Sylow numbers and nilpotent Hall subgroups.
Let \(\pi\) be a set of primes of a finite group \(G\). By the renowned Wielandt's theorem, if \(G\) has a nilpotent Hall \(\pi\)-subgroup then all Hall \(\pi\)-subgroups of \(G\) are conjugate and every \(\pi\)-subgroup is contained in some Hall \(\pi\)-subgroup of \(G\). In particular, all \(\pi\)-subgroups of \(G\) are nilpotent. The author presents a very interesting criterion for a finite group to possess nilpotent Hall \(\pi\)-subgroups. This is done by means of the Sylow numbers.NEWLINENEWLINE The main result claims that \(G\) has nilpotent Hall \(\pi\)-subgroups if and only if the three following conditions hold: (i) For any two different primes \(p,q\in\pi\), the prime \(p\) does not divide the number of Sylow \(q\)-subgroups of \(G\). (ii) If \(\{2,3\}\subseteq\pi\), then \(G\) does not have any composition factor isomorphic to \(\text{PSL}(2,q)\) with \((q^2-1)_{\{2,3\}}=24\). (iii) If \(\{2,7\}\subseteq\pi\), then \(G\) does not have any composition factor isomorphic to \(^2G_2(3^{2n+1})\) with \(n\not\equiv 3\pmod 7\). -- The proof is made first for the case \(|\pi|=2\) and then for the general case, and it depends on the recently completed classification of Hall subgroups of simple groups.
scientific article; zbMATH DE number 6221923