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I am trying to prove: for $H$ a subgroup of $G$ and $g$ an element of $G$

$N_{G}(gHg^{-1})=gN_{G}(H)g^{-1}$

I have started along these lines:

let $x$ be an element of $N_{G}(gHg^{-1})$

therefore $x$ is an element of $G$ such that

$x(gHg^{-1})x^{-1}=gHg^{-1}$

and then I'm confused as to how to go from here.

please help.

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    Thank you everyone! @mTurgeon you are like a super hero to me right now ;)2012-06-03

2 Answers 2

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We have the following chain of implications: $\begin{align*} x\in N_G(gHg^{-1}) &\iff x(gHg^{-1})x^{-1}=gHg^{-1}\\ &\iff g^{-1}xgHg^{-1}x^{-1}g=H \\ &\iff (g^{-1}xg)H(g^{-1}xg)^{-1}=H\\ &\iff g^{-1}xg\in N_G(H)\\ &\iff x\in gN_G(H)g^{-1}.\\ \end{align*}$


Added (for when you learn about group actions):

What is really happening here is that we have an action of $G$ on $\mathcal{P}(G)$ (the power set of $G$), where $g\in G$ acts by conjugation: $\begin{align*} G\times \mathcal{P}(G) &\to \mathcal{P}(G)\\ (g,H) &\mapsto g\cdot H:=gHg^{-1}.\\ \end{align*}$ Hence, what you are trying to prove is that the stabilizer of elements lying in a given orbit are conjugate: $ H =g\cdot H^\prime \Rightarrow \mathrm{Stab}(H)=g \mathrm{Stab}(H^\prime)g^{-1}.$

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    A complete $f$irm proof.2012-06-03
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You can rearrange your last equation to get $g^{-1}xgHg^{-1}x^{-1}g = H$. Remembering that $(ab)^{-1} = b^{-1}a^{-1}$ in a group, what does this say about the element $g^{-1}xg$?

The reverse inclusion should be more straightforward. Give it a shot!

Looking ahead, this is a special case of a fact about groups acting on sets: if $G$ acts on $S$ and $g \in G, s \in S$ then the stabilizers of $s$ and $gs$ are conjugate. Here $G$ acts on its set of subgroups by conjugation.