Show that every minimal normal subgroup of a finite solvable group is an elementary abelian $p$-group for some prime $p$.

I'm stuck on this one, any idea is appreciated.


First, note that $H$ is abelian. Indeed, we know that $G$ is solvable, and since $H\unlhd G$ this implies that $H$ is solvable. Now, $[H,H]$ is a characteristic subgroup of $H$, and thus normal in $G$, and so by assumption this implies that $H=[H,H]$ or $\{1\}=[H,H]$. If the former happened then $H$ would not be solvable (why?) and so the latter must happen, which says precisely that $H$ is abelian.

Let's show next that $|H|$ is divisible by only one prime. Let $p\mid |H|$, then $H$ has a $p$-Sylow subgroup $S$. Since $H$ is normal in $G$ and $S$ is characteristic in $H$ (because it's the unique Sylow subgroup since $H$ is abelian) we must have that $S$ is normal in $G$, by assumption this implies that $H=S$ so that $|H|=p^n$ for some $n$.

Finally, since $H$ is an abelian $p$-group we know that $pH$ is a proper (by Cauchy's theorem) of $H$ which is necessarily normal in $G$ (since it's characteristic in $H$), and so $pH=\{1\}$. This implies that $H=\mathbb{F}_p^n$ as desired.

  • $\begingroup$ Thank you so much for the both clear and instructive answer. $\endgroup$ – user112564 Dec 1 '13 at 6:33
  • $\begingroup$ @user112564 If you're happy with my answer you can upvote it and/or accept it. $\endgroup$ – Alex Youcis Dec 1 '13 at 6:34
  • $\begingroup$ @user I was just being lazy. Do you want me to change it? $\endgroup$ – Alex Youcis Dec 1 '13 at 8:31
  • $\begingroup$ @user Although, there is some logic in it. My observation that $pH$ is $0$ shows that $H$ must be a $\mathbb{F}_p$-vector space, and thus as a vector space isomorphic to $\mathbb{F}_p^n$. $\endgroup$ – Alex Youcis Dec 1 '13 at 8:31
  • $\begingroup$ @Alex Youcis I'm fine with that notation. It requires 15 reputations to vote up for an answer. I will vote up for your answer once I reach 15 reputations. $\endgroup$ – user112564 Dec 1 '13 at 8:36

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