he elements of finite order in the group of units 0: … · elements of finite order 2225 this...

. COMMUNICATIONS IN ALGEBRA, 17(9), 2223-2253 (1989)

HE ELEMENTS OF FINITE ORDER IN THE GROUP OF UNITS 0:T GROUP RINGS OF FREE PRODUCTS OF GROUPS

A.I. Lichtman

University of Wisconsin, ParksideKenosha, WI 53141

U.S.A.

and S.K. SehgalUniversity of Alberta

Edmonton, Alberta, T6G 2G 1Canada

1. Introduction. Let G = . * GJ be a free product of groups. \Ve studyJEJ

in this paper thc clements of finite ordcr in the group of units U( ZG) of

the integral group ring ZG. Let us denote by VI (ZG) the group of units of

augmentation one. Then V(ZG) = :tVI(ZG). We prove (Theorcm 1) that

if all the groups G j(j E J) are abelian and II E VI (ZG) ha$ order m then

one of the groups Gj(j E J) contain$ an element oj order m.

We consider then thc group ring KG, whcre K is an arbitrary ficld

and G is a free product of arbitrary groups and construct (see Theorcm 3) a

simple ring R,which contains KG and

R'::::'.KG 01":C 6

where C is the cartesian subgroup of G, ~ is the universal field of fractions

of the free group ring J{ C (see Cohn [2], Chapter 6). When all the groups

Gj(j E J) are finite and the set J is finite we show that R is in fact a matrix

2224 LICHTMAN AND SEHGAL

Gj(j E J) are finite abelian then we prove that every unit u of finite order

in U1(7LG) is conjugate in R to an element h of finite order in one of the

groups Gj. We conjecture that u and h are conjugate in fact in the group of

units of the rational group ring QQ.

Our proofs depend on the technical results of section 3. We consider

here the quotient groups Gjfk(C) (k = 1,2,...) where fk(C) is the kth

term of the lower central series and study the centralizer of the subgroup

fk-I(C)jfk(C) in Gjfk(G). The main result here is Theorem 2 which gives

a description of this centralizer. We hope that this theorem might be also of

independent interest.

2. The units of finite order in ZG, when G is a free product of

abelian groups.

Let (Hi,4>;) be an inverse system of finite p-groups and homomor-

phisms. Its inverse limit l~ Hi is called a pro-p-group (see [17], [19]). We

will consider in the sequel the Lazard-Zassenhaus-Jennings p-series Mi( C) of

the cartesian subgroup of * Gj. The inverse limit G = limGjMi(C) (i =JEJ ~

1,2, . . . ) is called a free pro-p-product (see [17], [19]); it contains the free

pro-p-group C = I~CjMi(C).

PROPOSITION 1. Let G be a group, H be a normal subgroup which is a

pro-p-group and let the index (G : H) be not divisible by p. Then H has

a complement S in G, i.e. there exists a subgroup S of G such that G is a

split extension, G = H )<IS.

ELEMENTS OF FINITE ORDER 2225

This extension therefore splits; there exists thus a subgroup 5i of Gi such

that Gi~Hi )<35i. Clearly, 5i':::!.G;jHi.

We pass now to the group Gi+l = G/Mi+1 (H) and obtain there a

complement 5i+1 of Hi+1' The only problem is to make sure that 5i+1

is picked in such a way that the homomorphism <Pi+1: Gi+1 -+ Gi with

the kernel Mi(H)/Mi+l(H) will map 5i+l onto 5i- This will imply that

the system of groups Gi = Hi )<35i is an inverse system with respect to

the system of homomorphisms <Pi and simultaneously the systems (Hi, <Pi)

and (5i, 4>i)(i = 1,2,...) are inverse systems.

limHi, 5 = lim 5i we will obtain that G-::::.H)<35.~

Since G = lim Gi, H -

Let thus 4>i+1(5i+d = Qi # 5i. Since the kernel of the homomorphism

4>i+lis a p-group we obtain that Qi-::::.5i+1-::::.G/H and hence Qi is a comple-

ment of Hi in Gi. Hence, (see [18], 9.1.2) Q, is conjugate to 5i. Let gi E Gi

be an element such that gi1 Qigi = 5i and gi+1 E G.+1 be an element such

that 4>i+l(gi+d = gi. We see that <Pi+1(g;+\5,+lgi+d = 5i and hence we

have found a complement of Hi+1 which was required. The proof is complete.

PROPOSITION 2. Let G be a split extension G = H )<35, where H is a pro-

p-group and 5 is a finite abelian p' -group. Let u be an element oE p' -order

m in U1(7LG). Then ml(5 : 1) and u = (1 + V)8, where 8 is an element oE

order m in U1(Z5). Furthermore, v E w(G, H), the ideal oETLG generated

by all the elements h - 1, (h E H).

PROOF: We point out first of all that the elements of finite order in the

set 1 + w(G, H) must be p-elements. Indeed, let x = 1 + w, w E w(G, H) and


an extension of a p-group if by a pi_group S. The image x of the element x

in Z[GJ has now a representation x = 1 + tV, tV E w(G, if) and it is known

([IJ (3.1» that such an element must be a p-element. This contradicts the

conditions x ~ 1 and xq = L

Now consider the homomorphism TLG -+ ?LS, induced by the group

homomorphism G -+ S. This homomorphism induces a homomorphism of

groups of units U(7LG) onto the group of units U(ZS). Clearly, the kernel of

this homomorphism is the subgroup U(1 + w(G, H», of units congruent to

one mod w(G,H). Furthermore, if T = 9P{U} then (T: 1) = m and hence

Tn (1 + w(G, H» = 1 because (1 + w(G, H» contains no p'-elements. This

implies that T is isomorphic to a subgroup of U(ZS); since 5 is abelian, T

is in fact isomorphic to a subgroup of S.

PROPOSITION 3. Let G be a group which is a split extension of a p-group

H by a p'_group S. Let

U = (1 + v)s with v E w(G, H), s E 5

be an element of order m in U1(7LG). Assume that p f m and let

z = 1 + v + vl+s + ... + vl+s+---+sm-2. (2.1)

Then z is invertible in ~ and

Z-lUZ = s. (2.2)

Furthermore, s is conjugate in G to an element from the support of u.


which belongs to [(G, (G] n TLG= [7LG,7LG]. It follows that a conjugate of

s appears in the support of u. For details of this argument, see [13, p. 342].

LEMMA 1. Let G be an arbitrary group, H <JG be a residually torsion free

nilpotent normal subgroup of G, K be an arbitrary commutative domain of

characteristic zero. Then the ideal 1 + w( G, H) contains no units of finite

order.

PROOF: The proof is reduced immediately to the case when K is a field.

Furthermore, the augmentation ideal w(K H) of K H is residually nilpotent

by Hartley's theorem (see [61 or [15]). Since w(G, H) is a free module over

w( K H) we obtain easily that00

nwn(G,H) = 0n=l

and a routine argument completes the proof.

THEOREM 1. Let G = . * Gj be a free product of abelian groups, u be aJEJ

unit of order m in 7LG. Then there exists an element h of order m in one of

the groups Gj. Furthermore, if all the groups Gj(j E J) are finite and J is

finite then this element h can be chosen in such a way that it is conjugate

to u in the group of units of every group ring <IXG /11,1;(C)).

PROOF: Consider first the case when all the groups Gj(j E J) are finite

and J is finite. Take an arbitrary prime p such that p f (Gj : 1), (j E J) and

p f m. Let once again G = Ii;::G/Mi(C),C = I~C/Mi(C) (i = 1,2,...).Then

,,~


and p t (G : C). Then by Proposition 1, G = H )<IS. By Proposition 2, u

can be written as

u = (1 + iJ)s,s E S,iJ E w(G,C), ord(s) = m. (2.3)

On the other hand, u is an element of ZG. Hence

u = al91 +a292 +...ak9k (ai E Z;9i E G; i = 1,2,...,k). (2.4)

Now consider the homomorphic images Gi - Gj Mi( C)~G j Mi( C).

Every group Gi is a split extension of a p-group Ci = CjMi(C) by a group

Si~S = IT G) , We write as usual Supp u = {91,92,...,9dand let, afterjEJ

renumeration,

0"1 = {91,92 ...,9kl}' 0"2= {gkl+I,...,9k2},...,O"r = {9kr-l+l,gkr-l+2,"',9k,J

be the partition of Supp u into subsets of elements which are conjugate in all

the homomorphic images Gi(i = 1,2,...). Let Ui be the image of U in ?LGi.

We pick io such that for every i > io the images of the subsets 0"1, 0"2,. . . , 0"r

in Gi give the partition of Supp Ui in Gi into classes of conjugate elements.

For every given i > io we consider the homomorphism of G onto Gi

and obtain from (2.3)

Ui = (1 + Vi)Si (2.5)

where Vi E W(Gi, Cd, Si E S. If now

z = 1 + iJ+ iJI+..+..2+...+..m-2 (2.1')

and Zi is the image of z in ?LGi then by Proposition 3 Zi is invertible in qJi

and

Z;lUiZi =Si (i = 1,2,...). (2.6)


NoW Proposition 3 implies that there exists a class, say (J"tsuch that 8i is

conjugate in Gi to an element of (J"t. It is important that in Gi+l' 8i+l is

conjugate to an element from the image of (J"t,because the image of this class

in Gi under the natural homomorphism Gi+1 -+ Gi is the only class which

contains the conjugates of 8i.

We obtain therefore that the class (J"tcontains elements whose images

in Gi (i > io) are conjugate to the image of the element 8; and we conclude

from (2.6) that these images are conjugate to the element Ui in the group

of units of the group ring Q(GjMi(C». Since 8 has order m, every element

of (J"thas order m in every homomorphic image Gi (i > io) and hence the

elements of (J"t have order m in * G)'. This implies that one of the groupsjEJ

Gij E J) contains an element h of order m which is conjugate to an element

of (J"tand the proof is completed for the case, when all the groups G j(j E J)

have finite orders.

To obtain the proof in the general case, when Gj(j E J) are arbitrary

abelian groups we point out, first of all, that a routine argument reduces it

to the case when all these groups are finitely generated and J is finite. Let

for every given j E J

Gj~Hj x Fj

where Hj is the torsion subgroup of Gj and Fj is free abelian. Let F be the

normal subgroup of G = .n Gj generated by all the subgroups Fj(j E J).)EJ

Thus, F is the kernel of the homomorphism

*G.-+ *H.=HjEJ) jEJ)

induced by the natural homomorphisms

l~;


The Kurosh Theorem implies that the normal subgroup F is isomorphic to

a free product of a free group and of groups which are conjugate in G to

some of the subgroups FjU E J), i.e. F is isomorphic to a free product of

residually torsion free nilpotent groups and hence it is residually torsion free

nilpotent by Malcev's theorem (see [12]). Lemma I therefore implies that

the subset 1+w( G, F) contains no elements of finite order and as in the proof

of Proposition 2 we conclude that the image of u under the homomorphism

TLG- TLH is a unit of order m in 7LH. Since H is a free product of finite

abelian groups HjU E J) and the truth of the assertion for this case has

already been established we obtain that there exists an element of order m

in one of the groups Hj ~ Gj and the proof is complete.

3. Let F be a noncyclic free group, fl(F) = F, f2(F),..., be the lower

central series of F. Lemmas 2-4 of this paragraph give some sufficient con-

ditions for an automorphism </>of F to induce a nonunit automorphism in

fk(F)jfk+I(F) (k = 1,2,...).

LEMMA 2. Let F be a free group with bases Xi, i E I, and <I>an automor-

phism of F. Assume that there exists a subset J ~ I, card J ~ 2, and an

element jo E J such that <I>(Xjo)= uxio' where io E I\J, € = xl, and u is a

word in xj,j E J. Then for every k,</>induces a nonunit automorphism of

fk(F)jfk+1(F).

PROOF: The assertion is obvious when k = 1; we suppose therefore that

k > 1 and consider a free nilpotent group F of class k + I with generators

Yj ,j E J, and define the homomorphism t/Jof F to F by

t/J(Xj) = Yj,j E J,t/J(Xio) = t/J(U)-E; t/J(Xi) = 1, i E I\{JU {io}}. (3.1)


We order now the set x j, j E J, in such a way that there exists an element

".. < XJ' . It follows that the commutator.., Jl 0

v = [xjo' X jl , . . . , , X j1], J

k

; is basic of the weight k and we see that t/J(v) f= 1. On the other hand, t/J

maps the element cP(v) to 1 and this implies that v f= cP(v) mod fHl(F),

which means that cP induces a nonunit automorphism of fk(F)jfHl(F).

LEMMA 3. Let F be a free group with free generators Xl, X2 and <i> the

automorphism of F which is defined by

cP(x d = X2X~I; cP( X2) = X~>,

where €l is an arbitrary integer and €2 = :f::1. Then the automorphism

induced by cPin fk(J)jfHl(F) is the identity if k = 2 and €2 = -1; and it

is nonidentity in other cases.

PROOF: The group f2(F)jf3(F) is an infinite cyclic group and it is gen-

erated by the commutator u = [Xl, X2]. We now have

cP(u) = cP([Xl,X2]) = [X2X~I,X~2] = [X2,X?]X;I. (3.2)

Since [X2,x.-l] == [X2,Xl]-1 (mod f3(F» (see e.g. [5], 11.1.) we ob-

tain from (3.2) that cP(u) == u(mod f3(F», when €2 = -1, and cP(u) ~

u (mod f3(F», when €2 = 1; i.e. cP induces the unit automorphism of

f2(F)jf3(F) in the first case and a nonunit automorphism in the second

Ii"'


We remark also that 4>induces a nonunit automorphism of the free

abelian group Fjr2(F) and we will now consider the case when k 2: 3. In

order to prove that 4>induces a nonunit automorphism of rk(F)jrk+I(F)

when k 2: 3, we consider the commutator v = [Xl, X2,..., X2], which is basic~ J

k

of weight k (with respect to the ordering Xl < X2); we will prove that the

elements v and

"'( ) [< 1 <2 < 2 ]VI='I'V = X2xI,xI ""XI

~ J~

k

are not equal mod rk+I(F). We will use the relation

VI == [X2, X?,..., x~2](mod rk+I(F» (3.3)

which follows easily from (3.2).

We will construct now a k nilpotent group U of class k + 1 with two

generators e, I and a homomorphism t/J of F to U such that t/J(v) =1=1

and t/J(vd = 1; this will imply that the elements v and v 1 are not equal

mod rk+I(F).

Let V be a free abelian group of rank k + 1 with generators e =

el, e2,..., ek+l. We consider Vasa free module over the ring of integers

7L and define the endomorphism 0 of the module V by O. el = e2, O. e2 =

e3,. ..,0. ek = ek+h O. ek+1 = O. We see that 0 is nilpotent of index k + 1

and ( = 1 - 0 is an automorphism of V.

Let U be a group which is a split extension of V by the infinite cyclic

group H with generator I, where I-leI = ( . e, e E V. One can check now

that (f, e] = 1-1e-l Ie = (1 - 0 . e = O. e and this implies easily that U is

nilpotent of class k + 1; we also see that U is generated by two elements e


We define now a homomorphism .,p of F onto U by

.,p(xd = e,.,p(X2) = f.

iWe have [f, e£2, e£2,. . . , e£2] = 1, which implies by (3.3) that .,p(VI) = 1; on, ,~

k

'the other hand [e,f,...,f] = (-O)k. e = (-l)kek. Thus, .,p(vd' v '

k

.,p(v) f- 1 and as it was remarked above the assertion follows.

= 1 and

COROLLARY. Let F be a non cyclic free group, R <JF. Then for every k

the centralizer in F offk(R)jfk+I(R) coincides with R.

This fact follows from Passi's theorem [14]. We obtain it here as a

consequence of Lemmas 2 and 3.

Indeed, let x E F\R, and FI = gp{x,R}. If F,/R has order r then by

the Schreier's theorem R has a free basis

xn,x-kYiXk(iEI), k=O,1,...,n-1;

when Fd R is infinite cyclic group, the elements

X-}YiX} (i E I);j 2: °

form a free basis of R. If (F1 : R) > 2 Lemma 2 implies now that x cannot

belong to the centralizer of fk(R)jfk+I(R); when (FI : R) = 2, we obtain

i this from Lemma 3. The corollary is proven.

LEMMA 4. Let F be a noncyclic free group with a system of generators

xi(i E I) and <Pbe an automorphism of F, which is defined by


Let for arbitrary k,;jJbe the automorphism Ofrk(F)jrk+I(F) induced by ljJ.

Then ;jJ= 1, when k is even and ljJ= -1, when k is odd.

PROOF: The basic commutators of weight k form a basis of rk(F)jrk+1 (F)

and the assertion follows from the identity

[fi~l ,fi~l,... ,fi~l] = (-1)k [fip !i2,..., iik] (mod rk+l(p)).

4. The centralizers of the factors of the lower central series of the

cartesian subgroup.

Throughout this paragraph G denotes the free product of the set of

groups Gj(j E J) and C is the cartesian subgroup of G. The group C is free

and we need an explicit basis of it. We may suppose that the set J is ordered

and then from results of Takahasi [20] it follows easily that a system of free

generators of C can be obtained by taking all elements of the form

u =[gr g +1 g ]

(9192...9r-l)-1, r ,...,.. (4.1)

when g.,. . ., gr,..., g.. (1 :S; r :S; s) are nonunit elements such that

gk E Gjk (k = 1,2,..., s), jl < h... < j... (4.2)

(We use the notations [x,y] = x-Iy-Ixy and yX = x-IyX).

We also will use a simple consequence of this result: if for some jo, A

is a subset of Gjo, then the set of commutators [h,g] (1 =I g E A,1 =I h E

G j , j =I jo) is a part of some system of free generators of C.

One can check also that any normal subgroup of the group G =G1 * G2 which contains properly the normal closure of the free factor G1


In order to avoid ,wme complications in the formulation~ of the results

we suppose in thi~ section that G =/: Z2 * Z2.

Let k be some fixed natural natural number. We denote by N the

centralizer in G of fk( C)jf HI (C).

PROPOSITION 3. Assume that for every) E J, N n Gj = 1. Then N = C.

PROOF: Kurosh's theorem implies that in this case N is a free group.

Since C is not cyclic and C <IN we now apply Corollary of Lemma 3 to

obtain that N = C.

PROPOSITION 4. Assume that N has a nontrivial intersection with some

of the subgroups G j. Then the following conditions hold simultaneously.

1) J contains only two element~, i.e. G = G1 * G2.

2) Let for clearness N n G1 f= 1. Then G2 f= Z2 implies G1 = Z2 and

G2 = Z2 implies G1 = ZJ.

The proof will be given in 4 steps.

Steg 1. Let for some )0 E J, N n G jo f= 1. In order to simplify the

notations we assume that )0 = 1 and let 1 =/: a E N n G 1. We prove that a

has order 2 or 3.

Indeed, let A = gp{a}. We have [h,g] E N n C for any 9 E A,

hE Gj(j f= 1). As it was remarked above all the commutators [h,g],g E A,

h E Gj (j =/:1), h f= 1, 9 f= 1 are a part of some free basis of C.

Suppose now that a has order more than 3; we apply the relation

,.


to obtain that in this case the conjugation by a induces an automorphism of

C which satisfies the conditions of Lemma 2 and hence a induces a nonunit

automorphism of rk(C)jrk+I(C), i.e. a f/.N. This contradiction shows that

a has order 2 or 3.

SteQ 2 We show that GI~Z2 or GI-:::=Z3.Indeed, let q E G1 \A. If A

has order 2 we consider the elements UI = [h,a], VI = [h,q], V2 = [h,qa],

and we obtain from (4.3) a-Iula = u;-I, a-Ivia = U;-IV2;if A has order 3

we consider the elements UI = [h,a], U2= [h,a2], VI = [h,q], V2= [h,qa]

d b . -1 -I -I -Ian 0 tron aUla = UI U2; a via = UI V2.

In both the cases Lemma 2 implies that a induces a nonunit auto-

morphism of rk(C)jrk+I(C), This contradiction shows that A = GI, i.e.

GI ~ Z2 or GI ~ Z3'

SteQ 3 We now prove that the set J contains precisely two elements,

I.e. G = GI * G2. Indeed, in the other case let GI1 G2, G3 be three different

subgroups from the family Gj,j E J, and g, h be nonunit elements from

G2 and G3 respectively; as above, 1 =1=a E N n GI. We take the elements

UI = [g,hI, U2 = a-I [g, h]a, and apply lemma 3 if a has order 2 and Lemma

2 if a has order 3. In both cases, a induces a nonunit automorphism of

rk(C)jrk+1(C), which contradicts the assumption a E N. Hence, card

J = 2, i.e. G = GI * G2.

SteQ 4 We prove that G2 =1= Z2 implies GI = Z2' Indeed, let hI =1= h2

be nonunit elements of G2. We suppose that GI = Z3. Then, as above, the el-

ements UI = [hI, a], u2 = [hl1 a2], vI = [h2, a], V2 = [h2, a2] are a part of some

free basis of C. Once more we obtain from (4.3) a-Iu2a = u;-I; a-Iv2a = V;I


LEMMA 5. Let G = GI * G2, wbere G1 = Z2 and G2 t= Z2, Z3. Tben

N = C if k is odd and N coincides witb tbe nonnal closure G1 if G1 is even.

PROOF: Let 1 t= a E G1. We see that the basis of G is given by all the

commutators [a, h], 1 t= h E G2, and

a-l[a,h]a = [a,ht1. (4.4)

We remark also that Proposition 4 implies that N n Gz = 1.

Let first k be an odd number. Lemma 4 together with (4.4) gives that

aft N; thus N n GI = 1;N n Gz = 1 and Proposition 4 implies that N = C.

If now k is even we see that GI ~ N and since N n Gz = 1 we obtain

that 81 = N and the assertion follows.

LEMMA 6. Let G = CI * Cz, wbere C1 is a cychc group of order 2 with

generator a and G2 is a cyclic group of order 3 with generator b; for a given

natural k let N be tbe centralizer ofrk(C)jrHI(C). Then N = C if k is

odd; N = G if k = 2, and N = G1, tbe normal closure of C 1, if k is an even

number more tban 2.

PROO F: The elements UI = [a,b]-I, Uz = [a,bZ]-1 are a basis for C and

it is easy to verify that

-I -1(

.1 2) b-1 b -I b-1 b -I

a uia = ui l = ,; UI = UZul; Uz = UI . (4.5)

When k is odd Lemmas 3 and 4 imply that a ri N, bN and we obtain

from Proposition 4 that N = C.

~..,,,


We recall that throughout this section it is assumed that G =I Zz * Zz.

Propositions 3 and 4, Lemmas 4 and 5 imply the following theorem.

THEOREM 2.For some natural k let N be the centralizer in G ofrk( C)jrk+l(C).

Then we have the following possibilities.

1) card J > 1 or card J = 2 but no one of the groups G1, Gz is Z2 or ZJ.

Then N = C.

2) card J = 2, G1 = Zz, Gz =I ZJ. Then N = C if k is odd, and N

coincides with the nonnal closure of the subgroup G1 if k is even.

3) card J = 2,G1 = Zz,Gz = ZJ. Then N = C if k is odd; N = G if

k = 2 and N coincides with the nonnal closure of G1 if k is an even

number more than 2.

COROLLARY. Let G = *G j. Assume that there exists ko such that for

all k 2: ko, the element 9 E G belongs to the centralizer ofrk(C)jrHl(C),

Then 9 E C.

Remark 1. It is well known that corollary of Theorem 2 is not true

when G = Zz * Zz.

Indeed, let G = gp{ a, b}, aZ = 1, bZ = 1. The subgroup R, generated

by the element t = ab is normal in G and a-I ta = t -1; we see also that

C = gp{ t2}. Thus, the centralizer of the unique nontrivial factor C =

r1(C)jr2(C) is R.

5. Let F be a free group, K F be its group ring over an arbitrary field K.

'T'1-~ _:_- T/r;" L_- LL- I '" " ~ ~


here we only point out that ~ can be obtained in the following way (Lewin

18]).

Let K < F> be the Malcev-Neumann power series of K F. Then ~ is

isomorphic to the subfield of K < F >, generated by K F.

We now give an account of the properties of ~ which will be needed in

our arguments.

1) Every automorphism of K F can be uniquely extended to an automor-

phism of~. (See Cohn [2], 6.5).

Let r i( F) be the i th term of the lower central series of F, then Fi =

Fjri(F) is a free nilpotent group of class i. Let <l>ibe the natural homo-

morphism K F -+ K Fi and ~i be the Ore field of fractions of ~,. We will

need the following property of the homomorphisms <l>i(i = 1,2,...). (See

Eisenbud and Lichtman [4]).

2) For every given i there exists a specialization Oi . ~ -+ ~i extending

the homomorphism <l>i,i. e. there exists a local subring ~ ;;2 T ;;2 K F

such that Td rad Ti'::::'.~iand (rad Ti) n K F = ker <l>i. Furthermore,

Tt <;::;T2 <;::;'..

andex>

UTi=~'i=1

The local subring Ti is called the domain of the specialization Oi.

3) If ( II ~i )j:F is an arbitrary ultraproduct of the fields ~i then K FiE[

generates in this ultraproduct a subfield, isomorphic to U.

It follows also from the relationship between the specializations and

'J~

iii


6. Let G be a group, containing a free normal subgroup F, KG be the group

ring of G, 6. be the uni versal field of fractions of K F. Clearly, the group

ring KG is isomorphic to an appropriate cross product K F * G j F; since the

automorphisms of K F are uniquely extended to the automorphisms of ~

(section 5) we can form the cross product R = 6. * G j F.

Now let i be given and Gi = Gjri(F). Consider the group ring KGi

and let Ri be its ring of fractions with respect to the subring K Fi where

Fi = Fjri(F). Since Gi is an extension of the free nilpotent group Fi by the

group (Gjri(F))j(FjC(F))'::::.GjF we conclude that Ri is isomorphic to an

appropriate cross product 6.i * G j F.

Now consider an arbitrary ultrafilter F on the set of indices I = 1,2,...

and form the ultraproduct (n Ri) j F. It is well known that there exists alEI

natural imbedding

KF ~ (n KFi)jF ~ (n 6.i)jF.lEI lEI

(6.1)

It is easy to verify that (6.1) defines an imbedding of suitable cross products

KG=(KF)*GjF~ ((nKFi)jF ) *GjF~ ((n6.i)jF ) * GjF. (6.2)lEI lEI

It is known also that the ultraproduct of fields (n 6.i) j F is a field; welEI

denote it by S. We obtain thus from (6.2) that KG is imbedded into the

cross product 5 * G j F in such a way that K F ~ S.

Now let G be an arbitrary group, F <IG be a free normal subgroup

and K be an arbitrary commutative field. We recall (see section 5) that the

subfield of 5 generated by K F is isomorphic to the universal field of fractions

of K F. We can now prove the following result.


isomorphic to a suitable cross product ~ * GIF. Furthermore,

R,::::~ 0KF KG. (6.3)

PROOF: Let gi (i E I) be a transversal of F in G. Since the elements

, gi( i E I) are linearly independent over S and ~ ~ S they are linearly

independent over~. Since ~ is G-invariant and ~ :::>K F we conclude

easily that the left vector supspace l: ~gi is closed also with respect to theiEf

multiplication; hence, it forms a subring R of (11 R;)/ F, which is isomorphiciEI

thus to .6. * G IF and to .6.0 K F KG.

COROLLARY. Let T be the domain of the specialization Ok . .6. -+ .6.k.

Then Tk and rad Tk and G generate in R subrings, isomorphic to suitable

cross products Tk * GIF and (rad Tk) * G / F; the quotient [jng

(Tk * GIF)/((rad Tk) * GIF) is isomorphic to a suitable cross product

(Tklrad Tk) * G/F'::::~k * G/F.

7. Throu ghout this section let G = * G) . Let G be the cartesian subgroupjEJ

of G and ~ be the universal field of fractions of KG. Proposition 5 implies

that R = .6. 0KC KG is a ring.

PROPOSITION 6. Assume that G =I- Z2 * Z2. Then the cen tralizer G (6.) of

~ in R is K.

.~


n

X = Lai9i,;=1

ai E 6. (7.1)

be an element which centralizes 6.. Since 6. is G-invariant we conclude

immediately that every sWllmand aigi in (7.1) must centralize 6.. But the

center of 6. is K (see Lewin [8]); the assertion will follow therefore if we

prove that for given elements 0 =/: a E 6. and 1 =/: 9 E G\C the relation

(ag)f = f(ag) (7.2)

is impossible when f is an arbitrary element of C.

To verify this statement we find a number ko such that for an arbi-

trary k ~ ko the local subring T"" which is the domain of the specialization

(h : 6. -+ 6.k, contains a and a-I. We consider the relation in the subring

Tk * G and its homomorphic image 6.k * G IC. (See Corollary of Proposition

5). Since a-I E Tk we obtain that its image a is nonzero in 6.k * GIF.

The subgroup fk-l(C)/fk(C) is in the center of the free nilpotent

group C/fk(C). Take now an arbitrary f E fk-l(C)/fk(C) and obtain

from (7.2) the following relation in 6.k * GI F

agf = fag. (7.3)

Since f is in the center of G k we obtain

agf = afg (7.3/)

or, because 0 =/: a E 6.k,

gf = fg. (7.4)


COROLLARY. Let G =/: Z2 * Z2' Then the center of R is the field K.

THEOREM 3. Let G = j~JGj be a free product of groups. Then the ring

R == 6. 0KC KG issimple. Ifall the groups Gj(j E J) are finiteand the set

J isfinitealso then R~Dnxn' where D isa skew fieldand n S; IT (Gj :1).}EJ

PROOF: Let A be a nonzero ideal in R. We will prove that A = R. Once

again let 9i( i E I) be a transversal of C in G. Pick a nonzero element x E A

which has the shortest representation of the form

n

X = L<Xi9ii=1

(Oi E 6.; i = 1,2,...,n). (7.5)

Multiplying, if necessary, x on the right by the element g;-1 0;-1 we can

assume that the firstsummand in the right side of (7.5) is 1. Pick an arbitrary

f E G and consider the element x - f-I X f. It is easy to verify that this

element has a representation of length S; n - 1 and hence itmust be equal

to zero. We obtain therefore

X=f-IXf

which implieseasily,via the uniqueness of the representation(7.5), that f

commutes witheverysummand °igi; since f is an arbitrary element of G and

KG generate 6. these elements commute with 6. and Proposition 6 implies

that x = 1, provided that G =/: Z2 *Z2. We have proved thus that R is simple

in all the cases except maybe the case when G = Z2 * Z2. But in this last

case (see Remark 1 in section 3) G = gp{a,t}, a-Ita = C1, where a2 = 1

and t has an infinite order, R coincides with the Goldie ring of fractions of

l'

2244LICHTMAN AND SEHGAL

Now let the set j be finite. In this case

{R: ~)l = (KG: KG) = II (Gj : 1).]Ei

Thus, R is a simple ring which has finite left dimension over its sub field ~j

this implies that R is a matrix ring of degree

n ~ .IIJ {Gj : 1)]E

over a field D and the assertion follows.

We will need in section 5 the following easy corollary of Theorem 3.

Let Kl 2 K be a field, containing K, ~l be the universal field of fractions

of K1G. Then ~1~K1 0K ~ and hence

~l 0K1C K1G~K1 0K (~0KC KG). (7.6)

We obtain therefore the following.

COROLLARY 1. Assmne that all the groups Gj(j E J) are finite and that

the set J is finite also. Let RI = ~l 0KC K1G. Then R1~{D.)nxn, where

DI is a field, isomorphic to Kl 0K D.

The proof follows from Theorem 3 and the relation (7.6).

COROLLARY 2. Assume that all the groups Gj(j E J) in Theorem 3 are

fiill te. Then R is a locally matrix ring.

PROOF: Follows immediately from the second part of Theorem 3.

ELEMENTS OF FINITE ORDER2245

imbed it into the ring ~ 0Q7 QJ'::::::.Dnxn.The ring Dnxn can be considered

thus as some ring of fractions of 7LG.

THEOREM 4. Assume that all the groups Gj(j E J) are finite abelian and

the set J is finite. Let u be a unit of finite order m in VI (Z G) and h be an

element of order m obtained in one of the groups G j in Theorem 1. Then

there exists an element c E GLn(D) such that C-Iuc = h i.e. the elements

u, hE VI (7LG) are conjugate in GLn(D).

We need first the following result which is a generalization of Corollary

1 of Theorem 1 in [10].

Let G be a finitely generated nilpotent-by-finite group, G be a torsion

free normal subgroup of G, K be a field such that char K f (G : G) and R

be the Goldie ring of fractions of KG.

PROPOSITION 7. Let

X(", (a = 1,2,...,k) (8.2)

be given non-zero elements of R and q -I char K be an arbitrary prime

number. Then there exists a G-invariant ideal A <::;KG and the ideal B =

A(KG)'::::::.A* GjG of KG such that

i) The quotient ring (K G)j A is a finite dimensional semisimple K -algebra,

generated by a finite q-group G, which is the image of G in (KG)/A.

Hence the quotient ring (KG)j B is isomorphic to a suitable cross prod-

uct (K[C)) * GjG and is a finite dimensional semisimple K-algebra,

generated by the finite group G, which is the image of G in (KG)/ B


subring KG). Clearly, tbe Jacobson radical of (KGh+B is tbe ideal

B(KGHB) and tbe quotient ring is isomorpbic to (KG)/ B.

PROOF: The assertion is proven in [10] for the case when G is torsion free

nilpotent, i.e. G = C. (See [10], Corollary of Theorem I). So, we give only

a sketch of the proof here.

The ring R contains a subfield ~ isomorphic to the field of fractions

of KG and if gl - 1, g2, . - - ,gr is a given transversal for G in G then this

transversal forms a basis of R over~. Let

k

x'" = L >..",fJgfJ13=1

(>""'13E~, a = 1,2,...r) (8.3)

be the representations of the elements (8.2) in this basis. Thus every nonzero

>-""13in (8.3) has a representation

>-""fJ = a"'fJb ~J(0 -=1=a "'13 , b"'fJ E KG). (8.4)

Now apply Corollary 1 of Theorem 1 in [10]. We find an ideal U s:;: KG such

that it contains none of the elements

-lbgi "'fJgi, (i=I,2,...,r) (8.5)

and such that the quotient ring (KG)jU is a simple K-algebra generated by

a finite group, the image of G in (KG)jU, and that the element

b = IIg;1 bOlfJgi (8.6)

is invertible in (KG)jUj hence all the elements (8.5) are also invertible. Let


Since 91 = 1,92,,'" 9r is a transversal for C in G we obtain that the ideal

A is G-invariant and hence B = A(KG) is an ideal of KG, B~A * GjG and

(KG)jB~«KC)jA) * GjC. (8.8)

I, Furthennore, since every quotient ring (KC)j9i-IU9i is isomorphic to the

simple algebra (KG)jU we conclude easily from (8.7) that

(KC)jA~(KC)jUil + (KC)jUi2 +... + (KC)jUi. (8.9)

where Uij = 9:;1 U9ij (j = 1,2,..., s) is the set of distinct elements among the

ideals 9;lU9i(i = 1,2,...,r). (See [9], Lemma 2.9 and note on page 577).

The representation (8.9) implies first of all that (KG)j A is a semisimple finite

dimensional algebra. Since the image of G in every (KC)jUij (j = 1,2,... , s)

is a finite q-group its image in (KG)jA is a finite q-group C, which generates

(KC)jA; this together with (8.8) completes the proof of the statement i).

Since the group C is torsion free nilpotent the ideal A is polycentral

(see [15], 11.3.12) and Roseblade's theorem ([151, 11.2.9) implies that the

ideal B = (A)KG has the weak AR-property; we obtain therefore from

P. Smith's theorem ([15], 12.2.10) that 1 + B is a right divisor set in KG.

Since the elements (8.5) are invertible in (KC)jA we obtain from (8.9) that

all the elements bOl{Jfrom (8.3) are invertible modulo the ideal A. This

implies easily that all the elements b;:J from (8.4) belong to the subring

(KGh+B ~ R and hence

)..0l{JE (KGh+B (a = 1,2,...,r; (3 = 1,2,...,k)

"i


Proof of Theorem 4. First of all, the Corollary of Theorem 3 implies

that KI 0K D contains no zero divisors for an abitrary extension KI 2 K.

Apply Corollary of Proposition 2 in [11] and obtain that there exist c}, C2 E

GLn(D) such that

UI = c~luCI E cQnxn, hi = C;I hC2 E cQnxn. (8.10)

It remains to prove that the elements UI, hi are conjugate in G Ln (CQ). Let

Ckl(1 :S k, R. :S n) be a given system of matrix units in the ring Dnxn':::!.

6 0/Or. qJ. Take in this ring the left 6-basis U Gj and represent by it the"t'-' j E J

elements

Ck(-I -I

(1:S k,R.:Sn);u;cI,cI ,C2,c2 . (8.11)

Let

Ao (a = 1,2,...,s) (8.12)

be all the nonzero coefficients, which occur in the representations of the

elements (8.11). Apply the results of [14] (see section 5) and find a ho-

momorphism <Pi : G ~ Gi = G/ri(G) and a specialization 8i : 6 ~ 6i,

extending <Pi,such that the domain Ti of 8i contains the elements

AO,-\;:;I (a = 1,2,...,s) (8.13)

as well as the subring KG. We have already pointed out (see the end of

section 5) that the subring Ti is G-invariant and Corollary of Proposition 5

implies that Ti and G generate a subring Ti * GIG. Furthermore, since Ti

contains all the elements (8.12) the subring Ti *GIG contains all the elements

(8.11). Now denote by .it the image of a subset X ~ Ti * GIG under the


Since the coordinates ,\.(0:' = 1,2,...,s) of the elements (S.12) are invert-

ible in Ti we conclude that the images of these elements in ~i . G IFare

nonzero and that the elements ek((1 :S k, e :S n) generate over Q a sub-

algebra Qnxn~Qnxn- We conclude from this that the elements UI a!1d hi

are conjugate in Qnxn if their images itl and hi are conjugate in Qnxn.

Furthermore, (S.lO) implies that

it I = cl1 itc I E Qn x n, hi = c:;1 hC2 E Qn)( n . (8.10' )

Since all the groups Gj(j E J) and the set J are finite we obtain from

(4.1) that the cartesian subgroup C is finitely generated and so is the group

Gi. We can therefore apply Proposition 7 to the system of elements

ek((1:Sk,e:Sn); it; cI,cll, C2, c;1 (8.11')

of the ring ~i.G IC, with a prime number q, which does not divide the orders

(Gj : 1)(j E J). We obtain an ideal B = A( QG;) such that the elements

(S.ll') belong to the subring (KGi)I+B ~ ~i. GIG and the relations (810')

imply for the images of these elements in the quotient ring

(~i)1+BI B( <QG;)I+B~(~i)1 B

the following relations:

UI = CllUCI E Q-::n, hI = C:;IhC2E Q-::n' (8.10ff)

Once again, because Qnxn~Qnxn we conclude that the elements UI and hI

are conjugate in Qnxn if and only if their images UI and hi are conjugate in

Qnxn.


GjCc:= IT G }', the algebra <Q[Gi]is the image of <QGunder the product ofjEJ

the homomorphisms: <Pi : <QG~ <Q(Gjf i(C))c:=<Q[Gd and tP : <QGi ~ <Q[G;]

and the homomorphism tP can be factored through the homomorphism tPk :

<QG ~ <Q(GjMk(C)), where k is an arbitrary natural number such that the

p-group Gi is a homomorphic image of CjA'h(C). Since the images of u and

h are conjugate in qCjMk(C)) (Theorem 1) their images u and h under the

product of homomorphisms

<QG--t <Q(GjM;(C)) ~ <QG; ~ qG;J

are conjugate in <Q[G;J.We see now from (8.10") that the conjugacy of the

elements UI and hI in Q--:::;nc:=Qnxn(and Theorem 4) will follow from the

following assertion.

LEMMA 8. Let R be a finite dimensional algebra over a field K of charac-

teristic zero, Knxn be a matrix subalgebra of R, which contains the unit. Let

GI and G2 be two subgroups ofGLn(K) which are conjugate in the group

of unit of R. Then these elements are conjugate in GLn(K).

PROOF: Consider the left regular K -representation p of the algebra R. It

is known (see [7] VI. 6.1.) that R is a free module over K n x n. If k is t he rank

of this module then the restriction of p to Knxn is equivalent to the repre-

sentation kT, where T is the regular representation of Knxn. Furthermore,

T = neT, where eT is the faithful irreducible representation of Knxn afforded

by its minimal left ideal I. Clearly, the isomorphism Knxn ~ eT(Knxn) maps

Knxn on the ring of the linear transformations of I and


Now let c E U(R) be an element such that

c-lGI C = G2. (8.15)

Pick an element u E G1 and v = c-luc E G2. We obtain from (8.15)

(p(C))-lp(U)p(C) = p(v)

and hence the traces of the elements p( u) and p( v) are equal

Tr(p(u)) = Tr(p(b)). (8.16)

Since p = kna we obtain from (8.16) that

Tr(a(u)) = Tr(a(v)). (8.17)

Since G1 and G2 are isomorphic we can consider a( Gd and a( G2) as two

representations of an abstract finite group G~G1~G2. Then (8.16) means

that these representations have the same character. Since char J( = 0 they

are conjugate in the group Aut(I) and (8.14) implies that G1 and G2 are

conjugate in GLn(K). This completes the proof.

ACKNOWLEDGEMENT

The work of the first author is supported by a grant from the Central

Research Fund of the University of Alberta and NSF Grant DMS 8802634.

The work of the second author is supported by NSERC Grant A 5300.


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Received: April 1988

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