If a group has no maximal subgroups then all elements are non-generators? Frattini subgroup characterization

Question

This question is the last leg of an exercise I've been working on in which we characterize the intersection of all maximal subgroups as the subgroup of all non-generators. I've already shown that if a group has a maximal subgroup then the intersection of all the maximal subgroups are precisely the non-generators. Also, I've shown that if every element of a group is a non-generator, then the group has no maximal subgroups. But I'm having difficulty with this claim:

If a group has no maximal subgroups, then all elements are non-generators.

All the proofs I've seen of this characterization assume that there are maximal subgroups, so I'm not even sure if the claim is true. Any help is appreciated.

ADDENDUM

Per request, here's my proof for the other three cases (none of which use Zorn's lemma):

Let $G$ be a group. Let's denote the intersection of all maximal subgroups of $G$ by $\Phi(G)$. If $G$ has no maximal subgroups, we set $\Phi(G)=G$. Let's denote the subgroup of all non-generators of $G$ by $N(G)$. We note that if $G$ is trivial, then $\Phi(G)=N(G)$. So we can forget about that case.

Case 1: $G$ has a maximal subgroup.

Let $g\in\Phi(G)$. Now let $X\subseteq G$ satisfy $\langle X, g\rangle=G$. Let $M$ be a maximal subgroup of $G$. Then we have that $M\subseteq\langle X, g\rangle=G$. Also, since $g\in\Phi(G)\subseteq M$, we have that $M\subseteq\langle X\rangle\subseteq G$. Thus $\langle X\rangle=M$ or $\langle X\rangle=G$ by maximality of $M$. Suppose $\langle X\rangle=M$. This would imply that $\langle X, g\rangle=M$ contradicting the assumption. Thus $\langle X\rangle=G$, and $g$ is a nongenerator. Thus $\Phi(G)\subseteq N(G)$.

Now let $g\in N(G)$. Let $M$ be an arbitrary maximal subgroup of $G$. Suppose $g$ were not in $M$. Then by maximality of $M$, $\langle M,g\rangle=G$. However, since $g$ is a nongenerator, this implies that $\langle M\rangle = M=G$. Contradiction. Thus $g\in M$. Since $M$ was arbitrary, $g\in\Phi(G)$.

Case 2: $G$ has no maximal subgroup.

We have that $\Phi(G)=G$. Thus we trivially have that $N(G)\subseteq\Phi(G)$.

We are left to prove that $\Phi(G)=G\subseteq N(G)$ which was the reason behind the question.

Answer 1

Suppose that $G$ has no maximal subgroups. We have to show that, for any $g \in G$ and $X \subseteq G$, $\langle g,X \rangle =G \Rightarrow \langle X \rangle = G$.

If not, then clearly $g \not\in \langle X \rangle$. By Zorn's Lemma, there is a subgroup $H$ of $G$ with $X \subseteq H$ that is maximal with respect to not containing $g$. Now if $H$ is not maximal in $G$, then it is contained in a larger proper subgroup $K$ of $G$. Then $g \in K$ by definition of $H$, contradicting $\langle g,X \rangle =G$. So $H$ is maximal in $G$, contrary to the assumption.

Answer 2

The point is that if something is not a non-generator, then it must be contained in a maximal subgroup. Indeed, if $\langle Y,x\rangle =G$, but $\langle Y\rangle $ is a proper subgroup, find a maximal subgroup $M$ w.r.t to containing $Y$ and not $x$. One can find such group, since $\langle Y\rangle$ doesn't contain $X$. Then whenever $M\subsetneq M'$, $x\in M'$. Since $Y\subseteq M$ already, $G\subseteq M'$, i.e. $M'=G$ so $M$ is maximal. This shows that if $G$ has no maximal subgroups, any element must be a non generator. If you think about it for a second, this is what we do to show that $\Phi(G)$ is a subset of the non-generators.