Let G be a group and Let A,B<=G
|A|+|B|>G
Show that G=AB
we only need to prove that $\displaystyle \forall g \in G: \ A \cap gB \neq \emptyset.$ so suppose, on the contrary, that $\displaystyle \exists g \in G: \ A \cap gB = \emptyset.$
then: $\displaystyle |G| \geq |A \cup gB|=|A|+|gB|=|A|+|B| > |G|,$ which is impossible! Q.E.D.