Kourovka Notebook Problem 21.150

Definitions and mathematical statement

An elementary abelian group is a vector space over a prime field ${\mathbb{F}}_p$, regarded additively; its rank is its vector-space dimension, finite or infinite. For a group $H$, let $Z(H)=\{h\in H:hu=uh\text{ for every }u\in H\}$ and let $H'=[H,H]$ be its derived subgroup.

The source considers an extension $1\to A\to G\to B\to1$, with $A$ and $B$ elementary abelian, and an element $a\in A$ with trivial stabilizer $C_B(a)$. Since $A$ is abelian, conjugation on $A$ factors through $G/A\cong B$, so that stabilizer is meaningful. To form the displayed subgroup $\langle a,B\rangle$ literally, however, a copy of $B$ inside $G$ is needed. In the split case $G=A\rtimes B$, as in the supplied counterexample, use its embedded complement and put $H=\langle a,B\rangle$. The proposed inequality is \[ C_B(a)=\{1\}\quad\Longrightarrow\quad \operatorname{rank}(Z(H)\cap H')\leq\operatorname{rank}(B). \] Here $H'\leq A$, so the left group is elementary abelian and its rank is defined. No restriction to finite ranks is imposed by the short statement. An arbitrary nonsplit-extension reading would require the meaning of $\langle a,B\rangle$ or a choice of lifts to be specified; the split case already falls under the intended inequality.

The supplied background records Monticone's negative answer in a letter of 21 March 2026. Let \[ R={\mathbb{F}}_3[x,y]/(x,y)^3,\qquad A=(R,+), \] and let $B\leq\operatorname{Aut}(A)$ be generated by multiplication by $1+x$ and by $1+y$. Form $G=A\rtimes B$, and take $a=1\in R$, the additive-group element represented by the constant polynomial. The source reports \[ B\cong C_3^2,\qquad C_B(a)=\{1_B\},\qquad Z(\langle a,B\rangle)\cap\langle a,B\rangle' =\langle x^2,xy,y^2\rangle_{{\mathbb{F}}_3}\cong C_3^3. \] Thus its reported ranks are 2 and 3, respectively. The stated quotient uses the cube of the ideal generated by $x,y$, not the ideal generated only by $x^3,y^3$.

Both registry and supplied literature triage dated 2026-08-17 classify the inequality solved negatively. The triage says the explicit example matches all intended hypotheses and reports an accompanying Lean formalization, a later joint paper and the original solution note. This draft reports those source claims without executing the formalization or independently checking the rank computation; proposed corrected bounds remain a separate question.

Short English statement

Does an element with trivial stabilizer in an elementary-abelian extension force the central derived subgroup of the generated group to have rank no larger than the quotient? The supplied characteristic-three split extension is reported to have quotient rank 2 and central derived rank 3, refuting the inequality.

Sources

• Local UnsolvedMath registry, permanent ID 2659, source code catalogue entry, "Kourovka Notebook Problem 21.150," from Kourovka Notebook, New Problems, 21st issue (2026), original problem 21.150; attributed to V. I. Zenkov. Includes supplied literature triage dated 2026-08-17.
• Pietro Monticone, "Solution of Kourovka Notebook Problem 21.150" (2026), associated in the background with a letter of 21 March 2026: https://kourovkanotebookorg.wordpress.com/wp-content/uploads/2026/03/solution_21_150-3.pdf.
• Wouter van Doorn, Elias Judin, Pietro Monticone, and Daniel Morrison, "On Some Problems from the Kourovka Notebook," arXiv:2607.17477 (2026): https://arxiv.org/abs/2607.17477.
• Pietro Monticone et al., Kourovka formalization repository, Problem 21.150 (2026): https://github.com/pitmonticone/Kourovka.