Prove $N_{G}(gHg^{-1})$=$gN_{G}(H)g^{-1}$ 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.
 A: 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.
A: 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:
$$ H =g\cdot H^\prime \Rightarrow \mathrm{Stab}(H)=g \mathrm{Stab}(H^\prime)g^{-1}.$$
