Every minimal normal subgroup of a finite solvable group is elementary abelian

For any group G a minimal normal subgroup is a normal subgroup M of G such that the only normal subgroups of G which are contained in M are M and 1. Prove that every minimal normal subgroup of a finite solvable group is an elementary abelian p-group for some prime p.


Let G be a finite solvable group and let N \leq G be a minimal normal subgroup.

If N=1, then the conclusion is trivial. Suppose N \neq 1.

Since G is solvable, N is solvable. Since N \neq 1, N^\prime < N is proper; now N^\prime is characteristic in N, hence normal in G. Since N is minimal, we have N^\prime = 1; thus N is abelian.

Recall from this previous exercise that N_p = \{ x \in N \ |\ x^p = 1 \} is characteristic in N, hence normal in G, for all primes p. Thus N_p \in \{1,N\} for all primes p. By Cauchy, some element of N has prime order, so that N_p \neq 1 for some prime p; in particular N_p = N. Thus N is an elementary abelian p-group.

Advertisements
Post a comment or leave a trackback: Trackback URL.

Comments

  • Steve  On March 6, 2011 at 10:38 pm

    You clearly meant to say in the last paragraph that:

    By Cauchy, some element of $N$ has prime order, so that $N^p \ne N$ for some prime $p$; in particular, $N^p=1$. Thus $N$ is an elementary abelian $p$-group.

    • nbloomf  On March 7, 2011 at 7:50 am

      There was a typo- that should have said N_p rather than N^p. Thanks!

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: