f01n1gThesetofnonnegativeintegersisdenotedbyNWedenotethesetof nitesequencesofelementsofasetXbySeqXandthelengthofasequences2SeqXbylhsTheemptysequenceisdenotedbyForst2SeqXlettheco ID: 98417
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2M.CENCELJ,K.EDAANDA.VAVPETICgeneratortoaperfectgroupPcanbeextendedtoahomomorphismfromagropegrouptoP.NotethatincasetheperfectgroupPhastheOreproperty([8],[6])thateveryelementinPisacommutator,intheaboveprocess()wecanmakeeverygeneratorinthechosenbasisofFnasinglecommutatoroftwobasiselementsofFn+1.ThegroupobtainedinthiswayistheminimalgropegroupM.ClearlyeverygropegroupadmitsmanyepimorphismsontoM.InthesequelweshowthatMadmitsanontrivialhomomorphismtoanothergropegroupGonlyifthelatteristhefreeproductG=MKwhereKisagropegroup.GropeswereintroducedbyStan'ko[9].Theyhaveanimportantroleingeometrictopology([3],formorerecentuseindimensionthe-orysee[5]and[4]).TheirfundamentalgroupswereusedbyBerrickandCasacubertatoshowthattheplus-constructioninalgebraicK-theoryislocalization[2].Recently[1]suchagrouphasappearedintheconstructionofaperfectgroupwithanonperfectlocalization.Intherstpartofthepaperwegiveasystematicdenitionofgropegroupsandprovesometechnicallemmas.Inthesecondpartweprovethattheminimalgropegroupadmitsnontrivialhomomorphismstoalmostnoothergropegroupthusprovingthatthereexistatleasttwodistinctgropegroups.2.SystematicdefinitionofgropegroupsandbasicfactsForeverypositiveintegernletn =f0;1;:::;n�1g.Thesetofnon-negativeintegersisdenotedbyN.WedenotethesetofnitesequencesofelementsofasetXbySeq(X)andthelengthofasequences2Seq(X)bylh(s).Theemptysequenceisdenotedby;.Fors;t2Seq(X)lettheconcatenationbest2Seq(X).Foranon-emptysetAletL(A)bethesetfa;a�:a2Ag,whichwecallthesetofletters.Weidentify(a�)�witha.LetW(A)=Seq(L(A)),whichwecallthesetofwords.ForawordWa0an,deneW�a�na�0.WewriteWW0foridentityinW(A)whileW=W0foridentityinthefreegroupgeneratedbyA.Forinstanceaa�=;butaa�6;.Weadopt[a;b]=aba�1b�1asthedenitionofacommutator.Todescribeallthegropegroupsweintroducesomenotation.AgropeframeSisasubsetofSeq(N)satisfying:;2Sandforeverys2Sthereexistsn�0suchthat2n =fi2N:si2Sgandwewite!(s)=2n�1.Ifthereisnoambiguitywewrite!=!(s).ForeachgropeframeSweinduceformalsymbolscSsfors2SanddeneESm=fcSs:lh(s)=m;s2SgandafreegroupFSm=hESmi. 4M.CENCELJ,K.EDAANDA.VAVPETICThesuccessivelettertoWin(C0)ishead(em+1;n[cs1])=cs10:::0,butin(C1)itishead(em+1;n[c�s1])=head(em+2;n[cs1!(s1)])=cs1!(s1)0:::0.ThusthesuccessivelettertoWinXisnotuniquelydetermined.However,ifXWYforsomeY,thecase(C1)cannotappear,sotheheadofYisuniquelydeterminedascs100.Similarly,theprecedinglettertoWisnotuniquelydetermined{therearefourpossibilities:tail(em+1;n[cs1])=tail(em+2;n[c�s1!(s1)])=tail(em+3;n[c�s1!(s1)0])=c�s1!(s1)0:::0incase(C1)IfXWYforsomeY,thereisnoprecedinglettertoWtail(em;n[ct])=tail(em+1;n[c�t!])=tail(em+2;n[c�t!0])=c�t!0:::0incase(C1)ifX=em+1;n[ZctcsY]forsomeZ;Ytail(em;n[c�t])=tail(em+1;n[c�t0])=c�t0:::0incase(C1)ifX=em+1;n[Zc�tcsY]forsomeZ;Y.However,theprecedinglettertoWdeterminesthesuccesivelettertoWuniquely:IftheprecedinglettertoWisc�s1!(s1)0:::0thenwearein(C1)Inallothercaseswearein(C0).Observation2.2.Alettercs002W(En)forlh(s)=mpossiblyappearsinem;n[W0]inthefollowingcases.Whenn=m+1,cs0appearsonceinem;n[cs]andalsoonceinem;n[c�s].Accordingtotheincreaseofn,cs00appearsinmanyparts.cs00appears2n�m�1-timesinem;n[cs]andalso2n�m�1-timesinem;n[c�s].Lemma2.3.ForawordW2W(Em)andnm,em;n[W]isreduced,ifandonlyifWisreduced.Lemma2.4.ForareducedwordV2W(En)andnm,V2FmifandonlyifthereexistsW2W(Em)suchthatem;n[W]V.Proof.Thesuciencyisobvious.Toseetheotherdirection,letWbeareducedwordinW(Em)suchthatem;n[W]=VinFn.ByLemma2.3em;n[W]isreduced.SinceeveryelementinFnhasauniquereducedwordinW(En)presentingitself,wehaveem;n[W]V.Lemma2.5.LetmnandAbeanon-emptywordinW(En).LetX0AY0andX1AY1bereducedwordsinW(En)satisfyingX0AY0;X1AY12Fm.(1)IfAisnotsmall,X0A=2FmandX1A=2Fm,thentheheadsofY0andY1arethesame.(2)LetX0beanemptyword.IfAisnotsmallandA=2Fm,theheadsofY0andY1arethesame. 6M.CENCELJ,K.EDAANDA.VAVPETICU62Fm.SinceUWU�;UW�U�2Fm,theheadofWandthatofW�isthesamebyLemma2.5(3),whichcontradictsthatWiscyclicallyreduced.Lemma2.9.LetXYandYXbereducedwordsinW(En)fornm.IfXYandYXbelongtoFm,thenbothofXandYbelongtoFm.Proof.Wemayassumen�m.Whenn�m,theheadofem;n[W]foranon-emptywordW2W(Em)iscs00orcsk00wherelh(s)=mandk+1=fi2N:si2Sgiseven.(Whenn=m+1,thereappearsno00.)SinceX�Y�2FmandX�Y�isreduced,thetailofXisoftheformc�s00orc�sk00.Weonlydealwiththeformercase.SupposethatX=2Fm.SinceXY2FmandXYisreduced,XZem+1;n[cs1c�s0]forsomeZ.ThisimpliesX�em+1;n[cs0c�s1]Z�,whichcontradictsthatX�Y�2FmandX�Y�isreduced.NowwehaveX;Y2Fm.Lemma2.10.LetmnandA;B;CinW(En)ande6=ABCA�B�C�2Fm.IfABCA�B�C�isareducedwordandatleastoneofA;B;Cisnotsmall,thenA;B;C2Fm.Proof.SinceABCA�B�C�6=e,atmostoneofA;B;Cisempty.WhenCisempty,theconclusionfollowsfromLemma2.6andthefactthatBAB�A�isalsoreducedandBAB�A�2Fm.NowweassumethatA;B;Carenon-empty.IfAisnotsmall,thenABCA�2FmandB�C�2FmbyLemma2.6.SinceBCiscycli-callyreduced,A2FmandBC2FmbyLemma2.8.TheconclusionfollowsfromLemma2.9.InthecasethatCisnotsmall,theargu-mentissimilar.TheremainingcaseiswhenAandCaresmall.ThenABCA�B�C�2FmandCBAC�B�A�2FmimplyAC,whichcontradictsthereducednessofABCA�B�C�.Lemma2.11.LetmnandA;B;CinW(En)ande6=ABCA�B�C�2Fm.IfABCA�B�C�isareducedwordandA;B;Caresmall,thenoneofA;B;Cisempty.AssumeCisempty.Thenthereexistscs2EmsuchthatsisbinarybranchedandeitherAem+1;n[cs0]andBem+1;n[cs1],orAem+1;n[cs1]andBem+1;n[cs0].Proof.SinceA;B;Caresmall,allthewordsA;B;Candtheirinversesmustbesubwordsofem+1;n[csi],i=0;1,orem+1;n[c�si],foranelementcs2Em,andinparticularthateitherABCA�B�C�=em;n(cs)=em+1;n[cs0cs1c�s0c�s1] 8M.CENCELJ,K.EDAANDA.VAVPETICProof.Weonlyshow(1.1).Thenon-emptinessofB0followsfromABA�B�6=e.SinceABandA�B�arereduced,AB0andA�B�0arecyclicallyreducedandhencethesecondstatementfollowsfromLemma2.9.Lemma3.3.LetA;B;C2W(En)bereducedwords(possiblyempty)suchthatABCA�B�C�6=eandABandCA�B�C�arereducedwords.Thenthefollowinghold:(2.1)IfBB0C�,thenAB0andA�CB�0C�arereducedwordsandAB0A�CB�0C�=ABCA�B�C�.InadditionifAB0A�;CB�0C�2Fm,thenAB;CA�B�C�2Fm.(2.2)IfCB�C0,thenAC0andA�B�C�0BarereducedwordsandAC0A�B�C�0B=ABCA�B�C�.InadditionifAC0A�;B�C�0B2Fm,thenAB;CA�B�C�2Fm.(2.3)IfBB0Z�andCZC0fornon-emptywordsB0andC0andB0C0isreduced,thenAB0C0A�ZB�0C�0Z�isreducedandAB0C0A�ZB�0C�0Z�=ABCA�B�C�.InadditionifAB0C0A�;ZB�0C�0Z�2Fm,thenAB;CA�B�C�2Fm.Proof.(2.1)Therstpropositionisobvious.LetB0XB1X�foracyclicallyreducedwordB1.Since(AX)B1(AX)�;(CX)B�1(CX)�2Fm,AX;CX;B12FmbyLemma2.8.NowAB=(AX)B1(CX)�2FmandCA�B�C�=(CX)(AX)�(CB�0C�)2Fm.Wesee(2.2)similarly.For(2.3)observethefollowing.SincethebothB0andC0arenon-empty,B0C0andB�0C�0arecyclicallyreduced.Hence,usingLem-mas2.8and2.9,wehave(2.3).Thenexttwolemmasarestraightforwardandweomittheproofs.Lemma3.4.LetA;B;C2W(En)bereducedwords(possiblyempty)suchthatABA�CB�C�6=eandABandA�CB�C�arereduced.Thenthefollowinghold:(3.1)IfAA0B,thenA0BandA�0CB�C�arereducedandA0BA�0CB�C�=ABA�CB�C�.InadditionifA0BA�0;CB�C�2Fm,thenABA�;CB�C�2Fm.(3.2)IfBB0A,thenAB0andCA�B�0C�arereducedandAB0CA�B�0C�=ABA�CB�C�.InadditionifAB0;CA�B�0C�2Fm,thenABA�;CB�C�2Fm.(3.3)IfBB0ZandAA0Zfornon-emptywordsA0andB0andB0A�0isreduced,thenA0ZB0A�0CZ�B�0C�isreduced.Inad-ditionifA0ZB0A�0;CZ�B�0C�2Fm,thenABA�;CB�C�2Fm. 10M.CENCELJ,K.EDAANDA.VAVPETIClh(B)+(k�1)lh(A0A1)+lh(A0),wehaveBA1A0X1.LetA0A0andB0A1,thenwehavetheconclusion.Iflh(B)�lh(X0)+lh(A),wehavek�0andB0;B1suchthatB0B1X0A,B(B0B1)kB0,andB1isnon-empty.(WeremarkthatB0maybeempty.)Sincelh(B1B0)=lh(AX1),wehaveB1B0AX1.NowBX0A(B0B1)k�1B0(B0B1)k�1B0AX1holds.LetA0AandB0(B0B1)k�1,thenwehavetheconclusion.InLemma3.7wehaveA�B�X0ABX�0=X1X�0=(A0)�(B0)�X0A0B0X�0.Lemma3.8.LetA;B;X;Y2W(En)bereducedwords(possiblyempty)suchthatXandYarenon-empty,Y�A�B�YX�ABX6=e,Y�A�B�YandX�ABXarereducedwords,andthereducedwordofY�A�B�YX�ABXiscyclicallyreduced.IfY�A�B�YX�ABX2Fm,then(1)Y�A�B�Y;X�ABX2Fm,or(2)Y�A�B�YX�ABXisequaltocsorc�sforsomessuchthatlh(s)=mandsisbinarybranched.Proof.IfYX�isreduced,thenY�A�B�YX�ABXiscyclicallyre-duced.ByanargumentanalyzingtheheadandthetailofY�andXwecanseeY�A�B�Y;X�ABX2Fm.Otherwise,inthecancellationofY�A�B�YX�ABXtheleftmostY�ortherightmostXisdeleted.SinceY�A�B�YX�ABX6=eandlh(Y�A�B�Y)=2lh(Y)+lh(AB)andlh(X�ABX)=2lh(X)+lh(AB),lh(X)6=lh(Y).Wesupposethatlh(X)�lh(Y),i.e.theheadofY�isdeleted.ThenwehaveXZYforanon-emptywordZ.WerstanalyzeareducedwordofA�B�Z�ABZ,whereA�B�isdeleted.TheheadpartofZ�ABisBA.ApplyingLemma3.7forX0Z�andX1repeatedly,wehavereducedwordsA0andB0suchthatZ�A0B0Zisreduced,Z�A0B0B0A0X1forsomeX1,A�0B�0Z�A0B0Z=A�B�Z�ABZ,A;B2hZ;A0;B0iandlh(B0)lh(Z).ItneveroccursthatthebothA0andB0areempty,butoneofA0andB0maybeempty.IfB0=;,interchangetheroleofA0andB0andbyLemma3.7wecanassumeB0isnon-emptyandlh(B0)lh(Z).FirstwedealwiththecaseA0isempty.SincetheleftmostB�0isdeletedinthereductionofB�0Z�B0Z,wehavenon-emptyZ0suchthatZZ0B�0andhaveareducedwordZ�0B0Z0B�0withZ�0B0Z0B�0=B�0Z�B0Z.SincetheleftmostY�isdeletedinthereductionofY�B�0Z�B0ZYandZ�0B0Z0B�0Yisreduced,Z�0B0Z0B�0iscyclically 12M.CENCELJ,K.EDAANDA.VAVPETICProof.Wehaven�msuchthatu;v2Fn.Itsucestoshowthelemmaincasethatthereducedwordfor[u;v]iscyclicallyreduced.For,supposethatwehavetheconclusionofthelemmaintheindicatedcase.Let[u;v]2Fmand[u;v]=XYX�whereXYX�isareducedwordandYiscyclicallyreduced.Thenwehave[X�uX;X�vX]=X�[u;v]X=Y.OntheotherhandX;Y2FmbyLemma2.8.BytheassumptionatleastoneofX�uXandX�vXdoesnotbelongtoFm.Since[u;v]isconjugatetoYinFm,wehavetheconclusion.Letu;v2Fnsuchthat[u;v]6=eandthereducedwordfor[u;v]iscyclicallyreduced.Thereexistacyclicallyreducednon-emptywordV02W(En)andareducedwordX2W(En)suchthatv=X�V0XandthewordX�V0Xisreduced.LetU0beareducedwordforuX�.SinceV0isacyclicallyreducedword,atleastoneofU0V0andV0U�0isreduced.WhenU0V0isreduced,thereexistk0andreducedwordsY;A;BsuchthatY�ABYisreduced,U0Y�AVk0andV0BA.WhenV0U�0isreduced,thereexistk0andreducedwordsY;A;BsuchthatY�ABYisreduced,U0Y�A(V�0)kandV0BA.Inthebothbasesuvu�1=Y�ABYandv=X�BAX.WeremarkthatABandBAarecyclicallyreduced.WeanalyzeareductionprocedureofY�ABYX�A�B�Xinthefol-lowing.(Case0):XandYareempty.InthiscasethebothAandBarenon-emptyandcorrespondstoLemma3.2.Using(1.1)and(1.2)alternatelyand(1.3)possiblyasthelaststepweobtainareducedwordofABA�B�.IfthereducedwordXYZX�Y�Z�satisesthatoneofX;Y;Zisnotsmall,by(1)and(2)ofLemma3.6andapplyingLemma3.2repeatedlywecanseeA;B2Fm.Otherwise,oneofX;Y;Zisemptyand[u;v]=csor[u;v]=c�sforsomebinarybranchedswithlh(s)=mbyLemma2.11.(Case1):Yisempty,butXisnon-empty.(Case2):Xisempty,butYisnon-empty.Inthesecasesargumentsaresymmetric,weonlydealwith(Case1).ThereispossibilitythatoneofAandBmaybeempty,thoughatleastoneofAandBisnon-empty.WeassumethatAisnon-empty.WetraceLemmas3.3,3.4,3.5togetareducedwordofABX�A�B�X.Thenweapplyoneof(2),(3)and(4)ofLemma3.6tothereducedwordandapplyingLemma3.2repeatedlywegetareducedword.ThenwehaveA;B2Fm,whichimpliesu;v2Fm,or[u;v]=csetc.asin(Case0).(Case3):ThebothXandYarenon-empty. 14M.CENCELJ,K.EDAANDA.VAVPETIC(b)Neitherh(cs0u0)norh(cs0u1)belongstoFn+k;(2)Foreveryv2Seq(2 )withlh(v)=kthereexistsu2Seq(2 )suchthatlh(u)=kandh(cs0u)isconjugatetodt0vord�t0vinFn+k.Wehaveshownthatthisholdswhenk=0.Supposethat(1)and(2)holdfork.Letlh(u)=kandh(cs0u)isconjugatetodt0vord�t0vetc.Then[h(cs0u0);h(cs0u1)]isconjugateto[dt0v0;dt0v1]or[dt0v1;dt0v0]inFn+k+1.Weclaimh(cs0u0)2Fn+k+1.Toshowthisbycontradiction,supposethath(cs0u0)=2Fn+k+1.ApplyLemma3.9toFn+k+1,thenwehave[h(cs0u0);h(cs0u1)]isaconjugatetodtord�twithlh(t)=n+k+1inFn+k+1,whichisimpossiblesince[h(cs0u0);h(cs0u1)]2[Fn+k+1;Fn+k+1].Similarlywehaveh(cs0u1)2Fn+k+1.Ontheotherhand,neitherh(cs0u0)norh(cs0u1)belongsto[Fn+k+1;Fn+k+1]byLemma3.10.Henceatleastoneofh(cs0u00)andh(cs0u01)doesnotbelongtoFn+k+1andconsequentlyneitherh(cs0u00)norh(cs0u01)be-longstoFn+k+1byLemma2.7.Henceh(cs0u0)isconjugatetodtord�twithlh(t)=n+k+1byLemma3.9.Similarly,h(cs0u1)isconjugatetodt0ord�t0withlh(t0)=n+k+1.Since[h(cs0u0);h(cs0u1)]isconjugateto[dt0v0;dt0v1]or[dt0v0;dt0v1]inFn+k+1,h(cs0u0)andh(cs0u1)areconjugatetodt0vjord�t0vjforsomej22 andforeachj22 theelementdt0vjisconjugatetoexactlyoneofh(cs0u0),h(cs0u1),h(cs0u0)�andh(cs0u1)�byLemma3.11.Hence(1)and(2)holdfork+1.Nowwehaveshowntheinductionstepandnishedtheproof.Remark3.12.ThoughtheconclusionofTheorem3.1israthersim-ple,embeddingsfromGS0intoGSmaybecomplicated.InparticularautomorphismsonGS0maybecomplicated,sincethefollowinghold:[dc�d�;dcd�c�d�]=dc�d�dcd�c�d�dcd�dcdc�d�=cdc�d�=[c;d]:References[1]BibliographyB.BadziochandM.Feshbach,`Anoteonlocalizationsofperfectgroups',PAMS133no.3(2004)693{697.[2]BibliographyA.J.BerrickandC.Casacuberta,`Auniversalspaceforplus-constructions',Topology1999.[3]BibliographyJ.W.Cannon,`Therecognitionproblem:Whatisatopologicalmanifold?',Bull.AMS84(1978)832{866.[4]BibliographyM.CenceljandD.Repovs,`Oncompactaofcohomologicaldi-mensiononeovernonabeliangroups',HoustonJ.Math.26(2000)527{536.[5]BibliographyA.N.DranishnikovandD.Repovs,`Cohomologicaldimensionwithrespecttoperfectgroups',TopologyandItsApplications74(1996)123{140.