The Frattini subgroup of a finite group is nilpotent

Let G be a finite group. Prove that \Phi(G) is nilpotent.

We will use Frattini’s Argument to show that every Sylow subgroup is normal.

Let P \leq \Phi(G) be a Sylow subgroup. By Frattini’s Argument, G = \Phi(G) N_G(P). By a lemma to a previous theorem, N_G(P) cannot be proper, so that N_G(P) = G and we have P \leq G normal. Then P \leq \Phi(G) is normal; thus all of the Sylow subgroups of \Phi(G) are normal, and thus \Phi(G) is nilpotent.

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  • Gwin  On December 16, 2010 at 10:35 am

    The link to the “previous theorem” does not show up. Perhaps the link is broken?

    • nbloomf  On December 16, 2010 at 10:42 am

      Fixed. Thanks!

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