The Frattini subgroup is a characteristic subgroup. 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? 
 A: HINT. If $H$ is a maximal subgroup of $G$, and $f\colon G\to K$ is a homomorphism with kernel contained in $H$, then $f(H)$ is a maximal subgroup of $f(G)$.
A: let $\alpha$ be an arbitrary automorphism of $G$. If $M$ be a maximal subgroup of $G$, then $\alpha(M)$ is a maximal subgroup, too. then $\alpha(\phi(G))=\alpha(\cap M)=\cap \alpha(M)=\cap M'=\phi(G)$, M' is a maximal subgroup. 
A: 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:
$\alpha(\Phi(G)) = \alpha\left(\bigcap\limits_{M \: maximal \: in \: G} \: M \right) = \bigcap\limits_{\alpha(M) \: maximal \: in \: G} \alpha(M) = \Phi(G)$.
