GroupTheoryandtheRubiksCubeJanetChen - PDF document

1 GroupTheoryandtheRubik'sCubeJanetChen AN
GroupTheoryandtheRubik'sCubeJanetChen ANotetotheReaderThesenotesarebasedona2-weekcoursethatItaughtforhighschoolstudentsattheTexasStateHonorsSummerMathCamp.Allofthestudentsinmyclasshadtakenelementarynumbertheoryatthecamp,soIhaveassumedinthesenotesthatreadersarefamiliarwiththeintegersmodnaswellastheunitsmodn.BecauseonegoalofthisclasswasacompleteunderstandingoftheRubik'scube,IhavetriedtousenotationthatmakesdiscussingtheRubik'scubeaseasyaspossible.Forexample,Ihavechosentouserightgroupactionsratherthanleftgroupactions.2 IntroductionHereissomenotationthatwillbeusedthroughout.Zthesetofintegers:::;�3;�2;�1;0;1;2;3;:::Nthesetofpositiveintegers1;2;3;:::Qthesetofrationalnumbers(fractions)RthesetofrealnumbersZ=nZthesetofintegersmodn(Z=nZ)thesetofunitsmodnThegoalofthesenotesistogiveanintroductiontothesubjectofgrouptheory,whichisabranchofthemathematicalareacalledalgebra(orsometimesabstractalgebra).Youprobablythinkofalgebraasaddition,multiplication,solvingquadraticequations,andsoon.Abstractalgebradealswithallofthisbut,asthenamesuggests,inamuchmoreabstractway!Ratherthanlookingataspecicoperation(likeaddition)onaspecicset(likethesetofrealnumbers,orthesetofintegers),abstractalgebraisalgebradonewithoutreallyspecifyingwhattheoperationorsetis.Thismaybetherstmathyou'veencounteredinwhichobjectsotherthannumbersarereallystudied!AsecondarygoalofthisclassistosolvetheRubik'scube.WewillbothdevelopmethodsforsolvingtheRubik'scubeandprove(usinggrouptheory!)thatourmethodsalwaysenableustosolvethecube.ReferencesDouglasHofstadterwroteanexcellentintroductiontotheRubik'scubeintheMarch1981issueofScienticAmerican.ThereareseveralbooksabouttheRubik'scube;myfavoriteisInsideRubik'sCubeandBeyondbyChristophBandelow.DavidSingmaster,whodevelopedmuchoftheusualnotationfortheRubik'scube,alsohasabookcalledNotesonRubik's'MagicCube,'whichIhavenotseen.Foranintroductiontogrouptheory,IrecommendAbstractAlgebrabyI.N.Herstein.Thisisawonderfulbookwithwonderfulexercises(andif

2 youarenewtogrouptheory,youshoulddolotsof
youarenewtogrouptheory,youshoulddolotsoftheexercises).Ifyouhavesomefamiliaritywithgrouptheoryandwantagoodreferencebook,IrecommendAbstractAlgebrabyDavidS.DummitandRichardM.Foote.3 1.FunctionsTounderstandtheRubik'scubeproperly,werstneedtotalkaboutsomedierentpropertiesoffunctions.Denition1.1.A functionor mapffromadomainDtoarangeR(wewritef:D!R)isarulewhichassignstoeachelementx2Dauniqueelementy2R.Wewritef(x)=y.Wesaythatyisthe imageofxandthatxisa preimageofy.NotethatanelementinDhasexactlyoneimage,butanelementofRmayhave0,1,ormorethan1preimage.Example1.2.Wecandeneafunctionf:R!Rbyf(x)=x2.Ifxisanyrealnumber,itsimageistherealnumberx2.Ontheotherhand,ifyisapositiverealnumber,ithastwopreimages,p yand�p y.Therealnumber0hasasinglepreimage,0;negativenumbershavenopreimages.vFunctionswillprovideimportantexamplesofgroupslateron;wewillalsousefunctionsto\translate"informationfromonegrouptoanother.Denition1.3.Afunctionf:D!Riscalled one-to-oneifx16=x2impliesf(x1)6=f(x2)forx1;x22D.Thatis,eachelementofRhasatmostonepreimage.Example1.4.Considerthefunctionf:Z!Rdenedbyf(x)=x+1.Thisfunctionisone-to-onesince,ifx16=x2,thenx1+16=x2+1.Ifx2Risaninteger,thenithasasinglepreimage(namely,x�1).Ifx2Risnotaninteger,thenithasnopreimage.Thefunctionf:Z!Zdenedbyf(x)=x2isnotone-to-one,sincef(1)=f(�1)but16=�1.Here,1hastwopreimages,1and�1.vDenition1.5.Afunctionf:D!Riscalled ontoif,foreveryy2R,thereexistsx2Dsuchthatf(x)=y.Equivalently,everyelementofRhasatleastonepreimage.Example1.6.Thefunctionf:Z!Rdenedbyf(x)=x+1isnotontosincenon-integersdonothavepreimages.However,thefunctionf:Z!Zdenedbyf(x)=x+1isonto.Thefunctionf:Z!Zdenedbyf(x)=x2isnotontobecausethereisnox2Zsuchthatf(x)=2.vExercise1.7.Canyounda...1....functionwhichisneitherone-to-onenoronto?2....functionwhichisone-to-onebutnotonto?3....functionwhichisontobutnotone-to-one?4....functionwhichisbothone-to-oneandonto?Denition1.8.Afunctionf:D!Riscalleda bijectionifitisboth

3 one-to-oneandonto.Equivalently,everyelem
one-to-oneandonto.Equivalently,everyelementofRhasexactlyonepreimage.Example1.9.Thefunctionf:Z!Zdenedbyf(x)=x+1isabijection.vExample1.10.IfSisanyset,thenwecandeneamapf:S!Sbyf(x)=xforallx2S.Thismapiscalledthe identity map,anditisabijection.vDenition1.11.Iff:S1!S2andg:S2!S3,thenwecandeneanewfunctionfg:S1!S3by(fg)(x)=g(f(x)).Theoperationiscalled composition.Remark1.12.Oneusuallywrites(gf)(x)=g(f(x))ratherthan(fg)(x)=g(f(x)).However,aslongasweareconsistent,thechoicedoesnotmakeabigdierence.WeareusingthisconventionbecauseitmatchestheconventionusuallyusedfortheRubik'scube.4 Exercises1.Whichofthefollowingfunctionsareone-to-one?Whichareonto?(a)f:Z!Zdenedbyf(x)=x2+1.(b)f:N!Ndenedbyf(x)=x2+1.(c)f:Z!Zdenedbyf(x)=3x+1.(d)f:R!Rdenedbyf(x)=3x+1.2.Supposef1:S1!S2andf2:S2!S3areone-to-one.Provethatf1f2isone-to-one.3.Supposef1:S1!S2andf2:S2!S3areonto.Provethatf1f2isonto.4.Letf1:S1!S2,f2:S2!S3,andf3:S3!S4.Provethatf1(f2f3)=(f1f2)f3.5.LetSbeaset.(a)Provethatthereexistsafunctione:S!Ssuchthatef=fandfe=fforallbijectionsf:S!S.Provethateisabijection.S.(b)Provethat,foreverybijectionf:S!S,thereexistsabijectiong:S!Ssuchthatfg=eandgf=e.6.Iff:D!RisabijectionandDisanitesetwithnelements,provethatRisalsoanitesetwithnelements.5 2.GroupsExample2.1.Togetanideaofwhatgroupsareallabout,let'sstartbylookingattwofamiliarsets.First,considertheintegersmod4.RememberthatZ=4Zisasetwith4elements:0,1,2,and3.Oneoftherstthingsyoulearnedinmodulararithmeticwashowtoaddnumbersmodn.Let'swriteanadditiontableforZ=4Z.+ 0123 0 01231 12302 23013 3012Now,we'regoingtorewritetheadditiontableinawaythatmightseemprettypointless;we'rejustgoingtousethesymbolinsteadof+foraddition,andwe'llwritee=0,a=1,b=2,andc=3.Then,ouradditiontablelookslike eabc e eabca abceb bceac ceabLet'sdothesamethingfor(Z=5Z),thesetofunitsmod5.Theunitsmod5are1,2,3,and4.Ifyouaddtwounits,youdon'tnecess

4 arilygetanotherunit;forexample,1+4=0,and
arilygetanotherunit;forexample,1+4=0,and0isnotaunit.However,ifyoumultiplytwounits,youalwaysgetaunit.So,wecanwritedownamultiplicationtablefor(Z=5Z).Hereitis:
1243 1 12432 24314 43123 3124Again,we'regoingtorewritethisusingnewsymbols.Letmeanmultiplication,andlete=1,a=2,b=4,andc=3.Then,themultiplicationtablefor(Z=5Z)lookslike eabc e eabca abceb bceac ceabNoticethatthisisexactlythesameasthetableforadditiononZ=4Z!Whyisitinterestingthatwegetthesametablesinthesetwodierentsituations?Well,thisenablesustotranslatealgebraicstatementsaboutadditionofelementsofZ=4Zintostatementsaboutmultiplicationofelementsof(Z=5Z).Forexample,theequationx+x=0inZ=4Zhastwosolutions,x=0andx=2.Withouralternatesetofsymbols,thisisthesameassayingthattheequationxx=ehassolutionsx=eandx=b.Ifwetranslatethisto(Z=5Z),thissaysthatthesolutionsofx
x=1in(Z=5Z)arex=1andx=4.Thatis,1and4arethesquarerootsof1in(Z=5Z),whichisexactlyright!Inmathematicallanguage,wesaythatZ=4Zwithadditionand(Z=5Z)withmultiplicationare\isomorphicgroups."Theword\isomorphic"meansroughlythattheyhavethesamealgebraicstructure;we'llgetintothislater.Fornow,let'sjustseewhata\group"is.vDenition2.2.A group(G;)consistsofasetGandanoperationsuchthat:6 1.Gisclosedunder.Thatis,ifa;b2G,thenab2G.Examples:Z=4Zisclosedunder+;afterall,wewrotedowntheadditiontable,whichtellsushowtoaddanytwoelementsofZ=4ZandgetanotherelementofZ=4Z.Similarly,(Z=5Z)isclosedundermultiplication.Zisclosedunder+:ifa;b2Z,thena+b2Z.Similarly,Zisalsoclosedunder�.Risclosedundermultiplication:ifwemultiplytworealnumbers,wegetarealnumber.Thesetofnegativenumbersisnotclosedundermultiplication:ifwemultiplytwonegativenum-bers,wegetapositivenumber.2.isassociative.Thatis,foranya;b;c2G,a(bc)=(ab)c.Examples:Additionandmultiplicationareassociative.Subtractionisnotassociativebecausea�(b�c)6=(a�b)�c.3.Thereisan\identityelement"e2Gwhichsatis

5 ;esg=eg=geforallg2G.Examples:

;esg=eg=geforallg2G.Examples:For(Z=4Z;+),0isanidentityelementbecauseg=0+g=g+0foranyg2Z=4Z.For((Z=5Z);
),1isanidentityelementbecauseg=1
g=g
1foranyg2(Z=5Z).For(Z;+),0isanidentityelementbecauseg=0+g=g+0foranyg2Z.For(R;
),1isanidentityelementbecauseg=1
g=g
1foranyg2R.4.Inversesexist;thatis,foranyg2G,thereexistsanelementh2Gsuchthatgh=hg=e.(hiscalledan inverseofg.)Examples:UsingtheadditiontableforZ=4Z,wecanndinversesofalltheelementsofZ=4Z.Forinstance,wecanseefromthetablethat1+3=3+1=0,so3istheinverseof1.Similarly,sincethetablefor(Z=5Z)isidentical,allelementsof(Z=5Z)haveinverses.For(Z;+),theinverseofn2Zis�nbecausen+(�n)=(�n)+n=0.For(R;
),noteveryelementhasaninverse|namely,0doesnothaveaninverse.However,ifx6=0,then1 xisaninverseofxbecausex
1 x=1 x
x=1.Example2.3.1.(Z=4Z;+)and((Z=5Z);
)aregroups.Infact,aswesaidearlier,theseshouldbethoughtofasthe\same"group,butwewon'tgointothisuntillater.2.(Z;+)isagroup.However,(Z;�)isnotagroupbecausesubtractionisnotassociative.3.(R;
)isnotagroupsince0doesnothaveaninverseundermultiplication.However,(R�f0g;
)isagroup.4.Thesetofnegativenumbersisnotclosedundermultiplication,sothesetofnegativenumberswithmultiplicationisnotagroup.5.Wecanconstructagroup(G;)whereGisasetwithjustoneelement.SinceGmusthaveanidentityelement,wewillcallthissingleelemente.Todenethegroupoperation,wejustneedtosaywhateeis.ThereisonlyonechoicesinceGhasonlyoneelement:eemustbee.Thisdenesagroupwhichiscalledthe trivial group.Asyoumightguess,thetrivialgroupisn'tveryinteresting.6.Soon,wewillseehowtomakethemovesofaRubik'scubeintoagroup!7 vTheexamplesofgroupswehaveseensofarallhaveanotherspecialproperty:foreveryg;h2G,gh=hg;thatis,theoperationiscommutative.Thisisnottrueofallgroups.Ifitistrueof(G;),wesay(G;)is abelian.Wewillsoonseeexamplesofnonabeliangroups.Now,wewillprovetwoimportantpropertiesofgroups.Le

6 mma2.4.Agrouphasexactlyoneidentityelemen
mma2.4.Agrouphasexactlyoneidentityelement.Proof.Let(G;)beagroup,andsupposeeande0areidentityelementsofG(weknowthatGhasatleastoneidentityelementbythedenitionofagroup).Then,ee0=esincee0isanidentityelement.Ontheotherhand,ee0=e0sinceeisanidentityelement.Therefore,e=e0becausebothareequaltoee0. Lemma2.5.If(G;)isagroup,theneachg2Ghasexactlyoneinverse.Proof.Letg2G,andsupposeg1;g2areinversesofG(weknowthereisatleastonebythedenitionofagroup);thatis,gg1=g1g=eandgg2=g2g=e.Byassociativity,(g1g)g2=g1(gg2).Sinceg1isaninverseofg,(g1g)g2=eg2=g2.Sinceg2isaninverseofg,g1(gg2)=g1e=g1.Therefore,g2=g1. Ingeneral,wewritetheuniqueinverseofgasg�1.However,ifweknowthatthegroupoperationisaddition,thenwewritetheinverseofgas�g.Exercises1.Whichofthefollowingaregroups?Proveyouranswer.(a)(f1g;
)(b)(S;+),whereSisthesetofnon-negativeintegersf0;1;2;3;:::g(c)(2Z;+),where2Zisthesetofevenintegers(d)(Z;
)(e)(Z;?)wherea?bisdenedtobea+b�1.(f)(Q�f0g;
)(g)(R;H)whereaHbisdenedtobe(a�1)(b�1).(h)(G;)whereGisthesetofbijectionsfromsomesetStoitselfanddenotescompositionoffunctions.2.Let(G;)beagroupanda;b;c2G.Prove:(a)Ifab=ac,thenb=c.(b)Ifba=ca,thenb=c.Thatis,wecancancelingroups.3.If(G;)isagroupandg2G,provethat(g�1)�1=g.4.If(G;)isagroupandg;h2Gsuchthatgh=e,provethathg=e.(Thatis,ifgh=e,thenhistheinverseofgandgistheinverseofh.)5.If(G;)isagroupandg;h2Gsuchthatgh=h,provethatgistheidentityelementofG.6.Let(G;)beanitegroup;thatis,(G;)isagroupandGhasnitelymanyelements.Letg2G.Provethatthereexistsapositiveintegernsuchthatgn=e(here,gnmeansgg


gwithncopiesofg).Thesmallestsuchintegerniscalledthe orderofg.8 7.Findtheorderof5in(Z=25Z;+).Findtheorderof2in((Z=17Z);
).8.If(G;)isagroupinwhichgg=eforallg2G,showthat(G;)isabelian.9.If(G;)isagroupwhereGhas4elements,showtha

7 t(G;)isabelian.10.Let(G;)bea
t(G;)isabelian.10.Let(G;)beanitegroup.Provethatthereisapositiveintegernsuchthatgn=eforallg2G.(ThisisdierentfromProblem6inthatyouneedtoshowthatthesamenworksforallg2G.)11.LetGbeasetandbeanoperationonGsuchthatthefollowingfourpropertiesaresatised:(a)Gisclosedunder.(b)isassociative.(c)Thereexistse2Gsuchthatge=gforallg2G.(Wecallea\rightidentity.")(d)Foreachg2G,thereexistsh2Gsuchthatgh=e.(Wecallha\rightinverse"ofg.)Provethat(G;)isagroup.9 3.TheRubik'sCubeandSubgroups3.1.CubenotationTheRubik'scubeiscomposedof27smallcubes,whicharetypicallycalled\cubies."26ofthesecubiesarevisible(ifyoutakeyourcubeapart,you'llndthatthe27thcubiedoesn'tactuallyexist).WhenworkingwiththeRubik'scube,it'shelpfultohaveasystematicwayofreferringtotheindividualcubies.Althoughitseemsnaturaltousethecolorsofacubie,itisactuallymoreusefultohavenameswhichdescribethelocationsofthecubies.Thecubiesinthecornersarecalled,appropriatelyenough,\cornercubies."Eachcornercubiehas3visiblefaces,andthereare8cornercubies.Thecubieswithtwovisiblefacesarecalled\edge"cubies;thereare12edgecubies.Finally,thecubieswithasinglevisiblefacearecalled\centercubies,"andthereare6centercubies.Now,let'snamethe6facesoftheRubik'scube.FollowingthenotationdevelopedbyDavidSingmaster,wewillcallthemright(r),left(l),up(u),down(d),front(f),andback(b).Theadvantageofthisnamingschemeisthateachfacecanbereferredtobyasingleletter.Tonameacornercubie,wesimplylistitsvisiblefacesinclockwiseorder.Forinstance,thecubieintheupper,right,frontcorneriswrittenurf.Ofcourse,wecouldalsocallthiscubierfuorfur.Sometimes,wewillcarewhichfaceislistedrst;inthesetimes,wewilltalkabout\orientedcubies."Thatis,theorientedcubiesurf,rfu,andfuraredierent.Inothersituations,wewon'tcarewhichfaceislistedrst;inthesecases,wewilltalkabout\unorientedcubies."Thatis,theunorientedcubiesurf,rfu,andfurarethesame.Similarly,tonameedgeandcentercubies,wewilljustlistthevisiblefacesofthecubies.Forinstance,thecubieinth

8 ecenterofthefrontfaceisjustcalledf,becau
ecenterofthefrontfaceisjustcalledf,becauseitsonlyvisiblefaceliesonthefrontofthecube.Wewillalsofrequentlytalkabout\cubicles."Thesearelabeledthesamewayascubies,buttheydescribethespaceinwhichthecubielives.Thus,iftheRubik'scubeisinthestartconguration(thatis,theRubik'scubeissolved),theneachcubielivesinthecubicleofthesamename(theurfcubielivesintheurfcubicle,thefcubielivesinthefcubicle,andsoon).IfyourotateafaceoftheRubik'scube,thecubiclesdon'tmove,butthecubiesdo.Notice,however,thatwhenyourotateafaceoftheRubik'scube,allcentercubiesstayintheircubicles.Finally,wewanttogivenamestosomemovesoftheRubik'scube.Themostbasicmoveonecandoistorotateasingleface.WewillletRdenoteaclockwiserotationoftherightface(lookingattherightface,turnit90clockwise).Similarly,wewillusethecapitallettersL,U,D,F,andBtodenoteclockwisetwistsofthecorrespondingfaces.Moregenerally,wewillcallanysequenceofthese6facetwistsa\move"oftheRubik'scube.Forinstance,rotatingtherightfacecounterclockwiseisamovewhichisthesameasdoingRthreetimes.Laterinthislecture,wewilldescribeanotationforthesemorecomplicatedmoves.Acoupleofthingsareimmediatelyclear.First,wealreadyobservedthatthe6basicmoveskeepthecentercubiesintheircubicles.Sinceanymoveisasequenceofthese6basicmoves,thatmeansthateverymoveoftheRubik'scubekeepsthecentercubiesintheircubicles(foraformalproof,seetheexampleafterProposition4.9).Also,anymoveoftheRubik'scubeputscornercubiesincornercubiclesandedgecubiesinedgecubicles;itisimpossibleforacornercubietoeverliveinanedgecubicleorforanedgecubietoliveinacornercubicle.Usingthesetwofacts,wecanstarttogureouthowmanypossiblecongurationstheRubik'scubehas.Let'slook,forinstance,attheurfcubicle.Theoretically,anyofthe8cornercubiescouldresideinthiscubicle.Thatleaves7cornercubiesthatcouldresideintheurbcubicle,6forthenextcornercubicle,andsoon.Therefore,thereare8
7
6
5
4
3
2
1=8!possiblepositioningsofthecornercubies.Noticethatacornercubiecantintoitscubiclein3dierentways.Forin

9 stance,ifared,white,andbluecubieliesinth
stance,ifared,white,andbluecubieliesintheurfcubicle,eitherthered,white,orbluefacecouldlieintheufaceofthecubicle(andthisdetermineswheretheother2faceslie).Sincethereare8cornercubiesandeachcanlieinitscubiclein3dierentways,thereare38dierentwaysthecornercubiescouldbeoriented.Therefore,thereare38
8!possiblecongurationsofthecornercubies.Similarly,sincethereare12edgecubies,thereare12!positionsoftheedgecubies;eachedgecubiehas2possibleorientations,giving212possibleorientations.So,thereare10 212
12!possiblecongurationsoftheedgecubies,givingatotalof212388!12!possiblecongurationsoftheRubik'scube.(Thisnumberisabout5:191020,or519quintillion!)Althoughthesecongurationsaretheoreticallypossible,thatdoesn'tmeanthatthesecongurationscouldreallyoccur.WewillsaythatacongurationoftheRubik'scubeis validifitcanbeachievedbyaseriesofmovesfromthestartingconguration.Itturnsoutthatsomeofthetheoreticallypossiblecongurationswehavecountedareactuallynotvalid.Therefore,wehavetwogoals:1.Demonstratethatsomecongurationsarenotvalid.2.Findasetofmovesthatcantakeusfromanyvalidcongurationbacktothestartconguration.3.2.MakingtheRubik'sCubeintoaGroupWecanmakethesetofmovesoftheRubik'scubeintoagroup,whichwewilldenote(G;).TheelementsofGwillbeallpossiblemovesoftheRubik'scube(forexample,onepossiblemoveisaclockwiseturnofthetopfacefollowedbyacounterclockwiseturnoftherightface).Twomoveswillbeconsideredthesameiftheyresultinthesamecongurationofthecube(forinstance,twistingafaceclockwiseby180isthesameastwistingthesamefacecounterclockwiseby180).Thegroupoperationwillbedenedlikethis:ifM1andM2aretwomoves,thenM1M2isthemovewhereyourstdoM1andthendoM2.Whyisthisagroup?Wejustneedtoshowthe4propertiesin[PS2,#11].Giscertainlyclosedundersince,ifM1andM2aremoves,M1M2isamoveaswell.Ifweletebethe\empty"move(thatis,amovewhichdoesnotchangethecongurationoftheRubik'scubeatall),thenMemeans\rs

10 tdoM,thendonothing."Thisiscertainlythesa
tdoM,thendonothing."ThisiscertainlythesameasjustdoingM,soMe=M.So,(G;)hasarightidentity.IfMisamove,wecanreversethestepsofthemovetogetamoveM0.Then,themoveMM0means\rstdoM,thenreverseallthestepsofM."Thisisthesameasdoingnothing,soMM0==e,soM0istheinverseofM.Therefore,everyelementofGhasarightinverse.Finally,wemustshowthatisassociative.Rememberthatamovecanbedenedbythechangeincongurationitcauses.Inparticular,amoveisdeterminedbythepositionandorientationitputseachcubiein.IfCisanorientedcubie,wewillwriteM(C)fortheorientedcubiclethatCendsupinafterweapplythemoveM,withthefacesofM(C)writteninthesameorderasthefacesofC.Thatis,therstfaceofCshouldendupintherstfaceofM(C),andsoon.Forexample,themoveRputstheurcubieinthebrcubicle,withtheufaceofthecubielyinginthebfaceofthecubicleandtherfaceofthecubielyingintherfaceofthecubicle.Thus,wewriteR(ur)=br.First,let'sinvestigatewhatasequenceoftwomovesdoestothecubie.IfM1andM2aretwomoves,thenM1M2isthemovewherewerstdoM1andthendoM2.ThemoveM1movesCtothecubicleM1(C);themoveM2thenmovesittoM2(M1(C)).Therefore,(M1M2)(C)=M2(M1(C)).Toshowthatisassociative,weneedtoshowthat(M1M2)M3=M1(M2M3)foranymovesM1,M2,andM3.Thisisthesameasshowingthat(M1M2)M3andM1(M2M3)dothesamethingtoeverycubie.Thatis,wewanttoshowthat[(M1M2)M3](C)=[M1(M2M3)](C)foranycubieC.Weknowfromourabovecalculationthat[(M1M2)M3](C)=M3([M1M2](C))=M3(M2(M1(C))).Ontheotherhand,[M1(M2M3)](C)=(M2M3)(M1(C))=M3(M2(M1(C))).So,(M1M2)M3=M1(M2M3).Thus,isassociative.Therefore,(G;)isindeedagroup.11 3.3.SubgroupsWecalculatedthattherearearound519quintillionpossiblecongurationsoftheRubik'scube(althoughthesearenotallvalid).Tryingtounderstandsuchalargenumberofcongurationsisnoeasytask!Itishelpfultorestricttheproblem;forinstance,insteadoflookingatallpossiblemovesoftheRubik'scube,wemightstartoutbylookingatthemoveswhichonlyinvolvet

11 wistsofthedownandrightfaces.Thisisagener
wistsofthedownandrightfaces.Thisisageneralphilosophyingrouptheory:tounderstandagroupG,weshouldtrytounderstandsmallpiecesofit.Denition3.1.AnonemptysubsetHofagroup(G;)iscalleda subgroupofGif(H;)isagroup.Theadvantageofstudyingsubgroupsisthattheymaybemuchsmallerand,hence,simpler;however,theystillhavealgebraicstructure.Example3.2.Agroupisalwaysasubgroupofitself.Also,thetrivialgroupisasubgroupofanygroup.However,thesesubgroupsaren'ttoointeresting!vExample3.3.Thesetofevenintegersisasubgroupof(Z;+):afterall,theevenintegersarecertainlyasubsetofZ,andweknowfrom[PS2,#1c]that(2Z;+)isagroup.vThefollowinglemmaoftenmakesiteasiertocheckifasubsetisactuallyasubgroup.Lemma3.4.Let(G;)beagroup.AnonemptysubsetHofGisasubgroupof(G;)i,foreverya;b2H,ab�12H.Proof.First,supposeHisasubgroup.Ifb2H,thenb�12Hsince(H;)isagroup.So,ifa2Haswell,thenab�12H.Conversely,supposethat,foreverya;b2H,ab�12H.First,noticethatisassociativesince(G;)isagroup.Leta2H.Then,e=aa�1,soe2H.Letb2H,Thenb�1=eb�12H,soinversesexistinH.Leta;b2H.Bythepreviousstep,b�12H,soa(b�1)�1=ab2H.Thus,Hisclosedunder.Therefore,(H;)isagroup,whichmeansthatHisasubgroupofG. 3.4.SimplifyingGroupNotationItiscommonpracticetowritegroupoperationsasmultiplication;thatis,wewriteghratherthangh,andwecallthisthe\product"ofgandh.Thestatement\letGbeagroup"reallymeansthatGisagroupundersomeoperationwhichwillbewrittenasmultiplication.WewillalsooftenwritetheidentityelementofGas1ratherthane.Finally,wewillusestandardexponentialnotation,sog2meansgg,g3meansggg,andsoon.Inparticular,wewilldothiswith(G;).Thatis,fromnowon,wewilljustcallthisgroupG,andwewillwritetheoperationasmultiplication.Forinstance,DRmeansthemoveDfollowedbythemoveR.Themovewhichtwiststherightfacecounterclockwiseby90isthesameasamovetwistingtherightfaceclockwisethreetimes,sowecanwritethismoveasR3.Exercises1.IfGisagroupandg;h2G,write(gh)�1intermsofg

12 �1andh�1.12 2.IfA;Baresubgroupsofa
�1andh�1.12 2.IfA;BaresubgroupsofagroupG,provethatA\BisasubgroupofG.3.LetGbeagroupandZ(G)=fz2G:zx=xzforallx2Gg.(ThisnotationmeansthatZ(G)isthesetofz2Gsuchthatzx=xzforallx2G.)ProvethatZ(G)isasubgroupofG.Z(G)iscalledthe centerofG.IfGisabelian,whatisZ(G)?4.RememberthatGisthegroupofmovesoftheRubik'scube.Provethatthisgroupisnotabelian.(Hint:picktwomovesM1andM2andlookattheir commutator[M1;M2],whichisdenedtobeM1M2M�11M�12.)5.LetC1andC2betwodierentunorientedcornercubies,andletC01andC02betwodierentunorientedcornercubicles.ProvethatthereisamoveoftheRubik'scubewhichsendsC1toC01andC2toC02.Sincewearetalkingaboutunorientedcubiesandcubicles,weonlycareaboutthepositionsofthecubies,nottheirorientations.(Forexample,ifC1=dbr,C2=urf,C01=dlb,andC02=urf,thenthemoveDsendsC1toC01andC2toC02.)6.LetGbeagroupandSbeasubsetofG.LetHbethesetofallelementsofGwhichcanbewrittenasaniteproductofelementsofSandtheirinverses;thatis,Hconsistsofallelementsoftheforms1


snwhereeachsiiseitheranelementofSortheinverseofanelementofS.ProvethatHisasubgroupofG.WecallHthe subgroup of G generated by SandwriteH=hSi.ItisthesmallestsubgroupofGcontainingS(doyouseewhy?).7.IfGisanabeliangroup,showthatfa2:a2GgisasubgroupofG.WhichpartofyourprooffailswhenGisnotabelian?8.Findallsubgroupsof(Z;+).9.Let(G;)beagroupandHbeanonemptynitesubsetofGclosedunder.ProvethatHisasubgroupofG.IsthestatementstilltrueifHisnotrequiredtobenite?10.TryscramblingyourRubik'scube.Howmanycubiescanyouputintherightposition?Therightorientation?11.BrainstormabitaboutwhatkindofstrategyyoushouldusetosolvetheRubik'scube.Whichcubiesshouldyoutrytosolverst?Whatkindofmovesshouldyoulookfor?13 4.GeneratorsLetGbeagroupandSbeasubsetofG.In[PS3,#6],wedenedthesubgrouphSi.Now,wewillgivesomepropertiesofhSi.First,let'slookatthespecialcasewhenhSiisjustG.Denition4.1.LetGbeagroupandSbeasubsetofG.WesaythatS generatesGorthatSisasetof generatorsofGifG=hSi;thatis,everyelementofGcanbewrittena

13 saniteproduct(underthegroupoperatio
saniteproduct(underthegroupoperation)ofelementsofSandtheirinverses.Example4.2.EveryelementofZcanbewrittenasanitesumof1'sor�1's,soZisgeneratedbyf1g.Thatis,Z=h1i.Forthesamereason,Z=h�1i.Ofcourse,it'salsotruethatZ=h1;2i.Ingeneral,therearemanypossiblesetsofgeneratorsofagroup.vExample4.3.EveryelementofZ=4Zcanbewrittenasanitesumof1's,soZ=4Z=h1i.EventhoughZ=h1iandZ=4Z=h1i,ZandZ=4Zarenotequal!hSionlymakessenseinthecontextofagivengroup.vExample4.4.EveryelementofGcanbewrittenasanitesequenceofturnsoftheRubik'scube,soG=hD;U;L;R;F;Bi.vYoumightthinkofgeneratorsasbeingthe\core"ofthegroup;sinceeveryelementofthegroupcanbewrittenintermsofthegenerators,knowledgeaboutthegeneratorscanoftenbetranslatedintoknowledgeaboutthewholegroup.Wewillmakethismoreprecisesoon.Denition4.5.AgroupGis cyclicifthereexistsg2GsuchthatG=hgi.Example4.6.ZandZ=4Zarecyclic.vFornitegroups,wecanevenrelaxthedenitionofgeneratorsslightlybyleavingouttheinversesofS.Toprovethis,weneedalemma.Lemma4.7.LetGbeanitegroupandg2G.Then,g�1=gnforsomen2N.Proof.Ifg=e,thenthereisnothingtoshow.So,supposeg6=e.By[PS2,#6],thereexistsapositiveintegermsuchthatgm=e.Sinceg6=e,m6=1,som�1.Letn=m�12N.Then,ggn=gm=e,sognistheinverseofgby[PS2,#4]. Lemma4.8.LetGbeanitegroupandSbeasubsetofG.Then,G=hSiieveryelementofGcanbewrittenasaniteproductofelementsofS.(Thatis,theinversesofSarenotnecessary.)Proof.IfeveryelementofGcanbewrittenasaniteproductofelementsofS,thenitisclearthatG=hSi.Conversely,supposeG=hSi.ThismeansthateveryelementofGcanbewrittenasaniteproducts1


snwhereeachsiiseitherinSortheinverseofanelementofS.ThebasicpointoftheproofisthattheinverseofanelementofScanalsobewrittenasaproductofelementsofSbythepreviouslemma.Tomakethiscompletelyrigorous,wewilluseinductiononn.Supposen=1.Eithers12Sors�112S.Ifs12S,thens1iswrittenasaproductofasingleelementofS.Ifs�112S,thens1canbewrittenasaniteproductofelementsofSbyLemma4.7.So,thebasecasei

14 strue.Now,supposethestatementistrueforal
strue.Now,supposethestatementistrueforallnaturalnumberssmallerthann;wewanttoshowthats1


sncanbewrittenasaniteproductofelementsofS.Bytheinductionhypothesis,s1:::;sn�1andsncanbothbewrittenasniteproductsofelementsofS.Therefore,theirproducts1


sncertainlycanaswell. Now,wewillseehowtotranslatepropertiesofgeneratorstothewholegroup.14 Proposition4.9.LetGbeanitegroupandSbeasubsetofG.Supposethefollowingtwoconditionsaresatised.1.EveryelementofSsatisessomepropertyP.2.Ifg2Gandh2GbothsatisfythepropertyP,thenghsatisesthepropertyPaswell.Then,everyelementofhSisatisesP.Beforeweprovethis,let'sseehowitmightbeused.Example4.10.LetS=fD;U;L;R;F;BgG.Then,everyM2Ssatisestheproperty\Mkeepsallcentercubiesintheircubicles."IfM1;M22Garesuchthattheykeepallcentercubiesintheircubicles,thenM1M2certainlykeepsallcentercubiesintheircubicles.SinceG=hSi,thepropositionsaysthateveryelementofGkeepsthecentercubiesintheircubicles.ThispropositionisextremelyusefulfortheRubik'scubebecauseitmeanswefrequentlyonlyneedtounderstandpropertiesofthe6basicmovesratherthanall51020possiblemoves.vNow,let'sproveProposition4.9.Proof.ByLemma4.8,anyelementofhSicanbewrittenass1


snwheren2NandeachsiisanelementofS.Wewillprovethepropositionbyinductiononn.Ifn=1,thens12SsatisespropertyPbyhypothesis.Supposeinductivelythats1


sn�1satisespropertyP.Then,theproduct(s1


sn�1)snisaproductoftwoelementssatisfyingpropertyP,soitsatisespropertyPaswell. Exercises1.IfHisasubgroupofagroupG,provethathHi=H.2.IfAisasubsetofB,provethathAiisasubgroupofhBi.Istheconversetrue?3.LetGbeanontrivialgroup(rememberthatthetrivialgroupisthegroupwithonlyoneelement,soanontrivialgroupisagroupwithatleast2elements).SupposethattheonlysubgroupsofGareGandf1g.ProvethatGiscyclicandnite,andprovethatthenumberofelementsinGisaprimenumber.4.Provethatanysubgroupofacyclicgroupiscyclic.5.RememberthatGisthegroupofmovesoftheRubik'scube,D2G

15 isaclockwisetwistofthedownface,andR2Gisa
isaclockwisetwistofthedownface,andR2Gisaclockwisetwistoftherightface.FindtheorderofD,R,andDR(remember,DRisthemovewhereyourstdoD,thenR;forthedenitionoforder,see[PS2,#6]).6.AgroupGis nitely generatedifthereexistsanitesubsetSofGsuchthatG=hSi.(a)Provethateverynitelygeneratedsubgroupof(Q;+)iscyclic.(b)Provethat(Q;+)isnotnitelygenerated.15 5.TheSymmetricGroupWhenwefoundthenumberofpossiblecongurationsoftheRubik'scube,weusedthefactthatanymovesendscornercubiestocornercubiclestodeducethatthereare8!possiblewaystopositionthecornercubies.Inthissection,wewilllaydownthemathematicalfoundationneededtounderstandthesepossibilities.Ratherthanjustlookingatcongurationsof8cubies,we'lllookatcongurationsofanynobjects.We'llcalltheseobjects1;2;:::;n,althoughthesenamesarearbitrary.Wecanthinkofarrangingtheseobjectsasputtingthemintonslots.Ifwenumbertheslots1;2;:::;n,thenwecandeneafunction:f1;2;:::;ng!f1;2;:::;ngbyletting(i)bethenumberputintosloti.Example5.1.Wecanputtheobjects1,2,3intheorder312.Here,3isintherstslot,1isinthesecondslot,and2isinthethirdslot.So,thisorderingcorrespondstothefunction:f1;2;3g!f1;2;3gdenedby(1)=3,(2)=1,and(3)=2.vWhatcanwesayabout?Lemma5.2.isabijection.Proof.Supposex6=y.Sinceanumbercannotbeinmorethanoneslot,ifx6=y,slotsxandymustcontaindierentnumbers.Thatis,(x)6=(y).Therefore,isone-to-one.Anynumbery2f1;2;:::;ngmustlieinsomeslot,sayslotx.Then,(x)=y.So,isonto. Ontheotherhand,if:f1;:::;ng!f1;:::;ngisabijection,thendenesanarrangementofthenobjects:justputobject(i)insloti.So,thesetofpossiblearrangementsisreallythesameasthesetofbijectionsf1;:::;ng!f1;:::;ng.Therefore,insteadofstudyingpossiblearrangements,wecanstudythesebijections.Denition5.3.The symmetric group on n lettersisthesetofbijectionsfromf1;2;:::;ngtof1;2;:::;ng,withtheoperationofcomposition.WewritethisgroupasSn.NotethatSnisagroupby[PS2,#1h].Let'sdoa

16 nexampletomakesurethatthegroupoperationi
nexampletomakesurethatthegroupoperationisclear.Example5.4.Let;2S3bedenedby(1)=3,(2)=1,(3)=2,(1)=1,(2)=3,and(3)=2.Then,()(1)=(3)=2,()(2)=(1)=1,and()(3)=(2)=3.v5.1.DisjointCycleDecompositionThereisamorecompactwayofwritingelementsofthesymmetricgroup;thisisbestexplainedbyanexample.Example5.5.Consider2S12denedby(1)=12(2)=4(3)=5(4)=2(5)=6(6)=9(7)=7(8)=3(9)=10(10)=1(11)=11(12)=8Wewillwrite"i7!j"(\imapstoj")tomean(i)=j.Then,17!12,127!8,87!3,37!5,57!6,67!9,97!10,107!127!4,47!277!7117!1116 Thisdatatellsuswhatdoestoeachnumber,soitdenes.Asshorthand,wewrite=(1128356910)(24)(7)(11):Here,(1128356910),(24),(7),and(10)arecalled cycles.Whenwritingthedisjointcyclede-composition,weleaveoutthecycleswithjustonenumber,sothedisjointcycledecompositionofis=(1128356910)(24).vNow,let'sactuallydenewhatacycleis.Denition5.6.The cycle(i1i2


ik)istheelement2Sndenedby(i1)=i2;(i2)=i3;:::;(ik�1)=ik;(ik)=i1and(j)=jifj6=irforanyr.The lengthofthiscycleisk,andthe supportofthecycleisthesetfi1;:::;ikgofnumberswhichappearinthecycle.Thesupportisdenotedbysupp.Acycleoflengthkisalsocalleda k-cycle.Denition5.7.Twocyclesandare disjointiftheyhavenonumbersincommon;thatis,supp\supp=?.Lemma5.8.Let;2Snbecycles.Ifandaredisjoint,then=.Proof.Leti2f1;:::;ng.Sincesupp\supp=?,thereareonlytwopossibilities:i62suppandi62supp.Inthiscase,(i)=iand(i)=i,so()(i)=(i)=iand()(i)=(i)=i.Otherwise,iisinthesupportofexactlyoneofand.Wemaysupposewithoutlossofgeneralitythati62suppandi2supp.Then,(i)=i,so()(i)=(i).Ontheotherhand,()(i)=((i)).Now,since(i)2sup

17 pandsupp\supp=?,(i)6
pandsupp\supp=?,(i)62supp.Therefore,((i))=(i).So,weagainhave()(i)=()(i).Therefore,()(i)=()(i)=iforalli,whichshowsthat=. Any2Sncanbewrittenasaproduct(underthegroupoperation,whichiscomposition)ofdisjointcycles.Thisproductiscalledthe disjoint cycle decompositionof.Inourexample,wegaveamethodforndingthedisjointcycledecompositionofapermutation.Wewritetheidentitypermutationas1.Example5.9.S2consistsoftwopermutations,1and(12).vExample5.10.Let;2S6bedenedby(1)=3(2)=5(3)=4(4)=1(5)=2(6)=6(1)=5(2)=4(3)=3(4)=2(5)=1(6)=6Incyclenotation,=(134)(25)and=(15)(24).Then,=(132)(45)and=(12)(345).Wecanalsoeasilycompute2=(143)and2=1.vDenition5.11.If2Snistheproductofdisjointcyclesoflengthsn1;:::;nr(includingits1-cycles),thentheintegersn1;:::;nrarecalledthe cycle typeof.5.2.Rubik'sCubeWecanwriteeachmoveoftheRubik'scubeusingaslightlymodiedcyclenotation.Wewanttodescribewhathappenstoeachorientedcubie;thatis,wewanttodescribewhereeachcubiemovesandwhereeachfaceofthecubiemoves.Forexample,ifweunfoldthecubeanddrawthedownface,itlookslike17 f f f l d d d r l d d d r l d d d r b b b Ifwerotatethisfaceclockwiseby90(thatis,weapplythemoveD),thenthedownfacelookslike l l l b d d d f b d d d f b d d d f r r r So,D(dlf)=dfrbecausethedlfcubienowlivesinthedfrcubicle(withthedfaceofthecubielyinginthedfaceofthecubicle,thelfaceofthecubielyingintheffaceofthecubicle,andtheffaceofthecubielyingintherfaceofthecubicle).Similarly,D(dfr)=drb,D(drb)=dbl,andD(dbl)=dlf.Ifwedothesamethingfortheedgecubies,wendD=(dlfdfrdrbdbl)(dfdrdbdl).Example5.12.CheckthatthedisjointcycledecompositionofRis(rfurubrbdrdf)(rurbrdrf).vExercises1.Let;2S5bedenedasfollows.(1)=3(2)=4(3)=5(4)=2(5)=1(1)=5(2)=3(3)=2(4)=4

18 ;(5)=1Findthecycledecompositionsofeachof
;(5)=1Findthecycledecompositionsofeachofthefollowingpermutations:,,2,and2.2.Let=(1128104)(213)(5117)(69).Find2and3.Whatistheorderof?3.Supposeisapermutationwithcycletypen1;:::;nr.Whatistheorderof?4.Let=(12)and=(23).Findand.(a)Whatistheorderof(12)(23)?Doesthisagreewithyouranswerto[PS5,#3]?(b)ForwhatnisSnabelian?5.WritealltheelementsofS3,thesymmetricgroupon3elements,usingdisjointcyclenotation.6.FindallsubgroupsofS3.7.RememberthatD,U,L,R,F,andBaredenedtobeclockwisetwistsofthedown,up,left,right,front,andbackfaces,respectively.WeshowedinclassthatDhasdisjointcycledecomposition(dlfdfrdrbdbl)(dfdrdbdl)andthatRhasdisjointcycledecomposition(rfurubrbdrdf)(rurbrdrf).WriteU,L,F,andBasproductsofdisjointcycles.8.Leta1;:::;ambedistinctelementsoff1;:::;ng,andlet=(a1a2)(a1a3)(a1a4)


(a1am).Findthedisjointcycledecompositionof.9.ShowthatSnisgeneratedbythesetof2-cyclesinSn.10.Let2Sn,andleta1;:::;ak)bedistinctelementsoff1;:::;ng.Provethat�1(a1a2


ak)=((a1)(a2)


(ak)).18 11.Let;02S6bedenedby=(132)(46)0=(254)(16)Find2S6suchthat0=�1.12.Let;02Sn.Provethatand0havethesamecycletypei0=�1forsome2Sn.Note:Anyelementoftheform�1iscalleda conjugateof,sothissaysthattheconjugatesofareexactlytheelementsofSnwiththesamecycletypeas.13.LetHbethesubgroupofGgeneratedbyD2andR2;thatis,H=hD2;R2i.HowmanyelementsdoesHhave?WhichoftheseelementsmightbehelpfulforsolvingtheRubik'scube?14.Let;2Sn.Supposesupp\supp=fxg;thatis,xistheonlynumberinf1;:::;ngwhichappearsinthecycledecompositionofbothand.Wedenethe commutatorofandtobe�1�1,andwewritethisas[;].Showthatsupp[;]hasatmost3elements.19 6.CongurationsoftheRubik'sCu

19 beAswesaidalready,acongurationofthe
beAswesaidalready,acongurationoftheRubik'scubeisdeterminedbyfourpiecesofdata:thepositionsofthecornercubiesthepositionsoftheedgecubiestheorientationsofthecornercubiestheorientationsoftheedgecubiesTherstcanbedescribedbyanelementofS8(i.e.,theelementofS8whichmovesthecornercubiesfromtheirstartpositionstothenewpositions).ThesecondcanbedescribedbyanelementofS12.Now,wewillseehowtounderstandthethirdandfourth.Thebasicideaistoxa\startingorientation"andasystematicwayofwritingdownhowagivenorientationdiersfromthisstartingorientation.Thsismostlyjustamatterofnotation.We'llstartwiththecornercubies.Eachcornercubiehas3possibleorientations,andwewillnumbertheseorientations0,1,and2.Let'sexplainwhatthesenumbersmean.ImaginethatyourRubik'scubeisinthestartconguration.Wearegoingtowriteanumberononefaceofeachcornercubicle,asfollows.Write:1ontheufaceoftheu cubicle2ontheufaceoftheurfcubicle3ontheufaceoftheubrcubicle4ontheufaceoftheulbcubicle5onthedfaceofthedblcubicle6onthedfaceofthedlfcubicle7onthedfaceofthedfrcubicle8onthedfaceofthedrbcubicleSo,eachcornercubiclenowhasexactlyonenumberedface.Eachcornercubiethushasonefacelyinginanumberedcubicleface.Labelthiscubieface0.Goingaroundthecubieclockwise,labelthenextface1,andthenlabelthenalface2.Example6.1.Ifwelookstraightatthedownfaceandunfoldthecube,thecubiefaceslooklikethis. f f f l d d d r l d d d r l d d d r b b b So,thecubiclenumberingsthatwecanseelooklikethis: 6 7 5 8 Therefore,thecubielabelslooklike 2 1 1 0 0 2 2 0 0 1 1 2 20 vNow,eachfaceofeachcornercubiehasanumberonit.IftheRubik'scubeisinanyconguration,wewilldescribetheorientationsofthecornercubieslikethis:foranyibetween1and8,ndthecubiclefacelabeledi;letxibethenumberofthecubiefacelivinginthiscubicleface.Wewritexfortheordered8-tuple(x1;:::;x8).Noticethatwecanthinkofeachxiascountingthenumberofclockwisetwiststhecubieiisawayfromhavingits0faceinthenumberedfaceofthecubicle.Butacubiethatis3twistsawayisorientedthesa

20 mewayasacubiethatis0twistsaway.Thus,wesh
mewayasacubiethatis0twistsaway.Thus,weshouldthinkofthexiasbeingelementsofZ=3Z.So,xisan8-tupleofelementsofZ=3Z;wewritex2(Z=3Z)8.Example6.2.IftheRubik'scubeisinthestartconguration,eachxiis0.Wealsowritex=0tomeanthateachxiis0.vExample6.3.Let'sseewhatthexiareafterweapplythemoveRtoacubeinthestartconguration.Inthestartconguration,therighthandfacelookslikethis: u u u f r r r b f r r r b f r r r b d d d Thecubiclenumbersonthisfaceare 2 3 7 8 Therefore,thelabelingofthecornercubieslookslikethis: 0 0 2 1 2 1 1 2 1 2 0 0 Ifwerotatetherightfaceofthecubeby90,thenthecubiefaceslooklike 1 2 0 2 1 0 0 1 2 0 2 1 ThecubiesontheleftfaceareunaectedbyR,sox1=0,x4=0,x5=0,andx6=0.Now,wecanseefromourdiagramsthatx2=1,x3=2,x7=2,andx8=1.So,x=(0;1;2;0;0;0;2;1).vWecandothesamethingfortheedgecubies.First,welabeltheedgecubiclesasfollows.Write:21 1ontheufaceoftheubcubicle2ontheufaceoftheurcubicle3ontheufaceoftheufcubicle4ontheufaceoftheulcubicle5onthebfaceofthelbcubicle6onthebfaceoftherbcubicle7ontheffaceoftherfcubicle8ontheffaceofthelfcubicle9onthedfaceofthedbcubicle10onthedfaceofthedrcubicle11onthedfaceofthedfcubicle12onthedfaceofthedlcubicleEachedgecubienowhasafacelyinginanumberedcubicleface;labelthiscubieface0,andlabeltheotherfaceofthecubie1.Then,letyibethenumberofthecubiefaceinthecubiclefacenumberedi.Thisdenesy2(Z=2Z)12.Thus,anycongurationoftheRubik'scubecanbedescribedby2S8,2S12,x2(Z=3Z)8,andy2(Z=2Z)12.So,wewillwritecongurationsoftheRubik'scubeasordered4-tuples(;;x;y).Example6.4.Thestartcongurationis(1,1,0,0).vExample6.5.Pretendthatyourcubeisinthestartconguration.Let(;;x;y)bethecongurationofthecubeafterwedothemove[D;R],whichisdenedtobeDRD�1R�1.Wewillwritedown,,x,andy.WeshowedinclassthatD=(dlfdfrdrbdbl)(dfdrdbdl)andR=(rfurubrbdrdf)(rurbrdrf).Therefore,D�1=(dbldrbdfrdlf)(dldbdrdf)andR�1=(rdfrbdrubrfu)(rfrdrbru).So,[D;R]=(dlfdfrlfdfrdfdlrdf)(drbbrubdrubrrbdrub)(dfdrbr)(

21 6.1)(Whenwritingthisdown,beverycarefulwi
6.1)(Whenwritingthisdown,beverycarefulwiththeorientations.)RememberthatisanelementofS12;wethinkofitasabijectionfromthesetof12unorientededgecubiestothesetof12edgecubicles.Itisdenedlikethis:ifCisanunorientededgecubieinthestartconguration,then(C)istheunorientededgecubiclewhereCislivinginthecurrentconguration.LikeanyelementofS12,canbewrittenindisjointcyclenotation.Inthisparticularexample,[D;R]movescubiedftocubicledr,cubiedrtocubiclebr,andcubiebrtocubicledf.Therefore,=(dfdrbr).Similarly,wethinkofasabijectionfromthesetof8unorientedcornercubiestothesetof8unorientedcornercubicles.Tond,wemustgureoutwhat[D;R]doestothepositionsofthecornercubies.Observethat[D;R]switchesthepositionsofthecubiesd anddfr,anditalsoswitchesthepositionsofdrbandbru.Therefore,=(drbbru)(d dfr).Recallthatwedenedxasfollows.Whenthecubewasinthestartconguration,wenumbered8cubiclefaceslikethis:1ontheufaceoftheu cubicle2ontheufaceoftheurfcubicle3ontheufaceoftheubrcubicle4ontheufaceoftheulbcubicle5onthedfaceofthedblcubicle6onthedfaceofthedlfcubicle7onthedfaceofthedfrcubicle8onthedfaceofthedrbcubicleWenumberedeachofthecorrespondingcubiefaces0.Startingfrom0onacornercubie,wewentclockwise22 andlabeledtheothertwofaces1and2.(Forexample,theufaceoftheu cubieislabeled0,sotheffaceis1andthelfaceis2.)Nowthatthecubeisnolongerinthestartconguration,wedenexitobethecubiefacenumberincubiclefacei.Inthestartposition,allofthenumberedcubiclefaceshavecubiefacesnumbered0.Sincethemove[D;R]doesnotaectcubiesu ,urf,ulb,ordbl,x1,x2,x4,andx5mustbe0.Tondx3,wewanttoseewhichcubiefaceisintheufaceoftheubrcubicle.Wecanseefrom(6.1)that[D;R]putsthebfaceofthedrbcubiethere;byournumberingscheme,thebfaceofthedrbcubieisnumbered2;therefore,x3=2.Similarly,x6=2,x7=0,andx8=2.Therefore,theordered8-tuplexis(0;0;2;0;0;2;0;2).Similarly,todeney,werstnumbered12edgecubiclefaces(whenthecubewasinthestartconguration):1ontheufaceoftheubcubicle2on

22 theufaceoftheurcubicle3ontheufaceoftheuf
theufaceoftheurcubicle3ontheufaceoftheufcubicle4ontheufaceoftheulcubicle5onthebfaceofthelbcubicle6onthebfaceoftherbcubicle7ontheffaceoftherfcubicle8ontheffaceofthelfcubicle9onthedfaceofthedbcubicle10onthedfaceofthedrcubicle11onthedfaceofthedfcubicle12onthedfaceofthedlcubicleThen,welabeledthecorrespondingcubiefaces0andtheotheredgecubiefaces1.Finally,wedenedytobethe12-tuple(y1;:::;y12)whereyiisthenumberinedgecubiclefacei.Since[D;R]onlyaectstheedgecubiesdf,dr,andbr,weknowrightawaythaty11,y10,andy6aretheonlyyithatmaybenonzero.Since[D;R]putsthebfaceofthebrcubicleinthedfaceofthedfcubicle,y11=0.Similarly,y10andy6areboth0.So,y=(0;0;0;0;0;0;0;0;0;0;0;0).YoumightwonderwhywebothertodenecongurationsoftheRubik'scubethisway,sinceitseemsmucheasiertogettheinformationdirectlyfrom(6.1).Themainreasonisthatwriting,,x,andyseparatelyallowsustorecognizepatternsmoreeasily(andprovethem!).vExercises1.SupposeyourRubik'scubeisintheconguration(;;x;y).IfyouapplythemoveDtothecube,itendsupinanewconguration(0;0;x0;y0).Writex0andy0intermsofxandy.DothesameforthemovesU,L,R,F,andB.Doyounoticeanypatterns?2.Writethecommutator[D;R]indisjointcyclenotation(becarefultokeeptrackoftheorientationsofthecubies).Whatistheorderof[D;R]?Canyound...(a)...amovewhichxesthepositions(butnotnecessarilyorientations)ofthebackcornercubiesandonefrontcornercubie?(b)...amovewhichleaves6cornercubiesxedandswitchestheother2?Tryusingthesemovestoputallthecornercubiesintherightpositions(butnotnecessarilyorienta-tions).Whatotherusefulmovescanyougetfrom[D;R]?3.(a)InthesubgrouphD;RiofG,ndamovewhichchangestheorientations(butnotpositions)oftwocornercubieswithoutaectinganyothercornercubies.Yourmovemaydoanythingtoedge23 cubies.(Hint:computeafewelementsofhD;Riandthinkaboutwhichpowersoftheseelementsaresimplest.Thentrytoputsomeofthesesimplethingstogether.)(b)Findamovewhichchangestheorientationsofthecubiesdbrandu butdoesnotaectanyotherco

23 rnercubies.24 7.GroupHomomorphismsInExam
rnercubies.24 7.GroupHomomorphismsInExample2.1,wesaidthatZ=4Zand(Z=5Z)shouldbethoughtofasthesamegroup.Afterall,theybothhave4elements,andadditioninZ=4Zbehavesexactlythesamewayasmultiplicationin(Z=5Z).Inordertounderstandthis,wereallyneedawaytotranslatebehaviorfromonegrouptoanother.Let'stakeanotherlookatExample2.1.WeobservedinthatexamplethattheadditiontableforZ=4Zandthemultiplicationtablefor(Z=5Z)werereallythesame.Essentially,ifwereplaced0,1,2,and3intheadditiontableforZ=4Zby1,2,4,3(respectively),thenwegotthethemultiplicationtablefor(Z=5Z).Anotherwayofthinkingaboutthisisthatwereallydenedafunctionf:Z=4Z!(Z=5Z)byf(0)=1,f(1)=2,f(2)=4,andf(3)=3.Thisfunctionhadaspecialproperty,though,whichwasthatittranslatedtheadditiontableforZ=4Ztothemultiplicationtablefor(Z=5Z).Howcanweexpressthis?Well,theideaisthat,foreverya;b2Z=4Z,theterma+bintheadditiontableforZ=4Zshouldcorrespondtothetermf(a)f(b)inthemultiplicationtablefor(Z=5Z).Thatis,f(a+b)=f(a)f(b)foralla;b2Z=4Z.Thisconditionexpressesthefactthatf\translates"theadditiontableforZ=4Ztothemultiplicationtablefor(Z=5Z).Wecangeneralizethisconditiontoanypairofgroups.Denition7.1.Let(G;)and(H;H)betwogroups.A homomorphismfromGtoHisamap:G!Hsuchthat(ab)=(a)H(b)foralla;b2G.Example7.2.Wecandeneamap:Z=4Z!(Z=5Z)bytaking(0)=1,(1)=2,(2)=4,and(3)=3.UsingtheadditiontableforZ=4Zandthemultiplicationtablefor(Z=5Z),wecancheckthat(a+b)=(a)(b)foralla;b2Z=4Z.Alternatively,observethat(x)=2xforallx,so(a+b)=2a+b=2a2b=(a)(b).vExample7.3.Wecandeneamapcorner:G!S8asfollows.AnymoveinGcertainlyrearrangesthecornercubiessomehow;thus,itdenesapermutationofthe8unorientedcornercubies.Thatis,anyM2Gdenessomepermtuation2S8.Letcorner(M)=.Thatis,corner(M)istheelementofS8whichdescribeswhatMdoestotheunorientedcornercubies.Forexample,weknowthat[D;R]hasdisjointcycledecomposition(dlfdf

24 rlfdfrdfdlrdf)(drbbrubdrubrrbdrub)(dfdrb
rlfdfrdfdlrdf)(drbbrubdrubrrbdrub)(dfdrbr).Therefore,corner(M)=(dlfdfr)(drbbru).Similarly,wecandeneahomomorphismedge:G!S12bylettingedge(M)betheelementofS12whichdescribeswhatMdoestothe12unorientededgecubies.Forexample,edge([D;R])=(dfdrbr).Finally,wecandenea\cube"homomorphismcube:G!S20,whichdescribespermutationsofthe20unorientededgeandcornercubies.Forexample,corner([D;R])=(dlfdfr)(drbbru)(dfdrbr).vExercises1.Let2S5bedenedby(1)=2,(2)=4,(3)=1,(4)=5,and(5)=3.Writeasaproductof2-cyclesinatleast3dierentways(thisispossibleby[PS5,#9]).Doyounoticeanypatternsinthewaysyouhavewritten?Howmany2-cyclesdidyouuse?2.Let:(Z;+)!(R�f0g;
)bedenedby(x)=2x.Provethatisahomomorphism.3.Findallhomomorphisms:(Z=4Z;+)!((Z=5Z);
).4.(a)FindamoveM2Gwhichswitchesthepositionsoftheurfandbdlcubieswithoutaectinganyothercornercubies.Whatiscorner(M)?Hint:Startwiththemoveyoufoundin[PS6,#2b].25 (b)LetC1andC2beanytwodistinctunorientedcornercubies.ProvethatthereissomemoveM2GwhichswitchesthepositionsofC1andC2withoutaectinganyothercornercubies.(Evenifyoudidn'tndthemovein[PS7,#4a],assumethatsuchamoveexists,andyoushouldbeabletodothisproof!)Whatiscorner(M)?5.Let:(G;)!(H;H)beahomomorphism.(a)Let1GbetheidentityelementofGand1HbetheidentityelementofH.Provethat(1G)=1H.(b)Provethat(g�1)=(g)�1forallg2G.(Firstmakesureyouunderstandexactlywhatthismeans!)6.Let:(G;)!(H;H)beahomomorphism.The imageofisdenedtobethesetim=f(g):g2Gg.ProvethatimisasubgroupofH.26 8.TheSignHomomorphismBy[PS5,#9],Snisgeneratedbythe2-cyclesinSn.Thatis,anypermutationinSncanbewrittenasaniteproductof2-cycles.However,anygivenpermutationofSncanbewrittenasaniteproductof2-cyclesininnitelymanyways,soitseemslikethereisnotmuchwecansayaboutthisproduct.SomepermutationsinSncanbewrittenasaproductofanevennumberof2-cycles;w

25 ecallthese even permutations.Otherpermut
ecallthese even permutations.OtherpermutationsinSncanbewrittenasaproductofanoddnumberof2-cycles;wecallthese odd permutations.Sofar,thereseemstobenoreasonthatapermutationcouldnotbebothevenandodd.However,itisinfacttruethatapermutationiseitherevenorodd,butnotboth.Unfortunately,adirectproofofthisfactisrathermessy;instead,wewillgiveaproofwhichisrelativelysimplebutusesanindirecttrick.Fixn,andletp(x1;:::;xn)beapolynomialinthenvariablesx1;:::;xn.Example8.1.Ifn=1,p(x1)isapolynomialinthevariablex1;thatis,p(x1)lookslikeamxm1+am�1xm�11+


+a0.So,p(x1)isasumoftermsthatlooklikeaxi1.Ifn=2,thenp(x1;x2)isasumoftermsthatlooklikeaxi1xj2.Ingeneral,p(x1;:::;xn)isasumoftermsthatlooklikeaxi11xi22


xinn.vIf2Sn,letpbethepolynomialdenedby(p)(x1;:::;xn)=p(x(1);:::;x(n)).Thatis,wesimplyreplacexibyx(i).Example8.2.Supposen=4,p(x1;x2;x3;x4)=x31+x2x3+x1x4,and2S4hascycledecomposition=(123).Then,(p)(x1;x2;x3;x4)=x3(1)+x(2)x(3)+x(1)x(4)=x32+x3x1+x2x4.vLemma8.3.Forany;2Sn,(p)=p.Thisstatementiseasytomisinterpret,sobeforewegiveaproof,let'sdoanexample.Example8.4.Asinthepreviousexample,letn=4,p(x1;x2;x3;x4)=x31+x2x3+x1x4,and=(123).Let=(13)(24).Weknowthatp=x32+x3x1+x2x4,so(p)=x3(2)+x(3)x(1)+x(2)x(4)=x34+x1x3+x4x2.Ontheotherhand,=(142),sop=x3()(1)+x()(2)x()(3)+x()(1)x()(4)=x34+x1x3+x4x2.vNow,wewillproveLemma8.3.Proof.Bydenition,(p)(x1;:::;xn)=p(x(1);:::;x(n)),so[(p)](x1;:::;xn)=p(x((1));:::;x((n))).Now,((i))=()(i),so[(p)](x1;:::;xn)=p(x()(1);:::;x()(n))=(p)(x1;:::;xn). Toproveourassertionaboutevenandoddpermutations,wewillapplyLemma8.3toaspecicpolynomial,namely
=Y1in(xi�xj):Example8.5.Ifn=3,
=(x1�x2)(x1�x3

26 )(x2�x3).vLemma8.6.Forany2Sn,

)(x2�x3).vLemma8.6.Forany2Sn,
=
.Example8.7.If=(132),then
=(x3�x1)(x3�x2)(x1�x2)=(x1�x2)(x1�x3)(x2�x3)=
.Ontheotherhand,if=(12),then
=(x2�x1)(x2�x3)(x1�x3)=�
.vNow,wewillproveLemma8.6.Asyoumightguessfromtheexamples,theideaistomatchtermsof
withtermsof
.Thatis,foreachtermxi�xjoftheproductfor
,eitherxi�xjoritsnegativeappearsintheproductfor
.27 Proof.Bydenition,
=Y1in(xi�xj);so
=Y1in(x(i)�x(j)):Inordertoshow
=
,wemustshowtwothings.First,foreachiandjwith1ijn,wemustshowthateitherx(i)�x(j)oritsnegativeappearsin
;thatis,eitherx(i)�x(j)oritsnegativehastheformxk�x`with1kln.Secondly,wemustshowthat,foreachiandjwith1ijn,eitherxi�xjoritsnegativeappearsin
.Since
and
havethesamenumberofterms,thesetwostatementstogetherprovethatthetermsof
and
matchup.Toprovetherststatement,allweneedtoshowisthateither(i)(j)or(j)(i);equivalently,weneedtoshowthat(i)6=(j)if1ijn.Thisistruebecauseisone-to-oneandi6=j.Toprovethesecondstatement,weneedtoshowthateitherxi�xjoritsnegativecanbewrittenasx(k)�x(`)with1k`n.Since2Sn,�12Sn;inparticular,�1isalsoabijection.Sincei6=j,�1(i)6=�1(j).Letkbethesmallerof�1(i)and�1(j),andlet`bethelarger.Then,1k`n,andxi�xjiseitherx(k)�x(`)oritsnegative. ByLemma8.6,wecandeneamap:Sn!f1gby
=()
.ByLemma8.3,
=(
)=[()
]=()
=()()
.Therefore,()=()().So,isahomomorphism.Wecallitthe sign homomorphism.Weclaimedatthebeginningthat()hadsomethingtodowiththenumberof2-cyclesinaproductdecom

27 positionof.Now,we'llprovethis.Theor
positionof.Now,we'llprovethis.Theorem8.8.Ifisa2-cycle,then()=�1.Proof.First,let=(12).Wecanwrite
=Y1in(xi�xj):Now,let'swritethetermswherei=1ori=2separately.Then,
=Y1n(x1�xj)Y2n(x2�xj)Y3in(xi�xj)=(x1�x2)Y2n(x1�xj)Y2n(x2�xj)Y3in(xi�xj)Therefore,
=(x(1)�x(2))Y2n(x(1)�x(j))Y2n(x(2)�x(j))Y3in(x(i)�x(j))=(x2�x1)Y2n(x2�xj)Y2n(x1�xj)Y3in(xi�xj)=�
Thus,wehaveprovedthestatementfor=(12).Wecouldgeneralizetheaboveargumenttoany2-cycle,butthereisaneasierway!Letbeany2-cycle.By[PS5,#12],isconjugateto(12).Thatis,=(12)�1forsome2Sn.Sinceisahomomorphism,()=()(12)()�1=(12)=�1. Sinceismultiplicative,if()=1,thenmustbeaproductofanevennumberof2-cycles.Similarly,if()=�1,thenmustbeaproductofanoddnumberof2-cycles.So,iseveni()=1,andisoddi()=�1.28 Exercises1.Let:(G;)!(H;H)beahomomorphism.IfSisasubsetofim,provethathSiisasubgroupofim.2.Provethatthehomomorphismcorner:G!S8isonto(equivalently,imcornerisS8).Whatdoesthistellyouaboutthepossiblepositionsofthecornercubies?3.SupposeyourRubik'scubeisintheconguration(;;x;y).IfyouapplythemoveM2GtotheRubik'scube,itendsupinanewconguration(0;0;x0;y0).Provethat0=corner(M)and0=edge(M).4.Let(;;x;y)beacongurationoftheRubik'scube.ProvethatthereisamoveM2Gwhichputsallofthecornercubiesinthecorrectpositions.5.Let:G!HbeahomomorphismandG0beasubgroupofG.Deneamap0:G0!Hby0(g)=(g).Provethat0isahomomorphism.Wecallthishomomorphismthe restriction of  to G0andwrite0=jG0.6.Findallhomomorphisms:Z!Z.Whichofthesehomomorphismsar

28 eisomorphisms?7.LetGbeagroupandleta2G.De
eisomorphisms?7.LetGbeagroupandleta2G.Deneamap:G!Gby(g)=a�1ga.Isahomomorphism?Isanisomorphism?8.WriteDR�1andD�1Rindisjointcyclenotation.Canyouusethesetondamovewhichchangestheorientationsoftwocornercubieswithoutaectinganyothercornercubies?9.Sofar,wehaveonlyusedtwistsofthe6facesD,U,L,R,F,andB.LetMRbeaclockwisetwist(lookingattherightface)ofthefacebetweentheleftandrightfaces.WriteMR,MRU,MRU�1,MRU2,andMR�1U2indisjointcyclenotation.Usetheseto...(a)...ndamoveinGwhichcycles3edgecubieswithoutaectinganyothercubies.(b)...ndamoveinGwhichchangestheorientationsof2edgecubieswithoutaectinganyothercubies.Note:RememberthatwedenedGtobethemovescomposedofsequencesofD;U;L;R;F;B;thatis,G=hD;U;L;R;F;Bi.Therefore,MRisnotanelementofG,soyouwillhavetorewriteyourmovestomakethemelementsofG.10.IsitpossiblefortheRubik'scubetobeinacongurationwhereexactlytwocubiesareinthewrongpositions?29 9.TheAlternatingGroupIntheprevioussection,wedenedwhatitmeantforanelementofSntobeevenorodd.Recallthat2Snisdenedtobeevenifitcanbewrittenasaproductofanevennumberof2-cycles,anditisdenedtobeoddifitcanbewrittenasaproductofanoddnumberof2-cycles.Example9.1.(12)(13)isevensinceitisaproductoftwo2-cycles.(12)isoddsinceitisaproductofone2-cycle.vWethenprovedthatanelementofSniseitherevenorodd,butnotboth.Thetoolweusedforthiswasthesignhomomorphism.Rememberthatthiswasahomomorphism:Sn!f1gsuchthat()=�1forany2-cycle.Sincethe2-cyclesgenerateSn,thispropertycharacterizesthehomomorphism.Infact,thewaywasactuallydenedisnolongerimportant!Sinceisahomomorphism,()=�1ifisodd,and()=1ifiseven.Example9.2.((12)(13))=1and((12))=�1.vExample9.3.(16342)=1because(16342)=(16)(13)(14)(12)iseven.vExample9.4.Ifisak-cycle,then()=(�1)k�1.Afterall,ifisak-cycle,thenwecanwrite=(a1a2:::ak)=(a1a2)(a1a3)


(a1ak).vThep

29 roductofanevenpermutationandanoddpermuta
roductofanevenpermutationandanoddpermutationisodd.Theproductoftwoevenpermutationsortwooddpermutationsiseven.Theinverseofanevenpermutationiseven,andtheinverseofanoddpermutationisodd.Therefore,wecandeneasubgroupofSnconsistingofalltheevenpermutations.Thisgroupiscalledthe alternating groupandisdenotedAn.Example9.5.IfM2Gisafacetwist(oneofD;U;L;R;F;B),thencube(M)isaproductoftwo4-cycles.A4-cycleisodd,soaproductoftwo4-cyclesiseven.Therefore,cube(M)iseven.SincethefacetwistsgenerateallofG,thismeansthatcube(M)isevenforallM2G.Thatis,cube(M)2A20forallM2G.Anotherwayofwritingthisistosaythatimcube(M)2A20.Now,cube(M)=corner(M)edge(M),soeithercorner(M)andedge(M)arebotheven,ortheyarebothodd.Thatis,corner(M)andedge(M)havethesamesign.SupposeyourcubeisinthestartcongurationandyoudothemoveMtoit.Then,itendsupinaconugration(;;x;y)where=corner(M)and=edge(M).Therefore,wehaveprovedthat,if(;;x;y)isavalidconguration,thenandhavethesamesign.vSinceAnconsistsofalloftheevenelementsofSn,Ancanalsobedescribedasf2Sn:()=1g.Thisdenitioncanbegeneralizedtoanyhomomorphism.Denition9.6.The kernelofahomomorphism:G!Hisdenedtobefg2G:(g)=1Hg,anditisdenotedbyker.Thatis,keristhepreimageof1HinG.Example9.7.Thekernelofthehomomorphismcube:G!S20consistsofallmovesoftheRubik'scubewhichdonotchangethepositionsofanyofthecubies.Thatis,kercubeconsistsofallmoveswhichonlyaecttheorientations,notthepositions,ofcubies.Asyoucanimagine,thisisausefulsettounderstand:ifyouhaveputallthecubiesintherightpositions,youwanttondmovesthatonlyaecttheorientationsofthecubies.vTheorem9.8.IfGandHaregroupsand:G!Hisahomomorphism,thenkerisasubgroupofG.30 Proof.ByLemma3.4,itsucestoshowthat,ifg;h2ker,thengh�12ker.So,letg;h2ker.Then,(gh�1)=(g)(h�1)sinceisahomomorphism=(g)

30 0;(h)�1by[PS7,#5b]=1H1�1Hsinceg;h2
0;(h)�1by[PS7,#5b]=1H1�1Hsinceg;h2ker=1HTherefore,gh�12ker. Example9.9.ThealternatinggroupAnisthekernelof:Sn!f1g.vExercises1.Let2S10bedenedby=(13)(2469)(149).Whatis()?Isevenorodd?2.If2Sn,provethat22An.3.If2Snhascycletypen1;:::;nr,whatis()?4.ProvethatAnisgeneratedbythesetof3-cyclesinSn.5.ProvethatSnisgeneratedby(12);(23);(34);:::;(n�1n).6.FindamoveM2Gwhichchangestheorientationsofthecubiesdfrandulbwithoutaectinganyothercornercubies.Whatiscorner(M)?Whatisastrategyforxingtheorientationsofallcornercubies?31 10.GroupActionsIftheRubik'scubeissomecongurationC=(;;x;y),thendoingamoveM2GputstheRubik'scubeinsomenewconguration.Let'swritethisnewcongurationasC
M.SupposetheRubik'scubestartsinthecongurationC.IfwedothemoveM1,thecongurationofthecubebecomesC
M1.IfwethendoanothermoveM2,thecongurationbecomes(C
M1)
M2.Ontheotherhand,whatwehavereallydoneisstartedwiththecongurationCandappliedthemoveM1M2,soanotherwaytowritethenewcongurationisC
(M1M2).Thatis,wehavejustshownthat(C
M1)
M2=C
(M1M2)forallcongurationsCandallmovesM1;M22G.Ifwedotheemptymove(theidentityelementeofG),thenthecongurationdoesnotchangeatall,soC
e=C.Thisisanexampleofamathematicalobjectcalleda\groupaction."Elementsofagroup(here,theelementsaremovesoftheRubik'scube)aectelementsofsomeset(thesetofcongurationsoftheRubik'scube).Wehaveactuallyusedgroupactionsalready;forinstance,tounderstandSn,westudiedhowelementsofSnaectedtheintegers1;:::;n.Togiveaformaldenition,werstneedsomenotation.IfS1andS2aretwosets,thenS1S2isthesetoforderedpairs(s1;s2)withs12S1ands22S2.Denition10.1.A (right) group actionofagroup(G;)ona(non-empty)setAisamapAG!A(thatis,givena2Aandg2G,wecanproduceanotherelementofA,whichwewritea
g)satisfyingthefollowingtwoproperties:1.(a
g1)
g2=a
(g1g2)forallg1;g22Ganda

31 2A.2.a
e=afora2A(here,eistheidentitye
2A.2.a
e=afora2A(here,eistheidentityelementofG).Thisisarightactionratherthanaleftactionbecauseweputtheelementsofthegroupontheright.Intherstcondition,a
g12A,so(a
g1)
g2makessense.Ontheotherhand,g1g22G,soa
(g1g2)alsomakessense.WhenwehaveagroupactionofGonasetA,wejustsay\GactsonA."Example10.2.ThegroupGactsonthesetofcongurations(;;x;y)oftheRubik'scube(weallowbothvalidandinvalidcongurations).vExample10.3.Snactsonthesetf1;:::;ng.Thegroupactionisdenedasfollows:giveni2f1;:::;ngand2Sn,leti
=(i).Tocheckthatthisreallyisagroupaction,observethati
()=()(i)=((i))=(i
)=(i
)
andi
1=1(i)=i.vExample10.4.Snactsonthesetofpolynomialsinthevariablesx1;:::;xn;infact,weusedthisactiontoprovetheexistenceofthesignhomomorphism.Namely,ifp(x1;:::;xn)wasapolynomial,wedenedanewpolynomialpbyp(x1;:::;xn)=p(x(1);:::;x(n)).Thiswasagainapolynomialinthevariablesx1;:::;xn.Weprovedthat(p)=p,anditisclearthatp1=p.Thus,ifwedenep
=p,wehaveagroupaction.vExample10.5.Thegroup(Z;+)actsonthesetRbya
g=g+aforg2Zanda2R.Afterall,(a
g1)
g2=(a
g1)+g2=(a+g1)+g2=a+(g1+g2)=a
(g1+g2)forallg1;g22Zanda2R.Moreover,a
0=0+a=0foralla2R.vExample10.6.Often,weareinterestedinthecasewhenthesetAisthegroupitself.Inthiscase,wesay32 thatthegroupactsonitself.Forinstance,wecandeneagroupactionasfollows:forg2Ganda2G,denea
g=ag,thenormalgroupmultiplicationofaandg(checkthatthisdenesagroupaction).WecallthistheactionofGonitselfbyrightmultiplication.vDenition10.7.IfGactsonasetA,thenthe orbitofa2A(underthisaction)isthesetfa
g:g2Gg.Example10.8.GactsonthesetofcongurationsoftheRubik'scube.TheorbitofthestartcongurationunderthisactionisexactlythesetofvalidcongurationsoftheRubik'scube.vExample10.9.InExample10.5,wesaidthat(Z;+)actsonthesetRbya
g=g+1forallg2Zanda2R.Thus,theorbitofaisthesetfa

32 +g:g2Zg,orthesetf:::;a�2;a�1;a;a+1
+g:g2Zg,orthesetf:::;a�2;a�1;a;a+1;a+2;:::g.Inparticular,a;a+1;a�1;:::allhavethesameorbit.Thereisadistinctorbitforeacha2[0;1).Therefore,wecanthinkofthesetoforbitsastheinterval[0;1).However,sincetheorbitof0isthesameastheorbitof1,wecouldalsothinkofthesetoforbitsas[0;1]with0and1viewedasthesamepoint.Onewaytovisualizethisistoimaginebendingtheinterval[0;1]aroundsothat0and1join|thisformsacircle!Thus,itisnaturaltothinkofthesetoforbitsofthisactionasformingacircle.vDenition10.10.Ifagroupactionhasonlyoneorbit,wesaythattheactionis transitive(orthatthegroupactstransitively).Example10.11.Gactsonthesetoforderedpairs(C1;C2)ofdierentunorientedcornercubies.Afterall,ifC1andC2aretwodierentunorientedcornercubies,applyingamoveM2GsendsthesecornercubiestotwodierentcornercubiclesC01andC02.Then,wecandenethegroupactionby(C1;C2)
M=(C01;C02).(Checkthatthisisagroupaction.)By[PS3,#5],thisactionistransitive.Inthesameway,Gactsonthesetoforderedtriples(C1;C2;C3)ofdierentunorientededgecubies.vWeoftenwanttoprovesomethingaboutallelementsofanorbit(forexample,wemightwanttoproveastatementaboutallvalidcongurationsoftheRubik'scube).Thefollowinglemmacanbeusefulinthesesituations.Lemma10.12.SupposeanitegroupGactsonasetA,andletSbeasetofgeneratorsofG.LetPbeapropertysuchthatthefollowingistrue:Whenevera2AsatisesPands2S,a
salsosatisesP.Then,ifa02AsatisesP,everyelementintheorbitofa0alsosatisesP.Proof.Let'sdeneanewpropertyQasfollows:sayg2GsatisespropertyQwhenthefollowingistrue:Whenevera2AsatisesP,a
galsosatisesP.Itsucestoshowthateveryg2GsatisespropertyQ.Afterall,thatwouldmeanthat,ifa02AsatisesP,thena0
gsatisesPforallg2G,whichisexactlywhatwewanttoshow.Byhypothesis,everyelementofSsatisespropertyQ.ByProposition4.9,allweneedtoshowisthat,ifg;h2GbothsatisfypropertyQ,thenghsatisespropertyQ.So,supposeg;h2GbothsatisfypropertyQ.ToshowthatghalsosatisespropertyQ,wewanttoshowth

33 at,ifa2AsatisesP,thena
ghalsosat
at,ifa2AsatisesP,thena
ghalsosatisesP.Supposea2AsatisesP.SincegsatisespropertyQ,a
gsatisespropertyP.SincehsatisespropertyQ,(a
g)
hsatisespropertyP.However,bythedenitionofagroupaction,(a
g)
h=a
gh.So,wehaveprovedthat,ifa2AsatisesP,thena
ghsatisesP.ThatmeansthatghsatisespropertyQ,whichnishesourproof. InthecaseoftheRubik'scube,wewilloftentrytoapplytheabovelemmatotheactionofthegroupGonthesetAofcongurations.Inparticular,ifweletS=fD;U;L;R;F;Bganda0bethestartconguration,thenwecanusethelemmatoprovethingsaboutallvalidcongurationsoftheRubik'scube.33 Exercises1.Provethatthegroup(nZ;+)actsonZbya
g=a+gforallg2nZanda2A.Whataretheorbitsofthisaction?Howmanydierentorbitsarethere?Doesthesetoforbitsremindyouofanythinginnumbertheory?2.LetGbeagroup.ProvethatGactsonGbya
g=g�1agforallg2Ganda2G.Wesaythat\Gactsonitselfbyconjugation."3.By[PS10,#2],SnactsonSnby
=�1forall2Snand2Sn.Whataretheorbitsofthisaction?(Youmightwanttorsttrytowriteouttheorbitsexplicitlyforn=3.)4.LetR2betheusualxy-plane,whichconsistsoforderedpairs(x;y)wherex;y2R.Provethatthegroup(R;+)actsonR2by(x;y)
r=(x+r;y)for(x;y)2R2andr2R.If(x;y)2R2,ndtheorbitof(x;y).Canyoudescribethesetoforbitsgeometrically?5.LetZ2bethesetoforderedpairs(z1;z2)wherez1;z22Z.Weaddtwoorderedpairsasfollows:(z1;z2)+(z3;z4)isdenedtobe(z1+z3;z2+z4).Provethat(Z2;+)isagroup,andprovethatthisgroupactsonR2by(x;y)
(z1;z2)=(x+z1;y+z2)forall(x;y)2R2and(z1;z2)2Z2.Canyoudescribethesetoforbitsgeometrically?6.LetAbethesetoforderedtriples(C1;C2;C3)whereC1,C2,andC3aredierentunorientededgecubies.Inclass,weexplainedhowGactsonA.Whatdoesitmean(intermsoftheRubik'scube)tosaythatthisactionhasonlyoneorbit?Convinceyourselfthattheactionreallyhasjustoneorbit.7.IfC1,C2,andC3areanythreedierentunorientededgecubies,provethatthereisamoveM2GsuchthatMdoesnotaectanycornercubiesand

34 ;edge(M)=(C1C2C3).8.SupposeyourRubik'scu
;edge(M)=(C1C2C3).8.SupposeyourRubik'scubeisinavalidconguration(e;;x;y)(thatis,allofthecornercubiesareintherightpositions).ProvethatisevenandthatthereisamoveM2Gwhichputsalloftheedgecubiesintherightpositions(withoutaectingthecornercubies).34 11.ValidCongurationsoftheRubik'sCubeNow,wewillputeverythingwehavelearnedtogethertogiveacharacterizationofthevalidcongurationsoftheRubik'scube.Theorem11.1.Aconguration(;;x;y)isvalidisgn=sgn,Pxi0(mod3),andPyi0(mod2).Therestofthissectionwillbedevotedtoprovingthistheorem.First,wewillshowthat,if(;;x;y)isvalid,thensgn=sgn,Pxi0(mod3),andPyi0(mod2).Intheprocess,wewillprovesomeslightlymoregeneralfactsthatwillbeusefulforprovingtheconverse.RecallthatGactsonthesetofcongurationsoftheRubik'scube.Thevalidcongurationsformasingleorbitofthisaction.So,itmakessensethatstatementswemakeaboutvalidcongurationscanbegeneralizedtootherorbits.Lemma11.2.If(;;x;y)and(0;0;x0;y0)areinthesameorbit,then(sgn)(sgn)=(sgn0)(sgn0).Proof.ByLemma10.12,itsucestoshowthat,if(0;0;x0;y0)=(;;x;y)
MwhereMisoneofthe6basicmoves,then(sgn)(sgn)=(sgn0)(sgn0).By[PS8,#3],0=corner(M)and0=edge(M).Therefore,(sgn0)(sgn0)=(sgn)(sgncorner(M))(sgn)(sgnedge(M)).IfMisoneofthe6basicmoves,thencorner(M)andedge(M)areboth4-cycles,sotheybothhavesign�1.Thus,(sgn0)(sgn0)=(sgn)(sgn). Corollary11.3.If(;;x;y)isavalidconguration,thensgn=sgn.Proof.ThisisadirectconsequenceofLemma11.2sinceanyvalidcongurationisintheorbitofthestartconguration(1;1;0;0). Lemma11.4.If(0;0;x0;y0)isinthesameorbitas(;;x;y),thenPx0iPxi(mod3)andPy0iPyi(mod2).Proof.InlightofProposition10.12,itsucestoshowthat,if(0;0;x0;y0)=(;;x;y)
Mwh

35 ereMisoneofthe6basicmoves,thenPx0iP
ereMisoneofthe6basicmoves,thenPx0iPxi(mod3)andPy0iPyi(mod2).Youshouldhavedonethisin[PS6,#1].Hereisatableshowingwhatx0andy0areif(0;0;x0;y0)=(;;x;y)
MandMisoneofthe6basicmoves.Ineachcase,itiseasytocheckthatPx0iPxi(mod3)andPy0iPyi(mod2).M x0andy0 D (x1;x2;x3;x4;x8;x5;x6;x7) (y1;y2;y3;y4;y5;y6;y7;y8;y10;y11;y12;y9)U (x2;x3;x4;x1;x5;x6;x7;x8) (y4;y1;y2;y3;y5;y6;y7;y8;y9;y10;y11;y12)R (x1;x7+1;x2+2;x4;x5;x6;x8+2;x3+1) (y1;y7;y3;y4;y5;y2;y10;y8;y9;y6;y11;y12)L (x4+2;x2;x3;x5+1;x6+2;x1+1;x7;x8) (y1;y2;y3;y5;y12;y6;y7;y4;y9;y10;y11;y8)F (x6+1;x1+2;x3;x4;x5;x7+2;x2+1;x8) (y1;y2;y8+1;y4;y5;y6;y3+1;y11+1;y9;y10;y7+1;y12)B (x1;x2;x8+1;x3+2;x4+1;x6;x7;x5+2) (y6+1;y2;y3;y4;y1+1;y9+1;y7;y8;y5+1;y10;y11;y12)Asanexample,we'llseehowtondx0whenMisthemoveR.Thecubiclesoftherighthandfacelooklikethis:35 u u u f r r r b f r r r b f r r r b d d d Thecubiclesarelabeledlikethis: 2 3 7 8 Therefore,iftheRubik'scubeisintheconguration(;;x;y),thecubiesontherightfacearelabeledlikethis: x2 x3 x2+2 x2+1 x3+2 x3+1 x7+1 x7+2 x8+1 x8+2 x7 x8 Ifwerotatethisfaceby90clockwise,thenthecubieslooklike: x7+1 x2+2 x7 x7+2 x2+1 x2 x8 x8+1 x3+2 x3 x8+2 x3+1 Thus,x0=(x1;x7+1;x2+2;x4;x5;x6;x8+2;x3+1).So,Px0i=Pxi+6Pxi(mod3). Corollary11.5.If(;;x;y)isavalidconguration,thenPxi0(mod3)andPyi0(mod2).Proof.ThisisadirectconsequenceofLemma11.4sinceanyvalidcongurationisintheorbitofthestartconguration(1;1;0;0). Thus,wehaveprovedonedirectonofTheorem11.1.Now,wewillprovetheconverse.Supposesgn=sgn,Pxi0(mod3),andPyi0(mod2).Wewanttoshowthatthereisaseriesofmoveswhich,whenappliedto(;;x;y),givesthestartconguration;thatis,iftheRubik'scubeisintheconguration(;;x;y),itcanbesolved.TheideaoftheproofisbasicallytowritedownthestepsrequiredtosolvetheRubik'scube.Thus,wewillprovethesefourfacts:1.If(;;x;y)isacongurationsuchthatsgn=sgn,Pxi0(mod3),

36 andPyi0(mod2),thenthereisamoveM2Gsu
andPyi0(mod2),thenthereisamoveM2Gsuchthat(;;x;y)
Mhastheform(1;0;x0;y0)withsgn0=1,Px0i0(mod3),andPy0i0(mod2).Thatis,wecanputallthecornercubiesintherightpositions.2.If(1;;x;y)isacongurationwithsgn=1,Pxi0(mod3),andPyi0(mod2),thenthereisamoveM2Gsuchthat(1;;x;y)
Mhastheform(1;0;0;y0)withsgn0=1andPy0i0(mod2).Thatis,wecanputallthecornercubiesintherightorientations(andpositions).3.If(1;;0;y)isacongurationwithsgn=1andPyi0(mod2),thenthereisamoveM2Gsuchthat(1;;0;y)
Mhastheform(1;1;0;y0)withPy0i0(mod2).Thatis,wecanputalltheedgecubiesintherightpositions(withoutdisturbingthecornercubies).4.If(1;1;0;y)isacongurationwithPyi0(mod2),thenthereisamoveM2Gsuchthat(1;1;0;y)
M=(1;1;0;0).Thatis,wecansolvethecube!36 Beforeprovingthese,let'spointoutausefulfact.Supposethat(;;x;y)satisessgn=sgn,Pxi0(mod3),andPyi0(mod2).Then,Lemma11.2and11.4showthat,forany(0;0;x0;y0)inthesameorbitas(;;x;y),sgn0=sgn0,Px0i0(mod3),andPy0i0(mod2).Thus,forexample,intherststatementabove,ifwecanprovethatthereisamoveM2Gsuchthat(;;x;y)
Mhastheform(1;0;x0;y0),itisautomaticthatsgn0=1,Px0i0(mod3),andPy0i0(mod2).Therefore,tonishtheproofofTheorem11.1,itsucestoprovethefollowingfourpropositions.Proposition11.6.If(;;x;y)isacongurationsuchthatsgn=sgn,Pxi0(mod3),andPyi0(mod2),thentheorbitof(;;x;y)containssomecongurationoftheform(1;0;x0;y0).Proposition11.7.If(1;;x;y)isacongurationwithsgn=1,Pxi0(mod3),andPyi0(mod2),thentheorbitof(1;;x;y)containssomecongurationoftheform(1;0;0;y0).Proposition11.8.If(1;;0;y)isacongurationwithsgn=1andPyi0(mod2),thentheorbitof(1;;0;y)containssomecongurationoftheform(1;1;0;y).Proposition11.9.If(1;1;0;y)isacon

37 2;gurationwithPyi0(mod2),thentheorb
2;gurationwithPyi0(mod2),thentheorbitof(1;1;0;y)con-tainsthestartconguration(1;1;0;0).Wewillprovetheseinorder.So,wewanttorstshowthatwecanputallthecornercubiesintherightpositions.Lemma11.10.Thehomomorphismcorner:G!S8isonto.Proof.By[PS5,#9],S8isgeneratedbythesetSof2-cyclesinS8.ItsucestoshowthatSimcorner.Afterall,ifSimcorner,thenS8=hSihimcorneriby[PS4,#2].By[PS7,#6],imcornerisagroup,sohimcorneri=imcornerby[PS4,#1].So,wewanttoshowthatevery2-cycleinS8isintheimageofcorner.In[PS6,#2b],youshouldhavefoundamovewhichswitchesjust2cornercubiesandleavestheothercornercubiesxed.OnesuchmoveisM0=([D;R]F)3,whichhasdisjointcycledecomposition(dbrurb)(druf)(brrf)(dflf).Then,corner(M0)=(dbrurb).So,weatleastknowthat(dbrurb)liesintheimageofcorner.LetC1andC2beanypairofcornercubies.By[PS3,#5],thereexistsamoveM2GwhichsendsdbrtoC1andurbtoC2.Let=corner(M).Then,(dbr)=C1and(urb)=C2.Sincecornerisahomomorphism,corner(M�1M0M)=corner(M)�1corner(M0)corner(M)=�1(dbrurb)=((dbr)(urb))by[PS5,#10]=(C1C2)Therefore,(C1C2)2imcorner,whichnishestheproof. ProofofProposition11.6.ByLemma11.10,thereexistsamoveM2Gsuchthatcorner(M)=�1.By[PS8,#3],(;;x;y)
M=(1;0;x0;y0)forsome02S12,x02(Z=3Z)8,andy02(Z=2Z)12. Next,wewillproveProposition11.7.Thebasicideafororientingallofthecornercubiescorrectlywastousemoveswhichchangetheorientationsofjust2cubies.First,wemustshowthatsuchmovesexist.Lemma11.11.IfC1andC2areanytwocornercubies,thereisamoveM2Gwhichchangestheorientations(butnotpositions)ofC1andC2andwhichdoesnotaecttheothercornercubiesatall.Moreover,thereissuchamoveMwhichrotatesC1clockwiseandrotatesC2counterclockwise.Proof.AsintheproofofLemma11.10,thepointistorstndasinglemoveM0whichchangestheorientationsof2cubiesandthenconjugateM0tondothermovesthatchangetheorientationsof2cubies.In[PS6,#

38 3],youshouldhavefoundsuchamove;onepossib
3],youshouldhavefoundsuchamove;onepossibilityisM0=(DR�1)3(D�1R)3,whichhasdisjointcycledecomposition(dfrrdffrd)(drbrbdbdr)(dfdrfrurbrdbdl).Then,corner(M0)=1and corner(M0)=(dbrrdbbrd)(drfrfdfdr).So,ifC1=dbrandC2=drf,thelemmaistrue.37 Now,wewillconjugatethismove.By[PS3,#5],thereexistsM2GwhichsendsdbrtoC1anddrftoC2.LetM0=M�1M0M.Byapplying[PS5,#10]tond corner(M0),weseethatM0changestheorientationsofC1andC2anddoesnotaecttheothercornercubies.Specically,M0rotatesC1clockwiseandrotatesC2counterclockwise. ProofofProposition11.7.SupposethattheRubik'scubeisinacongurationwhereatleasttwocornercubiesC1andC2havethewrongorientation.ByLemma11.11,thereisamovewhichrotatesC1clockwise,rotatesC2counterclockwise,anddoesnotaecttheothercornercubies.Byapplyingthismoveonceortwice,wecanensurethatC1hasthecorrectorientation.SincethismovedoesnotaectanycornercubiesbesidesC1andC2,theRubik'scubenowhasonefewercornercubiewithanincorrectorientation.Doingthisrepeatedly,weendupwithaconguration(1;0;x0;y0)wherethereisatmostonecornercubiewiththeincorrectorientation.Thatis,atleast7ofthex0iare0.ByLemma11.4,Px0iPxi0(mod3),soitmustbethecasethatthelastx0iisalso0,sothecongurationoftheRubik'scubeis(1;0;0;y0). Next,wewanttoproveProposition11.9;thatis,wewanttoxthepositionsoftheedgecubies.TheideaoftheproofisverysimilartotheoneweusedtoproveProposition11.7.Recallthat,inthatcase,werstprovedthatcorner:G!S8isonto.Inthiscase,weonlywanttousemovesthatdon'taectthecornercubies,sincewehavealreadydonealotofworktogetthecornercubiesintherightpositionsandorientations.Therefore,insteadoflookingdirectlyatedge,wewilllookattherestrictionofedgetoker corner(see[PS8,#5]).Lemma11.12.Theimageofedgejker corner:ker corner!S12containsA12.Proof.By[PS9,#4],A12isgeneratedbythesetof3-cyclesinA12.BythesameargumentasintheproofofLemma11.10,itsucestoshowthatevery3-cycleisintheimageofedgejker corner.AsintheproofofLemma

39 11.10,thestrategyistouseconjugatesofasin
11.10,thestrategyistouseconjugatesofasinglemovetoprovethis.Youshoudlhavefoundamovein[PS8,#9a]thatdoesnotaectanycornercubiesbutcycles3edgecubies.OnesuchmoveisM0=LR�1U2L�1RB2,whichhasdisjointcycledecomposition(ubufdb).Then,M02ker corner,andedge(M0)=(ubufdb).By[PS10,#6],ifC1,C2,andC3areany3cornercubies,thereisamoveMoftheRubik'scubewhichsendsubtoC1,uftoC2,anddbtoC3.Then,by[PS5,#10],M0=M�1M0Mhasdisjointcycledecomposition(C1C2C3),soM02ker cornerandedge(M0)=(C1C2C3).Therefore,(C1C2C3)2imedgejker corner,whichcompletestheproof. Remark11.13.Infact,theimageofedgejker corner:ker corner!S12isexactlyA12,whichyoucanproveusingCorollary11.3.Now,Proposition11.8followsdirectlyfromLemma11.12.(TheproofisexactlythesameideaastheproofofProposition11.6.)Finally,wemustproveProposition11.9.ThisisquitesimilartoProposition11.7;rst,weneedananalogofLemma11.11.Lemma11.14.IfC1andC2areanytwoedgecubies,thereisamoveM2Gwhichchangestheorientations(butnotpositions)ofC1andC2andwhichdoesnotaecttheothercubiesatall.Proof.In[PS8,#9b],youshouldhavefoundamovewhichswitchestheorientationsof2edgecubieswithoutaectinganyothercubies.OnesuchmoveisLR�1FLR�1DLR�1BLR�1ULR�1F�1LR�1D�1LR�1B�1LR�1U�1(thismoveisdescribedmoreeasilyas(MRU)4(MRU�1)4).CallthismoveM0;ithasdisjointcycledecom-position(fuuf)(buub).By[PS10,#6],Gactstransitivelyonthesetoforderedtriples(C1;C2;C3)whereC1,C2,andC3aredierentedgecubies.Inparticular,ifC1andC2areanytwodierentedgecubies,thereexistsM2GsendinguftoC1andubtoC2.By[PS5,#10],MM0M�1changestheorientationsofC1andC2anddoesnotaecttheothercubiesatall. 38 Now,theargumentweusedtoproveProposition11.7provesProposition11.9aswell.ThiscompletestheproofofTheorem11.1.Remark11.15.Earlier,wecalculatedthattherewere212388!12!possiblecongurationsoftheRubik'scube;now,Theorem11.1tellsusthatonly1 12ofthosearevalid.Ofcourse,thismeanstherearestillmorethan41019validcongurations,nosmallnu