On primitive representations of soluble groups of finite rank (Q2736062)

The following conjecture underlies much of the work of this paper: Let \(G\) be a linear group of finite rank over a field of characteristic zero. If \(k\) is a field of characteristic zero and \(kG\) has an irreducible module \(M\) such that \(C_G(M)=1\) and \(M\) is not induced from any subgroup of infinite index in \(G\), then \(G\) is polycyclic-by-finite. (Here, we say that \(G\) has `finite rank' \(r\) if every finitely generated subgroup of \(G\) can be generated by \(r\) elements, with \(r\) best possible.)NEWLINENEWLINENEWLINEThe main theorem presented here as evidence in favour of this conjecture is this: Let \(G\) be a non-polycyclic soluble group of finite rank satisfying the ascending chain condition for normal subgroups, let \(k\) be a field of characteristic zero and let \(M\) be an irreducible \(kG\)-module with \(C_G(M)=1\). Then there is a subgroup \(H\) of \(G\) whose torsion-free rank is less than that of \(G\), and a \(kH\)-submodule \(U\) of \(M\), such that \(M=U\otimes_{kH}kG\).NEWLINENEWLINENEWLINEThe proof requires a considerable amount of intricate analysis of modules over soluble and nilpotent groups of finite torsion-free rank building on earlier work of C.~J.~B. Brookes, the reviewer, the author himself and others. This yields a number of results of independent interest, the statements of which are too technical to be reproduced here.