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PROCEEDINGS OF THEAMERICAN MATHEMATICAL SOCIETYVolume 124, Number 4, April 1996

CENTRAL UNITS OF INTEGRAL GROUP RINGSOF NILPOTENT GROUPS

E. JESPERS, M. M. PARMENTER, AND S. K. SEHGAL

(Communicated by Ronald Solomon)

ABSTRACT. In this paper a finite set of generators is given for a subgroup offinite index in the group of central units of the integral group ring of a finitelygenerated nilpotent group.

In this paper we construct explicitly a finite set of generators for a subgroup offinite index in the centre Z(U (ZG)) of the unit group U (ZG) of the integral groupring ZG of a finitely generated nilpotent group G. Ritter and Sehgal [4] did thesame for finite groups G, giving generators which are a little more complicated.They also gave in [2] necessary and sufficient conditions for Z(U(ZG)) to be trivial;recall that the units ::i:G are called the trivial units. We first give a finite set of

generators for a subgroup of finite index in Z (U(ZG)) when G is a finite nilpotentgroup. Next we consider an arbitrary finitely generated nilpotent group and provethat a central unit of ZG is a product of a trivial unit and a unit of ZT, where Tis the torsion subgroup of G. As an application we obtain that the central unitsof ZG form a finitely generated group and we are able to give an explicit set ofgenerators for a subgroup of finite index.

1. FINITE NILPOTENTGROUPS

Throughout this section G is a finite group. When G is Abelian, it was shown in[1] that the Bass cyclic units generate a subgroup of finite index in the unit group.Using a stronger version of this result, also proved by Bass in [1], we will constructa finite set of generators from the Bass cyclic units when G is finite nilpotent.

Our notation will follow that in [6]. The following lemma is proved in [1].

Lemma 1. The images of the Bass cyclic units of ZG under the natural homo-morphism j : U(ZG) -+ K1 (ZG) generate a subgroup of finite index.

Let L denote the kernel of this map j, and B the subgroup of U(ZG) generatedby the Bass cyclic units. It follows that there exists an integer m such that zm E LBfor all z E Z(U(ZG)), and so we can write zm = Ib1b2 . .. bk for some 1 ELandBass cyclic units bi.

Received by the editors August 4, 1994.1991 Mathematics Subject Classification. Primary 16U60, 20CO5, 20CO7; Secondary 20C10,

20C12.This work is supported in part by NSERC Grants OGPOO36631, A8775 and A5300, Canada,

and by DGICYT, Spain.

@1996 American Mathematical Society

1007

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1008 E. JESPERS, M. M. PARMENTER, AND S. K. SEHGAL

Next, let Zi denote the i-th centre of G, and suppose from now on that G isnilpotent of class n. For any x E G and Bass cyclic unit bE Z(x), we define

b(l) = b

and for 2 ~ i ~ n

b(i) = II b(i-l)'gEZi

where ag = g-lag for a E ZG. Note that by induction b(i) is central in Z(Zi, x) andindependent ofthe order of the conjugates in the product expression. In particular,

b(n) E Z(U(ZG)).Recall again that if z E Z(U(ZG)), then zm = lb1b2... bk for some l ELand

Bass cyclic unitsbi. Since Kl (ZG) is Abelian, we can write

zmIZ21IZ31...IZnl = (lb1b2... bk)lz2!1Z31..'IZnl- l IT blz21Iz31'.'IZnl- 1 l<i<k i- l IT - - bIZ31'..IZnl- 2 l:S:i:S:ki(2)

- If ITl:S:i:S:kbien)

for some it E L

for some l2 E Lfor some If E L.

Since each bien) is in Z(U(ZG)), we conclude that If E L n Z(U(ZG)). But weshall show next that L n Z(U(ZG)) is trivial, so If E :f:Z(G). The argument usesthe same idea as in [3, Lemma 3.2].

For every primitive central idempotent e in the rational g~oup algebra QG, thesimple ring QGe has a reduced norm which we denote by nre. Further, denote

me = v[QGe: Z(QGe)]

and let

r = IIme.e

Now let If E L n Z(U(ZG)). By definition of K1(ZG) this means that a suitablematrix

If

1

1

is a product of commutators. Therefore Ife has reduced norm one. Since Ife is alsocentJ;al, we obtain

(ife)me = nr(lf e)e = e.

Hence

['r = 1.

So If is a torsion central unit, and therefore is trivial [7, Corollary 1.7, page 4].Since Z(U(ZG)) is finitely generated (see, e.g., [2]), (Z(U(ZG)))mIZ21IZ31"'IZnl is

of finite index. But we have just seen that the latter subgroup is contained in the

subgroup generated by :f:Z(G) and {b(n) I b a Bass cyclic unit}. We have proved

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CENTRAL UNITS OF INTEGRAL GROUP RINGS OF NILPOTENT GROUPS 1009

Proposition 2. Let G be a finite nilpotent group of class n. Then

(b(n) I b a Bass cyclic)

is of finite index in Z(U(ZG)).

Remark. Note that our method for constructing generators for a subgroup of finiteindex in Z(U(ZG)) can be adapted for some other classes of finite groups G. Forexample if G = D2n = (a,b I xn = 1, y2 = 1, yx = xn-ly), the dihedral groupof order 2n, then the only nontrivial Bass cyclic units b of ZD2n belong to Z(x).It follows that bbY = bYb is central. Our proof now remains valid and yields that(bbY I b a Bass cyclic in Z(x» is of finite index in Z(U(ZD2n)).

2. FINITELY GENERATED NILPOTENT GROUPS

We will now consider central units of an integral group ring of an arbitraryfinitely generated nilpotent group G. The torsion subgroup of G is denoted T.First we show that central units of ZG have the following decomposition.

Proposition 3. Let G be a finitely generated nilpotent group. Every u E Z (U (ZG))can be written as u = rg, r E ZT, g E G.

Proof Let F = G/T. Since T is finite and F acts on the set of primitive centralidempotents of QT by conjugation, by adding the idempotents in an orbit of thisaction we obtain

QT = EB(QT)ei = EBRi'

where ei are primitive central idempotents of QG. Then QG is the crossed product

(*) QG = QT * F = (EB Ri) * F = EB Ri * F.

Decompose u as a sum of elements in (*):

u = $. (

tUj/j

), 0 =1= Uj E Ri' /j E G, for each j.

t J=1

We assume that we have put together the Uj'S with the same /jT E G/T, namelyfor k =1=j, IkT =1=/jT.

We claim that n = 1. Let us denote by - the projection of QG onto Ri * F.Then since U is central we have QTu = uQT, which implies QTuj/j = uj/jQTfor all j. It follows that Uj is not a zero divisor provided ~ has only one simple(artinian) component, and so Uj is a unit. The only time Uj can be a nonunit iswhen it has some zero components in the simple components of Ri. However, bythe construction of ~, these latter components can be moved to any other placeby conjugating suitably. But they must stay put due to the facts that F is orderedarid u is central. It follows that Uj is a unit for all j. Hence, working in ~ * F and

using again that F is ordered, it follows by a classical argument that 11= L:j ujljis simply equal to unl n as claimed.

Changing notation, we write

U = EBal, a E Ri' lEG.

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1010 E. JESPERS, M. M. PARMENTER, AND S. K. SEHGAL

Let k = IAut(T)I, so fk commutes with T for f E G. Hence

uk = Ef)(af)k = Ef),Bfk, ,BE ~

(note that the number of summands in uk is the same as the number of summandsin u, because each a is a unit in Ri), and thus

Uk = (uk)ff = Ef) (,Bfk)g = Ef) ,Btfk, t E T.The last step followsfrom the fact that conjugation will preserve the order on the

fT's in the ordered group F. Since (Jk)ff = tfk, we can choose k large enough sothat all the fk commute with each other and with T. Thus we may assume that

uk=Ef),Bfk.

Again, we put together all ,B with the same fkT. In other words, we assume thatuk = fIJ,Bfk with all fkT different. Note that these new valuesof,B all lie in ZT.Furthermore, we now obtain for each t E T,

uk = (uk) t = Ef)(,B)tfk,

and thus ,Bt = ,B. So the ring R generated by all the ,B is commutative. Again, ifnecessary, replacing k by a high enough power, we may assume that the group Agenerated by all the fk in the summation of uk is a torsion-free Abelian group, andthus a free Abelian group. Consequently

uk ERA,

the commutative group ring of A over R. Let N = Rad(R) be the set of nilpotentelements of R. Now ZT has only trivial idempotents [6, Theorem 2.20, page 25].Hence since R ~ ZT and since idempotents of Rj N can be lifted to R, it followsthat RjN also has only trivial idempotents. Therefore [6, Lemma 3.3, page 55]together with an inductive argument tells us that (RjN)A has only trivial units.It follows that

Uk = ,Bfk + nilpotent elements.

But as each ,Bis a sum of units in various Ri, it followsthat the last term must bezero. Hence uk = ,Bfk, and thus all f's in the original decomposition of u = fIJiafwere in the same coset of T. Thus u = rf as required. 0

We give two important consequences of the last result. We say that Z(U(ZG))is trivial if it contains only trivial units.

Corollary 4. Let G be a finitely generated nilpotent group. If Z(U(ZT)) is trivial,then Z(U(ZG)) is trivial.

Proof. Let u E Z(U(ZG)) be nontrivial. Then the support of u contains two differ-ent elements, say x and y. Since finitely generated nilpotent groups are residuallyfinite, there exists a finite factor GjN = G so that xi- y in G (see [5, page 149]).Hence u has in its support at least two different elements, and thus u is of infiniteorder ([7, Corollary 1.7, page 4]). By Proposition 3 we write u = rg, r E ZT,9 E G. Since u is central, r commutes with g. It then follows easily that r, andhence also r, is of infinite order. Moreover, there exists a positive integer n suchthat (gn, T) = 1. Consequently it follows from un = rngn that rn commutes withT. Thus rn is a nontrivial unit of Z(U(ZT)). 0

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CENTRAL UNITS OF INTEGRAL GROUP RINGS OF NILPOTENT GROUPS 1011

Corollary 5. Let G be a finitely generated nilpotent group. Then Z(U(ZG)) isfinitely generated. Furthermore, (Z(U(ZG)) n Z(U(ZT))) Z(G) is of finite index inZ(U(ZG)).

Proof. Let S = Z(U(ZG)) n Z(U(ZT)). First we show that Z(U(ZG))ISZ(G)is a torsion group of bounded exponent. Indeed, let u E Z(U(ZG)). Because ofProposition 3 write u = rg, with r E U(ZT) and 9 E G. Considering the natu-ral epimorphism ZG --? Z(G/T) and using the fact that Z(U(Z(GIT))) is trivialbecause GIT is ordered, it follows that gT E Z(GIT). Hence (gk, T) = 1 andgl E Z(G) for k = IAut(T)1 and 1= kiTI. Now since u is central, rand 9 commute.Therefore

Ul = rlgl and rl E S.

Consequently ul E SZ(G), and the claim follows.'" As a subgroup of the finitely generated group Z(U(ZT)), the group S is itselffinitely generated. Hence so is SZ(G). Since the torsion subgroup of Z(U(ZG))is finite (see for example [6, page 46]), the above claim now easily yields thatZ(U(ZG)) is indeed finitely generated.' 0

We will now construct finitely many generators for the central units of any finitelygenerated nilpotent group.

Let n be the nilpotency class of T and h the Hirsch number of G IT. Let k =~Aut(T)I. Further let Xl,'" , Xh be elements of G such that for each 1 S; i S; h thegroup Gi = (T, Xl, . . . , Xi) is normal in G and Gd Gi-l ~ Z, where Go = T. Forany generator b(n) described in Proposition 2 define

(0)b(n) = b(n)

and for 1 S; i S; hj

b(i) - II (b(i-l) )Xi

(n) - (n)'O~j<k

Since each b(n)is in Z(U(ZT)), the order of the conjugates in the product expression

for b~~)is unimportant. It follows by induction that b~~) is in Z(U(ZGi)). In

particular, b~~~E Z(U(ZG)).

Theorem 6. Let G be a finitely generated nilpotent group. Suppose n is the nilpo-tency class of T and h is the Hirsch number of GIT. Then

(b~~~Ib a Bass cyclic of ZT) Z(G)

is of finite index in Z(U(ZG)).

Proof. Because of Corollary 5 the group SZ(G) with S = Z(U(ZG)) n Z(U(ZT))is of finite index in Z(U(ZG)). Let at,... ,ap be a finite set of generators for S.By Proposition 2 there exists a positive integer m such that all aT', . . . ,a; are in(b(n) Ib a Bass cyclic in ZT). For simplicity, write a = aT'. Then

a = IIb(n),

where the product runs over a finite number of Bass cyclic units of ZT. Since a isin Z(U(ZG)), and using the notation introduced above, we obtain

k k-la =aaXl...axl .

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I" As each b(n) is central in ZT, this implies

~k - IIb(l)'-'" - (n)"

Continuing this process one obtains that~kh - IIb(h)'-'" - (n)'

Since the group generated by alkh, . . . , a~kh is of finite index infollows.

E. JESPERS, M. M. PARMENTER, AND S. K. SEHGAL

S, the result0

Note that Corollary 4 can now also be obtained as an easy consequence of The-orem 6.

We now give an example showing that the converse of Corollary 4 does not hold.

Example. Let G = (a, x I aX = a3, a8 = 1). Clearly G is a nilpotent group withT = (a), acyclic group of order 8. From Higman's result (see [6]) it follows thatZ(U(ZT)), modulo the trivial units, is a free Abelian group of rank 1. We nowshow that Z(U(ZG)) contains only trivial units. For this suppose u is a nontrivialcentral unit in ZG. By Proposition 3, we can write u = rxi for some integer iand r E U(ZT)). We know from the above that r is of infinite order, and since rcommutes with x, it must be in Z(U(ZG)).

Because the only Bass cyclic unit, up to inverses, in ZT is

b = (1 + a + a2)4 - lOa, a = 1 + a +... + a7,

Proposition 2 yields that

rk = bl,

for some nonzero integers k, l. Observe, however, that bX = b-1. Since bl = rk iscentral in ZG, we obtain bl = b-I, contradicting the fact that b is of infinite order.

REFERENCES

1. H. Bass, The Dirichlet Unit Theorem, Induced Characters and Whitehead Groups of FiniteGroups, Topology, 4 (1966), 391-410. MR 33:1341

2. J. Ritter and S.K. Sehgal, Integral group rings with trivial central units, Proc. Amer.Math.Soc., 108 (1990), 327-329. MR 90d:16009

3. J. Ritter and S.K. Sehgal, Construction of Units in Integral Group Rings of Finite NilpotentGroups, Trans. Amer. Math. Soc., 324 (2)(1991), 603-621. MR 91h:20008

4. J. Ritter and S.K. Sehgal, Units of group rings of solvable and Frobenius groups over largerings of cyclotomic integers, Journal of Algebra, 158 (1993), 116-129. MR 95d:16045

5. Derek J.S. Robinson, A course in the theory of groups, Springer Verlag, 1982. MR 84k:200016. S. K. Sehgal, Topics in group rings, Marcel Dekker, New York, 1978. MR 80j:160017. S. K. Sehgal, Units in integral group rings, Longman Scientific and Technical, Essex, 1993.

MR 94m:16039

(E. Jespers and M. M. Parmenter) DEPARTMENT OF MATHEMATICS AND STATISTICS, MEMORIALUNIVERSITY OF NEWFOUNDLAND, ST. JOHN'S, NEWFOUNDLAND, CANADA A1C 5S7

E-mail address:ejespers~albert.math.mun.caE-mail address:mparmen~lato.ucs.mun.ca

(S. K. Sehgal) DEPARTMENT OF MATHEMATICS, UNIVERSITY OF ALBERTA, EDMONTON, AL-

BERTA, CANADA T6G 2G 1

E-mail address:ssehgal~schur.math.ualberta.ca