Finite-by-torsion-free abelian groups (or compact abelian groups with finitely many components)

Question

Here's a question I should know the answer to but don't:

Suppose $1\to F \to G \to G/F \to 1$ is a short exact sequence of abelian groups with $F$ finite and $G/F$ torsion-free. Must the sequence split?

This is not true if you merely assume that $F$ is torsion. A counterexample is given by YCor here: https://mathoverflow.net/a/314536/20598.

Equivalently, suppose $G$ is a compact abelian group with finitely many components. Then does $G_0 \to G \to G/G_0$ split? This is not true without assuming there are finitely many components, as YCor's example shows, and it's also not true for nonabelian groups, as Max's answer here shows: https://math.stackexchange.com/a/954539/23805 (though I think it's true whenever $G/G_0$ is cyclic).

Answer 1

Here’s a quick homological proof.

Suppose $F$ is finite and $H$ torsion free. Then $F\cong\text{Hom}(F,\mathbb{Q}/\mathbb{Z})$, so $$\text{Ext}^1(H,F)\cong\text{Ext}^1\left(H,\text{Hom}(F,\mathbb{Q}/\mathbb{Z})\right) \cong\text{Hom}\left(\text{Tor}_1(H,F),\mathbb{Q}/\mathbb{Z}\right), $$ which is zero since torsion free abelian groups are flat.