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Let me address a slightly more general question:

To what extent is a closed, connected $M^3$ determined by its fundamental group?

Following the Geometrization Theorem, we have a complete answer. There are only two ways in which a closed $3$--manifold $M$ can fail to be determined by its fundamental group:

1. $M$ is a lens space, or a connected sum of something with a lens space. It is well-known that lens spaces are not determined up to homeomorphism by their fundamental groups. Note that in this case, you would see a ${Z}/p$ free factor in your group $G$, so it can't arise in the context of your question.

2. $M = N_1 \sharp N_2$, where each $N_i$ is orientable, and each is chiral (fails to have an orientation-reversing symmetry). In this case, reversing the orientation on one factor would produce $M' = N_1 \sharp \overline{N_2}$, which is not homeomorphic to $M$ but has the same fundamental group. This also cannot arise in your context, for exactly the reasons that Igor outlined, and because $S^1 \times S^2$ does have an orientation-reversing symmetry.

Finally, let me point out that although Geometrization may seem like an overly big hammer for this question, in fact one needs the Poincare conjecture. For, if there existed a fake $3$--sphere, one could take the connect sum with that manifold without altering the group.

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Let me address a slightly more general question:

To what extent is a closed, connected $M^3$ determined by its fundamental group?

Following the Geometrization Theorem, we have a complete answer. There are only two ways in which a closed $3$--manifold $M$ can fail to be determined by its fundamental group:

1. $M$ is a lens space, or a connected sum of something with a lens space. It is well-known that lens spaces are not determined up to homeomorphism by their fundamental groups. Note that in this case, you would see a ${Z}/p$ free factor in your group $G$, so it can't arise in the context of your question.

2. $M = N_1 \sharp N_2$, where each $N_i$ is orientable, and each is chiral (fails to have an orientation-reversing symmetry). In this case, reversing the orientation on one factor would produce $M' = N_1 \sharp \overline{N_2}$, which is not homeomorphic to $M$ but has the same fundamental group. This also cannot arise in your context, for exactly the reasons that Igor outlined, and because $S^1 \times S^2$ does have an orientation-reversing symmetry.

Finally, let me point out that although Geometrization may seem like an overly big hammer for this question, in fact one needs the Poincare conjecture. For, if there existed a fake $3$--sphere, one could take the connect sum with that manifold without altering the group.