Let $\displaystyle H$ and $\displaystyle K$ be subgroups of the group $\displaystyle G$, and let $\displaystyle a,b\in G$. Show that either $\displaystyle aH\cap bK=\emptyset$ or else $\displaystyle aH\cap bK$ is a left coset of $\displaystyle H\cap K$.

I've been working on this one for over a day, now, and I'm still stumped. Assistance would therefore be much appreciated!

EDIT: I finally solved it, and it is indeed extremely simple. I can't believe I missed it for so long.

**Proof:** Suppose $\displaystyle aH\cap bK\neq\emptyset$, and let $\displaystyle x\in aH\cap bK$. Then there are $\displaystyle h\in H$, $\displaystyle k\in K$ with $\displaystyle x=ah=bk$. So $\displaystyle a=bkh^{-1}$, and therefore $\displaystyle aH\cap bK=bkh^{-1}H\cap bK=bkH\cap bK=bkH\cap bkK=bk(H\cap K)$. The conclusion follows. $\displaystyle \blacksquare$