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1ApplicationofTheFundamentalHomomorphismTheoremofGroupLIQian-qianLIUZhi-gangYANGLi-ying(DepartmentofMathematicsandComputerScience,GuangxiTeachersEducationUniversity,NanningGuangxi530001,P.R.China)Abstract:Thefundamentalhomomorphismtheoremisveryimportantconsequenceingrouptheory,byusingitwecanresolvemanyproblems.InthispaperweresearchesmainlyaboutthefundamentalhomomorphismtheoremappliedtodirectproductsofgroupsandgroupofinnerautomorphismsofagroupG.Keyword:TheFundamentalHomomorphismTheorem;DirectProducts;InnerAutomorphismsMR(2003)SubjectClassification:16WChineseLibraryClassification:O153.3Documentcode:AIntherealmofabstractalgebra,groupisoneofthebasicandimportantconcept,haveextensiveapplicationinthemathitselfandmanysideofmodernsciencetechnique.ForexampleTheoriesphysics,Quantummechanics,Quantumchemistry,Crystallographyapplicationareclearcertifications.Sothat,afterwestudyabstractalgebracourse,godeepintoagroundoftheoriesofresearchtohavethenecessityverymuchmore.Inthecontentsofgroup,thefundamentalhomomorphismtheoremisveryimportanttheorem,wecanuseitprovemanyproblemsaboutgrouptheory,inthispapertoproveseveralconclusionsasfollowwiththefundamentalhomomorphismtheorem:Thesecontentsareallstandardifwenottothespecialprovisionandexplained.Definition1.LetNbeasubgroupofagroupGwithsymbolN≤G,wesayNisthenormalsubgroupofGifoneofthefollowingconditionshold.Tosimplifymatters,wewriteNG.(1)aNNaforanyaG;(2)NaNa1wheneveranyaG;(3)Nana1foreveryNnandanyaG.Definition2.ThekernelofagrouphomomorphismfromGtoagroupGwithidentityeisthesetexGx.ThekernelofisdenotedbyKer.Definition3.Let12,,,mGGGbeacollectionofgroups.Theexternaldirectproductof12,,,mGGG,广西自然科学基金(0447038)资助项目2writtenas12mGGG,isthesetofallm-tuplesforwhichtheitscomponentisanelementofiG,andtheoperationiscomponentwise.Insymbols12mGGG=12{(,,,)}miiggggG,where1212(,,,)(,,,)mmggggggisdefinedtobe1122(,,,)mmggggggNoticethatitiseasilytoverifythattheexternaldirectproductofgroupsisitselfagroup.[4]Definition4.LetGbeagroupandHbeasubgroupofG.ForanyGa,thesetHhahaHiscalledtheleftcosetofHinGcontaininga.AnalogouslyHhhaHaiscalledtherightcosetofHinGcontaininga.Lemma1.[1](Thefundamentalhomomorphismtheorem)LetbeagrouphomomorphismfromGtoG.ThentheN=KeristhenormalsubgroupofG,and()GNG.Tosimplifymatters,wecallthetheoremastheFHT.Lemma2.[2]LetbeagrouphomomorphismfromGtoG.Thenwehavethefollowingproperties:(1)IfHisasubgroupofG,then()HHisasubgroupofG;(2)IfHisanormalinG,then()HHisanormalinG;(3)IfNisasubgroupofG,then1()NNisasubgroupofG;(4)IfNisanormalsubgroupofG,then1()NNisanormalsubgroupofGLemma3.[3]LetbeahomomorphismfromagroupGtoagroupG,andNG,1()NN.ThenNGNG.Lemma4.[4]LetHbeasubgroupofGandletabelongtoG,then:(1)HaHifandonlyifHa;(2)HaaHifandonlyifHaaH1.Byusingtheabovelemmaswecanobtainthefollowingmainlyresults.Theorem1.LetGandHbetwogroups.SupposeJGandKH,then()()JKGHandKHJGKJHG.Proof.Firstwewillprove()()JKGH.3ForanyHGhg,andeveryKJkj,.Wehave:11111,,.,,,,hkhgjghgkjhghgkjhg.SinceGJandHK,wecangetKJhkhgjg11,,i.e.KJhgkjhg1,,,.Thus()()JKGH.WemakeuseoftheFHTtoprovethatKJHGisisomorphictoKHJG.ThereforewemustlookforagrouphomomorphismfromHGontoKHJGanddeterminethekernelofit.Infactonecandefinecorrespondence:fGHGJGKdefinedby(,)(,)fxyJxKy.Clearly,KHJGKyJx,,theremustbeHGyx,tosatisfyKyJxyxf,,.Thus,fisonto.BecauseofJG,wehavexJJxforGx,similarly,yKKyforHy.WhenHGyxHGyx',',,,thereare','',',yyxxyxyx.ForanyHGyxyx2211,,,,wehave11221212((,)(,))(,)fxyxyfxxyy=1212(,)JxxKyy==12121212(,)(,)xJJxyKKyJxJxKyKy=11221122(,)(,)(,)(,)JxKyJxKyfxyfxy.Hence22112211,,,,yxfyxfyxyxf.ThereforefisgroupahomomorphismfromHGontoKHJGandKJ,istheidentityofKHJG.ForanyHGyx,,thenKyJxyxf,,,accordingtothepropertyofcoset,wecanget:KJKyJxyxf,,,ifandonlyifJxandKy,i.e.ker,,fxyxJyK=KJ.Nowletwelookatourproof:()()JKGH,fisagrouphomomorphismfromHGontoKHJGandthekerneloffisKJ.AccordingtotheFHT,wecangetKHJGKJHG.Theorem2.LetisagrouphomomorphismfromGontoG.IfHGandkerH,then4HGHGwhereHhhHH.Proof:AccordingtoLemma2.[2](2),weknowGH.ToestablishHGHG,wefirstlyneedtoconstructamappingfandprovefisagrouphomomorphismfromGontoHG.Wegivethemapping:fHGGdefinedbyHggfwhereg=()g.ForHGHg,sinceisasurjectionfromGtoG,wemustbefoundGgsuchthatHggf.Thusfisonto.ForarbitraryGba,,()()fababHabHHaHbHfafbThereforefisagrouphomomorphism.WewillnowshowHfker,infactweknowthatHisidentityofHG,accordingtoLemma4,wecangetthatforgH,then()fggHgHH,saykergf,sothatkerHf.Ontheotherhand,kergf,()fggHgHH,thatistosay,()gH.Moreover1()kergHHH,becauseofkerH,thereforekerfH.ThatisHfker.AccordingtotheFHT,wecanobtainHGHG.Theorem1andTheorem2applyExercise1andExercise2.Exercise1.3isnormalsubgroupof16U,4isanormalsubgroupof8Z.Sothatforany16Uxand8Zy,forafunctionf:yxyxf4,3,wehave4/3/1616:8438ZU
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