I am wondering (but I may be wrong) if one can say that the homology of a K(G,n) is finite dimensional for G finitely generated abelian group. I looked up on the internet for some counterexample and I didn't find any. What makes me believe it could be true it's the way we constructed the K(G,n) in class: we took a system of generators of G, say $\alpha_1, \dots, \alpha_m$ and we called $X_n = \bigvee_{1,\dots,m} S^n$. Then we considered the relations between these generators and attached $S^{n+1}$'s to $X_n$ using the maps these relations define. The subgroup of relations is finitely generated because it is a subgroup of a finitely generated abelian group. What I am wondering now is if the fact that I am attaching a finite number of cells it's enough to say that the $\pi_{n+1}(X_{n+1})$ is finitely generated. If it is I can continue inductively saying that in each dimension $K(G,n)$ has a finite number of cells and so it has finite dimensional homology.
EDIT: For finite dimensional homology I meant that in each degree: $H_m(K(G,n),G)$ is a finitely generated abelian group.
Thank you all!
Here is a sketch of an argument that $H_m(K(G,n))$ is finitely generated for $G$ a finitely generated abelian group.
The class of finitely generated abelian groups forms a Serre class. So by induction on $n$ using the Serre spectral sequence for the path-loop fibration, it suffices to prove this for $n = 1$.
By the structure theorem for finitely-generated abelian groups and the Kunneth theorem, it suffices to check this for $G = \mathbb{Z}$ and $G = \mathbb{Z}/p^r$. In fact, it suffices to show this only for $G = \mathbb{Z}$, because we have short exact sequences $$0 \to \mathbb{Z} \to \mathbb{Z} \to \mathbb{Z}/p^r \to 0,$$ and we can run the Serre spectral sequence argument again.
But $K(\mathbb{Z}, 1) \simeq S^1$, and it's clear that the homology groups of $S^1$ are finitely generated.
This argument is essentially an application of Serre's mod $\mathcal{C}$ theory.