Q:

Proof that every subgroup of $\mathbb Z \oplus \mathbb Z $ is finitely generated with covering spaces?

I'm ultimately concerned with a proof showing that every subgroup of a finitely generated abelian group is also finitely generated by covering spaces. 
These thoughts are premature, but my curiosity could not stop me from asking:
Can we show that every subgroup of  $\pi_1(T) \cong \mathbb Z^2$ is a finitely generated abelian group?
We know that subgroups of $\pi_1(T)$ are in correspondence with normal covering spaces (since they are all normal.)
Taking the universal cover $p:X \to T$, we can see that anu subgroup $H$ acts on the quotient by the group action $X/H$. 
The problem here, is that I don't know at all what kind of restriction being abelian puts on the structure of a covering space so there is no way for me to proceed here.

On the other hand, we could study instead the subgroup $\langle(1,0) \rangle \subset \mathbb Z^2$, and try to look at a quotient by it (which will be finitely generated abelian) and use the correspondence/iso theorems to lift back up. If there is a nice geometric interpretation of this procedure, I am interested, but otherwise I know this proof already.

A:

The argument is easy for finite index subgroups: if $H \subseteq \mathbb{Z}^2$ is finite index, then it corresponds to a finite cover $X$ of $T^2$, which is necessarily a closed surface having Euler characteristic $0$ (by the multiplicativity of Euler characteristic in covers); said another way, it's necessarily a closed surface having a flat metric. Either way the conclusion is that by the classification of surfaces, $X$ must itself be homeomorphic to $T^2$, hence $H \cong \mathbb{Z}^2$.
(Presumably there is a more geometric argument here which avoids the classification and continues to apply to $T^n, n \ge 3$.) 
For infinite index subgroups we need a classification of noncompact surfaces with flat metrics. The classification should be that such a surface is either $T^2$, a flat cylinder $S^1 \times \mathbb{R}$, or $\mathbb{R}^2$ itself, which corresponds to a subgroup $H$ isomorphic to $\mathbb{Z}^2, \mathbb{Z}$, or the trivial group. But I don't actually know how to prove this off the top of my head. 

