Groups with faithful irreducible projective unitary representations. (Q2836444)

A consequence of Schur's lemma is that a finite group has a faithful irreducible complex representation only if its center is cyclic. A general criterion was given by Gaschütz in 1954. He showed that a finite group has a faithful irreducible representation over an algebraically closed field of characteristic zero if and only if the direct product of all minimal normal subgroups is generated by a single conjugacy class of the group. In a beautiful earlier paper in the Commentarii Mathematici Helvetici [83, No. 4, 847-868 (2008; Zbl 1154.22005)], the authors had -- apart from other things -- generalized Gaschütz's criterion to the case of faithful irreducible unitary representations of countably infinite groups.NEWLINENEWLINE In the present paper, the authors build on their earlier effort by considering projective unitary representations. Once again, the groups considered are countable as they use measure theory and direct integrals. Indeed, the criteria are invalid at times when the group is uncountable. Not surprisingly, the notion of capability plays a role -- a group is capable if it is isomorphic to the quotient group of a group by its center. The discussion depends on the study of multipliers (or \(2\)-cocyles) with values in the unit circle group \(T\). Another notion which plays a role is that of a group having the property FAb -- a group in which every normal subgroup generated by a single conjugacy class has a finite Abelianization.NEWLINENEWLINE One of the main results asserts: Let \(\Gamma\) be a countable group and \(\zeta\colon\Gamma\times\Gamma\to T\) be a \(2\)-cocycle. Then, if \(\Gamma\) has an irreducible projective faithful unitary representation corresponding to the cocycle \(\zeta\), then so does the minisocle of \(\Gamma\) (the subgroup generated by all finite, minimal normal subgroups). Further, the latter condition holds if and only if it holds for the subgroup generated by all finite Abelian normal subgroups. Finally, if \(\Gamma\) has FAb, then all three conditions are equivalent.NEWLINENEWLINE The paper is very pleasant to read with a wealth of examples and remarks.