Embedding torsion-free abelian groups into $\mathbb Q^n$?

Glass' Partially Ordered Groups states without proof:

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

1. 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)?
2. 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.
• i think the natural interpretation would put no bounds on the cardinality of the vectorspace Jun 20 '13 at 13:45
• To answer your second question, it means that the embedding (the injective map into the $\mathbb Q$-vector space) is only assume to be a group homomorphism: it may preserve order, or it may not. Jun 20 '13 at 14:20
• – user26857
Jun 23 '13 at 16:57

Let $G$ be an abelian torsion-free group. Then $G$ is a torsion-free $\mathbb Z$-module and consider $S=\mathbb Z-\{0\}$. Then we have a canonical homomorphism $\varphi$ from $G$ to the module of fractions $S^{-1}G$ given by $\varphi(x)=x/1$. Since $G$ is torsion-free we have that $\varphi$ is injective, so $G$ is embedded in $S^{-1}G$. But $S^{-1}G$ is an $S^{-1}\mathbb Z=\mathbb Q$-module, that is, a $\mathbb Q$-vector space.

• Why not directly $G \otimes \mathbb Q$? If OP is familiar with localization, he's probably familiar with tensor product too. Jun 20 '13 at 14:28
• I'm not very familiar with modules, but when you say "$\mathbb Q$-module", you mean that the scalars are rational and the "vectors" are elements of $G$, right? I thought the proof was $G \hookrightarrow \mathbb Q^n$, but maybe I don't understand what is meant by "rational vector space". Jun 21 '13 at 13:33
• No, the "vectors" are elements of $S^{-1}G$, that is, a kind of fractions of the form $g/n$ with $g\in G$ and $n\in\mathbb Z$, $n\neq 0$. Furthermore, a rational vector space is not necessarily finitely dimensional. Or when you say $\mathbb Q^n$ you simply assume that the dimension of the space is $n$, and this is why you have such troubles with countability.
– user26857
Jun 21 '13 at 14:54
• But since your homomorphism is $x\mapsto x/1$ the "relevant" vectors are just $G$, yes? Jun 21 '13 at 15:11

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.

"As a group" means that there is an injective group homomorphism $$G \hookrightarrow V$$ where $$V$$ is a $$\mathbb Q$$-v.s, thought of just as a group via addition.

As a very minor variant on YACP's description of the actual embedding, if $$G$$ is torsion-free then the natural map $$G \to \mathbb Q\otimes_{\mathbb Z} G$$ is injective, and the target is a $$\mathbb Q$$-vector space. (This is the same construction as in YACP's answer, just decribing localization at $$\mathbb Z \setminus \{0\}$$ as tensoring with $$\mathbb Q$$.)

Also, there is nothing in the statement which suggests that the vector space should be finite dimensional, and in general it won't be.