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The Frattini subgroup $\Phi(G)$ is the intersection of all maximal subgroups of $G$. (If there are none, then $\Phi(G)=G$.) We say that an element $g\in G$ is a nongenerator if whenever $\langle X\cup \{g\}\rangle = G$, we have $\langle X\rangle = G$ for subsets $X\subseteq G$.

If $G$ is finite, show that $\Phi(G)$ is the set of nongenerators of $G$.

So I want to start by assuming that $g\in G$ is in all maximal subgroups of $G$, and $\langle X\cup\{g\}\rangle = G$, and to prove that $\langle X\rangle = G$. At first I thought if $\langle X\cup\{g\}\rangle = G$, then $\langle X\rangle$ might be a maximal subgroup itself, but it seems that this is not necessarily the case. (For example $G$ is a cyclic group of order $4$ generated by $g$, and $X$ is the trivial subgroup.)

Paul S.
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2 Answers2

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Let $g\in\Phi(G)$ and suppose that $g$ is not a non-generator for the group. Then there exists a subset $X$ of $G$ such that $G=\langle X,g\rangle$ and $G\neq \langle X\rangle$. Then $g\notin \langle X\rangle$, otherwise $\langle X,g\rangle=\langle X\rangle=G$, contradicting our assumptions. Let $\mathscr{S}$ be the set of all subgroups of $G$ containing $\langle X\rangle$ and not containing $g$. First of all, $\mathscr{S}$ is not the empty set, because $\langle X\rangle$ belongs to it. Moreover $(\mathscr{S},\subseteq)$ is an ordered set. If we take a chain of elements of $\mathscr{S}$, and do their insiemistic union, this union also belongs to $\mathscr{S}$ and evidently is an upper bound for the elements in the chain. By Zorn's Lemma we get that $\mathscr{S}$ has a maximal element, call it $M$. Suppose $M<H\leq G$. Then $g\in H$ and $H=G$ (since $G=\langle X,g\rangle\leq \langle H,g\rangle=\langle H\rangle =H$). It follows that $M$ is a maximal subgroup of $G$, not containing $g$ , and this is absurd since we've taken $g$ in $\Phi(G)$. So $g\in \Phi(G)\leq M$, thus $G=\langle g,X\rangle=M$, absurd. We conclude that $g$ must be a non-generator of the group.

Conversely, suppose that $g$ is a non-generator and that $g$ is not an element of Frattini subgroup of $G$. Then there exists a maximal subgroup $M$ of $G$ that does not contain the element $g$. It follows that $M\neq \langle g,M\rangle$ and so, by maximality of $M$ that $G=\langle g,M\rangle$. But $g$ is a non-generator, so that $G=M$, contradicting hypothesis saying that $M$ is a maximal (hence proper) subgroup of $G$

Federica Maggioni
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    A LaTeX tip: < and > mean "less than" and "greater than", and produce spacing correct for that meaning only. When you want angle brackets, you need to use \langle and \rangle. – Zev Chonoles May 25 '13 at 19:24
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    @ZevChonoles thank you very much, i'll edit in some minutes – Federica Maggioni May 25 '13 at 19:25
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    Appealing at Zorn's lemma for finite groups? ;-) – egreg May 25 '13 at 21:25
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    @egreg whoops, i actually didn't see "finite" in the question :-), well, the proof works in the general case – Federica Maggioni May 25 '13 at 21:33
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    You last paragraph is not really a proof by contradiction. You're effectively proving that if $x\notin \Phi(G)$; then $x$ is not a non-generator by taking $M=\langle M\rangle <\langle M,x\rangle=G$ for $M$ is maximal, and $x\notin M$. – Pedro Mar 14 '14 at 22:44
  • The first paragraph isn't a proof by contradiction either: Scratch the "Let $g\in \Phi(G)$" part and you've effectively proven that if $g$ is not a generator, then it is not an element of the Frattini subgroup. I was going to ask a question looking exactly for a verification of this (direct!) argument I took some time in coming up with. In particular the correct use of Zorn's lemma. +1 to you. – Pedro Mar 14 '14 at 22:47
  • As a stylistic (?) sidenote, saying $\langle X\rangle \subseteq H$ for a subgroup of $G$ is the same as saying $X\subseteq H$. – Pedro Mar 14 '14 at 22:54
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    +1 for giving the proof in the general case. In fact I have spent the last hour or so learning about Frattini subgroups for the first time, and most of the references I found restrict the hypothesis in this result, to the extent that I spent a while searching for counterexamples in the general case. – Pete L. Clark Aug 25 '15 at 04:35
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    Surely there's a way to prove this that doesn't assume choice? – Little Narwhal Aug 28 '21 at 13:31
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Suppose that $\langle X\cup\{g\}\rangle=G$, but $\langle X\rangle\neq G$. Because $\langle X\rangle$ is not all of $G$, it is contained in some maximal subgroup $M\subset G$...

but so is $\{g\}$, so you'd have to have $\langle X\cup\{g\}\rangle\subseteq M$, which is a contradiction.

Zev Chonoles
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