Show that a nonempty subset $\displaystyle H$ of a group $\displaystyle G$ is a subgroup of $\displaystyle G$ if and only if $\displaystyle ab^{-1} \in H$ for all $\displaystyle a, b \in H$.
Clearly this condition must hold if your set is a subgroup (as subgroups are closed under multiplication).
If this condition holds then note that you have the identity in your set (take $\displaystyle a=b$), and that if $\displaystyle g\in H$ then $\displaystyle g^{-1} \in H$ (take $\displaystyle a=1$, $\displaystyle b=g$).
So the only thing to prove is that if $\displaystyle g, h\in H$ then $\displaystyle gh\in H$. However, as $\displaystyle h^{-1}\in H$, take $\displaystyle a=g$ and $\displaystyle b=h^{-1}$...and your done!