Torsion Freeness of the Unit Group of 1 + Δ(G)Δ(A)*

Abstract

Let ℤG be the integral group ring of a finite group G and U=U(ℤG) its group of units. For a normal subgroup N ⊲ G we denote by Δ(G, N) the kernel of the natural map ℤG → ℤ(G/N). In particular, for N = G, Δ(G, G) is the augmentation ideal Δ(G) which is the kernel of the augmentation map e. Let us denote by U 1 the units of augmentation one. It is a classical result of G. Higman that if A is an abelian group then (UℤA) is the direct product ±A × U 2 (ℤA) where U 2(ℤA) = u ∈U(ℤA) : u ≡ 1 mod (ΔA)2. Furthermore, ±A is the torsion subgroup of U(ℤA), namely, U 2(ℤA) is torsion free. An extension of this result is that if A ⊲ G and Uℤ(G/A) is trivial (equivalently G/A is abelian of exponent 2, 3, 4 or 6) then again U(ℤG) = ±G × U(1 + Δ(G)Δ(A)) with the second factor torsion free. So it was asked by Dennis [2] if the embedding ±G →U(ℤA) splits and if the normal complement is torsion free. An affirmative answer to this question also has an affirmative answer to the (ISO), the isomorphism problem. As it turns out, it was proved by Cliff-Sehgal-Weiss [1] that the answer is affirmative if G/A is abelian of odd order. The normal complement arises from ideals related to Δ(G)Δ(A). The unit group U(1 + Δ(G)Δ(A)) was proved to be torsion free for all metabelian groups G. Roggenkamp-Scott [3] answered the question of Dennis in the negative by providing counterexamples for even metabelian groups. In spite of this, the following was proposed as Problem 28 in [4].