If $S$ is a subset of $G$ with $|S| > \frac{|G|}{2}$ , then given $g \in G$ prove that $\exists a,b \in S$ so that $ab^{-1} = g$

Question

I was trying the following problem :

Let $G$ be a finite group. If $S$ is a subset of $G$ with $|S| > \frac{|G|}{2}$ , then given $g \in G$ prove that $\exists a,b \in S$ so that $ab^{-1} = g$ .

My attempt:

Let $S^{/} $ denote the subgroup of $G$ generated by $S$, then since $|S| > \frac{|G|}{2}$ , by Lagrange's theorem, $S^/ = G$ , then again since $G$ is finite , I guess it follows that such $a,b$ exist.

Is my argument alright? Thanks in advance for help!

Answer 1

Hint:

$|gS| = |S| \stackrel{|S| > \frac{|G|}{2}}{\Rightarrow} gS \cap S \neq \emptyset$