Kourovka Notebook Problem 21.136

Definitions and mathematical statement

A profinite group is a compact Hausdorff topological group that is an inverse limit of finite discrete groups. Conjugacy in $G$ is the relation $x\sim y$ when $y=g^{-1}xg$ for some $g\in G$; the conjugacy class of $x$ is $x^G=\{g^{-1}xg:g\in G\}$.

An element has infinite order if $x^n\ne1$ for every positive integer $n$. Put \[ \mathcal C_\infty(G)=\{x^G:x\in G\text{ has infinite order}\}. \] Since element order is constant on a conjugacy class, this is a well-defined set of classes. Let $\mathfrak c=2^{\aleph_0}$ denote the cardinality of the real numbers.

If $|\mathcal C_\infty(G)|<\mathfrak c$, must $G$ be torsion? Here torsion means that every individual element has some finite order, with no common exponent bound required. Equivalently, does the existence of even one infinite-order element force at least continuum many distinct conjugacy classes consisting of infinite-order elements?

The hypothesis counts classes, not elements, so a single class may itself be infinite or uncountable. The inequality is strict and is a comparison of cardinals, without an assumption about the continuum hypothesis. No finite-generation, metrizability, or countability condition is imposed on the profinite group. All conjugations and all element orders are taken in the given group $G$.

Short English statement

If a profinite group has any infinite-order element, must it have at least continuum many conjugacy classes of such elements?

Sources

• S1 ULAM.AI and the UnsolvedMath Contributors, supplied problems.json, record 2645 (KOU-21.136), 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 21.136. Expanded formulation of the supplied catalogue statement. https://arxiv.org/pdf/1401.0300v47.