$H,N(H)$ are subgroups of $G$ show that $H\lhd N(H)$

Question

Let $G$ be a group and $H$ subgroup of $G$, $N(H):=\{g\in G; gHg^{-1}=H\}$

$N(H)$ is also subgroup of $G$.

I need to prove that $H$ is a normal subrgoup in $N(H)$

Attempt:

$H\lhd N(H) \iff nhn^{-1}\in N(H)$ for all $n\in N(H),h\in H$

Let $z\in N(H),h\in H,g\in G$

$zhz^{-1}\in ? H $

$\iff (gHg^{-1})h(gHg^{-1})^{-1} \in H$

Is it true? how can I continue?

Answer 1

Take $z\in N(H)$ and $h\in H$ (there is no need to take a $g\in G$), what you want to do is to show that $zhz^{-1}\in H$. Remark that $zhz^{-1}\in zHz^{-1}$ (by definition of the conjugation) now use the fact that $z\in N(H)$ to conclude that $zhz^{-1}$ is actually in $H$ and you are done (I think you have everything right except the fact that you take some $g\in G$ that you don't need to take and that blurs your attempt).

Answer 2

@Nehorai, now you are enlightened (:-)), you might want to prove (another definition of the normalizer) that $N_G(H)$ is the largest subgroup of $G$ in which $H$ is normal.