# Groups with three conjugacy classes that define an ordering

Consider the following property for a group $$(\mathcal{G},\cdot,1)$$:

There are exactly three conjugacy classes $$\{1\}$$, $$\mathcal{C}_1$$, $$\mathcal{C}_2$$ in $$\mathcal{G}$$, and we have $$\mathcal{C}_1 \mathcal{C}_1 \subseteq \mathcal{C}_1$$ and $$\mathcal{C}_2=\mathcal{C}^{-1}_1$$.

Note that the only finite groups with exactly three conjugacy classes are the cyclic group of order $$3$$ and the symmetric group of order $$6$$. Those do not satisfy the property above. So any such group must be infnite, hence also non-abelian.

In fact, given such a group $$\mathcal{G}$$, define $$x if and only if $$y \cdot x^{-1} \in \mathcal{C}_1$$. This defines a linear order on $$\mathcal{G}$$ for which it is bi-ordered, i.e. with $$\forall x,y,z \in \mathcal{G},\ y>1 \Longleftrightarrow x \cdot y \cdot z> x \cdot z$$. So it is perhaps best to think of those groups as linearly bi-ordered groups where any two strictly positive elements are conjugated.

I suspect not so many such groups are known. So my question is: are there known examples of such groups? Or better yet: have they been studied to some extent?

One could also expect such groups to be related to the first-order theory $$T$$ of linearly bi-ordered non-trivial groups with trivial center. The question would be: does any algebraically closed model of $$T$$ satisfy the property above? Indeed I believe that László Fuchs proved that in the case of partially bi-ordered, lattice ordered groups, the algebraically closed models have the property that any two strictly positive elements are conjugated. In that case however there may still be more conjugacy classes than three because not every element must be positive or negative.

• As far as I'm aware, this is an open problem: see Problem 3.31 of arxiv.org/abs/0906.2621. – shane.orourke Apr 8 at 14:43
• In fact, you can even say that such a group cannot be finitely generated, since f.g. bi-orderable groups are indicable (i.e., admit an epimorphism onto $\mathbb Z$), hence they must contain infinitely many conjugacy classes. – Ashot Minasyan Apr 8 at 15:10
• What do you mean by an "algebraically closed model of $T$"? – Alex Kruckman Apr 8 at 15:15
• @shane.orourke : Thanks for this article, a lot of related and more general problems there! Strangely the reference for Problem 3.31 is the "Black Swamp Problem Book" which I cannot find, so I'll try to find another reference and post it here. – nombre Apr 8 at 15:28
• @AlexKruckman I mean a model $\mathcal{M}$ of $T$ such that if $\mathcal{N}\supseteq \mathcal{M}$ is a model of $T$ and $\varphi[x]$ is a conjunction of equations which has a solution in $\mathcal{N}$, then it has a solution in $\mathcal{M}$. So it's a little weaker than being existentially closed. – nombre Apr 8 at 15:31