Progression-free sets in finite abelian groups

Let G be a finite abelian group. Write 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|<2|G|/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 ε>0 are fixed, |2G| is large enough, and a subset A⊆G satisfies |A|?(1/k+ε)|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.