Pure subgroup implies direct factor in torsion-free abelian group that is finitely generated as a module over the ring of integers localized at a set of primes

Statement

Consider the following:

Then, $H$ must be a direct factor.

Proof

By Fact (1), $G/H$ is also torsion-free. It is also finitely generated over $\mathbb{Z}[\pi^{-1}]$. Thus, by Fact (2), it is free as a $\mathbb{Z}[\pi^{-1}$-module, and we can thus find a complement to $H$ in $G$.