Show that the Frattini subgroup is the set of nongenerators of $G$.

Question

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 $G$.

Proof: Let $N$ be the set of nongeneratos of $G$. We need to show that $\Phi(G)=N$.

(1) $\Phi(G)\subseteq N$

$\Phi(G)=\cap \mathcal{H}$ where $\mathcal{H}$ is maximal subgroup. We need to show that all $g\in \Phi(G)$ is a nongenerator of $G$. We will do this by contradiction.

i.e. We will show that there exists $g\in \Phi(G)$ such that $g$ is not generator.

i.e. We will show that there exists $X$, $\langle X\cup\{g\}\rangle=G$ and $\langle X\rangle\neq G$.

At this point all the definitions have me confused can some clear things up for me and help me finish the proof?

Answer 1

Hint: Assume $X$ does not generate $G$, but $X \cup \{g\}$ does. Let $H$ be the subgroup generated by $X$, so that $g \not\in H$. In general, you can use Zorn's lemma to find a maximal subgroup $K \supseteq H$ such that $g \not\in K$. If $G$ is finite (as in your question), Zorn's lemma isn't needed: you can take $K \supseteq H$ to be subgroup of maximal order such that $g \not\in K$. Either way: is $K$ a maximal subgroup of $G$?