Is the following arithmetic on conjugation of a set correct? I am given a set $H$ subset of a group $G$ closed under the binary operation and satisfying that if $g \in G$ then $g^2 \in H$. And I am to prove that $H$ is a normal subgroup of $G$ and that $G/H$ is Abelian.
What I have done is:
Take an an arbitrary $g \in G$. 
We were told that for all $g \in G$, $g^2 \in H$ so $(gHg^{-1})^2 \subseteq H$
However $(gHg^{-1})^2 = gHg^{-1}gHg^{-1} = gH^2g^{-1} = gHg^{-1}$
I am not sure if my argument above is correct because I am not sure if my approach is mathematically correct.
 A: I don't think you can get $(gHg^{-1})^{2}\subset H$, because an arbitrary element of $A^{2}$ is $ab$ where $a,b\in A$. Also, you should prove $H$ is a subgroup first, that is, prove $H$ is closed under taking inverses and $H\neq \emptyset$. And then prove it is normal. 
A: Like highlighted in Delong's answer $(gHg^{-1})^2\subseteq H$ does not follow directly from the fact that $g^2\in H$ for all $g\in G$, because the most general element of $(gHg^{-1})^2$ is not a square but it is element of the form $gh_1h_2g^{-1}$ where $h_1,h_2$ are two elements of $H$. 
Proving that $H$ is a group has already been dealt with.
Hint : $gHg^{-1}=H$ is equivalent to $gH=Hg$. Try to write any element $gh$ where $h\in H$ as some squares and elements of $H$ with a $g$ at the end. 

 $gh=ghghh^{-1}g^{-1}=\underbrace{(gh)^2h^{-1}(g^{-1})^2}_{\in H}g$

For the remaining part (abelianity of $G/H$) Jalex Stark's comment is the good path. The hypothesis tells you that $G/H$ is of exponent $2$ or $1$ and any group of this kind needs to be abelian.
