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(iv) ⇒ (v): Let P be a Sylow p-subgroup of G and let N = P Φ(G)(which is a subgroup of G, sinceΦ(G) G by Lemma 7.17). Let x ∈ Nand g ∈ G. Thenx −1 x g =[x, g] ∈ G ′ Φ(G) N.Hence x g ∈ N for all x ∈ N and g ∈ G, soN G. Now P is a Sylowp-subgroup of N (since it is the largest possible p-subgroup of G, so iscertainly largest amongst p-subgroups of N). Apply the Frattini Argument(Lemma 6.35):G =N G (P ) N=N G (P ) P Φ(G)=N G (P )Φ(G) (as P N G (P )).From this we deduce that G =N G (P ): for suppose N G (P ) ≠ G. ThenN G (P ) M < G for some maximal subgroup M of G. By definition,Φ(G) M, soN G (P )Φ(G) M