If $G = \langle x,y,z\rangle$ and $x \in \Phi(G)$, the intersection of all maximum subgroups of $G$, then $G = \langle y,z\rangle$

Question

If $G = \langle x,y,z\rangle $ and $x \in \Phi(G)$, the intersection of all maximum subgroups of $G$, then $G = \langle y,z\rangle $.

I have already proved that $\Phi(G) \triangleleft G$.

So i'm basically trying to show that under the given conditions, $x \in (y,z)$.

I know that $zxz^{-1} \in \Phi(G)$ and $yxy^{-1} \in \Phi(G)$... But i guess I'm a little lost

https://math.stackexchange.com/questions/402292/frattini-subgroup-is-set-of-nongenerators — the_fox, Jul 21 '19 at 17:38

score 2 · Accepted Answer · answered Jul 21 '19 at 17:22

Suppose $x\not\in \langle{y,z}\rangle$.

By Zorn's lemma, we can extend $\langle{y,z}\rangle$ to a subgroup of $A$ of $G$, such that $A$ is maximal among all subgroups of $G$ containing $y$ and $z$, but not $x$.

Since $x\not\in A$, $A$ is a proper subgroup of $G$.

Suppose $A$ is not a maximal subgroup of $G$.

Then there must be a subgroup $B$ of $G$ such that $A\subset B\subset G$, where all inclusions are strict.

By choice of $A$, we must have $x\in B$, but then, since $y,z\in A$, we get $x,y,z\in B$, so $B=G$, contradiction.

Hence $A$ must be a maximal subgroup of $G$.

But then $x\in A$, contradiction.

Therefore $x\in\langle{y,z}\rangle$.

If $G = \langle x,y,z\rangle$ and $x \in \Phi(G)$, the intersection of all maximum subgroups of $G$, then $G = \langle y,z\rangle$

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