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By Bieri R.

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Let f ∈ F be such that f C ∈ γω (F/C). Then the hypothesis implies that R∩S [ RV∩ S , f C] = 1. Furthermore, since the action of F/RS on [R, S] is assumed to be faithful, it follows that f ∈ RS and consequently f ∈ C. Observe that, in the case R = S, C/V is exactly the centre of R/V . 19). Let V ⊂ R and let C/V be the centre of R/V . Then γω (F/V ) = 1 implies that γω (F/C) = 1. R∩S The investigation of the modules [R, S] is also motivated from topological considerations, particularly in view of the relationship of such modules with the second homotopy modules which we discuss next.

An ideal I in the group algebra k[G], where k is a field, is said to be controlled by a normal subgroup H of G if I = (k[H] ∩ I)k[G]. α = 0}. Observe that if a k[G]-module M embeds in a free k[G]-module, then Annk[G] (M ) is an annihilator ideal in k[G]. We need a result, due to M. e. group algebras which do not have any nonzero two-sided ideal with square zero. If k is a field of characteristic 0, then the group algebra k[G] is always semi-prime ([Pas77b]). 53 (Smith [Smi70]). In a semi-prime group algebra k[G], in particular if char(k) = 0, the annihilator ideals are controlled by the normal subgroup δ + (G).

Let us now have some examples of linear groups for which residual nilpotence can be proved by the above-mentioned method. Consider subgroups of the Picard group of indices 12 and 24 respectively. The presentations of these groups can be found in [Wie78, Kru86] and [Bru84]. 29 Let G1 = x, y | [x, yxy −2 xy] = 1 . This group has a faithful representation f1 : G1 → PSL2 (Z[i]) given by x −→ 10 11 , y −→ i 1+i −1 −1 , and the group f1 (G1 ) has index 12 in PSL2 (Z[i]) [Bru84]. Let a = ϕ2 (f1 (x)), b = ϕ2 (f1 (y)).

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