Confusion about the top homology group of a compact manifold.

Question

I know that if the manifold is compact, then all of its homology groups are finitely generated. But on the other hand, we know (for example Hatcher 3.26) that if the manifold is closed and orientable, then its top homology group $H_{n}(M; G) \cong G$. Both have relevant proofs. But how can they both be correct at the same time? If $H_{n}(M; G)$ is finitely generated and $G$ is not, then in the latter case, there would be contradiction, wouldn't it?

Answer 1

I know that if the manifold is compact, then all of its homology groups are finitely generated.

This is true for homology groups with coefficients in $\mathbb{Z}$. As you have observed, it's obviously not true with arbitrary coefficients (there's no need to use nontrivial facts about $H_n$ to see that; just look at $H_0(M;G)$ which is trivially isomorphic to $G$ if $M$ is connected).

(More generally, if $R$ is a Noetherian ring, then $H_i(M;R)$ will be finitely generated as an $R$-module. This does not necessarily mean it is finitely generated as a group.)