU

= 2}1

= x-'}/

and p(1) = ItGI = (G : H ) . If a E A - {l},then {U

E

G~u-'uu= a-', u #

U,

O(U) =

2) = A3

and therefore p(u) = IAl. Now choose y E G - H G with maximal p ( y ) . Then

( { ( a , b ) ( ~E bG - H G , O ( U ) = ~ ( b=) 2}( JG- H G J (G : H ) 2 - I{(a,b)lubE A', O ( U ) = ~ ( b =) 2}1 IGI - lAGl (G : H ) 2 ((G : H ) lAl(lA1- 1)(G : H)) [GI - 1 - ( [ A ]- 1)(G : H ) (G : H ) - 1 - lAl(lAl - 1) > IHI - (IAI - 1) - ( 2 1 4 t S)(lAl t 6) - 1 - lAl(lAl - 1) 2 1 4 - (I4 - 1) -

+

Since p(y) > 0, then by (iv) we may assume that t-lyt = y-'. Step 7 . P u t K = Cc(y), M = N ~ ( l i )k, = IKI,n = ( M : K ) and Mo the set of all involutions in M . It is clear that t E Mo. Because K fl H G = 1, we have (IKI,[ H I )= 1. In particular, k is odd and, for u E Mo, CK(U) = 1. Hence any element of K is a commutator of the form X - ~ U ~with U 2 E K and so u inverts K. It follows that K is an abelian group and for any a E K - (1) and any involution u of Mo, I< = c G ( a ) , U 'a?/,= (8)

Applications of Characters

1202

It follows from (8) that Ii < t > d M , MO = K t , M = KCn,l(t) and C,v(t) contains precisely one involution t . Since t E M n A and As consists of involutions, we have C M ( ~ =)M n H = M n ( A U As) = M n A . Therefore M is the semidirect product

Because ( ( I { ( ,IAI) = 1, it follows from (8) and (9) that

Step 8 . The set of all involutions inverting y coincides with K t . Therefore, by Step 6,

and

Now consider the set

L = Uv&t4oMCG(V) We wish to find the number of cosets of M and the number of involutions in L. If w E Mo, then M = KCM(V),K 4 M , K n C M ( V = ) 1, and IC,v(v)l = n. Because v E t M and each subgroup of A is normal in Cc(t), we have Cn,l(w) d CG(D).Hence, applying (lo), we have cM(v)

= M n g - ' ~ g for a~ g E CG(V) - CM(V)

Moreover, for u,w E

Mo,u # w, MCG(Un ) MCc(v)= M ,

since M

# Mz = M g ( z E c G ( U ) , g E C G ( V )implies )

and u = v since Cn,l(u)contains only one involution. It follows from the above that for all v E Mo,

4 The Brauer-Suzuki-Wall theorem

1203

and L consists of

cosets with respect to M (since lM0l = \.lit1 = k). The set C G ( ~-)C M ( ~has ) precisely IAl involutions (these are the elements of As). Hence L - M contains precisely klAl involutions and L contains precisely kIAI k involutions. Step 9. Here we show the existence of Q satisfying (a) - (d) under the assumption that L = G. So assume that L = G. Then

+

(G : H ) = k(IAI t 1) (13) Invoking (4) and (13), we see that [ A ] 1 divides (21AI t & ) ( [ A [ 6). Hence, taking into account that the even integer 1AI is at least 6, we obtain :

+

6 = 1, k = 2lAl+ 1 = q

+

(14)

Because [MI = kn, IHI = 21AI it follows from (12), (13) and (14) that n = IAl. Now put Q = li and B = A. Then it is easy to verify that Q satisfies (a) - (d). Step 10. Here we show the existence of Q satisfying (a) - (d) under the assumption that L # G. By (lo), for any z E G - M ,the set {Mzglg E K } contains precisely k cosets modulo M . Moreover, if g E K ,u E Mo,we have MCG(u)g= MCG(g-'ug) L. Therefore G - L contains at least Ic cosets modulo M . Accordingly,

( G : M ) L ( w - k + l ) + k = - 2klAl + 1 n n Now any coset M z E G- L contains at most one involution. Indeed, if v1,02 are distinct involutions in Ms,then 1 # I = 0102 E M and vi'tv1 = z-'. From (10) and (9) it follows that C M ( Z n) MO # 8. Hence either o ( z ) is even and q centralizes an involution of < z >or v1 inverts CG(Z)and therefore centralizes C M ( Zn) Mo.It follows easily from (10) and (9) that M z = Mvl C L , a contradiction. Thus the number of involutions in G - L is at most IG - LI/IMI. Therefore (G : H ) - k(lAl+ 1) I (G M ) - ILI/IMI

= ( G : M ) - ( Un - k + l )

Applications of Characters

1204

which implies

(-

(G: H ) - (G : M) 5 k(lAl+ 1) - 2k'A' - k + 1) n We now claim that

k = /A\ - 1, = 2 , 6 = -1

(17)

Indeed, if IM 5 IHI, then

This implies (17) by applying (11) and the fact that k is odd and 1AI is even. Assume that [MI> [ H I , i.e. l M l / l H l - 1 > 0. Then, by (15) and (16), we have n

(fi

- 1)

<

(G : M ) ( I M ( / I H (- 1)

whence k < IAI t 2. It follows from (11) that 6 = -1 and k 2 1A1 - 1. If k = IAI 1, then, since (IKl,IHI) = 1, it follows from (4)that IA1 1 divides (21AI - l)(IAl- l ) , which is impossible. Thus k = IA[ - 1. Moreover, since n = / A n M I , n divides IAl. Since any nonidentity element of K under the conjugation by any element of A n M - { 1) is not fixed, we see that n divides k - 1 = IAI - 2. Therefore n divides IAI - ([A]- 2). Hence n = 2 and (17) is established. It follows from (4) and (17) that

+

+

lKl = IAl - 1, I N G ( K ) = ~ 2(IAI - 1) Now put Y = (G - H G ) - KG. Since HG = A' and A - {l},K - (1) are trivial intersection sets, we have

4 The Brauer-Suzuki-Wall theorem

1205

If x E Y , then ICG(Z)~ is coprime to llil and [HI ( K G and H G contain centralizers of their nonidentity elements). Since l C ~ ( x ) divides l [GI, we have

where m is odd. It follows that ' consists of two conjugacy classes of G an( IcG(x)l = 21AI - 1. It is clear that CG(Z)C Y U (1). Because Y is a union of two conjugacy classes, we see that CG(Z)is a p-group for some prime p . Since IGl = IHI IIil-lC~(x)l, it follows that, for all x E Y , C c ( x ) is a Sylow psubgroup of G. If x E Y and x is conjugate to x-l, then there exists an involution inverting Z. By (iv], we may assume that this involution is t. Now the group C G ( X < ) t > has [CG(Z)~ involutions inverting x. Thus

and so as I<, we can choose CG(Z). Then

or

ICG(Z)~ = k = (A1 - 1 But this is impossible, since ICG(Z)~ = 2lAl - 1. Thus no element of Y is conjugate to its inverse. Let S be a nontrivial p-subgroup of G, P a Sylow p-subgroup of N G ( S ) and let T # p be a prime dividing I N G ( S ) ~Denote . by R a Sylow r-subgroup of N G ( S ) . Because there are no involutions which centralize or invert an is odd. element of Y , we see that I N G ( S ) ~ Moreover, since I< - (1) and A - (1) are trivial intersection sets, the indices of K and A in their normalizers are equal to two and since TI IKI or TI IAI, we see that N = N x ( R ) ,X = N G ( S ) ,is conjugate to a subgroup of K or A. Thus N is an abelian group and so, by Theorem 26.1.15, R has a normal complement in X. Since T was arbitrary, it follows that P Q X . Hence the intersection of two distinct Sylow p-subgroups of G is trivial. Moreover, P R 4 X ,X = P N and N is conjugate to a subgroup of K or A.

Applications of Characters

1206

Denote by Q a Sylow p-subgroup of G. Since Q - { 1) is a trivial intersection set, IQI - 1 C G ( X ) Q and IQ n xGI = 2

for all

3

E Q - {l},we see that

where The subgroup B is conjugate to a subgroup of A or K . Since IKI = \A1 - 1, we see that B is conjugate to K . It is now easily verified that Q satisfies (a) - (4. Step 11. Here we treat the case where IAl = 2. Clearly H =< t > x < s > is a Sylow 2-subgroup of G. Setting N = N G ( H ) , we see that N = H < b > where o(b) = 3. We may assume that G # N (otherwise, for q = 3 conditions (a) - (d) are fulfilled). Let ~1,212E Hs be involutions. Then 01 centralizes ~1212and so 01 E H . Hence, if Ha: # H then Ha: contains at most one involution. Note also that H b U Hb2 has no involutions and that G contains precisely (1/4)IGI involutions. This easily implies that each Hs in G - N contains exactly one involution. Let u be an arbitrary involution. If H b u has no involutions, then Hbu N and u E N. On the other hand, if v E Hbu is an involution, then u inverts the element wu of order 3 and H b C N . Thus, for any involution u E G - N , 2 = u-'Nu n N has order 3 and is inverted by u. Now { u } = H u n N G ( Z ) , for otherwise N G ( Z )n H # 1. Hence

C G ( Z )= 2, N G ( Z ) = Z < u

>, INc(2)l= 6

Each subgroup of order 3 of N is inverted by precisely 3 involutions. Note also that N contains 4 subgroups of order 3. Therefore G - N contains precisely 12 involutions, while G contains 15 involutions. Thus 1GI = 15 !HI = 60. It is now clear that for q = 5 conditions (a) - (d) are fulfilled. Step 12. Here we treat the case IAl = 4. We apply the same argument as in the previous step. Put N = NG(< t > x < s >). Then one easily verifies that N 2 Sq. Assume that NG(< b >) C N , where b is an element of order 3 in N . Then counting the number of involutions in cosets of N , we see that [GI = 8 3 . 7 and that for q = 7 conditions (a) - (d) are fulfilled.

-

5 Applications to U ( Z G )

1207

Assume that N G ( < b >) $ N .lThen N G ( < b >) = C ~ ( b ) uwhere , u is an involution which inverts Cc(b). Put q = ICc(b)l,q 2 9. One easily verifies that N G ( < b >)x, x !$ N G ( < b >), contains at most two involutions. Denote by ,s the number of cosets NG( < b >)z containing precisely rn involutions. Since, by the structure of H , any involution of N G ( < b >) commutes with four involutions outside of N c ( < b >) and since the number of involutions in N G ( < b >) is q, we have s2 = 2q. Hence

ltGl = (1/8)(GI = q

+ 292 + si = 5q + si

(G : N G ( < b >)) = IG1/2q = 1 t ~2 t si t SO = 1 t 2q t s1 t so It follows that 0 5 s1 5 ( l / q - 4)q(19 - 2q). Thus q = 9, s1 = 0 and . the conditions (a) - (d) are fulfilled for Q = Cc(b). IGl = 8 9 ~ 5 Now Step 13. Completion of the proof. Here we use the following grouptheoretic fact due to Zassenhaus (1936). Let G be a 2-transitive permutation group of degree q t 1 which has no nonidentity permutation fixing 3 points. If the stabilizer B of two points is abelian of order (1/2)(q- 1)and is inverted by some involution in G, then G E PSL2(q) for some odd prime power q. Let Q and B satisfy (a) - (d). For each x E G , put Q" = z-lQz. Consider the permutation representation n of G given by

Then one easily verifies that n(G) Z G and that either q = 3, G Z PSL2(3) or n(G) satisfies the conditions of Zassenhaus mentioned above. This completes the proof of the theorem. W

5

Applications to U ( Z G )

Let G be a finite abelian group. A number of results of algebraic topology (see Milnor (1966, Theorem 12.8 and Corollary 12.9)) and unitary I<-theory (see Novikov (1970,p. 502)) suggest the importance of finding out precisely what units exist within the ring ZZG.' In this long section, we provide some insight into the problem by using characters.

Applications of Characters

1208

A. Preliminary results. Let A be an abelian group. A system elements of A is called independent if ayl

a;2

- - - aik = 1

{al,a2,.

implies each u l i = 1

..,ak} of

nonidentity

(n;E 21; )

An infinite system M of elements of A is said to be independent if every finite subsystem of M is independent. Owing to Zorn’s lemma, we can select an independent system that contains only elements of infinite order and is maximal with respect t o this property. The cardinality of this system r ( A ) is an invariant of A , called the torsion-free rank of A. For convenience of reference, we next quote some standard facts pertaining to algebraic number theory. An algebraic number field is a finite field extension of the field Q of rational numbers. Let Ii be an algebraic number field. Then the integral closure of Z in K is also called the ring of algebraic integers of I<. Denote by cn a primitive n-th root of 1 over Q .

Proposition 5.1. Let R be the ring of algebmic integers of an algebraic number field K . Then (i) R is a free 211 -module with ( R : Z ) = (K : Q ). (ii) Supjwse K = Q ( E ~ ) . Then R = Z[E,]and ( R : Z ) = ( K : Q ) = cp(n), where cp is the Euler function. Proof. See Janusz (1973) or Lang (1970). H Proposition 5.2. (Dirichlet’s Unit Theorem). The unit group of the ring of algebraic integers of an algebraic number field K is a direct product of a cyclic group of finite order and a free abelian gmup of rank r1 t 1-2 - 1. Here r1 is the number of real roots and r2 the number of pairs of complex conjugate roots of the defining equation of I< over Q . Proof. See Janusz (1973). U Corollary 5.3.

For any given n > 2,

> XF < - ~ n> X F


if n is even if n is odd

5 Applications to U ( Z G )

1209

where F is a free abelian group of rank ( 1 / 2 ) p ( n )- 1. Proof. This is a direct consequence of Proposition 5.2. It is a consequence of Corollary 5.3 that there exist T

= (1/2)(p(n)- 1

units ul,.. . , u r of Z[E,]such that any unit u of Z[E,]can be written uniquely in the form u = &;I u;2 . . .u:

, the torsion subgroup of U = U ( Z [ E ~ ]The ). with ni E z1; and 6 E< f ~ >, set (211,. ,.,u,} is called a fundamental system of units of U . The determination of such a system requires deep results and delicate computations. In fact, with the exception of some particular values of n, no specific system of fundamental units is known. Proposition 5.4. Let n be a prime power. Then the units ( 1 < a < n / 2 , ( a ,n ) = 1 )

( 1 - E : ) / ( 1 - E,)

are independent and genemte a subgroup of finite index in U(Z[E,]). Proof. See Washington (1982). W Proposition 5.5.

Let n be a prime power with p(n) 5 66. Then

( 1- E:)/( 1 - E

~ )

( 1 < a < n / 2 , ( a , n) = 1 )

is a fundamental system of units of U ( Z [ E , ] ) .

+

Proof. It suffices to show that the class number h i of Q ( E ~ E ; ' ) is equal to 1 (see Washington (1982)). The latter being true by virtue of a result due to van der Linden (1982), the required assertion follows. We next quote a deep result due to Franz (1935). If we write e(E) = 1 - E if E # 1 and e(E) = 1 if E = 1.

E

is a root of 1 , then

Proposition 5.6, (Franz independence lemma). Assume m > 1, let and let g,,, =< E~ >.

ambe the set of all primitive m-th mots of 1 over Q

Applications of Characters

1210

If

a,, E E

am,are

integers such that (i) a, = a c - ] , (ii) x u c = 0 , and (iii)

for all linear CC -characters

x of g,,,,

n e(x(W

=1

,€Om

then all a, = 0. Proof. See Franz (1935). As a preliminary to our next result, it will be convenient to record the following notation : an is the set of all primitive n-th roots of 1 over Q . !I!a is the group of all n-th roots of 1 over Q

.

Q n = Q (En).

Lemma 5.7. Let n = ps, where p is a prime, and let N = NQ ,/Q *. (i) Assume p [ s, 1 # E E !I!#,and 6 E !DP. Then N ( ~ - E )= ( l - ~ ) ~ - ' N(1- ~ 6 ) = (1 - E P ) ( ~ - c)-'

(ii) Assume PIS,

and let 1 #

N(1-

E)

=

{

E

E

Qn.

(1-E)P

1 - 9

Then if

EE!Ps

if

E$!%

Proof. In both cases, the first formula is simply a question of degree :

For the proof of the second formula, we require the following identity : 1

-en

=

n

(1 -OX)

8

aroot of I

X€Qn

To prove (l),we set X = 8-1 in nxEQn(X - A) = X n- 1 to infer that

r n -1

=

n n

(e-I

-A)

A€'€",

=

&Qn

=

e-n

d-l(l-eX)

n (i-ex)

(1)

5 Applications to U ( Z G )

1211

Now multiply by en and we obtain ( 1 ) . Returning to the second formula of (i), we note that

( & 6 ) P = &P E Q s It follows that the conjugates of €6 are all &A, X E

N(1 -&a)

=

n

(1 - E X )

A€@,

= [A€..

n

(1 - E X )

1

QP.

Therefore

(l-&)-l

= (1 - EP)( 1 - &)'I

using (1). In case (ii)) E P E Q and \vp C Q s. Hence, if E 4 Q s, the conjugates of E are all E X , X E \vp. Applying ( l ) ,we therefore deduce that

n

N(1-€) =

(1-EX)

A€O,

= 1-&P as required.

Lemma 5.8. Let R C S be commutatiwe rings such that S is integral over R . Then T E R is a unit of R if and only if T is a unit of S . Proof. If T E R is a unit of R, then obviously versely, assume that T € R is a unit of s. Say T-"

+ T ~ T - ( ~ - ' +)

6 .

+

T,

=0

T

(Ti

is a unit of S. ConE R)

exhibits the fact that r - l is integral over R. Multiplying this equation by P , we obtain T(TI T 2 f t * * * t TnTn-l) = -1,

+

thus completing the proof.

B. Torsion units. Throughout, G denotes a finite group and R a commutative ring. A unit u of RG is said to be trivial if u = rg for some T E U ( R ) ,g E G. Our aim

Applications of Characters

1212

here is to exhibit a class of rings R for which all central units of finite order in RG are trivial.

Lemma 5.0. Let E ~ , E z , . . , ~ tbe n-th roots of unity over Q and let a l , . . . ,at be positive rational numbers such that a1 - at = 1 . If

+ - +

a = al&l t ... t at&t is an algebraic integer, then either cr = 0 or E I = ~2 = - - - = € 1 .

Et.

Proof. Observe first that la1 5 1 and la1 = 1 if and only if ~1 = - = It therefore suffices to show that if a # 0, then la1 = 1. Assume by

way of contradiction that a # 0 and that la1 < 1. Let E be a primitive n-th root of unity over Q Then, for any u E G d ( Q ( E ) / Q ), In(cr)I < 1. Hence INQ (a)[< 1, contrary to the fact that the norm of cr is a nonzero integer. o the lemma is true.

(e)q

.

Lemma 5.10. Let (Y be an algebraic number and let n be a natural number such that ncr is an algebraic integer. If (a1 = a,a2,. , a t } is the set of all Q -conjugates of a , then either a is an algebraic integer or in the ring Z [ a l ,a2,.. . ,at]at least one prime divisor of n is a unit.

..

Proof. Let E;, 1 5 i 5 t , be the elementary symmetric function of t variables and degree i. If

t ' ' t at

f ( X )= X t t

*

(a; E Q )

is the irreducible monic polynomial satisfied by a , then Assume that a is not an algebraic integer. Then there exists i E { 1 , . . . ,t } such that E;(cr1,cr2, ...,at) = a / b for some integers a and b with 6 > 1. Hence, denoting by p a prime divisor of 6 , we see that a l p E Z [ q ,0 2 , . , , a t ] .Furthermore, p divides n because

.

E ; ( n a l ,na2,.

. . ,nat) = na(a/b) E z

Finally, since ( a , p ) = 1, there exist c,d E 23 such that

uc

+ dp = 1

5 Applications to U ( Z G )

Therefore as desired. W

1213

1 1 a - = -(ac+ d p ) = c . - t d E P P P

Z[Cul,rrZ .,at] ,..

The following result for the case where R is the ring of algebraic integers was established by G. Higman (1940).

Theorem 5.11. (Saksonov (1971)). Let R be an integral domain of characteristic 0 and let G be Q finite group. Assume that no prime dividing u9g is Q unit of finite order in the order of G is a unit of R. If u = CgEG RG with u1 # 0 , then u = u1 - 1. In particular, all central units of finite order in RG are trivial.

Proof. Let z = CgeG zgg E RG be a central unit of finite order. Then zt # 0 for some t E G and therefore u = zt-' is a unit of finite order with u1 = zt # 0. Hence it suffices to show that if urn = 1 and u1 # 0, then 21 = 211 * 1. To prove this assertion, let F be the quotient field of R and let p be the regular matrix representation of RG. Denote by x the character of p. Since urn = 1, p ( u ) is similar to diag(E1,. . . , E ~ ) ,n = IG(,where each E ; is an m-th root of 1 over F . Hence

x ( u ) = u1 - n= ~1

+. - . + E ~

By looking at the ~ ( u " )where , ( p , r n ) = 1, it is also clear that the set {Pl = U l r P Z , .

..

, P S I

of all Q -conjugates of u1 belongs to R. Thus Z[Pl,P2,...,PSl

GR

Because nu1 is an algebraic integer, Lemma 5.10 ensures that u1 is also an algebraic integer. But then, by Lemma 5.9, ~1 = - = E" which implies that p(u) = diag(E, . . . , E ) = P(E l), where E = E I . Thus u = E - 1 as desired.

-

-

C. The isomorphism class of U ( Z G ) .

For any commutative ring R and any integer n 2 1, we write R" for the direct product of n copies of R. As before, E~ denotes a primitive n-th root of 1 over Q .

Applications of Characters

1214

Lemma 5.12. Let G be a finite abelian group of exponent n and, for any dln, let ad be the number of cyclic subgroups of order d in G. Then the integral closure of ZG'in Q G can be identified with

n

z[&dIud

dln

Proof. We keep the notation of the proof of Theorem 19.4.2. For any given subring R of F , the isomorphism of Theorem 19.4.2 (ii) induces an isomorphism

RG[zx,

7

3

* *

7

zx3]

+

n

R[&dIad

dln

of R-algebras. Thus, if we identify RG with its image, then ndlnR[~dIad can be regarded as an integral ring extension of RG. In particular, by integrally closed taking F = Q , R = Z3 and applying the fact that Z [ Eis ~ (Proposition 5.1 (ii)), we infer that the integral closure of Iu: in Q G can Z [ ~ d ] ~ d . be identified with

ndln

Lemma 5.13. Let S be a subring of a commutative ring R. Assume that the additive group R+ of R is finitely generated and that the torsionfree ranks of S+ and R+ are equal. Then the torsion-free ranks of U ( R )and U ( S ) are also equal. Proof. The hypothesis on ranks of S+ and R+ ensures that S+ has finite index, say m, in R+ and so mR E S . Since R+ is finitely generated, we also have (R+ : m R + ) < 00. Therefore it suffices to show that for any Z7Y

E

W),

- y E mR

implies x-'y E U(S) Because x - y E mR implies both 1 - x-'y and y-'x - 1 E m R C S , the result follows. W 2

The following result is essentially due to G. Higman (1940) (see also R.G. Ayoub and C. Ayoub (1969)).

Theorem 5.14. Let G be a finite abelian group. Then

U ( Z G )= f G x F where F is a free abelian group of rank 1

-(\GI - 21 t m 2

+ 1)

5 Applications to U ( Z G )

1215

Here m (respectively, I) is the number of cyclic subgroups of G of order 2 (respectively, the number of cyclic subgroups of G).

Proof. Let a d be the number of cyclic subgroups of order d in G and let n be the exponent of G. By Lemma 5.12, we have

n

ZG

Z[&dlad

= R, ,say

dln

and by Corollary 19.4.3,

Since, by Proposition 5.1 (ii), ( Z [ E: ~ Z]) = (Q( ~ d :) Q ), it follows that and R+ have the same torsion-free rank, equal to [GI. the groups (ZG)+ Because R+ is free, we may apply Lemma 5.13 to conclude that U ( m ) and U ( R )have the same torsion-free rank. Furthermore, by (1) and Proposition 5.2, U(2zG)is finitely generated. Hence, by Theorem 5.11,

U ( Z ) = fG x F for some free abelian group F . By the foregoing, we are left to verify that 1 r a n k ( F ) = -(]GI - 21 2

+ m + 1)

By Corollary 5.3, if d > 2 then the torsion-free rank of u ( z [ & d ] ) is equal to (1/2)(p(d) - 1. By the preceding paragraph, ranL(F) = rank(U(R)). Thus

1

= -(IGl 2 as desired.

- 21 + TTZ + l ) ,

Applications of Characters

1216

D. Effective construction of units of ZG. Throughout this subsection, we employ the following notation : n = the exponent of a finite abelian group G. ad = the number of cyclic subgroups of G of order d. r = (1/2)(IGI - 21 t m t l), where m and 1 are given by Theorem 5.14. E~ = a primitive n-th root of 1 over Q . All characters of G are assumed to be linear (c -characters. Thus any character x of G is a group homomorphism

Recall that the augmentation map is a homomorphism

given by

a u g ( C w) = g€G

c

zg

(% f

g€G

z1

We denote by V(=) the subgroup of U ( = ) consisting of units u with aug(u) = 1. For convenience, we shall identify any character x of G with its extension to the ring homomorphism

In particular, the principal character will be identified with the augmentation map. As we mentioned earlier, a number of results of algebraic topology suggest the importance of finding out precisely what units exist within the ring ZG. Recall that, by Theorem 5.14, there exists a system of r units

uI,u2, ...,u, in Z G such that every unit of Z G is represented uniquely in the form

We call (u1, . .. ,u r } a f u n d a m e n t a l system of units in 2 % ' and refer to each u;as a f u n d a m e n t a l u n i t , The problem we have alluded to earlier, may be stated as follows :

5 Applications to U ( Z G )

1217

Problem. Given a finite abelian group G, find a specific fundamental system of units in 1zG. This is an exceptionally difficult problem. For example, if G is cyclic of prime order p , then as we shall see below the given problem is equivalent to one of finding a specific fundamental system of units of the subgroup of U ( Z [ E ~consisting ]) of all units which are congruent to 1 modulo E~ - 1. The latter is, needless to say, a formidable task. In general, the main difficulty seems to be the fact that effective calculations in U ( = ) are intimately ). into account that, in general, no connected with those in U ( Z [ E ~ ]Taking specific fundamental system of units of Z[E,]is known, any optimism in solving the corresponding problem for U ( Z G )must be guarded.

.

Lemma 5.15. Let X I , ~ 2 , . . ,xs be a full set of representatives of equivalence classes of Q -conjugate characters of G. (i) If x,y E ZGG, then x = y if and only if xi(.)

= xi(y) for all i E {I,. . .,s}

(ii) A n element x E ZG' is a unit if and only if for all i E (1,. . . ,s}, x ; ( x ) is a unit of

Z[E,]

Proof. (i) Direct consequence of Theorem 19.4.2. (ii) Owing to Lemmas 5.12 and 5.8, x is a unit of Z G if and only if for all i E { 1,. . .,s } , x ; ( x ) is a unit in Imx;. Since Z[E,]is integral over Imxi, it follows from Lemma 5.8 that xi(%) is a unit in I m x ; if and only if xi(%) is a unit of Z[E,]. So the lemma is true. 1 Turning to the case where G is cyclic of prime order p > 2, we now prove the following result.

Theorem 5.16. Let G be a cyclic group of prime order p > 2 and let

be the homomorphism which sends g to E ~ Then . (i) K e r X = ~ ( t 1 gt t gp-'). (ii) A n element u E ZG is a unit if and only if aug(u) = f l

and

x(u)E U ( Z [ E ~ ] )

Applications of Characters

1218

(iii) The induced homomorphism x* : U ( Z G ) U(ZCJ[E~]) is injective and x*(V (Z G ) ) is the subgroup of V (Z [ E ~ consisting ]) of all units which are congruent to 1 modulo ( E ~ 1). (iv) The indices of x*(V(ZG)) and x * ( U ( Z G ) )in U ( Z [ E ~ are ]) p -1 and (1/2)(p - l), respectively. --f

Proof. (i) This statement follows from the equality

and the fact that { l , c p , .. . , E ; - ~ } is a %basis for Z [ E ~ ] . (ii) We first note that { l ~x} ,is a full system of nonconjugate characters. Hence the desired conclusion is a consequence of Lemma 5.15 (ii). (iii) If u E U ( Z G )is such that ~ ( u=) 1, then x(u2)

= aug(u2) = 1

= l&)

so u2 = 1 by Lemma 5.15 (i). Because, by Theorem 5.14, f G is the torsion subgroup of U(ZE),the assumption that p is odd forces TA = fl, whence u = 1. Thus x* is injective. Let F be the field of p elements. We next claim that

Indeed, since I)-

v-2

2

the ring Z [ E ~ ]-/ 1) (E coincides ~ with its prime subring. On the other hand, p E (cp - 1) by Lemma 30.1.13,proving (1). By (ii), we have

Because

P-1 i=O

i=O

5 Applications to U ( Z G )

1219

we see that each element of x*(V(ZG)) is congruent to 1 modulo ( E ~ 1). Conversely, let u = Cyzi A;&;(& E Z ) be a unit of Z [ Esuch ~ ] that u z l(mod(Ep- 1))

Because p E (2),

( E ~

l), we have ( c p - 1) n P-2

C Xi

(2)

Z = pZ . On the other hand, by

1(mod (

E ~

1))

i=O

Thus A0

+- A1 + ... + Ap-2

=1

+ kp

for some k E 23

Setting 5

= (A,

- k) * 1 + (A, - k)g +

* * *

t (Ap-z - k)gp-2 - kgp-l

it follows that ~ ( 5=) u and aug(s) = 1. Hence u E x*(V(ZG)), proving (iii). (iv) Since U ( E ) = {*1} x V ( W ) ,it suffices to prove that the index of x*(V(ZG))in U(Z[E,]) is equal t o p - 1. By (iii), x*(V(W)) is the kernel of the natural homomorphism V(Z[EPl)

+

~(W,I/(EP

- 1))

Hence, by (l),it suffices to show that this homomorphism is surjective. Now a typical element of V ( Z [ E , ] / ( E ~1 ) ) is a i- ( E -~ 1) where a is a positive integer such that 1 5 a < p . On the other hand, the element &; - 1 -= 1t E p + Ep - 1

and is mapped to a t (

E ~

* * *

+ &;-I

E U(Z[Ep])

l),as desired.

Corollary 5.17. Let G be a cyclic group of prime order p > 3 , let r = (1/2)(p - 3 ) and let ~ 1 , ...,u, be units of 1zG. Then (211,. ..,u,} is a fundamental system if and only if the image of < f G , 111,. . . ,u, > in U ( Z [ c p ] is ) a subgroup of index ( 1 / 2 ) ( p- 1). Proof. By Theorem 5.14, U ( Z ) = f G x F where F is a free abelian group of rank T . Hence { u ~ ,...,u,} is a fundamental system of units if and only if

U ( ZG)=< f G , ~

. . ,U, >

1 , .

Applications of Characters

1220

Thus the desired conclusion follows by virtue of Theorem 5.16 (iii), (iv).

We close by providing three examples which illustrate the interplay between the unit group of Z G and that of Z [ E ~ ] . Example 5.18. (Kaplansky, unpublished). Let G be a cyclic group of order 5 with generator g . Then

Proof. We first observe that the torsion-free rank of U ( Z [ E ~is] )1 and that ~5 1 is a fundamental unit (see Proposition 5.5). On the other hand, the element

+

21

= (g t 1)2 - (1 t 9 t g 2 t g 3 t g 4 ) = -g(g2 t g3 - 1)

is mapped to ( E S + 1)2 and has augmentation -1. Hence u is a unit (Theorem 5.16 (ii)) such that the image of < f G , u > is a subgroup of U ( Z [ E ~ ] of ) index 5 2 . Thus, by Theorem 5.16 (iv), this index must be equal to 2. Now apply Corollary 5.17. The next example deals with a more involved situation.

Example 5.19. Let G be a cyclic group of order 7 with generator g .

Then

Proof. By Proposition 5.5, { 1+&7, 1+&7+&;} is afundamental system of units of U ( Z [ E ~ ] ) Hence .

is also a fundamental system of units of U ( Z [ E ~ ] ) Observe . that the element u1

+

+

= (9 + - (1 t 9 t !I2 g3 t g4 g5 t g6) = g ( 2 t 29 - 9 3 - 94 - g5)

5 Applications to U ( Z G )

1221

is mapped to ( ~ t7 1)3and has augmentation -1. Hence, by Theorem 5.16 (ii), u1 is a unit. Similarly, the element

is mapped to

= (1 t E7)-l( 1 t E7

+

2 1 E7)-

and has augmentation -1. Thus, by Theorem 5.16 (ii), 212 is also a unit. Let V be the image of < fG,211, u2 > in U = U(Z[&7]). Because

it follows that ( U : V) 5 3. Therefore, by Theorem 5.16 (iv), ( U : V) = 3. Now apply Corollary 5.17.

Example 5.20. Let G be a cyclic group of order 8 with generator g. Then U ( Z G ) = ~ G

-+

is 1 and Proof. We first observe that the torsion-free rank of U(Z[&8]) that 1 &g t E: is a fundamental unit (see Proposition 5.5). Let

+

be the homomorphism which sends g to &8. Taking into account that { ~ , E ~ , E ; , E ; } is a Z b a s i s of Z [ & g ] , it is clear that

t~

~t g 7 ( )1~ ~E i

z31

(3)

We claim that the image of U(2G)is a proper subgroup of U ( Z [ E ] ) .To prove the claim, it suffices to verify that no unit in 2%'is mapped to 1 E8

+

+

&*:

Applications of Characters

1222

Assume by way of contradiction that x(z) = 1 + &8 + &; for some z E U(ZG). Then, by (3), there exist A; E Z such that 2

= 1t g t g2 t Xo(1 t g4) t h ( g t g5) t A2(g2 g6) t X3(g3 9')

+

+

Observe that {XI = ~ G , x J , x ~ , x } where , x2(g) = -1 and x3(g) = i, is a full set of nonconjugate characters of G. Thus x ; ( s )is a unit for i = 1,2,3. The latter is equivalent to the following system of equations :

It is routine to verify that this system does not admit integral solutions,

which gives a desired contradiction. This substantiates OUT claim. Setting 21

= (1tgtg2)-(g+g5)-2(g2tg6)-(g3tg7) = -g(g6 t 295 t g4 - g2 - - 1)

it follows that Therefore, by Lemma 5.15 (ii), '(L is a unit. Furthermore, x(< f G , u >) is a subgroup of U(Z[&g]) of index 2. Because the image of U ( = ) is a proper subgroup of U ( Z [ E ~ we ] ) , deduce that

We are therefore left to verify that the kernel of the induced homomorphism U ( Z G )-+ U(Z[&g]) is contained in fG. To this end, we employ Theorem 19.4.2 (i) to deduce that the map

ZG

+

1z x

z x z[i]x z q E 8 1

is an injective homomorphism. Since the unit group of the first three factors is finite, it follows that any 9 E U(zU=) for which x ( v ) = 1 is of finite order. This implies v E fG and so the proof is complete.

5 Applications to U ( Z G )

1223

E. Cyclic groups. Throughout, we employ the following notation : Qn is the set of all primitive n-th roots of 1 over Q . Q n is the group of n-th roots of 1 over Q . Q n ( X )= n,,,,(X - E ) , the n-th cyclotomic polynomial e(E)=

{

1;E

if if

E #

~

=

l

where E is a root of 1. By a character of Q n , we mean a homomorphism x : Q n --t In this subsection, we examine the following problem.

Qn.

Problem. Given a finite abelian group G, find a system of independent units of Z G of infinite order which generates a subgroup of finite index.

Our goal is to present a theorem, due to Bass (1966), which solves this problem for cyclic groups. The proof is based on the application of the following important result. Theorem 5.21. (Bass (1966)). Let m 2 1 be an integer. Assume that a,(& E \urn) an? integers such that

(1)

a, = a,-1

and such that, for all chamcters x of

n 4x(aUc Qm,

=1

eE@m

Then a, = 0 for all

E

#

1.

Proof. We wish to show that conditions (1) and (2) imply that a, = 0 if E @ !Dm and that & ' a, a, = 0. Once this is accomplished, the result will follow by virtue of Proposition 5.6. A remark that will be used implicitly is the following. If x is the identity map, then (2) becomes

Suppose that x has trivial kernel. Then (3) implies (2) since an automorphism of the field Q = Q ( E ~ ) Em , E am.

x extends to

Applications of Characters

1224

The theorem being vacuous for m = 1, we argue by induction on m. For the sake of clarity, we divide the rest of the proof into three steps. Step 1. Our aim here is to prove

by showing that

and that

=0

(6)

To establish ( 5 ) , write m = ks and choose a character QErl, and such that ImX = !Pa. Setting

x such that K e r x =

Ifplm, pprime, then

be =

C

ac

S€@,

ag

(0 E qS)

x(c)=e

it follows that ( 5 ) is equivalent to be = 0, for all 0 # 1. But be, 6 E Q 8 , satisfy the hypothesis of the theorem. Thus, by induction, be = 0 for all e + 1. To prove (6), write rn = p"q with (p,q) = 1 and put N = NQ / Q. It is a m consequence of Lemma 30.1.3 that if

Applying N to (3), we may thus neglect all but prime power order Segregating the latter according to the relevant prime, we deduce that

E'S.

a, = 0. If i > 1, aPpi is a union of nontrivial cosets Therefore Cy=lri CcEQp, a, = 0. Thus (6) is of Q p , so it follows from ( 5 ) (with k = p ) that CcEQp,

5 Applications to U ( Z G )

verified. Step 2.

1225

Assume that m = pt, ( p , t ) = 1, p prime. Here we prove that a,

= 0 for all 1 #

E

E

\Et

(7)

To this end, put N = NQ m / t .~ Applying N to (3) and using Lemma 5.7 (i) and Lemma 30.1.13, we obtain 1 =

n

N(e(~))"c

r€Qm

n

E

E

E

#

e(E)(P-l)ac

n

[~(EP)~(E)-']"s~

6€@p

\Et

1

-

Pd

where

and

It follows from (6) that d = 0. By ( 5 ) (with k = p), we also have

Thus

1=

e(E)'C,

E & # E

c,P

= pa,p - a,

Qt

I

We next claim that to prove (7), it suffices to verify the hypothesis of the theorem for t and ce. For then we can appiy induction and conclude cb = 0 for E 1, i.e. a, = p e p for 1 # E E Qi. Selecting T such that pr = l(modt)

+

Applications of Characters

1226

and setting q = p', it follows that E B = E , whence a, = qa,q = qa,. This proves that a, = 0 and substantiates our claim. That t and c, satisfy (1) is obvious. To prove (2), suppose x is a character of 9t. If K e r x = 1, we can apply the remark of the first paragraph of the proof. So assume that K e r x = Q k , k > 1. Extend x to a character of !ljm = XPt x !Pp by the identity map on ! l j p . By (2), we then have 1 =

n

e(x(W

EEQm

n

[e(X(E)Yc

= { EE@t-*k

n

]

l-I e ( V 6

[Ec'@k6e@p

n

6E@p

1

e(X(E)6)Qc61

Since k > 1, we have &Qk a,6 = 0(6 E QP) by ( 5 ) ; hence the second factor is 1. We now apply N to the first factor and calculate as above :

(since e(1) = 1). This establishes (2) for the C,(E E Q,)and so (7) is proved. Step 3. Let m = p t , plt, p prime. Here we complete the proof by showing that a, = 0 for all 1 # E E Qt (8) Again, we apply N = to (3), this time applying Lemma 5.7 (ii) :

N~

m/Q

t

Collecting exponents in the second factor leads to sums over nontrivial cosets of Q p , and these sums are zero by ( 5 ) . Hence EEQt

5 Applications to

U(ZG)

1227

and these data clearly satisfy (1). For ( 2 ) , suppose x is a character of Q t and write rn = pns, ( p ,s) = 1. Extending x to a character of Q m , it follows from (2) that

If K e r x n Qpn # 1, then we can collect exponents modulo Q p to eliminate the second factor. Otherwise, we apply N to the last equation and calculate as above, applying Lemma 5.7 (ii). Hence, by induction, we see that pa, = 0 for all 1 # E E qt, proving (8). The following observation now completes the proof. For any E E Q m am,we can choose a prime p , m = p t , so that E E Q t . Then a, = 0, b y virtue of (7) and (8). Thus the theorem reduces to Proposition 5.6, by applying (4) and (2). Theorem 5.22. (Bass(1966)). Let G be a cyclic group of order n > 2 , let m be a multiple of y ( n ) and, for each dJn,d > 2, let Qd be a fixed element of order d in G. Put

Sd = {g:ll < p and, for any s = g$ E s d , put

< d/2

and

( p , d ) = 1)

Then Udln,d>2{"slS E S d ) is a system of independent units of ZG of infinite order which genemtes a subgroup of finite index in U ( Z G ) .

Proof. By assumption, cp(n)lm,and so cp(d)lmfor all dln. Since ( p , d ) = 1, we also have E l ( m o d d ) . Thus d l l - p m and therefore us E ZG. Let x be a linear character of G . If x(gd) = 1, then

If X(gd) = E

#

1, then Ed- 1 - - -0 E-1

Applications of Characters

1228

Thus X('1LS)

= (1 t E

t

=

* * *

(-)1l--&&P

E U(Z[E])

(9)

Hence, by Lemma 5.15 (ii), a, E U ( Z G ) . Since G is cyclic, G has exactly one subgroup of order dln. Hence, by the proof of Theorem 5.14, the torsion-free rank of U ( Z G )is

Since [Sdl = (1/2)(p(d) - 1, we are therefore left to verify that the units us are independent. To prove independence, suppose we have a relation

We must show that all as = 0. Our aim is to apply Theorem 5.21, so we must construct the necessary data. Denote by @d(G)the set of all elements of order d in G . If g E G, then g E @d(G)for a unique d)n. We now define the integer 6, as follows : if d 5 2 set 6, = 0; otherwise exactly one of the following four cases occurs : (a) g E sd,(b) 9-' E s d , (c) = gd, (d) 9-l = gd- Define

bg = and

[

mag

(-m>&s,,

as

if if

g Esd g = 9d

-

bg-1 = bg

Now choose an isomorphism xo : G iJn and put b, = b, if E = xo(g). We claim that n and the be(& E S,) satisfy conditions (1) and (2) of Theorem 5.21; if sustained, it will follow that b, = 0 for all E # 1. In particular, for any s E s d , bX(+) = mas = 0 and thus a, = 0, as required. To substantiate our claim, we first observe that condition (1) follows from the construction. Hence, we are left to verify that for any character x of $,

5 Applications to U ( Z C )

1229

Let u: be the image of us under the automorphism of Z G which sends each element of G to its inverse. Then, by (lo), we have

and the product of this with (10) implies

Applying the character 0 = x o xo of G to the above, we find that

and this, by (9))is

the last equality being true by the definition of the 6,. This proves (11) and hence the result.