On the number of conjugates defining the solvable radical of a finite group.
The authors want to characterize the solvable radical of a finite group by finding the smallest integer \(k\) such that any \(k\) conjugates of \(G\) generate a solvable subgroup. It is clear that \(4\) is a lower bound (e.g., consider \(g\) to be a transposition in \(S_n\), \(n\geq 5\)). They also consider a similar problem of the subgroup generated by \(k\) commutators of the form \([g,b_i]\), \(1\leq i\leq k\). In fact, Flavell, in a preprint has shown that \(10\) conjugates suffice to characterize the solvable radical. His proof does not use the classification of finite simple groups. \textit{Guest} [A solvable version of the Baer-Suzuki theorem and generalizations, USC Ph.D. Thesis] has proved an analog of the Baer-Suzuki theorem -- more precisely, he has shown that if \(G\) is a finite group, and \(x\) has prime order at least \(5\), then \(x\) is in the solvable radical of the finite group if and only if every two conjugates of \(x\) generate a solvable group. Using this and other results in the thesis, it is not hard to show that \(4\) conjugates do suffice to test whether an element in a finite (or linear) group is in the solvable radical.
scientific article; zbMATH DE number 5080176
scientific article