Every torsion-free divisible abelian group admits a total order compatible with the group operation.
I am looking for a proof of this result. I found a proof which goes like this: Call the group $G$.
Take a maximally independent {$r_a$} and totally order it.
Every element of $G$ can be written as a finite linear combination of {$r_a$} with rational coefficients.
Let $r\in G$. Write $\displaystyle{r = \sum_{i=1}^k c_i r_{a_i} }$ where $c_i\neq 0$ and $a_1
Declare an element $r$ to be positive iff $c_k>0$ with respect to the expression in #2.
What I don't understand:
- What is meant by a maximally independent subset? Is there an implication that a torsion-free divisible abelian group is a free $\mathbb{Q}$-module?
- How would an arbitrary ordering of the "generating set" result in an order compatible with the group operation?
Any pointers will be greatly appreciated. I can add more details if needed.