## 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.