Kourovka Notebook Problem 15.54

Definitions and mathematical statement

A group is periodic if every element has finite order. An involution is an element of order exactly two. A group is locally cyclic if each of its finitely generated subgroups is cyclic, and a $2$-group here means a group in which every element has order a power of $2$, including $2^0=1$.

Let $G$ be any periodic group, and let $i\in G$ be an involution such that its centralizer \[ C_G(i)=\{g\in G:gi=ig\} \] is a locally cyclic $2$-group. Define \[ S_i=\{g\in G:\operatorname{ord}(g)\text{ is odd and }i^{-1}gi=g^{-1}\}. \] The identity belongs to $S_i$, since its order is $1$. Must $S_i$ be a subgroup of $G$ for every such pair $(G,i)$? No finiteness assumption on $G$ is imposed beyond periodicity.

Short English statement

In a periodic group, suppose an involution has a locally cyclic centralizer consisting of elements of $2$-power order. Must all odd-order elements that this involution inverts, including the identity, together form a subgroup?

Sources

• Local UnsolvedMath registry, permanent ID $701142$, exact source code catalogue entry, title Kourovka Notebook Problem 15.54; supplied source record, upstream revision d371ee917ac9647fbb06cd6fc529deec6f3dda82.
• Attributed in the supplied record to V. D. Mazurov, proposed in 2002: The Kourovka Notebook (Unsolved Problems in Group Theory), 15th issue (2002), Problem 15.54. The record also identifies the 21st issue edition (2026), arXiv:1401.0300v48. The supplied collection description names E. I. Khukhro and V. D. Mazurov as editors.
• The registry labels the problem open. This is a supplied source label, not an independently verified assessment of the mathematical literature. The record was updated on 2026-10-16; this timestamp is registry metadata, not a supplied literature-verification date. No source was independently consulted for this draft.