TITLE: Let $A,B$ be two subsets of a finite group $G$. If $|A|+|B|>|G|$, show that $G=AB$
QUESTION [7 upvotes]: Let $A,B$ be two subsets of a finite group $G$. If $|A|+|B|>|G|$, show that $G=AB$. My attempt is : Since $|A|+|B|>|G|$, there exists one common element in both sets $A$ and $B$, say $g$. Then since $G$ is a group, by closure, $g^2 \in G$, which implies that $G \subset AB$. Let $a \in A$, $b \in B$. Then I get stuck at proving another inclusion.

REPLY [13 votes]: Take any $g \in G$.  Let $A^{-1} = \lbrace  a^{-1}, a \in A \rbrace$. Then $\vert A^{-1}g \cap B \vert  \gt 0$ by easy counting. Let $b = a^{-1} g$ for some $a \in A$.  Then $ab = a  a^{-1} g = g$.