Small triple-representation set bounds the sumset via multiplicity

Let $G$ be an additive commutative group and let $A, B, S \subseteq G$ be finite sets, and let $M : \mathbb{N}$. Suppose every sum $a + b$ with $a \in A$ and $b \in B$ admits at least $M$ representations

Then

The proof is double counting: the fibres of the map $S \times S \times S \to G$, $(s_1,s_2,s_3) \mapsto s_1 - s_2 + s_3$, over distinct sums $a+b \in A+B$ are pairwise disjoint and each has size at least $M$, while they all sit inside $S \times S \times S$, which has $|S|^3$ elements.

Note that $S \times S \times S$ is the Cartesian cube of $S$, and the image of the map above is the triple-representation set $S - S + S$. Neither is a dilate of $S$: no scaling is applied to $S$, and the three coordinates range independently.

In the Balog-Szemeredi-Gowers project this is the final counting step: the Tao-Vu injection supplies a uniform multiplicity $M$ of such triple representations for every sum in $A' + B'$, and this lemma converts that into the honest sumset bound $|A' + B'| \le |S|^3/M$.