For a finite subgroup, $G$, I am trying to show that the Frattini subgroup, $$\Phi(G) = \bigcap M \Longleftrightarrow \{g \in G \mid g\text{ is non-generator}\}.$$ Here $M$ is any maximal subgroup of $G$.
I am studying for an exam. Could someone please confirm if my reasoning is all right?
($\Longrightarrow$)
Let $g \in \bigcap M$. Now, choose any $X \subset G$, with $g \notin X$ such that, $G = \langle X, g \rangle$.
Now consider, $\langle X \rangle$. If $G = \langle X \rangle$, we are done, because this would mean that $g$ is a non-generator. If not, that is, if $\langle X \rangle \subsetneq G$, then there is a maximal subgroup $M$, in $G$, such that, $\langle X \rangle \subset M \subsetneq G$. Now if $g \notin \langle X \rangle$, then we have $\langle X \rangle \subset \langle X, g \rangle \subset M$.
But $\langle X, g \rangle$ generates $G$ by assumption, and since $M$ is a maximal subgroup, this containment cannot hold. Thus $G = \langle X \rangle$, and it follows that $g$ is a non-generator.
($\Longleftarrow$)
Let $g$ be a non-generator. Let $g \in X$ with $G = \langle X \rangle$. Now consider any arbitrary maximal subgroup, $M$, of $G$. We have the following containment tower: $M \subset \langle M, g \rangle \subset \langle X \rangle$. The first containment cannot be proper unless $G = \langle M, g \rangle$, and the second containment must be proper because $g$ is a non-generator. This forces $M = \langle M, g \rangle$; whence we have $g \in M$.
Since $M$ was an arbitrary maximal subgroup, $g \in \bigcap M$, and thus $g \in \Phi(G)$.