Lower Bounds for the Finite Quotients of Conder's Group on the Cubic Tree (Kourovka Notebook Problem 16.46)

Problem 16.46 of the Kourovka Notebook, posed by M. Conder and going back to his paper of 1992, concerns the free product \(G=G_3*_{G_1}G_4\) of two of the Djoković–Miller groups that act arc-transitively on the cubic tree, amalgamated along the modular group \(G_1\), and the normal subgroup \(K\) of index \(8\) of \(G\), which has a presentation with six generators and \(22\) relators: does \(K\) have a non-trivial finite quotient? Conder found no proper subgroup of finite index. We do not decide the question. We prove, without computation, that \(K\) is perfect, that in every non-trivial image of \(K\) the generators keep their orders and the symmetric group \(S_4\) embeds, and that \(K\) is the amalgamated free product of four copies of \(L=S_4*_{D_8}S_4\) over a common subgroup \(C_3*C_3\), glued along its four sign automorphisms; a non-trivial finite quotient of \(K\) gives one of the subgroup \(J\) generated by two of these copies. By exhaustive computer searches we show that \(J\), and hence \(K\), has no non-trivial action on at most \(36\) points, so \(K\) has no subgroup of index between \(2\) and \(36\), and, with GAP, that no non-abelian finite simple group of order less than \(10^8\) receives a non-trivial homomorphism from \(J\). Hence every non-trivial finite quotient of \(K\) has order at least \(10^8\). Jambor (2015) proved, by a machine computation, that \(K\) has no quotient \(\mathrm{PSL}_2(q)\) or \(\mathrm{PGL}(2,q)\); together with a search of the twelve other simple groups of order between \(10^8\) and \(10^9\), every non-trivial finite quotient of \(K\) has order at least \(10^9\). In contrast, the amalgam of two copies differing by the twist \((A,B)\)\({}\mapsto(A^{-1},B^{-1})\) maps onto \(A_6\) and \(\mathrm{PSL}(2,7)\). The bounds are computer-assisted; the question itself remains open. This is an unrefereed note.