Let $\Phi(G)$ be the set of non-generators of $G$, that is the groups of elements $g\in G$ such that $\langle X,g\rangle=G \implies \langle X\rangle=G$
We call such set the Frattini Subgroup of $G$
A simple observation is that the contra-positive gives us another way of defining $\Phi(G)$ since:
( $\langle X,g\rangle=G \implies \langle X\rangle=G) \iff ( \langle X\rangle\neq G\implies\langle X,g\rangle\neq G)$
I must show that:
$\Phi(G)$ coincides with the intersection of all maximal proper subgroups $M<G$ (under the statement that if $G$ has no maximal elements then $\Phi(G)=G$).
Then conclude that $\Phi(G)$ is a subgroup.
What I've done is:
Consider the case where $G$ has maximal subgroups
Let $r\in \Phi(G)$, $M$ an arbitrary maximal proper subgroup.
Then $\langle M\rangle=M\neq G$ (since $M$ is proper)
therefore $\langle M,r\rangle\neq G$ (since $r\in\Phi(G)$)
By the maximality of M and $M\subseteq \langle M,r\rangle$ we get that $\langle M,r\rangle=M$ which implies $r\in M$
but $M$ was arbitrary therefore $r\in \bigcap M$
So $\Phi(G)\subseteq\bigcap M$
For the other side I consider this unfinished proof:
Let $g\in G$ and $X\subset G$ for which $\langle X\rangle\neq G$ but $\langle X,g\rangle=G$, now I consider the set $\mathcal{H}:=\{ H<G $ such that $ \langle X\rangle\leq H, g\notin H\}$ I know the set is non-empty since $\langle X \rangle\in\mathcal{H}$...
But im having problems with the part where you find the maximal element on $\mathcal{H}$, the proof I read began with the statement that the union of a chain in $\mathcal{H}$ is an element of it, for a finite chain I can see that but I dont get the countable case.
Then it goes to say that the union is an upper bound therefore by Zorn's Lemma we get the maximal element.
Im fine with that but wanted to know how would I choose a maximal element on the finite case, do I just pick the biggest on my chain? Why should that be a maximal subgroup?
I tried searching books for this part but they go over more refined results that I don't cover in my course.