Let $G$ be a group and $H$ subgroup of $G$, $N(H):=\{g\in G; gHg^{-1}=H\}$
I need to prove that $N(H)$ is subgroup of $G$.
It's almost the same question like :How to show $\forall g \in G, gHg^{-1} = H \Leftrightarrow \forall g \in G, gHg^{-1} \subseteq H$?
But I need the inverse direction of the answer.
Attempt: The inverse of $gHg^{-1}$ is:
$$\begin{align}(gHg^{-1})^{-1}&=(g^{-1})^{-1}Hg^{-1}=gHg^{-1}=H \\&\implies (gHg^{-1})^{-1}=gHg^{-1} \\&\implies \text{the inverse of element is the element}\end{align}$$
Is it true?