If $G$ is a group with a (left) linear order, does every (left) partial order on $G$ extend to a (left) linear order?

The answer is affirmative on abelian groups, where being torsion-free is necessary and sufficient both for having a linear order and for partial orders to extend to linear orders (Fuchs, 1950, since any partial order on a torsion-free abelian group extends to a normal order in his terminology).

  • $\begingroup$ Groups in which every bi-invariant partial order extends to a bi-invariant order are called $O^*$-group in Gupta, Narain; Rhemtulla, Akbar On ordered groups. Algebra Univ. 1 (1971/72), 129-132. Possibly more information can be found in Rhemtulla's book, but I haven't yet had access to it. $\endgroup$
    – YCor
    Feb 16, 2020 at 19:39

1 Answer 1


There are necessary and sufficient conditions in the literature for a (left) partial order $\le$ on $G$ to extend to a (left) linear order $\le^{\ast}$ on $G$. This shows in particular, that not every partial left order extends to a linear left order in the non-abelian case, even though the group is orderable.
In the paper "Right-orderability of groups" by Richard Kaye (1998) these conditions are called "a sort of mini completeness/soundness theorem". In the paper "Compactness of the space of left orders" (arXiv) of Dabkowska, Dabkowski, Harizanov, Przytycki and Veve, these conditions are referred to as Conrad's theorem (P. F. Conrad, Right-ordered groups, Mich. Math. J. 6(3), 1959, pp. 267–275.) This paper also gives an explicit example of a partial order on the fundamental group of the Klein bottle that does not extend, even though the group is orderable.

Remark: I have edited the answer to include Alexander's useful comments, which helped to clarify the answer.

  • $\begingroup$ Thanks for these references! I may be missing something obvious, but I don't off-hand see how to use these necessary and sufficient conditions to prove or disprove that the extensibility of partial orders is equivalent to right orderability. $\endgroup$ Sep 17, 2013 at 12:04
  • $\begingroup$ On page 164 Ohnishi constructs a partial order which cannot be made into a linear one. $\endgroup$ Sep 17, 2013 at 13:41
  • $\begingroup$ First, Ohnishi works with biorders, not just right (or left) orders. Second, he shows that if his condition (III) is false, then such a partial order can be constructed, yes. But to give a negative answer to a question like mine (i.e., the biorderability analogue of my question), one would then need to show that there is a biorderable group where (III) fails. Maybe there is an obvious case, but I'm not very familiar with the area. $\endgroup$ Sep 17, 2013 at 16:24
  • $\begingroup$ I see. I thought that not every partial left order on $G$ extends to a linear left order on $G$ in the non-abelian case. I will check again (there is Conrad's theorem, see arxiv.org/pdf/math/0606264.pdf, section $4$). $\endgroup$ Sep 17, 2013 at 18:05
  • $\begingroup$ Thanks! That last reference gives a solution. Right after stating Conrad's theorem it gives an example of a partial order on the fundamental group of the Klein bottle that does not extend, even though the group is orderable. You might want to edit your answer and incorporate this. $\endgroup$ Sep 18, 2013 at 22:14

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