Kourovka Notebook Problem 21.18

Definitions and mathematical statement

Let $G$ be a finite group and let $A_1,A_2,A_3\subseteq G$. Assume the ordered multiplication map \[\mu:A_1\times A_2\times A_3\to G,\qquad(a_1,a_2,a_3)\mapsto a_1a_2a_3\] is bijective. Thus every $g\in G$ has exactly one expression $g=a_1a_2a_3$ with $a_i\in A_i$; in particular $|G|=|A_1||A_2||A_3|$. The factors are subsets, not assumed subgroups, and no identity-containment or commutativity assumption is imposed.

Let $H=\langle A_2\rangle$ be the least subgroup of $G$ containing $A_2$. The question is whether every such exact factorization satisfies \[|A_2|\mid |H|,\] equivalently whether $|H|=m|A_2|$ for some positive integer $m$.

The supplied discussion attributes the original question to G. M. Bergman, Question 8 of the 2020 note. It reports the corresponding endpoint conclusions $|A_1|\mid|\langle A_1\rangle|$ and $|A_3|\mid|\langle A_3\rangle|$. For $i=1,3$, let $H_i$ be the least subgroup containing $A_2$ that is closed under conjugation by every member of $A_i$; because $G$ is finite this can be expressed as $aH_ia^{-1}=H_i$ for all $a\in A_i$. The source reports $|A_2|\mid|H_i|$ as well. These closure groups may be larger than $\langle A_2\rangle$ and do not supply the requested middle-factor conclusion.

The source is Kourovka Notebook issue 21 (2026), Problem 21.18. The literature triage checked 2026-08-17 reports that the official maintained update records a negative solution by Mikhail Kabenyuk's evolving arXiv preprint: a finite exact three-factor factorization exists with $|A_2|\nmid|\langle A_2\rangle|$. The registry and triage label it solved / SOLVED-IN-LITERATURE. The review expressly bases classification on the authoritative update and attributed primary paper while requiring independent extraction and checking of the concrete finite witness before downstream theorem-level use; the abstract does not expose this specialization and no witness is supplied in the record. No concrete group or counterexample is invented here.

Short English statement

In a finite group uniquely factored as three ordered subset factors, must the size of the middle factor divide the order of the subgroup it generates? The supplied Kourovka update reports no, while requiring a separate check of the concrete counterexample in the cited evolving preprint.

Sources

• Local UnsolvedMath registry, permanent ID 2527, source code catalogue entry, "Kourovka Notebook Problem 21.18"; supplied issue-21 (2026) statement, Bergman attribution, and negative-answer qualifications.
• George M. Bergman, "A note on factorizations of finite groups," Journal of the Iranian Mathematical Society 1 (2020), 157--161, Question 8: https://math.berkeley.edu/~gbergman/papers/gp_factzn.pdf. Supplied primary source for the question and endpoint divisibility results.
• Mikhail Kabenyuk, "Factorizations of finite groups," arXiv:2102.08605 (2021; later revisions): https://arxiv.org/abs/2102.08605. Supplied primary preprint credited for the negative answer.
• Kourovka Notebook, issue 21, Problem 21.18, checked 2026-08-17: https://alglog.org/21tkt.pdf. Supplied authoritative maintained update recording the negative answer.
• Supplied literature triage checked 2026-08-17, SOLVED-IN-LITERATURE; it reports no OCR defect, requires a separate concrete-witness check before theorem-level use, and cautions that dated triage does not exclude later or unindexed results.