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

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 $$p$$.

• The standard definition of $G^p$ is not the set of $p$-th powers but the subgroup of $G$ generated by the $p$-th powers. So $G^p$ is always a subgroup. But having said that, with your definition it will still work, because the images of the elements $x^p$ in $G/[G,G]$ (which is abelian) do form a subgroup. Commented Sep 18, 2015 at 22:03
• Oh, thanks for that. Commented Sep 18, 2015 at 22:07
• But would $G'\subset \Phi(G)$? Commented Sep 18, 2015 at 22:16

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.