3

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?

Guest
  • 1,657
  • 3
    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 $N$.) – Andreas Caranti Jul 25 '22 at 11:02
  • 2
    A minimal normal subgroup of a group is characteristically simple. Now see this post – Derek Holt Jul 25 '22 at 11:22

1 Answers1

1

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

Shaun
  • 47,747
markvs
  • 19,896