Transitive projective planes and insoluble groups. (Q2790581)

There is a long standing conjecture that a finite projective plane admitting an automorphism group \(G\) which acts transitively on the points has to be Desarguesian. The best result in this direction is due to \textit{W. M. Kantor} [J. Algebra 106, 15-45 (1987; Zbl 0606.20003)]. He showed that if \(G\) acts primitively and \(x\) is the order of the plane, then either it is Desarguesian or \(x^2+x+1\) is a prime. In the latter \(G\) is regular or a Frobenius group of order dividing \((x^2+x+1)(x+1)\) or \((x^2+x+1)x\). This result uses the classification of the finite simple groups.NEWLINENEWLINE In [Adv. Geom. 7, No. 4, 475-528 (2007; Zbl 1139.51014)], the author proved that a transitive group \(G\) on a non Desarguesian plane does not posses components. This was the first step for a proof that a transitive group on a non Desarguesian plane has to be solvable. The paper under review now is the second step. The first result is that the Sylow 2-subgroup has to be cyclic or a generalized quaternion group. This then will be used to show that in the nonsolvable case \(G/O(G)\) has a subgroup of index at most two, which is isomorphic to \(SL_2(5)\). Furthermore the order \(x\) of the plane is a square \(u^2\) and any prime divisor of \(|F(G)|\) divides \(u^2+u+1\) or \(u^2-u+1\). A nice corollary is that a finite projective plane, which admits a transitive automorphism group on the points, also admits a transitive group of odd order. (In the Desarguesian case this is the Singer cycle.) The proof of the results of this paper relies on the results mentioned above and so in particular on the classification of the finite simple groups. -- This paper also contains quite a few results about subgroups of \(GL_n(q)\) which might be of independent interest. The author claims that for proving solvability of \(G\) if \(G\) acts transitively on the points of a non Desarguesian plane one needs much more number theory.