According to the Kneser-Milnor prime decomposition theorem for 3-manifolds, any compact, connected, orientable 3-manifold $M$ is diffeomorphic to $S^3 / \Gamma\_1$ # $\cdots$ # $S^3/ \Gamma\_n$ # $(S^2 \times S^1)\_1$ # $\cdots$ # $(S^2 \times S^1)\_r$ # $K( \pi\_1,1)$ # $\cdots$ # $K( \pi\_m,1)$, where # is the connect sum, and $\Gamma\_i$ is a non-trivial finite subgroup of $SO(4)$ acting orthogonally to $S^3$ (so that the result is a spherical space form).


I suppose my question boils down to if $\pi\_1(M)=\mathbb{Z} \times \cdots \times \mathbb{Z}$, k times, does that uniquely identify a manifold as a $(S^2 \times S^1)\_1$ # $\cdots$ # $(S^2 \times S^1)\_k$, or can the various quotient manifolds or aspherical factors create a more complicated topology without changing the fundamental group?



Is a compact, connected, orientable 3-manifold with $\mathbb{Z}^K$ fundemental group uniquely determined?
NO, since the three-torus $T^3$ does not have this form.


**EDIT** if the OP really means a free product of $\mathbb{Z}$s, so the free group $F\_k,$ then the answer is YES. It is a fact (see Hempel's book, chapter 7) that every splitting of the fundamental group of $M^3$ as a free product comes from a connected sum decomposition. On the other hand, a prime three manifold is either a $K(\pi, 1)$ or $S^2 \times S^1.$ In the former case, its fundamental group cannot be $\mathbb{Z}$


