# Prove using group action

(1)If $\,H ≤ G\,$, the factor group $\,N_G(H)⁄C_G(H)\,$ is isomorphic to a subgroup of $\,\operatorname{Aut}(H)\,$ using in the proof group action

(2)Let $\,H\leq G\,$ . The centralizer of $\,H\,$ is the set $\,C_G(H):=\{g∈G:hg=gh\;\; ∀h∈H\}\,$.Show that $\,C_G(H)\leq N_G(H)$

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(1) $N(H)$ operates on $H$ by conjugation and precisely $C(H)$ opererates trivially (i.e. is the kernel of the corresponding homomorphism $N(H)\to \operatorname{Aut}(H)$

(2) should be both obvious and actually shown before (1).

EDIT: After (justified) complaints about the use of the word "obvious", let us rewrite to obtain something to answer (1) and (2) at the same time:

Recall that if $\psi\colon A\to B$ is a group homomorphism, then $\ker\psi\lhd A$ and $A/\ker\psi\cong\operatorname{im}\phi<B$.

If $g\in N_G(H)$ and $h\in H$, then by definition $ghg^{-1}\in H$. Since conjugation-with-$g$ is an automorphism of $G$ and leaves $H$ invariant, it is also an automorphism of $H$. Thus we have a group homomorphism $$\begin{matrix}\Phi\colon &N_G(H)&\to&\operatorname{Aut}(H)\\&g&\mapsto&(h\mapsto ghg^{-1})\end{matrix}$$ What is the kernel of $\Phi$? We have $g\in\ker\Phi$ iff $h=ghg^{-1}$ for all $h\in H$ or equivalently $hg=gh$ for all $h\in H$, i.e. $\ker\Phi = C_G(H)$, thus showing both (1) and (2).

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what is obvious – Dania Nabeel Hadieh Nov 6 '12 at 18:59