Show simple item record

dc.contributor.authorJespers, Eric 
dc.contributor.authorRío Mateos, Ángel del 
dc.contributor.authorRuiz Marín, Manuel 
dc.date.accessioned2009-05-28T08:28:15Z
dc.date.available2009-05-28T08:28:15Z
dc.date.issued2002-09
dc.identifier.citationJESPERS, Eric, RÍO, Ángel, de, RUIZ, Manuel. Groups generated by two bicyclic units in integral group rings. Journal of Group Theory, 5 (4): 493–511, 17 Septiembre 2002. ISSN 1433-5883es
dc.identifier.issn1433-5883
dc.description.abstractIn [5] Ritter and Sehgal introduced the following units, called the bicylic units, in the unit group U(ZG) of the integral group ring ZG of a finite group G: ¯a;g = 1 + (1 ¡ g)abg; °a;g = 1 + bga(1 ¡ g); where a; g 2 G and bg is the sum of all the elements in the cyclic group hgi. It has been shown that these units generate a large part of the unit group of ZG. Indeed, for most finite groups G, the bicyclic units together with the Bass cyclic units generate a subgroup of finite index in U(ZG) [3, 6]. The Bass cyclic units are only needed to cover a subgroup of finite index in the centre and the group B generated by the bicyclic units contains a subgroup of finite index in a maximal Z-order of each non-commutative simple image Mn(D) of the rational group algebra QG. In particular, if n > 1, then B contains a subgroup of finite index in SLn(O), where O is a maximal order in D; and hence B contains free subgroups of rank two. A next step in determining the structure of U(ZG) is to investigate relations among the discovered generators. Presently this is beyond reach. Hence a more realistic goal is to study the structure of the group generated by two bicyclic units. In [4] Marciniak and Sehgal proved that if ¯a;g is a non trivial unit in ZG (here G is not necessarily finite) then the group h¯a;g; °a¡1;g¡1i is free of rank 2. Clearly, bicyclic units are of the form 1 + a with a2 = 0. Salwa, in [7], used the ideas of Marciniak and Sehgal to prove that if x and y are two elements of an additively torsion-free ring such that x2 = y2 = 0 and xy is not nilpotent then h(1 + x)m; (1 + y)mi is free of rank 2 for some positive integer m. In particular, if b1 and b2 are two bicyclic units and (b1 ¡ 1)(b2 ¡ 1) is not nilpotent, then hbm 1 ; bm 2 i is free of rank 2 for some positive integer m. In this paper we investigate the minimum positive integer m so that hbm 1 ; bm 2 i is free provided that b1 and b2 are two bicyclic units so that (b1 ¡1)(b2 ¡1) is not nilpotent. We prove the following theorem which indicates that if b1 and b2 are of the same type then frequently m = 1.es
dc.description.sponsorshipThe first author has been partially supported by the Onderzoeksraad of Vrije Universiteit Brussel and the Fonds voor Wetenschappelijk Onderzoek (Vlaanderen) and the second by the D.G.I. of Spain and Fundación Séneca of Murcia.We would like to express our gratitude to Victor Jiménez for some helpful conversation on inequality .es
dc.formatapplication/pdf
dc.language.isoenges
dc.publisherWalter de Gruyteres
dc.rightshttp://www.reference-global.com/doi/abs/10.1515/jgth.2002.018es
dc.titleGroups generated by two bicyclic units in integral group ringses
dc.typeinfo:eu-repo/semantics/articlees
dc.subject.otherEconomía Aplicadaes
dc.subjectBicylic unitses
dc.subjectBass cyclic unitses
dc.subjectStructure of the groupes
dc.subjectDihedral groupes
dc.subjectUnidades bicíclicases
dc.subjectUnidades cíclicas de Basses
dc.subjectEstructuras de grupoes
dc.subjectGrupo diédricoes
dc.identifier.urihttp://hdl.handle.net/10317/999
dc.identifier.doi10.1515/jgth.2002.018


Files in this item

This item appears in the following Collection(s)

Show simple item record