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.