# 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.

Notice that only the last step used something specific about abelian groups. The same argument shows that a (nonabelian) group is totally orderable if and only if all its finitely generated subgroups are, and likewise for other ordered structures (e.g., semigroups or rings).

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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
Well, the argument is similar, but I would not say it’s quite the same. Your argument goes by embedding the whole group into something rich enough that it turns out to be easily orderable, whereas here we reduce the problem to subgroups that are poor enough to be easily orderable. Essentially, one looks at the group as the direct limit of its finitely generated subgroups. (This does not literally work, as the orders on the subgroups are not canonically chosen. The purpose of compactness here is to make these choices in a consistent way.) –  Emil Jeřábek Jun 18 '13 at 12:35
Thanks. Now the basic difference between the two approaches, far beyond the wording, is clear to me too. –  Salvo Tringali Jun 18 '13 at 12:52
I have added a note on generalization to other structures. I think this also illustrates the difference between the two proofs. –  Emil Jeřábek Jun 18 '13 at 13:24
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