Closed 3-manifolds with free abelian fundamental groups Which free abelian groups can be realized as the fundamental group of a closed 3-manifold?  The only one I can come up with is $\mathbb{Z}$, which is the fundamental group of $S^1 \times S^2$.  For the application I have in mind, the key case is $\mathbb{Z}^2$.  Here it is easy if you allow boundary (just take $T^2 \times [0,1]$), but I don't see how to do it without the boundary.
 A: John Hempel, in his book $3$-manifolds, shows that if $G$ is a finitely generated abelian group which is a subgroup of the fundamental group of a closed $3$-manifold, then $G$ is one of $\mathbb Z$,  $\mathbb Z\oplus\mathbb Z$,   $\mathbb Z\oplus\mathbb Z \oplus\mathbb Z$, $\mathbb Z_p$ or $\mathbb Z\oplus\mathbb Z_2$. This is theorem 9.13 in the book.
He also proves, in theorem 9.14, that an abelian group which is not finitely generated and a subgroup of the fundamental group of a $3$-manifold, then it is isomorphic to a subgroup of $\mathbb Q$ (and proposes, as an exercise, to show that all such groups in fact occur) 
A: Only $\mathbb Z$ and $\mathbb Z^3$ (for $T^3$) are free abelian groups that appear as fundamental groups of $3$-manifolds. Hopefully the following is an approximative proof.
The manifold must be prime (otherwise the $\pi_1$ is not ableian), hence it is $K(\pi,1)$.
Hence  its cohomology are just cohomology of the group $\mathbb Z^n$. So the can not get 
$\mathbb Z^2$ since $H^3(\mathbb Z^2)=0$, and we can not get $\mathbb Z^n$ with $n>3$ since $H^n(\mathbb Z^n)=\mathbb Z$. 
A: (I assume all occuring 3-manifolds to be orientable and closed)
A manifold with a free abelian fundamental group cannot be a connected sum of non-trivial 3-manifolds since its fundamental group is not a free product. A prime manifold is either $S^1\times S^2$ or irreducible (Hatcher's notes on 3-manifolds, 1.4). By 3.9 in the same source, an irreducible 3-manifold $M$ with infinite fundamental group is a $K(\pi,1)$. If $\pi = \mathbb{Z}^n$, then $M$ needs therefore to be homotopy equivalent to $(S^1)^n$. This can only be if $n=3$. 
