# 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.

• Hint: Prove that if $g: S^1\times R\to S^1\times R$ is an isometry such that $g(S^1\times \{0\})\cap S^1\times \{0\}= \emptyset$, then the group $<g>$ acts cocompactly on $S^1\times R$. Dec 1 '17 at 0:06
• @moishe so yours is to go by way of method $2$? Dec 1 '17 at 0:10
• That's right. The way you use "abelian" is by arguing that every subgroup is normal in $Z^2$ and, hence, $H$ acts on $R^2/<g>$ for every $g\in H$. But the proof is completely topological. Dec 1 '17 at 0:18

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.
• Totally foolish. I meant higher dimensions, $(S^1)^n$ Dec 4 '17 at 2:27
• @Andres: here's one idea. $T^n$ is a compact connected abelian Lie group, and so all of its finite covers are also compact connected abelian Lie groups (and its infinite covers are at least connected abelian Lie groups). These are known to all be isomorphic to finite products of copies of $S^1$ in the finite cover case and finite products of copies of $S^1$ and $\mathbb{R}$ in the infinite case, but the proof uses the classification of discrete subgroups of $\mathbb{R}^n$ (they are all $\mathbb{Z}^k$ for some $k \le n$) so we might as well talk about that and ignore covering space theory. Dec 4 '17 at 5:55