# If $G$ is a finitely generated abelian group, then $G \cong \mathbb{Z}^r/\phi(\mathbb{Z}^s)$

Let $G=\langle x_1,x_2,\ldots,x_r \rangle$ be a finitely generated abelian group. Show that there exist positive integers $r, s$ such that $G \cong \mathbb{Z}^r/\phi(\mathbb{Z}^s)$, where $\phi: \mathbb{Z}^s \rightarrow \mathbb{Z}^r$ is a group homomorphism.

I used the fact that if $G$ is a finitely generated abelian group, then there is an onto homomorphism $\psi: \mathbb{Z}^r \rightarrow G$ defined by $(a_1,a_2, \ldots , a_r) \mapsto a_1x_1+a_2x_2+ \ldots +a_rx_r$ (For this problem the group operation will be written additively). By the First Isomorphism Theorem, $G \cong \mathbb{Z}^r/\ker(\psi)$. So I am tempted to show that $\ker(\psi)=\phi(\mathbb{Z}^s)$ for some integer $s$. In that case $\phi: \mathbb{Z}^s \rightarrow \mathbb{Z}^r$ is the linear transformation that maps $\mathbb{Z}^s$ onto $\ker(\psi)$. But how do I show such a transformation exists? Am I on the right track?

• Are you allowed to use the fact that any subgroup of a free group is free? – Yeldarbskich Nov 25 '15 at 6:57
• No since we haven't talked about free groups. – user112358 Nov 25 '15 at 6:59

This is a good approach, but you need to show that any subgroup of $\mathbb{Z}^r$ is finitely generated.