5
$\begingroup$

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.

$\endgroup$
14
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.