Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|cite|improve this question

marked as duplicate by draks ..., Arkamis, amWhy, rschwieb, Norbert Nov 7 '12 at 17:21

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

It is obvious by definition. Please have a try. – 23rd Nov 7 '12 at 14:36
$C_{G}(H)$ := {$g \in G | ghg^{-1}=h, \forall h \in H$} 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
@draks: Not a duplicate. The question you refer to is focused on the subgroup property; this one is about normality. – Harald Hanche-Olsen Nov 7 '12 at 15:56
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 at 23:19

1 Answer 1

up vote 3 down vote accepted

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

share|cite|improve this answer
+1 Nice, short and accurate. – DonAntonio Nov 7 '12 at 15:51

Not the answer you're looking for? Browse other questions tagged or ask your own question.