Showing that a subgroup $H\subseteq G$ is contained in its normalizer $N_G(H)$ 
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 ?
 A: 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 $H\subseteq N_G(H)$.
A: First we have if $h$ $\in$ $H$ then $Hh=H$ and $hH=H$,it concludes that $\forall h \in H$,$hHh^{-1}=H$
thus ,which show $H\subseteq N_G(H)$,and $H$ is a group ,therefore 
$H \leq N_G(H)$.
A: You wrote 

I try to use the statement '$H \leq G \iff ab^{-1} \in H\,\,\forall a,b \in H$'.

This statement proves that $H$ is a subgroup of $G$. Since we already know that $H$ is a subgroup of $G$, the only thing left to show is that $H$ is contained in $N_G(H)$. 
Let $h\in H$ and we will show that $h\in N_G(H)$. That is, by definition of $N_G(H)$, we will show that $hHh^{-1}=H$. Indeed, since $H$ is a subgroup, we have for all $a \in H$ : $aH=H$ and $Ha=H$ (because a subgroup is closed under multiplication). So $hHh^{-1}=H$. 
