Let $H$ be a subgroup of $G$. Define $N(H) = \{ a \in G | aHa^{-1}=H \}$. Show $H \subset N(H)$. Let $H$ be a subgroup of $G$. Define $N(H) = \{ a \in G | aHa^{-1}=H \}$. Show $H \subset N(H)$.


I was able to show $N(H) $ is a subgroup of $G$ but now I am unsure about showing the relation $H \subset N(H)$.

My attempt:

Let $x \in H$. Let $h_1 \in H$ be some element in $H$. 

Then $x h_1 x^{-1} = (x h_1) x^{-1} = h_2 x^{-1}$ where $x h_1 = h_2 \in H$. Now $h_2 x^{-1}= h_3 \in H$.

So we have shown for any $h_1 \in H$, $\exists h_3 \in H$ s.t. $x h_1 x^{-1} = h_3$. Hence, $x H x^{-1} = H$ and thus $x \in N(H) \implies H \subset N(H)$.

Did I make any mistake or leave out any necessary detail?



Edit: as per the comment below, I have incorrectly claimed that $xHx^{-1} = H$ whereas I had only shown $xHx^{-1} \subset H$. So for the other part, 

Let $h \in H$. Consider $x^{-1}hx = (x^{-1}h)x = h_1 x $ where $h_1 = x^{-1}h \in H$. Similarly $h_1x = h_2 \in H$.

So $x^{-1}hx = h_2 \implies h = x h_2 x^{-1} \in xHx^{-1} \implies H \subset xHx^{-1}$.

Therefore, $H = xHx^{-1}$ and the rest of the proof follows from here.
see that $a\in H\implies aH=Ha=H\implies aHa^{-1}=H$ so,   $a\in H\implies a\in N(H)$ hence $H\subset N(H)$.
see that $a\in H\implies aH=Ha=H\implies aHa^{-1}=H$ so,   $a\in H\implies a\in N(H)$ hence $H\subset N(H)$.