Progression-free sets in finite abelian groups

Let G be a finite abelian group. Write 2G {2g: g G} and denote by rk(2G) the rank of the group 2G. Extending a result of Meshulam, we prove the following. Suppose that A G is free of “true” arithmetic progressions; that is, a1+a3=2a2 with a1,a2,a3 A implies that a1=a3. Then A<2G/rk(2G). When G is of odd order this reduces to the original result of Meshulam. As a corollary, we generalize a result of Alon and show that if an integer k 2 and a real >0 are fixed, 2G is large enough, and a subset A G satisfies A (1/k+ )G, then there exists A0 A such that 1 A0 k and the elements of A0 add up to zero. When G is of odd order or cyclic this reduces to the original result of Alon.