# Finitely Generated Abelian Group (torsion)

I've been working on the following question, but haven't made much progress at all... so any help would be greatly appreciated.

Prove that if $G$ is a finitely generated abelian group, then $G \cong G_{\text{tors}} \times \mathbb{Z}^r$.

We can't use the Fundamental Theorem of Finitely Generated Abelian Groups for this, since this problem is supposed to help us fill in some of the gaps that the professor left out.

## 1 Answer

I'm assuming you already know that each finitely generated torsion-free abelian group is isomorphic to $\mathbb{Z}^r$ for some $r$. Then you can proceed like this:

1. $G/G_{\text{tors}}$ is torsion-free and finitely generated. Therefore, $G/G_{\text{tors}} \simeq \mathbb{Z}^r$ for some $r$. So we have a surjective homomorphism $\varphi: G \to \mathbb{Z}^r$ such that $\ker \varphi = G_{\text{tors}}$.
2. There exists a homomorphism $\psi: \mathbb{Z}^r \to G$ such hat $\varphi \circ \psi = \text{Id}_{\mathbb{Z}^r}$. To build such a $\psi$ you can just pick a basis $(a_1, \ldots, a_r)$ in $\mathbb{Z}^r$ and choose $\psi(a_i)$ to be any element in the set $\varphi^{-1}(a_i)$.
3. Now it is easy to show that $G = \ker \varphi \oplus \operatorname{im} \psi$. Since $\ker \varphi = G_{\text{tors}}$ and $\operatorname{im} \psi \simeq Z^r$, we have the desired isomorphism.