If $S$ is a simple subgroup of a finite group $G$ such that $NS=SN$ where $N$ is any subnormal subgroup of $G$, then $S\subseteq N_G(N)$

Question

I’m looking for a proof of the following statement, or any hint/help on how to attempt one:

If $S$ is a simple subgroup of a finite group $G$ such that $NS=SN$ where $N$ is any subnormal subgroup of $G$, then $S\subseteq N_G(N)$ for every subnormal subgroup $N$ of $G$.

Answer 1

Let $N \lhd \lhd G$. We prove by induction on the subnormal depth of $N$ in $G$ that $S \le N_G(N)$. This is clear for subnormal depth $1$, i.e. when $N \lhd G$.

Otherwise, there exists $K \lhd G$ with $N \le K$ such that the subnormal depth of $N$ in $K$ is less than in $G$. If $S \le K$ then the result follows by induction, and otherwise, since $S$ is simple, we have $S \cap K = 1$.

Now $SN=NS$ implies that $NS$ is a subgroup of $G$ and so $N^S \le NS \cap K = N(S \cap K) = N$, and hence $S \le N_G(N)$.