# Minimal normal subgroups are either Abelian or a product of non-Abelian simple normal subgroups

In order to prove that the generalized fitting group of a non-trivial group is non-trivial, I'm required to prove the following:

Let $$N$$ be a minimal normal subgroup of a finite group $$G$$. Then $$N$$ is either Abelian or a product of non-Abelian simple normal subgroups of $$N$$.

My attempt: Let $$E$$ be a minimal normal subgroup of $$N$$. If $$E$$ is Abelian, then $$E\subseteq F(N)$$ and so $$F(N)\neq 1$$. Now $$F(N)$$ is a normal subgroup of $$G$$ contained in $$N$$, so by minimality of $$N$$, we have $$F(N)=N$$. Thus, $$N$$ is nilpotent and hence $$Z(N)\neq 1$$. Again using minimality of $$N$$, we get $$N=Z(N)$$ and so $$N$$ is Abelian.

Now, suppose that $$E$$ is not Abelian. If $$E$$ is simple, then using minimality of $$N$$, one can see that $$N=\prod_{g\in G} E^g$$ and so $$N$$ is a product of non-Abelian simple normal subgroups.

How to treat the case when $$E$$ is non-Abelian and non-simple?

• It is sharper than that. The conclusion is that either $N$ is Abelian, or it is the direct product of non-Abelian, pairwise isomorphic simple groups. (These factors are then automatically normal in their direct product $N$.) Jul 25 at 11:02
• A minimal normal subgroup of a group is characteristically simple. Now see this post Jul 25 at 11:22

You are almost done. Take any simple non-Ab subgroup $$H$$. Apply all automorphisms. The images form a characteristic direct product $$N_0$$, so $$N=N_0$$.
• I don't get it. $H$ is a subgroup of $G$ or $E$ or $N$? $N_0$ is characteristic in which group? Jul 25 at 12:07