# Prove that the centralizer is a normal subgroup. [duplicate]

Possible Duplicate:
Prove that the center of a group is a normal subgroup

Suppose that $H$ is a normal subgroup of $G$. Prove that $C_{G}(H)$ is a normal subgroup of $G$, where $C_{G}(H)$ is the centralizer of $H$ in $G$.

I have proved that $C_{G}(H)$ is a subgroup but how do I prove that it is normal - is this not obvious by the definition of a centralizer?

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## marked as duplicate by draks ..., Emily, amWhy, rschwieb, NorbertNov 7 '12 at 17:21

$C_{G}(H)$ := {$g \in G | ghg^{-1}=h, \forall h \in H$} and we want to show that $C_{G}(H)$ is normal i.e. that $fC_{G}(H)f^{-1} \subseteq C_{G}(H)$ for all $f \in G$. Can we just say that $fgf^{-1}=f(gf^{-1})=f(f^{-1}g)=(ff^{-1})g=g \subseteq G$ and this is true for all $f \in G$ hence $C_{G}(H)$ is normal in G. – Mark Nov 7 '12 at 14:52
Nooo.. $gf^{-1}\ne f^{-1}g$, unless we know sth like $g\in H$ and $f\in C_G(H)$, but this was not assumed here. – Berci Nov 7 '12 at 15:04
Just another way of seeing normality: If $H$ is normal, then $G$ acts on $H$ by conjugation with kernel $C_G(H)$, so as $C_G(H)$ is the kernel of the permutation representation in $S_H$ it is normal. – Stefan Oct 25 '15 at 23:19
Let $x\in C_G(H)$, and $g\in G$ arbitrary. Then, as $xh=hx$ for all $h\in H$, and $H$ is normal, thus $g^{-1}hg\in H$, we have $$gxg^{-1}\cdot h=gx(g^{-1}hg)g^{-1} = g(g^{-1}hg)xg^{-1} = h\cdot gxg^{-1} .$$