TITLE: Torsion-free divisible group
QUESTION [5 upvotes]: I have some confusion.
Let $G$ be a torsion-free divisible abelian group. Then, $G$ is a $\mathbb Q$-vector space. If $G$ has a finite dimension as a vector space then can we write $G$ as a finite direct sum of copies of $\mathbb Q$ ?

REPLY [7 votes]: You are correct. Since $G$ (a torsion-free divisible group) is divisible given any $y\in G$ and $m\in\mathbb{Z}$ there exists a solution $x$ to $mx=y$. Since $G$ is torsion-free this solution is unique. Thus, we can define $\displaystyle \frac{1}{m}y:=x$. One then shows that this defines a $\mathbb{Q}$-vector space structure on $G$, and so $G\cong \mathbb{Q}^{\oplus\lambda}$ for some cardinal $\lambda$. Saying that the dimension of $G$ over $\mathbb{Q}$ is finite is just saying that $\lambda$ is finite, and so we get exactly what you wanted.