(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 $\,C_G(H)\leq N_G(H)$


(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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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