TITLE: The Frattini subgroup is a characteristic subgroup.
QUESTION [6 upvotes]: I am revising for my Group Theory exam and am stuck on the following question;
The Frattini subgroup $\Phi(G)$ of a group $G$ is defined to be the intersection of all maximal subgroups of $G$. Prove that $\Phi(G)$ is a characteristic subgroup of G.
Why is this the case?

REPLY [2 votes]: We need this Lemma 1:

Automorphism maps a maximal subgroup to another maximal subgroup.

The proof is here.
Let $\alpha:G \rightarrow G \in Aut(G)$.
Let:
$\Phi(G) = \bigcap\limits_{M \: maximal \: in \: G} \: M$
be the Frattini subgroup of $G$.
By Lemma 1 $\alpha(M)$ is also another maximal subgroup of $G$. We could just quit here by noticing that $Aut(G)$ maps maximal subgroups of $G$ to another maximal subgroups so their intersection stays the same, which means Frattini subgroup is a characteristic subgroup of $G$. However I want to make it more clear by writing it more formal:
$\alpha(\Phi(G)) = \alpha\left(\bigcap\limits_{M \: maximal \: in \: G} \: M \right) = \bigcap\limits_{\alpha(M) \: maximal \: in \: G} \alpha(M) = \Phi(G)$.