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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?

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I'd prove that if $P/N$ is elementary abelian, then $N$ is the intersection of the maximal subgroups of $G$ that contain it.

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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.

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  • $\begingroup$ @Feryll: In fact when $P/N$ is elementary abelian so it is a vector space over $\mathbb Z_p$ $\endgroup$ – mrs Apr 5 '13 at 17:06
  • $\begingroup$ @DonAntonio Does $P^p$ need to be a group? Because the $p$-power map shouldn't necessarily be a homomorphism on an arbitrary $p$-group. $\endgroup$ – Feryll Apr 5 '13 at 17:26
  • $\begingroup$ @Feryll, that's how the subgroup generated by all $p-$th power of elements of $\,P\,$ is usually denoted $\endgroup$ – DonAntonio Apr 5 '13 at 18:49

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