Exact sequence of 4 abelian groups, 3 of them being free

Question

Let $G$ be a finitely generated abelian group that fits in an exact sequence of the form

$$0 \to \mathbb{Z} \to \mathbb{Z}^n \stackrel{f}{\to} G \stackrel{g}{\to} \mathbb{Z}^m \to 0$$

for some $n \geq 1$ and $m \geq 0$. What can we say about $G$?

Answer 1

Since $\mathbb{Z}^m$ is free, the surjection $G\to \mathbb{Z}^m$ splits. Thus, $G$ is a direct product of $\mathrm{Im}(f)$ and $\mathbb{Z}^m$, So the only question is what is $\mathrm{Im}(f)$.

Every subgroup of $\mathbb{Z}^n$ is free, and moreover, given any subgroup $H$ of $\mathbb{Z}^n$, either $H$ is trivial or there exists a basis $x_1,\ldots,x_n$ of $\mathbb{Z}^n$, an integer $r$, $1\leq r\leq n$, and positive integers $d_1,\ldots,d_r$ such that $d_1|\cdots|d_r$ and $H$ is (freely) generated by $d_1x_1,\ldots,d_rx_r$.

Here, $r$ must be equal to $1$, so there exists an integer $d\gt 0$ such that $\mathbb{Z}^n/H\cong (\mathbb{Z}/d\mathbb{Z})\oplus \mathbb{Z}^{n-1}$. Thus, $$G \cong \frac{\mathbb{Z}}{d\mathbb{Z}}\oplus \mathbb{Z}^{n+m-1},$$ for some $d\geq 1$.