From Thompson to Baer-Suzuki: a sharp characterization of the solvable radical. (Q975094)

The classical Baer-Suzuki theorem states that the nilpotent radical of a finite group \(G\) coincides with the collection of \(g\in G\) such that for every \(x\in G\) the subgroup generated by \(g\) and \(g^x\) is nilpotent. The following problem is parallel to the Baer-Suzuki theorem: what is the minimal number \(k\) such that \(g\) belongs to the solvable radical \(R(G)\) if and only if \(\langle g^{x_1},\dots,g^{x_k}\rangle\) is solvable for every \(x_1,\dots,x_k\) in \(G?\) In a previous paper the authors proved that \(R(G)\) coincides with the collection of \(g\) in \(G\) with the property that the subgroup generated by every four conjugates of \(g\) is solvable [J. Pure Appl. Algebra 213, No. 2, 250-258 (2009; Zbl 1163.20010)]. However one can expect a precise analogue of the Baer-Suzuki theorem to hold for the elements of prime order greater than \(3\) in \(R(G)\). The main result of the paper confirms this expectation: an element \(g\) of prime order \(l>3\) belongs to the solvable radical \(R(G)\) of a finite group \(G\) if and only if for every \(x\in G\) the subgroup \(\langle g,g^x\rangle \) is solvable. This implies that a finite group \(G\) is solvable if and only if in each conjugacy class of \(G\) every two elements generate a solvable subgroup. The proof uses the classification of finite simple groups: a standard argument reduces the proof to the case of almost simple groups. -- The authors add the information that similar results were independently proved in forthcoming works by Guest, Guralnick and Flavell.