TITLE: Embedding torsion-free abelian groups into $\mathbb Q^n$?
QUESTION [7 upvotes]: Glass' Partially Ordered Groups states without proof:

Every torsion-free abelian group can be embedded into a rational vector space (as a group).


Can someone link me to a proof of this? It seems to me like it's probably false: $\mathbb Q^n$ is countable, so how can an uncountable group be embedded (unless the space has uncountable dimension)?
Does "embedded as a group" mean "embedded in a way that doesn't preserve order"? I'm not sure what the "as a group" modifier indicates.

REPLY [5 votes]: An alternative proof based on divisible groups:

Any abelian group can be embedded into a divisible group.

Let $G$ be a group. It can be written as the quotient of a free group, say $F/N$. If $G$ is abelian, the commutator subgroup $F'$ is included in $N$, so $G$ is a quotient of the abelianization $F^{ab} \simeq \bigoplus\limits_{i \in I} \mathbb{Z}$, say $F^{ab}/M$. Clearly, $F^{ab}$ can be embedded into $Q=\bigoplus\limits_{i \in I} \mathbb{Q}$, so $G=F^{ab}/M$ is embedded into $Q/M$. But $Q/M$ is divisible as the quotient of a divisible group.

Any torsion-free abelian group can be embedded into a torsion-free divisible group.

Using the previous embedding $G=F^{ab}/M \hookrightarrow Q/M$, it is clear that for all $x \in Q/M$ there exists $n \in \mathbb{Z}$ such that $nx \in G$. Therefore, if $G$ is torsion-free, $Q/M$ also.

Any torsion-free divisible group is a $\mathbb{Q}$-vector space.

Therefore, $G \hookrightarrow Q/M$ is an embedding from $G$ into a $\mathbb{Q}$-vector space.