If $G$ is abelian such that $mG=G$ for some $m\in\Bbb{Z}$ then every short exact sequence splits

Question

Suppose $G$ is an abelian group with $mG=G$ for some $m\in \mathbb{Z}$. Show that every short exact sequence

$$0\to G\xrightarrow{f}E\xrightarrow{g}\Bbb Z_m\to 0$$

splits.

I know that ${\rm Ext}_\mathbb{Z}^1(\mathbb{Z}_m,G)=0$, then I deduce the long exact sequence

$$0\to{\rm Hom}_{\Bbb Z}(\Bbb Z_m, G)\xrightarrow{{\rm Hom}_{\Bbb Z}(G, f)}{\rm Hom}_{\Bbb Z}(\Bbb Z_m, E)\xrightarrow{{\rm Hom}_{\Bbb Z}(\Bbb Z_m, g)}{\rm Hom}_{\Bbb Z}(\Bbb Z_m, \Bbb Z_m)\to 0$$

I'm not sure how to put all together to get that the sequence splits.

Answer 1

Your proof (plus the hint in the comments) is more succint, but we can also do it explicitly:

Pick some lift $\widetilde{1}$ in $E$ of the element $1 \in \mathbb{Z}_m$. Now $m\widetilde{1}$ is in the kernel of $g$, so it's in the image of $f$; say it's $f(\alpha)$. By hypothesis, there is $\beta \in G$ with $m\beta = \alpha$.

The claim is then that the map $\phi: \mathbb{Z}_m \to E$ given by $\phi(j) = j(\widetilde{1} - f(\beta))$ is a splitting of this sequence. You need to check that $\phi(m) = 0$ (else the map wouldn't be well-defined) and that $g \phi$ is the identity map, both of which are straightforward.