3.7J.A.Beachy13.7HomomorphismsfromAStudyGuideforBeginner'sbyJ.A.Beachy - PDF document

3.7J.A.Beachy13.7HomomorphismsfromAStudyGuideforBeginner'sbyJ.A.Beachy,asupplementtoAbstractAlgebrabyBeachy/Blair21.FindallgrouphomomorphismsfromZ4intoZ10.Solution:AsnotedinExample3.7.7,anygrouphomomorphismfromZnintoZkmusthavetheform([x]n)=[mx]k,forall[x]n2Zn.Underanygrouphomomorphism:Z4!Z10,theorderof([1]4)mustbeadivisorof4andof10,sotheonlypossibilitiesare(([1]4)=1or(([1]4)=2.Thus([1]4)=[0]10,whichdenesthezerofunction,orelse([1]4)=[5]10,whichleadstotheformula([x]4)=[5x]10,forall[x]42Z4.22.(a)FindtheformulasforallgrouphomomorphismsfromZ18intoZ30.Solution:AsnotedinExample3.7.7,anygrouphomomorphismfromZ18intoZ30musthavetheform([x]18)=[mx]30,forall[x]182Z18.Sincegcd(18;30)=6,thepossibleordersof[m]30=([1]18)are1;2;3;6.Thecorrespondingchoicesfor[m]30are[0]30(order1),[15]30(order2),[10]30and[20]30(order3),and[5]30and[25]30(order6).(b)Chooseoneofthenonzeroformulasinpart(a),andnameit.Find(Z18)andker(),andshowhowelementsof(Z18)correspondtoequivalenceclassesof.Solution:Forexample,consider([x]18)=[5x]30.Theimageofconsistsofthemultiplesof5inZ30,whichare0;5;10;15;20;25.Wehaveker()=f0;6;12g,andusingProposition3.7.9tondtheequivalenceclassesof,weadd1,2,3,4,and5,respectively,tothekernel.Wehavethefollowingcorrespondence:f[0]18;[6]18;[12]18g !([0]18)=[0]30,f[3]18;[9]18;[15]18g !([3]18)=[15]30,f[1]18;[7]18;[13]18g !([1]18)=[5]30,f[4]18;[10]18;[16]18g !([4]18)=[20]30,f[2]18;[8]18;[14]18g !([2]18)=[10]30,f[5]18;[11]18;[17]18g !([5]18)=[25]30.23.(a)ShowthatZ11iscyclic,withgenerator[2]11.Solution:AnelementofZ11canhaveorder1,2,5,or10.Since22461and251061,itfollowsthat[2]11cannothaveorder2or5,soitmusthaveorder10.(b)ShowthatZ19iscyclic,withgenerator[2]19.Solution:SinceZ19hasorder18,theorderof[2]is2,3,6,or18.Theelement[2]19isageneratorforZ19becauseithasorder18,since22=461,23=861,and26(23)2761.(c)CompletelydetermineallgrouphomomorphismsfromZ19intoZ11.Solution:Anygrouphomomorphism:Z19!Z11isdeterminedbyitsvalueonthegenerator[2]19,andtheorderof([2]19)mustbeacommondivisorof18and10.Thustheonlypossibleorderof([2]19)is1or2,soeither([2]19)=[1]11or([2]19)=[10]11=[1]11.Intherstcase,([x]19)=[1]11forall[x]192Z19,andinthesecondcase([2]n19)=[1]n11,forall[x]19=[2]n192Z19. 3.7J.A.Beachy224.Dene:Z4Z6!Z4Z3by([x]4;[y]6)=([x+2y]4;[y]3).(a)Showthatisawell-denedgrouphomomorphism.Solution:Ify1y2(mod6),then2y12y2isdivisibleby12,so2y12y2(mod4),andthenitfollowsquicklythatisawell-denedfunction.Itisalsoeasytocheckthatpreservesaddition.(b)Findthekernelandimageof,andapplyTheorem3.7.8.Solution:If([x]4;[y]6)belongstoker(),theny0(mod3),soy=0ory=3.Ify=0,thenx=0,andify=3,thenx=2.ThustheelementsofthekernelKare([0]4;[0]6)and([2]4;[3]6).Itfollowsthatthereare24=2=12equivalenceclassesdeterminedby.Theseareinone-to-onecorrespondencewiththeelementsoftheimage,somustmapZ4Z6ontoZ4Z3.Thus(Z4Z6)==Z4Z3:25.Letnandmbepositiveintegers,suchthatmisadivisorofn.Showthat:Zn!Zmdenedby([x]n)=[x]m,forall[x]n2Zn,isawell-denedgrouphomomorphism.Solution:First,notethatisawell-denedfunction,sinceif[x1]n=[x2]ninZn,thennj(x1x2),andthisimpliesthatmj(x1x2),sincemjn.Thus[x1]m=[x2]m,andso([x1]n)=([x2]n).Next,isahomomorphismsincefor[]n;[]n2Zn,wehave([]n[]n)=([]n)=[]m=[]m[]m=([]n)([]n).26.Forthegrouphomomorphism:Z36!Z12denedby([x]36)=[x]12,forall[x]362Z36,ndthekernelandimageof,andapplyTheorem3.7.8.Solution:Thepreviousproblemshowsthatisagrouphomomorphism.Itisev-identthatmapsZ36ontoZ12,sinceifgcd(x;12)=1,thengcd(x;36)=1.ThekernelofconsistsoftheelementsinZ36thatarecongruentto1mod12,namely[1]36;[13]36;[25]36.ItfollowsthatZ36==Z12.27.ProvethatSLn(R)isanormalsubgroupofGLn(R).Solution:LetG=GLn(R)and=SLn(R).Theconditionweneedtocheck,thatgxg1foralln2andallg2G,translatesintotheconditionthatifPisanyinvertiblematrixanddet(A)=1,thenPAP1hasdeterminant1.Thisfollowsimmediatelyfromthefactthatdet(PAP1)=det(P)det(A)det(P1=det(P)det(A)1 det(P)=det(A);whichyoumayrememberfromyourlinearalgebracourseasthepropositionthatsimilarmatriceshavethesamedeterminant.Alternatesolution:Hereisaslightlymoresophisticatedproof.NotethatSLn(R)isthekernelofthedeterminanthomomorphismfromGLn()intoR(Example3.7.1inthetextshowsthatthedeterminantdenesagrouphomomorphism.)TheresultthenfollowsimmediatelyfromProposition3.7.4(a),whichshowsthatthekernelofanygrouphomomorphismisanormalsubgroup. 3.7J.A.Beachy3ANSWERSANDHINTS28.Proveordisproveeachofthefollowingassertions:(a)ThesetofallnonzeroscalarmatricesisanormalsubgroupofGLn(R).Answer:ThissetisthecenterofGLn(R),andsoitisanormalsubgroup.(b)ThesetofalldiagonalmatriceswithnonzerodeterminantisanormalsubgroupofGLn(R).Answer:Thissetisasubgroup,butitisnotnormal.31.Dene:Z15!Z15by([x]15)=[x]315,forall[x]152Z15.Findthekernelandimageof.Answer:Thefunctionisanisomorphism.32.Dene:Z15!Z5by([x]15)=[x]35.Showthatisagrouphomomorphism,andnditskernelandimage.Answer:Thefunctionisontowithker()=f[1]15;[11]15g.34.HowmanyhomomorphismsaretherefromZ12intoZ4Z3?Answer:Thereare12,since[1]12canbemappedtoanyelementofZ4Z3.36.Dene:R!Cbysetting(x)=eix,forallx2R.Showthatisagrouphomomorphism,andndker()andtheimage(R).Answer:Theimageistheunitcircleinthecomplexplane,andker()=f2kjk2Zg.37.LetGbeagroup,withasubgroupHG.Dene(H)=fg2GjgHg1=Hg.(b)LetG=S4.Find(H)forthesubgroupHgeneratedby(1;2;3)and(1;2).Answer:Inthisexample,(H)=H.38.FindallnormalsubgroupsofA4.Answer:f(1)g,=f(1);(1;2)(3;4);(1;3)(2;4);(1;4)(2;3)g,andA4arethenormalsubgroupsofA4.