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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?

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    $\begingroup$ 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 $N$.) $\endgroup$ Jul 25 at 11:02
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    $\begingroup$ A minimal normal subgroup of a group is characteristically simple. Now see this post $\endgroup$
    – Derek Holt
    Jul 25 at 11:22

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You are almost done. Take any simple non-Ab subgroup $H$. Apply all automorphisms. The images form a characteristic direct product $N_0$, so $N=N_0$.

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  • $\begingroup$ I don't get it. $H$ is a subgroup of $G$ or $E$ or $N$? $N_0$ is characteristic in which group? $\endgroup$
    – Guest
    Jul 25 at 12:07
  • $\begingroup$ @Guest make your guess. $\endgroup$
    – markvs
    Jul 26 at 0:00

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