Kourovka Notebook Problem 21.80

Definitions and mathematical statement

A left order on a group $G$ is a strict total order $<$ on its elements such that $x<y$ implies $ax<ay$ for all $a,x,y\in G$. A group is left-orderable if it admits at least one such order; the order is not required to be invariant under multiplication on the right. Finite generation means that some finite subset $S\subseteq G$ generates $G$ using products and inverses.

For $g\in G$, its conjugacy class is $g^G=\{h^{-1}gh:h\in G\}$. The identity always forms a conjugacy class by itself. Having exactly one nontrivial conjugacy class means \[ G\ne\{1\},\qquad \forall a,b\in G\setminus\{1\}\ \exists h\in G: h^{-1}ah=b. \] Thus $G$ has exactly two conjugacy classes, one equal to $\{1\}$ and one equal to its complement.

Does there exist a group satisfying all three requirements: finite generation, left-orderability, and this two-class condition? Left-orderability already excludes elements of nontrivial finite order, so any example must be infinite and torsion-free. The conjugating element may depend on the ordered pair $(a,b)$, and no bound on its word length is part of the question. The target concerns conjugacy inside the proposed group itself.

Short English statement

Does a finitely generated left-orderable group with exactly two conjugacy classes exist?

Sources

• S1 ULAM.AI and the UnsolvedMath Contributors, supplied problems.json, record 2589 (KOU-21.80), Kourovka Notebook - New Problems, Issue 21. Catalogue identity and source transcription; CC BY 4.0. https://huggingface.co/datasets/ulamai/UnsolvedMath.
• S2 The Kourovka Notebook, 21st edition, version 47 (30 September 2026), new problems, Problem 21.80. Expanded formulation of the supplied catalogue statement. https://arxiv.org/pdf/1401.0300v47.