## The Frattini subgroup is characteristic

Let $G$ be a group and denote by $\Phi(G)$ the Frattini subgroup of $G$. Prove that $\Phi(G) \leq G$ is characteristic.

We begin with some lemmas. Let $\mathcal{S}(G)$ denote the set of subgroups of $G$; recall that $\mathsf{Aut}(G)$ acts on $\mathcal{S}(G)$ by application: $\alpha \cdot H = \alpha[H]$.

Lemma 1: Let $G$ be a group. If $\mathcal{O}$ is an orbit of the action of $\mathsf{Aut}(G)$ on $\mathcal{S}(G)$, then $\bigcap \mathcal{O}$ is characteristic in $G$. Proof: Let $\alpha$ be an automorphism of $G$. Now $\alpha[\bigcap_{H \in \mathcal{O}} H]$ $= \bigcap_{H \in \mathcal{O}} \alpha[H]$ $= \bigcap_{H \in \mathcal{O}} H$, since $\alpha$ permutes the elements of $\mathcal{O}$. $\square$

Lemma 2: The set $\mathcal{M} \subseteq \mathcal{S}(G)$ of all maximal subgroups of $G$ is a union of orbits under the action of $\mathsf{Aut}(G)$. Proof: It suffices to show that if $M$ and $H$ are maximal and nonmaximal subgroups of $G$, respectively, then no automorphism carries $M$ to $H$. Suppose to the contrary that $\alpha$ is such an automorphism and that $H < K < G$. Then $G > \alpha^{-1}[K] > M$, a contradiction. $\square$

Now to the main result. We see that $\Phi(G) = \bigcap \mathcal{M}$ is an intersection of characteristic subgroups, hence characteristic.