# Baer-Specker group is not free abelian

From Groupprops

## Statement

The Baer-Specker group, which is defined as the external direct product of countably many copies of the group of integers, is *not* a free abelian group.

## Facts used

- A slight variant of 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 (we need to relax the condition of finite generation)
- Homomorphism from Baer-Specker group to group of integers that is zero on restricted direct product is zero

## Proof

Proof follows by combining (1) and (2).