Every torsion-free divisible abelian group admits an order compatible with the group operation 
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 $a_1<a_2<...<a_k$

*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.
 A: Here is the answer to your first question.
Suppose $G$ is a torsion-free divisible abelian group. Let $n$ be an integer and $y \in G$. Since $G$ is divisible, there exists $x \in G$ such that $y = nx$. And since $G$ is torsion-free, this $x$ is unique, because if $y = nx = nx'$, then $n(x-x') = 0$ hence $x-x'$ is a torsion element, so it must be equal to $0$. Hence we may define an action of $\mathbb{Q}$ on $G$ by defining $(a/b) \cdot x$ to be the (unique) element $y$ of $G$ satisfying $by = ax$. Hence $G$ becomes a $\mathbb{Q}$ -module, i.e. a vector space over $\mathbb{Q}$.
A: To add to the previous answer, note that the set of $r_a$ is a basis for that vector space, and the representation of an element $x \in G$ as $x = \sum\limits_{i=1}^k c_i r_{a_i}$ with $a_1 < \dots < a_k$ and all coefficients non-zero is unique.
For the second question, let $H = \{0_G\} \cup \{\sum\limits_{i=1}^k c_i r_{a_i}:c_k>0\}$, with notation as in the question. Call $c_k$ the leading coefficient in this representation. Since $G$ is Abelian, to show that $H$ is the positive cone for a compatible total order on $G$ you need only check that $H$ is closed under addition and that if $0_G \ne x \in G$, then exactly one of $x$ and $-x$ is in $H$. 
To see that $H$ is closed under addition, let $x = \sum\limits_{i=1}^k c_i r_{a_i}$ and $y = \sum\limits_{i=1}^j d_i r_{a'_i}$, where $a_1 < \dots < a_k$, $a'_1 < \dots < a'_j$, all $c_i$ and $d_i$ are non-zero rationals, and $c_k,d_j > 0$. Then $x+y$ is a rational combination of $r_{a_1}, r_{a_2}, \dots, r_{a_k}, r_{a'_1}, r_{a'_2}, \dots, r_{a'_j}$, possibly with some zero coefficients, and its leading coefficient is (i) $c_k$ if $a'_j < a_k$; (ii) $d_j$ if $a_k < a'_j$; and (iii) $c_k + d_j > 0$ if $a_k = a'_j$. In every case the leading coefficient is positive, so $x+y \in H$.
Now suppose that $0_G \ne x \in G$, say $x = \sum\limits_{i=1}^k c_i r_{a_i}$. 
Then $-x = \sum\limits_{i=1}^k (-c_i) r_{a_i}$, with leading coefficient $-c_k$, and obviously exactly one of $c_k$ and $-c_k$ is positive.
A: The proof through modules is nice, but I'll record the (in my opinion easier) proof for general torsion-free abelian groups.
In this proof, rather than using a basis to first figure out all the relations and then picking the whole order at once (knowing we won't get into trouble because we know no new relations will pop up), we just order bigger and bigger finite sets and take a limit in the Cantor topology of $\{-1,1\}^G$.
Let $\Gamma$ be a group. A left order on $\Gamma$ is a total order $<$ such that $\forall a, b, c \in \Gamma: a < b \iff ca < cb$.
A group is left orderable if it admits a left order. For a property of groups P, we say a group is locally P if its finitely-generated
subgroups are P.

If $\Gamma$ is not torsion-free, it is not left-orderable.

Proof. Let $\Gamma$ be a non-torsion-free group and suppose $g^n = \mathrm{id}$ for $n > 0$. Suppose for a contradiction that $\Gamma$ is orderable, 
let $<$ be the order, and suppose that $g$ is positive (the negative case is symmetric). Then $g^i < g^{i+1}$ for all $i$
by setting $a = \mathrm{id}, b = g, c = g^i$ in the definition of left-orderable. By transitivity we have $\mathrm{id} < g < \cdots < g^{n-1} < g^n$,
so by transitivity of the $<$ we have $g < g^n = \mathrm{id}$, a contradiction. Thus $\Gamma$ cannot be orderable, concluding the proof.
The monograph Groups, Orders, and Dynamics by B. Deroin, A. Navas and C. Rivas begins with the following fact:

A group is left-orderable if and only if it is locally left-orderable.

This is deduced from the following characterization, which itself is proved by an easy compactness argument: a group $\Gamma$ is left-orderable if and
only if for every finite family $\mathcal{G}$ of non-identity elements, there exists a choice of (compatible) exponents $\eta : \mathcal{G} \to \{−1, +1\}$ such
that $\mathrm{id}$ does not belong to the semigroup generated by the elements $g^{\eta(g)}, g \in \mathcal{G}$.
In fact, you can make this slightly more local: a group is left orderable if and only if for every finite subset $\mathcal{G}$ you can find $\eta : \mathcal{G} \to \{-1, 1\}$ with the following properties
$$ \forall g, h \in \mathcal{G}: g^{\eta(g)} h^{\eta(h)} \neq \mathrm{id},$$
$$\forall g, h \in \mathcal{G}: (gh \in \mathcal{G} \wedge \eta(g) = \eta(h)) \implies \eta(gh) = \eta(g).$$
The proof is the same.

An abelian group is left-orderable if and only if it is torsion-free.

Proof: A non-torsion-free group is never left-orderable by the first lemma above.
Finite subgroups of torsion-free abelian groups are of course torsion-free finitely-generated abelian groups, thus isomorphic to $\mathbb{Z}^d$ for some $d$
by the fundamental theorem of abelian groups. Clearly $\mathbb{Z}^d$ is orderable, for example by the lexicographic order. We conclude by the previous lemma.
