Automorphisms fixing subnormalsubgroups of polycyclic groups

Abstract

We denote by $\rm{Aut}_{sn}(G)$ the set of all automorphisms that fix every subnormal subgroup of $G$ setwise. In their paper [5], Franciosi and de Giovanni began the study of $\rm{Aut}_{sn}(G)$. Other authors have also considered the structure of $\rm{Aut}_{sn}(G)$ under various restrictions on the structure of $G$ (Robinson [11], Cossey [2], Dalle Molle [4]). The inner automorphisms in $\rm{Aut}_{sn}(G)$ are precisely the inner automorphisms induced by elements of $\omega(G)$, the Wielandt subgroup of $G$. Recall that the Wielandt subgroup of a group $G$ is the set of all elements of $G$ that normalise each subnormal subgroup of $G$ and that $\zeta(G)$, the centre of $G$, is contained in $\omega;(G)$. Thus $\rm{Aut}_{sn}(G)\cap;\rm{Inn}(G)$ is isomorphic to $\omega(G)/\zeta(G)$ and some of the results obtained indicate that the structure of $\rm{Aut}_{sn}(G)$ is controlled by the structure of $\omega(G)/\zeta(G)$; for example, Robinson [11, Corollary 3] shows that, for a finite group $G,\rm{Aut}_{sn}(G)$ is insoluble if and only if $\omega(G)$ is insoluble. We shall prove a result of a similar nature here. One of the main results (Theorem B) of Franciosi and de Giovanni [5] is that, for a polycyclic group $G$, $\rm{Aut}_{sn}(G)$ is either finite or abelian. We shall show that $\rm{Aut}_{sn}(G)$ can indeed be infinite, but only if $\omega(G)/\zeta(G)$ is infinite.