Let $H$ be a subgroup of $G$ and $x,y \in G$. Show that $x(Hy)=(xH)y.$ 
Let $H$ be a subgroup of $G$ and $x,y \in G$. Show that $x(Hy)=(xH)y.$

I have that $Hy=\{hy \mid h \in H\}$ so wouldn't $x(Hy)=\{x hy \mid h \in H\}$? If so there doesn't seem to be much to be shown since if this holds I suppose that $(xH)y=\{x h y \mid h \in H\}$ would also hold and these two are clearly the same sets? Am I misinterpreting the set $x(Hy)$? Should this be $\{xhy \mid h \in H, y \in G\}$ for fixed $y$?

If $w\in x(Hy),$ then for some $v\in Hy,$ $w=xv.$ Since $v\in Hy,$ for some $h\in H,$ $v=hy.$ So $w = xhy.$ So for some $u\in xH,$ $w=uy.$ Thus $w\in (xH)y.$
Therefore $x(Hy)\subseteq (xH)y.$
The inclusion in the other direction can be shown in the same way.
