For any finitely-generated abelian group, show M(G,n) exists

Question

Let $G$ be a finitely-generated abelian group. Prove there is a CW complex $M(G,n)$ which has $\tilde H_k(M(G,n))$ equals $G$ if $k=n$ or zero otherwise.

Here's what I have so far:

By the fundamental theorem for finitely generated abelian groups, $G \cong \mathbb{Z}^k \times \mathbb{Z}/p_1^{n_1}\times \dots \times \mathbb{Z}/p_k^{n_k}$ for primes $p_i$.

$S^n$ is an $M(\mathbb{Z},n)$.

I'm thinking the CW complex I'm looking for will be a wedge sum with $k$ $S^n$ and other spaces.

How can I continue from here?

Answer 1

You're right that using wedge sums is a good idea. In fact, it should be clear that using wedge sums and using $S^n$, you only need to prove that $M(\mathbb Z/p^k\mathbb Z, n)$ exists for all $k,n$.

Hint: use the fact that $\mathbb Z/p^k\mathbb Z$ is the homology of $\mathbb Z\overset{p^k}\to \mathbb Z$, together with cellular homology.