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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 $H = G.$

Shaun
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Andrew
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  • I don't think this does matter. – Andrew Apr 23 '21 at 17:10
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    My infinite group theory knowledge is very rusty, however the assumption that "Let $G$ be a group with finitely generated Frattini subgroup" implies that $G$ might be infinite, and in particular it may not have maximal subgroups (so your assumption that $H$ is contained in a maximal subgroup may not hold). For finite groups, your proof certainly works, but I am not sure about infinite groups. – Levent Apr 23 '21 at 17:14
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    What Levent identifies is a gap: you need to show that there must exist a maximal subgroup that contains $H$ when $G$ is infinite. Also, I think that after "since $M$ is maximal", rather than $\Phi(G)\subset G$ you mean $\Phi(G)\subset M$. – Arturo Magidin Apr 23 '21 at 17:19
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    You can check this question to see why Zorn's lemma does not imply the existence of maximal subgroups for infinite groups. – Levent Apr 23 '21 at 17:22
  • The definition of Frattini subgroup says that if G does not have maximal subgroup then the Frattini subgroup Φ(G) is just G. In this case there is nothing to prove I think – Andrew Apr 23 '21 at 17:23
  • Quite the opposite: if $\Phi(G)$ were equal to $G$, then the conclusion you want would be false, since $H\Phi(G)=G$ for all $H$, including $H={e}$. But if $G$ had no maximal subgroups, then $\Phi(G)=G$, which under you assumptions means $G$ is finitely generated. But it turns out that finitely generated groups do have maximal subgroups, so the initial assumption cannot hold. You still need to show the existence of one that contains $H$, though. – Arturo Magidin Apr 23 '21 at 17:31
  • The point is: the claim that there must exist a maximal subgroup containing $H$ is clear for finite groups, but it is not immediate for infinite groups (and is false in general for arbitrary infinite groups). That means that it must be justified in the situation at hand using your other hypotheses (that $\Phi(G)$ is finitely generated), and this is not being done here. Notice that you never used that $\Phi(G)$ is finitely generated, at least not explicitly, which should be a red flag. – Arturo Magidin Apr 23 '21 at 17:33
  • You should fix your link, and it's not the same link. You are linking to question 4113796, I'm linking to question 779590. There's no need to apologize for that, but I hope you are seeing the issue we are pointing out. It's almost certain you can fill the gap and fix the proof, but as presented it is definitely gappy. – Arturo Magidin Apr 23 '21 at 17:35
  • @Arturo Yes, my logic is loose. Then with the evidence you provided which I think it works. – Andrew Apr 23 '21 at 17:35
  • @Andrew: There's going to be problems. You really need to show that "every proper subgroup is contained in a maximal subgroup", and that may not be true even if $G$ does have maximal subgroups. E.g., $\mathbb{Z}_{p^{\infty}}\times C_p$ has maximal subgroups, but any subgroup containing ${e}\times C_p$ is not contained in any maximal subgroup. (Of course, here $\Phi(G)$ is the Prufer $p$-group, so again it violates the hypotheses of your problem; but I think it's going to be tricky.) – Arturo Magidin Apr 23 '21 at 17:44
  • @Arturo Magidin One more question, I seem not so happy about using Zorn's lemma in solving problems, since all I know the proof uses Axiom of Choise, so is there any elementary proof of Zorn's lemma that avoid the AoC which make me happy to accept? – Andrew Apr 23 '21 at 17:44
  • @Andrew: Zorn's Lemma is equivalent to the Axiom of Choice. Any proof of Zorn's Lemma that does not use the Axiom of Choice explicitly is using it implicitly; e.g., by using the Well-Ordering principle. If you are going to do algebra/group theory, you better come to terms with the Axiom of Choice sooner rather than later. – Arturo Magidin Apr 23 '21 at 17:45
  • @Arturo By the link you provided I think we also can generate a chain from H with a maximal element – Andrew Apr 23 '21 at 17:52
  • @Andrew: But the link assumes $G$ is finitely generated; you are only assuming $\Phi(G)$ is finitely generated. That guarantees that $G$ has some maximal subgroups (because otherwise you would have $\Phi(G)=G$, and then you could conclude $G$ is finitely generated, contradiction the assumption that $G$ has no maximal subgroups at all). But it does not guarantee that it has maximal subgroups that contains a given a subgroup: I gave you an example of an infinite group with $\Phi(G)\neq G$ but where not every proper subgroup is contained in a maximal one. – Arturo Magidin Apr 23 '21 at 17:54
  • Which book are you referring to? – Shaun Dec 22 '21 at 23:32

1 Answers1

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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.

Arturo Magidin
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