The unit group of 1 + A(G)A(A) is torsion-free

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1 J. Group Theory 6 (2003), Journal of Group Theory cg de Gruyter 2003 The unit group of 1 + A(G)A(A) is torsion-free Zbigniew Marciniak and Sudarshan K. Sehgal* (Counicated by I. B. S. Passi) Abstract. Let A be an abelian noral subgroup of a finite group G. Then all units of the integral group ring 7lG which are of the for 1+ J with 0 ""J E d( G)d(A), where d denotes the augentation ideal, are of infinite order. 1 Introduction Let TLGbe the integral group ring of a group G. Graha Higan asked in 1940, in [4], the following question. The IsoorphisProble. Suppose that the rings TLGand TLHare isoorphic. Does it follow that the groups G andh are isoorphic? Today we know that for soe classes of groups the answer is positive; see [9, p. 207] for an account. On the other hand, Hertweck [3] has recently constructed two (large) non-isoorphic finite groups with isoorphic integralgroup rings. However, the situation is still far fro being copletely understood. Higan hiself positively solved this proble for finite abelian groups A, by identifying + A with the set of torsion eleents in the group O//(7LA)of units of 7LA. He proved that O//(7LA) = :t:a x all(1 +.1(A)2), where.1(a) is the kernel of the augentation hooorphis e: TLA-4 7L defined by e(~ ngg) = ~ ng. Higan proved that the subring 1 +.1(A)2 has no torsion units, hence the subgroup A < O//(7LA)has a torsion-free noral copleent. Higan's description of units in 7lA was generalized in [1] for groups G with a noral abelian subgroup A such that G/A is abelian of exponent 2, 3, 4 or 6. This tie O//(7LG) = +G. all(1+.1(g).1(a)), *This research was supported by NSERC Canada and Polish KBN Grant No. 2PO3AO ~~-

2 224 Zbigniew Marciniak and Sudarshan K. Sehgal where ~(G)~(A) is the two-sided ideal in 7lG generated by all eleents of the for (g - I)(a - I) with g E G, a E A. Again, the subring I + ~(G)~(A) has no torsion units. It is easy to show that in general, if a finite group G has a torsion-free noral copleent in the unit group rj/i(71g)and 7lG ~ 7lH then G ~ H. Hence it was asked by Dennis [2]whether the ebedding G --*rj/i(7lg)always splits and the noral copleent arising as the kernel of the splitting hooorphis is torsionfree. Roggenkap and Scott [7] constructed counter-exaples to this even for etabelian groups. On the other hand, Cliff, Sehgal and Weiss [I] gave an affirative answer for etabelian groups A <J G --* G/ A with IG/A Iodd. Again, noral copleents arise fro ideals related to ~(G)~(A). Moreover, they proved that the group rj/i(1+ il(g)il(a)) is torsion-free for any eta beiian group. Therefore the following was proposed as Proble 28 in [9]. Proble. Let A be a noral abelian subgroup of a finite group G. Is the unit group rj/i(1+ ~(G)il(A)) torsion-free? An easy induction arguent proves that this is true for nilpotent groups G; see [9, p. 182]. In [6] an affirative answer is provided for groups which are cyclic-bynilpotent. In this paper we give a positive answer to this proble for arbitrary finite groups G. Main Theore. If A is a noral abelian subgroup of a finite group G then the unit group rj/i(1+ ~(G)il(A)) is torsion-free. Reduction to a single prie case It is enough to prove that 1 + ~(G)~(A) contains no units of prie order p, for all pries p. Hence pick a prie p and suppose that for soe u E I + ~(G)~(A) we have up = I. We proceed by induction on the order of G. Let B be the pi_part of A. The subgroup B is noral in G and if B =1=1 then, by induction, u aps to 1 under the ap TLG--* 7l[G/ B]. Hence u E I + 7LG~(B). It is proved in [1, Lea 3.1] that if p does not divide IBI then the group rj/i(1+ 7LG~(B)) has no p-torsion. Hence u ust be equal to 1. Therefore the proble reduces to the case when A is a p-group. It reains to prove the following assertion. Let G be a finite group with a noral abelian p- o subgroup A. If a unit u E I + ~(G)~(A) satisfies up = I then u = 1. The atrix representation Let us choose a syste {Xl, X2,...,x} of right coset representatives of A in G with Xl = 1. Then we have TLG = EB xi7la and u = 1+ L Xi(Xi with (XiE 7LA. i=l i=l Here (XiE il(a), as fro u - 1 E il( G)~(A) c 7LG~(A) it follows that each eleent (Xi ap to 0 under the natural projection TLG--* 7L[G/A]

3 The unit group of 1 + ~(G)~(A) is torsion-free 225 Left ultiplication by u defines an endoorphis of 7LG,treated as a free right. 7LA-odule. The atrix of this endoorphis with respect to the basis {Xj} has the for U = 1+ M, where I is the identity x atrix and M has all entries in ~(A). More precisely, al a2 ax2 I M= I a3... ax} I a... ax I The (i, j)-entry of M is equal to cija~(j)' where cij E A is a value of the co cycle describing the extension A <J G -t Gj A and aj is a perutation of the set {I,...,}. Because A is abelian, the conjugation action of G on A factors to an action of K = Gj A and extends linearly to 7LA. In particular, if Zj E K is the iage of Xj under the natural ap G -t Gj A then at = at. In this notation, the trace of M is equal to LZEK ai. Our basic tool is Weiss' Rigidity Theore [10].This states that the atrix U can be conjugated in GL(ZpA) to a diagonal atrix diag(al,'",a) with all aj E A. Here Zp denotes the ring of p-adic integers. As the atrix U originated fro a unit of 7LG,the eleents aj are closely related. In fact, we have Lea 1. All eleents aj E A are conjugate in G and hence of the sae order. Proof We prove the stateent by showing that for each i there exists an eleent Zj E K such that aj = afi. We start by writing I + al = fji fjs where the support of each fjj is contained in a single K-orbit and the supports of different fj/s are contain~d in different K-orbits. Pick one elel1!entbj E supp(fjj for each 1 ~ j ~ s and define bj = LZEK bj. Then LZEK fjj = (fjj)bj. Becausetraces of conjugate atrices coincidewe have L aj = + L af = L(l + al)z = L (fjj)bj. j=1 ZEK ZEK j=1 As the supports of bj are pairwise disjoint, we ust have (fjj) ~ 0 for each j. But these integers add up to (1 + ad = 1. It follows that precisely one (fjjo)is different fro 0 and hence equal to 1. Thus as desired. L aj j=1 ZEK = (fjjo )bjo = L bfo, s

4 226 Zbigniew Marciniak and Sudarshan K. Sehgal Reark. We have just proved that the unit u has non-zero trace for at ost one conjugacy class of G contained in A. If we were able to prove that the traces vanish on all classes contained in G\A we would know that u has the unique trace property: only one trace is not zero. Fro [5] it would then follow that the Zassenhaus Conjecture holds for u: the unit is conjugate in <QGto a group eleent, which then ust be trivial. However, even though we know that u ay have non-zero traces on classes of p- eleents only [5], it is hard to control the because we ake no assuptions about G. Therefore we chose a different path. Proof of the Main Theore The atrix M has one additional property. Lea 2. Let w = (1,...,1) E?lA. Then all coordinates ofw. M belong to L\(A)2. Proof We have U L ai = L(Xi - l)ai E L\(G)L\(A) i=1 i=l and u - 1 E L\(G)L\(A), and hence L ai E L\( G)L\(A) n?la = L\(A)2. i=1 But 2:::1 ai is the first coordinate of w. M. For the j-coordinate we obtain the su 2:::1 cija;(i)' where aj is a perutation of the index set {I,...,}. However J Z z. 2 cija:(i) == a:(i) odl\(a) and hence the above su is congruent to (2:::1 ai)zje L\(A)2. Notice that if p-i UP = diag(al,'",a) for a atrix P E GL(ZpA) then we also have p-i M P = diag( al - 1,...,a - 1). Consider the vector v = w. P. By Lea 2, the vector v. diag(al - 1,..., a - 1) = w. p. p-i MP = (w. M)P has all entries in ~(A)2, where ~(A) is the augentation ideal of ZpA. If v = (VI,..., v) then for all i we have

5 The unit group of 1 + Ll(G)Ll(A) is torsion-free 227 ~ 2 Vi(ai - 1) E L1(A), ~ 2 e(vi)(ai - 1) = (e(vi)- Vi)(ai - 1) + Vi(ai - 1) E L1(A), ~ 2 e(vi)(ai - 1) E L1(A), where e(vi) is the augentation of Vi E 7LpA. Notice that at least one of the nubers e(vi) ust be invertible in 7Lp.Otherwise we have e*(v) E (p7lp) and hence W = e*(w)= e*(v). e*(p -I ~ ) E (p7lp) as well, which iplies I E p7lp, a contradiction. Suppose that Vio,the io-coordinate of V, has augentation invertible in 7Lp.Then aio - I E A(A)2. It follows now fro [8, Theore 1.9 and Proposition 1.14] that aio = 1. We recall the short arguent for the reader's convenience. Lea 3. If A is abelian p-group and a E A satisfies a - I E A(A)2 then a = 1. Proof If a =1=I then we can find a projection n : A -+ C = <c : cpn = I> such that n(a) = ck =1=1. As A(C) is generated by (c - 1), we get k 2 ~ C - I = (c - 1) 13 for soe 13E 7LpC. Notice that if (c - I)y = 0 then y = d. (I + c cp"-l) for soe d E 7Lp.Hence 1+ c ck-i = (c - 1)13+ d. (I + c cpn-i). When we copare the augentations, we get k = dpn. However k rtpnzp, as ck =1=1. Now we can conclude the proof of the Main Theore. We already know that aio = 1. By Lea 1 all reaining eleents ai are trivial. Therefore M = 0, U = I and u = 1, as desired. Corollary. Let a finite group G have an abelian noral subgroup A such that the group ring 7L[GjA] has trivial units only. Then G has a torsion-free noral copleent in 0Zt(7LG). The above result was proved in [1], in the case when the factor group G/A was abelian. The sae proof applies here; copare [9, Lea 30.6]. The Main Theore takes care of the only issing link: the torsion-freeness of 0Zt(l+ L1(G)L1(A)) in the general case.

6 228 Zbigniew Marciniak and Sudarshan K. Sehgal References [I] G. H. Cliff, S. K. Sehgal and A. R. Weiss. Units in integral group rings of etabelian groups. J. Algebra 73 (1981), [2] R. K. Dennis. The structure of the unit group of group rings. In Ring Theory II (Marcel Dekker, 1977), pp [3] M. Hertweck. A counterexaple to the isoorphis proble for integral group rings. Ann. of Math. (2) 154 (2001), [4] G. Higan. Units in group rings. D. Phil. thesis. University of Oxford (1940). [5] Z. Marciniak, J. Ritter, S. K. Sehgal and A. R. Weiss. Torsion units in integral group rings of soe etabelian groups, II. J. Nuber Theory 25 (1987), [6] M. Parenter and S. K. Sehgal. Torsion freeness of the unit group of 1 + ~(G)~(A). To appear. [7] K, W. Roggenkap and L. L. Scott. Units in etabelian group rings: nonsplitting exaples for noralized units. J. Pure Appl. Algebra 27 (1983), [8] S. K. Sehgal. Topics in group rings (Marcel Dekker, 1978). [9] S. K. Sehgal. Units in integral group rings (Longan, 1993). [10] A. A. Weiss. Rigidity of p-adic torsion. Ann. of Math. (2) 127 (1988), Received 27 February, 2002 Z. Marciniak, Institute of Matheatics, Warsaw University ul. Banacha 2, Warsaw, Poland S. K. Sehgal, Departent of Matheatical and Statistical Sciences, University of Alberta, Edonton, Alberta, T6G 2G 1 Canada