Frattini subgroup is nilpotent in finite

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Statement

Verbal statement

The Frattini subgroup of any finite group is nilpotent.

Symbolic statement

Let $G$ be a finite group and $\mathrm{\Phi }(G)$ denote the intersection of all maximal subgroups of $G$ (the so-called Frattini subgroup of $G$).

Definitions used

Frattini subgroup

The Frattini subgroup of a (here, finite) group is the intersection of all its maximal subgroups.

Nilpotent group

A finite group is nilpotent if every Sylow subgroup of it is normal.

Generalizations

There is a somewhat more general version of this result: the Frattini subgroup of a finite group (or more generally, a group where every subgroup is contained in a maximal subgroup) has the property of being an ACIC-group: any automorph-conjugate subgroup in it is characteristic. For full proof, refer: Frattini subgroup is ACIC.

Proof

We are given a finite group $G$, and $\mathrm{\Phi }(G)$ is the Frattini subgroup. We need to show that for any Sylow subgroup $P$ of $\mathrm{\Phi }(G)$, $P$ is normal in $\mathrm{\Phi }(G)$.

In fact, we shall show that $P$ is normal in $G$.

Here's the idea. By applying Frattini's argument and the fact that $\mathrm{\Phi }(G)◃G$, we have $\mathrm{\Phi }(G)N_G(P)=G$. Now if $N_G(P)\ne G$, it is contained in a maximal subgroup $M$ of $G$. Since $\mathrm{\Phi }(G)$ is contained in every maximal subgroup, $\mathrm{\Phi }(G)N_G(P)\le M$, leading to a contradiction.