2
$\begingroup$

A linearly ordered group is a group $(G,\dot,1)$ together with a linear ordering $<$ on $G$ with $f<g \Longrightarrow (f h<g h$ and $h f < h g)$ for all $f,g,h \in G$.

A commutative-transitive group (or CT-group) is a group $(G,\dot,1)$ in which two elements which commute with a third, non-trivial one, commute together. Equivalently, this is a group in which centralizers are commutative.

I'm having a hard time finding linearly ordered groups whose underlying group is not a CT-group, but I see no reason why any linearly ordered group should be a CT-group. Are there ways to construct linearly ordered non CT-groups?

$\endgroup$
2
  • 1
    $\begingroup$ To be sure I understand what you are asking, from what you wrote in the 2nd to last sentence I infer that you can neither prove nor disprove that linear orderability implies CT. From that, I was expecting you to ask for constructions of linearly ordered groups that are not CT. Nonetheless, you are asking for constructions of linearly ordered groups that are CT, correct? $\endgroup$
    – Lee Mosher
    Commented Aug 19, 2022 at 16:29
  • 1
    $\begingroup$ @LeeMosher Your intuition was good, and my attention is not. I meant non CT-groups. $\endgroup$
    – nombre
    Commented Aug 19, 2022 at 16:32

1 Answer 1

Reset to default
2
$\begingroup$

Such groups are often called "bi-orderable".

Torsion-free nilpotent groups are all bi-orderable.

But a non-abelian nilpotent group is never commutative-transitive.

Hence, every non-abelian torsion-free nilpotent group yields an example (bi-orderable, not commutative-transitive).

$\endgroup$
1
  • 1
    $\begingroup$ Thanks, that's a nice answer. $\endgroup$
    – nombre
    Commented Aug 20, 2022 at 15:15

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .