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This question arose from my attempt at understanding the answer in this post - the comments below it, to be precise. Everything revolves around the following problem:

Let $S$ be a simple, nonabelian and subnormal subgroup of a group $G$. Show that $S^G$, the normal closure of $S$ in $G$ is a minimal normal subgroup of $G$ .

The question (from Isaacs' book "Finite Group Theory") comes with a hint. It says:

HINT: Work by induction on $|G|$ to conclude $S \subset \operatorname{Soc}(H)$ whenever $S \subset H$. Deduce that each conjugate of $S$ in $G$ is a minimal normal subgroup of $S^G$. Then, apply the previous problem to $S^G$.


I managed to do everything, except for the part in italics. What I did was (briefly) as follows:

  1. $S \lhd \lhd G \implies S \lhd \lhd S^G$. If $S^G = G$, then $S = G$, which means the result is trivial. Therefore, we can assume $S^G \neq G$
  2. Using induction, it was simple to prove $S \subset \operatorname{Soc}(H)$ if $H \neq G$. In particular, $S \subset \operatorname{Soc}(S^G) \implies S^G = \operatorname{Soc}(S^G)$
  3. Using a previous problem, $S^G$ is a direct product of minimal normal subgroups (and of simple groups)

In the aforementioned post, user @Stefan4024 stated:

Since $S \lhd \lhd S^G$, and $S^G$ is a product of minimal normal subgroups, then $S$ is a minimal normal subgroup of $S^G$ (I took $K = S^G$)

I just can’t see why this follows so immediately.

Continuing my previous reasoning, I managed to find, by subnormality, a normal subgroup $S \lhd \lhd K \lhd S^G$ (which is non-abelian) and this yielded, by Exercise $2A.6$, a minimal normal subgroup $X$ of $S^G$ such that $X \subset K$. It follows, both by simplicity and from the fact that minimal normal subgroups normalize all subnormal subgroups, that either $S = X$ or $S \cap X = 1$. And I couldn't rule out this last case...

Is there a simpler way to prove the statement in the last block quote? If not, how do I continue from my arguments?

Thanks in advance!

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The only normal subgroups of a direct product of nonabelian simple groups are direct products of some of the direct factors. Hence these are also the only subnormal subgroups. So its subnormal subgroups are all normal.

Since $S$ is simple and subnormal in $S^G$, $S$ must be one of these direct factors, so it is indeed a minimal normal subgroup of $S^G$.

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  • $\begingroup$ Thank you so much! So simple, yet I didn't know this result... Is there a simple proof or reference? $\endgroup$
    – Gauss
    Commented Jan 4, 2023 at 23:21
  • $\begingroup$ Is there a simple proof of what? $\endgroup$
    – Derek Holt
    Commented Jan 4, 2023 at 23:23
  • $\begingroup$ I was referring to your statement about normal subgroups of a direct product of nonabelian simple groups, but I now realize it comes straight from the fact that a normal subgroup of a direct product is either central or intersects one of the factors non-trivially $\endgroup$
    – Gauss
    Commented Jan 4, 2023 at 23:28

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