Classification of Certain Families of Finite P-Groups

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Classification of Certain Families of Finite P-Groups
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english
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Brennan, Joseph P
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Doctorate ( Ph.D.)
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University of Florida
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Mathematics
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Turull, Alexandre
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Keating, Kevin
Sin, Peter
Crew, Richard M
Ray, Greg B

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algebra -- classification -- group -- p-group
Mathematics -- Dissertations, Academic -- UF
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Mathematics thesis, Ph.D.
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In 1999 Simon Blackburn published a classification of finite groups of prime powered order for which the derived subgroup is of prime order. A n-generalized Blackburn group is defined to be a group of prime powered order which is of nilpotence class 2 and for which the derived subgroup is elementary abelian of rank n. In 1949 George Szekeres presented a determination of finite p-groups possessing an abelian maximal subgroup. Following Dr. Szekeres' construction, in 1974 Sam Conlon presented a classification of finite p-groups possessing an abelian maximal subgroup which have cyclic centers, refereed to here as Conlon groups. This dissertation has two parts. First, I present a classification for the subclass of 2-generalized Blackburn groups which have an odd prime powered order and a center of index p3. The isomorphism classes for such groups of order p5 are enumerated and a general algebraic construction is presented. Second, I present a separate classification for Conlon groups and two separate constructions for Conlon groups.
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by Joseph P Brennan.
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Thesis (Ph.D.)--University of Florida, 2012.
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Adviser: Turull, Alexandre.
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CLASSIFICATIONOFCERTAINFAMILIESOFFINITEP-GROUPSByJOSEPHP.BRENNANADISSERTATIONPRESENTEDTOTHEGRADUATESCHOOLOFTHEUNIVERSITYOFFLORIDAINPARTIALFULFILLMENTOFTHEREQUIREMENTSFORTHEDEGREEOFDOCTOROFPHILOSOPHYUNIVERSITYOFFLORIDA2012

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c2012JosephP.Brennan 2

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ACKNOWLEDGMENTS Iwouldliketobeginbyacknowledgingmyadviser,AlexandreTurull.ThroughhispatienceanddirectionIwasabletobecomeamathematician.Iwillalwaysbegratefultotheexcellentadviceandconstantharassmentofmylovingparents:KevinBrennan(father),KarenPerrin(step-mother),IleenRoot(mother),andRichardRoot(step-father).Thankstomycommittee:PeterSin,KevinKeating,RichardCrew,andGregRay.Finally,forVanessaFreeze-Sumner,thankyouforbeingbymysidethroughitall;IamveryexcitedaboutourupcomingBinghamtonadventure. 3

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TABLEOFCONTENTS page ACKNOWLEDGMENTS .................................. 3 ABSTRACT ......................................... 5 CHAPTER 1INTRODUCTION ................................... 6 1.1GroupTheory .................................. 6 1.2ParticularClassesofp-Groups ........................ 9 1.2.1SimonBlackburn ............................ 9 1.2.2GeorgeSzekeresandSamConlon .................. 10 1.3DissertationOutline .............................. 11 2MATHEMATICALPRELIMINARIESFORCHAPTERS3AND4 ......... 13 2.1GeneralGroupTheory ............................. 13 2.2CharacteristicSubgroups ........................... 14 2.3NilpotentGroups ................................ 15 2.4GroupProducts ................................. 16 2.5VectorSpaceProperties ............................ 17 3GENERALIZEDBLACKBURNGROUPS ..................... 19 3.1Propertiesofn-generalizedBlackburngroups ................ 19 3.2AbelianMaximalSubgroup .......................... 21 3.3TheClassicationSet ............................. 26 3.4TheClassicationMap ............................. 30 3.5TheMainTheorem ............................... 39 3.6EnumerationofD(p5) ............................. 44 3.6.1ZpZpZpZp=A12A(p4) ................... 46 3.6.2Zp2ZpZp=A22A(p4) ...................... 47 3.6.3Zp2Zp2=A32A(p4) ......................... 54 4CONLONGROUPS ................................. 59 4.1Preliminaries .................................. 59 4.2TheConlonClass ............................... 62 4.3ConlonType .................................. 67 4.4ConlonGroupConstruction .......................... 75 4.5FurtherResults ................................. 85 REFERENCES ....................................... 90 BIOGRAPHICALSKETCH ................................ 91 4

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AbstractofDissertationPresentedtotheGraduateSchooloftheUniversityofFloridainPartialFulllmentoftheRequirementsfortheDegreeofDoctorofPhilosophyCLASSIFICATIONOFCERTAINFAMILIESOFFINITEP-GROUPSByJosephP.BrennanMay2012Chair:AlexandreTurullMajor:MathematicsIn1999SimonBlackburnpublishedaclassicationofnitegroupsofprimepoweredorderforwhichthederivedsubgroupisofprimeorder.An-generalizedBlackburngroupisdenedtobeagroupofprimepoweredorderwhichisofnilpotenceclass2andforwhichthederivedsubgroupiselementaryabelianofrankn.In1949GeorgeSzekerespresentedadeterminationofnitep-groupspossessinganabelianmaximalsubgroup.FollowingDr.Szekeres'construction,in1974SamConlonpresentedaclassicationofnitep-groupspossessinganabelianmaximalsubgroupwhichhavecycliccenters,refereedtohereasConlongroups.Thisdissertationhastwoparts.First,Ipresentaclassicationforthesubclassof2-generalizedBlackburngroupswhichhaveanoddprimepoweredorderandacenterofindexp3.Theisomorphismclassesforsuchgroupsoforderp5areenumeratedandageneralalgebraicconstructionispresented.Second,IpresentaseparateclassicationforConlongroupsandtwoseparateconstructionsforConlongroups. 5

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CHAPTER1INTRODUCTIONMathematiciansandscientistshaverelieduponclassication,theactofcollectingobjectsintoacommoncategorybasedonsharedcharacteristics,asameanstoorderourchaoticuniverseforgenerations.Biologistsclassifyorganismsintokingdoms,phylum,classes,orders,families,generas,andspecies.Chemistsclassifyandorderelementsbasedonatomicweights,theabundanceofprotonsandneutronspresentinthenucleusofanatom,andelectronstates.Astronomersclassifytheheavensbycelestialbodies,starsystems,galaxies,andnebulae.Mathematiciansdealmainlywithabstractionsofrealityandhaveroughlyclassiedtheseabstractionsintofoundations,analysis,geometry,andalgebra;thoughtherearenumeroussubclassicationsandthedivisionsbetweenthebranchesareoftenblurred.Theworkinthisdissertationrelatestoalgebra,thestudyofoperationsandrelations;specicallythestudyofgroups. 1.1GroupTheoryGroupsrepresentanabstractionofsymmetryfoundinthestudyofnumbertheory,geometry,algebraicequations,andmanyothersubjects.GroupswereoriginallyconceivedbyNielsHenrikAbelin1824andEvaristeGaloisin1830toaidinthesearchforaformularepresentingtherootsofalgebraicequations.Whiletherehasbeenmanydenitionsofthisabstraction,todayagroupisuniversallydenedasin[ 7 ]: Denition1.1.1. Agroupisanorderedpair(G,)whereGisasetandisabinaryoperationonGsatisfyingthefollowingaxioms: 1) Associativity:(ab)c=a(bc),foralla,b,c2G. 2) ThereexistsanelementeinG,calledtheidentityofG,suchthatforalla2Gwehaveae=ea=a. 3) Foreacha2Gthereisanelementa)]TJ /F5 7.97 Tf 6.59 0 Td[(1ofG,calledtheinverseofa,suchthataa)]TJ /F5 7.97 Tf 6.59 0 Td[(1=a)]TJ /F5 7.97 Tf 6.58 0 Td[(1a=e. 6

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Thisrelativelysimpledenitionofagroupmasksthecomplexityandvarietyofstructuresthatcanbelabeledasgroups,andamajorgoalofgrouptheoristshasbeentheclassicationofgroupsbyisomorphism.Anobviouswaytoclassifyanitegroupisbyit'ssize.Forexample,givenaprimep,thecyclicgroupistheuniquegroup,uptoisomorphism,oforderp.Therearenumeroustableslistinggroupsofsmallorderuptoisomorphism;suchastheSmallGroupsLibraryinGAP,acomputeralgebrasystem,whichallowsaccesstomostgroupsoforderlessthan1000[ 8 ].However,groupsaretoonumerousandcomplextohopetoclassifybyorderalone.Groupclassicationsusuallyfocusonstructureratherthansize.Anexampleofastructureclassicationwouldbetheclassicationofnitely-generatedabeliangroups,groupsconstructedbyanitenumberofelementswhichcommutewitheachother.Thesegroupswereconsideredearlyinthehistoryofgrouptheory;consequentially,theywerealsoclassiedearly,beingthedirectsumofnitecyclicgroups,correspondingtoatorsionsubgroup,andinnitecyclicgroups,correspondingtothefreerank.Duringthelatterhalfofthetwentiethcenturythegrouptheoreticalcommunitywascommittedtothesearchforandclassicationofnitesimplegroups,whicharegroupswhosepropernormalsubgroupsaretrivial.In1955theChevalleygroupswereintroducedtothemathematicalcommunity,whichledtoinnitefamiliesconsistingofsimplegroupsofLietype.In1963ThompsonandFeitpublishedtheOddOrderTheorem[ 12 ],allowinggrouptheoriststofocusongroupsofevenordergroup.In1972DanielGorensteinorganizedthegrouptheoreticalcommunityinanattempttoclassifythenitesimplegroups,andhewasabletoclaimsuccess,althoughprematurely,in1983.Uponcloserinspection,signicantholesappearedintheproofoftheclassicationandwererepaired.Notably,in2004MichaelAschbacherandStephenSmithpublisheda1,221pageclassicationoftheoverlookedquasi-thingroups[ 14 ].Today,mostgrouptheoristsbelievethattheclassicationofnitesimplegroupsiscorrect.Thenallistofnitesimplegroupsconsistsof 7

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1) Cyclicgroupsofprimeorder.(Theonlyp-groups.) 2) Alternatinggroupsofdegreeatleast5. 3) FamiliesofsimplegroupsofLietype. 4) The26sporadicsimplegroups.Theclassication,spreadbetweenhundredsofjournalarticleswithmorethanahundredauthors,iscurrentlybeingorganizedbyDr.Gorenstein(deceased),Dr.Lyons,andDr.Solomon[ 9 ].Themonumentalachievementoftheclassicationofnitesimplegroupscertainlytookadvantageofthepresenceofp-groupscontainedinsimplegroups.However,non-trivialp-groupshavenon-trivialcenters,hencethenon-abelianp-groupsareautomaticallynon-simple.Therefore,theclassicationofnitesimplegroupsdoesnotgiveinsightintotheclassicationofp-groups.Acompleteclassicationofnitep-groupsisconsideredtobeunobtainablebygrouptheoristsduetothesheerquantityofisomorphismclassesforgroupsofaspecicorder.In[ 5 ],theauthorsenumeratetheisomorphismclassesofgroupsoforderlessthan2001usingvariouscomputeralgorithms.Notably,thereare49,910,529,484isomorphismclassesofgroupsoforderatmost2000,amongthem49,487,365,422are2-groupsoforder1,024.Infact,manygrouptheoristsbelievethat2-groupsasymptoticallymakeupallthenitegroups[ 11 ].Moreover,thenumberofisomorphismclassesforgroupsoforderpn,representedbythefunctionf(pn),isboundedp2 27n3)]TJ /F7 7.97 Tf 6.58 0 Td[(O(n2)f(pn)p2 27n3+O(n8=3)([ 6 ],[ 13 ]).Earlyinthestudyofp-groups,1897,WilliamBurnsideclassiedthegroupsoforderdividingp4[ 10 ].Today,thegroupsoforderdividingp7havebeenclassiedandthegroupsoforderdividingp6canbeaccessedfromthesmallgrouplibraryinGAP.However,classifyingisomorphismclassesofp-groupsbyorderisviewedasinefcient,sothefocushasbeenondescribingclassesofnitep-groupsrelatedtotheirinternalstructure. 8

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1.2ParticularClassesofp-GroupsInthisSection,twofamiliesofp-groupsthathavesuccessfullybeenclassiedandarerelevanttothisdissertationarediscussed.WenowbrieypresenttheworkofSimonBlackburn,ontherstofthesefamilies,andtheworkofGeorgeSzekeresandSamConlon,onthesecondofthesefamilies. 1.2.1SimonBlackburnIn1999,SimonBlackburnidentiedtheisomorphismclassesoftheclassPconsistingofnitep-groupswithderivedsubgroupoforderp.BlackburnconstructedabijectionbetweentheisomorphismclassesofgroupsoforderpninPandthesetSn,denedbelow.Notably,thenumberofisomorphismclassesrelyonlyuponnandnottheprimep.NowwewilldenetheclassicationsetSn.AssumeG2P.SincejG0j=p,Ghasnilpotenceclass2andG0Z(G).SinceG0hasexponentp,itfollowsthatG0(G)Z(G).ThegroupV=G=Z(G)canbeviewedasavectorspaceoverFpsince(G)Z(G).Themap:G=Z(G)G=Z(G)!G0denedforx,y2Gby(xZ(G),yZ(G))=[x,y]isanon-degeneratealternatingbilinearform.Equippedwiththisform,Visasymplecticspace.AmaintoolofBlackburn'sclassicationistheclassicationoftheorbitsoftheagsofVundertheactionofthegeneralsymplecticgroup.TheisomorphismclassesisthendeterminedbytheorbitofaspecicaginG=Z(G)andthestructureofZ(G). Denition1.2.1. Letnbeapositiveinteger.LetSnbethesetoftriples(,e,A)suchthat: 1) =(k1,...,kr)isapartitionofsomeintegercwherec
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a) X(i,j)2Jc+1i,j=n)]TJ /F3 11.955 Tf 11.96 0 Td[(c 2. b) Forallintegersi,denemitobethenumberofpartsofofsizei.Thenforallk2f2,...,c+2g,X1iki,k+Xkjc+2k,j8>><>>:mk)]TJ /F5 7.97 Tf 6.58 0 Td[(1ifke.1ifk=e+1.me)]TJ /F4 11.955 Tf 11.95 0 Td[(1ifk=e+2.mk)]TJ /F5 7.97 Tf 6.58 0 Td[(2ifk>e+2.ThedenitionofAisrelatedtotheclassicationofagorbitsinVundertheactionofthegeneralsymplecticgroup.BlackburnconstructedabijectionbetweentheisomorphismclassesofgroupsoforderpninPandthesetSn. 1.2.2GeorgeSzekeresandSamConlonIn1949[ 2 ],GeorgeSzekeresprovidedthetoolstoclassifyallthenon-abeliangroupsGwithanormalabeliansubgroupAwithcyclicG=AsuchthatnoprimedivisorofjAjoccurstothesecondpowerinjG=Aj.LetPbeanon-abelianp-groupenumerablebySzekeres'techniquesandletAbeanabelianmaximalsubgroupofP.SinceAismaximalinanilpotentgroup,AisnormalinP.Forx2PnA,denethemapx:A!Atobetheautomorphismdenedbyconjugationbyx.DenetheA-endomomorphisms,x=x)]TJ /F4 11.955 Tf 11.96 0 Td[(IdAx=IdA+x+2x+...+p)]TJ /F5 7.97 Tf 6.58 0 Td[(1x.Szekeresusesx,x,andxptodetermineinvariantsfortheisomorphismclassesofnitenon-abelianp-groupscontaininganabelianmaximalsubgroup.In1974,SamConlon[ 4 ],[ 3 ]explicitlyclassiedaclassofnon-abelianp-groupsusingSzekeres'tools.Thesearethenon-abelianp-groupswhichhaveanabelianmaximalsubgroupandcycliccenter,referredtohereasConlongroups.HisclassicationresultwasimplicitinSzekeres'workandhegivesaproofofthisclassicationusingwreathproducts,centralproducts,andthefactthatthesegroupscanbeviewedasnitelinearp-groupsofdegreep. 10

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1.3DissertationOutlineThisdissertationstudiesindetailtwofamiliesofp-groups.ThisrstfamilyinvolveswhatIcalln-generalizedBlackburngroupsandachievestheircompleteclassicationinthecasewheren=2andthecenterofthegrouphasindexp3.Thesecondfamilyisthefamilyofgroupsthatcontainacycliccenterandanabelianmaximalsubgroup,referredtohereasConlongroups.HerewegiveanewproofoftheclassicationofConlongroups,provideaconstruction,andinvestigatetheirconjugacyclassesandanumberoftheircharacteristicsubgroupsincludingtheircentersandFrattinisubgroup.Inthisdissertationwedenean-generalizedBlackburngrouptobeanitep-groupwithnilpotenceclass2andelementaryabelianderivedsubgroupoforderpn(Denition 3.0.6 ).Chapter3pertainstoasubclassofthegeneralizedBlackburngroups;namely,theclassGBp(2,n)]TJ /F4 11.955 Tf 12.33 0 Td[(3,n)consistingofgeneralizedBlackburngroupswhoseelementaryabelianderivedsubgrouphasrank2andwhosecenterhasindexp3.ForP2GBp(2,n)]TJ /F4 11.955 Tf 10.94 0 Td[(3,n),thevectorspaceP=Z(P)overFpisofdimension3andmaybeequippedwiththenon-degeneratealternatingbilinearcommutatormap.InSection2,P=Z(P)isshowntopossessauniquetotally-isotropicsubspaceofdimensiontwo.ItthenfollowsthatPhasauniqueabelianmaximalsubgroup(Theorem 3.2.7 ).InSection3,aclassicationsetDisdened(Denition 3.3.14 )andinSection4aclassicationmapf4isconstructed(Denition 3.4.26 ).InSection5,anelementofGBp(2,n)]TJ /F4 11.955 Tf 12.15 0 Td[(3,n)isconstructedusinginvariantsfoundinD(Denition 3.5.3 )anditisshownthatf4providesabijectionbetweenDandtheisomorphismclassesofGBp(2,n)]TJ /F4 11.955 Tf 11.96 0 Td[(3,n)(Theorem 3.5.6 ).InSection6,thesetD(p5)isshowntohavep+8elements.AmainresultfromtheinvestigationofgroupsinGBp(2,n)]TJ /F4 11.955 Tf 13.09 0 Td[(3,n)isthattheypossessanabelianmaximalsubgroup.Therefore,isomorphismclassesofgroupsinGBp(2,n)]TJ /F4 11.955 Tf 12.65 0 Td[(3,n)areabletobeenumeratedviathetechniquesofSzekeres.IbegantostudyConlongroupsinanefforttofamiliarizemyselfwithSzekeres'techniques;thisinvestigationhasyieldedaseparateapproachtotheirclassication.Myapproachis 11

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moredirectthanConlon'soriginal,usingthegroupsthemselvesratherthanviewingthemassubgroupsofknowngroups.Thegoalwastondaproofthatwouldeasilyexplicitlyextendtofurtherclassesofp-groupswhereSzekeres'techniquesapply.InChapter4,Conlongroupsaredenedtobethenon-abelianp-groupswithcycliccenterandanabelianmaximalsubgroup(Denition 4.2.1 ).InSection2,connectionsbetweentheindexofthecenterandthenilpotenceclassofConlongroupsareexplored(Theorem 4.2.8 ).InSection3,aclassicationsetisdenedandrepresentedbyatripleofintegersandreferredtoasConlontriples(Denition 4.3.14 ).InSection4,twoisomorphicConlongroupsareconstructed(Denitions 4.4.9 and 4.4.11 )andabijectionisconstructedbetweenConlontriplesandtheisomorphismclassesofConlongroups(Theorem 4.4.15 ).Finally,inSection5,furtherresultspertainingtoelementorder,conjugacyclasses,andthestructureofcertaincharacteristicsubgroupsareshown. 12

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CHAPTER2MATHEMATICALPRELIMINARIESFORCHAPTERS3AND4Wenowwillremindthereaderofsomestandarddenitions,constructions,andresultsthatwillbeusedthroughoutChapters3and4(see[ 7 ]).ThroughoutthisChapter,pwilldenoteaprimenumberandGanitegroup.Throughoutthispaper,functionsandabeliangroupswillberepresentedadditively,whilegeneralgroupswillberepresentedmultiplicatively. 2.1GeneralGroupTheoryAnitegroupisap-groupifitsorderisapowerofp.AsubgroupNofGisnormal,denotedN/G,ifforallg2Gg)]TJ /F5 7.97 Tf 6.59 0 Td[(1Ng=fg)]TJ /F5 7.97 Tf 6.59 0 Td[(1ngjn2NgN.AgroupGiscalledAbelianifgh=hgforallg,h2G.TheexponentofanitegroupGistheminimumnumberrsuchthatr>0andgr=1forallg2G.Anitep-groupGiscalledelementaryabelianifGisabelianandhasexponentdividingp.Denotetherankofanelementaryabelianp-grouptobethemultiplicityoftheprimepintheorderofG.ThefreegroupF(S)withgeneratingsetSisdenedtobethegroupofallreducedwordsonthealphabetSwhereconcatenationofwordsfollowedbyreductionistheassociatedgroupoperation. Theorem2.1.1(UniversalPropertyofFreeGroups[ 7 ]). LetGbeagroup,Saset,and:S!Gasetmap.Let:S!F(S)betheinclusionsetmap.Thereexistsauniquegrouphomomorphism:F(S)!Gsuchthat=.AconsequenceoftheabovetheoremisthatanygroupGisthehomomorphicimageofsomefreegroupF(S).ApresentationofagroupGconsistsofapair(S,R),whereG=hSiandRisasetofwordsinF(S)suchthatthenormalclosureofhRiinF(S)isthekernelofasurjectivehomomorphism:F(S)!G. 13

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Theorem2.1.2(vonDyck'sTheorem). LetSbeaset,RasetofreducedwordsonS,andGagroupwithpresentation(S,R).LetHbeagroupandlet:S!Hbeaninclusionsetmap.IfHissuchthatH=h(S)iandHsatisestherelationsof(R),thenthereexistsasurjectivegrouphomomorphism :G!H.WeconcludethisSectionwithtwousefulcombinatorialidentities. Theorem2.1.3([ 15 ]). Forallnonnegativeintegersnandk,wehave 1) )]TJ /F7 7.97 Tf 5.62 -4.38 Td[(nk+)]TJ /F7 7.97 Tf 11.12 -4.38 Td[(nk+1=)]TJ /F7 7.97 Tf 5.61 -4.38 Td[(n+1k+1. 2) )]TJ /F7 7.97 Tf 5.48 -4.38 Td[(kk+)]TJ /F7 7.97 Tf 5.48 -4.38 Td[(k+1k+...+)]TJ /F7 7.97 Tf 5.61 -4.38 Td[(nk=)]TJ /F7 7.97 Tf 5.61 -4.38 Td[(n+1k+1. 2.2CharacteristicSubgroupsAsubgroupHofGischaracteristic,denotedHcharG,if(H)=HforeveryautomorphismofG.Characteristicsubgroupsareinvariantunderautomorphisms.Therefore,iftwogroupsarethoughttobeisomorphic,thenanisomorphismbetweenthegroupsshouldextendtoabijectionbetweenthesetsofcharacteristicsubgroups.ThecenterofagroupGisdenotedZ(G),whereZ(G)=fg2Gjga=agforalla2Gg.NotethatZ(G)charG.Thecommutatoroftwoelementsg,h2Gisdened[g,h]=g)]TJ /F5 7.97 Tf 6.59 0 Td[(1h)]TJ /F5 7.97 Tf 6.59 0 Td[(1gh. Theorem2.2.1(CommutatorCalculus[ 7 ]). LetGbeagroupandletx,y,z2G. 1) xy=x[x,y]. 2) [x,y]=[y,x])]TJ /F5 7.97 Tf 6.59 0 Td[(1. 3) [x,yz]=[x,z][x,y]zand[xy,z]=[x,z]y[y,z].Thecommutatormeasurestheextenttowhichagroupoperationfailstobecommutative.Noticethat[g,h]=1ifandonlyifgh=hg.ThecommutatorsubgroupofagroupGisdenotedbyG0,whereG0=h[g,h]jg,h2Gi.NotethatG0charG. 14

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AssumeGisanitep-group.TheithomegasubgroupofG,denotedi(G),isdenedi(G)=Dg2Gjgpi=1E.TheithagemosubgroupofG,denoted0i(G),isdened0i(G)=Dgpijg2GE.Notethatbothi(G)and0i(G)arecharacteristicinGforallnumbersi.TheFrattinisubgroupistheintersectionofallmaximalsubgroupsofG,andisdenoted(G).Notethat(G)charG. Theorem2.2.2([ 7 ]). IfGisap-group,thentheFrattinisubgroupofGisgeneratedbypthpowersandcommutatorsofit'selements.Specically,(G)=GpG0.Furthermore,thequotientG=(G)isanelementaryabeliangroup. 2.3NilpotentGroupsGrouptheoristsusecommutatorstomeasurehowcloseagroupistobeingAbelian.Letthe0thcenterofagroupGbedenotedbyZ0(G),whereZ0(G)=f1g.Fori1,theithcenterGisdenotedbyZi(G),whereZi(G)=fg2Gj[g,h]2Zi)]TJ /F5 7.97 Tf 6.58 0 Td[(1(G)forallh2Gg.NotethatZi(G)charGforallnumbersiandZ1(G)isthecenterofG.Theuppercentralseriesisdenedasf1g=Z0(G)/Z1(G)/.../Zi(G)/...AgroupGissaidtobenilpotentiftheuppercentralseriesreachesthetop,ie,ifthereexistsanintegeri0suchthatZi(G)=G.Ifiisthesmallestsuchinteger,thanGissaidtohavenilpotenceclassi.Notethatanon-trivialabeliangrouphasanilpotenceclassof1. Theorem2.3.1([ 7 ]). Notablepropertiesofnilpotentgroups: 15

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1) IfGisanitep-group,thenGisnilpotent. 2) Everymaximalsubgroupofanilpotentgroupisnormal. 3) IfGisnilpotentofclass2,thenthederivedsubgroupiscontainedinthecenter. 2.4GroupProductsGroupproductsarerulesthatcombinetwogroupsintoanewgroup. Denition2.4.1. Let(G1,1)and(G2,2)begroups.ThedirectproductG1G2isthegroupconsistingofthesetf(g1,g2)jg12G1,g22G2gandanoperationdenedcomponentwise,forg1,h12G1andg2,h22G2,(g1,g2)(h1,h2)=(g11h1,g22h2). Denition2.4.2. Let(G1,1)and(G2,2)begroupsandlet:G2!Aut(G1)beagrouphomomorphism.ThesemidirectproductG1oG2isthegroupconsistingofthesetf(g1,g2)jg12G1,g22G2gandanoperationdened,forg1,h12G1andg2,h22G2,(g1,g2)(h1,h2)=(g11(g2)(h1),g22h2).Notably,theorderofGHandGoHbothequaljGjjHj.Thesubgroup~G=f(g,1H)jg2GgofGoHisnormalandisomorphictoG.Thesubgroup~H=f(1G,h)jh2HgofGoHisisomorphictoH.Furthermore,ifistrivial,thenGHandGoHareequalasgroups.ThefollowingtheoremwillbeusedingroupconstructionsinChapters3and4: Theorem2.4.3. LetHandKbenitegroupssuchthatthereexistsagrouphomomor-phism:K!Aut(H).IfGisagroupsuchthatthereexistsgrouphomomorphismsH:H!G(injective)andK:K!GsatisfyingK(k)H(h)K(k)]TJ /F5 7.97 Tf 6.59 0 Td[(1)=H((k)(h))forallh2H,k2K.Thenthereexistsauniquegrouphomomorphism:HoK!Gsuchthat((h,1))=H(h)and((1,k))=K(k)forallh2H,k2K. 16

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2.5VectorSpacePropertiesForthisSection,FwilldenoteaeldandFpwilldenotetheeldwithp-elements.Anyelementaryabelianp-groupofrankrsatisesthepropertiesofanr-dimensionalvectorspaceoverFp. Denition2.5.1(TensorProduct[ 7 ]). LetVandWbevectorspacesoverF.LetGbethefreeabeliangrouponthecartesianproductVW.LetKbethesubgroupofGgeneratedbyallelementsoftheform: 1) (v1+v2,w))]TJ /F4 11.955 Tf 11.96 0 Td[((v1,w))]TJ /F4 11.955 Tf 11.96 0 Td[((v2,w); 2) (v,w1+w2))]TJ /F4 11.955 Tf 11.96 0 Td[((v,w1))]TJ /F4 11.955 Tf 11.95 0 Td[((v,w2); 3) (fv,w))]TJ /F4 11.955 Tf 11.95 0 Td[((v,fw);forv,v1,v22V,w,w1,w22W,andf2F.ThequotientgroupG=KisthetensorproductofVandW,denotedVFW.Thecoset(v,w)+KisdenotedvFw. Theorem2.5.2. LetVbeann-dimensionalvectorspaceoverFpandletFbeanalgebraicallyclosedeldsuchthatF=Fp.ThenFFpVisisomorphictoVasFpvectorspaces.Furthermore,iffeigni=1isabasisforV,thenf1Fpeigni=1isabasisforFFpV. Denition2.5.3(ExteriorTensorProduct[ 7 ]). LetVbeavectorspaceoverF.TheexteriortensorproductofVwithitself,denotedV^FV,isthevectorspaceVFVmodulothesubspaceIspannedbythesetoftensorsfvFvjv2Vg.Denotev^FwforthecosetvFw+I.IfVisavectorspaceoverFp,thendim(V^FpV)=)]TJ /F5 7.97 Tf 5.48 -4.38 Td[(dimV2. Denition2.5.4. LetR,S,TbevectorspacesoverF.Amap:RS!Tisbilinearif 1) (f1r1+f2r2,s)=f1(r1,s)+f2(r2,s). 2) (r,f1s1+f2s2)=f1(r,s1)+f2(r,s2).forallf1,f22F,r,r1,r22R,ands,s1,s22S.Amap:RR!Tisnondegenerateif(r,s)=0forallr2Rimpliesthats=0.Amap:RR!Tisalternatingif(r,r)=0forallr2R.Dene:VV!V^FVtobethecanonicalmap(v,w)=v^Fw.Thismapisalternatingandbilinear 17

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Theorem2.5.5(UniversalPropertyoftheExteriorProduct[ 7 ]). LetVandWbevectorspacesoverFandlet:VV!V^FVbethecanonicalmap.If:VV!Wisanalternatingbilinearmap,thenthereexistsauniquevectorspacehomomorphism:V^FV!Wsuchthat=. 18

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CHAPTER3GENERALIZEDBLACKBURNGROUPSAsoutlinedintheintroduction,SimonBlackburnin[ 1 ]presentedaclassicationofthenitep-groupswithprimeorderedderivedsubgroups. Denition3.0.6. Ap-groupisageneralizedBlackburngroupifithasnilpotenceclass2andhasanelementaryabelianderivedsubgroup.Specically,ap-groupwillbecalledan-generalizedBlackburngroupifitisageneralizedBlackburngroupanditsderivedsubgrouphasorderpn.ThisgeneralizationofthegroupsconsideredbyBlackburnisnaturalasitensuresP=Z(P)isasymplecticvectorspace,forPageneralizedBlackburngroup,equippedwiththealternatingbilinearcommutatormap.Underthisdenition,theworkofBlackburnrelatestothe1-generalizedBlackburngroups.Myresearchinvolves2-generalizedBlackburngroups,specicallytheclass: Denition3.0.7. DeneGBp(r,s,t)tobetheclassofr-generalizedBlackburngroupsoforderpt,podd,whosecenterisoforderps.TheprimaryachievementofthisChapteristheclassicationandconstructionofgroupscontainedinGBp(2,n)]TJ /F4 11.955 Tf 12.04 0 Td[(3,n)(Denition 3.5.3 andTheorem 3.5.6 ).ThisChapterwillconcludewithanenumerationofthep+8isomorphismclassesofGBp(2,2,5).ThroughoutthisChapter,letpbeanoddprimeandnanumber. 3.1Propertiesofn-generalizedBlackburngroups Lemma3.1.1. LetPbeageneralizedBlackburngroup.ThenP0(P)Z(P). Proof. SincePhasnilpotenceclass2,itfollowsfromTheorem 2.3.1 thatP0Z(P).ByTheorem 2.2.2 ,theFrattinisubgroupofap-groupPisgeneratedbythepth-powersandcommutatorsofallelements,implyingP0(P).Letx,y2P,usingLemma 2.2.1 wehavethat[xp,y]=[x,y]p=1wheretherstequalityholdssinceP0Z(P)and 19

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thesecondequalityholdssinceP0iselementaryabelian,hencehasexponentp.Wenowhavethatelementsgenerating(P)commutewithalltheelementsofP,therefore(P)Z(P). Denition3.1.2. LetPbeageneralizedBlackburngroup.DenethemapP:P=Z(P)P=Z(P)!P0byP(xZ(P),yZ(P))=[x,y]forx,y2P. Lemma3.1.3. IfPbeisageneralizedBlackburngroup,thenPisawell-dened,non-degeneratealternatingbilinearmap. Proof. Letx,y,r,s2Pbesuchthat(xZ(P),yZ(P))=(rZ(P),sZ(P)).Itfollowsthatx)]TJ /F5 7.97 Tf 6.59 0 Td[(1randy)]TJ /F5 7.97 Tf 6.59 0 Td[(1sareelementsofZ(P)andP(xZ(P),yZ(P))=x)]TJ /F5 7.97 Tf 6.59 0 Td[(1y)]TJ /F5 7.97 Tf 6.59 0 Td[(1xy=(x)]TJ /F5 7.97 Tf 6.59 0 Td[(1r)r)]TJ /F5 7.97 Tf 6.58 0 Td[(1(y)]TJ /F5 7.97 Tf 6.59 0 Td[(1s)s)]TJ /F5 7.97 Tf 6.58 0 Td[(1xy=r)]TJ /F5 7.97 Tf 6.59 0 Td[(1s)]TJ /F5 7.97 Tf 6.59 0 Td[(1x(x)]TJ /F5 7.97 Tf 6.59 0 Td[(1r)y(y)]TJ /F5 7.97 Tf 6.59 0 Td[(1s)=r)]TJ /F5 7.97 Tf 6.59 0 Td[(1s)]TJ /F5 7.97 Tf 6.59 0 Td[(1rs=P(rZ(P),sZ(P)).Hence,Piswell-dened.Forx,y,z2PP((xy)Z(P),zZ(P))=[xy,z]=[x,z]y[y,z]=[x,z][y,z]=P(xZ(P),zZ(P))P(yZ(P),zZ(P)),andP(xZ(P),(yz)Z(P))=[x,yz]=[x,z][x,y]z=[x,y][x,z]=P(xZ(P),yZ(P))P(xZ(P),zZ(P)).Therefore,Pisbilinear.Forx2P,P(xZ(P),xZ(P))=[x,x]=18x2P.Hence,Pisalternating.Assumethatx2PissuchthatP(xZ(P),yZ(P))=1forally2P.Thisimpliesthat[x,y]=1forally2Panditfollowsthatx2Z(P).Therefore,theradicalofPistrivialandPisnon-degenerate. 20

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Lemma3.1.4. LetAbeanabelianp-group.IfisanautomorphismofAsuchthatIm()]TJ /F4 11.955 Tf 11.95 0 Td[(IdA)ker()]TJ /F4 11.955 Tf 11.96 0 Td[(IdA)andIm()]TJ /F4 11.955 Tf 11.95 0 Td[(IdA)iselementaryabelian,thenIdA++...+p)]TJ /F5 7.97 Tf 6.59 0 Td[(1=pIdA. Proof. SinceIm()]TJ /F4 11.955 Tf 11.96 0 Td[(IdA)ker()]TJ /F4 11.955 Tf 11.96 0 Td[(IdA),wehavethat()]TJ /F4 11.955 Tf 11.96 0 Td[(IdA)=()]TJ /F4 11.955 Tf 11.95 0 Td[(IdA).Itfollowsthatfori2N,i)]TJ /F4 11.955 Tf 11.95 0 Td[(IdA=i)]TJ /F5 7.97 Tf 6.59 0 Td[(1()]TJ /F4 11.955 Tf 11.95 0 Td[(IdA)+...+()]TJ /F4 11.955 Tf 11.96 0 Td[(IdA)+()]TJ /F4 11.955 Tf 11.96 0 Td[(IdA)=i()]TJ /F4 11.955 Tf 11.95 0 Td[(IdA).WenowhaveIdA++...+p)]TJ /F5 7.97 Tf 6.58 0 Td[(1=pIdA+()]TJ /F4 11.955 Tf 11.96 0 Td[(IdA)+...+(p)]TJ /F5 7.97 Tf 6.59 0 Td[(1)]TJ /F4 11.955 Tf 11.96 0 Td[(IdA)=pIdA+p(p)]TJ /F4 11.955 Tf 11.96 0 Td[(1) 2()]TJ /F4 11.955 Tf 11.95 0 Td[(IdA)=pIdA.ThelastequalityisduetoIm()]TJ /F4 11.955 Tf 11.95 0 Td[(IdA)beingelementaryabelian. 3.2AbelianMaximalSubgroupForP2GBp(2,n)]TJ /F4 11.955 Tf 13.17 0 Td[(3,n),thequotientP=Z(P)iselementaryabeliansince(G)Z(P)(Lemma 3.1.1 ).ItfollowsthatVP=P=Z(P)isavectorspaceoverFpequippedwiththenon-degeneratealternatingbilinearmapP.SinceZ(P)hasindexp3,Vphasdimension3.WenowrestrictourattentiontothevectorspacesoverFpwithnon-degeneratealternatingbilinearmaps. Denition3.2.1. LetVbeavectorspaceoverFpandletA2GL(V).LetFbeanalgebraicallyclosedeldsuchthatF=Fp.Dene V=FFpV.Denoteby A:V^FpV!V^FpVthelinearmapsatisfying A(v^Fpw)=A(v)^FpA(w)forallv,w2V.Denoteby A: V! Vthelinearmapsatisfying A(fFpv)=fFpA(v)forallf2F,v2V. Lemma3.2.2. LetVbeann-dimensionalvectorspaceoverFpandletfeigni=1beabasisforV.Thereexistsavectorspaceisomorphism: V^Fp V! V^FpVsuchthat 21

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for1ijn,((1Fpei)^F(1Fpej))=1Fp(ei^Fpej). Proof. Let: V V! V^F Vbethemapdened((v,w))=v^Fwforallv,w2 V.Letfeigni=1beabasisforV,thenf1Fpeigni=1isabasisfor V(Theorem 2.5.2 ).Thereexistsauniquebilinearmap: V V! V^FpVinducedby(1Fpei,1Fpej)=1Fp(ei^Fpej)forintegers1ijn.Letf2Fandv=nXi=1aiei2V.Recallei^Fpei=0andei^Fpej=)]TJ /F4 11.955 Tf 9.3 0 Td[((ej^Fpei)forintegers1ijn.Now,(fFpv,fFpv)=f2 1Fp nXi=1aiei!,1Fp nXj=1ajej!!=f2nXi=1nXj=1aiaj(1Fpei,1Fpej)=f2nXi=1nXj=1aiajFp(ei^Fpej)=0.Thereforeisanalternatingmap.ByLemma 2.5.5 thereexistsauniquevectorspacehomomorphism: V^F V! V^FpVsuchthat=and((1Fpei)^F(1Fpej))=1Fp(ei^Fpej)forintegers1ijn.Sincef(1Fpei)^Fp(1Fpej)j1i
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A(f2)= A(e1^Fpe3)= A(e1)^Fp A(e3)=(ac)f2+(ab3)f3. A(f3)= A(e1^Fpe2)= A(e1)^Fp A(e2)=(ab)f3.Now,thematrixrepresentationfor Aonthisbasisis266664bc00a2cac0a2b3)]TJ /F3 11.955 Tf 11.96 0 Td[(a3bab3ab377775.ByLemma 3.2.2 A= A)]TJ /F5 7.97 Tf 6.59 0 Td[(1.Wehavedet( A)=det( A)=det( A)]TJ /F5 7.97 Tf 6.58 0 Td[(1)=(bc)(ac)(ab)=(abc)2.Wenowhaveestablishedthatdet( A)=det(A)2.Nowdet(A)6=0forA2GL(V),whichimpliesthatdet( A)6=0.Therefore A2GL(V^FpV). Denition3.2.4. LetVbea3-dimensionalvectorspaceoverFp.Dene:GL(V)!GL(V^FpV)by(A)= AforA2GL(V). Lemma3.2.5. LetVbea3-dimensionalvectorspaceoverFp.isahomomorphismsuchthat 1) jker()j=2. 2) Im()=f2GL(V^FpV)j9y2Fpsuchthaty2=det()g. 3) Im()istransitiveonthe1-dimensionalsubspacesofV^FpV. Proof. LetA,B2GL(V).Forv,w2V A B(v^Fpw)= A(B(v)^FpB(w))=AB(v)^FpAB(w)= AB(v^Fpw)Since A B= ABforelementarywedgesofV^FpVandtheelementarywedgesgenerateV^FpV, A B= ABforallelementsofV^FpV.Therefore(A)(B)=(AB)andisahomomorphism.LetA2ker(),thisimpliesthat A=IdV^FpV.Letfe1,e2,e3gbeabasisfor Vchosensothatthematrixrepresentationfor AoverFis266664aa2a30bb300c377775.Deningf1=e2^Fe3,f2=e1^Fe3,f3=e1^Fe2,wehavethatff1,f2,f3gisabasisfor V^F V.Now A= A)]TJ /F5 7.97 Tf 6.59 0 Td[(1(Lemma 3.2.2 ),sothematrixrepresentationfor Aoverthebasis 23

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f1Fpf1,1Fpf2,1Fpf3gis266664bc00a2cac0a2b3)]TJ /F3 11.955 Tf 11.95 0 Td[(a3bab3ab377775.Now A=Id V^FpV,since A=IdV^FpV.Wehavea,b,c6=0anda2,a3,b3=0.Nowdet( A)=1since1=det( A)=det( A)2.Sincebc,ac,ab=1anddet( A)=abc,wehavea=b=c=det( A).So A=Id VimplyingA=IdVorA=)]TJ /F4 11.955 Tf 11.29 0 Td[(IdV,hencejker()j=2.LetI=f2GL(V^FpV)j9y2Fpsuchthaty2=det()g.Let2ImandletA2GL(V)besuchthat A=(A)=.ByLemma 3.2.3 ,det()=det(A)2,hence2IandImI.NowjIm()j=jGL(V)j=2sinceisa2-to-1homomorphism.Therefore,jIj=jGL(V^FpV)j=2sincehalfthedeterminantsoftheautomorphismsofanitedimensionalvectorspaceoveraniteeldareaperfectsquare.GL(V)=GL(V^FpV)sinceVandV^FpVare3-dimensionalvectorspacesoverFp.Therefore,jIm()j=jIjandIm=I.Since12Fpisasquare,anyautomorphismofV^FpVwithdeterminant1canbeinducedbyanautomorphismofV.Hencethereare2orbitsofelementsinV^FpVofGL(V);f0gandfvjv6=0g.ThisimpliesthatGL(V)istransitiveon1-dimensionalsubspacesofV^FpV. Lemma3.2.6. LetVbea3-dimensionalvectorspaceoverFp.LetS1,S2besubspacesofdimension1ofV^FpVandsuppose!:(V^FpV)=S1!(V^FpV)=S2isanisomorphism.ThenthereexistsanA2GL(V)suchthatforallr2V^FpVwehave A(r)+S2=!(r+S1). Proof. ByLemma 3.2.5 ,thereexistsB2GL(V)suchthat B(S2)=S1and = Bj(V^FpV)=S2:(V^FpV)=S2!(V^FpV)=S1.Now !2GL((V^FpV)=S1). 24

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LetS1=hf1iandchooseabasisff1,f2,f3gforV^FpV.Let !berepresentedby264b11b12b21b223752GL((V^FpV)=S1)withdet( !)=b.Let2GL(V^FpV)beamapthatinduces !with(f1)=bf1.Nowcanberepresentedby266664b000b11b120b21b223777752GL(V^FpV).Wehavethatdet()=b2.ByLemma 3.2.5 thereexistsC2GL(V)suchthat C=.Wehaveforallr2V C(r)+S1= !(r+S1)) B)]TJ /F8 5.978 Tf 5.76 0 Td[(1 C(r)+S2=!(r+S1).Since B)]TJ /F8 5.978 Tf 5.76 0 Td[(1 C= B)]TJ /F8 5.978 Tf 5.75 0 Td[(1CandB)]TJ /F5 7.97 Tf 6.59 0 Td[(1C2GL(V),wehavethelemma. ReturningourattentiontoGBp(2,n)]TJ /F4 11.955 Tf 12.57 0 Td[(3,n)wearenowinapositiontostateandprovethisSection0smaintheorem. Theorem3.2.7. IfP2GBp(2,n)]TJ /F4 11.955 Tf 12.58 0 Td[(3,n),thenthereexistsauniqueabelianmaximalsubgroupMofP.Furthermore,Z(P)M. Proof. Bydenition,Z(P)hasindexp3inP.ByLemma 3.1.1 ,(P)Z(P).Therefore,VP=P=Z(P)canbeviewedasa3-dimensionalvectorspaceoverFp.ByTheorem 3.1.3 ,P:VPVP!P0denedby(xZ(P),yZ(P))=[x,y]forallx,y2Pisanon-degeneratealternatingbilinearmap.ByLemma 2.5.5 ,thereexistsalineartransformation:VP^FpVP!P0suchthat=.SincetheimageofPisasetofgeneratorsofP0,theimageofisthegeneratorsofP0wheredim(P0)=2whenviewedasavectorspaceoverFp.ByLemma 3.2.6 ,abasisfe1,e2,e3gcanbechosenforVPsuchthatker()=e2^Fpe3.Denethemap:P!P=Z(P)=VPby(x)=xZ(P)forx2PandletH=he2,e3iVP.DeneMtobethepreimageofHinPandnotethatZ(P)M.SinceZ(P)hasindexp3inPandHhasorderp2,itfollowsthatMhasindexpinP. 25

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AssumeJVP,dim(J)=2andJ6=H.SinceJ6H,thereexists1,2,32Fpwith16=0suchthatj1=1e1+2e2+3e32J.Sincedim(J)=2thereexists2,32Fp,notbothzero,suchthatj2=2e2+3e32J.Now,(j1,j2)=((j1,j2))=(12(e1^Fpe2)+13(e1^Fpe3)+(23)]TJ /F13 11.955 Tf 11.96 0 Td[(32)(e2^Fpe3))=12(e1^Fpe2)+13(e1^Fpe3)6=0,sinceatleastoneof12or13isnotzero.ThereforeJisnottotallyisotropic.Hence,Histheuniquetotallyisotropic2-dimensionalsubspaceofVP.Letx,y2M,andletxZ(P)=a12e2+a13e3andyZ(P)=a22e2+a23e3.Sinceei^Fpei=0foranyintegeriandker()=e2^Fpe3,wehavethat(e2,e3)=(e3,e2)=0.Therefore,[x,y]=(xZ,yZ)=(a12e2+a13e3,a22e2+a23e3)=a12a22(e2,e2)+a12a23(e2,e3)+a13a22(e3,e2)+a13a23(e3,e3)=0.Hence,allelementscommuteandMisabelian.LetNbeanabelianmaximalsubgroupofP.IfZ(P)6NthenwehaveP=Z(P)N,sinceNisamaximalsubgroupofP.SinceNisabelian,itfollowsthatPisabelian,acontradiction.WenowhavethatZ(P)NandJ=(N)Visa2-dimensionalsubspace.SinceNisabelian,wehave[a,b]=0fora,b2N.ItfollowsthatJisatotallyisotropic2-dimensionalsubspaceofV.HenceJ=HandN=M.ThereforeMistheuniqueabelianmaximalsubgroupofPandZ(P)M. 3.3TheClassicationSetThegoalofthisSectionistodeneasetofalgebraicinvariantsthatwillbeinbijectionwiththeisomorphismclassesofGBp(2,n)]TJ /F4 11.955 Tf 12.75 0 Td[(3,n).ThiswillbeachievedbydeningthreeintermediatesetsA,B,andC. Denition3.3.1. LetA(pn)beasetofrepresentativesfortheisomorphismclassesofabeliangroupsAoforderpnwhichpossess: 26

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1) AsubgroupDwhichiselementaryabelianoforderp2. 2) AsubgroupZoforderpn)]TJ /F5 7.97 Tf 6.59 0 Td[(2thatcontainsDandwhosequotientinAiselementaryabelian. Denition3.3.2. LetA2A(pn)]TJ /F5 7.97 Tf 6.59 0 Td[(1).DeneS(A)tobethesetofpairs(D,Z)where 1) DZA. 2) Diselementaryabelianoforderp2. 3) Zisoforderpn)]TJ /F5 7.97 Tf 6.59 0 Td[(3andA=Ziselementaryabelian. Lemma3.3.3. IfA2A(pn)and(D,Z)2S(A),then01(A)ZandD1(A). Proof. Bydenition,AmustcontainasubgroupZofindexp2whosequotientinAiselementaryabelian,hence01(A)Z.Bydenition,Diselementaryabelian,henceD1(A). Lemma3.3.4. LetA2A(pn)]TJ /F5 7.97 Tf 6.59 0 Td[(1).TheautomorphismsofAactonS(A). Proof. Let(D,Z)2S(A)andleta2Aut(A).Sinceaisanautomorphism;a(D)iselementaryabelianoforderp2,a(Z)isoforderpn)]TJ /F5 7.97 Tf 6.59 0 Td[(3.SinceDZAwehavea(D)a(Z)a(A)=A.Therefore(a(D),a(Z))2S(A). Denition3.3.5. ForA2A(pn)]TJ /F5 7.97 Tf 6.58 0 Td[(1)xR(A)tobeasetofrepresentativesfortheorbitsoftheactionofAut(A)onS(A).DeneB(pn)tobethesetoftriples(A,D,Z)whereA2A(pn)]TJ /F5 7.97 Tf 6.59 0 Td[(1)and(D,Z)2R(A). Denition3.3.6. Let(A,D,Z)2B(pn)anddenotebyAut(AjD,Z)thegroupofautomorphismsofAthatxthepair(D,Z).DeneC(A,D,Z)tobethesetofautomor-phismsofAwhere 1) ker()]TJ /F4 11.955 Tf 11.95 0 Td[(IdA)=Z. 2) Im()]TJ /F4 11.955 Tf 11.96 0 Td[(IdA)=D. Lemma3.3.7. Let(A,D,Z)2B(pn).ThenAut(AjD,Z)actsonC(A,D,Z).Further-more,p=IdAfor2C(A,D,Z). 27

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Proof. Let2C(A,D,Z)andleta2Aut(AjD,Z).Sinceaisanautomorphism,aa)]TJ /F5 7.97 Tf 6.59 0 Td[(12Aut(A).SinceaisanautomorphismofAxingZ,ker(aa)]TJ /F5 7.97 Tf 6.59 0 Td[(1)]TJ /F4 11.955 Tf 11.95 0 Td[(IdA)=ker(a()]TJ /F4 11.955 Tf 11.95 0 Td[(IdA)a)]TJ /F5 7.97 Tf 6.59 0 Td[(1)=a(ker()]TJ /F4 11.955 Tf 11.95 0 Td[(IdA))=a(Z)=Z.Similarly,sinceaisanautomorphismofAxingD,Im(aa)]TJ /F5 7.97 Tf 6.59 0 Td[(1)]TJ /F4 11.955 Tf 11.95 0 Td[(IdA)=Im(a()]TJ /F4 11.955 Tf 11.96 0 Td[(IdA)a)]TJ /F5 7.97 Tf 6.59 0 Td[(1)=a(Im()]TJ /F4 11.955 Tf 11.95 0 Td[(IdA))=a(D)=D.Thereforeaa)]TJ /F5 7.97 Tf 6.59 0 Td[(12C(A,D,Z);henceAut(AjD,Z)actsonC(A,D,Z).Leta2Aand2C(A,D,Z).Wehavethat(a)=a+dforsomed2D,sinceIm()]TJ /F4 11.955 Tf 12.5 0 Td[(IdA)=D.Also,(z)=zforz2Z,sinceker()]TJ /F4 11.955 Tf 12.5 0 Td[(IdA)=Z.Therefore,sinceDZ,wehave(a)p=p)]TJ /F5 7.97 Tf 6.59 0 Td[(1(a+d)=p)]TJ /F5 7.97 Tf 6.59 0 Td[(2(a+2d)=...=a+pd=a,withthelastequalityholdingduetoDbeingelementaryabelian.Thereforep=IdAforall2C(A,D,Z). Denition3.3.8. For(A,D,Z)2B(pn)let,2C(A,D,Z).DenearelationonC(A,D,Z)whereifthereexistsanautomorphisma2Aut(AjD,Z)andanintegermnotdivisiblebyp,suchthatm=aa)]TJ /F5 7.97 Tf 6.59 0 Td[(1. Lemma3.3.9. Let(A,D,Z)2B(pn).TherelationisanequivalencerelationonC(A,D,Z). Proof. Letr,s,t2C(A,D,Z).Sincer1=IdArId)]TJ /F5 7.97 Tf 6.58 0 Td[(1A,wehaverr;henceisreexive.Assumethatrs.Thisimpliesthatthereexistsm2Z,p-m,anda2Aut(AjD,Z)suchthatrm=asa)]TJ /F5 7.97 Tf 6.59 0 Td[(1.Fromelementarynumbertheory,thereexistsanintegerl2Z,p-l,suchthatml=1(modp).ByLemma 3.3.7 ,wehaverp=IdA.Therefore,r=rml=(rm)l=(asa)]TJ /F5 7.97 Tf 6.59 0 Td[(1)l=asla)]TJ /F5 7.97 Tf 6.59 0 Td[(1. 28

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Wenowhavesl=a)]TJ /F5 7.97 Tf 6.58 0 Td[(1rawitha)]TJ /F5 7.97 Tf 6.58 0 Td[(12Aut(AjD,Z).Thereforesrandissymmetric.Assumethatrsandst.Thisimpliesthatthereexistsl2Z,p-l,anda2Aut(AjD,Z)suchthatrl=asa)]TJ /F5 7.97 Tf 6.58 0 Td[(1andthereexistsm2Z,p-m,andb2Aut(AjD,Z)suchthatsm=btb)]TJ /F5 7.97 Tf 6.58 0 Td[(1.Wenowhavep-mlandbyLemma 3.3.7 ,rml=(asa)]TJ /F5 7.97 Tf 6.59 0 Td[(1)m=asma)]TJ /F5 7.97 Tf 6.59 0 Td[(1=(ab)t(ab))]TJ /F5 7.97 Tf 6.59 0 Td[(1.Sincea,b2Aut(AjD,Z),ab2Aut(AjD,Z),wehavert;henceistransitive.ThereforeisanequivalencerelationonC(A,D,Z). Denition3.3.10. For(A,D,Z)2B(pn)xR(A,D,Z)tobeasetofrepresentativesfortheequivalenceclassesoftherelationonC(A,D,Z).DeneC(pn)tobethesetofquadruples(A,D,Z,)where(A,D,Z)2B(pn)and2R(A,D,Z). Denition3.3.11. Let(A,D,Z,)2C(pn).Denethehomologyof(A,D,Z,)tobethegroupH(A,D,Z,)=CA()=(IdA++...+p)]TJ /F5 7.97 Tf 6.59 0 Td[(1)(A). Lemma3.3.12. Let(A,D,Z,)2C(pn).ThenH(A,D,Z,)=Z=01(A). Proof. Leta2CA(),implying(a)=aandhence()]TJ /F4 11.955 Tf 12.91 0 Td[(IdA)(a)=0.Recallthatker()]TJ /F4 11.955 Tf 12.21 0 Td[(IdA)=Zbydenition,implyingthata2Z;henceCA()Z.Letb2Z.Sinceker()]TJ /F4 11.955 Tf 11.96 0 Td[(IdA)=Z,(b)=()]TJ /F4 11.955 Tf 11.96 0 Td[(IdA+IdA)(b)=()]TJ /F4 11.955 Tf 11.95 0 Td[(IdA)(b)+IdA(b)=b.Wehaveb2CA(),henceCA()=Z.ByLemma 3.1.4 ,IdA++...+p)]TJ /F5 7.97 Tf 6.59 0 Td[(1=pIdAItfollowsthatIm(IdA++...+p)]TJ /F5 7.97 Tf 6.58 0 Td[(1)=01(A).Therefore,H(A,D,Z,)=Z=01(A). Corollary3.3.13. If(A,D,Z,)2C(pn),thenAut(AjD,Z)actsonH(A,D,Z,). 29

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Proof. ByLemma 3.3.12 ,H(A,D,Z,)=Z=01(A).Since01(A)ischaracteristicinAandAut(AjD,Z)xesZ,wehavethatAut(AjD,Z)actsonH(A,D,Z,). Denition3.3.14. For(A,D,Z,)2C(pn)xR(A,D,Z,)tobeasetofrepre-sentativesfortheorbitsoftheactionofAut(AjD,Z)onH(A,D,Z,).Finally,de-netheclassicationset,D(pn),tobethesetofallpentuples(A,D,Z,,r)where(A,D,Z,)2C(pn)andr2R(A,D,Z,). 3.4TheClassicationMapInthisSection,theclassicationmapF:GBp(2,n)]TJ /F4 11.955 Tf 13.11 0 Td[(3,n)!D(pn)willbeconstructed.SimilartoSection4,Fwillbeconstructedinthreestepswiththemapsf1:GBp(2,n)]TJ /F4 11.955 Tf 10.77 0 Td[(3,n)!A(pn)]TJ /F5 7.97 Tf 6.59 0 Td[(1),f2:GBp(2,n)]TJ /F4 11.955 Tf 10.77 0 Td[(3,n)!B(pn),andf3:GBp(2,n)]TJ /F4 11.955 Tf 10.77 0 Td[(3,n)!C(pn). Denition3.4.1. Denethemapf1:GBp(2,n)]TJ /F4 11.955 Tf 11.96 0 Td[(3,n)!A(pn)]TJ /F5 7.97 Tf 6.59 0 Td[(1)byf1(P)=A2A(pn)]TJ /F5 7.97 Tf 6.58 0 Td[(1)forP2GBp(2,n)]TJ /F4 11.955 Tf 11.95 0 Td[(3,n)ifthereexistsanisomorphism:M!A,whereMistheabelianmaximalsubgroupofP.DeneI1(P)=f:M!f1(P)janisomorphismg. Theorem3.4.2. Themapf1iswell-dened.IfP,Q2GBp(2,n)]TJ /F4 11.955 Tf 12.26 0 Td[(3,n)areisomorphic,thenf1(P)=f1(Q). Proof. LetP2GBp(2,n)]TJ /F4 11.955 Tf 12.54 0 Td[(3,n).ByTheorem 3.2.7 andLemma 3.1.1 ,thereexistsauniqueabelianmaximalsubgroupMofPanditsatisesP0(P)Z(P)M.NowMisabelianoforderpn)]TJ /F5 7.97 Tf 6.58 0 Td[(1withanelementaryabeliansubgroupP0oforderp2andasubgroupoforderpn)]TJ /F5 7.97 Tf 6.58 0 Td[(3,Z(P),whosequotientinMiselementaryabelian.Therefore,thereexistsanisomorphism:M!AforauniquelydeterminedA2A(pn)]TJ /F5 7.97 Tf 6.59 0 Td[(1).Hencef1iswell-dened.LetPandQbeinGBp(2,n)]TJ /F4 11.955 Tf 13 0 Td[(3,n)suchthatP=Q.Let:P!Qbeanisomorphism.LetM(resp.N)betheuniqueabelianmaximalsubgroupofP(resp.Q). 30

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Then,duetotheiruniqueness,(M)=NandinducesanisomorphismfromMtoN.Hence,wehavef1(P)=f1(Q). Lemma3.4.3. LetP2GBp(2,n)]TJ /F4 11.955 Tf 12.34 0 Td[(3,n),letMbetheabelianmaximalsubgroupofP,andletA=f1(P).Forany2I1(P),thepair((P0),(Z(P)))2S(A). Proof. ByTheorem 3.2.7 andLemma 3.1.1 ,P0(P)Z(P)M,hencewehave(P0)(Z(P))(M)=A.SinceP0iselementaryabelianoforderp2andisinjective,(P0)iselementaryabelianoforderp2.SinceZ(P)isoforderpn)]TJ /F5 7.97 Tf 6.59 0 Td[(3andisinjective,(Z(P))isoforderpn)]TJ /F5 7.97 Tf 6.58 0 Td[(3.Since(P)Z(P)M,wehavethatP=Z(P)iselementaryabelian,henceM=Z(P)iselementaryabelianwhichimpliesthat(M)=(Z(P))iselementaryabelian.Wenowhavethat((P0),(Z(P)))2S(A). Denition3.4.4. LetP2GBp(2,n)]TJ /F4 11.955 Tf 12.39 0 Td[(3,n),letMbetheabelianmaximalsubgroupofP,andletA=f1(P).DeneS(A,P)tobethesubsetofS(A)consistingofthepairs((P0),(Z(P)))where2I1(P). Lemma3.4.5. AssumePandQinGBp(2,n)]TJ /F4 11.955 Tf 13.03 0 Td[(3,n)aresuchthatP=Q.Ifr2S(f1(P),P)ands2S(f2(Q),Q),thenthereexistsa2Aut(f1(P))suchthatar=s. Proof. SinceP=Q,f1(P)=f1(Q)=AforsomeA2A(pn)]TJ /F5 7.97 Tf 6.58 0 Td[(1)(Theorem 3.4.2 ).Letr=(rd,rz)2S(A,P)andlet2I1(P)satisfy((P0),(Z(P)))=r.Lets=(sd,sz)2S(A,Q)andlet2I1(Q)besuchthat((Q0),(Z(Q)))=s.Let:P!Qbeanisomorphism,notingthat(P0)=Q0and(Z(P))=Z(Q)sinceboththederivedsubgroupandthecenterarecharacteristicinagroup.Leta=)]TJ /F5 7.97 Tf 6.59 0 Td[(1.Wehavethata2Aut(A)=Aut(f1(P))andar=((P0),(Z(P)))=((Q0),(Z(Q)))=s. Lemma3.4.6. IfP2GBp(2,n)]TJ /F4 11.955 Tf 12.03 0 Td[(3,n)andA=f1(P),thenS(A,P)isafullorbitofS(A)undertheactionofAut(A). 31

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Proof. ByLemma 3.4.5 ,S(A,P)iscontainedinanorbitofS(A)undertheactionofAut(A).Letr2S(A,P)anda2Aut(A).Assume2I1(P)satises((P0),(Z(P)))=randlet=a.Wehavethat((P0),(Z(P)))=ar.Sinceaandareisomorphisms,wehavethat2I1(P).Wenowhavear2S(A,P).ItfollowsthatS(A,P)isafullorbitofS(A)undertheactionofAut(A). Denition3.4.7. Denethemapf2:GBp(2,n)]TJ /F4 11.955 Tf 11.96 0 Td[(3,n)!B(pn)byf2(P)=(A,D,Z)2B(pn)forP2GBp(2,n)]TJ /F4 11.955 Tf 11.95 0 Td[(3,n),wheref1(P)=Aand((P0),(Z(P)))=(D,Z)2R(A)forsome2I1(P).DeneI2(P)=f2I1(P)j((P0),(Z(P)))2R(f1(P))g. Theorem3.4.8. Themapf2iswell-dened.IfP,Q2GBp(2,n)]TJ /F4 11.955 Tf 12.57 0 Td[(3,n)aresuchthatP=Qthenf2(P)=f2(Q). Proof. AssumeP2GBp(2,n)]TJ /F4 11.955 Tf 12.46 0 Td[(3,n).PismappedtoauniqueelementAofA(pn)]TJ /F5 7.97 Tf 6.59 0 Td[(1)andbyLemma 3.4.5 elementsofS(A,P)arecontainedinthesameorbitundertheactionofAut(A).Therefore,Pismappedtoauniqueelement(A,D,Z)ofB(pn)where(D,Z)2R(A)istherepresentativefortheorbitS(A,P).AssumeP,Q2GBp(2,n)]TJ /F4 11.955 Tf 12.44 0 Td[(3,n)aresuchthatP=Q.ByLemma 3.4.2 f1(P)=f1(Q),thereforeletA=f1(P)=f1(Q).ByLemma 3.4.5 S(A,P)andS(A,Q)areinthesameorbitundertheactionofAut(A).Itfollowsthatf2(P)=f2(Q). Denition3.4.9. AssumethatP2GBp(2,n)]TJ /F4 11.955 Tf 12.25 0 Td[(3,n)andthatMistheabelianmaximalsubgroupofP.Forx2Pdene x:M!Mby x(m)=xmx)]TJ /F5 7.97 Tf 6.59 0 Td[(1. Lemma3.4.10. AssumethatP2GBp(2,n)]TJ /F4 11.955 Tf 12.4 0 Td[(3,n)andthatMistheabelianmaximalsubgroupofP.Thenforx2P, 1) x2Aut(M). 2) Ifx=2M,thenker( x)]TJ /F4 11.955 Tf 11.96 0 Td[(IdM)=Z(P). 3) Ifx=2M,thenIm( x)]TJ /F4 11.955 Tf 11.96 0 Td[(IdM)=P0. 32

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Proof. SinceMisnormalinP, x(M)=xMx)]TJ /F5 7.97 Tf 6.59 0 Td[(1=Manditfollowsthat x2Aut(M).Assumex2PnM.Form2M,( x)]TJ /F4 11.955 Tf 12.86 0 Td[(IdM)(m)=mxm)]TJ /F5 7.97 Tf 6.59 0 Td[(1x)]TJ /F5 7.97 Tf 6.58 0 Td[(1=[m)]TJ /F5 7.97 Tf 6.58 0 Td[(1,x)]TJ /F5 7.97 Tf 6.58 0 Td[(1]2P0implyingthatIm( x)]TJ /F4 11.955 Tf 13.1 0 Td[(IdM)P0.Since x(z)=zforallz2Z(P),wehave( x)]TJ /F4 11.955 Tf 11.96 0 Td[(IdM)(Z(P))=1andZ(P)ker( x)]TJ /F4 11.955 Tf 11.96 0 Td[(IdM).Assumethata2Missuchthat( x)]TJ /F4 11.955 Tf 11.96 0 Td[(IdM)(a)=1,then[a,x)]TJ /F5 7.97 Tf 6.58 0 Td[(1]=a)]TJ /F5 7.97 Tf 6.59 0 Td[(1ax=( x)]TJ /F4 11.955 Tf 11.96 0 Td[(IdM)(a)=1.SinceP=hM,xiandMisabelian;wehavethata2Z(P).Wenowhavethatker( x)]TJ /F4 11.955 Tf 11.95 0 Td[(IdM)=Z(P)andbyanorderargumentthatIm( x)]TJ /F4 11.955 Tf 11.95 0 Td[(IdM)=P0. Denition3.4.11. AssumethatP2GBp(2,n)]TJ /F4 11.955 Tf 12.09 0 Td[(3,n)andthatMistheabelianmaximalsubgroupofP.Denethemap:P!Aut(M)by(x)= xforx2P. Lemma3.4.12. AssumethatP2GBp(2,n)]TJ /F4 11.955 Tf 12.4 0 Td[(3,n)andthatMistheabelianmaximalsubgroupofP.ThenisagrouphomomorphismwithkernelM. Proof. Leta,b2P.Form2M, ab(m)=abmb)]TJ /F5 7.97 Tf 6.58 0 Td[(1a)]TJ /F5 7.97 Tf 6.59 0 Td[(1= a(bmb)]TJ /F5 7.97 Tf 6.58 0 Td[(1)= a b(m).Thisimplies(ab)= ab= a b=(a)(b)andisagrouphomomorphism.Letm2M.Forn2M,sinceMisabelianitfollowsthat m(n)=mnm)]TJ /F5 7.97 Tf 6.58 0 Td[(1=n=IdM(n).Wehave(m)=IdMandMker().Assumethatx2PnM.ByLemma 3.4.10 ker( x)]TJ /F4 11.955 Tf 11.95 0 Td[(IdM)6=M,whichimplies(x)6=IdM.Itfollowsthatker()=M. Denition3.4.13. AssumethatP2GBp(2,n)]TJ /F4 11.955 Tf 12.09 0 Td[(3,n)andthatMistheabelianmaximalsubgroupofP.SinceisagrouphomomorphismwithkernelM,itdeterminesaninjectivehomomorphism :P=M!Aut(M). 33

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Lemma3.4.14. AssumethatP2GBp(2,n)]TJ /F4 11.955 Tf 12.86 0 Td[(3,n),thatMistheabelianmaximalsubgroupofP,andthat(A,D,Z)=f2(P).For2I2(P)and g2P=M, g6=M,wehave ( g))]TJ /F5 7.97 Tf 6.59 0 Td[(12C(A,D,Z). Proof. Let g2P=Mandlet= ( g))]TJ /F5 7.97 Tf 6.59 0 Td[(1.Sinceisanisomorphismand ( g)2Aut(M),itfollowsthat2Aut(A).Now)]TJ /F4 11.955 Tf 11.95 0 Td[(IdA=( ( g))]TJ /F5 7.97 Tf 6.58 0 Td[(1)]TJ /F4 11.955 Tf 11.96 0 Td[(IdM=( ( g))]TJ /F4 11.955 Tf 11.95 0 Td[(IdM))]TJ /F5 7.97 Tf 6.59 0 Td[(1.Since(Z(P))=Zandker( ( g))]TJ /F4 11.955 Tf 13.06 0 Td[(IdM)=Z(P)(Lemma 3.4.10 ),wehavethatker()]TJ /F4 11.955 Tf 11.86 0 Td[(IdA)=Z.Since(P0)=DandIm( ( g))]TJ /F4 11.955 Tf 11.86 0 Td[(IdM)=P0(byLemma 3.4.10 ),wehaveIm()]TJ /F4 11.955 Tf 11.95 0 Td[(IdA)=D.Itnowfollowsthat2C(A,D,Z). Denition3.4.15. AssumethatP2GBp(2,n)]TJ /F4 11.955 Tf 12.55 0 Td[(3,n),thatMistheabelianmaximalsubgroupofP,andthat(A,D,Z)=f2(P).Dene(, g)=( g))]TJ /F5 7.97 Tf 6.59 0 Td[(12C(A,D,Z)where2I2(P)and g2P=M, g6=M.DeneC(A,D,Z,P)tobethesubsetofC(A,D,Z)consistingofmaps(, g). Lemma3.4.16. AssumeP,Q2GBp(2,n)]TJ /F4 11.955 Tf 12.5 0 Td[(3,n)andthatM(resp.N)istheabelianmaximalsubgroupofP(resp.Q).Let P:P=M!Aut(M)and Q:Q=N!Aut(N)beasinDenition 3.4.13 .Letx2PnMandy2QnN.IfP=Q,thenthereexistsn2Z,p-n,andanisomorphism:M!Nsatisfying((P0),(Z(P)))=(Q0,Z(Q))suchthat Q( y)n= P( x))]TJ /F5 7.97 Tf 6.59 0 Td[(1. Proof. Assume:P!Qisanisomorphism.SinceMandNaretheuniqueabelianmaximalgroupsofPandQ(resp.),(M)=N.Let=jMandnote:M!Nisanisomorphism.Since(P0,Z(P))(resp.(Q0,Z(Q)))arecharacteristicinP(resp.Q)andarecontainedinM(resp.N),((P0),(Z(P)))=(Q0,Z(Q)).Sincex=2Mandyp2N,thereexistsn2Z,p-n,andb2Msuchthat(x)=byn.Letr2Zandm2M,( P( xr)(m))=(xrmx)]TJ /F7 7.97 Tf 6.58 0 Td[(r)=(byn)r(m)(byn))]TJ /F7 7.97 Tf 6.59 0 Td[(r= 34

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Q( yn)(b)... Q( y(r)]TJ /F5 7.97 Tf 6.59 0 Td[(1)n)(b) Q( yrn)((m)) Q( y(r)]TJ /F5 7.97 Tf 6.59 0 Td[(1)n)(b)]TJ /F5 7.97 Tf 6.59 0 Td[(1)... Q( yn)(b)]TJ /F5 7.97 Tf 6.59 0 Td[(1)= Q( yrn)((m)).Therefore P( x))]TJ /F5 7.97 Tf 6.59 0 Td[(1= Q( yn)= Q( y)n. Lemma3.4.17. AssumethatPandQareinGBp(2,n)]TJ /F4 11.955 Tf 12.46 0 Td[(3,n)andthatM(resp.N)istheabelianmaximalsubgroupofP(resp.Q).AssumeP=Q,thenf2(P)=f2(Q)=(A,D,Z)(Theorem 3.4.5 ).Ifr2C(A,D,Z,P)ands2C(A,D,Z,Q)thenrs. Proof. Let2I2(P)besuchthatr=(, x)forsome x2P=M, x6=M.Let2I2(Q)besuchthats=(, y)forsome y2Q=N, y6=N.Since x6=Mand y6=N,wecanassumex2PnM(resp.y2QnN)issuchthatxM= x(resp.yN= y)).ByLemma 3.4.16 thereexistsn2Z,p-n,andanisomorphism:M!Nsatisfying((P0),(Z(P)))=(Q0,Z(Q))suchthat Q( y)n= P( x))]TJ /F5 7.97 Tf 6.59 0 Td[(1.Now,sn=( Q( y))]TJ /F5 7.97 Tf 6.59 0 Td[(1)n= Q( y)n)]TJ /F5 7.97 Tf 6.59 0 Td[(1= P( x))]TJ /F5 7.97 Tf 6.59 0 Td[(1)]TJ /F5 7.97 Tf 6.59 0 Td[(1=)]TJ /F5 7.97 Tf 6.59 0 Td[(1r)]TJ /F5 7.97 Tf 6.58 0 Td[(1)]TJ /F5 7.97 Tf 6.58 0 Td[(1.Lettinga=)]TJ /F5 7.97 Tf 6.58 0 Td[(12Aut(A),wehavesn=ara)]TJ /F5 7.97 Tf 6.59 0 Td[(1.Hence,rs. Lemma3.4.18. AssumethatP2GBp(2,n)]TJ /F4 11.955 Tf 12.5 .01 Td[(3,n)andthat(A,D,Z)=f2(P).ThenC(A,D,Z,P)isafullequivalenceclassforonC(A,D,Z). Proof. ByLemma 3.4.17 ,C(A,D,Z,P)iscontainedinanequivalenceclassofonC(A,D,Z).LetMbetheabelianmaximalsubgroupofP.Assumer2C(A,D,Z,P)ands2C(A,D,Z)aresuchthatsr.Assumen2Z,p-n,anda2Aut(AjD,Z)aresuchthats=arna)]TJ /F5 7.97 Tf 6.58 0 Td[(1.Let2I2(P)and g2P=M, g6=M,besuchthatr= ( g))]TJ /F5 7.97 Tf 6.58 0 Td[(1.Wehaves=arna)]TJ /F5 7.97 Tf 6.59 0 Td[(1=a( ( g))]TJ /F5 7.97 Tf 6.59 0 Td[(1)na)]TJ /F5 7.97 Tf 6.59 0 Td[(1=(a) ( gn)(a))]TJ /F5 7.97 Tf 6.59 0 Td[(1.Sincep-nandMisofindexp,wehavethat gn2P=Missuchthat gn6=M.Sincea2Aut(AjD,Z)and2I2(P),wehavethataisanisomorphismfromMtoAsuchthat 35

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(a(P0),a(Z(P)))=(D,Z).Itfollowsthata2I2(P),implyingthats2C(A,D,Z,P).ThereforeC(A,D,Z,P)isafullequivalenceclassforonC(A,D,Z). Denition3.4.19. Denethemapf3:GBp(2,n)]TJ /F4 11.955 Tf 11.95 0 Td[(3,n)!C(pn)byf3(P)=(A,D,Z,)2C(pn)forP2GBp(2,n)]TJ /F4 11.955 Tf 11.96 0 Td[(3,n),wheref2(P)=(A,D,Z)and(, g)=forsome2I2(P)and g2P=M,whereMistheabelianmaximalsubgroupofP, g6=M.DeneI3(P)=f2I2(P)j(, g)=g. Theorem3.4.20. Themapf3iswell-dened.IfP,Q2GBp(2,n)]TJ /F4 11.955 Tf 12.41 0 Td[(3,n)aresuchthatP=Q,thenf3(P)=f3(Q). Proof. LetP2GBp(2,n)]TJ /F4 11.955 Tf 12.82 0 Td[(3,n).ByTheorem 3.4.8 ,f2(P)=(A,D,Z)isuniquelydetermined.ByLemma 3.4.18 ,C(A,D,Z,P)isafullequivalenceclassofonC(A,D,Z).Let2R(A,D,Z)betherepresentativefortheequivalenceclassC(A,D,Z,P),thenf3(P)=(A,D,Z,)2C(pn)andisuniquelydetermined.Hence,f3iswell-dened.LetPandQbeinGBp(2,n)]TJ /F4 11.955 Tf 12.43 0 Td[(3,n)suchthatP=Q.ByTheorem 3.4.8 ,f2(P)=f2(Q)=(A,D,Z).ByLemma 3.4.17 ,C(A,D,Z,P)andC(A,D,Z,Q)arethesameequivalenceclass.Wenowhavef3(P)=f3(Q). Lemma3.4.21. AssumethatP2GBp(2,n)]TJ /F4 11.955 Tf 11.91 0 Td[(3,n),thatMistheuniqueabelianmaximalsubgroupofP,andthat(A,D,Z,)=f3(P).If2I3(P),thenthereisaunique g2P=Msuchthat ( g))]TJ /F5 7.97 Tf 6.59 0 Td[(1=. Proof. Assume g, h2P=Maresuchthat ( g))]TJ /F5 7.97 Tf 6.59 0 Td[(1== ( h))]TJ /F5 7.97 Tf 6.58 0 Td[(1.Thisimplies ( gh)]TJ /F5 7.97 Tf 6.59 0 Td[(1)=IdM.Wenowhave gh)]TJ /F5 7.97 Tf 6.59 0 Td[(1=Mbydenitionof .Implying g= h. 36

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Denition3.4.22. AssumethatP2GBp(2,n)]TJ /F4 11.955 Tf 12.55 0 Td[(3,n),thatMistheabelianmaximalsubgroupofP,andthat(A,D,Z,)=f3(P).For2I3(P),let g2P=M, g6=M,betheuniqueelementsuchthat ( g))]TJ /F5 7.97 Tf 6.59 0 Td[(1=.Dener(P)=f(hp)jh2 gg. Lemma3.4.23. AssumethatP2GBp(2,n)]TJ /F4 11.955 Tf 12.69 0 Td[(3,n)andthat(A,D,Z,)=f3(P).If2I3(P),thenr(P)2H(A,D,Z,). Proof. LetMbetheabelianmaximalsubgroupofP.Let g2P=M, g6=M,betheuniqueelementsuchthat ( g))]TJ /F5 7.97 Tf 6.59 0 Td[(1=,byLemma 3.4.21 .ByLemma 3.1.4 ,forg2PnMandm2M,(IdM+ g+...+ p)]TJ /F5 7.97 Tf 6.59 0 Td[(1g)(m)=mpByLemma 3.1.1 ,(P)Z(P)andinap-group,(P)=P0Pp.Itfollowsthatxp2Z(P)foranyx2P.Forx2 g,x=gmforsomem2M.Now,xp=(gm)p= g(m) g2(m)... gp)]TJ /F8 5.978 Tf 5.76 0 Td[(1(m)mgp=gp(IdM+ g+...+ p)]TJ /F5 7.97 Tf 6.59 0 Td[(1g)(m)=gpmpItfollowsthatfxpjx2 gg=gp01(M)2Z(P)=01(M).Since:M!Aisanisomorphismand01(M)(resp.01(A))ischaracteristicinM(resp.A),(01(M))=01(A).Bydenition(Z(P))=Z,anditfollowsthatr(P)2Z=01(A)=H(A,D,Z,). Denition3.4.24. LetP2GBp(2,n)]TJ /F4 11.955 Tf 13.02 0 Td[(3,n)andlet(A,D,Z,)=f3(P).DeneH(A,D,Z,,P)tobethesubsetofH(A,D,Z,)consistingofelementsr(P)for2I3(P). Lemma3.4.25. LetPandQbeinGBp(2,n)]TJ /F4 11.955 Tf 13.04 0 Td[(3,n).Iff3(P)=(A,D,Z,),thenH(A,D,Z,,P)isafullorbitofH(A,D,Z,)undertheactionofAut(AjD,Z,).Moreover,ifQissuchthatP=Q,thenH(A,D,Z,,P)=H(A,D,Z,,Q). 37

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Proof. Let2I3(P),andnotethatbydenitionI3(P)=Aut(AjD,Z).Itfollowsthatar(P)2H(A,D,Z,)foranya2Aut(AjD,Z).WenowhavethatH(A,D,Z,,P)isafullorbitundertheactionofAut(AjD,Z).LetM(resp.N)betheabelianmaximalsubgroupofP(respQ).Assume:P!Qisanisomorphism,then((M),(P0),(Z(P)),(01(M)))=(N,Q0,Z(Q),01(N))ItfollowsthatI3(Q))]TJ /F5 7.97 Tf 6.59 0 Td[(1=I3(P).Wehave P( g))]TJ /F5 7.97 Tf 6.59 0 Td[(1= Q( h)forsome g2P=M, g6=M,and h2Q=N, h6=N.Hence,H(A,D,Z,,P)=H(A,D,Z,,Q). Denition3.4.26. DenethemapF:GBp(2,n)]TJ /F4 11.955 Tf 11.96 0 Td[(3,n)!D(pn)byF(P)=(A,D,Z,,r)2D(pn)forP2GBp(2,n)]TJ /F4 11.955 Tf 11.96 0 Td[(3,n),wheref3(P)=(A,D,Z,)andforsome2I3(P),r(P)=r.DeneI4(P)=f2I3(P)jr(P)=rg. Theorem3.4.27. ThemapFiswell-dened.IfP,Q2GBp(2,n)]TJ /F4 11.955 Tf 12.41 0 Td[(3,n)aresuchthatP=Q,thenF(P)=F(Q). Proof. LetP2GBp(2,n)]TJ /F4 11.955 Tf 12.57 0 Td[(3,n).ByTheorem 3.4.8 ,f3(P)=(A,D,Z,)isuniquelydetermined.ByLemma 3.4.25 ,C(A,D,Z,P)isafullorbitofH(A,D,Z,)undertheactionofAut(AjD,Z).Letr2R(A,D,Z,)betherepresentativefortheorbitH(A,D,Z,,P),thenf3(P)=(A,D,Z,,r)2D(pn)andisuniquelydetermined.Hence,Fiswell-dened. 38

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LetPandQbeinGBp(2,n)]TJ /F4 11.955 Tf 12.28 0 Td[(3,n)suchthatP=Q.ByTheorem 3.4.20 ,f3(P)=f3(Q)=(A,D,Z,).ByLemma 3.4.25 ,H(A,D,Z,,P)=H(A,D,Z,,Q)arethesamefullorbitofH(A,D,Z,)onAut(AjD,Z).WenowhaveF(P)=F(Q). 3.5TheMainTheoremWenowhaveaclassicationsetDandaclassicationmapF.ThisSectionwillprovideaconstructionforgroupsinGBp(2,n)]TJ /F4 11.955 Tf 11.96 0 Td[(3,n)andprovethatFisabijection. Denition3.5.1. Let(A,D,Z,,r)2D(pn).Denethegrouphomomorphism :Z!Aut(A)by (x)=xforallx2Z.DenethegroupAo Zandlet(resp.)betheembeddingofA(resp.Z)inAo Z.Letr02randdeneI=h()]TJ /F3 11.955 Tf 9.3 0 Td[(p)(r0)i. Lemma3.5.2. Let(A,D,Z,,r).ThenIisnormalinAo Z. Proof. ByLemma 3.3.7 ,hasorderp.ByLemma 3.3.12 ,r02r2Z=01(A).Bydenitionker()]TJ /F4 11.955 Tf 12.02 0 Td[(IdA)=Z,implying(z)=zforz2Z.Therefore, (x)(r0)=x(r0)=r0forallx2Z.Forany(a)(x)2Ao Z,(a)(x)(r0)()]TJ /F3 11.955 Tf 9.3 0 Td[(p)=(a+ (x)(r0))(x)]TJ /F3 11.955 Tf 11.96 0 Td[(p)=(r0+ ()]TJ /F3 11.955 Tf 9.3 0 Td[(p)(a))()]TJ /F3 11.955 Tf 9.3 0 Td[(p+x)=(r0)()]TJ /F3 11.955 Tf 9.3 0 Td[(p)(a)(x).Therefore,(r0)()]TJ /F3 11.955 Tf 9.3 0 Td[(p)2Z(Ao Z),whichimpliesthatIZ(Ao Z).Hence,IisnormalinAo Z. Denition3.5.3. Let(A,D,Z,,r)2D(pn)andassumethenotationofDenition 3.5.1 .WenowdenethegroupG(A,D,Z,,r)=Ao Z I. Lemma3.5.4. Let(A,D,Z,,r)2D(pn).ThenG(A,D,Z,,r)isanelementofGBp(2,n)]TJ /F4 11.955 Tf 11.96 0 Td[(3,n)suchthatF(G(A,D,Z,,r))=(A,D,Z,,r). Proof. LetG=G(A,D,Z,,r).Let ,,,andIbedenedasinDenition 3.5.1 .Let:Ao Z!Gbethesurjectivehomomorphismdened(g)=gIforg2Ao Z.Recallthattheembedding:A!Ao Zisinjective.Let =:A!G. 39

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Fora,b2Aandassume (a)= (b).)((a))=((b)))(a)I=(b)I)(a)]TJ /F3 11.955 Tf 11.96 0 Td[(b)2I.Since(c)2Iifandonlyifc=0,itfollowsthata=band isinjectiveand (A)isnormalinG.Let=:Z!G.Now,(1)=(1)I6=Iand(1)p=(p)=(p)I=(p)(r0)()]TJ /F3 11.955 Tf 9.3 0 Td[(p)I=(r0)I= (r0)2 (A).Sinceissurjectiveandh(A),(1)i=Ao Z,wehavethath(A),(1)i=h (A),(1)i=G.Since isinjective,j (A)j=pn)]TJ /F5 7.97 Tf 6.58 0 Td[(1.Since(1)p2 (A),wehavethatjGj=pn.Let=)]TJ /F4 11.955 Tf 12.21 0 Td[(IdA.Bydenitionof,Im()=DZ=ker(),whichimplies2=0.Forx2Zanda2A, (x)(a)=x(a)=(IdA+)x(a)=a+x(a).Ifa,b2Aandx,y2Z,then[(a)(x)I,(b)(y)I]=(a+ (x)(b)+ (y)()]TJ /F3 11.955 Tf 9.29 0 Td[(a))]TJ /F3 11.955 Tf 11.96 0 Td[(b)I=(x(b))]TJ /F3 11.955 Tf 11.96 0 Td[(y(a))ISinceIm()=D,itfollowsthatG0(D)I.Since[(1)I,(m)I]=((m))I,itfollowsthat(D)IG0.ThereforeG0=(D)I.Since(a)2Iifandonlyifa=0fora2A,itfollowsthat(D)I=D.SinceDiselementaryabelianoforderp2,wehavethatG0=Diselementaryabelianoforderp2.Leta2Aandx2Z.If(a)(x)I2Z(G),thenforallb2Aandy2Z[(a)(x)I,(b)(y)I]=I 40

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)(x(b))]TJ /F3 11.955 Tf 11.96 0 Td[(y(a))I=I.Since(c)2Iifandonlyifc=0forc2A,itfollowsthatx(a))]TJ /F3 11.955 Tf 12.42 0 Td[(y(b)=0forallb2Aandy2Z.SinceIm()=DandDiselementaryabelian,wehavethatpjxand)]TJ /F3 11.955 Tf 9.3 0 Td[(y(a)=0.Sinceker()=Z,wehavea2Z.ItfollowsthatZ(G)(Z)I.Letz2Z.Foranyx2Zandm2Awehave((m)(x))(z)=(m+ (x)(z))(x)=(z+x(z))(m)(x)=(z)((m)(x)).Therefore,(Z)IZ(G)and(Z)I=Z(G).Since(a)2Iifandonlyifa=0fora2A,itfollowsthat(Z)I=Z.SinceDZwehaveG0=(D)I(Z)I=Z(G).SinceZ(G)=Z,wehavethatZ(G)isoforderpn)]TJ /F5 7.97 Tf 6.59 0 Td[(3.SincejGj=pn,wehavethatZ(G)hasindexp3inG.WenowhavethatG0iselementaryabelianofrank2,Z(G)hasindexp3,andG0=(D)I(Z)IZ(G),henceGhasnilpotenceclass2.ThereforeG2GBp(2,n)]TJ /F4 11.955 Tf 12.9 0 Td[(3,n).Since(A)Iisanabeliansubgroupoforderpn)]TJ /F5 7.97 Tf 6.59 0 Td[(1,wehavethat(A)IistheabelianmaximalsubgroupofG.LetM=(A)I.SinceMistheabelianmaximalsubgroupofGanditisisomorphictoA,wehavef1(G)=A.Let:M!Abetheisomorphismdened((a)I)=afora2Aandnotethat2I1(G).Wehave((G0),(Z(G)))=(((D)I),((Z)I))=(D,Z).Hencef2(G)=(A,D,Z)and2I2(G).Let g=((1)I)M,wehave g2G=M, g6=M.Fora2A, ( g))]TJ /F5 7.97 Tf 6.59 0 Td[(1(a)= ( g)((a)I)=((1)(a)()]TJ /F4 11.955 Tf 9.3 0 Td[(1)I)= 41

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=(( (1)(a)I)=(((a))I)=(a).Itfollowsthat ( g))]TJ /F5 7.97 Tf 6.58 0 Td[(1=.Hencef3(G)=(A,D,Z,)and2I3(G).Notethat(p)I=(p)()]TJ /F3 11.955 Tf 9.3 0 Td[(p)(r0)I=(r0)I.Wenowhaver(G)=f(hp)jh2 gg=f((ap)(p)I)ja2Ag=f((r0+ap)I)ja2Ag=fr0+apja2Ag=r001(A)=r.Therefore,F(G)=(A,D,Z,,r)and2I4(G). Lemma3.5.5. Letpbeanoddprimeandnapositiveinteger. D(pn)isnon-emptyifandonlyifA(pn)]TJ /F5 7.97 Tf 6.58 0 Td[(1)isnon-empty. A(pn)isnon-emptyifandonlyifn4. Proof. IfA(pn)]TJ /F5 7.97 Tf 6.59 0 Td[(1)isnon-empty,thenbydenitionB(pn),C(pn),andD(pn)mustbeempty.AssumethatA(pn)]TJ /F5 7.97 Tf 6.59 0 Td[(1)isnon-emptyandletA2A(pn)]TJ /F5 7.97 Tf 6.59 0 Td[(1).BydenitionofA(pn)]TJ /F5 7.97 Tf 6.59 0 Td[(1),S(A)andhenceB(pn)arenon-empty.Let(A,D,Z)2B(pn).SinceA=ZandDareelementaryabelianoforderp2thereexistsanisomorphism:A=Z!D.SinceZisnormal,thereexistsasurjectivehomomorphism:A!A=Zsuchthat(a)=aZfora2A.Hence, =:A!Aisagrouphomomorphismsuchthatker( )=ZandIm( )=D.Denethegrouphomomorphism= +IdA,notingker()]TJ /F4 11.955 Tf 11.96 0 Td[(IdA)=ker( )=ZandIm()]TJ /F4 11.955 Tf 11.95 0 Td[(IdA)=Im( )=D.Assumea2ker(),then(a)= (a)+IdA(a)= (a)+a=0) (a)=)]TJ /F3 11.955 Tf 9.3 0 Td[(a.Thereforea2Im( )=D.SinceDZ,wehavea2ker( )=Zanditfollowsthata=0.Hence2Aut(A)anditfollowsthat2C(A,D,Z).WenowhavethatC(pn)andR(A,D,Z,)arenon-empty.ItfollowsfromdenitionthatD(pn)isnon-empty.AssumeAisanabelianp-groupcontainingasubgroupZoforderatleastp2whosequotientisorderp2.ThisassumptionimpliesthatjAj=jA=ZjjZjp4.SuchanAis 42

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thebasicrequirementforinclusioninthesetA(pn),thereforeA(pn)isemptyforn3.Recallthatsubgroupsofelementaryabeliangroupsareelementaryabelianandthequotientsofelementaryabeliangroupsareelementaryabelian.Hencetheelementaryabelianp-groupEofrankn4containsanelementaryabeliansubgroupZofindexp2andrankatleast2.Therefore,thereexistsA2A(pn)suchthatE=Aforn4. Theorem3.5.6(MainTheorem). Letpbeanoddprime.LetGBp(2,n)]TJ /F4 11.955 Tf 12.61 0 Td[(3,n)betheclassofp-groupsoforderpnandnilpotenceclass2withelementaryabelianderivedsubgroupofrank2whosecenterhasindexp3.Let(A,D,Z,,r)2D(pn)(Denition 3.3.14 )andletG=G(A,D,Z,,r)(Denition 3.5.3 ). ThegroupGisanelementofGBp(2,n)]TJ /F4 11.955 Tf 12.04 0 Td[(3,n)andF(G)=(A,D,Z,,r)underF,theclassicationmap(Denition 3.4.26 ). LetP2GBp(2,n)]TJ /F4 11.955 Tf 13 0 Td[(3,n).IfF(P)=(A,D,Z,,r),thenPisisomorphictoG(A,D,Z,,r). Forn0,GBp(2,n)]TJ /F4 11.955 Tf 11.96 0 Td[(3,n)isnotemptyifandonlyforn5. Proof. ByLemma 3.5.4 ,thegroupG=G(A,D,Z,,r)isanelementofGBp(2,n)]TJ /F4 11.955 Tf 11.79 0 Td[(3,n)andF(G)=(A,D,Z,,r).AssumethatP2GBp(2,n)]TJ /F4 11.955 Tf 12.25 0 Td[(3,n)issuchthatF(P)=(A,D,Z,,r)andletMbetheabelianmaximalsubgroupofP.Let2I4(P)andlet g2P=M, g6=M,betheuniqueelementsuchthat(, g)=(Lemma 3.4.21 ).Bydenition,r=r001(A)=(gp)01(A),forsomeg2 g.Denethegrouphomomorphismg:Z!Pbyg(z)=gzforz2Z.Recallthatthegrouphomomorphism :Z!Aut(A)isdened (z)=zforz2Z.Fora2Aandz2Z,g(z))]TJ /F5 7.97 Tf 6.59 0 Td[(1(a)g()]TJ /F3 11.955 Tf 9.3 0 Td[(z)=gz)]TJ /F5 7.97 Tf 6.59 0 Td[(1(a)g)]TJ /F7 7.97 Tf 6.59 0 Td[(z= ( gz)()]TJ /F5 7.97 Tf 6.59 0 Td[(1(a))= ( g)z()]TJ /F5 7.97 Tf 6.59 0 Td[(1(a))=)]TJ /F5 7.97 Tf 6.59 0 Td[(1(z(a))=)]TJ /F5 7.97 Tf 6.58 0 Td[(1( (z)(a)). 43

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ByTheorem 2.4.3 thereexistsauniquegrouphomomorphism)-277(:Ao Z!Psuchthat\((a))=)]TJ /F5 7.97 Tf 6.58 0 Td[(1(a)and\((z))=x(z).)]TJ /F1 11.955 Tf 10.1 0 Td[(issurjectivesinceIm()]TJ /F5 7.97 Tf 6.59 0 Td[(1)=M,g(1)=g,andP=hM,gi.Assume(a)(z)2ker(\forsomea2A,z2Z,implying\((a)(z))=)]TJ /F5 7.97 Tf 6.59 0 Td[(1(a)gz=1)gz=)]TJ /F5 7.97 Tf 6.59 0 Td[(1()]TJ /F3 11.955 Tf 9.3 0 Td[(a)2M.Wenowhavethatpjz,sinceg2PnM,MisofindexpinP,andgz2M.Wealsohavea=(g)]TJ /F7 7.97 Tf 6.58 0 Td[(bp)forsomeintegerb.Since\(((bp))(g)]TJ /F7 7.97 Tf 6.58 0 Td[(bp))=1foranyintegerb,itfollowsthatI=h()]TJ /F3 11.955 Tf 9.3 0 Td[(p)(r0)i=ker(\.HenceG=Ao Z=IisisomorphictoP.ByLemma 3.5.5 ,D(pn)isnon-emptyifandonlyifn5.SinceeveryelementofGBp(2,n)]TJ /F4 11.955 Tf 13.28 0 Td[(3,n)isisomorphictoangroupG(A,D,Z,,r)arisingfromsome(A,D,Z,,r)2D(pn),weareabletoconcludethatGBp(2,n)]TJ /F4 11.955 Tf 12.11 0 Td[(3,n)isnotemptyifandonlyifn5. WenowareabletoconcludethatthesetDdescribesacompletelistofisomorphismclassesfortheclassGBp(2,n)]TJ /F4 11.955 Tf 11.96 0 Td[(3,n). 3.6EnumerationofD(p5)InthisSectionresultspertainingtotheenumerationoftheclassicationsetDwillbepresented.Notably,itisshownthatD(p5)hasp+8elements.Thisimpliesthattherearep+8isomorphismclassesofthe2-generalizedBlackburngroupsoforderp5whosecenterhasindexp3.ThroughoutthisSection,assumethedenitionsandnotationofSection3. Lemma3.6.1. ForA2A(p4)(Denition 3.3.1 ),anyelement(D,Z)2S(A)(Denition 3.3.2 )issuchthatD=Z.Moreover,A(p4)contains3elements: 44

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1) A1=ZpZpZpZp. 2) A2=Zp2ZpZp. 3) A3=Zp2Zp2. Proof. BytheFundamentalTheoremofFinitelyGeneratedAbelianGroups,theisomorphismclassesofabelianp-groupsoforderpnisinbijectionwiththepartitionsofn.Therefore,A(p4)hasatmost5elements.AssumeA2A(p4)andlet(D,Z)2S(A).SincejDj=p2=jZjandDZ,wehavethatD=Z.AssumebywayofcontradictionthatA2A(p4)isisomorphictoZp4.SinceDisdenedtobeaanelementaryabelianp-subgroupofrank2,Acontainsatleastp2)]TJ /F4 11.955 Tf 12.34 0 Td[(1elementsoforderp.ThiscontradictstheisomorphismwiththecyclicgroupZp4,asitcontainsp)]TJ /F4 11.955 Tf 11.95 0 Td[(1elementsoforderp.AssumebywayofcontradictionthatA2A(p4)isisomorphictoZp3Zp.Since01(Zp3Zp)=f(pa,0)ja2Zp3g=Zp2hasindexp2inZp3Zpand01(A)Z(Lemma 3.3.3 ),itfollowsthatZ=01(A).Bydenition,Zmustcontainanelementaryabelianp-subgroupofrank2.ThisisacontradictionsinceZ=01(A)wasshowntobecyclicoforderp2.Sincesubgroupsandquotientsofanelementaryabeliangroupareelementaryabelian,thegroupZpZpZpZpisisomorphictoanelementofA(p4).ThesubgroupH=f(pa,b,0)ja2Zp2,b2ZpgofG=Zp2ZpZpiselementaryabelianoforderp2andG=Hiselementaryabelianoforderp2.Therefore,GisisomorphictoanelementofA(p4).ThesubgroupH=f(pa,pb)ja,b2Zp2gofG=Zp2Zp2iselementaryabelianoforderp2andG=Hiselementaryabelianoforderp2.Therefore,GisisomorphictoanelementofA(p4).WenowhaveidentiedthethreeelementsA1,A2,andA3ofA(p4). 45

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3.6.1ZpZpZpZp=A12A(p4) Lemma3.6.2. LetA12A(p4)beisomorphictoZpZpZpZp.ThenR(A1)(Denition 3.3.5 )hasauniqueelement(D1,Z1).Moreover,B(p5)hasauniqueelement,(A1,D1,Z1),arisingfromA1. Proof. SinceA1isanelementaryabelianp-group,itmaybeviewedasa4-dimensionalvectorspaceoverFp.ByLemma 3.6.1 ,anypair(D,Z)2S(A1)issuchthatD=Z.SinceDiselementaryabelian,Dmaybeviewedasa2-dimensionalvectorsubspaceofA1.Let(D1,Z1)and(D2,Z2)beelementsofS(A1).Sincethe2-dimensionalsubspacesofa4-dimensionalvectorspaceformanorbitundertheactionofthegenerallineargroup,thereexistsan2Aut(A)suchthat(D1)=D2.Therefore,thereisauniqueelement(D1,Z1)2R(A1)andthelemmafollows. Lemma3.6.3. Let(A1,D1,Z1)2B(p5).ThenR(A1,D1,Z1)hasauniqueelement1andC(p5)hasauniqueelement,(A1,D1,Z1,1),arisingfromA1.Furthermore,fornon-trivialelementsv,w2Z1,thereexists2Aut(A1jD1,Z1)suchthat(v)=w. Proof. Let,2C(A1,D1,Z1)(Denition 3.3.6 ).SinceD1=Z1,thisimpliesthatIm()]TJ /F4 11.955 Tf 11.96 0 Td[(IdA1)=ker()]TJ /F4 11.955 Tf 11.96 0 Td[(IdA1)=Im()]TJ /F4 11.955 Tf 11.95 0 Td[(IdA2)=ker()]TJ /F4 11.955 Tf 11.95 0 Td[(IdA2)=Z1areoforderp2.Letv,w2Z1benon-trivial.Ifc2ker()]TJ /F4 11.955 Tf 12.48 0 Td[(IdA1),then(c)=candifc2ker()]TJ /F4 11.955 Tf 12.23 0 Td[(IdA1),then(c)=c.Choosee22Im()]TJ /F4 11.955 Tf 12.23 0 Td[(IdA1)suchthatfv,e2ggeneratesIm()]TJ /F4 11.955 Tf 11.18 0 Td[(IdA1).Choosee3,e42A1suchthat()]TJ /F4 11.955 Tf 11.19 0 Td[(IdA1)(e3)=vand()]TJ /F4 11.955 Tf 11.18 0 Td[(IdA1)(e4)=e2.Nowfe1,e2,e3,e4ggeneratesA1andwehave(v)=v(e3)=v+e3(e2)=e2(e4)=e2+e4. 46

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Similarly,choosef22Im()]TJ /F4 11.955 Tf 12.35 0 Td[(IdA1)andf3,f42A1suchthatfw,f2,f3,f4ggeneratesA1and(w)=w(f3)=w+f3(f2)=f2(f4)=f2+f4.Dene2Aut(A1)by(v)=wand(ei)=fi,itfollowsthat=)]TJ /F5 7.97 Tf 6.59 0 Td[(1.Therefore,thereisauniqueequivalenceclassunderinC(A1,D1,Z1).ItfollowsthatR(A1,D1,Z1)hasauniqueelement1andbyLemma 3.6.2 thereisauniqueelementinC(p5),(A1,D1,Z1,1),arisingfromA1. Lemma3.6.4. Let(A1,D1,Z1,1)2C(p5).ThenR(A1,D1,Z1,1)(Denition 3.3.14 )hastwoelements.Moreover,therearetwoelementsinD(p5)arisingfromA1. Proof. ByLemma 3.3.12 ,H(A1,D1,Z1,1)=Z1=01(A1).Now01(A1)istrivialsinceA1iselementaryabelian;thereforeH(A1,D1,Z1,1)=Z1.ByLemma 3.6.3 ,Aut(A1jD1,Z1)istransitiveonthenon-trivialelementsofZ1.Therefore,R(A1,D1,Z1,1)=f1A1,rgforsomenon-trivialrinZ1.ByLemma 3.6.3 ,wehavetwodistinctelementsofD(p5)arisingfromA1:(A1,D1,Z1,1,1A1)and(A1,D1,Z1,1,r). 3.6.2Zp2ZpZp=A22A(p4) Lemma3.6.5. LetA22A(p4)beisomorphictoZp2ZpZp.ThenR(A2)(Denition 3.3.5 )hasauniqueelement(D2,Z2).Furthermore,B(p5)hasauniqueelement,(A2,D2,Z2),arisingfromA1.Moreover,foranyx,y21(A2)n01(A2)thereexists2Aut(A2)suchthat(x)=y. Proof. Itfollowsthat01(A2)=Zpand1(A2)=ZpZpZpsinceA2=Zp2ZpZp.Assumea2A2generates01(A2).Givenanyx21(A2)n01(A2),thesubgroupha,yiiselementaryabelianoforderp2inA2and,sinceitcontains01(A2),itsquotientinA2iselementaryabelianoforderp2.Itfollowsthat(ha,yi,ha,yi)2S(A2).Infact,givenanypair(D,Z)2S(A2),wehavethat01(A2)D=Z1(A2)(Lemma 3.6.1 47

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Lemma 3.3.3 ).Wemayconcludethat(D,Z)2S(A2)ifandonlyifthereexistssomex21(A2)n01(A2)suchthatD=ha,xi.Let(D1,Z1),(D2,Z2)2S(A2).Bytheaboveparagraphthereexistsx,y21(A2)n01(A2)suchthatD1=ha,xiandD2=ha,yi.Assumethatfa,x,rggenerates1(A2)andfa,y,sggenerates1(A2).Sinceagenerates01(A2),thereexistssome^a2A2suchthatp^a=a.Wenowhavethath^aihxihri=A2=h^aihyihsi.Denethemap:A2!A2bylinearcombinationsof(^a)=^a(x)=y(r)=s.SinceA2=h^aihxihri,iswell-dened.Now,forintegersiandj,(1^a+2x+3r)+(1^a+2x+3r)=1^a+2y+3s+1^a+2y+3s=(1+1)^a+(2+2)y+(3+3)s=((1+1)^a+(2+2)x+(3+3)r)=(1^a+2x+3r+1^a+2x+3r).Therefore,isagroupendomorphism.SinceA2=h^aihyihsi,wehavethatissurjective,anditfollowsthatisanautomorphism.Thelemmanowfollows. Denition3.6.6. Let(A2,D2,Z2)2B(p5).Denethegrouphomomorphisms 1) :Aut(A2jD2,Z2)!Aut(Z2)by()=jZ2for2Aut(A2jD2,Z2). 2) :Aut(A2jD2,Z2)!Aut(A2=Z2)by()= for2Aut(A2jD2,Z2)where (a+Z2)=(a)+Z2fora2A2.Moreover,denethemapf:Aut(A2jD2,Z2)!Aut(Z2)Aut(A2=Z2)byf()=((),())for2Aut(A2jD2,Z2).Finally,denetheset=f:A2=Z2!Z2jisanisomorphismg. 48

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Denition3.6.7. Let(A2,D2,Z2)2B(p5).Lete1beanelementofA2suchthatpe2generates01(A2).Lete2beanelementofZ2n01(A2)suchthatZ2=hpe1,e2i.Lete3beanelementof1(A2)nZ2suchthat1(A2)=hpe1,e2,e3i.SinceA2hasorderp4,wehavethatA2isgeneratedbyhe1,e2,e3i.NowZ2(resp.A2=Z2)canbeviewedasavectorspaceoverZpwithbasisB1=fpe1,e2g(resp.B2=fe3+S,e1+Sg). Lemma3.6.8. Let(A2,D2,Z2)2B(p5).Then 1) Themapfisagrouphomomorphism. 2) Theimageoffisthesetofpairs(,)2Aut(Z2)Aut(A2=Z2)where(01(A2))=01(A2),(1(A2)=Z2)=1(A2)=Z2,andtheeigenvaluesoftheactionofon01(A2)andtheactionofon(A2=Z2)=(1(A2)=Z2)areequal. Proof. Leta,b2Aut(A2jD2,Z2).Sinceandaregrouphomomorphisms,(ab)=(a)(b)and(ab)=(a)(b).Therefore,f(ab)=((ab),(ab))=((a),(a))((b),(b))=f(a)f(b),andfisagrouphomomorphism.Let(,)2Aut(Z2)Aut(A2=Z2)besuchthat(01(A2))=01(A2),(1(A2)=Z2)=1(A2)=Z2,andistheeigenvalueoftheactionofon01(A2)andtheactionofon(A2=Z2)=(1(A2)=Z2).Since01(A2)=hpe1iandZ2=hpe1,e2i,thereexistsintegersa1anda2,wherep-a2,suchthat(pe1)=pe1(e2)=a1pe1+a2e2.Since1(A2)=Z2=he3+Z2iandA2=Z2=he3+Z2,e1+Z2i,thereexistsintegersb1andb2,p-b1,suchthat(e3+Z2)=b1e3+Z2(e1+Z2)=b2e3+e1+Z2.Denethemapa:A2!A2bylinearcombinationsofa(e1)=e1+b2e3a(e2)=(a1p)e1+a2e2a(e3)=b1e3. 49

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Sincefe1,e2,e3ggeneratesA2,themapaiswell-dened.Nowforx=2Xi=1xieiandy=3Xj=1yieiwherexiandyjareintegers,a(x+y)=a 3Xi=1(xi+yi)ei!=[(x1+y1)+a1p(x2+y2)]e1+[a2(x2+y2)]e2+[b2(x1+y1)+b1(x3+y3)]e3=(x1+y1)[e1+b2e3]+(x2+y2)[a1pe1+a2e2]+(x3+y3)[b1e3]=a(x1e1+x2e2+x3e3)+a(y1e1+y2e2+y3e3)=a(x)+a(y).Thereforeaisanendomorphism.SincetheprojectionofaontoA2=Z2isandtherestrictionofaontoZ2is,bothbijections,wehavethataissurjective.SinceA2isanitegroup,wehavethataisanautomorphismofA2.SinceajZ2=,a(Z2)=Z2anditfollowsthata2Aut(A2jD2,Z2).Since(a)=and(a)=,wehavethat(,)2Im(f).Leta2Aut(A2jD2,Z2).Since01(A2)ischaracteristicinA2and01(A2)Z2(Lemma 3.3.3 ),wehavethat(a)(01(A2))=01(A2).Since1(A2)ischaracteristicinA2andZ2=D21(A2),wehavethat(a)(1(A2)=Z2)=1(A2)=Z2.SinceA2isgeneratedbyfe1,e2,e3g,thereexistintegersisuchthata(e1)=3Xi=1iei.Nowa(pe1)isageneratorof01(A2)since01(A2)=hpe1i.Itfollowsthatp-1anda(pe1)=1(pe1).Thereforepe1isaneigenvectorwitheigenvalue1fortheactionofaonZ2.Theelemente1+1(A2)isaneigenvectorwitheigenvalue1fortheactionofaonA2=1(A2).Since(A2=Z2)=(1(A2)=Z2)isisomorphictoA2=1(A2),itfollowsthate1+1(A2)isaneigenvectorwitheigenvalue1fortheactionofaon(A2=Z2)=(1(A2)=Z2).Thelemmanowfollows. Lemma3.6.9. Let(A2,D2,Z2)2B(p5).DeneanactionofAut(A2jD2,Z2)Zponby((,n),)=n()())]TJ /F5 7.97 Tf 6.59 0 Td[(1wheren2Zp,2Aut(A2jD2,Z2),2. 50

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RecallthatZ2andA2=Z2maybeviewedas2-dimensionalvectorspacesoverZp,henceelementsofmayberepresentedaselementsofGL2(p).AsetofrepresentativesfortheorbitsofundertheactionofAut(A2jD2,Z2)ZpisR=8><>:0B@10011CA9>=>;[8><>:0B@01101CA9>=>;. Proof. Let2Aut(A2jD2,Z2).Similarto,theisomorphisms()and()maybeviewedaselementsofGL2(p).Let(dij)(resp.(eij))bethematrixof()(resp.())withrespecttothebasisB1(resp.B2).ByLemma 3.6.8 ,d21=0=e21andd11=e22.Assume2hasmatrixrepresentation(aij).Since()and()areisomorphisms,wehavethate116=0,d116=0,andd226=0.Theactionof(,n)2Aut(A2jD2,Z2)Zponisn(a)(a))]TJ /F5 7.97 Tf 6.59 0 Td[(1=n0B@d11d1,20d221CA0B@a11a12a21a221CA0B@e11e120d111CA)]TJ /F5 7.97 Tf 6.59 0 Td[(1=n0B@d11a11+d12a21 e11d11e11a12+d12e11a22)]TJ /F3 11.955 Tf 11.96 0 Td[(d11e12a11)]TJ /F3 11.955 Tf 11.96 0 Td[(d12e12a21 d11e11d22a21 e11d22e11a22)]TJ /F3 11.955 Tf 11.96 0 Td[(d22e12a21 d11e111CA.Let06=a2Zp.Assumethereexists(,n)2Aut(A2jD2,Z2)Zpsuchthat0B@0aa01CA=n(a)0B@10011CA(a))]TJ /F5 7.97 Tf 6.59 0 Td[(1=n0B@d11d120d221CA0B@10011CA0B@e11e120d111CA)]TJ /F5 7.97 Tf 6.59 0 Td[(1=n0B@d11 e11d12e11)]TJ /F3 11.955 Tf 11.96 0 Td[(d11e12 d11e110d22 d111CA.Thisimpliesa=0,acontradiction.Therefore,0B@0aa01CAisnottheintheorbitof0B@10011CAundertheactionofAut(A2jD2,Z2)Zp. 51

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WenowcanconcludethattheelementsofRareindistinctorbitsundertheactionofAut(A2jD2,Z2)Zp.ItremainstoshowthateveryorbitisrepresentedinR.Assume2hasamatrixrepresentationsuchthata21=0.Sinceisanisomorphism,wehavethata11,a216=0.Let(,n)2Aut(A2jD2,Z2)Zpbesuchthatn=1,()=0B@d11d11 a220)]TJ /F3 11.955 Tf 9.3 0 Td[(d11a12 a221CA()=0B@d11a1100d111CA.Wehavethatn()())]TJ /F5 7.97 Tf 6.58 0 Td[(1=0B@10011CA.Assume2hasamatrixrepresentationsuchthata216=0.Let(,n)2Aut(A2jD2,Z2)Zpbesuchthatn=a21 det()(a)=0B@d11)]TJ /F3 11.955 Tf 9.3 0 Td[(d11a11 a120e12det() a121CA(a)=0B@e12e12a220d111CA.Wehavethatn()())]TJ /F5 7.97 Tf 6.58 0 Td[(1=0B@01101CA.WenowhavethatRisasetofrepresentativesfortheactionofAut(A2jD2,Z2)Zpon. Denition3.6.10. Let(A2,D2,Z2)2B(p5).For2C(A2,D2,Z2)(Denition 3.3.6 ),denetheisomorphism )]TJ /F4 11.955 Tf 11.95 0 Td[(IdA2:A2=Z2!Z2by )]TJ /F4 11.955 Tf 11.95 0 Td[(IdA2(a+Z2)=()]TJ /F4 11.955 Tf 12.74 0 Td[(IdA2)(a)fora+Z22A2=Z2.Denethemap:C(A2,D2,Z2)!by()= )]TJ /F4 11.955 Tf 11.96 0 Td[(IdA2for2C(A2,D2,Z2). Lemma3.6.11. Let(A2,D2,Z2)2B(p5)and2C(A2,D2,Z2).For2Aut(A2jD2,Z2)andanintegern,p-n,wehave(n)=n()()]TJ /F5 7.97 Tf 6.59 0 Td[(1)=()()())]TJ /F5 7.97 Tf 6.59 0 Td[(1.Moreover,issurjective. 52

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Proof. Let2Aut(A2jD2,Z2)and2C(A2,D2,Z2).Fora+Z22A2=Z2,()]TJ /F5 7.97 Tf 6.59 0 Td[(1)(a+Z2)= )]TJ /F5 7.97 Tf 6.58 0 Td[(1)]TJ /F4 11.955 Tf 11.96 0 Td[(IdA2(a+Z2)=()]TJ /F4 11.955 Tf 11.95 0 Td[(IdA2))]TJ /F5 7.97 Tf 6.59 0 Td[(1(a)= )]TJ /F4 11.955 Tf 11.96 0 Td[(IdA2()]TJ /F5 7.97 Tf 6.58 0 Td[(1(a)+Z2)=()()]TJ /F5 7.97 Tf 6.59 0 Td[(1)(a+Z2)=()()()]TJ /F5 7.97 Tf 6.59 0 Td[(1)(a+Z2),wherethelastequalityfollowssinceIm(())=Z2.Therefore,()]TJ /F5 7.97 Tf 6.58 0 Td[(1)=()()())]TJ /F5 7.97 Tf 6.58 0 Td[(1.Leta2A2andlet2C(A2,D2,Z2).SinceIm()]TJ /F4 11.955 Tf 10.76 0 Td[(IdA2)=D2=Z2,thereexistssomez2Z2suchthat(a)=a+z.Sinceker()]TJ /F4 11.955 Tf 12.13 0 Td[(IdA2)=Z2,wehavethatn(a)=a+nz.Itfollowsthat(n)]TJ /F4 11.955 Tf 11.96 0 Td[(IdA2)(a)=nz=n()]TJ /F4 11.955 Tf 11.95 0 Td[(IdA2)(a))(n)=n().Let2.Let:A2!A2=Z2bethenaturalprojection.Sinceker()=Z2andisanisomorphism,wehavethatker()=Z2.Sinceissurjectiveandisanisomorphism,wehavethatIm()=Z2.Assumea2A2issuchthata2ker(+IdA2).Thisimpliesthat(a)=)]TJ /F3 11.955 Tf 9.3 0 Td[(aanda2Im()=Z2.Sinceker()=Z2,wehavethat(a)=0A2.Therefore,a=0A2.Wenowhavethat+IdA2isaninjectiveendomorphismofnitegroupA2,hence+IdA2isanautomorphism.Wecanconcludethat+IdA22C(A2,D2,Z2)and(+IdA2)=.Therefore,issurjective. Lemma3.6.12. Let(A2,D2,Z2)2B(p5).ThereexiststwodistinctelementsofR(A2,D2,Z2)(Denition 3.3.10 )andtwoelementsofC(p5)arisingfromA2. Proof. Assume,2C(A2,D2,Z2)suchthat,implyingthatthereexists2Aut(A2jD2,Z2)andanintegern,p-n,suchthatn=)]TJ /F5 7.97 Tf 6.59 0 Td[(1.Therefore,byLemma 3.6.11 ,n()=(n)=()]TJ /F5 7.97 Tf 6.58 0 Td[(1)=()()())]TJ /F5 7.97 Tf 6.59 0 Td[(1.Sinceisawell-denedsurjectivemap,wecanconcludethatifandonlyif()and()areinthesameorbitofundertheactionofAut(A2jD2,Z2)Zp. 53

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ByLemma 3.6.9 ,thereare2orbitsforundertheactionofAut(A2jD2,Z2)Zp.Therefore,R(A2,D2,Z2)hastwouniqueelementsandthelemmafollows. Lemma3.6.13. TherearefourelementsofD(p5)(Denition 3.3.14 )arisingfromA2. Proof. Assumethat(A2,D2,Z2,)2C(p5).ThenH(A2,D2,Z2,)=Z2=01(A2)(Lemma 3.3.12 ).BydenitionZ21(A2)andbyLemma 3.6.5 theautomorphismsofA2aretransitiveontheelementsof1(A2)n01(A2).Therefore,R(A2,D2,Z2,)=f01(A2),r+01(A2)gforsomenon-trivialelementofZ2n01(A2).ByLemma 3.6.12 ,therearetwodistinctelementsinC(p5)arisingfromA2.Bytheaboveparagraph,eachofthisgivesrisetotwoelementsofD(p5).Thelemmanowfollows. 3.6.3Zp2Zp2=A32A(p4) Lemma3.6.14. LetA32A(p4)beisomorphictoZp2Zp2.ThenR(A3)(Denition 3.3.5 )hasauniqueelement(D3,Z3).Furthermore,B(p5)hasauniqueelement,(A3,D3,Z3),arisingfromA3.Moreover,D3=Z3=01(A3)=1(A3). Proof. Assume(D,Z)2S(A3)andlet:Zp2Zp2!A3beanisomorphism.Wehavethat01(A3)=f(pa,pb)ja,b2Zp2g=1(A3)iselementaryabelianoforderp2.Since01(A3)Z(Lemma 3.3.3 )andD=Zisoforderp2(Lemma 3.6.1 ),wehavethat1(A3)=01(A3)=D=Z.Since01(A3)ischaracteristicinA3,wehavethat((D),(Z))=(D,Z)forall2Aut(A3).Itfollowsthatthereisauniqueelement(D3,Z3)inR(A3)andthelemmafollows. Denition3.6.15. Let(A3,D3,Z3)2B(p5).SinceA3=Z3andZ3areelementaryabelianoforderp2,theyareisomorphic.Let:A3!A3=Z3bethenaturalprojection.Dene :A3=Z3!Z3=01(A3)tobethemap (a+Z3)=pafor(a+Z3)2A3=Z3. Lemma3.6.16. Let(A3,D3,Z3)2B(p5).Thenthemap isawell-denedisomorphism. 54

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Proof. Assumea+Z3=b+Z3.Thisimpliesthat(a)]TJ /F3 11.955 Tf 11.95 0 Td[(b)2Z3=01(A3)andthereexistsc2A3suchthatpc=a)]TJ /F3 11.955 Tf 11.96 0 Td[(b.Now, (a+Z3)=pa=p(b+pc)=pb+p2c= (b+Z3),wherep2c=0A3sinceA3hasexponentp2.Wenowhavethat iswell-dened.Fora,b2A3, ((a)+(b))= (a+b)=p(a+b)=pa+pb= (a)+ (b),therefore isagrouphomomorphism.Since01(A3)=1(A3),wehavethatker( )=Z3=01(A3),anditfollowsthat isanisomorphism. Denition3.6.17. Let(A3,D3,Z3)2B(p5).For2Aut(A3=Z3),denethemap:Aut(A3jD3,Z3)!Aut(A3=Z3)by()= )]TJ /F5 7.97 Tf 6.58 0 Td[(1jZ3 .For2C(A3,D3,Z3),bydenitionker()]TJ /F4 11.955 Tf 12.05 0 Td[(IdA3)=Z3=Im()]TJ /F4 11.955 Tf 12.06 0 Td[(IdA3).Hencethereexistsanisomorphism )]TJ /F4 11.955 Tf 11.96 0 Td[(IdA3:A3=Z3!Z3where )]TJ /F4 11.955 Tf 11.96 0 Td[(IdA3(a+Z3)=()]TJ /F4 11.955 Tf 11.96 0 Td[(IdA3)(a)for(a+Z3)2A3=Z3.Denethemap:C(A3,D3,Z3)!Aut(A3=Z3)by()= )]TJ /F5 7.97 Tf 6.59 0 Td[(1 )]TJ /F4 11.955 Tf 11.96 0 Td[(IdAfor2C(A3,D3,Z3). Lemma3.6.18. Let(A3,D3,Z3)2B(p5)andlet2C(A3,D3,Z3).For2Aut(A3jD3,Z3)andanintegern,p-n,wehavethat(n)=n()()]TJ /F5 7.97 Tf 6.59 0 Td[(1)=()()()]TJ /F5 7.97 Tf 6.58 0 Td[(1).Moreover,issurjective. Proof. SinceIm()]TJ /F4 11.955 Tf 12.69 0 Td[(IdA3)=Z3=1(A3),fora2A3thereexistsb2A3suchthat(a)=a+pb.Sinceker()]TJ /F4 11.955 Tf 12.22 0 Td[(IdA3)=Z3=01(A3),wehavethat(pb)=pb.Therefore 55

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n(a)=a+npb.Now,for(n)((a))= )]TJ /F5 7.97 Tf 6.58 0 Td[(1 n)]TJ /F4 11.955 Tf 11.95 0 Td[(IdA3(a+Z3)= )]TJ /F5 7.97 Tf 6.59 0 Td[(1(n)]TJ /F4 11.955 Tf 11.95 0 Td[(IdA3)(a)= )]TJ /F5 7.97 Tf 6.59 0 Td[(1(npb)=nb+Z3=n(b+Z3)=n( )]TJ /F5 7.97 Tf 6.59 -.01 Td[(1(pb))=n( )]TJ /F5 7.97 Tf 6.59 -.01 Td[(1(n)]TJ /F4 11.955 Tf 11.95 0 Td[(IdA3)(a))=n( )]TJ /F5 7.97 Tf 6.58 0 Td[(1 )]TJ /F4 11.955 Tf 11.95 0 Td[(IdA3(a+Z3))=n()((a)).Therefore,(n)=n().Let2Aut(A3jD3,Z3).Bydenition,)]TJ /F5 7.97 Tf 6.58 0 Td[(1=())]TJ /F5 7.97 Tf 6.58 0 Td[(1.Nowfora2A3,()]TJ /F5 7.97 Tf 6.58 0 Td[(1)(a)= )]TJ /F5 7.97 Tf 6.58 0 Td[(1 )]TJ /F5 7.97 Tf 6.58 0 Td[(1)]TJ /F4 11.955 Tf 11.96 0 Td[(IdA3(a+Z3)= )]TJ /F5 7.97 Tf 6.58 0 Td[(1()]TJ /F4 11.955 Tf 11.96 0 Td[(IdA3))]TJ /F5 7.97 Tf 6.58 0 Td[(1(a)=() )]TJ /F5 7.97 Tf 6.58 0 Td[(1 )]TJ /F4 11.955 Tf 11.95 0 Td[(IdA3()]TJ /F5 7.97 Tf 6.59 0 Td[(1(a)+Z3)=()()()]TJ /F5 7.97 Tf 6.59 0 Td[(1)(a).Therefore,()]TJ /F5 7.97 Tf 6.59 0 Td[(1)=()()()]TJ /F5 7.97 Tf 6.59 0 Td[(1).Let2Aut(A3=Z3).Sinceker()=Z3and, areisomorphisms,wehavethatker( )=Z3.Since,,and aresurjective,itfollowsthatIm( )=Z3.Assumethata2A3issuchthata2ker( +IdA3),implyingthat (a)=)]TJ /F3 11.955 Tf 9.3 0 Td[(a.SinceIm( )=Z3,wehavethata2Z3.Sinceker( )=Z3,wehavethata=0A3.Therefore, +IdA3isaninjectivegroupendomorphism,implyingthat +IdA32Aut(A3).Wenowhavethat +IdA32C(A3,D3,Z3)and( +IdA3)=,anditfollowsthatissurjective.Then:A3!A3=Z3isanisomorphismand 2Aut(A3). Lemma3.6.19. Let(A3,D3,Z3)2B(p5)andlet,2C(A3,D3,Z3).Now,ifandonlyifthereexistsanintegern,p-n,and2Aut(A3=Z3)suchthatn()=())]TJ /F5 7.97 Tf 6.59 0 Td[(1. Proof. Assume,implyingthatthereexists2Aut(A3jD3,Z3)andanintegern,p-n,suchthatn=)]TJ /F5 7.97 Tf 6.59 0 Td[(1.ByLemma 3.6.18 wehaven()=(n)=()]TJ /F5 7.97 Tf 6.59 0 Td[(1)=()()()]TJ /F5 7.97 Tf 6.58 0 Td[(1).Letting=()2Aut(A3=Z3),wehaven()=())]TJ /F5 7.97 Tf 6.58 0 Td[(1. 56

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Assumethereexists2Aut(A3=Z3)andanintegern,p-n,suchthatn()=())]TJ /F5 7.97 Tf 6.59 0 Td[(1.Since= )]TJ /F5 7.97 Tf 6.59 0 Td[(1 2Aut(A3jD3,Z3),wehavethat()=.Nowwehave(n)=()]TJ /F5 7.97 Tf 6.59 0 Td[(1)byLemma 3.6.18 .Therefore, )]TJ /F5 7.97 Tf 6.58 0 Td[(1 n)]TJ /F4 11.955 Tf 11.95 0 Td[(IdA3= )]TJ /F5 7.97 Tf 6.58 0 Td[(1 )]TJ /F5 7.97 Tf 6.58 0 Td[(1)]TJ /F4 11.955 Tf 11.96 0 Td[(IdA3)n=)]TJ /F5 7.97 Tf 6.59 0 Td[(1.Wecannowconcludethat. Lemma3.6.20. DenethenormalsubgroupD=fnIdj06=n2ZpgofGL2(p)andthecanonicalsurjectivegrouphomomorphism:GL2(p)!PGL2(p)by()=Dfor2GL2(p).Let,2GL2(p),()isconjugateto()inPGL2(p)ifandonlyifthereexistsA2GL2(p)and06=n2Zpsuchthatn=AA)]TJ /F5 7.97 Tf 6.58 0 Td[(1. Proof. Let,2GL2(p)besuchthatthereexistsA2GL2(p)and06=n2Zpwithn=AA)]TJ /F5 7.97 Tf 6.59 0 Td[(1.Sincen)]TJ /F5 7.97 Tf 6.59 0 Td[(1=nId2D,wehavethat()=(n).Therefore,()=(n)=(AA)]TJ /F5 7.97 Tf 6.59 0 Td[(1)=(A)()(A))]TJ /F5 7.97 Tf 6.58 0 Td[(1.Let,2GL2(p)andassumethatthereexistsa2PGL2(p)suchthat()=A()A)]TJ /F5 7.97 Tf 6.59 0 Td[(1.Sinceissurjective,thereexistsA2GL2(p)suchthat(A)=a.Therefore,(BB)]TJ /F5 7.97 Tf 6.59 0 Td[(1)=a()a)]TJ /F5 7.97 Tf 6.59 0 Td[(1=().Thisimplies()]TJ /F5 7.97 Tf 6.59 0 Td[(1BB)]TJ /F5 7.97 Tf 6.59 0 Td[(1)=D,hence)]TJ /F5 7.97 Tf 6.59 0 Td[(1BB)]TJ /F5 7.97 Tf 6.59 0 Td[(1=nIdforsome06=n2Z.Thereforen=BB)]TJ /F5 7.97 Tf 6.59 0 Td[(1. Lemma3.6.21. Let(A3,D3,Z3)2B(p5).ThereisabijectionbetweentheelementsofR(A3,D3,Z3)(Denition 3.3.10 )andtheconjugacyclassesofPGL2(p).Furthermore,thereexistsp+2elementsofC(p5)arisingfromA3. Proof. SinceA3=Z3iselementaryabelianoforderp2,Aut(A3=Z3)canbeidentiedwithGL2(p).ByLemma 3.6.19 ,Lemma 3.6.20 ,andsinceissurjective,theequivalenceclassesofR(A3,D3,Z3)areinbijectionwiththeconjugacyclassesofPGL2(p).By 57

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Lemma 3.6.14 ,(A3,D3,Z3)istheuniqueelementofB(p5)arisingfromA3.Thereexistsp+2elementsinR(A3,D3,Z3)andp+2elementsofC(p5)arisefromA3,sincethereexistsp+2conjugacyclassesofPGL2(p)foranoddprimep. Lemma3.6.22. Thereexistsp+2distinctelementsofD(p5)(Denition 3.3.14 )arisingfromA3. Proof. For(A3,D3,Z3,)2C(p5),thenH(A3,D3,Z3,)=Z3=01(A3)(Lemma 3.3.12 ).ByLemma 3.6.14 ,Z3=01(A3)istrivial.Therefore,thereexistsauniqueelementofR(A3,D3,Z3,)foreach(A3,D3,Z3,)2C(p5).ByLemma 3.6.21 ,thereexistsp+2elementsofC(p5)arisingfromA3,thereforethereexistsp+2distinctelementsofD(p5)arisingfromA3. WearenowabletoconcludethemaintheoremforthisSection: Theorem3.6.23. Letpbeanoddprime.Thereexistsp+8isomorphismclassesamongthegroupsof2-generalizedBlackburngroupsoforderp5whosecenterhasindexp3. Proof. ByTheorem 3.5.6 ,theelementsofD(pn)andtheisomorphismclassesofGBp(2,n)]TJ /F4 11.955 Tf 12.72 0 Td[(3,n)areinbijection.ByLemma 3.6.1 thereareonlythreeisomorphismclassesofabeliangroupsadmittedintoA(p4)andbyLemma 3.6.4 ,Lemma 3.6.13 ,andLemma 3.6.22 theseclassesgiverisetop+8elementsofD(p5). 58

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CHAPTER4CONLONGROUPSConlongroupsaredenedtobethenon-abelianp-groupswithcycliccenterandanabelianmaximalsubgroup(Denition 4.2.1 ).InthisChapterwegiveanewproofoftheclassicationofConlongroups(Theorem 4.4.15 ),providetwoconstructionsofConlongroups,andinvestigatetheirconjugacyclassesandanumberoftheircharacteristicsubgroupsincludingtheircentersandFrattinisubgroup.ThroughoutthisChapter,pwilldenoteaprimenumber. 4.1Preliminaries Denition4.1.1. DeneGptobetheclassofnilpotentgroupswithanormalabeliansubgroupofindexp. Denition4.1.2. LetG2GpandletAbeanormalabeliansubgroupofindexpinG.Letx2Gandnotethatxp2A.Denethemapx:A!Atobetheautomorphismdenedbyconjugationbyx.DenetheA-endomomorphismsx=x)]TJ /F4 11.955 Tf 11.96 0 Td[(IdAx=IdA+x+x2+...+xp)]TJ /F8 5.978 Tf 5.75 0 Td[(1. Lemma4.1.3. LetG2Gp,letAbeanormalabeliansubgroupofindexpinG,andletx2G.Thenx=p)]TJ /F5 7.97 Tf 6.59 0 Td[(1Xr=0pr+1rx. Proof. Webeginbyprovingxr=rXi=0riixforaninteger0
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=r)]TJ /F5 7.97 Tf 6.59 0 Td[(1Xi=0r)]TJ /F4 11.955 Tf 11.95 0 Td[(1i(ix+i+1x)=r)]TJ /F5 7.97 Tf 6.59 0 Td[(1Xi=0r)]TJ /F4 11.955 Tf 11.95 0 Td[(1iix+r)]TJ /F5 7.97 Tf 6.59 0 Td[(1Xi=0r)]TJ /F4 11.955 Tf 11.96 0 Td[(1ii+1x=0x+r)]TJ /F5 7.97 Tf 6.59 0 Td[(1Xi=1r)]TJ /F4 11.955 Tf 11.96 0 Td[(1iix+r)]TJ /F5 7.97 Tf 6.59 0 Td[(1Xi=0r)]TJ /F4 11.955 Tf 11.96 0 Td[(1i)]TJ /F4 11.955 Tf 11.96 0 Td[(1ix+rx=0x+r)]TJ /F5 7.97 Tf 6.58 0 Td[(1Xi=1r)]TJ /F4 11.955 Tf 11.95 0 Td[(1i+r)]TJ /F4 11.955 Tf 11.96 0 Td[(1i)]TJ /F4 11.955 Tf 11.95 0 Td[(1ix+rx=0x+r)]TJ /F5 7.97 Tf 6.59 0 Td[(1Xi=1riix+rx=rXi=0riix.Theidentitynowfollowsfrominduction.Usingtheaboveidentityandthecombinatorialidentityp)]TJ /F5 7.97 Tf 6.59 0 Td[(1Xk=rkr=pr+1(Lemma 2.1.3 ),wehavex=IdA+x+...+xp)]TJ /F8 5.978 Tf 5.76 0 Td[(1=0Xi=00iix+1Xi=01iix+...+p)]TJ /F5 7.97 Tf 6.59 0 Td[(1Xi=0p)]TJ /F4 11.955 Tf 11.95 0 Td[(1iix=p)]TJ /F5 7.97 Tf 6.58 0 Td[(1Xr=0p)]TJ /F5 7.97 Tf 6.58 0 Td[(1Xk=rkrrx=p)]TJ /F5 7.97 Tf 6.59 0 Td[(1Xr=0pr+1rx.Thelemmanowfollows. Lemma4.1.4. LetG2Gp,letAbeanormalabeliansubgroupofindexpinG,andletx2GnA.Thenxandxarenilpotentmutuallyannihilatinghomomorphismssuchthat 1) Im(x)Z(G). 2) x(z)=zpifz2Z(G). 3) ker(x)=Z(G). Proof. Leta2A,sinceAisabelian,xx(a)=xaax...axp)]TJ /F8 5.978 Tf 5.76 0 Td[(1=xax)]TJ /F5 7.97 Tf 6.59 0 Td[(1xaxx)]TJ /F5 7.97 Tf 6.59 0 Td[(1...axp)]TJ /F8 5.978 Tf 5.76 0 Td[(1x)]TJ /F5 7.97 Tf 6.59 0 Td[(1x=axp)]TJ /F8 5.978 Tf 5.75 0 Td[(1a...axp)]TJ /F8 5.978 Tf 5.76 0 Td[(2x=x(a)x.Therefore[x,x(a)]=1.SinceP=hA,xiandx(a)2Aisabelian,itnowfollowsthatx(a)2Z(P). 60

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Letz2Z(G),thenx(z)=zzx...zxp)]TJ /F8 5.978 Tf 5.75 0 Td[(1=zpx(z)=[z,x]=1.Itfollowsthatx(z)=zpforz2Z(GandZ(G)ker(x).SinceG=hA,xiandAisabelian,ifx(a)=[a,x]=1fora2Athena2Z(G).Wenowhavethatker(x)=Z(G).SinceGisnilpotentandxisacommutatormap,itfollowsthatxisnilpotent.ByLemma 4.1.3 wehavex=p)]TJ /F5 7.97 Tf 6.58 0 Td[(1Xr=0pr+1rx,hencexisnilpotent.Sincexp2A,wehavepx=IdA.Now,xx=(x)]TJ /F4 11.955 Tf 11.96 0 Td[(IdA)(IdA+x+2x+...+p)]TJ /F5 7.97 Tf 6.59 0 Td[(1x)=px)]TJ /F4 11.955 Tf 11.96 0 Td[(IdA=(IdA+x+x2+...+xp)]TJ /F8 5.978 Tf 5.75 0 Td[(1)(x)]TJ /F4 11.955 Tf 11.96 0 Td[(IdA)=xx.Hencexx=xx=px)]TJ /F4 11.955 Tf 11.95 0 Td[(IdA=0A. Lemma4.1.5. LetG2Gp,letAbeanormalabeliansubgroupofindexpinG,andletx2GnA.Foraninteger1j
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Therefore,xj=x+xj)]TJ /F8 5.978 Tf 5.76 0 Td[(1(x+IdA).Byourinductionhypothesis,xj=x+ j)]TJ /F5 7.97 Tf 6.58 0 Td[(1Xi=1j)]TJ /F4 11.955 Tf 11.96 0 Td[(1iix!(x+IdA)=j)]TJ /F5 7.97 Tf 6.59 0 Td[(1Xi=0j)]TJ /F4 11.955 Tf 11.95 0 Td[(1ii+1x+j)]TJ /F5 7.97 Tf 6.59 0 Td[(1Xi=1j)]TJ /F4 11.955 Tf 11.96 0 Td[(1iix=j)]TJ /F5 7.97 Tf 6.59 0 Td[(1Xi=1j)]TJ /F4 11.955 Tf 11.95 0 Td[(1i)]TJ /F4 11.955 Tf 11.95 0 Td[(1+j)]TJ /F4 11.955 Tf 11.95 0 Td[(1iix+jx=jXi=1jiix. 4.2TheConlonClass Denition4.2.1. DeneaConlongrouptobeanitenon-abelianp-groupPwithcycliccenterandanabelianmaximalsubgroup.DeneCptobetheclassofConlongroupsoforderdivisiblebyp. Lemma4.2.2. TheclassCpisasubclassofGp. Proof. Sincep-groupsarenilpotentandthemaximalsubgroupsofap-grouparenormalwithindexp(Theorem 2.3.1 ),wehavethatCpisasubclassofGp. InthisSection,thegroupstructureofConlongroupswillbeexplored.TheresultsfromSection1nowholdfortheclassCp. Lemma4.2.3. IfP2CpandAisanabelianmaximalsubgroupofP,thenZ(P)AP.Furthermore,ifx2PnA,thenxp2Z(P). Proof. SinceAisamaximalsubgroupofap-group,AhasindexpinP.Assumethereexistsx2PnAsuchthatx2Z(P),itfollowsthatPisabeliansinceP=hA,xi,acontradiction.HenceZ(P)AP.Letx2PnA,thenxp2Z(P),sincexp2A,Aisabelian,andP=hA,xi. Lemma4.2.4. LetP2CpandletAbeanabelianmaximalsubgroupofP.Then 1) x=yforx,y2PnA. 2) (ax)p=x(a)xpforx2PnAanda2A. 62

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Proof. Letx2PnAandletibeanintegersuchthatp-i.Fora2A,xi(a)=aaxi...axi(p)]TJ /F8 5.978 Tf 5.75 0 Td[(1).Sinceaxp=ax0=a,wecanidentifyfaxjgp)]TJ /F5 7.97 Tf 6.58 0 Td[(1j=0withZpviathemapaxj7!j.Sincei:Zp!Zpdenedi(n)=inforn2Zpisanautomorphism,wemayviewaxi,...,axi(p)]TJ /F8 5.978 Tf 5.75 0 Td[(1)asapermutationofax,...,axp)]TJ /F8 5.978 Tf 5.76 0 Td[(1.SinceAisabelianandnormalinP,wehavexi(a)=aaxi...axi(p)]TJ /F8 5.978 Tf 5.76 0 Td[(1)=aax...axp)]TJ /F8 5.978 Tf 5.76 0 Td[(1=x(a).Forb2Aandanintegerj,axjb=axjsinceAisabelian.SinceAhasindexpinP=hx,Ai,anyy2PnAmaybewritteny=xjbwithjaninteger,p-j,andb2A.Fora2Ay(a)=xjb(a)=aaxjb...a(xjb)p)]TJ /F8 5.978 Tf 5.76 0 Td[(1=aaxj...axj(p)]TJ /F8 5.978 Tf 5.75 0 Td[(1)=xj(a)=x(a)and(ax)p=axax)]TJ /F5 7.97 Tf 6.59 0 Td[(1x2ax)]TJ /F5 7.97 Tf 6.59 0 Td[(2x3...ax)]TJ /F5 7.97 Tf 6.59 0 Td[((p)]TJ /F5 7.97 Tf 6.59 0 Td[(1)xp=aax)]TJ /F8 5.978 Tf 5.76 0 Td[(1...ax)]TJ /F8 5.978 Tf 5.76 0 Td[((p)]TJ /F8 5.978 Tf 5.75 0 Td[(1)xp=x)]TJ /F8 5.978 Tf 5.76 0 Td[(1(a)xp=x(a)xp.Therefore,thelemmafollows. Denition4.2.5. LetP2CpandletAbeanabelianmaximalsubgroupofP.DeneA=xforsomex2PnA.NoticethatAiswell-denedduetoLemma 4.2.4 Lemma4.2.6. P2Cphasnilpotenceclass2ifandonlyifjP:Z(P)j=p2. Proof. AssumejP:Z(P)j=p2.SinceP=Z(P)isabelianandZ(P)6=P,wehavethatPhasnilpotenceclass2.AssumePhasclass2,thenP0Z(P)(Theorem 2.3.1 ).LetZ1=1(Z(P)),andnotethatZ1hasorderpsinceZ(P)iscyclic.Fora2Aandx2PnAwehavex(a)2Z(P)sinceP0Z(P).Hencex(a)p=A(x(a))=1andx(a)2Z1for 63

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a2A.SincejAj=jker(x)jjIm(x)jandker(x)=Z(P)andIm(x)=Z1,wehavethatjA:Z(P)j=pandhencejP:Z(P)j=p2. Lemma4.2.7. IfP2Cphasnilpotenceclassgreaterthan2,thenZ(P)hasindexpinthesecondcenter,Z2(P). Proof. AssumebywayofcontradictionthatAZ2(P)=P.ByLemma 4.2.3 ,Z(P)A.Dene P=P=Z(P).Byourassumption, AZ( P)= P,thereforeP=Z2(P).ThisimpliesthatPhasnilpotenceclass2,acontradiction.WenowhavethatZ2(P)A,moreoverZ2(P)=fa2Ajx(a)2Z(P)g.LetP2=hZ2(P),xiP.SinceAisabelian,Z(P2)=Z(P)iscyclicandZ2(P)isanabelianmaximalsubgroupofP2,henceP2Cp.BydenitionP2hasnilpotenceclass2andbyLemma 4.2.6 jP2:Z(P)j=jP2:Z(P2)j=p2.SinceZ2(P)ismaximalinP2,wehavejZ2(P):Z(P)j=p. Theorem4.2.8. LetP2Cp,letkbethenilpotenceclassofP,andletzbesuchthatjP:Z(P)j=pz.Thenk=z. Proof. Assumebywayofcontradictionthatthelemmaisfalse.AmongallcounterexampleschooseP2Cptohaveminimalclassk.Let P=P=Z(P).Assume Pisabelian,thenPhasclass2andbyLemma 4.2.6 jP:Z(P)j=p2,hencePisnotacounterexample.Therefore Pisnon-abelianandPhasclassgreaterthan2.LetAbeanabelianmaximalsubgroupofPandlet A=A=Z(P) P.SinceZ(P)A,wehavethat Aisabelianmaximalin P.ItnowfollowsfromLemma 4.2.7 thatZ( P)hasorderp,henceiscyclic.Therefore P2Cp.Bydenition Phasclassk)]TJ /F4 11.955 Tf 12.56 0 Td[(1,henceisnotacounterexample.Itfollowsthatj P:Z( P)j=pk)]TJ /F5 7.97 Tf 6.58 0 Td[(1.SincejZ( P)j=pwehavej Pj=pkanditfollowsthatjP:Z(P)j=pk,contradictingourchoiceofcounterexample. 64

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Corollary4.2.9. LetP2CphavenilpotenceclasskandletZi=Zi(P),theithcenterofP.LetAbeanabelianmaximalsubgroupofP.Then 1) jPj=pk+lwherejZ(P)j=pl. 2) ZiAfor1i
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Proof. SincePisofclasskandZk)]TJ /F5 7.97 Tf 6.59 0 Td[(1
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Proof. AssumeIm(A)=Z(P)andletx2PnA.ByLemma 4.2.4 ,fora2Atheelementax2PnAand(ax)p=A(a)xp.Sincexp2Z(P)byLemma 4.2.3 ,thereexistsb2AsuchthatA(b)=x)]TJ /F7 7.97 Tf 6.59 0 Td[(p.Thereforey=bx2PnAissuchthatbp=1.AssumeIm(A)=Z(P)pandletx2PnA.Everyelementy2PnAcanbewritteny=axjfora2Aandanintegerj,p-j.IfxpgeneratesZ(P),thenyp=A(a)xjpgeneratesZ(P)sincep-j.Ifxp2Z(P)p,thenyp=A(a)xjp2Z(P)pandthereexistsa2AsuchthatA(a)=x)]TJ /F7 7.97 Tf 6.59 0 Td[(jpandtheelementz=axj2PnAissuchthatzp=1. Lemma4.2.13. LetP2Cp,letAbeanabelianmaximalsubgroupofP,letPhavenilpotenceclassk,andletZi=Zi(P)betheithcenterofP. 1) IfIm(A)=Z(P),thenA(a)generatesZ(P)foreverya2AnZk)]TJ /F5 7.97 Tf 6.59 0 Td[(1. 2) IfIm(A)=Z(P)p,thenthereexistsa2AnZk)]TJ /F5 7.97 Tf 6.58 0 Td[(1suchthatA(a)=1. Proof. AssumeIm(A)=Z(P),byCorollary 4.2.11 thereexistsa2AnZk)]TJ /F5 7.97 Tf 6.58 0 Td[(1suchthatA(a)generatesZ(P).ByCorollary 4.2.9 elementsofAnZk)]TJ /F5 7.97 Tf 6.59 0 Td[(1areoftheformb=ajzwherez2Zk)]TJ /F5 7.97 Tf 6.58 0 Td[(1andjisaninteger,p-j.SinceA(z)2Z(P)pandp-j,wehavethatA(b)isageneratorofZ(P).AssumeIm(A)=Z(P)p.Letx2AnZk)]TJ /F5 7.97 Tf 6.58 0 Td[(1andletb2Z(P)besuchthatb)]TJ /F7 7.97 Tf 6.58 0 Td[(p=A(x).Lettinga=xb2AnZk)]TJ /F5 7.97 Tf 6.59 0 Td[(1wehaveA(a)=1. 4.3ConlonTypeInthisSection,thedenitionofaConlonTypeforaConlongroupwillbedeveloped.ItwillbeshownthateveryConlongrouphasauniqueConlontype. Denition4.3.1. LetP2CpandletAbeanabelianmaximalsubgroupofP.SupposePhasnilpotentclasskandZ(P)hasorderpl.WerefertoasetC=fx1,...,xk,y,x1gasapreA-ConlonsetforPif 1) PisgeneratedbyC. 2) AisgeneratedbyCnfx1g. 3) [xj,x1]=xj)]TJ /F5 7.97 Tf 6.58 0 Td[(1for2jk. 4) Z(P)isgeneratedbyy. 67

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5) x1=ypl)]TJ /F8 5.978 Tf 5.75 0 Td[(121(Z(P)). Denition4.3.2. Foraprimep,letFpbetheeldofpelementsandletFpbethemultiplicativegroupofFp.Let:Z!Fpbethesurjectiveringhomomorphism. Denition4.3.3. LetP2CpandletAbeanabelianmaximalsubgroupofP.SupposePhasnilpotentclasskandZ(P)hasorderpl.LetCbeapreA-ConlonsetforP. 1) CissaidtohaveConlontype0,ifxp1=1andA(xk)=1. 2) CissaidtohaveConlontype1,ifxp1=yandA(xk)=1. 3) CissaidtohaveConlontypemform2Fp,ifxp1=1andA(xk)=ynwherem= n.Furthermore,wesaythatCisanA-ConlonsetforPifCisapreA-ConlonsetforPandChasaConlontype. Lemma4.3.4. LetP2CpandletAbeanabelianmaximalsubgroupofP. 1) IfIm(A)=Z(P),thenthereexistsanA-ConlonsetoftypemforP,forsomem2Fp. 2) IfIm(A)=Z(P)pandthereexistsx2PnAsuchthatxp2Z(P)p,thenthereexistsanA-Conlonsetoftype0forP. 3) IfIm(A)=Z(P)pandthereexistsx2PnAsuchthatxpgeneratesZ(P),thenthereexistsanA-Conlonsetoftype1forP. Proof. AssumePhasnilpotenceclasskandjZ(P)j=pl.LetZi=Zi(P)betheithcenterofP.AssumeIm(A)=Z(P).ByLemma 4.2.11 thereexistsa2AnZk)]TJ /F5 7.97 Tf 6.59 0 Td[(1suchthatA(a)generatesZ(P).ByLemma 4.2.12 ,thereexistsx12PnAsuchthatxp1=1.Letxk)]TJ /F7 7.97 Tf 6.58 0 Td[(i=ix1(a)for0i
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AssumeIm(A)=Z(P)pandx2PnAissuchthatxp2Z(P)p.ByLemma 4.2.12 ,thereexistsx12PnAsuchthatxp1=1.ByLemma 4.2.13 ,thereexistsa2AnZk)]TJ /F5 7.97 Tf 6.59 0 Td[(1suchthatA(a)=1.Letxk)]TJ /F7 7.97 Tf 6.59 0 Td[(i=ix1(a)for0i
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AssumeIm(A)=Z(P)p.ByLemma 4.2.12 ,eithereveryx2PnAissuchthatxpgeneratesZ(P)oreveryx2PnAissuchthatxp2Z(P)p.Assumethereexistsx2PnAsuchthatxpgeneratesZ(P),byLemma 4.3.4 thereexistsanA-Conlonsetoftype0forP.Assumethereexistsx2PnAsuchthatxp2Z(P)p,byLemma 4.3.4 thereexistsanA-Conlonsetoftype1forP. Lemma4.3.7. LetP2CphavenilpotenceclasskandletAbeanabelianmaximalsubgroupofP.AssumethereexistsanA-ConlonsetoftypemforP,wherem2Fp.ThereexistsanA-ConlonsetoftypenforPifandonlyifn2m(Fp)k)]TJ /F5 7.97 Tf 6.59 0 Td[(1. Proof. Assumefx1,...,xk,y,x1gisanA-ConlonsetoftypemforPandf^x1,...,^xk,^y,^x1gisanA-ConlonsetoftypenforP.Letr,s2Zbesuchthatm= randn= s,A(xk)=yr,andA(^xk)=^ys.Bothxkand^xkareelementsofAnZk)]TJ /F5 7.97 Tf 6.59 0 Td[(1(P)(Lemma 4.3.5 ).SinceAisgeneratedbyxkandZk)]TJ /F5 7.97 Tf 6.59 0 Td[(1(P)(Lemma 4.2.9 ),wehave^xk=xikzwherep-iandz2Zk)]TJ /F5 7.97 Tf 6.59 0 Td[(1(P).Sincex1and^x1areelementsofPnA,wehave^x1=axj1wherep-j,a2A.Sincex1isacommutatormapandPhasnilpotenceclassk,wehavethatjx1istrivialforjk.Usingthisfact,thedenitionofpreA-ConlonsetsandLemma 4.1.5 ,wehave^ypl)]TJ /F8 5.978 Tf 5.76 0 Td[(1=^x1=k)]TJ /F5 7.97 Tf 6.59 0 Td[(1^x1(^xk)=k)]TJ /F5 7.97 Tf 6.59 0 Td[(1axj1(xikz)= jXi=1jiix1!k)]TJ /F5 7.97 Tf 6.58 0 Td[(1(xikz)= j1k)]TJ /F5 7.97 Tf 6.59 0 Td[(1k)]TJ /F5 7.97 Tf 6.58 0 Td[(1x1!(xikz)=k)]TJ /F5 7.97 Tf 6.59 0 Td[(1x1(xikz)jk)]TJ /F8 5.978 Tf 5.75 0 Td[(1=xijk)]TJ /F8 5.978 Tf 5.76 0 Td[(11=ypl)]TJ /F8 5.978 Tf 5.76 0 Td[(1ijk)]TJ /F8 5.978 Tf 5.76 0 Td[(1.UsingthedenitionofConlonexponents,^ys=A(^xk)=A(xikz)=yriA(z).Sincez2Zk)]TJ /F5 7.97 Tf 6.58 0 Td[(1(P),wehavethatA(z)2Z(P)p.Itfollowsthat^ypl)]TJ /F8 5.978 Tf 5.76 0 Td[(1s=ypl)]TJ /F8 5.978 Tf 5.76 0 Td[(1ri.Wenowhavethatypl)]TJ /F8 5.978 Tf 5.75 0 Td[(1sijk)]TJ /F8 5.978 Tf 5.76 0 Td[(1=ypl)]TJ /F8 5.978 Tf 5.75 0 Td[(1ri,implyingthatypl)]TJ /F8 5.978 Tf 5.76 0 Td[(1i(r)]TJ /F7 7.97 Tf 6.59 0 Td[(sjk)]TJ /F8 5.978 Tf 5.76 0 Td[(1)=1.Sinceypl)]TJ /F8 5.978 Tf 5.75 0 Td[(121(Z(P))andp-i,wehaversjk)]TJ /F5 7.97 Tf 6.59 0 Td[(1(modp),implyingm2n(Fp)k)]TJ /F5 7.97 Tf 6.58 0 Td[(1. 70

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Assumefx1,...,xk,y,x1gisanA-ConlonsetoftypemforPandletn2m(Fp)k)]TJ /F5 7.97 Tf 6.59 0 Td[(1.Letr,s2Zbesuchthatm= randn= s,implyingrsjk)]TJ /F5 7.97 Tf 6.58 0 Td[(1(modp)forsomej2Z.Notethatp-r,p-s,andp-j.Let^xk=xsk,^x1=xj1,and^y=yr.Let^xa)]TJ /F5 7.97 Tf 6.59 0 Td[(1=[^xa,^x1]for2ak.LetC=f^x1,...,^xk,^y,^x1g,wenowshowthatCisanA-ConlonsetoftypenforP.Sincep-sandxk2AnZk)]TJ /F5 7.97 Tf 6.58 0 Td[(1(Lemma 4.3.5 )wehave^xk2AnZk)]TJ /F5 7.97 Tf 6.58 0 Td[(1.Similarly,sincep-jandx12PnAwehave^x12PnA.Itfollowsthat^xk)]TJ /F7 7.97 Tf 6.59 0 Td[(i=i^x1(^xk)for1ik.ByLemma 4.2.10 thesetCnf^x1ggeneratesAandsince^x12PnAwehavethatCgeneratesP.Sincep-r,wehavethat^ygeneratesZ(P).RecallingLemma 4.1.5 wehave^x1=^xk)]TJ /F8 5.978 Tf 5.76 0 Td[(11(^xk)=k)]TJ /F5 7.97 Tf 6.59 0 Td[(1x1(xk)sjk)]TJ /F8 5.978 Tf 5.76 0 Td[(1=xsjk)]TJ /F8 5.978 Tf 5.76 0 Td[(11=ypl)]TJ /F8 5.978 Tf 5.75 0 Td[(1sjk)]TJ /F8 5.978 Tf 5.76 0 Td[(1.Sinceypl)]TJ /F8 5.978 Tf 5.76 0 Td[(121(Z(P))andrsjk)]TJ /F5 7.97 Tf 6.59 0 Td[(1(modp)wehave^x1=ypl)]TJ /F8 5.978 Tf 5.76 0 Td[(1r.Wenowhave^ypl)]TJ /F8 5.978 Tf 5.75 0 Td[(1=(yr)pl)]TJ /F8 5.978 Tf 5.75 0 Td[(1=^x1,anditfollowsthatCisapreA-ConlonsetforP.Since^xp1=xpj1=1andA(^xk)=A(xsk)=yrs=^ys,wehavethatCisanA-ConlonsetoftypenforP. Denition4.3.8. LetP2CphavenilpotenceclasskandletAbeanabelianmaximalsubgroupofP. 1) AhasConlontype0ifthereexistsanA-Conlonsetoftype0forP. 2) AhasConlontype1ifthereexistsanA-Conlonsetoftype1forP. 3) AhasConlontypemform2Fp=(Fp)k)]TJ /F5 7.97 Tf 6.59 0 Td[(1ifthereexistsanA-ConlonsetoftyperforPwherem=r(Fp)k)]TJ /F5 7.97 Tf 6.59 0 Td[(1. Theorem4.3.9(A-ConlonSetUniqueness). LetP2CpandletAbeanabelianmaximalsubgroupofP.ThenAhasauniqueConlontype. 71

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Proof. ByLemma 4.3.6 ,thereexistsanA-ConlonsetforeveryabelianmaximalsubgroupAofP.Therefore,thereexistsaConlontypeforeachabelianmaximalsubgroupA.AssumeAhasConlontypemwherem2Fp=(Fp)k)]TJ /F5 7.97 Tf 6.58 0 Td[(1.LetCbeanA-ConlonsetoftyperforPwherem=r(Fp)k)]TJ /F5 7.97 Tf 6.58 0 Td[(1.Bydenition,A(xk)generatesZ(P),wehavethatIm(A)=Z(P).SinceeverypreA-ConlonsetforPhasxk2AnZk)]TJ /F5 7.97 Tf 6.58 0 Td[(1(P)(Lemma 4.2.12 )andA(a)6=1foralla2AnZk)]TJ /F5 7.97 Tf 6.59 0 Td[(1(P)(Lemma 4.2.13 ),therearenoA-Conlonsetsoftype0or1forP.ByLemma 4.3.7 thereexistsanA-ConlonsetoftypenforP,wheren2Fp,ifandonlyifr2n(Fp)k)]TJ /F5 7.97 Tf 6.58 0 Td[(1.Therefore,AhasauniqueConlontypem.AssumeAhasConlontype1.AssumeCisanA-Conlonsetoftype1forP.BydenitionwehaveIm(A)=Z(P)psinceA(xk)=1,AisgeneratedbyxkandZk)]TJ /F5 7.97 Tf 6.59 0 Td[(1(P)(Lemma 4.3.5 andLemma 4.2.9 ),andA(Zk)]TJ /F5 7.97 Tf 6.59 0 Td[(1(P))=Z(P)p(Lemma 4.2.11 ).Sincexp1generatesZ(P)theredoesnotexisty2PnAsuchthatyp=1(Lemma 4.2.12 ).Hence,therearenoA-Conlonsetsoftype0ormforP,wherem2Fp.AssumeAhasConlontype0.IfAalsohadConlontype1orm,wherem2Fp=(Fp)k)]TJ /F5 7.97 Tf 6.58 0 Td[(1,thentheaboveparagraphswouldbecontradicted.Hence,anabelianmaximalsubgroupAhasauniqueConlontype. WenowhaveauniquedescriptionoftheabelianmaximalsubgroupsofConlongroups.AttentionisnowdrawntothenumberofabelianmaximalsubgroupsofaConlongroupandthevarietyofA-Conlontypespossibleforeach. Lemma4.3.10. LetP2Cp.IfPhasnilpotenceclassgreaterthan2,thenthereexistsauniqueabelianmaximalsubgroupofP. Proof. AssumethatPhasnilpotenceclassk>2.LetAandBbemaximalsubgroupsofPsuchthatA6=B.AssumebywayofcontradictionthatbothAandBareabelian.ByLemma 4.2.9 ,thek)]TJ /F4 11.955 Tf 12.96 0 Td[(1thcenterofP,Zk)]TJ /F5 7.97 Tf 6.58 0 Td[(1(P)6=Z(P),iscontainedinbothAandB.Letx2PnAbesuchthatx2B.Forxk2AnZk)]TJ /F5 7.97 Tf 6.59 0 Td[(1(P),Lemma 4.2.10 givesZk)]TJ /F5 7.97 Tf 6.59 0 Td[(1(P)=x(xk),...,k)]TJ /F5 7.97 Tf 6.58 0 Td[(1x(xk),Z(P).Sincex2PnA,wehavex=2Zk)]TJ /F5 7.97 Tf 6.59 0 Td[(1(P)A, 72

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hencex2BnZk)]TJ /F5 7.97 Tf 6.58 0 Td[(1(P).Bydenition,[x(xk),x]=2x(xk)6=1anditfollowsthatBisnotabelian,acontradiction. Lemma4.3.11. LetP2Cp.IfPhasnilpotenceclass2,thenPhasp+1abelianmaximalsubgroups. 1) IfjPj6=8,theneitherPhasp+1abelianmaximalsubgroupsofConlontype0,orPhas1abelianmaximalsubgroupofConlontype1andpabelianmaximalsubgroupsofConlontypem,wherem2Fp=Fp. 2) IfjPj=8,theneitherPhas3abelianmaximalsubgroupsofConlontype1,orPhas1abelianmaximalsubgroupofConlontype0and2abelianmaximalsubgroupsofConlontypem,wherem2F2=F2. Proof. AssumethatPhasnilpotenceclass2.ByTheorem 4.2.8 ,Z(P)hasindexp2inP.LetAbeanabelianmaximalsubgroupofPandletfx1,x2,y,x1gbeapreA-ConlonforP.SinceZ(P)hasindexpinAandxp12Z(P)(Lemma 4.2.3 ), P=P=Z(P)iselementaryabelianoforderp2.AnymaximalabeliansubgroupforPcontainsZ(P)(Lemma 4.2.3 ),hencethereisabijectionbetweentheabelianmaximalsubgroupsofPandthepropernon-trivialsubgroupsof P,ofwhichtherearep+1.LetBj=Dxj2x1,Z(P)Efor0j
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AssumejPj6=8,implyingp6=2orjZ(P)j6=p.AssumeAhasConlontype0,implyingthereexistsanA-ConlonsetCforPoftype0.Bydenitionandequation 4 ,wehavexp2=x(p2)1andBj(xj2x1)=x)]TJ /F4 11.955 Tf 6.58 -1.17 Td[((p2)1,henceIm(Bj)=Z(P)pandx22PnBjissuchthatxp22Z(P)p.ByLemma 4.3.4 thereexistsaBj-Conlonsetoftype0forPfor0j
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Theorem4.3.13. IfP2CpthenPhasauniqueConlontype. Proof. LetP2Cp.AssumePhasnilpotenceclassk,wherek2.ByLemma 4.3.10 ,PhasauniqueabelianmaximalsubgroupAofP.ByTheorem 4.3.9 AhasauniqueConlontype.ThereforePhasauniqueConlontypeassociatedwiththeConlontypeofit'suniqueabelianmaximalsubgroup.AssumePhasnilpotenceclass2.ByLemma 4.3.11 ,Phasp+1abelianmaximalsubgroupsandregardlessoforder,PeitherhasanabelianmaximalsubgroupofConlontype1oranabelianmaximalsubgroupofConlontype0.ByTheorem 4.3.9 abelianmaximalsubgroupshaveauniqueConlontype,implyingPeitherhasConlontype1or0. Denition4.3.14. Letpbeaprime.Foranintegeri2,denethesetM(p,i)=8><>:f0,1gi=2.f0,1g[Fp=(Fp)i)]TJ /F5 7.97 Tf 6.59 0 Td[(1i3.LetQpbethesetofConlontriples(k,l,m),wherek2andl1areintegers,andm2M(p,k). Lemma4.3.15. ForP2Cp,thereexistsauniquetriple(k,l,m)2Qp,where 1) Phasnilpotenceclassk. 2) jZ(P)j=pl. 3) PhasConlontypem. Proof. LetP2Cp.SincePisanon-trivialp-group,jZ(P)j=plwithl1.SincePisanon-abelianp-group,Phasnilpotenceclassk,wherek2.ByTheorem 4.3.13 ,PhasauniqueConlontypem2Mk.Therefore,Paffordsauniquetriple(k,l,m)2Qp. 4.4ConlonGroupConstruction Denition4.4.1. Dene Ctobethesetofquadruples(p,k,l,m)suchthat 1) pisaprime. 2) k2isaninteger. 75

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3) l1isaninteger. 4) 0mpisaninteger.InthisSection,twoisomorphicgroupswillbeconstructedusingtheset C.ThesegroupswillbeshowntolieinCpandwillbeusedinthemainclassicationtheorem. Denition4.4.2. Let(p,k,l,m)2 C.DeneFA(p,k,l,m)bethefreeabeliangroupgeneratedbythedistinctvariablesfYl,X2,...,Xkg.Weset 1) X1=pl)]TJ /F5 7.97 Tf 6.59 0 Td[(1Yl. 2) Xj=0forj0.. 3) 1=plYl. 4) i=p)]TJ /F5 7.97 Tf 6.59 0 Td[(1Xj=0pj+1Xi)]TJ /F7 7.97 Tf 6.59 0 Td[(jfor2i
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Denition4.4.4. For(p,k,l,m)2 CletRA(p,k,l,m)bethenormalclosureofh1,...,kiinFA(p,k,l,m).DeneA(p,k,l,m)=FA(p,k,l,m)=RA(p,k,l,m)andlet:FA(p,k,l,m)!A(p,k,l,m)bethenaturalprojection.Sety=(Y1)andxj=(Xj)for2jk. Lemma4.4.5. If(p,k,l,m)2 C,thenA(p,k,l,m)hasorderpk+l)]TJ /F5 7.97 Tf 6.59 0 Td[(1. Proof. LetMbethekk-matrixforf1,...,kgwithrespecttothebasisfYl,X2,...,Xkg.NowMhasentries:Mi,j=8>>>>>>>>>>>>>>>>><>>>>>>>>>>>>>>>>>:pl(i,j)=(1,1).0i>j.)]TJ /F7 7.97 Tf 14.76 -4.37 Td[(pi)]TJ /F7 7.97 Tf 6.58 0 Td[(j+10i)]TJ /F3 11.955 Tf 11.95 0 Td[(jp)]TJ /F4 11.955 Tf 11.96 0 Td[(1andj6=1.0i)]TJ /F3 11.955 Tf 11.95 0 Td[(jpand(i,j)6=(k,1).)]TJ /F7 7.97 Tf 5.48 -4.38 Td[(pipl)]TJ /F5 7.97 Tf 6.58 0 Td[(1j=1and2ip.0(i,j)=(k,1)andm2f0,pg.)]TJ /F3 11.955 Tf 9.3 0 Td[(m(i,j)=(k,1)and1mp)]TJ /F4 11.955 Tf 11.95 0 Td[(1.SinceMi,j=0fori>j,Misalowertriangularmatrix.SinceMi,i=pif1
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1) \(1)=\(plYl)=plYl=1. 2) \(i)=)]TJ /F14 11.955 Tf 29.59 20.45 Td[( p)]TJ /F5 7.97 Tf 6.59 0 Td[(1Xj=0pj+1Xi)]TJ /F7 7.97 Tf 6.59 0 Td[(j!=p)]TJ /F5 7.97 Tf 6.59 0 Td[(1Xj=0pj+1(Xi)]TJ /F7 7.97 Tf 6.59 0 Td[(j+Xi)]TJ /F7 7.97 Tf 6.59 0 Td[(j+1)=i+i)]TJ /F5 7.97 Tf 6.59 0 Td[(1for2i<>:h(0,p)i0mp)]TJ /F4 11.955 Tf 11.95 0 Td[(1.h(y,p)im=p. Lemma4.4.8. N(p,k,l,m)Z(S(p,k,l,m))for(p,k,l,m)2 C. Proof. Let(a,b)2S=S(p,k,l,m).Since(a,b)(0,p)=(a+(b)(0),b+p)=(0+a,p+b)=(0+(p)(a),p+b)=(0,p)(a,b) 78

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and(a,b)(y,0)=(a+(b)(y),b+0)=(a+y,b+0)=(y+a,0+b)=(y,0)(a,b)wehavethat(0,p),(y,0),and(y,p)areelementsofZ(S).ItfollowsthatN(p,k,l,m)Z(S). Denition4.4.9. Let(p,k,l,m)2 CandN=N(p,k,l,m).DeneQ(p,k,l,m)=S(p,k,l,m)=N.Set 1) y=(y,0)N. 2) xj=(xj,0)Nfor1jk. 3) x1=(0,)]TJ /F4 11.955 Tf 9.3 0 Td[(1)N. Theorem4.4.10. Let(p,k,l,m)2 CandletD=f x1..., xk, y, x1g.LetQ=Q(p,k,l,m),then 1) Qhasorderpk+l. 2) Z(Q)iscyclicoforderpl. 3) Q2Cp. 4) Qhasnilpotenceclassk.Moreover,DisaConlonsetforQoftype0ifm=0, m2Fpif1mp)]TJ /F4 11.955 Tf 12.14 0 Td[(1,and1ifm=p. Proof. LetA=A(p,k,l,m)andN=N(p,k,l,m).Leta,b2Aandassumethat(a,0)N=(b,0)N.Then(a,0)(b,0))]TJ /F5 7.97 Tf 6.58 0 Td[(1=(a,0)()]TJ /F3 11.955 Tf 9.3 0 Td[(b,0)=(a)]TJ /F3 11.955 Tf 12 0 Td[(b,0)2N,whichimpliesthata=b.Itfollowsthat A=f(a,0)Nja2AgisisomorphictoA.SinceA/S(p,k,l,m),wehavethat A/Q.SinceS(p,k,l,m)isgeneratedbyf(y,0),(x2,0),...,(xk,0),(0,)]TJ /F4 11.955 Tf 9.3 0 Td[(1)g,wehavethatQisgeneratedbyD.Therefore,Q= Aiscyclicgeneratedby x1 Aand( x1 A)p= x1p A=(0,)]TJ /F4 11.955 Tf 9.3 0 Td[(1)pN A=(0,)]TJ /F3 11.955 Tf 9.3 0 Td[(p)N A. 79

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Ifm2f0,pg,then(0,)]TJ /F3 11.955 Tf 9.3 0 Td[(p)N A=(0,0)N A= A.If1m
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2) Ifm=p,thenDhasConlontype1since x1p=(0,)]TJ /F3 11.955 Tf 9.3 0 Td[(p)N=(y,0)N= yand A( xk)=p)]TJ /F5 7.97 Tf 6.58 0 Td[(1Yj=0 xk)]TJ /F7 7.97 Tf 6.58 0 Td[(j(pj+1)= p)]TJ /F5 7.97 Tf 6.59 0 Td[(1Xj=0pj+1xk)]TJ /F7 7.97 Tf 6.59 0 Td[(j,0!N=(0,0)N. 3) If1mp)]TJ /F4 11.955 Tf 12.09 0 Td[(1,thenDhasConlontype m2Fpsince x1p=(0,)]TJ /F3 11.955 Tf 9.3 0 Td[(p)N=(0,0)Nand A( xk)=p)]TJ /F5 7.97 Tf 6.58 0 Td[(1Yj=0 xk)]TJ /F7 7.97 Tf 6.58 0 Td[(j(pj+1)= p)]TJ /F5 7.97 Tf 6.58 0 Td[(1Xj=0pj+1xk)]TJ /F7 7.97 Tf 6.58 0 Td[(j,0!N=(my,0)N= ym.Hence,thetheoremfollows. WenowhaveaconstructionforConlongroupsarisingfromfreeabeliangroups.Next,apresentationwillbegivenforConlongroups.Thispresentationisusedinthemainclassicationtheorem. Denition4.4.11. Let(p,k,l,m)2 C.LetF(p,k,l,m)bethefreegroupgeneratedbythedistinctvariablesfS,R1,...,Rk,R1g.Weset 1) j=[Rj,R1]R)]TJ /F5 7.97 Tf 6.59 0 Td[(1j)]TJ /F5 7.97 Tf 6.59 0 Td[(1for2jk. 2) zj=[S,Rj]for2jkandj=1. 3) i,j=[Ri,Rj]for1i,jk. 4) Ifm
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Proof. SinceQ(p,k,l,m)isgeneratedbyf x1,..., xk, y, x1g,thereexistsasurjectivehomomorphism:F(p,k,l,m)!Q(p,k,l,m)where 1) (S)= y. 2) (Rj)= xjfor1jkorj=1.LetN=N(p,k,l,m).Wehavethat 1) For2jk,(j)=[ xi, x1] xi)]TJ /F5 7.97 Tf 6.58 0 Td[(1)]TJ /F5 7.97 Tf 6.59 0 Td[(1=[(xi,0),(0,)]TJ /F4 11.955 Tf 9.3 0 Td[(1)]()]TJ /F3 11.955 Tf 9.3 0 Td[(xi)]TJ /F5 7.97 Tf 6.58 0 Td[(1,0)N=()]TJ /F3 11.955 Tf 9.3 0 Td[(xi+(1)(xi))]TJ /F3 11.955 Tf 11.95 0 Td[(xi)]TJ /F5 7.97 Tf 6.59 0 Td[(1,0)N=(0,0)N. 2) Since y2Z(Q(p,k,l,m)),(zj)=[ y, xj]=(0,0)Nfor1jkandj=1. 3) For1i,jk,(i,j)=[ xi, xj]=(0,0)N. 4) Ifm
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Proof. SincescommuteswitheveryelementinC=C(p,k,l,m)andspl)]TJ /F8 5.978 Tf 5.75 0 Td[(1=r1,wehavethathr1,siZ(C).Since[r2,r1]=r1and[r2,r1]=1,wehavethathr1,r2,siZ2(C).Proceedingsimilarly,since[ri,r1]=ri)]TJ /F5 7.97 Tf 6.59 0 Td[(1for3ikand[ri,rj]=1for1i,jk,wehavehr1,...,ri,siZi(C)forik.Sincehr1,...,rk)]TJ /F5 7.97 Tf 6.59 0 Td[(1,si2Zk)]TJ /F5 7.97 Tf 6.59 0 Td[(1(C)and[ri,r1]=ri)]TJ /F5 7.97 Tf 6.58 0 Td[(12Zk(C)forik,wehaver12Zk(C),henceC=hr1,...,rk,s,r1iZk(C).WenowhavethatCisnilpotent.LetB=hr1,...,rk,siandnotethatBisabelian.Bydenition,r1sr)]TJ /F5 7.97 Tf 6.58 0 Td[(11=sandr1rjr)]TJ /F5 7.97 Tf 6.58 0 Td[(11=rj)]TJ /F5 7.97 Tf 6.58 0 Td[(1rj,hencer1Br)]TJ /F5 7.97 Tf 6.58 0 Td[(11B.SinceBisabelianandCisgeneratedbyfr1,...,rk,s,r1g,wehavethatzBz)]TJ /F5 7.97 Tf 6.59 0 Td[(1Bforeveryz2C.HenceBisnormalinC.Regardlessofm,rp12Z(C)B,hencejC:Bj=psinceC=hB,r1i.WenowhavethatBisanormalabeliansubgroupofCofindexp.ThereforeC2Gp.ByLemma 4.1.4 ,wehavemutuallyannihilatingnilpotentA-endomorphismsr1andr1.Bydenition,wemayviewrk)]TJ /F7 7.97 Tf 6.59 0 Td[(i=ir1(rk)for0i
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hencejBjpk+l)]TJ /F5 7.97 Tf 6.59 0 Td[(1.SincejC:Bj=p,wehavejCjpk+l.SinceQ(p,k,l,m)hasorderpk+l(Lemma 4.4.12 )andQ(p,k,l,m)isahomomorphicimageofC(Theorem 4.4.10 ),itfollowsthatCisisomorphictoQ(p,k,l,m). Denition4.4.14. Foraprimep,anintegerk2,andaninteger0mp,denemp,k=8>>>><>>>>:0m=0. m(Fp)k)]TJ /F5 7.97 Tf 6.59 0 Td[(11mp)]TJ /F4 11.955 Tf 11.96 0 Td[(1.1m=p. Theorem4.4.15(ClassicationTheorem). Foranyquadruple(p,k,l,m)2 C(Denition 4.4.1 ),thegroupsQ(p,k,l,m)(Denition 4.4.9 )andC(p,k,l,m)(Denition 4.4.11 )areisomorphicConlongroupsaffordingtheConlontriple(k,l,mp,k)(Denition 4.3.14 ).Furthermore,ConlongroupsPandQareisomorphicifandonlyiftheConlontriplesforPandQareequal. Proof. Let(p,k,l,m)2 C.ThegroupQ(p,k,l,m)isaConlongroupisomorphictoC(p,k,l,m)(Theorem 4.4.13 )andaffordingtheConlontriple(k,l,mp,k)(Theorem 4.4.10 ).LetR2CpaffordtheConlontriple(k,l,mp,k).LetD=fx1,...,xk,y,x1gbeaConlonsetoftypemp,kforR.DisageneratingsetforRthatsatisestherelationsofC(p,k,l,m).BythevonDyckTheorem 2.1.2 ,thereexistsasurjectivehomomorphism:C(p,k,l,m)!R.ByTheorem 4.4.13 and 4.4.10 ,jC(p,k,l,m)j=pk+landbyTheorem 4.2.8 jPj=pk+l,henceisanisomorphism.AssumePandQareConlongroupsaffordingtheConlontriple(k,l,m). 1) Ifm=0,letn=0. 2) Ifm=1,letn=p. 3) Ifm2Fp=(Fp)k)]TJ /F5 7.97 Tf 6.59 0 Td[(1,letnbesuchthat1np)]TJ /F4 11.955 Tf 11.95 0 Td[(1wherem= n(Fp)k)]TJ /F5 7.97 Tf 6.59 0 Td[(1.Noticethatnp,k=m.Fromabove,bothPandQareisomorphictoC(p,k,l,n),thereforeP=Q. 84

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AssumeP=Q.ItfollowsthatPandQhavethesamenilpotenceclasskandjZ(P)j=jZ(Q)j.ByTheorem 4.3.9 ,P(resp.Q)hasauniqueConlontypem2Mk(resp.m02Mk).SincethereexistsanabelianmaximalsubgroupAofPofConlontypem,thereexistsanabelianmaximalsubgroupBofQofConlontypem0.SinceQhasauniqueConlontype,wehavem=m0.ItfollowsthatbothPandQaffordtheConlontriple(k,l,m). 4.5FurtherResultsUsingtheclassication,furtherresultscanbedeterminedregardingthestructureofConlongroups.Notably, 1) TheorderofeveryelementinaConlonset,henceeveryelementinaConlongroup,canbedetermined. 2) TheFrattinisubgroupofaConlongroup,thecentersofaConlongroup,andanabelianmaximalsubgroupofaConlongroupcanbedescribedbyaConlonset. 3) TheDerivedsubgroupofaConlongroupcanbedescribedbyaConlonset. 4) Theconjugacyclassescanbeenumerated. Lemma4.5.1. LetP2CpaffordtheConlontriple(k,l,m)andletAbeanabelianmaximalsubgroupofP.Letfx1,...,xk,y,x1gbeaA-ConlonsetoftypemforP. 1) For1i
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)xpi=p)]TJ /F5 7.97 Tf 6.59 0 Td[(1Yj=1x)]TJ /F4 11.955 Tf 6.59 -1.16 Td[((pj+1)i)]TJ /F7 7.97 Tf 6.59 0 Td[(j.(4)ByLemma 4.2.12 ,xk2AnZk)]TJ /F5 7.97 Tf 6.58 0 Td[(1(P)andbyLemma 4.2.10 ,x121(Z(P)).Sincel1 p)]TJ /F5 7.97 Tf 6.59 0 Td[(1m=1foranyprimep,wehavethattheorderofx1=p1where1=l1 p)]TJ /F5 7.97 Tf 6.58 0 Td[(1m.Assumexjhasorderrj=lj p)]TJ /F5 7.97 Tf 6.59 0 Td[(1mforeveryj1then(Zi(P))=hypi. 86

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3) Ifi1and(Zi(P))=f1gifl=1.ByLemma 4.5.1 ,xpp=x)]TJ /F5 7.97 Tf 6.58 0 Td[(11.Itfollowsthat(Zp(P))=hx1,(Zp)]TJ /F5 7.97 Tf 6.59 0 Td[(1(P))i.Assume(Zj(P))=hxi)]TJ /F7 7.97 Tf 6.58 0 Td[(p+1,(Zj)]TJ /F5 7.97 Tf 6.58 0 Td[(1(P))iforallpj
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issuchthatap=xi)]TJ /F7 7.97 Tf 6.58 0 Td[(p+1.Thereforehxi)]TJ /F7 7.97 Tf 6.58 0 Td[(p+1,(Zi)]TJ /F5 7.97 Tf 6.59 0 Td[(1(P))i(Zi(P)).ElementsofZi(P)nZi)]TJ /F5 7.97 Tf 6.59 0 Td[(1(P)areoftheformxaizbwherep-aandz2Zi)]TJ /F5 7.97 Tf 6.59 0 Td[(1(P),notice(xaizb)p=zpbx)]TJ /F7 7.97 Tf 6.59 0 Td[(ai)]TJ /F7 7.97 Tf 6.59 0 Td[(p+1p)]TJ /F5 7.97 Tf 6.59 0 Td[(2Yj=1x)]TJ /F7 7.97 Tf 6.58 0 Td[(a(pj+1)i)]TJ /F7 7.97 Tf 6.59 0 Td[(j.Sincepj)]TJ /F7 7.97 Tf 5.48 -4.37 Td[(pjfor2jp)]TJ /F4 11.955 Tf 11.96 0 Td[(1wehavep)]TJ /F5 7.97 Tf 6.59 0 Td[(2Yj=1x)]TJ /F7 7.97 Tf 6.59 0 Td[(a(pj+1)i)]TJ /F7 7.97 Tf 6.59 0 Td[(j2(Zi)]TJ /F5 7.97 Tf 6.59 0 Td[(1).Wenowhavethat(xaizb)p2hxi)]TJ /F7 7.97 Tf 6.59 0 Td[(p+1,(Zi)]TJ /F5 7.97 Tf 6.58 0 Td[(1(P))i,anditfollowsthat(Zi(P))hxi)]TJ /F7 7.97 Tf 6.58 0 Td[(p+1,(Zi)]TJ /F5 7.97 Tf 6.59 0 Td[(1(P))i.Wenowhave(Zi(P))=hxi)]TJ /F7 7.97 Tf 6.58 0 Td[(p+1,(Zi)]TJ /F5 7.97 Tf 6.59 0 Td[(1(P))iforpi
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Proof. SincePaffords(k,l,m),jZ(P)j=plandPhasnilpotenceclassk.Ifz2Z(P),thenza=zforalla2P.Therefore,thereareplconjugacyclassesinZ(P),eachwith1element.ByLemma 4.2.9 ,jZi(P)j=pl+i)]TJ /F5 7.97 Tf 6.59 0 Td[(1for1ik)]TJ /F4 11.955 Tf 12.09 0 Td[(1andZi(P)AforsomeabelianmaximalsubgroupAofP.Moreover,jA:Zk)]TJ /F5 7.97 Tf 6.59 0 Td[(1(P)j=pandjAj=pk+l)]TJ /F5 7.97 Tf 6.59 0 Td[(1.Forx2PnA,ax=aifandonlyifa2Z(P).Fora2A)]TJ /F3 11.955 Tf 12.57 0 Td[(Z(P)wehaveA=StabP(a),henceaiscontainedinaconjugacyclassofsizejPj)-248(jStabP(a)j=p.SinceconjugationinducesagroupautomorphismandZi(P)ischaracteristic,theconjugacyclasscontainingz2Zi(P)for2ik)]TJ /F4 11.955 Tf 11.97 0 Td[(1iscontainedinZi(P))]TJ /F3 11.955 Tf 11.97 0 Td[(Zi)]TJ /F5 7.97 Tf 6.59 0 Td[(1(P).Sincetherearepl+i)]TJ /F5 7.97 Tf 6.59 0 Td[(2(p)]TJ /F4 11.955 Tf 11.96 0 Td[(1)elementsinZi(P))]TJ /F3 11.955 Tf 10.26 0 Td[(Zi)]TJ /F5 7.97 Tf 6.59 0 Td[(1(P),therearepl+i)]TJ /F5 7.97 Tf 6.59 0 Td[(3(p)]TJ /F4 11.955 Tf 10.26 0 Td[(1)conjugacyclassesinZi(P))]TJ /F3 11.955 Tf 10.26 0 Td[(Zi)]TJ /F5 7.97 Tf 6.59 0 Td[(1(P),eachwithpelements.Similarly,sinceAisnormal,itcontainstheconjugacyclassofanya2A.Therefore,therearepl+k)]TJ /F5 7.97 Tf 6.59 0 Td[(1(p)]TJ /F4 11.955 Tf 12.06 0 Td[(1)conjugacyclassesinA)]TJ /F3 11.955 Tf 12.06 0 Td[(Zk)]TJ /F5 7.97 Tf 6.59 0 Td[(1(P),eachwithpelements.Forx2PnA,StabP(x)=hZ(P),xi,hencejStabP(x)j=pl+1.Therefore,xiscontainedinaconjugacyclasswithpk)]TJ /F5 7.97 Tf 6.59 0 Td[(1.Bytheaboveparagraph,conjugacyclassescontainingelementsofPnAarecontainedinPnA.Sincetherearepk+l)]TJ /F5 7.97 Tf 6.59 0 Td[(1(p)]TJ /F4 11.955 Tf -442.17 -23.91 Td[(1)elementsofPnA,therearepl(p)]TJ /F4 11.955 Tf 12.98 0 Td[(1)conjugacyclassesinPnA,eachwithpk)]TJ /F5 7.97 Tf 6.58 0 Td[(1elements. 89

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REFERENCES [1] S.R.Blackburn,Groupsofprimepowerorderwithderivedsubgroupofprimeorder,J.Algebra,219(1999),no.2,pp.625657. [2] G.Szekeres,Determinationofacertainfamilyofnitemetabeliangroups,Trans.Amer.Math.Soc.,66,(1949),pp.143. [3] S.B.Conlon,Finitelinearp-groupsofdegreepandtheworkofG.Szekeres,ProceedingsoftheInternationalConferenceonRepresentationsofAlgebras,CarletonUniv.,Ottawa,Ont.,(1974),PaperNo.8,8pp.CarletonMath. [4] S.B.Conlon,P-groupswithanabelianmaximalsubgroupandcycliccenter,J.Austral.Math.Soc.Ser.A,22(1976),no.2,pp.221233. [5] H.U.Besche,B.Eick,andE.A.O'Brien,Amillenniumproject:constructingsmallgroups,Internat.J.AlgebraComput.,12(2002),no.5,pp.623644. [6] C.Sims,Enumeratingp-groups,Proc.LondonMath.Soc.(3),15(1965),pp.151166. [7] D.S.DummitandR.M.Foote,Abstractalgebra.Thirdedition,JohnWileyandSons,Inc.,2004,pp.xii+932. [8] TheGAPGroup,GAPGroups,Algorithms,andProgramming,Version4.4.12,2008, http://www.gap-system.org [9] D.Gorenstein,R.Lyons,andR.Solomon,TheClassicationoftheFiniteSimpleGroups,MathematicalSurveysandMonographs,40,(1994),Providence,R.I.:AmericanMathematicalSociety,ISBN978-0-8218-0334-9. [10] W.Burnside,Theoryofgroupsofniteorder,CambridgeUniversityPress,(1897). [11] A.Mann,SomeQuestionsAboutp-Groups,J.Austral.Math.Soc.(SeriesA),67,(1999),pp.356-37. [12] W.FeitandJ.G.Thompson,SolvabilityofGroupsofOddOrder,PacicJournalofMathematics13,(1963),pp.775-1029. [13] G.Higman,Enumeratingp-groups.I:Inequalities,Proc.LondonMath.Soc.(3),10,(1960),pp.23-30. [14] M.AschbacherandS.D.Smith,TheClassicationofQuasithinGroups.I.StructureofstronglyquasithinK-groups,MathematicalSurveysandMonographs,111,AmericanMathematicalSocietyProvidence,RI(2004),pp.xiv+477,ISBN:0-8218-3410-X. [15] M.Bona,AWalkThroughComibinatorics:AnIntroductiontoEnumerationandGraphTheory,WorldScienticPublishingCompany,(2002),pp.xviii+406,ISBN:981-02-4900-4. 90

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BIOGRAPHICALSKETCH JosephBrennanwasborninTampa,FloridaandisanalumniofTampaCatholicHighSchool.JosephhascompletedhisBachelorofScienceandMasterofScienceinmathematicsattheUniversityofFlorida.AresidentofGainesvilleandastudentattheUniversityofFloridasince2003,hehasbeenamemberoftheproductionsstaffattheReitzUnion,tutoredstellarstudentathletesattheUniversityAthleticAssociation,educatedstudentsattheFlagshipUniversityoftheFloridaHigherEducationSystem,andhehasgreatlyenjoyednineexcitingfootballseasons.Anavidoutdoorsman,hehasenjoyedthebestcyclingenvironmentinthesoutheastandtraveledtoTurkeyandtheAppalachianTrailforextendedhikingexcursions. 91