# Why is the fundamental group of a prime, reducible 3-manifold $\mathbb{Z}$?

I've read in a paper that if $M$ is a prime, reducible $3$-manifold, then $\pi_{1}(M) \cong \mathbb{Z}$. Can anyone explain why this is true?

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en.wikipedia.org/wiki/… –  Brandon Carter Jul 10 '12 at 19:30
I'm probably assuming $M$ to be orientable.
Reducible means that there is an essential sphere $S \subseteq M$. Two cases happen : either that sphere disconnects $M$, and $M$ is decomposable as a nontrivial connected sum (so it is not prime) or $M$ is homeomorphic to $S^2 \times S^1$.
So, $S^2 \times S^1$ is the only prime reducible $3$-manifold.
The same technique of proof works in the non-orientable case. Either way, take a loop intersecting this non-separating sphere once transversally, then an open neighborhood is either a (punctured) $S^2\times S^1$ or its nonorientable cousin. –  user641 Jul 10 '12 at 20:18