Let $G$ be a finite group and $S$ is a subset such $|S|>|G|/2$. Then prove that for any $g\in G$ there exist elements $a,b\in S$ such that $g=ab^{-1}$.
My idea is if $G$ has the order 2 then it is trivial. If $|G|\geqslant3$ then $S$ has elements $a,b$ such that $ab^{-1}=g$ for some $g\in G$, but I cannot proceed like this. Any Hint, please?
Hint:
Consider the cardinality of $S$ and $gS$.
can you say anything about their intersection?