Kourovka Notebook Problem 21.92

Definitions and mathematical statement

Let $G$ be a finite group and $p$ a prime. Over an algebraically closed field $k$ of characteristic $p$, let $\ell_p(G)$ be the number of isomorphism classes of simple $kG$-modules. A module is simple if it is nonzero and has no proper nonzero $G$-invariant linear subspace. This number is also the number of irreducible $p$-Brauer characters of $G$.

A $p'$-subgroup of $G$ means a subgroup $H\le G$ whose order is not divisible by $p$. For any finite group $H$, write $k(H)$ for the number of its conjugacy classes, where conjugation is performed by elements of $H$ itself. Define \[ M_p(G)=\max\{k(H):H\le G,\ p\nmid|H|\}. \] This maximum exists because $G$ has finitely many subgroups and the trivial subgroup is allowed.

The conjecture is the universal inequality $\ell_p(G)\le M_p(G)$. Equivalently, for each pair $(G,p)$, at least one $p'$-subgroup $H$ should have as many internal conjugacy classes as $G$ has irreducible $p$-Brauer characters. The subgroup is allowed to depend on both $G$ and $p$. It need not be normal, and the assertion does not require a subgroup whose order equals the full $p'$-part of $|G|$. Repeated modular character degrees still represent distinct characters in $\ell_p(G)$.

Short English statement

Can some subgroup of order prime to $p$ always supply at least as many conjugacy classes as there are irreducible $p$-Brauer characters of the whole finite group?

Sources

• S1 ULAM.AI and the UnsolvedMath Contributors, supplied problems.json, record 2601 (KOU-21.92), 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 21.92. Expanded formulation of the supplied catalogue statement. https://arxiv.org/pdf/1401.0300v47.