My question is similar to the following two questions: Exact sequence of abelian groups, special case and Show $0 \to A\mathop \to B\mathop \to C \to 0$ is split when $C$ is a free Abelian group, but neither have been answered.
I have an idea of how to do this proof:
Let $0\to A\to B\to\mathbb{Z}\to 0$ be a short exact sequence of abelian groups with $f:A\to B$ and $g:B\to\mathbb{Z}$. I want to use the splitting lemma, which states that if there exists a map $j:\mathbb{Z}\to B$ with $gj=1_{\mathbb{Z}}$, then the sequence is split.
Because this is a short exact sequence, we have that $B/A\cong\mathbb{Z}$. Then because $\mathbb{Z}$ is cyclic, so is $B/A$, and in particular $b+A$ for some $b\in B$ generates all of $B/A$. I then want to use this fact somehow to construct a map from $\mathbb{Z}\to B$ which composes with $g$ to give the identity in $\mathbb{Z}$, but I can't quite figure out the construction.
Take $b_0\in B$ whose image in $\Bbb Z$ is $1$, check that $b_0+A$ generates $B/A$ and define $j$ by $j(1) = b_0$. Since $1$ generates $\Bbb Z$, this defines $j$ completely. Since $g(b_0) = 1$, we have $gj(1) = 1$ thus $gj = 1_{\Bbb Z}$.