On the Wielandt length of a finite supersoluble group (Q2708150)

The Wielandt subgroup of a group \(G\) is the subgroup consisting of those elements which normalize each subnormal subgroup of \(G\). It is denoted by \(\omega(G)\). Wielandt showed that in a finite group \(G\), \(\omega(G)\) contains any minimal normal subgroup of \(G\), hence \(\omega(G)\neq 1\). The ascending Wielandt series of \(G\) is defined inductively as follows: \(\omega_0(G)=1\), and for \(i\geq 1\), \(\omega_i(G)/\omega_{i-1}(G)=\omega(G/\omega_{i-1}(G))\).NEWLINENEWLINENEWLINEIn a finite group \(G\) there is a smallest \(n\) such that \(\omega_n(G)=G\), this \(n\) is the Wielandt length of \(G\).NEWLINENEWLINENEWLINEThe author studies the Wielandt subgroup of a semidirect product of two finite groups of coprime order to obtain the following: If \(G\) is a finite supersoluble group and \(n\) is the maximum of the nilpotency classes of the Sylow subgroups of \(G\), then \(G\) has Wielandt length bounded by \(n+1\). Moreover this bound is best possible.