If $G$ is a $p$-group then $\Phi(G)=G'G^p$

Question

Okay this problem is quite the confounding one for me.

If $G$ is a $p$-group then it follows that $\Phi(G)=G'G^p$.

Where:

1. $\Phi(G)$- Frattini subgroup (which in this case is the intersection of all subgroups of index $p$)
2. $G'$ commutator subgroup
3. $G^p=\{x^p:x\in G\}$

I am having trouble tackling a couple sub-problems with this problem.

1. Why is $G'G^p$ a subgroup? In general $G^p$ isn't a subgroup, so why does $G'G^p$ become a subgroup?

2. While I understand that $G^p\subset \Phi(G)$, why would $G'G^p\subseteq M_i$ for all $i$ where $\{M_i\}$ is a collection of subgroups of index p?

   Alot of my problems, center around $G^p$ not being a subgroup in general. But even if I were to prove that $G'G^p$ is a normal subgroup, I would still have to prove (2.) which would be easy if it is true that every maximal class is conjugate to one another, but I don't know that (or maybe it isn't true).

Is there a way for me to see that $G'G^p$ is a normal subgroup and that it is contained in every subgroup of index $p$.

Answer 1

For a finite $p$-group $G$, any maximal subgroup $M<G$ is normal of index $p$.

Proof. $G$ has nontrivial center so pick a central subgroup $Z$ of order $p$. If $Z\subseteq M$ then $M/Z<G/Z$ is normal of index $p$ by induction hypothesis, and then $M<G$ is normal of index $p$. Otherwise if $Z\not\subseteq M$ then the whole group $G=MZ$ normalizes $M$, and $G$ has $|Z|$ cosets of $M$ in it.

Because $G/M\cong C_p$, the $p$-power map on $G/M$ is the zero map so $G^p\subseteq M$, and the quotient $G/M$ is abelian so $G'\subseteq M$. Therefore $G^pG'\subseteq M$ for every maximal $M$. Both $G^p$ and $G'$ are normal so we know that $G^pG'$ is normal too.