Suppose that $A$ is a finitely generated abelian group with even rank and without $2$-torsion. Does there exist a compact symplectic $6$-manifold $(M,\omega)$ such that $A$ Is isormorphic to $H^{3}(M,\mathbb{Z})$?

(The reason for excluding $2$-torsion is the fact that a (closed, orientable, connected) smooth $6$-manifold without torsion in $H^{3}$ has an almost complex structure hence a hope to be symplectic).

Edit 1: Someone pointed out to me a very simple solution to the original question. Let $2k = rank(A)$, by Gompf we may find a compact symplectic $4$-manifold N such that $\pi_{1}(N) = H_{1}(N,\mathbb{Z}) = \mathbb{Z}^{k} \oplus T$ where $T$ is the torsion sub-group of $A$. Then by the Kunneth short exact sequence $H^{3}(N \times S^2,\mathbb{Z}) \cong A$. This works for any finitely generated abelian group of even rank without the assumption on $2$-torsion.

We can even (by finding appropriate algebraic surface with $\pi_{1} = T$) stay in the category of smooth projective 3-folds.

**Question** Same question as above but we insist $\pi_{1}(M) = \{1\}$. What about in the category of (smooth) projective 3-folds?

For any closed hyperbolic 3-manifold $M$ and any fg abelian group $A$, there is a finite-sheeted cover $M_0\to M$ so that $A$ is a summand of $H_1(M_0)$.$\endgroup$ – HJRW Jun 24 '17 at 19:13