## 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.

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### 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!