Rational linear groups with finitely many automorphism orbits

The automorphisms of a group permute its elements, and a group is highly symmetric when they leave only finitely many orbits. For groups of rational matrices, does this symmetry force the group to be close to soluble? Two independent proposed proofs, both produced with AI assistance in September 2026, say yes and even give virtual nilpotence with an explicit bound; neither has yet been confirmed by mathematicians.

Problem

For a group $G$, let $\omega(G)$ be the number of orbits of the full automorphism group $\mathrm{Aut}(G)$ acting on $G$. (Automorphisms here are arbitrary abstract group automorphisms, not necessarily induced by conjugation of matrices.)

Let $G \le GL(n, \mathbb{Q})$ be a group of invertible rational $n \times n$ matrices with $\omega(G) < \infty$. Must $G$ be virtually soluble, that is, have a soluble subgroup of finite index?

No finite generation is assumed.

Background

The problem was posed by A. Dantas and E. de Melo in the 21st issue of the Notebook (2026). Groups with few automorphism orbits have been studied for both finite and infinite groups; for instance, Bastos, Dantas and de Melo (2020) proved that a soluble group of finite rank with $\omega(G) < \infty$ contains a torsion-free characteristic nilpotent subgroup $K$ with $G = K \rtimes H$ for a finite group $H$. The question asks whether a linearity hypothesis over $\mathbb{Q}$ similarly controls the structure, without assuming solubility.

What is known

• Status in the Notebook (marked ♯). The Notebook marks this problem with ♯: artificial-intelligence systems have proposed solutions that mathematicians have not yet confirmed, so the problem is to be regarded as unsolved. Readers are invited to confirm the solutions or find mistakes.
• The proposed solutions. Two preprints (A. Jayadevan, 14 September 2026; M. Kinyon and J. Urban, 17 September 2026) answer yes, and more strongly claim that $G$ has a torsion-free normal nilpotent subgroup $U$ of class at most $n - 1$ with $[G : U] \le (2n+1)^{d}$, where $d \le n^2$ is the dimension of the span of $G$. The argument: finitely many orbits force every element's eigenvalues to be roots of unity, so traces lie in $\{-n, \dots, n\}$; then $U$ is the intersection of $G$ with $1 + R$, where $R$ is the trace radical of the rational algebra spanned by $G$. Jayadevan's preprint mentions an accompanying Lean formalisation.
• Related results. For soluble groups of finite rank, see Bastos, Dantas and de Melo (2020) above.

What would count as progress

• Verify or refute the proposed solutions: check the two short arguments line by line (the root-of-unity step, the nilpotence of the trace radical, and the index count), and, if available, the Lean formalisation and whether its statement matches the problem.
• Sharper index bounds, or examples showing how large $[G : U]$ must be for given $n$.
• Extensions to other fields, such as number fields or fields of positive characteristic, where the trace argument needs to change.

References

1. R. Bastos, A. C. Dantas and E. de Melo, Soluble groups with few orbits under automorphisms, Geom. Dedicata 209 (2020), 119–123; arXiv:1908.01375.
2. A. Jayadevan, Bounded-index nilpotent subgroups of rational linear groups with finitely many automorphism orbits, preprint (2026), as cited in the Notebook.
3. M. Kinyon and J. Urban, Results on problems from the Kourovka Notebook, preprint, version 4 (2026), entry for Problem 21.40.

Source

Imported from The Kourovka Notebook 21.40, curated by E. I. Khukhro and V. D. Mazurov. Open according to the source as of 2026-10-20. The statement above is written in our own words; see the source for the original wording, history and discussion.