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.


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


2 Answers 2


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).

  • $\begingroup$ 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! $\endgroup$ Commented Jun 18, 2013 at 11:57
  • 2
    $\begingroup$ 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.) $\endgroup$ Commented Jun 18, 2013 at 12:35
  • $\begingroup$ Thanks. Now the basic difference between the two approaches, far beyond the wording, is clear to me too. $\endgroup$ Commented Jun 18, 2013 at 12:52
  • $\begingroup$ I have added a note on generalization to other structures. I think this also illustrates the difference between the two proofs. $\endgroup$ Commented Jun 18, 2013 at 13:24
  • 1
    $\begingroup$ 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 $a,b\in G$. $\endgroup$ Commented Jun 18, 2013 at 15:43

A group $G$ is not bi-orderable if and only if for some finite subset $J$ of $G\smallsetminus\{1\}$, for every $e\in \{-1,1\}^J$ there exists $n\ge 1$, group elements $c_1,\dots,c_n$, and a function $s:\{1,\dots,n\}\to J$ such that $$\prod_{i=1}^nc_is(i)^{e(s(i))}c_i^{-1}=1_G.$$

Indeed, first clearly if there is such $J$, then if by contradiction there is a bi-invariant total order, there exists $e$ such that $s^{e(s)}>1$ for all $i$, and then since being $>1$ is stable under conjugation and taking products, we get a contradiction.

The converse is where compactness takes place. Suppose that the condition fails. That is, for every finite subset $J$ there exists $e\in\{-1,1\}^J$ such that the condition fails: $1_G$ is not in the subsemigroup generated by the union of all conjugates of $\{s^{e(s)}:s\in J\}$. By compactness, there exists a function $e:G\smallsetminus\{1\}\to\{-1,1\}$ that satisfies the same condition. Then

  • $e(g^{-1})=-e(g)$ for all $g\neq 1$, because $g^{e(g)}(g^{-1})^{e(g^{-1})}\neq 1$),

  • $e(g)=e(hgh^{-1})$ for all $g\neq 1,h$, because $g^{e(g)+e(ghg^{-1})}=g^{e(g)}h^{-1}(hgh^{-1})^{e(hgh^{-1})}h\neq 1$;

  • $e(g)=1,e(h)=1$ imply $e(gh)=1$, because $(gh)^{1+e(gh)}=g^{e(g)}h^{e(h)}(gh)^{e(gh)}\neq 1$

    So defining $S=\{g\neq 1:e(g)=1\}$, we have $S$ stable under conjugation and product, and $G=1\sqcup S\sqcup S^{-1}$. Thus defining $g<h$ if $g^{-1}h\in S$ defines a bi-invariant (strict) total order.

It is clear that the above criterion is "local" and in particular a group satisfies it iff it all its finitely generated subgroups do.

Note that similarly (and more simply) we have the classical:

A group $G$ is not left-orderable if and only if for some finite subset $J$ of $G\smallsetminus\{1\}$, for every $e\in \{-1,1\}^J$ there exists $n\ge 1$ and a function $s:\{1,\dots,n\}\to J$ such that $$\prod_{i=1}^ns(i)^{e(s(i))}=1_G.$$

The interest of such criteria appears for instance considering orderability of ultraproducts (of possibly non-orderable groups, e.g., of finite groups). Namely, define for a given subset $J$, $n_J$ and $n'_J$ as the smallest integer $n$ in the above criterion (for left-, resp. bi-orderability), and $\infty$ otherwise. Define $\mu_G(k)=\inf_Jn_J$ and $\mu'_G(k)=\inf_Jn'_J$ where $J$ ranges over finite subsets of $G$ of cardinal $\le k$. So the above criteria say that $G$ is non-left-orderable (resp. non-bi-orderable) if and only if $\mu_G(k)<\infty$ (resp. $\mu'_G(k)<\infty$) for some $k$. By the way also note that $\mu_G(1)=\infty$ iff $G$ is torsion-free, and otherwise $\mu_G(1)$ equals the smallest prime $p$ such that $G$ has an element of order $p$.

Then ($\bullet$) an ultraproduct $\prod^\omega G_i$ is left-orderable (resp. bi-orderable) if and only if $\forall k,\lim_{i\to\omega}\mu_{G_i}(k)=\infty$ (resp. $\forall k,\lim_{i\to\omega}\mu'_{G_i}(k)=\infty$). (And it is torsion-free iff $\lim_{i\to\omega}\mu_{G_i}(1)=\infty$.)

Note that this makes the failure of left orderability appear as a generalization of torsion.

In turn an application of the latter is the following

Proposition. A pseudofinite group is bi-orderable iff it is torsion-free.

(A group is pseudofinite if it is elementary equivalent to some ultraproduct of finite groups. Examples of non-left-orderable torsion-free groups are mentioned at several places on MO, e.g., here)

Proof of the nontrivial implication: every nonprincipal ultraproduct of $C_p$ for $p$ prime is a torsion-free abelian group, and hence for every $k$, we have $\mu'_k(C_p)\to\infty$ when $p$ prime tends to infinity (an explicit estimate should be doable but I don't need it). Write $u_p=\min_{p'\ge p}\mu'_k(C_p)$, it tends to infinity too. Next, I claim that for every finite solvable group $G$ with smallest prime divisor $p$ of $|G|$, we have $\mu'_k(G)\ge u_p$. Indeed, considering a subset $J$ as in the definition, considering the subgroup generated by $J$ and passing to a cyclic quotient, we see that there exists a prime divisor $q$ of $|G|$ such that $ \mu'_k(G)\ge\mu'_k(C_q) \;(\ge u_p)$.

Now let $G$ be pseudofinite and torsion-free, hence it is elementary equivalent to some ultraproduct $U=\prod^\omega G_n$ of (a sequence of) finite groups, and hence $U$ is torsion-free as well. Since it is torsion-free, the smallest prime divisor $|p_n|$ of $G_n$ $\omega$-tends to infinity. In particular, $\omega$-a.e., $|G_n|$ is of odd order, hence solvable by the Feit-Thompson theorem [if one wishes to be self-contained, assume $G$ pseudo-(finite solvable)]. Hence, by the previous paragraph, for every $k$, the number $\mu_k(G_n)$ $\omega$-tends to infinity. So, by the above criterion ($\bullet$) $U$ is bi-orderable. Since the criterion given at the beginning also shows that being bi-orderable is invariant under elementary equivalence, we deduce that $G$ is bi-orderable.

  • $\begingroup$ Note: Rhemtulla (Proc AMS 1973) proved that a group that is residually-$p$ for infinitely many primes $p$ is bi-orderable. The above proposition is a generalization of this. Indeed, every group that is residually (finite with no element of prime order $\le n$) for every $n$, embeds into a torsion-free ultraproduct of finite groups. The argument is in spirit the same, and actually Rhemtulla provides a quantitative bound for $\mu'_k(C_p)=\mu_k(C_p)$. $\endgroup$
    – YCor
    Commented Feb 16, 2020 at 17:53
  • $\begingroup$ Another note for record: a group $G$ for which $\mu'_1(G)=\infty$ (i.e., for every $g\neq 1$ the semigroup generated by the conjugacy class of $g$ does not contain $1$) is called an $R^*$-group by Fuchs (On Orderable Groups, in Proc. Internat. Conf. Theory of Groups, Austral. Nat. Univ., Canberra, 89-98 (1965)) and such groups are considered in N. Gupta, A. Rhemtulla, On ordered groups. Algebra Univ. 1 (1971/72), 129-132. $\endgroup$
    – YCor
    Commented Feb 16, 2020 at 18:33

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