Frattini Subgroup of p-Groups Letting $P$ be a $p$-group and $\Phi(P)$ be the Frattini subgroup of $P$ (the intersection of all maximal subgroups), the challenge is "Prove that $P/N$ is elementary abelian implies $\Phi(P)≤N$" (from Dummit and Foote 6.1.26b). The previous exercise has already established $P/\Phi(P)$ is elementary abelian since $P'≤\Phi(P)$ and $\langle~x^p~|~x∈P~\rangle≤\Phi(P)$.
The first step I took here was proving $P/N$ is elementary abelian if and only if $\langle~x^p,~P'~|~x∈P~\rangle ≤ N$, so it seemed natural to prove that $\Phi(P)=\langle~x^p,~P'~|~x∈P~\rangle$. I believe $\Phi(P)/P'=(P/P')^p$ (where $(P/P')^p$ is the image of the $p$-power map) would imply $\Phi(P)P'=\langle~x^p~|~x∈P~\rangle P'=\langle~x^p,~P'~|~x∈P~\rangle$, which would imply $\Phi(P)=\langle~x^p,~P'~|~x∈P~\rangle$ as desired, but I can't find much way to proceed further. The $\supseteq$ direction is clear, but I can't find the reverse containment. Am I going at all the right way about this?
 A: I'd prove that if $P/N$ is elementary abelian, then $N$ is the intersection of the maximal subgroups of $G$ that contain it.
A: A (huge) hint: prove that under the given hypothesis, we have that
$$\Phi(P)=P^p[P,P]$$
With the above, and with the part $\,P^p\le N\,$ that you almost got already,  what is left is $\,P'\le N\,$ , and the abelian part does this job.
A: I am not sure about my answer.
If a group is an elementary abelian group, then its Frattini subgroup is 1. (I am not sure about this part).
Now we assume that $x \in \Phi (P)$ and $x \notin N$. As a result, $xN \neq 1N $ and by the above statement, it couldn't be a nanogenerator. Thus, there exits a subgroup in $P /N$ that is in the format of $I/N$ ($N \subseteq I)$.
$$<xN,I / N> = P /N $$
This means that $<x,I>$ should contain at least one element in each $pN$ for all $p \in P$. We want to conclude that $<x,I>=P$, and this is a contradiction. For each $p \in P$, there exists $n \in N$:
$$ pn \in <x,I>$$
We know that $I$ should contains $N$. So, $\; p \in <x,P>$.
