Sumset of a subset of a group

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!

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$\newcommand{\Size}[1]{\lvert #1 \rvert}$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 $g = s_{1} s_{2} \in S \cdot S$.
So $g{s}^{-1}$ is in $S$... If $S$ were a subgroup I'd agree that we're done, but I don't see why we are in the case where $S$ is just a subset... Can you explain? – user160371 Jan 23 at 15:11
$gs^{-1} \in S$ is equivalent to $g \in S + S$, which is what you are trying to prove. I am guessing that by $S+S$ you mean $\{ st : s,t \in S \}$. – Derek Holt Jan 23 at 15:18