Introduction to Abstract Algebra (Math 113)

1 IntroductiontoAbstractAlgebra(Math113) +and ,CosetsandLagrange (ProofsOmitted).. , , (ProofsOmitted).. ,themostlikelyresponsewillbe: Somethinghorribletodowithx,yandz .Ifyou reluckyenoughtobumpintoamathematicianthe nyoumightgetsomethingalongthelinesof: Algebraistheabstractencapsulationofourin tuitionforcomposition .Bycomposition, , , ,evenasalittlebaby. N:={1,2, }, +and . Z:={.. 2, 1,0,1,2,..}, +and .AdditiononZhasparticularlygoodpropertie s, 2Q:={ab|a,b Z,b =0}, +and .Thistime,multiplicationishasparticularl ygoodproperties, ,therealnumbersandthenC, +and , +and +b=b+aforalla,b Q,ora (b+c)=a b+a cforalla,b,c (setswithextrastructure),ofwhichZandQare definitivemembers.

2 (Z,+) Groups(Z,+, ) Rings(Q,+, ) FieldsInlinearalgebratheanalogousideais( Rn,+,scalarmultiplication) ,alltheprotonsonEarth,everythoughtyou veeverhad,N,Z,Q,R, +and .Inthiswholecourse, ,butitiscrucialweallunderstandthefollowi ng:3 IfPandQaretwostatements,thenP :xodd x = IfP QandQ PthenwewriteP Q,whichshouldbereadasPistrueifandonlyifQ istrue. Thesymbol shouldbereadas forall . Thesymbol shouldbereadas thereexists .Thesymbol !shouldbereadas thereexistsunique .LetSandTbetwosets. IfsisanobjectcontainedinSthenwesaythatsi sanelement, Z.

3 (orsize)by|S|. {NotationforelementsinS|Propertieswhichs pecifiesbeinginS}.Theverticalbarshouldbe readas suchthat .Forexample,ifSisthesetofallevenintegert henS={x Z|2dividesx}.Wecanalsousethecurlybracket notationforfinitesetswithoutusingthe| ,thesetSwhichcontainsonly1,2and3canbewri ttenasS={1,2,3}. IfeveryobjectinSisalsoanobjectinT, TandT S S= T. IfS TthenT\S:={x T|x/ S}.T\SiscalledthecomplimentofSinT. T. T. S T={(a,b)|a S,b T}.Wecallthisnewsetthe(cartesian) .WesaythatSandTaredisjointifS T= .Theunionoftwodisjointsetsisoftenwritten asS!

4 (orfunction) :f:S Tx$ f(x) ,f:N Na$ Z,T=Z,f:Z Z Z(a,b)$ a+ ,observethatcalculusisjustthestudyofcert ainclassesoffunctions(continuous,differe ntiableorintegrable) ,andf:S (x)=x, x (x)=f(y) x=y x,y T,thereexistsx Ssuchthatf(x)= ,SandTaresetsandg:R Sandf:S Taremapsthenwemaycomposethemtogiveanewfu nction:f g:R g=IdTandg f= ,inZwecouldsaythatx,y Zarerelatedifx , S Ssatisfying:1.(x,y) U (y,x) U.(Thisiscalledthesymmetricproperty.)2. x S,(x,x) U.(Thisiscalledthereflexiveproperty.) ,y,z S,(x,y) Uand(y,z) U (x,z) U.(Thisiscalledthetransitiveproperty.)

5 IfU S Sisanequivalencerelationthenwesaythatx,y Sareequivalentifandonlyif(x,y) ,wewritex [x]:={y S|y x} [x]. [x]ifandonlyif[y]=[x]. [x] {Xi} {Xi}formsapartitionofSifeachXiisnon-empt y, , +and +and inthefollowingsettheoreticway:+:Z Z Z(a,b)# a+b :Z Z Z(a,b)# a bHereare4elementarypropertiesthat+satisf ies: (Associativity):a+(b+c)=(a+b)+c a,b,c Z (Existenceofadditiveidentity)a+0=0+a=a a Z. (Existenceofadditiveinverses)a+( a)=( a)+a=0 a Z (Commutativity)a+b=b+a a,b satisfy: (Associativity):a (b c)=(a b) c a,b,c Z (Existenceofmultiplicativeidentity)a 1=1 a=a a Z.

6 (Commutativity)a b=b a a,b +and interactbythefollowinglaw: (Distributivity)a (b+c)=(a b)+(a c) a,b,c llsimplifythenotationformultiplicationto a b= +and :Givena Q\{0}, b Qsuchthatab=ba= , (andQ) :a,b Zsuchthatab=0 eithera=0orb= :CancellationLaw:Fora,b,c Z,ca=cbandc =0 a= ,b c Zsuchthatb= |bandsaythataisadivisor(orfactor) (or 1) ,b ,denotedHCF(a,b), ,b ZaresaidtobecoprimeifHCF(a,b)= (300BC),whichI ,b Z,ifb>0then !q,r Zsuchthata=bq+rwith0 r< ,b Z, u,v Zsuchthatau+bv=HCF(a,b).Inparticular,aandbarecoprimeifanonlyifthereexistu,v Zsuchthatau+bv= ,b |ab p|aorp| ,a,greaterthan1canbewrittenasaproductofprimes:a= ,letc ,hencec=c1c2wherec1,cc N,c1<candc2< (uptoordering) | sLemmaweknowthatp1| ,sop1= < :1=qr+ , + > ,cisdivisiblebyatleastoneprime, (d )=c Qcanbewrittenuniquely(uptoreordering)int heform:a=p 11 p nn;piprimeand i Qcanbewrittenuniquelyintheform:a= , , (+and ) +and whichsatisfythesameelementaryproperties?

7 ,ifa Z, !q,r Zsuchthata=qm+rand0 r< :a b aandbhavethesameremaindermodulom m|(a b) ! ,b Zarecongruentmodulom m|(a b).Thiscanalsobewritten:a , []:Z Z/mZa+ [a](1) {[0],[1],..[m 1]}.Thefollowingresultallowsustodefine+a nd , a,b,a ,b Z:[a]=[a ]and[b]=[b ] [a+b]=[a +b ]and[ab]=[a b ]. [a] [b]=[a b] a,b Z[a]+[b]=[a+b] a,b , ,hence+and +and onZ/mZisliftedfromZ,hencetheysatisfythee ightel-ementaryproperitesthat+and [0] Z/mZbehaveslike0 Z:[0]+[a]=[a]+[0]=[a], [a] Z/mZ;and[1] Z/mZbehaveslike1 Z:[1] [a]=[a] [1]=[a], [a] [a] Z/mZisnon-zeroif[a] =[0].

8 Eventhough+and onZ/mZsharethesameelementarypropertieswi th+and onZ, ,noticethat[1]+[1]+[1]+ +[1](mtimes)=[m]=[0]Hencewecanadd1(inZ/m Z)toitselfandeventuallyget0(inZ/mZ).Also observethatifmiscompositewithm=rs,wherer <mands<mthen[r]and[s]arebothnon-zero( =[0])inZ/mZ,but[r] [s]=[rs]=[m]=[0] N,a Zthecongruenceax 1modmhasasolution(inZ) u,v Zsuchthatau+mv= [a] [x]=[1] [a] Z/mZhasamultiplicativeinverseif [x] Z/mZsuchthat[a] [x]=[1].Hencewededucethattheonlyelements ofZ/mZwithmuliplicativeinversearethosegi venby[a], {1,2, ,m 1} :Therearenaturallyoccuringsets(otherthan ZandQ)whichcomeequippedwithaconceptof+an d , llsee, : :G G (a,b)=a b a,b ,ifG=Zthen+and ,togetherwithafixedbinaryoperation ,weoftenwrite(G, ).

9 ,togetherwithabinaryoperation ,suchthatthefollowinghold:1.(Associativi ty):(a b) c=a (b c) a,b,c (Existenceofidentity): e Gsuchthata e=e a=a a (Existenceofinverses):Givena G, b Gsuchthata b=b a= :(Z,+),(Q,+),(Q\{0}, ),(Z/mZ,+),and(Z/mZ\{[0]}, ) (Z, ) ,agroupisamonoidinwhicheveryelementisinv ertible.(Z, ) , b=b a a,b nmatriceswithrealentries,denotedGLn(R), (G, )iscalledAbelianifitalsosatisfiesa b=b a a,b (Z,+).NoticealsothatanyvectorspaceisanAb eliangroupunderit (afunction). (G, )and(H, ) ,fromGtoH,isamapofsetsf:G H,suchthatf(x y)=f(x) f(y) x,y :G same ( :G G) (G, ),(H, )and(M, ) :G Handg:H :G ,y (x y)=g(f(x) f(y))=gf(x) gf(y).

10 (G).Thisisanaloguestothecollectionofn (G, ) ,e e =e . (G, ) Ghas2inverses,b,c :(a b)=ec (a b)=c e(c a) b=c(associativityandidentity)e b=cb=cThefirstpropositiontellsusthatweca nwritee Gwecanwritea 1 , Zanda G,wewritear= a a a(rtimes),ifr>0e,ifr=0a 1 a 1 a 1( rtimes),ifr< ,b,c c=a b c=bandc a=b a c= 1 G, (G, )and(H, )betwogroupsandf:G GandeH f(eG)=eH. f(x 1)=(f(x)) 1, x GProof. f(eG) eH=f(eG)=f(eG eG)=f(eG) f(eG).Bythecancellationlawwededucethatf( eG)=eH. Letx (eG)=f(x x 1)=f(x) f(x 1)andeH=f(eG)=f(x 1 x)=f(x 1) f(x).