Generating triples of conjugate involutions for finite simple groups (Q6633364)

Let \(G\) be a finite non-abelian simple group. By [\textit{G. Malle} et al., Geom. Dedicata 49, No. 1, 85--116 (1994; Zbl 0832.20029)], if \(G \not \simeq U_{3}(3)\), then \(G\) is generated by three involutions. Moreover, \(U_{3}(3)\) is a true exception in that at least four involutions are needed to generate it (see [\textit{A. Wagner}, Boll. Unione Mat. Ital., V. Ser., A 15, 431--439 (1978; Zbl 0401.20038)]). The work of many authors shows that most finite simple groups are generated by two elements of orders \(2\) and \(3\) and it has been conjectured that the only exceptions are the groups: \(A_{6}\), \(A_{7}\), \(A_{8}\), \(M_{11}\), \(M_{22}\), \(M_{23}\), \(\mathsf{McL}\), \(^{2}\!B_{2}(2^{2n+1})\), \(S_4(2^{n})\), \(S_{4}(3^{n}\), \(L_{3}(4)\), \(U_{3}(3)\), \(U_{3}(5)\), \(U_{4}(3)\), \(U_{5}(2)\), \(D_{4}(2)\), \(D_{4}(3)\).\N\NIn the paper under review, the authors address the problem of seeing when \(G\) can be generated by three involutions that are conjugate to each other. They prove that, in the hypothesis of the conjecture, only the groups \(U_{3}(3)\) and \(A_{8}\) are not generated by three conjugate involutions.