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Lemma 0.2.7 (Freiman). Let $G$ be an abelian group, written multiplicatively, and let $A$ be a finite subset of $G$. Suppose that $A$ is not contained in any proper coset of $G$. Then either $A \cdot A^{-1}=G$ or $|A \cdot A| \geqslant \frac{3}{2}|A|$.
Proof. Suppose that $|A \cdot A|<\frac{3}{2}|A|$; we show that $A \cdot A^{-1}$ is closed under multiplication. Indeed, let $w, x, y, z \in A$. The set $\{a \in A: w a \in z A\}$ has size greater than $\frac{1}{2}|A|$, since $|A \cdot A|<\frac{3}{2}|A|$. Similarly the set $\{a \in A: x a \in y A\}$ has size greater than $\frac{1}{2}|A|$. Therefore these two sets intersect, and we have $a, a_z, a_y \in A$ such that $w a=z a_z$ and $x a=y a_y$. Hence
$$
\left(w x^{-1}\right)\left(y z^{-1}\right)=w a a^{-1} x^{-1} y z^{-1}=z a_z a_y^{-1} y^{-1} y z^{-1}=a_z a_y^{-1} .
$$
Therefore $A \cdot A^{-1}$ is a subgroup of $G$ (the other axioms are trivial), and as $A$ is not contained in any proper coset of $G$ we conclude that $A \cdot A^{-1}$ must be the whole of $G$. $_\blacksquare$
This lemma is sharp: consider the case where $A$ is the union of a subgroup and a single non-trivial coset of the same subgroup. |
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