Xjyuk

Is there an orientable closed compact $3$-manifold such that its fundamental group is $mathbbZ$?

How about $mathbbZ^2$?

If $M$ is a closed oriented $3$-manifold with $Bbb Z^2$ fundamental group, then pass to the universal cover $tildeM$ and note that $pi_2(tildeM) = 0$ as otherwise by sphere theorem there would be an embedded sphere contained in $tildeM$, which by the covering map descends to a homotopically nontrivial embedded sphere in $M$, i.e., $M$ is not irreducible. Unless $M = S^2 times S^1$, it's not prime either, forcing $M$ to be a connected sum. But that forces $pi_1$ to be a free product which $Bbb Z^2$ isn't.

By Hurewicz theorem $pi_3(tildeM) = H_3(tildeM)$, also zero as $tildeM$ is noncompact $3$-dimensional. All the higher homology groups, hence the higher homotopy groups by Hurewicz, are subsequently zero. This implies $tildeM$ is contractible, hence $M$ is a $K(pi, 1)$-space, but as $pi_1 = Bbb Z^2$ here and $K(Bbb Z^2, 1)$ is homotopy equivalent to $T^2$, $M$ must be homotopy equivalent to $T^2$. But $Bbb Z = H_3(M) neq H_3(T^2) = 0$ so that can't happen. There are no closed oriented $3$-manifolds with $Bbb Z^2$ fundamental group.

$S^1 times S^2$ is the unique closed oriented $3$-manifold with fundamental group $Bbb Z$ by following a same trail of arguments as above: if $pi_2(tildeM) neq 0$ then there's a homotopically nontrivial embedded sphere in $M$. If you put aside the case $M = S^2 times S^1$, that forces $M$ to be non-prime, i.e., a free product, which again $Bbb Z$ isn't. The rest of the argument to show there are no other cases is identical.

Here's a sketch of an argument for why $S^2 times S^1$ is the unique prime non-irreducible closed oriented 3-manifold. Suppose $M$ is prime non-irreducible, then there's a homotopically nontrivial sphere $S$ in $M$ that doesn't bound a ball. Take an embedded closed $epsilon$-neighborhood $S times [-1, 1]$ of $S$ in $M$ and let $gamma$ be an arc from $S times -1$ to $S times 1$ which doesn't intersect $S$; this exists because $M setminus S$ is not disconnected ($M$ is prime!). Take union of $S times -1, 1$ with an embedded unit normal bundle of $gamma$ (which has to be diffeomorphic to $[0, 1] times S^1$ fixing the boundary, i.e., an "orientation-preserving tube", because $M$ is oriented) to obtain another embedded sphere in $M$ whose interior is union of a tubular neighborhood of $S$ and a tubular neighborhood of $gamma$ which deformation retracts to $S^2 vee S^1$. This new embedded sphere has to bound a $3$-ball in the exterior, as $M$ is prime. This gives a CW-decomposition of $M$ as a $D^3$ attatched to $S^2 vee S^1$, and it's not too hard to check this is indeed $S^2 times S^1$.