Prove $H \leq G$ implies $N_G(H) / C_G(H) \cong B \leq \mathrm{Aut}(H)$ where $B$ is some subgroup of $\mathrm{Aut}(H).$

I did see this question:

For $H \leq G$, showing that $N_G(H)/C_G(H) \leq \text{Aut}(H)$

which has been helpful. My question now is with the Proposition 13 that the question makes reference to:

$\textbf{Proposition 13: }$ Let $H$ be a normal subgroup of the group $G$. Then $G$ acts by conjugation on $H$ as automorphisms of $H$. More specifically, the action of $G$ on $H$ by conjugation is defined for each $g \in G$ by $$h\mapsto ghg^{-1} \text{ for each $h \in H$.}$$ For each $g \in G$, conjugation by $g$ is an automorphism of $H$. The permutation representation afforded by this action is a homomorphism of $G$ into $\mathrm{Aut}(H)$ with kernel $C_G(H)$. In particular, $G/C_G(H)$ is isomorphic to a subgroup of $\mathrm{Aut}(H)$.

Specifically, my question is what do they mean by

"The permutation representation afforded by this action..."?

I see that the conjugation map stated in the proposition above is an automorphism of $H$ and so it just permutes the elements of $H$ but I don't know what the permutation representation is or means.

My goal is, I think at least for now, to prove this proposition and the referred question above states that the result I need to show follows from there.

  • $\begingroup$ Just ignore that formulation, you already know what is intended: $g\mapsto(h\mapsto ghg^{-1})$ is a homomorphism $G\to\operatorname{Aut}(H)$ and the kernel of ths is $C_G(H)$. $\endgroup$ – Hagen von Eitzen May 16 '13 at 6:16
  • $\begingroup$ Okay awesome, I will complete the proof from here thanks very much! $\endgroup$ – name May 16 '13 at 6:21
  • $\begingroup$ So your question is not what appears in the title, uh? Please do edit your question and title to make this clear...\ $\endgroup$ – DonAntonio May 16 '13 at 6:35
  • $\begingroup$ Sorry. I am open to other approaches to solving the initial statement I made about proving the existence of such a $B$. I am just going with what has been said to lead to a solution so far. I think I am on the way to finishing this up but other ways of solving it are welcome. I will post my solution below when I finish working this out $\endgroup$ – name May 16 '13 at 6:40

$\textbf{Proof of Proposition 13}$: As $H$ is normal in $G$, $ghg^{-1}\in gHg^{-1}=H$ and so each $ghg^{-1}\in H$. If $h,k \in H$ then $$f_g(hk)=g(hk)g^{-1} \text{ and } f_g(h)f_g(k)=(ghg^{-1})(gkg^{-1})=g(hk)g^{-1}$$ so $f_g(hk)=f_g(h)f_g(k)$ so $f$ is a homomorphism. If $h \in H=gHg^{-1}$ then $h=gh'g^{-1}$ for some $h' \in H$ and so $$f_g(h')=gh'g^{-1}=h$$ so $f_g$ is a surjection. We have that \begin{align*} \mathrm{Ker}(f_g) &= \{ h \in H: f_g(h)=e_H \} \\ &= \{ h \in H: ghg^{-1}=e_H \} \\ &= \{ h \in H: gh=e_Hg \} \\ &= \{ h \in H: gh=ge_H \} \\ &= \{ h \in H: h=e_H \} \\ &= \{ e_H \} \end{align*} so $f_g$ is injective. We conclude that $f_g$ is an isomorphism and as $f_g:H \rightarrow H$, we have $f_g$ is an automorphism.

Let $f$ be as above. If $g,k \in G$ then $$f(gk)=f_{gk} \text{ and } f(g) \circ f(k) = f_g \circ f_k.$$ For any $h \in H$, $f_{gk}(h)=(gk)h(gk)^{-1}$ and $$(f_g \circ f_k)(h)=f_g( f_k(h) ) = f_g(khk^{-1})=g(khk^{-1})g^{-1}=(gk)h(gk)^{-1}=f_{gk}(h)$$ so $f_{gk} = f_g \circ f_k$ meaning $$f(gk)=f(g) \circ f(k).$$ Therefore $f$ is a homomorphism. We have then \begin{align*} \mathrm{Ker}(f) &= \{ g \in G : f(g)=\mathrm{id}_H \} \\ &= \{ g \in G: f_g = \mathrm{id}_H \} \end{align*} and so if $g \in \mathrm{Ker}(f)$ then for all $h \in H$, $f_g(h)=ghg^{-1}=h$ and so $gh=hg$ meaning that $g \in C_G(H)$. Further if $g \in C_G(H)$ then $f_g(h)=ghg^{-1}=hgg^{-1}=h=\mathrm{id}_H$ so $g \in \mathrm{Ker}(f)$. We have shown then $$\mathrm{Ker}(f)=C_G(H).$$ Therefore, we can take the function $j: G \rightarrow \mathrm{Im}(f)$ by $j(g)=f(g)$. This function $j$ is a surjective homomorphism and so by the first isomorphism theorem we have that since $\mathrm{Ker}(f)=\mathrm{Ker}(j)$ that $$G/C_G(H) \cong \mathrm{Im}(f)$$ and we know that $\mathrm{Im}(f) \leq \mathrm{Aut}(H)$ proving what was needed.

$\textbf{Proof of desired thing: }$ We have that $H$ is a normal subgroup of $N_G(H)$ and so from above $N_G(H)/C_{N_G(H)} \cong B \leq \mathrm{Aut}(H)$ and further $C_{N_G(H)}(H)=C_G(H)$ proving the statement.

  • 1
    $\begingroup$ A shortcut: to show that each $f_{g}$ is a bijection, note that it has inverse $f_{g^{-1}}$. $\endgroup$ – Andreas Caranti May 16 '13 at 9:50

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