A question related to Frattini subgroup I am doing a question which need the following statement as a lemma:
Statement: If $G$ is a group with a finitely generated Frattini subgroup $\Phi(G)$, then the only subgroup $H$ of $G$ such that $H\Phi(G) = G$ is $H = G$.
The proof of this statement on my book is indirect and using the following theorem:
Theorem: The Frattini subgroup of a nontrivial group $G$ is the set of all elements $x$ in $G$ with the property that, whenever a set $K\cup \{x\}$ generates $G$, $K$ generates $G$.
However, my proof of the statement above is rather simply and more directly. So I am afraid there would be some mistake in my proof so I want someone to check it. Thanks!
My proof of the statement: Suppose there exist a maximal subgroup $M$ such that $H \subset M \subset G.$ Then since $M$ is maximal, $\Phi(G) \subset G$, so $G = H\Phi(G) \subseteq M \subset G$. So $H$ is a maximal subgroup not equal to $G$ or $H = G$. The former case is impossible since if it does then $G = H\Phi(G) = H$, a contradiction. So finally we have $H = G.$
 A: This is a little tricky. For your argument to work, you need to show the existence of a maximal subgroup that contains $H$, which is not immediate. One cannot guarantee it for arbitrary infinite groups, even ones that have some  maximal subgroups (e.g., $\mathbb{Z}_{p^{\infty}}\times C_p$ has maximal subgroups, so $\Phi(G)\neq G$, but not every subgroup is contained in a maximal subgroup).
So there's an argument missing in your proof:  that given $H$ such that $H\Phi(G)=G$, if $H\neq G$ then there must exist a maximal subgroup of $G$ that contains $H$.
To that end, let $\mathscr{S}$ be the collection of all proper subgroups of $G$ that contain $H$. This is nonempty, since we are assuming that $H\neq G$. To apply Zorn's Lemma to $\mathscr{S}$, let $\{K_i\}_{i\in I}$ be a chain of elements of $\mathscr{S}$. Being a chain of subgroups, we know that $\cup K_i$ is a subgroup; they all contain $H$, so $\cup K_i$ contains $H$. The issue is proving that $\cup K_i\neq G$.
Here is where we use that $\Phi(G)$ is finitely generated: let $m_1,\ldots,m_k$ be a finite generating set for $\Phi(G)$. If $\cup K_i = G$, then for each $i$ there exists $j_i\in I$ such that $m_i\in K_{j_i}$. Since there are finitely many indices, there exists $N\in I$ such that $\Phi(G)\subseteq K_N$. But then $K_N=HK_N$ (since $H\subseteq K_N$ by construction), and $G = H\Phi(G)\subseteq HK_N = K_N\neq G$, a contradiction. Thus, $\cup K_i\neq G$ and hence lies in $\mathscr{S}$.
Applying Zorn's Lemma, we conclude that $\mathscr{S}$ has maximal elements, and thus that $G$ has maximal subgroups  that contain $H$, as desired. $\Box$

That said: you objected to the proof you saw being "indirect"; the proof here is, in my view, even more convoluted: it is a proof by contradiction which relies on applying Zorn's Lemma, and the obvious proof that Zorn's Lemma applies is itself an argument by contradiction: an argument by contradiction sitting inside an argument by contradiction. Not the most direct way of doing things.
