How to prove centraliser of a subgroup is normal? Notations
$C_{G}(H) = \{g \in G \mid hg = gh, \forall h \in H\}$
Claim : If $G,H \le S_n$ and $G$ normalises $H$, then $C_{G}(H)$ is normal in $G$.
Question1  : To prove $C_{G}(H)$ is normal in $G$, I need to prove that for all $g \in G$, $C_{G}(H)g = gC_{G}(H)$. One direction is $C_{G}(H)g \subseteq gC_{G}(H)$, let $g_1 \in C_{G}(H)$ now $c = g_1g$ and $c \in C_{G}(h)$.
Question 2 : Is proving $g^{-1}C_{G}(H)g = C_{G}(g^{-1}Hg)$, for $g \in G$ sufficient to prove the claim? Please explain how?
 A: Note that $C_G(g^{-1}Hg)=C_G(H)$ as $g\in G$ normalises $H$ by hypothesis.
Thus $g^{-1}C_G(H)g=C_G(g^{-1}Hg)$ implies
$g^{-1}C_G(H)g=C_G(H)$ which implies the normality.
The statement $g^{-1}C_G(H)g=C_G(g^{-1} Hg)$ can be proved as follows:
$a\in g^{-1}C_G(H)g$ iff $gag^{-1}\in C_G(H)$ iff $hgag^{-1}h^{-1}
=gag^{-1}$ for all $h\in H$ iff $hga(hg)^{-1}=gag^{-1}$ for all $h\in H$
iff $kak^{-1}=gag^{-1}$ for all $k\in Hg$.
As $G$ normalizes $h$, $Hg=gH$. Then
$a\in g^{-1}C_G(H)g$ iff  $kak^{-1}=gag^{-1}$ for all $k\in gH$
iff $ghah^{-1}g^{-1}=gag^{-1}$ for all $h\in H$ iff $hah^{-1}=a$
for all $h\in H$ iff $a\in C_G(H)$.
A: That's not quite what the centralizer is: rather, $C_G(H) = \{g \in G: ghg^{-1}= h \forall h \in H\}$. Now just unwrap the definitions: suppose that $g\in C_G(H)$, and take any $k \in G$. It suffices to show that $kgk^{-1} \in C_G(H)$. But then for any $h\in H$, we have $kgk^{-1}hkg^{-1}k^{-1}=kg(k^{-1}hk)g^{-1}k^{-1}$. What can you say about $k^{-1}hk$ using the fact $G$ normalizes $H$? Then how can you simplify this expression?
