Is there some simple proof that $\mathbb{Z}$ is not isomorphic to the fundamental group of any compact Kähler manifold? This follows from the main result of <https://arxiv.org/abs/0709.4350> which states that any $3$-manifold group which is Kähler must be finite. The simplest non-finite 3-manifold group is $\mathbb{Z} = \pi\_{1}(S^{2} \times S^{1})$. So I was wondering, is there an argument to rule out this particularly simple group, without applying to a lot of machinery? (also historically when were we first able to rule this group out?) 



Why is $\mathbb{Z}$ not a Kähler group?
If $X$ is a compact Kähler manifold, then $h^{p,q}(X) = h^{q,p}(X)$ and $b\_k(X) = \sum\_{p+q=k}h^{p,q}(X)$, so in particular, $b\_1(X) = h^{1,0}(X) + h^{0,1}(X) = 2h^{1,0}(X)$ is even. Now, 


$$b\_1(X) = \operatorname{rank} H^1(X; \mathbb{Z}) = \operatorname{rank} \operatorname{Hom}(\pi\_1(X), \mathbb{Z}).$$ 


If $\pi\_1(X) = \mathbb{Z}$, $b\_1(X) = 1$ which is not even, so $X$ is not Kähler. More generally, if $\pi\_1(X)$ is abelian, then it must have even rank.


Note, there are non-compact Kähler manifolds with fundamental group $\mathbb{Z}$, for example $X = \mathbb{C}^\*$.


