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.(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.

2 Thesymbol shouldbereadas forall . Thesymbol shouldbereadas thereexists .Thesymbol !shouldbereadas thereexistsunique .LetSandTbetwosets. IfsisanobjectcontainedinSthenwesaythatsi sanelement, Z. (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! (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.

3 (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.)If U 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. (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 !

4 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? ,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].

5 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, ). ,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.

6 (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). (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).Hencef(x 1)=(f(x)) ,CosetsandLagrange sTheoremInlinearalgebra, (G, ) ,y H x y H x 1 N,thenthesubsetmZ:={ma|a Z}isasubgroupof(Z,+).

7 ,K Gsubgroups H K ,Ksubgroups,e Hande K e H ,y H K x y Handx y K x y H H K x 1 Handx 1 K x 1 H (G, )beagroupandletH :Givenx,y G,x y x 1 y :1.(Reflexive)e H x 1 x H x G x x2.(Symmetric)x y x 1 y H (x 1 y) 1 H y 1 x H y x3.(Transitive)x y,y z x 1 y,y 1 z H (x 1 y) (y 1 z) H x 1 z H x Gtheequivalenceclass(orleftcoset)contain ingxequalsxH:={x h|h H} y x 1 y H x 1 y=hforsomeh H y=x h {Equivalenceclasscontainingx} xH y=x hforsomeh H x 1 y H y {Equivalenceclasscontainingx}. ,y G,xH=yH x 1 y x y x 1 y xH yH= , {e} , (G, )beagroupandH (G:H)=|G/H|, N,thesubgroupmZ (ofsets) :H xHh x ,h H, (h)= (g) x h=x g h= xH h Hsuchthatg=x h g= (h).Nowlet (G, )beafinitegroupandH x G,|xH|=|H|. (G, )beafinitegroupandH |H|divides|G|. , (clearlynon-uniqueingeneral)x |H|.HencewehavepartitionedGintosubsetsea chofsize|H|.Weconcludethat|H|divides|G|. (G, ) {e}. |H| |H|=1or|H|= {e}.InthesecondcaseH= (X) (X) GisasubgroupsuchthatX Hthengp(X) (X) (Z,+)andX={1} (X)= generated (G, )isfinitelygeneratedif X Gfinitesuchthatgp(X)= (Q\{0}, ) (G, )issaidtobecyclicif x Gsuchthatgp({x})=G, {xn|n Z}.

8 Bytheaboveobservations(Z,+)and(Z/mZ,+) ({x}) Gisnon-trivialandbyLagrange ({x}). (notnecessarilycyclic).Forr,s Zandx G,xrxs=xr+s=xs+r= ({x}) ,G =(Z,+) |G|=m N,thenG =(Z/mZ,+) ({x}),thenG={ x 2,x 1,e,x,x2 }.Assumeallelementsinthissetaredistinct, thenwecandefineamapofsets: :G Zxn nThen, a,b Z, (xa xb)= (xa+b)=a+b= (xa)+ (xb)so ,(G, )isisomorphicto(Z,+). a,b Z,b>asuchthatxa= (b a)=e x 1=x(b a 1) G={e, ,xb a 1}. Nsuchthatxm= {e,x, ,xm 1} |G|= :G Z/mZxn [n]forn {0,..m 1}Thisisclearlyasurjection,henceabijecti onbecause|G|=|Z/mZ|= a,b {0,..,m 1}weknow (xa xb)= (xa+b)=[a+b]=[a]+[b]= (xa)+ (xb) (G, )isisomorphicto(Z/mZ,+). (G, ) =(Z,+).LetH Nminimalsuchthatm H(m =0).HencemZ={ma|a Z} n Hsuchthatn/ ,n=qm+r,r,q Zand0<r<m r ({m})=mZ (G, ) =(Z/mZ,+).LetH ,choosen Nminimalandpositivesuchthat[n] ({[n]}) (G, ) Nsuchthatmdivides|G|. |G|=dweknowthatG =(Z/dZ,+). ({[n]}) Z/dZisasecondsubgroupofordermthenbytheab oveproofweknowthattheminimaln Nsuchthat[n] Hmustben= ({[n]}).Let(G, )beagroup(notnecessarilycyclic)andx ({x}) |gp({x})|< wesaythatxisoffiniteorderanditsorder,wri ttenord(x)equals|gp({x})|.

9 Gisoffiniteorder,thenord(x)=minimalm Nsuchthatxm= (G, )beafinitegroupandx (x)divides|G|andx|G|= (x)=|gp({x})|.Therefore,byLagrange stheorem,ord(x)mustdivide|G|.Alsonotetha tbydefinitionxord(x)= |G|=x(ord(x) |G|ord(x))=e|G|ord(x)= ,denoted (S), , f,g (S)ands S(f g)(s)=f(g(s)). , x S,e(s)= := ({1,2,..,n}).IfSisanysetofcardinalitynth en (S)isisomorphictoSymn,theisomorphismbein ginducedbywritingabijectionfromSto{1,2,. .,n}. (S)wecanthinkabout as moving (S)naturally acts (G, ) (G, )onSwemeanamap: :G S Ssuchthat1. x,y G,s S, (x y,s)= (x, (y,s))2. (e,s)=s20 Iftheactionofthegroupisunderstoodwewillw ritex(s)= (x,s) x G,s :(1)becomes(x y)(s)=x(y(s)) x,y G,s Sand(2)becomese(s)=s s (S)onS: : (S) S S(f,s) f(s) (G, ) : :G G G(x,y) x yProperty(1)holdsas (2)holdsbecausee x=x x :G S S(g,s) s s S,g : :G G G(x,y) x 1 y xProperty(1)holdsbecauseofassociativityo f andthat(g h) 1=h 1 g (2) :G S Gnaturallygivesrisetoamap: g:S Ss& g(s)Observethatproperty(1)of beinganactionimpliesthat g h= g h g,h (2)tellisthat e= g,ifwecanfindaninversefunction, g 1isinverseto givesrisetoamapofsets: :G (S)g gProposition.

10 ,property(1)of beinganaction h g= h g h,g :G (S). :G (S)g' :G (S).Cayley (G).Inparticularif|G|=n N, :G (G) ,s G, g(s)=g ,g G,suppose h= s=g s s G h= (G, )beagroup,togetherwithanaction t g Gsuchthatg(s)= , (G, )beagroup,togetherwithanaction ,andwewriteOrb(s):={t S| g Gsuchthatg(s)=t} Sfortheequivalenceclasscontainings (s) (G, )beagroup,togetherwithanaction , istransitiveifgivens,t S, g Gsuchthatg(s)= (S) (thegroupwithoneelement) (e)={e}. (G, )beagroup,togetherwithanaction (s)={g G|g(s)=s} GForthisdefinitiontomakesensewemustprove thatStab(s) (s) (s)=s e Stab(s) ,y Stab(s) (x y)(s)=x(y(s))=x(s)=s x y Stab(s). Stab(s) x 1(s)=x 1(x(s))=(x 1 x)(s)=e(s)=s x 1 Stab(s)ThuswemayformtheleftcosetsofStab( s)inG:G/Stab(s):={xStab(s)|x G}.RecallthatthesesubsetsofGaretheequiva lenceclassesfortheequivalencerelation:Gi venx,y G,x y x 1 y Stab(s), ,y GthenxStab(s)=yStab(s) x(s)=y(s). x 1y Stab(s).Hencex 1y(s)= (s)=y(s).Wededucethatthereisawelldefined map(ofsets): :G/Stab(s) Orb(s)xStab(s) x(s)Proposition.