TITLE: Is stabilizer subgroup Normal
QUESTION [6 upvotes]: Let $G\le S_n$, and set $\Omega = \{1,2,3,\ldots,n\}$. Let $\Delta \subseteq \Omega$, Stabilizer subgroup of $\Delta$ is defined as 
$$Stab_{\Delta} = \{g \in G  \mid \Delta^g = \Delta \}$$
Is $Stab_{\Delta}$ Normal subgroup of $G$?
My Attempt : 
To prove normal we need to show that if any $h\in Stab_{\Delta} $ then $ghg^{-1} \in Stab_{\Delta}$
$$x_i^g = x_j$$
$$(x_i^g)^h = (x_j)^h $$
$$(x_i^{gh})^{g^{-1}} = (x_{l})^{g^{-1}}$$

REPLY [7 votes]: The answer is in general no.
Take $n=3$ and $G=S_3$. The stabilizer of $\{1, \,2\} \subset \{1, \,2, \,3\}$ is the order two subgroup generated by $(1 \,2)$, which is obviously not normal in $S_3$.