Frattini subgroup of a finite group is nilpotent, the

The Frattini subgroup of a finite group is nilpotent (http://planetmath.org/NilpotentGroup).

Proof.

Let $\mathrm{\Phi }(G)$ denote the Frattini subgroup of a finite group $G$.Let $S$ be a Sylow subgroup of $\mathrm{\Phi }(G)$.Then by the Frattini argument, $G=\mathrm{\Phi }(G)N_G(S)=⟨\mathrm{\Phi }(G)\cup N_G(S)⟩$.But the Frattini subgroup is finite and formed of non-generators,so it follows that $G=⟨N_G(S)⟩=N_G(S)$.Thus $S$ is normal in $G$, and therefore normal in $\mathrm{\Phi }(G)$.The result now follows, as any finite group whose Sylow subgroups are all normal is nilpotent (http://planetmath.org/ClassificationOfFiniteNilpotentGroups).∎

In fact, the same proof shows that for any group $G$,if $\mathrm{\Phi }(G)$ is finite then $\mathrm{\Phi }(G)$ is nilpotent.