{"title":"Character induced subgroups","authors":"R. Merris, W. Watkins","doi":"10.6028/JRES.077B.010","DOIUrl":null,"url":null,"abstract":"One easil y sees that Cx is the norm al s ubgroup of C co ns is ting of those ele me nts represe nted by scalars in {A (g )}. Moreove r, A = xlx(l) is a linear c haracte r on Cx whi ch is in vari a nt under conjugation b y eleme nts of C . we call C x the subgroup induced by X. THEOREM 1 : We have X (1) 2 :s:; [G : Gx], with equality if and only if X is the only irreducible character of G whose restriction to Gx contains A as a component. PROOF: Let '11.(; be the charac te r of C induced b y A. The n, b y the Froben ius Reciprocity Theore m, X occurs in '11.(; exactly x(l) tim es. Moreover , if'Y/ is a n irreduc ible charac te r on C whose restri ction to C x conta ins A, th en 'Y/ E '11. (;. But, the degree of '11.. (; is [G : G xl W e might point out th a t if'Y/ is an irredu c ible charac ter on C such that A E 'Y/ I Cx, then 'Y/ I Cx = 'Y/ (1) '11.[8 , p. 53J. In partic ular , Cx C CT) . COROLLARY 1: Let g E G be arbitrary. Then A can be extended to a character of < Gx, g), the group generatedbyG x and g.lfX(I )2 = [ G: Gx > 1,thenA cannot beextendedtoG. PROOF: T he firs t s ta te me nt follows because Gx is normal and A is invariant. The second follows from Theore m l. W e now give another proof of Theore m 1 whic h leads to an appa re ntl y diffe re nt case of equality. First, define th e support of X to be supp X = {g E C : X(g ) 0/= O}. TH EOREM 1': We have X (l) 2 :s:; [G : Gx], with equality if and only if supp X = Gx.","PeriodicalId":166823,"journal":{"name":"Journal of Research of the National Bureau of Standards, Section B: Mathematical Sciences","volume":"27 1","pages":"0"},"PeriodicalIF":0.0000,"publicationDate":"1973-07-01","publicationTypes":"Journal Article","fieldsOfStudy":null,"isOpenAccess":false,"openAccessPdf":"","citationCount":"5","resultStr":null,"platform":"Semanticscholar","paperid":null,"PeriodicalName":"Journal of Research of the National Bureau of Standards, Section B: Mathematical Sciences","FirstCategoryId":"1085","ListUrlMain":"https://doi.org/10.6028/JRES.077B.010","RegionNum":0,"RegionCategory":null,"ArticlePicture":[],"TitleCN":null,"AbstractTextCN":null,"PMCID":null,"EPubDate":"","PubModel":"","JCR":"","JCRName":"","Score":null,"Total":0}
引用次数: 5
Abstract
One easil y sees that Cx is the norm al s ubgroup of C co ns is ting of those ele me nts represe nted by scalars in {A (g )}. Moreove r, A = xlx(l) is a linear c haracte r on Cx whi ch is in vari a nt under conjugation b y eleme nts of C . we call C x the subgroup induced by X. THEOREM 1 : We have X (1) 2 :s:; [G : Gx], with equality if and only if X is the only irreducible character of G whose restriction to Gx contains A as a component. PROOF: Let '11.(; be the charac te r of C induced b y A. The n, b y the Froben ius Reciprocity Theore m, X occurs in '11.(; exactly x(l) tim es. Moreover , if'Y/ is a n irreduc ible charac te r on C whose restri ction to C x conta ins A, th en 'Y/ E '11. (;. But, the degree of '11.. (; is [G : G xl W e might point out th a t if'Y/ is an irredu c ible charac ter on C such that A E 'Y/ I Cx, then 'Y/ I Cx = 'Y/ (1) '11.[8 , p. 53J. In partic ular , Cx C CT) . COROLLARY 1: Let g E G be arbitrary. Then A can be extended to a character of < Gx, g), the group generatedbyG x and g.lfX(I )2 = [ G: Gx > 1,thenA cannot beextendedtoG. PROOF: T he firs t s ta te me nt follows because Gx is normal and A is invariant. The second follows from Theore m l. W e now give another proof of Theore m 1 whic h leads to an appa re ntl y diffe re nt case of equality. First, define th e support of X to be supp X = {g E C : X(g ) 0/= O}. TH EOREM 1': We have X (l) 2 :s:; [G : Gx], with equality if and only if supp X = Gx.