If $H\leq G$ and $x^{2}\in H$ for all $x\in G$, show that $H\lhd G$ and $G/H$ is abelian. I have already showed that $H$ is normal in $G$. But I can't show that $G/H$ is abelian. 
What I have tried was taking $X,Y\in G/H$ and show that $XYX^{-1}Y^{-1}=1_{G/H}$. If $X=Hg_1$ and $Y=Hg_2$ then $XYX^{-1}Y^{-1}=Hg_1g_2g_1^{-1}g_2^{-1}$. How can I show that the element $g_1g_2g_1^{-1}g_2^{-1}\in H$. Maybe another approach would be easier.

Hint For each $x,y \in G$ you have $xyxy\in H, xx yy \in H$ and hence
$$Hxyxy=H=Hxxyy$$
