Kourovka Notebook Problem 21.96

Definitions and mathematical statement

The order of an element $g$ of a group $G$ is the least positive integer $m$ with $g^m=1$, when one exists. A group is periodic, or torsion, if every element has finite order. It is locally finite if every subgroup generated by finitely many elements is finite; equivalently, each finite subset lies in a finite subgroup.

An involution is an element of order exactly two. For $x\in G$, its centralizer is the subgroup $C_G(x)=\{g\in G:gx=xg\}$. Suppose $G$ is periodic, contains an involution, and satisfies \[ \forall x\in G\text{ of even order},\qquad C_G(x)\text{ is locally finite}. \] Must $G$ itself be locally finite?

More explicitly, the centralizer hypothesis says that whenever $x$ has even order and $a_1,\ldots,a_r$ all commute with $x$, the subgroup $\langle a_1,\ldots,a_r\rangle$ is finite. The desired conclusion removes the requirement that these generators have a common even-order element in their centralizers. It asks for finiteness of every finitely generated subgroup, not for finiteness of $G$ as a whole. No common upper bound on element orders or on the sizes of these finite subgroups is assumed. Centralizers are tested for all even-order elements, not only involutions.

Short English statement

Must a periodic group with an involution be locally finite when every even-order element has a locally finite centralizer?

Sources

• S1 ULAM.AI and the UnsolvedMath Contributors, supplied problems.json, record 2605 (KOU-21.96), 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 21.96. Expanded formulation of the supplied catalogue statement. https://arxiv.org/pdf/1401.0300v47.