Math-1212 · mathematics · Group theory

Math-1212

Kourovka Notebook Problem 21.40

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Statement

Definitions and mathematical statement

For a positive integer $n$, let $\operatorname{GL}_n(\mathbb Q)$ be the group of invertible $n$ by $n$ matrices with rational entries, under matrix multiplication. Let $G$ be any subgroup, not assumed finitely generated. Its abstract automorphism group $\operatorname{Aut}(G)$ acts on its elements by $\alpha\cdot g=\alpha(g)$.

Suppose this action has finitely many orbits. In explicit terms, some finite list $g_1,\ldots,g_t$ has the property that for every $g\in G$ there are $i$ and $\alpha\in\operatorname{Aut}(G)$ with $g=\alpha(g_i)$. The automorphisms in this condition need not be induced by conjugation inside $\operatorname{GL}_n(\mathbb Q)$, and the number $t$ is not prescribed uniformly.

A group $L$ is soluble if its derived series $L^{(0)}=L$, $L^{(j+1)}=[L^{(j)},L^{(j)}]$ reaches $1$ after finitely many steps. A group is virtually soluble if it contains a soluble subgroup of finite index. The question asks whether every $G$ as above is virtually soluble: \[ |G/\operatorname{Aut}(G)|<\infty \quad\Longrightarrow\quad \exists L\le G:\ [G:L]<\infty,\ L\text{ soluble}, \] where the quotient notation on the left denotes the set of automorphism orbits, not a quotient group. The possible dimension, index and derived length may vary with $G$. The conclusion concerns one soluble subgroup of finite index in the whole group, not just solubility of selected finitely generated subgroups.

The editors mark this entry with an unverified AI solution, which their preface expressly distinguishes from a confirmed solution. The mathematical question is therefore retained without endorsing that proposed argument.

Short English statement

Is every rational linear group with finitely many orbits of elements under its abstract automorphisms virtually soluble?

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