Kourovka Notebook Problem 21.114

Definitions and mathematical statement

For any finite group $H$, let $H'=[H,H]$ be its commutator subgroup, generated by $a^{-1}b^{-1}ab$ with $a,b\in H$. Its abelianization $H/H'$ is the largest abelian quotient of $H$, and its order is the index $[H:H']$.

A finite group $G$ is weakly ab-maximal if \[ [H:H']\le[G:G']\qquad\text{for every subgroup }H\le G. \] The inequality is weak, so subgroups may have abelianizations of the same order as that of $G$. The subgroups being tested need not be normal, and each is abelianized using its own commutator subgroup.

Define $G^{(0)}=G$ and $G^{(j+1)}=[G^{(j)},G^{(j)}]$. When $G^{(d)}=1$ for some $d$, its derived length is the least such nonnegative integer; otherwise assign derived length infinity.

Do weakly ab-maximal finite groups have bounded derived length? Precisely, is there an absolute integer $D\ge0$ such that \[ G^{(D)}=1\qquad\text{for every weakly ab-maximal finite group }G? \] The same $D$ must work simultaneously for all orders, all primes dividing those orders, and all numbers of generators. A bound depending on $|G|$ or on $[G:G']$ would not answer this uniform question. The hypothesis compares only the sizes of abelian quotients of subgroups, while the conclusion bounds the number of iterations of the derived operation.

Short English statement

Is there one universal upper bound on the derived lengths of all finite groups whose subgroups have no larger abelianization?

Sources

• S1 ULAM.AI and the UnsolvedMath Contributors, supplied problems.json, record 2623 (KOU-21.114), 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 21.114. Expanded formulation of the supplied catalogue statement. https://arxiv.org/pdf/1401.0300v47.