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.

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