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In Neukirch/Schmidt/Wingberg, "Cohomology of Number Fields", the following sentence is part of the proof of Proposition 3.5.7. (here $E$ is a finite group, and $\Phi(E)$ is the Frattini subgroup):

Let $N$ be a minimal normal subgroup of $E$ and assume that $N\subset\Phi(E)$. Since $\Phi(E)$ is nilpotent, the group $N$ is abelian.

I don't understand why this follows. If $N$ were minimal normal in $\Phi(E)$, it would certainly be true, but I don't see why that should be the case.

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  • $\begingroup$ Any minimal normal subgroup of any finite group is a direct product of isomorphic simple groups. If these simple groups were nonabelian, then they would not be nilpotent. $\endgroup$ – Derek Holt Oct 25 '15 at 17:06
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Since $N$ is nilpotent, its center $Z(N)$ is non-trivial. Since the center is a characteristic subgroup of $N$, it follows that $Z(N)$ is normal in $G$. By minimality of $N$ and the fact that $Z(N) \neq 1$, we conclude that $Z(N) = N$, so $N$ is abelian.

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