J. Group Theory 6 (2003), 223-228 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* (Communicated by I. B. S. Passi) Abstract. Let A be an abelian normal subgroup of a finite group G. Then all units of the inte- gral group ring 7lG which are of the form 1+ J with 0 ""J E d( G)d(A), where d denotes the augmentation ideal, are of infinite order. 1 Introduction Let TLGbe the integral group ring of a group G. Graham Higman asked in 1940, in [4], the following question. The IsomorphismProblem. Suppose that the rings TLGand TLHare isomorphic. Does it follow that the groups G andH are isomorphic? Today we know that for some 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-isomorphic finite groups with isomorphic integralgroup rings. However, the situation is still far from being completely understood. Higman himself positively solved this problem for finite abelian groups A, by identifying + A with the set of torsion elements 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 augmentation homomorphism e: TLA-4 7L defined by e(~ ngg) =~ ng. Higman proved that the subring 1 + .1(A)2 has no torsion units, hence the subgroup A < O//(7LA) has a torsion-free normal complement. Higman's description of units in 7lA was generalized in [1] for groups G with a normal abelian subgroup A such that G/ A is abelian of exponent 2, 3, 4 or 6. This time O//(7LG) = +G. all(1 + .1(G).1(A)), *This research was supported by NSERC Canada and Polish KBN Grant No. 2PO3AO0218. -- ~~-