A new solvability criterion for finite groups. (Q2880388)

\textit{J. G. Thompson}, in his fundamental ``N-group'' paper of 1968 [Bull. Am. Math. Soc. 74, 383-437 (1968; Zbl 0159.30804)] has shown that a finite group is solvable if and only if every pair of its elements generates a solvable group. The authors of this interesting paper prove that solvability of a finite group is guaranteed by a seemingly weaker condition than the solvability of all its \(2\)-generated subgroups.NEWLINENEWLINE In Theorem A it is proved that if \(G\) is a finite group, then the following conditions are equivalent:NEWLINE{\parindent=7mmNEWLINE\begin{itemize}\item[(1)]\(G\) is solvable;NEWLINE\item[(2)]for all \(x,y\in G\), there exists an element \(g\in G\) such that \(\langle x,y^g\rangle\) is solvable;NEWLINE\item[(3)]for all \(x,y\in G\) of prime power order, there exists an element \(g\in G\) such that \(\langle x,y^g\rangle\) is solvable.NEWLINENEWLINE\end{itemize}} NEWLINEA fundamental tool in the proof of Theorem A is the following result (Theorem B): Let \(G\) be a finite nonabelian simple group. Then there exist distinct prime divisors \(a\), \(b\) of \(|G|\) such that, for all \(x,y\in G\) with \(|x|=a\) and \(|y|=b\), the subgroup \(\langle x,y\rangle\) is nonsolvable.NEWLINENEWLINE The methods used to derive Theorem A from Theorem B, together with a recent result of Guralnick and Malle, lead the authors to prove the following generalization of Theorem A: Let \(\mathfrak X\) be a family of finite groups that is closed under taking subgroups and quotients and forming extensions. Then a finite group is in \(\mathfrak X\) if and only if, for every pair of conjugate classes \(\mathcal C\) and \(\mathcal D\) of \(G\), there exist \(x\in\mathcal C\) and \(y\in\mathcal D\) for which \(\langle x,y\rangle\in\mathfrak X\).NEWLINENEWLINE Another nice consequence of Theorem A is Corollary D: Let \(G\) be a finitely generated linear group. Then \(G\) is solvable if and only if for all \(x,y\in G\), there exists \(g\in G\) such that \(\langle x,y^g\rangle\) is solvable.