# Simple subgroups and the Wielandt subgroup

My goal is to prove the following:

Let $$G$$ be a finite group and let $$S \leq G$$ be a simple subgroup. Suppose that $$SH = HS$$ for all subnormal subgroups $$H$$ of $$G$$. Show that $$S$$ is contained in the Wielandt subgroup $$\omega(G)$$ of $$G$$

Note: $$\omega(G) = \bigcap N_G(H)$$, where $$H$$ runs over the subnormal subgroups of $$G$$.

Here's my attempt:

We know (from this paper by Wielandt) that $$\omega(G)$$ contains every minimal normal subgroup of $$G$$ (as stated in the Encyclopedia of Math on the Wielandt subgroup). So it would suffice to prove that $$S \subset \operatorname{Soc}(G)$$, since this is the subgroup generated by such minimal normal subgroups.

Suppose, then, that this is not true. Then, we must have, by the simplicity of $$S$$, $$S\cap \operatorname{Soc}(G) = 1$$, which implies $$S \cap M = 1$$ for all minimal normal subgroups $$M$$ of $$G$$.

This feels like it can't happen, but I don't really know how to prove it. I haven't yet used $$SH = HS$$, so that's probably where to go next, but I just don't see where this hypothesis comes in... I know, from this post, that it wouldn't be possible if $$S$$ were nonabelian and subnormal, but I don't think it's necessarily the case here.

I don't actually need (or even really want) a complete solution, just a hint of what to do now, or if there is a better approach.

What we will prove is the following for a single subnormal subgroup first.

Proposition Let $$G$$ be a finite group and let $$S \leq G$$ be a simple subgroup. Suppose that $$H$$ is a subnormal subgroup of $$G$$ with $$SH = HS$$. Then $$S$$ normalizes $$H$$.

Proof Note that from $$SH=HS$$ it does not follow immediately that $$S$$ normalizes $$H$$. What is true is that it implies that $$SH$$ is in fact a subgroup of $$G$$ and that is what we will use.

Since all groups are finite we can apply induction on the possible triples $$(G,H,S)$$ satisfying the conditions of the proposition. We choose $$|G|$$ as small as possible. If $$|G|=1$$ then the proposition is trivially true. Now consider the triple $$(HS,H,S)$$. This is a valid triple since $$S \subseteq HS$$ is simple, and $$H \lhd \lhd HS$$. Hence, if $$|HS| \lt |G|$$, then by induction, $$S \subseteq N_{HS}(H) \subseteq N_G(H)$$ and we are done.

We can assume that $$G=HS$$. Now $$H \cap S \lhd \lhd S$$, and since $$S$$ is simple we get, either $$H \cap S=1$$ or $$H \cap S=S$$. The latter case yields $$S \subseteq H$$, implying $$G=HS=H$$ and the statement becomes trivially true. So we need to deal with the case $$H \cap S=1$$. We can assume that $$H$$ is not normal, otherwise $$S \subseteq N_G(H)=G$$.

Hence, we can find in the normal series running from $$H$$ to $$G$$ a subgroup $$H_1$$ with $$H \lhd H_1 \lhd \lhd G$$, and $$H \neq H_1$$. By Dedekind's Modular Law $$H(H_1 \cap S)=H_1 \cap HS= H_1 \cap G=H_1$$. So $$H_1 \neq H$$ implies $$H_1 \cap S \gt 1$$, and since $$S$$ is simple and $$H_1 \lhd \lhd G$$, we conclude that $$H_1 \cap S=S$$, so $$S \subseteq H_1$$. But also $$H \subseteq H_1$$, whence $$G=HS \subseteq H_1$$, that is $$G=H_1$$. But then $$H \lhd G$$, a final contradiction.$$\square$$

Corollary Let $$G$$ be a finite group and let $$S \leq G$$ be a simple subgroup. Suppose that $$SH = HS$$ for all subnormal subgroups $$H$$ of $$G$$. Then $$S$$ is contained in the Wielandt subgroup $$\omega(G)=\bigcap_{H \lhd \lhd G} N_G(H)$$.

Note This is the solution to Problem 2A.4 from I.M. Isaacs, Finite Group Theory, p.54. See also here.

Added January 20th 2023. Here is a shorter proof, without induction that I learned from @mesel. Credits go to him!

Let $$H \lhd \lhd G$$, say $$H=H_0 \lhd H_1 \lhd \cdots \lhd H_r=G$$ be a normal series running from $$H$$ to $$G$$. We can assume that $$S \nsubseteq H$$, hence there is a minimal $$i \in \{1,2, ..., r\}$$ subject to $$S \nsubseteq H_i$$ and $$S \subseteq H_{i+1}$$. Note that since $$S$$ is simple, $$H_i$$ is subnormal, and $$S \nsubseteq H_i$$, we must have $$S \cap H_i=1$$.

Now we consider the normal closure of $$H$$ by $$S$$: $$H^S=\langle s^{-1}hs: s \in S, h\in H\rangle$$. Since $$HS=SH$$ this subgroup $$H^S$$ lies in $$HS$$. But $$H_i \lhd H_{i+1}$$, $$S \subseteq H_{i+1}$$, hence $$S$$ normalizes $$H_i$$, so $$H^S \subseteq H_i^S=H_i$$. We conclude that $$H^S \subseteq HS \cap H_i=H(S \cap H_i)=H$$ (here we applied Dedekind's Modular Law), that is $$S \subseteq N_G(H)$$. $$\square$$

• Thank you for such an elegant proof - the last bit ($H \cap S = 1$) is the one that got me. I'll keep this idea in mind for future reference! Commented Jan 19, 2023 at 20:50
• Thanks! You learned something! Keep going! Commented Jan 19, 2023 at 22:00