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In the following article https://www.math.cornell.edu/~hatcher/Papers/3Msurvey.pdf Hatcher mentioned that there is only one prime closed 3-manifold with infinite cyclic fundamental group, which is $S^1\times S^2$.

This implies that the 3-torus $T^3$ can be decomposed as a direct sums of three $S^1\times S^2$. Is it true that $T^3=S^1\times S^2\# S^1\times S^2\# S^1\times S^2$?

This also implies that the 3-dimensional mapping torus $M^3$: $T^2 \to M^3 \to S^1$ (with $S^1$ as the base manifold and $T^2=S^1\times S^1$ as the fiber) can be decomposed as a direct sums of $S^1\times S^2$. How such decomposition is achieved?

In general, I would like to to know the decomposition of 3-dimensional mapping torus into connected sum. In particular the connected sum decomposition of a torus bundle $E^3$: $T^2 \to E^3 \to S^1$.

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    $\begingroup$ $T^3$ is covered by $R^3$. It follows that $\pi_2(T^3)$ is trivial. I guess it follows that $T^3$ is prime. Or did I misunderstood something? $\endgroup$
    – Marc Kegel
    Commented Mar 19, 2017 at 14:25
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    $\begingroup$ The torus $T^3$ is irreducible (hence prime) and thus can not be split into the connect sum you alluded to $\endgroup$
    – T. Amdeberhan
    Commented Mar 19, 2017 at 16:30
  • $\begingroup$ The fundamental group of a cartesian product of spaces is the direct product of their fundamental groups. On the other hand, the fundamental group of a connect sum of spaces is the free product of their fundamental groups. So, the connect sum of copies of $S^2 \times S^1$ has a free fundamental group, not an abelian one. $\endgroup$
    – Sam Nead
    Commented Apr 16, 2019 at 15:47

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The first part of this answer is merely an elaboration on Marc Kegel and T. Amdeberhan's comments, and addresses the question before the edit. As they say, $T^3$ is prime by virtue of being irreducible (a stronger condition) -- since $\pi_2(T^3)$ is trivial, every embedded 2-sphere is null-homotopic in $T^3$ and it follows that it bounds a ball. This wikipedia page may be helpful for you.

While it is true that $S^1\times S^2$ is the only prime closed 3-manifold with fundamental group $\mathbb{Z}$ (and all infinite cyclic groups are isomorphic to $\mathbb{Z}$), this does not imply that $T^3$ is not prime. (On the off chance this helps, note that $\mathbb{Z}^3$ is not a cyclic group).

In the language of Hatcher's paper, $T^3$ is a prime 3-manifold of type III: it is $K(\mathbb{Z}^3,1)$. Note also that $T^3$ is a Seifert manifold, thus the following statement in Hatcher's paper is consistent with this:

The only Seifert manifold that is not prime is $\mathbb{RP}^3\#\mathbb{RP}^3$, the sum of two copies of real projective 3-space.

I believe the edit of your question is addressed in the answers to this question. Namely, torus bundles are irreducible, so they admit no nontrivial connected sum decomposition.

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    $\begingroup$ I am not sure that there is something called the infinite cyclic group (just as there is no the algebraic closure of $\mathbb{Q}$). Given two infinite cyclic groups, you do not have a unique isomorphism between them. $\endgroup$
    – user137767
    Commented Apr 16, 2019 at 10:03
  • $\begingroup$ Fair enough, I will edit. $\endgroup$
    – j.c.
    Commented Apr 16, 2019 at 15:13

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