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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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    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.2015-10-25

1 Answers 1

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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 $gxg^{-1}\cdot h=gx(g^{-1}hg)g^{-1} = g(g^{-1}hg)xg^{-1} = h\cdot gxg^{-1} .$