[Math] Showing that a subgroup $H\subseteq G$ is contained in its normalizer $N_G(H)$

Question

Suppose $G$ is a group and $H \leq G.$ Define a set $$N_G(H)=\lbrace a \in G:aHa^{-1}=H \rbrace.$$ Show that $H \leq N_G(H)$.

I try to use the statement '$H \leq G \iff ab^{-1} \in H\,\,\forall a,b \in H$'. But I got stuck. Can anyone guide me ?

Answer 1

Note that $$aHa^{-1}=\{aha^{-1}\mid h\in H\}.$$ From now on, let's suppose that $a$ is an element of $H$.

For any $h\in H$, we have that $aha^{-1}\in H$ because subgroups are closed under taking inverse and products. Thus $aHa^{-1}\subseteq H$.

On your own, now show that any element $g\in H$ is equal to $aha^{-1}$ for some $h\in H$, thereby demonstrating that $H\subseteq aHa^{-1}$.

Thus, for any $a\in H$, we have that $aHa^{-1}=H$, which shows that $H\subseteq N_G(H)$.