Why do torsion-free abelian groups admit linear orders? I have read a theorem that says that every torsion-free abelian group admits a linear order. The proof used tensor products and so was above my head. I tried to find another proof on the web and I found one which I also didn't understand, because it uses model theory of which I know nothing. It's here. Wikipedia gives a reference to a Levi paper which is supposed to prove this statement but I was unable to find it and can't know if it's my level mathematics.
My question is: is there a proof of this fact which can be understood by a student with group-theory knowledge just a bit above the first course?
 A: This proof won't be to everybody's taste, but let me offer it as an illustration of the power of ultraproducts and non-standard arguments.
First let's consider a finitary version of the question, and see if it's any easier: why do finitely generated torsion-free abelian groups admit linear orders? This is not hard to answer. Any such group is isomorphic to a direct product of infinite cyclic groups, and such a group admits a linear order, for example the lexicographic order.
Next, for an arbitrary torsion-free abelian group $G$ let its finitely generated subgroups $G_i$ be indexed by $I$. Take a non-principal ultrafilter $\mathcal{D}$ on $I$ that contains all subsets of $I$ of the form $E_x=\{i\in I:x\in G_i\}$ where $x$ ranges through $G$. (Since the sets $E_x$ have the finite intersection property it follows from Zorn's Lemma that such an ultrafilter exists.)
Now take the ultraproduct $\Pi_{i\in I}G_i/\mathcal{D}$ of its finitely generated subgroups $G_i$. One can now check that 
(1) the map $g\mapsto(g_i)_{i\in I}/\mathcal{D}$, where $g_i=g$ if $g\in G_i$ and $g_i=1$ otherwise, is a well-defined injective group homomorphism. (The choice of $\mathcal{D}$ is crucial here.)
(2) The ultraproduct is linearly ordered. (The orders on the $G_i$ induce a linear order on $G$.)
(3) The linear order on the ultraproduct is compatible with the group operation.
The embedded image of $G$ therefore inherits a suitable linear order, so we are done.
One of the morals of this story is that one can often embed an object in an ultraproduct of its finitary subobjects (the details are typically similar to those above), and such an embedding often allows one to reduce the argument to the finitary case. 
A: Another solution is to use divisible groups to show the following statement (see my answer here):

Theorem: Any torsion-free abelian group is embeddable into a $\mathbb{Q}$-vector space.

Because any subgroup of a totally orderable group is itself totally orderable, it is sufficient to prove that a direct sum of totally orderable groups is itself totally orderable in order to conclude:

Property: Let $(I, \leq)$ be a well-ordered set and $\{ (G_i, \leq_i) : i \in I \}$ be a collection of totally ordered groups. Then the lexicographic order with respect to $\leq_i$ totally orders $G= \bigoplus\limits_{i \in I } G_i$.

A: Are you used to Zorn's lemma proofs? Because here is a quick one:
Let $A$ be a torsion-free abelian group. A sub-monoid of $A$ is a subset $P$ containing $0$ and closed under addition. We'll say that $P$ is pointed if it does not contain both $x$ and $-x$ for any nonzero $x \in A$. Clearly, the union of an ascending collection of pointed sub-monoids is itself a pointed sub-monoid. So, by Zorn's lemma, there is a maximal element pointed sub-monoid of $A$.
Claim 1: For any $x \in A$, and any positive integer $k$, if $kx \in P$ then $x \in P$.
Proof: Let $Q = \{ x  \in A : kx \in P\ \text{for some } k \in \mathbb{Z}_{>0} \}$. Note that $Q$ is a sub-monoid: It clearly contains $0$ and, if $k_1 x_1 \in P$ and $k_2 x_2 \in P$, then $k_1 k_2 (x_1+x_2) \in P$. We also claim that $Q$ is pointed: If there is $x \neq 0$ such that $k x \in P$ and $\ell(-x) \in P$ then $k \ell x \in P$ and $- k \ell x \in P$. Since $A$ is torsion-free, $k \ell x \neq 0$, contradicting that $P$ is pointed.
So $Q$ is a pointed monoid containing $P$ and, by the maximality of $P$, we have $Q=P$. In other words, if $kx \in P$ for $k>0$, then $x \in P$, as desired. $\square$
Claim 2: For any $x$ in $A$, either $x$ or $-x \in P$. 
Proof: Suppose that $-x \not \in P$.
Let $P' = \{ y+nx : y \in P, n \in \mathbb{Z}_{\geq 0} \}$. Clearly $P'$ is a sub-monoid of $A$. We claim that $P'$ is pointed. If not, then we have $y_1 + n_1 x = - (y_2 + n_2 x)$ for some $y_1$, $y_2 \in A$ and some $n_1$, $n_2 \geq 0$ with $y_1+n_1 x \neq 0$. 
So $y_1 + y_2 =- (n_1+n_2) x$. If $n_1=n_2=0$, then $y_1 = - y_2$ and, since $P$ is pointed, $y_1=y_2=0$. In this case, $y_1+n_1 x = 0$, while we assumed otherwise.
Alternatively, suppose that $n_1+n_2>0$. We know $y_1 +y_2 \in P$ and by claim 1, the identity $y_1 + y_2 =- (n_1+n_2) x$ then forces $-x \in P$, a contradiction.
So we now know that $P'$ is pointed. So $P'=P$ and $x \in P$, as desired. $\square$
With $P$ as above, define $y \leq z$ if $z-y \in P$. This is reflexive (as $0 \in P$) and transitive (as $P$ is closed under addition.) It is anti-symmetric (because $P$ is pointed) and a total order by the claim. 
So we have equipped $A$ with a linear order. As $P$ is closed under addition, the order is compatible with the group structure, meaning that $w \leq x$ and $y \leq z$ means $w+y \leq x+z$. 
A: The group $(\mathbb{Z}^n,+)$ (sum component-wise), is torsion free and is abelian, right? Also, $(\mathbb{Z}^n,\leq)$ (component-wise too) is a conditional complete lattice (but not a total order). What am I missunderstanding?
