If $S \subset T \subset X$, then is $G_S$ normal in $G_T$, where the groups are the $S, T$ preserving symmetries of $X$?

Question

Suppose that $S \subset T \subset X$ and $G_S = \{ \varphi \in \text{Sym}_X : \varphi S = S, \ \varphi(x) = \text{id}(x) \ \forall x \notin S\}$, sim $G_T$. Then $G_S \leqslant G_T$ since $\varphi(x) = x$ when $x \in T\setminus S$ and $\varphi S = S$ by definition. Then, further, is $G_S$ normal in $G_T$?

Answer 1

This is not in general true.

For example if $G=S_4$, $S=\{1,2\}$, $T=\{1,2,3\}$, $X=\{1,2,3,4\}$. This gives $G_S=\{1,(1,2)\}$ and $(1,2,3)\in G_T$, but $(1,2,3)G_S(1,2,3)^{-1}=\{1,(2,3)\}$.

In general $\phi G_S\phi^{-1}=G_{\phi(S)}$ so $G_S$ is normal in $G_T$ if and only if for all $\phi\in G_T$ you have $G_S=G_{\phi(S)}$.