Sumset of a subset of a group

Question

I am interested in the following which I believe is known:

Let $S$ be a subset of a finite group $G$ containing more than half of $G$'s elements. Then $S+S = G$.

I have been looking but can not find a reference. Does anyone know of one? Or know a nice proof?

Thanks!

Answer 1

$\newcommand{\Size}[1]{\lvert #1 \rvert}$Let $g \in G$, and write $G$ multiplicatively.

By the inclusion-exclusion principle $$ \Size{A \cup B} = \Size{A} + \Size{B} - \Size{A \cap B}, $$ the sets $A = g S^{-1}$ and $B = S$ must have a non-trivial intersection...

[EDIT: the ending] so there are $s_{1}, s_{2} \in S$ such that $g s_{2}^{-1} = s_{1}$ or $g = s_{1} s_{2} \in S \cdot S$.