THE AUTOMORPHISM TOWER OF A CENTERLESS GROUP MOSTLY WITHOUT CHOICE

Thomas Simon

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Niveau: Supérieur, Master
ar X iv :m at h. LO /0 60 62 16 v 1 9 Ju n 20 06 THE AUTOMORPHISM TOWER OF A CENTERLESS GROUP (MOSTLY) WITHOUT CHOICE ITAY KAPLAN AND SAHARON SHELAH Abstract. For a centerless group G, we can define its automorphism tower. We define G?: G0 = G, G?+1 = Aut (G?) and for limit ordinals G? = ??

• group


• ?g


• zf ? ?a


• without choice


• smallest ordinal


• power ≤


• ?a


• given any centerless

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• Thomas

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• Publié par: cidur
• Nombre de lectures: 17
• Langue: English

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THEAUTOMORPHISMTOWEROFACENTERLESSGROUP (MOSTLY)WITHOUTCHOICE ITAYKAPLANANDSAHARONSHELAH Abstract. Foracenterlessgroup G ,wecandefineitsautomorphismtower.Wedefine G α : G 0 = G , G α +1 = Aut ( G α )andforlimitordinals G δ = α<δ G α .Let τ G bethe S ordinalwhenthesequencestabilizes.Thomas’celebratedtheoremsays τ G < 2 | G | + and
� .erom IfweconsiderThomas’prooftoosettheoretical,wehavehereashorterproofwithlittle settheory.However,settheoreticallywegetaparalleltheoremwithouttheaxiomof choice. Weattachtoeveryelementin G α ,the α -thmemberoftheautomorphismtowerof G ,a uniquequantifierfreetypeover G (whishisasetofwordsfrom G ∗h x i).Thissituation isgeneralizedbydefining“( G,A )isaspecialpair”. 1. Introduction background.GivenanycenterlessgroupG,wecanembedGintoitsautomorphism groupAut(G).SinceAut(G)isalsowithoutcenter,wecandothisagain,andagain. Thuswecandefineanincreasingcontinuousseries h G α | α ∈ ord i -Theautomorphism tower.Thenaturalquestionthatrises,iswhetherthisprocessstops,andwhen.Wedefine τ G =min { α | G α+1 =G α } . In1939(see[3])WeilantprovedthatforfiniteG,τ G isfinite.Butthereexistexamples ofcenterlessinfinitegroupssuchthatthisprocessdoesnotstopinanyfinitestage.For example-theinfinitedihedralgroupD ∞ =hx,y|x 2 =y 2 =1isatisfiesAut(D ∞ ) = ∼ D ∞ . Sothequestionremainedopenuntil1984,whenSimonThomas’celebratedwork(see[4]) provedthatτ G ≤ 2 |G| + .Helater(see[2])improvedthistoτ G <2 |G| + .
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� ThesecondauthorwouldliketothanktheUnitedStates-IsraelBinationalScienceFoundationforpartial supportofthisresearch.Publication882. 1

THEAUTOMORPHISMTOWEROFACENTERLESSGROUP(MOSTLY)WITHOUTCHOICE2 Foracardinalκwedefineτ κ asthesmallestordinalsuchthatτ κ >τ G forallcenterless groupsGofcardinality ≤ κ.AsanimmediateconclusionfromThomas’theoremwehave τ κ <(2 κ ) + . WealsodefinethenormalizertowerofH-asubgroupofagroupG-inG: hnor G α (H)|α∈ordibynor 0 G (H)=H,nor G α +1 (H)=nor(nor G α (H))andnor δ G (H)= { nor G α (H) | α<δ } forδlimit.Let S 1+ α  τ G,H =minα  nor G (H)=nor G α (H).Thisconstructionturnsouttobeveryuseful, thankstothefollowing: Foracardinalκ,letτ κ nlg bethesmallestordinalsuchthatτ κ nlg >τ Aut ( A ) ,H ,foreverystruc- tureofcardinality≤κandH≤Aut( A )ofcardinality≤κ. In[1],Just,ShelahandThomas,foundaconnectionbetweentheseordinals:τ κ ≥τ κ nlg . Inthispaperwedealwithanupperboundofτ κ ,butthereareconclusionregardinglower boundsaswell,andtheinequalityaboveisusedtoprovetheexistenceofsuchlowerbounds byfindingstructureswithlongnormalizertowers.In[4],Thomasprovedthatτ κ ≥ κ + , andin[1]theauthorsfoundthatonecannotproveinZFCabetterexplicitupperbound forτ κ then(2 κ ) + (usingsettheoreticforcing).In[5],Shelahprovedthatifκisstronglimit singularofuncountablecofinalitythenτ κ >2 κ (usingresultsfromPCFtheory). ItremainsanopenquestionwhetherornotthereexistsacountablecenterlessgroupG suchthatτ G ≥ω 1 . Inasubsequentpaperweplantoprovethatτ κ nlg ≤ τ κ istrueevenwithoutchoice. Results.Ourmaintheorem:(ofcourse,Thomas’didnotneedtodistinguishGand ω> G) Theorem1.1.(ZF)τ |G| <θ P( ω> G) foracenterlessgroupG.Thatis,thereisanordinalα andafunctionfrom ω> Gontoitsuchthatτ G <α.Moreover,τ G′ <αforeverycenterless groupG ′ suchthat|G ′ |≤|G|. Thisisanessentiallytheorem3.16. Wedealwithfindingτ G withoutchoice,anddiscoverthatThomas’theoremstillholds. WeprovethatgivenacertainalgebraicpropertyofGandasubsetA((G,A)isspecial

THEAUTOMORPHISMTOWEROFACENTERLESSGROUP(MOSTLY)WITHOUTCHOICE3 -seedefinition3.6)wecanreducethebound.Alongthewaywegiveadifferentproofof thetheoremwithoutchoiceinconclusion3.14(ThomasusedFodor’slemmainhisproof asyoucanseeinsection5,anditisknownthatitsnegationisconsistentwithZF).Then weconcludethatifV ′ isasubclassofVwhichisamodelofZFsuchthatP(κ)∈V ′ , thenτ κ <θ P( κ ) foreveryκandsoτ ℵ 0 <θ R .(seeconclusion3.19) �
V ′ L [ R ] Moreover,wegiveadescriptivesettheoreticapproachtofindingτ ℵ 0 insection4. Finally,wereturntotheaxiomofchoice,toseethatwecanimprovetheboundforcertain groupsthatholdaweakeralgebraicproperty((G,A)isweaklyspecial-seedefinition5.4). Anoteaboutreadingthispaper.Howshouldyoureadthispaperifyouarenotinter- estedintheaxiomofchoicebutonlyonthenewandsimpleproofofThomas’Theorem? Youcanreadonlysection3,andinthere,you: Startwithdefinition3.6.Continuetoclaim3.8,whichisverysimple.Thenconclusion 3.10isasimpleapplicationofthatlemma.Claim3.12Isaveryimportantsteptowards 3.13,andthenfinallyconclusion3.14wrapsitup. Notation1.2. (1)ForagroupG,itsidentityelement,willbedenotedase=e G . (2)ifA ⊆ Gthen h A i G isthesubgroupgeneratedbyAinG.Similarly,ifx ∈ G , h A,x i G isthesubgroupgeneratedbyA ∪{ x } . (3)Thelanguageofastructureisitsvocabulary. (4)Vwilldenotetheuniverseofsets;V ′ willdenoteatransitiveclasswhichisamodel .FZfo 2. TheNormalizerTowerWithoutChoice Definition2.1. (1)ForagroupGandasubgroupH ≤ G,wedefinenor G α (H)foreveryordinalnumber :ybα • nor 0 G (H)=H. •nor α G +1 (H)=nor G (nor G α (H)).

THEAUTOMORPHISMTOWEROFACENTERLESSGROUP(MOSTLY)WITHOUTCHOICE4 • nor δ G (H)= { nor G α (H) | α<δ } ,forδlimit. S  (2)Wedefineτ G n , lg H =τ G,H =minα  nor G α+1 (H)=nor G α (H). (3)Forasetk,wedefineτ | n k l | g asthesmallestordinalα,suchthatforeverystructure A ofpowerk A k≤|k|,τ Aut ( A ) ,H <αforeverysubgroupH≤Aut( A )=Gofpower |H|≤|k|.Notethatτ | n k l | g =sup{τ G,H +1|forsuchGandH}. (4)Foracardinalnumberκ,defineτ κ nlg similarly. Remark2.2.Notethatτ | n k l | g iswelldefined(inZF)sincewecanrestrictourselvestostruc- tureswithlanguagesofpower ≤ n<ω | k | n anduniversecontainedink.Seeobservation P .3.2 Observation2.3. (1)(ZF)Foranystructure A whoseuniverseis | A | =Athereisastructure B such :taht • A , B havethesameuniverse(i.e.A=| B |). • A , B havethesameautomorphismgroup(i.e.Aut( A )=Aut( B )). •thelanguageof B isoftheformL B ={R a¯ |a¯∈ ω> A}whereeachR a¯ isalg(a¯) placerelation. (2)(ZFC)If A isinfinitethenthelanguageof B hascardinalityatmost | A | . Proof.Define B asfollows:itsuniverseis| A |.ItslanguageisL={R a¯ |a¯∈ n A,n<ω} whereR a ¯ B =o(a¯),whichisdefinedbyo(a¯)={f(a¯)|f∈Aut( A )}-theorbitofa¯under Aut( A ).  Definition2.4.ForasetA,wedefineθ A =θ(A)tobethefirstordinalα>0suchthat thereisnofunctionfromAontoα. Remark2.5. (1)ZFC ⊢ θ A = | A | + (2)ZF ⊢ θ A isacardinalnumber,andifAisinfinite(i.e.thereisaninjectionfromω intoA)thenθ A >ℵ 0 .

THEAUTOMORPHISMTOWEROFACENTERLESSGROUP(MOSTLY)WITHOUTCHOICE5 (3)Usually,weshallconsiderθ A V ′ whereV ′ isatransitivesubclassofVwhichisa modelofZF. Claim2.6.(ZF)IfGisagroup,H≤Gasubgroupthenτ G,H <θ G . Proof.Ifτ G,H =0itisclear.ifnot,defineF:G → τ G,H byF(g)=αiffg ∈ nor G α+1 (H) \ nor G α (H),andifthereisnosuchα,F(g)=0.Bydefinitionofτ G,H ,F isonto.Fromthedefinitionofθ,τ G,H <θ G .  Wecandoevenmore: Claim2.7.(ZF)τ | n k l | g <θ P( ω> k) . Proof.Let B k = { ( A ,f,x)  A isastructure,L A ⊆ ω> k, | A |⊆ k,f:k → k k, x∈G=Aut( A )andH≤G,H=image(f)} LetF: B k → τ | n k l | g bethefollowingmap:F( A ,f,x)=αiffx ∈ nor G α+1 (H) \ nor G α (H), andifthereisnosuchα,F(G,H,x)=0(whereG=Aut( A ),andH=image(f)).Since Fisontoτ | n k l | g ,itsenoughtoshowthatthereisaonetoonefunctionfrom B k to P ( ω> k). Butx∈G,hencex⊆k×kandf∈ kk k⊆k×k×k,and A isaseriesofsubsetsof ω> k,i.e.
� >ω afunctionin k P( ω> k)⊆ ω> k×P( ω> k),andwecanencodesuchafunctionasamember ofP( ω> k).(How?defineaninjectivefunctionf 1 : ω> k× ω> k→ ω> k,usingthedefinable injectivefunctioncd:ω × ω → ω.Then,definetheencodingf 2 : ω> k ×P ( ω> k) →P ( ω> k) usingf 1 ).Henceitisclear.  Claim2.8.AssumethatV ′ isatransitivesubclassofVwhichisamodelofZF,G∈V ′ ′′ agroup,H∈V ′ asubgroupthenτ G V ,H =τ G V ,H <θ G V . ′ Proof.Byclaim2.6,itremainstoshowthatτ G V ,H =τ G V ,H .Byinductiononα ∈ V ′ ,one ′ canseethat(nor G α (H)) V =(nor G α (H)) V (theformulathatsaysthatxisinnor G′ (H ′ )is boundedintheparametersG ′ andH ′ ). 

THEAUTOMORPHISMTOWEROFACENTERLESSGROUP(MOSTLY)WITHOUTCHOICE6 Itisalsotruethatτ | n k l | g ispreservedinV ′ ,foreveryk∈V ′ ,suchthatP( ω> k)∈V ′ : Claim2.9.AssumethatV ′ isatransitivesubclassofVwhichisamodelofZF.  V ′  V (1)IfP( ω> k)∈V ′ thenτ | n k l | g =τ | n k l | g <θ P V ( ′ ω> k ) . (2)Ifk=κacardinalnumberandP(κ)∈V ′ thenτ κ nlg =τ κ nlg <θ P V ( κ ) �
V ′ �
V ′ Proof.(2)followsfrom(1),aswehaveanabsolutedefinablebijectioncd: ω> κ → κ. Forasetk ∈ V ′ ,suchthat P ( ω> k) ∈ V ′ let A k ={(G,H)|Thereisastructure A ,with| A |⊆k,suchthat G=Aut( A )andH≤G,|H|≤|k|} ′ Itisenoughtoprovethat( A k ) V =( A k ) V ,becausebydefinition  nlg  V [ n V o τ |k| =τ G,H +1  (G,H)∈(A k ) ′ on [ =τ G,H +1  (G,H) ∈ ( A k ) V  nlg  V ′ ′ =τ |k| <θ P V ( ω> k ) ′′ Soletusprovetheaboveequality:(A k ) V ⊆(A k ) V ,sinceif(G,H)∈(A k ) V and A ∈V ′ ′ astructuresuchthatG=Aut( A )then A ∈Vand(Aut( A )) V =(Aut( A )) V ,because ′ (Aut( A )) V ⊆ k k⊆P(k×k)∈V ′ .So(G,H)∈(A k ) V ,aswitnessedbythesame structure. V Ontheotherhand,suppose(G,H) ∈ ( A k ).Solet A beastructureonksuchthat G=Aut( A ).Byobservation2.3,wemayassumethatL A = { R a ¯ | a¯ ∈ ω> k } ,andeachR a ¯ isalg(a¯)placerelation(Thisisnotnecessary,itjustmakesitmoreconvenient).Define
� X A =a¯ˆb ¯  lg(a¯)=lgb ¯ ∧b ¯ ∈R a ¯ A .Observethat: •X A ∈V ′ ,asX A ⊆ ω> k. • A canbedefinedusingX A :itsuniverseisk,andforeacha¯∈ ω> k,R a¯ =
� b ¯  lgb ¯ =lg(a¯)∧a¯ˆb ¯ ∈X A .

THEAUTOMORPHISMTOWEROFACENTERLESSGROUP(MOSTLY)WITHOUTCHOICE7 Soinconclusion, A ∈ V ′ ,andsoG ∈ V ′ asbefore.InadditionH ∈ V ′ ,becauseH istheimageofafunctionin kk k,and kk k ⊆P (k × k × k) ∈ V ′ .Bydefinition
�
� ′ V (G,H)∈(A k )andwearedone.  3. TheAutomorphismTowerWithoutChoice Definition3.1.ForacenterlessgroupG,wedefinetheseries h G α | α ∈ ord i : • G 0 =G. •G α+1 =Aut(G α ) •G δ =∪{G α |α<δ}forδlimit. Remark3.2.SinceGiscenterless,thismakessense-G = ∼ Inn(G)≤Aut(G),andAut(G) isagainwithoutcenter.SoweidentifyGwithInn(G),andsoG α ≤ G α +1 .Thisseriesis thereforemonotoneandcontinuous. Definition3.3. (1)Defineanordinalτ G byτ G =min { α | G α+1 =G α } .Weshallshowbelowthatτ G iswelldefined. (2)Forasetk,wedefineτ |k| tobethesmallestordinalαsuchthatα>τ G forall groupsGwithpower≤|k|. (3)Foracardinalnumberκ,defineτ κ similarly. Definition3.4.ForagroupG(notnecessarilycenterless)andasubsetA,wedefinean equivalencerelationE G,A byxE G,A yifftp qf (x,A,G)=tp qf (y,A,G)wheretp qf (x,A,G)= {σ(z,a¯)|a¯∈ n A,n<ω,σaterminthelanguageofgroups(i.e.aword) withparametersfromA, zisit’sonlyfreevariableandG | =σ(x,a¯)=e } Remark3.5.

THEAUTOMORPHISMTOWEROFACENTERLESSGROUP(MOSTLY)WITHOUTCHOICE8 (1)NotethatxE G,A yiffthereisanisomorphismbetween h A,x i G and h A,y i G taking xtoyandfixingA. (2)TherelationE G,A isdefinableandabsolute(sincetp qf (x,A,G)isabsolute-the formuladefiningitisbounded). Definition3.6.Wesay(G,A)isaspecialpairifA ⊆ G,GisagroupandE G,A = { (x,x) | x ∈ G } (i.e.theequality). Example3.7. (1)IfG= h A i G then(G,A)isspecial. (2)IfGiscenterlessthen(Aut(G),G)isspecial(seeclaim3.8),soingeneral,the converseof(1)isnottrue. (3)ThereisagroupGwithcentersuchthat(Aut(G),Inn(G))isspecial,e.g. Z /2 Z , tub (4)IfGisnotcenterlessthen(2)isnotnecessarilytrue,evenifthe | Z(G) | =2: Itisenoughtofindagroupwhichsatisfiestheseproperties: (a)Z(G)={a,e}wherea6=e. (b)H i ≤Gfori=1,2aretwodifferentsubgroupsofindex2. (c)Z(G)=Z(H i )fori=1,2 Letπbethehomomorphismπ:G→Aut(G)takinggtoi g (i g (x)=gxg −1 ).Then inn(G)=image(π  ).Wewishtofindx 1 6 =x 2 ∈ Aut(G)withx 1 E inn ( G ) ,Aut ( G ) x 2 .  agg ∈ /H Sodefinex i (g)= i .Sincex i 2 =id,x i π(g)x i −1 =π(x i (g))=π(g)  gg ∈ H i andthefactthatx i ∈ /Inn(G)(becauseZ(G)=Z(H i ))itfollowsthat tp qf (x 1 ,Inn(G),Aut(G))=tp qf (x 2 ,Inn(G),Aut(G)).Nowwehavetocon- structsuchagroup.Noticethatitisenoughtofindacenterlessgroupsatisfying onlythelasttwoproperties,sincewecantakeit’sproductwith Z /2 Z .Sotake G=D ∞ = h a,b | a 2 =b 2 =e i ,andH a =kerϕ a whereϕ a :G → Z /2 Z takesato1 andbto0.InthesamewaywedefineH b ,andfinish. Thefollowingisthecrucialclaim:

THEAUTOMORPHISMTOWEROFACENTERLESSGROUP(MOSTLY)WITHOUTCHOICE9 Claim3.8.AssumeG 1 E G 2 ,C G 2 (G 1 )= { e } andthat(G 1 ,A)isaspecialpair.Then (G 2 ,A)isaspecialpair. Proof.FirstweshowthatC G 2 (A)= { e } .Supposethatx ∈ C G 2 (A),soxax −1 =aforall a∈A.Sinceconjugationbyx(i.e.themaph7→xhx −1 inG 1 )isanautomorphismofG 1 , (asG 1 isanormalsubgroupofG 2 ),itfollowsfrom(G,A)beingaspecialpair(byremark 3.5,clause(1))thatitmustbeid.Hence,x∈C G 2 (G 1 ),butweassumedC G 2 (G 1 )={e} hencex=e. NextassumethatxE G 2 ,A ywherex,y ∈ G 2 andweshallprovex=y.Thereisan isomorphismπ: h x,A i G 2 →h y,A i G 2 takingxtoyandfixingA.Wewishtoshowthat x=y,soitisenoughtoshowthatx −1 π(x)∈C G 2 (A).Thisisequivalenttoshowing x −1 π(x)aπ(x −1 )x=a,i.e.x −1 π(xax −1 )x=a,i.e.π(xax −1 )=xax −1 (remember thatπ(a)=a)foreverya∈A.Butxax −1 isanelementofG 1 (asG 1 E G 2 ),and π: h xax −1 ,A i G 1 →h π(xax −1 ),A i G 1 mustbeidbecause(G 1 ,A)isaspecialpair,andwe aredone.  Note3.9.IfGiscenterlessthenG E Aut(G),andC Aut(G) (G)= { e } . Conclusion3.10.AssumeGiscenterlessand(G,A)isaspecialpairthen: (1)(G α ,A)isaspecialpairforeveryα ∈ ord. (2)C G α (A)= { e } foreveryα. Proof.(2)followsfrom(1).Prove(1)byinductiononα.Forlimitordinal,itsclearfrom thedefinitions,andforsuccessors,thepreviousclaimfinishesthejobusingtheabove note.  Conclusion3.11.Letγbeanordinal,Gacenterlessgroupthen: (1)C G γ (G)={e}. (2)nor G γ G β =G β +1 ,forβ<γ.
� (3)nor β G γ (G)=G β forβ ≤ γ.