# Frattini subgroup of a finite group is nilpotent, the

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

###### Proof.

Let $\Phi(G)$ denote the Frattini subgroup of a finite group $G$. Let $S$ be a Sylow subgroup of $\Phi(G)$. Then by the Frattini argument, $G=\Phi(G)N_{G}(S)={\langle\Phi(G)\cup N_{G}(S)\rangle}$. But the Frattini subgroup is finite and formed of non-generators, so it follows that $G={\langle N_{G}(S)\rangle}=N_{G}(S)$. Thus $S$ is normal in $G$, and therefore normal in $\Phi(G)$. 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 $\Phi(G)$ is finite then $\Phi(G)$ is nilpotent.

Title Frattini subgroup of a finite group is nilpotent, the FrattiniSubgroupOfAFiniteGroupIsNilpotentThe 2013-03-22 13:16:44 2013-03-22 13:16:44 yark (2760) yark (2760) 16 yark (2760) Theorem msc 20D25