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

Solution 1:

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.

Solution 2:

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.

Solution 3:

"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.