How to show $\forall g \in G, gHg^{-1} = H \Leftrightarrow \forall g \in G, gHg^{-1} \subseteq H$? The following note is not obvious to me. I tried a couple steps, but was not able to proceed:

Note: $\forall g \in G, gHg^{-1} = H \Leftrightarrow \forall g \in G, gHg^{-1} \subseteq H$

It is direct that
$$\forall g \in G, gHg^{-1} = H \Rightarrow \forall g \in G, gHg^{-1} \subseteq H.$$
Now we show that
$$\forall g \in G, gHg^{-1} \subseteq H \Rightarrow \forall g \in G, gHg^{-1} = H.$$
It is equivalent to say that we want to show that
$$\forall g \in G, gHg^{-1} \subseteq H \Rightarrow \forall h \in H, \forall g \in G, h \in gHg^{-1}.$$
 A: Suppose that the following holds:
$$
\forall g \in G,\ gHg^{-1} \subseteq H
$$
For a fixed $g$, apply the property above to $g^{-1}$ to get $g^{-1}Hg \subseteq H$. Multiply by $g$ from the left and by $g^{-1}$ from the right to get $H \subseteq gHg^{-1}$. Thus, equality holds and we have $gHg^{-1} = H$ as desired.
A: Hint: Note that for any $g \in G$ you have $gHg^{-1} \subseteq H$ and $g^{-1}Hg \subseteq H$.
A: You have  : $gHg^{-1}\subseteq H$ for all $g\in G$
You want : $gHg^{-1}=H$ for all $g\in G$
Let us do it for each $g\in G$ step by step....
fix $g\in G$, You want to show $gHg^{-1}=H$
As $g\in G$, you have is $gHg^{-1}\subseteq H$
As $g^{-1}\in G$, you have is $g^{-1}H(g^{-1})^{-1}\subseteq H$ 
i.e., $g^{-1}Hg \subseteq H$ 
i.e., $Hg\subseteq gH$ (??)
i.e., $H\subseteq gHg^{-1}$ (??)
So, it is given that $gHg^{-1}\subseteq H$ and now we have seen that $H\subseteq gHg^{-1}$ 
These combine to give $gHg^{-1}=H$.
As fixed $g\in G$ is arbitrary, we see that $gHg^{-1}=H$ for all $g\in G$
I believe you can easily see  
$ gHg^{-1} = H$ for all  $g\in G$ implies $gHg^{-1} \subseteq H$ for all  $g\in G$.
So, combining all these, we conclude that :
$ gHg^{-1} = H$ for all  $g\in G$ if and only if $gHg^{-1} \subseteq H$ for all  $g\in G$.
P.S : It is very obvious (at least for me) to think that result is "Not obvious"... 
