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By J. D. Dixon, M. P. F. Du Sautoy, A. Mann, D. Segal

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6 that G»+i = {a:p | x G Gi}, and then by induction that Gj = {xp | x G G}. Since Gi is a subgroup this implies that Gi = Gp . 5 show that Gi = {a\ , . . ,a^ ). Thus we have (iii). 4 (ii). Finally, taking G{ in place of G and k + 1 in place of i, in (iii), we get = {/~ 1+fc I y e G} = Gi+k , giving (ii). 8 Corollary. If G = ( a i , . . ,a^) is a powerful p-group then G = (cLi)... e. G is the product of its cyclic subgroups (a*). Proof Say G e > G e +i = 1. Arguing by induction on e, we may suppose that G = ( o i ) .

The exact converse of this statement is false (see Exercise 3); our second major theorem is nevertheless a sort of converse: it shows that in anyfinitep-group G, there is a powerful normal subgroup of index bounded by a function of rk(G). The proof requires some preparation. 10 Definition For a finite p-group G and a positive integer r, F(G, r) denotes the intersection of the kernels of all homomorphisms of G into GL r (F p ). Since the image of any homomorphism of a p-group G into GLr (Fp) is a p-group, and every p-subgroup of GL r(Fp) is conjugate to a subgroup of the lower uni-triangular group U r (F p ), we could equally well define V(G,r) as the intersection of the kernels of all homomorphisms of G into Ur(Fp).

Since N/[N, V] is elementary abelian, we have d(M/[N,V\) = d(N/[N,V]) - 1 < r - 1; as [N, V] is cyclic it follows that d(M) < r. Hence, by the inductive hypothesis, [M, V] < Mp = 1. Thus M is central in JV, and as N/M is cyclic it follows that N is abelian. Then N is an Fp-vector space of dimension at most r, so the conjugation action of V on N must be trivial. Thus [N, V] = 1, in contradiction to the initial assumption. Now suppose p = 2. As above, we reduce to the case where iV4 = 1 and \[N, W]\ = 2.

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