Kourovka Notebook Problem 21.81

Definitions and mathematical statement

For an integer $n\ge3$, let $F_n=\langle x_1,\ldots,x_n\rangle$ be the free group on $n$ generators: arbitrary choices of $n$ elements of any group extend uniquely to a homomorphism from $F_n$. Let $\Gamma$ be a finite simple group, meaning a nontrivial finite group with no normal subgroups other than $1$ and itself. Define \[ \mathcal N_n(\Gamma)=\{N\trianglelefteq F_n:F_n/N\cong\Gamma\}. \] These are exactly the kernels of surjective homomorphisms $F_n\to\Gamma$. The automorphism group $\operatorname{Aut}(F_n)$ acts on this set by sending $N$ to $\alpha(N)$.

The conjecture is that this action is transitive for every such $n$ and $\Gamma$. Explicitly, for any $N_1,N_2\in\mathcal N_n(\Gamma)$, there should exist $\alpha\in\operatorname{Aut}(F_n)$ with $\alpha(N_1)=N_2$.

Equivalently, given any two epimorphisms $\varphi,\psi:F_n\to\Gamma$, there should be $\alpha\in\operatorname{Aut}(F_n)$ and $\beta\in\operatorname{Aut}(\Gamma)$ such that $\psi=\beta\circ\varphi\circ\alpha$. Epimorphisms with the same kernel differ by an automorphism of their common quotient, explaining the appearance of $\beta$. Accordingly the proposed transitivity is on quotient kernels, with no chosen identification of the quotient with $\Gamma$. The stated range begins at rank three and imposes no upper bound on the order of $\Gamma$.

Short English statement

For every finite simple group, are all its quotient kernels in a free group of rank at least three equivalent under free-group automorphisms?

Sources

• S1 ULAM.AI and the UnsolvedMath Contributors, supplied problems.json, record 2590 (KOU-21.81), 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 21.81. Expanded formulation of the supplied catalogue statement. https://arxiv.org/pdf/1401.0300v47.