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Here is question similar to another one asked today (Proof of Pollard's inequality).

When working in $\mathbb{Z}/n\mathbb{Z}$ for composite $n$ Cauchy-Davenport and Pollard's inequalities may not hold.

Is it true however that if $A, B$ satisfy $|A+B| \geq |A|+|B|-1$, then also Pollard's inequality holds for $A, B$ and all $t$?

I am aware of the works of Green, Ruzsa, Hamidoune, Serra and more recently Grynkiewicz on extending Pollard's inequality to abelian groups, but such a conclusion does not seem to follow immediately from their results.

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  • $\begingroup$ Maybe, we should ask something more, say, $|A'+B'|\geq |A'|+|B'|-1$ for all non empty subsets $A'\subset A$, $B'\subset B$? $\endgroup$
    – Fedor Petrov
    Commented Mar 28, 2015 at 7:42

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This is wrong; here is a simple counterexample. Fix a finite subgroup $H$ with $|H|>2$ and a group element $g\notin H$, and let $A:=H$ and $B:=H\cup\{g\}$. Then $A+B=H\cup(g+H)$, whence $|A+B|=2|H|=|A|+|B|-1$. Also, the set $A{\stackrel2+} B$ of all those group elements with at least two representations in $A+B$ is precisely the subgroup $H$. Therefore, $$ |A+B|+|A{\stackrel2+} B| = 3|H| < 2(2|H|-1) = 2(|A|+|B|-2). $$

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