# Frattini subgroup is set of nongenerators

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.)

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 $G$

• 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. May 25, 2013 at 19:24
• @ZevChonoles thank you very much, i'll edit in some minutes May 25, 2013 at 19:25
• Appealing at Zorn's lemma for finite groups? ;-) May 25, 2013 at 21:25
• @egreg whoops, i actually didn't see "finite" in the question :-), well, the proof works in the general case May 25, 2013 at 21:33
• 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, 2014 at 22:44

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.

• Not every proper subgroup of G is necessarily contained in a maximal subgroup: take Q for example which has no maximal subgroups. Aug 28, 2021 at 13:18
• Nevermind I didn't see the finite specification of the question. Aug 28, 2021 at 13:19