$G'G^p=\Phi(G)$

Question

Given a $p$-group $|G|=p^n$, consider $G'=[G,G]$ and define $G^p:=\langle g^p\;:\;g\in G\rangle$: then we have that $G'G^p=\Phi(G)$, where $\Phi(G)$ is the Frattini subgroup, defined as the intersection of all maximal subgroups of $G$.

There are no hints, and I'm really stuck. Since now, no good idea came to my mind, otherwise I'd wrote it here. Any help will be appreciated!

Answer 1

Hints: Show that $G^{\prime}G^p\leq\Phi(G)$ and that $G^{\prime}G^p\geq\Phi(G)$.

• One direction: Maximal subgroups are normal of index $p$ (use Nilpotency).
• Other direction: $G/G^{\prime}G^p$ is an elementary abelian $p$-group. What is its Fratinni subgroup?