How to show $\forall g \in G, gHg^{-1} = H \Leftrightarrow \forall g \in G, gHg^{-1} \subseteq H$?

Question

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 $$\forall g \in G, gHg^{-1} \subseteq H \Rightarrow \forall h \in H, \forall g \in G, h \in gHg^{-1}.$$

Answer 1

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.

Answer 2

Hint: Note that for any $g \in G$ you have $gHg^{-1} \subseteq H$ and $g^{-1}Hg \subseteq H$.

Answer 3

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"...