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

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Since every subgroup of finite index in G contains a normal subgroup of finite index, this proves the theorem. In the course of the above proof, we saw that 3>(G) = GP[G, G]. 20 Corollary. If G is a finitely generated pro-p group, then $(G) = Gp[G,G] and Pi+i(G) = Pi{Gy[Pi{G),G] for each I Proof We have already established the first claim. Put G^ = Pi (G) for each i. 14 show that each Gi is a finitely generated pro-p group, and that <&(G;) is open in G{. By the first part, applied with d in place of G, we have = [Gi,Gi]Gpi<[Gi,G}G?.

18, G/K is a finite p-group; consequently M < G, and the result of the first paragraph shows that M is open in G. 7, and our inductive hypothesis shows that K is open in M. Thus K is open in G. Since every subgroup of finite index in G contains a normal subgroup of finite index, this proves the theorem. In the course of the above proof, we saw that 3>(G) = GP[G, G]. 20 Corollary. If G is a finitely generated pro-p group, then $(G) = Gp[G,G] and Pi+i(G) = Pi{Gy[Pi{G),G] for each I Proof We have already established the first claim.

G] < Z(G), and it follows that for any given x € N and gG G, the map w i-> [x, #, w] is a homomorphism from G into Z(G). *»*]'=[^^^] p(p " i)/2 • j=0 j=0 Hence 0 = II faglfagix3] j=p-i p-i = [x,g]PY[[x,g,xj] since [x,g,x^] £ 1(G) for each j since [N,G]P = 1. Thus [NP,G] = 1, giving the result. Powerful p-groups 39 Case 2: p = 2. We may now assume that [TV, G] < TV4 and that [TV2, G, G] = [TV2, G}2 = (TV2)4 = l . For x G TV and g G G we have so TV4 < Z(G). Since TV has exponent dividing 8, TV4 is generated by elements of order 2, hence (TV4)2 = 1.

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