The question is: Prove that if $H \unlhd G$, then $C_{G}(H) \unlhd G$, where $C_{G}(H)$ is the centralizer of the group $H$ in $G$.
What I need to show: $gC_{G}(H)=C_{G}(H)g$, which is equivalent to (1) $gC_{G}(H) \subseteq C_{G}(H)g$ and (2) $C_{G}(H)g \subseteq gC_{G}(H)$.
(1) Let $gc \in gC_{G}(H)$, where $c \in C_{G}(H)$. Note that $gc=gcg^{-1}g$. So it's enough to show that $gcg^{-1} \in C_{G}(H)$. Note that $c=gh=hg$ (because $c \in C_{G}(H)$). And note that
$c=gh=hg \iff gcg^{-1}=gghg^{-1}=ghgg^{-1}$. And $ghgg^{-1}=ghe=gh$. The last one ($gh$) is equal to $c$ by construction. Then, $gcg^{-1}=ghgg^{-1}=gh=c$. Therefore, $gcg^{-1} \in C_{G}(H)$.
My doubt is: is this proof right? Am I missing something? Can I really use the same $g$'s in both cases: $g\in G$ and $g \in C_{G}(H)$?. I think so, cus it's the same group $G$ in both, but I need some validation here.
I know the proof of (2) is similar to the (1). So if (1) is right, (2) will also be right.
Best Answer
If $ch=hc$ for every $h\in H$ and $g\in G$, then $(gcg^{-1})h= gc(g^{-1}hg)g^{-1}=g(g^{-1}hg)cg^{-1}=h(gcg^{-1})$ because $g^{-1}hg\in H$. Hence $gC_G(H)g^{-1}\subseteq C_G(H)$ and $C_G(H)$ is normal.