A counterexample to the unit conjecture for group rings

On noncommutative bounded factorization domains and prime rings

“…We write supp(f ) := { s ∈ S : a s = 0 } for the support of f . In the noncommutative setting, in general, the problem of characterising (semi)group algebras that are domains is a very hard one, known as Kaplansky's zero-divisor conjecture [Pas77, Chapter 13], as is the related characterisation of units (with a recent counterexample to the unit conjecture given by Gardam [Gar21]). To avoid these issues, we work with right orderable monoids.…”

Complex group rings and group C$$^*$$-algebras of group extensions

“…Theorem 3.4 with N = {e}). Example 3.9 Let P be Passman's fours group [5,11], which is a non-split extension 1 → Z 3 → P → Z/2Z × Z/2Z → 1. Note that P is torsion-free.…”

“…First of all, we notice that G is not a unique product group because P is not (cf. [11]). Indeed, there are non-empty finite subsets A, B of P witnessing that P is not a unique product group.…”

“…We emphasize that it is not known whether every unit in C[P] is trivial (cf. [11]). By Theorem 3.2, C[P] ∼ = C[Z 3 ] × ( S, ω) (Z/2Z × Z/2Z), meaning that we may study the units of C[P] inside the crossed product on the right-hand side.…”

“…The unit problem has been answered affirmatively for special classes of groups (see e.g. [5,26]), but in 2021 G. Gardam [11] gave an example of a group ring K [P] possessing non-trivial units, where K is the field of two elements and P is Passman's fours group. Building on Gardam's ideas, further counterexamples to the unit conjecture in positive characteristic were provided by A. G. Murray [23].…”