# Every abelian torsion-free group is strictly totally orderable (via the compactness theorem)

Let $\mathbb G = (G, +)$ be a group. We say that $\mathbb G$ is strictly totally orderable (others would say bi-orderable) if there exists a total order $\preceq$ on $G$ such that $x+z \prec y + z$ and $z + x \prec z + y$ for all $x,y,z \in G$ with $x \prec y$. It is not difficult to give a direct proof of the fact that if $\mathbb G$ is abelian and torsion-free then it is strictly totally orderable (Proof. There is a group embedding of $\mathbb G$ into a divisible group, and then into $(\mathbb Q^\kappa,+)$ for $\kappa := |G|$); the result is credited to F. W. Levi [1]. However, an exercise in Hodges' Model Theory asks for a proof of the same result by the compactness theorem, a proof which I wasn't able to reconstruct. So the questions are:

Q1. Could you mention an article or a book where such a proof can be found? Q2. Would you sketch such a proof here?

Thanks in advance for any help.

References.

[1] F. W. Levi, Arithmetische Gesetze im Gebiete diskreter Gruppen, Rend. Circ. Mat. Palermo 35 (1913), 225–236.

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A compactness argument which Hodges may have had in mind can go as follows. Since a subgroup of a totally ordered group is also a totally ordered group, it suffices to embed the given abelian torsion-free group $G$ into a totally ordered group, i.e., to show that the theory of totally ordered abelian groups is consistent with the diagram of $G$. By the compactness theorem, it is enough to show that this is true for any finite subset of the diagram. This finite subset only mentions finitely many constants from $G$, hence it suffices to show that every finitely generated subgroup of $G$ is totally orderable. However, every finitely generated abelian torsion-free group is isomorphic to $\mathbb Z^n$ for some $n\in\omega$, which can be given e.g. the lexicographic order.
So essentially, the model theoretic proof, if I'm not missing anything, is sort of a "rewording" of the same argument given in the OP (I omitted some details, but it should be clear how to conclude once that the problem has been embedded into $\mathbb Q^\kappa$), right? Still, interesting and quite instructive. Thank you! –  Salvo Tringali Jun 18 '13 at 11:57
It may be worth noting that, although the proof given by Emil appears to use compactness for first-order logic, it really uses compactness only for propositional logic. The point is that the relevant first-order sentences are either quantifier-free (the diagram of $G$) or universal (the axioms for ordered groups). Replacing the latter by all their instances, we get just propositional combinations of atomic sentences $a\preceq b$ for $a,b\in G$. –  Andreas Blass Jun 18 '13 at 15:43