Every torsion free divisible abelian group admits an 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 $G$.

1. Take a maximally independent {$r_a$} and totally order it.

2. Every element of $G$ can be written as a finite linear combination of {$r_a$} with rational coefficients.

3. Let $r\in G$. Write $\displaystyle{r = \sum_{i=1}^k c_i r_{a_i} }$ where $c_i\neq 0$ and $a_1<a_2<...<a_k$

4. Declare an element to be positive iff $c_k>0$ with respect to the expression in #2.

What I don't understand:

1. 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.
2. 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.

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What do you mean by an order compatible with the operation, do you mean : i) if a>0 and b>c then ab>ac and a,b>0, then a*b>0? –  gary Jul 3 '11 at 6:28
@gary: What I mean is, if $a>b$, then $a+c>b+c$ for any group element $c$. The use of $0$ and $*$ in your statement suggests that you are thinking of two operations, like in a ring. –  B M Jul 3 '11 at 15:47

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 $0_G \ne x \in H$, 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 $a_1, a_2, \dots, a_k, a'_1, a'_2, \dots, a'_j$, 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 $0_G \ne x \in H$, 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.

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Thanks, Brian for a very detailed resolution of my questions. I needed to read up on positive cones but I think I understand the complete proof now. –  B M Jul 3 '11 at 21:27

Here is the answer to your first question.Suppose $G$ is a torsion free divisible 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 $(a/b) \cdot x$ to be (the unique) element $y$ of $G$ satisfying $by = ax$. Hence $G$ becomes a $\mathbb{Q}$ -module, i.e. a vector space over $\mathbb{Q}$.

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Thank you algebra_fan. This is quite elegant and interesting in itself. –  B M Jul 3 '11 at 21:28