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# Maximal Subgroups of Finite Groups

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Title:
Maximal Subgroups of Finite Groups
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1 online resource (57 p.)
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english
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Lauderdale, Lindsey K
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University of Florida
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## Thesis/Dissertation Information

Degree:
Doctorate ( Ph.D.)
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University of Florida
Degree Disciplines:
Mathematics
Committee Chair:
TURULL,ALEXANDRE
Committee Co-Chair:
KEATING,KEVIN P
Committee Members:
SIN,PETER K
ROBINSON,PAUL L
MAZYCK,DAVID W

## Subjects

Subjects / Keywords:
group -- maximal
Mathematics -- Dissertations, Academic -- UF
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Mathematics thesis, Ph.D.
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## Notes

Abstract:
Finite group theory is a topic that has held the attention of mathematicians for over a hundred years. This is in part due to its applications throughout multiple branches of science including biology, chemistry and physics. Investigating finite groups is a classical problem, with many open questions remaining. A major component to understanding finite groups is comprehending their structure. One way to investigate the structure of finite groups is to study their maximal subgroups. In this dissertation, we study the number of maximal subgroups is a finite group $G$, denoted $|m(G)|$. When $G$ is a cyclic group, an elementary calculation proves that $|m(G)|=|\pi(G)|$, where $\pi(G)$ denote the set of primes which divide $|G|$. We further prove that if $G=P_1 \times P_2 \times \cdots \times P_n$, then $|m(G)|=|m(P_1)|+|m(P_2)|+\cdots+|m(P_n)|,$ where $\pi(G) = \{p_1, p_2, \ldots, p_n\}$ and $P_i \in \syl_{p_i}(G)$ for each $i \in \{1,2, \ldots, n\}$. We proceed by turning our attention to bounding the number of maximal subgroups in an arbitrary finite group. First, we consider the upper bound for the number of maximal subgroups in a finite group. Upper bounds on the number of maximal subgroups have gradually been sharpened by other mathematicians through making further and further assumptions on the group. We make more assumptions on the structure of a finite group and then improve the existing upper bound for the number of the maximal subgroups in a finite solvable group. To narrow the possible number of maximal subgroups in a finite group, we must consider the lower bound. We prove multiple lower bounds for the number of maximal subgroups in an arbitrary noncyclic finite group. In general, for a noncyclic group $G$, $|m(G)| \geq |\pi(G)|+p$ where $p \in \pi(G)$ is the smallest prime that divides $|G|$. If $G$ has a noncyclic Sylow subgroup and $q \in \pi(G)$ is the smallest prime such that $Q \in \syl_q(G)$ is noncyclic, then $|m(G)| \geq |\pi(G)|+q$. We conclude by producing two new lower bounds for $|m(G)|$, both of which consider all of the primes in $\pi(G)$.
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In the series University of Florida Digital Collections.
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Includes vita.
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Statement of Responsibility:
by Lindsey K Lauderdale.
Thesis:
Thesis (Ph.D.)--University of Florida, 2014.
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Full Text

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c 2014Lindsey-KayLauderdale

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ACKNOWLEDGMENTS IwouldliketothankmyadvisorDr.AlexandreTurull,whorstsparkedmyinterest ingrouptheory.Hisendlesspatienceandsupportwereinvaluabletothecompletionof thisdissertation.Iamalsogratefulforthesupport,questionsandsuggestionsofmy committeemembersDr.KevinKeating,Dr.DavidMazyck,Dr.PaulRobinsonandDr. PeterSin. 4

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TABLEOFCONTENTS page ACKNOWLEDGMENTS ................................. 4 LISTOFTABLES ..................................... 6 ABSTRACT ........................................ 7 CHAPTER 1DISSERTATIONOUTLINE ............................. 9 2ABRIEFBACKGROUNDOFFINITEGROUPTHEORY ........... 11 2.1Introduction ................................... 11 2.2MaximalSubgroups ............................... 14 2.3NotationandSomeDenitions ......................... 17 2.4OntheNumberofMaximalSubgroups .................... 18 3UPPERBOUNDSFORTHENUMBEROFMAXIMALSUBGROUPS .... 20 3.1Introduction ................................... 20 3.2UpperBoundsProvenbyOthers ....................... 21 3.3PreliminaryResults ............................... 22 3.4ProofofMainTheorem ............................. 23 4LOWERBOUNDSFORTHENUMBEROFMAXIMALSUBGROUPS .... 25 4.1Introduction ................................... 25 4.2PreliminaryResults ............................... 25 4.3ProofofMainTheorem ............................. 29 4.4ASpecialCase ................................. 35 5SOMEFURTHERLOWERBOUNDSFORTHENUMBEROFMAXIMAL SUBGROUPS ..................................... 38 5.1Introduction ................................... 38 5.2Denitions .................................... 40 5.3PreliminaryResults ............................... 41 5.4 | m ( G ) | for p -SolvableGroups ......................... 43 5.5 | m ( G ) | forNonsolvableGroups ........................ 46 5.6ProofofMainTheorems ............................ 51 5.7Remarkson ( G ) ................................ 52 5.8RemarksonPropertyB ............................ 54 REFERENCES ....................................... 56 BIOGRAPHICALSKETCH ................................ 57 5

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LISTOFTABLES Table page 5-2Valuesof ( G ). .................................... 54 6

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AbstractofDissertationPresentedtotheGraduateSchool oftheUniversityofFloridainPartialFulllmentofthe RequirementsfortheDegreeofDoctorofPhilosophy MAXIMALSUBGROUPSOFFINITEGROUPS By Lindsey-KayLauderdale May2014 Chair:AlexandreTurull Major:Mathematics Finitegrouptheoryisatopicthathasheldtheattentionofmathematiciansforover ahundredyears.Thisisinpartduetoitsapplicationsthroughoutmultiplebranchesof scienceincludingbiology,chemistryandphysics.Investigatingnitegroupsisaclassical problem,withmanyopenquestionsremaining.Amajorcomponenttounderstanding nitegroupsiscomprehendingtheirstructure.Onewaytoinvestigatethestructureof nitegroupsistostudytheirmaximalsubgroups. Inthisdissertation,westudythenumberofmaximalsubgroupsinanitegroup G ,denoted | m ( G ) | .When G isacyclicgroup,anelementarycalculationprovesthat | m ( G ) | = | ( G ) | ,where ( G )denotesthesetofprimeswhichdivide | G | .Wefurtherprove thatif G = P 1 P 2 " P n ,then | m ( G ) | = | m ( P 1 ) | + | m ( P 2 ) | + + | m ( P n ) | where ( G )= { p 1 ,p 2 ,...,p n } and P i # Syl p i ( G )foreach i # { 1 2 ,...,n } Weproceedbyturningourattentiontoboundingthenumberofmaximalsubgroups inanarbitrarynitegroup.First,weconsidertheupperboundforthenumberof maximalsubgroupsinanitegroup.Upperboundsforthenumberofmaximalsubgroups havegraduallybeensharpenedbyothermathematiciansthroughmakingfurtherand furtherassumptionsonthegroup.Wemakemoreassumptionsonthestructureofa nitegroupandthenimprovetheexistingupperboundforthenumberofthemaximal subgroupsinanitesolvablegroup. 7

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Tocontinuetonarrowtherangeofthepossiblenumberofmaximalsubgroupsina nitegroup,wemustconsiderthelowerbound.Weprovemultiplelowerboundsforthe numberofmaximalsubgroupsinanarbitrarynoncyclicnitegroup.Ingeneral,fora noncyclicgroup G | m ( G ) | $| ( G ) | + p where p # ( G )isthesmallestprimethatdivides | G | .If G hasanoncyclicSylowsubgroupand q # ( G )isthesmallestprimesuchthat Q # Syl q ( G )isnoncyclic,then | m ( G ) |$ | ( G ) | + q .Weconcludebyproducingtwonew lowerboundsfor | m ( G ) | ,bothofwhichconsideralloftheprimesin ( G ). 8

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CHAPTER1 DISSERTATIONOUTLINE Thisrstchaptergivesanoutlineoftheremainingchaptersinthisdissertation. ThemaincontentbeginsinChapter 2 withsomebasicdenitionsandanaturalwayof classifyinggroupsbytheirstructuralproperties.Weproceedwithsomeresultswhich provideafoundationforthestudyofmaximalsubgroupsofnitegroups,thefocusofthis dissertation.Afterabriefintroductiontomaximalsubgroups,wecalculatethenumber ofmaximalsubgroupsofsomebasicnitegroups.However,withminimalassumptions onagroupwecannotalwayscalculatetheexactnumberofmaximalsubgroups.Thus weattempttonarrowtherangeforthepossiblenumberofmaximalsubgroupsinanite groupbyconsideringbothupperandlowerbounds. Chapter 3 focusesontheupperboundsforthenumberofmaximalsubgroups.Here weonlyconsiderthenumberofmaximalsubgroupsinnitesolvablegroups.Several upperboundsforthenumberofmaximalsubgroupshavebeencalculatedbyother mathematicians,eachoneimprovingtheupperboundthatwasprovenbefore.Inthelast sectionofChapter 3 ,wemakesomefurtherassumptionsonastructureofanitesolvable grouptoimprovetheupperboundforthenumberofthemaximalsubgroups. Tocontinuetonarrowtherangeforthepossiblenumberofmaximalsubgroupsof agivennitegroup,wenextconsiderthelowerbound.BothChapter 4 andChapter 5 considerthelowerboundsforthenumberofmaximalsubgroupsinanitegroup.In2013, themaincontentofChapter 4 appearedinVolume101ofArchivderMathematikand wastitled,Lowerboundsonthenumberofmaximalsubgroupsinanitegroup'(see[ 6 ]). Thischapterfocusesonproducingalowerboundforthenumberofmaximalsubgroupsin anitegroupandthenproceedstoimprovetheboundbymakingminimalassumptionson thestructureofthegroup.Weconcludethischapterbyprovidinganexampleofagroup whichachievesthestatedlowerbounds. 9

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InChapter 5 ,weimprovethepreviouslystatedlowerboundsforthenumberof maximalsubgroupsbymakingfurtherassumptionsonthestructureofthegroup.In particular,thelowerboundisimprovedbypartitioningthesetofprimeswhichdividethe orderofthegroupintothreesets.Again,weconcludethechapterbyprovidingexamples ofgroupswhichachievethestatedlowerbounds.ThemaincontentofChapter 5 is currentlysubmitted. 10

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CHAPTER2 ABRIEFBACKGROUNDOFFINITEGROUPTHEORY 2.1Introduction Groupsaremathematicalobjectswhichmeasuresymmetry.Theyarestudied throughoutmanyareasofmathematics,aswellasmultiplebranchesofscience.Agroupis denedbelow: Denition2.1.1. Agroupisanorderedpair( G, % ),where G isasetand % isabinary operationon G satisfyingthefollowingaxioms: (i) ( a % b ) % c = a % ( b % c ),forall a,b,c # G (ii) thereexistsanelement e # G suchthatforall a # G wehave a % e = e % a = a (iii) foreach a # G thereisanelement a 1 of G suchthat a % a 1 = a 1 % a = e Thissimpledenitionofagroupcanyieldextremelydi!cultquestionswhicharise ingrouptheory.Inthisdissertation,westudynitegroupsonlyandpartofnitegroup theoryattemptstoclassifynitegroups.Oneofthesimplestwaysofclassifyingnite groupsisbytheirorder,orthenumberofelementsinthegroup.Alistofsmallgroups, classiedbyorder,canbefoundintheprogramGroups,Algorithms,andProgramming (GAP)[ 4 ].Amorenaturalwaytoclassifygroupsisbytheirstructuralproperties.These structuralpropertiesformahierarchy,asseenbelow: CyclicGroups & AbelianGroups & NilpotentGroups & SolvableGroups & AllGroups Thesimpleststructuralpropertyisseenincyclicgroups.Acyclicgroupisagroup whichisgeneratedbyasingleelement.Everynitecyclicgroupisisomorphicto Z n where n isapositiveinteger.Allcyclicgroupsareabeliangroups,thatisforall a,b # G wehave a % b = b % a .Abeliangroupsarecompletelyclassiedandtheclassicationcanbe seeninthefollowingtheorem: Theorem2.1.2. Let G beaniteabeliangroup.Then G = Z n 1 Z n 2 " Z n m 11

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forsomeintegers n 1 ,n 2 ,...,n m satisfyingthefollowingconditions: (1) (i) n i $2 forall i # { 1 2 ,...,m } (ii) n j +1 | n j for j # { 1 2 ,...,m ( 1 } (2) Theexpressionin(1)isunique,uptoisomorphism. Proof. Seeforexample[ 3 ,Theorem3.22]. Thenextcategoryofgroupsisnilpotentgroups.Nilpotentgroupsenjoythefollowing equivalentproperties: Theorem2.1.3. Let G beanitegroup.Suppose p 1 ,p 2 ,...,p n arethedistinctprimes dividingtheorderof G andlet P i beaSylow p i -subgroupof G for i # { 1 2 ,...,n } .Then thefollowingareequivalent: (1) G isnilpotent (2) G = P 1 P 2 " P n (3) Forall i # { 1 2 ,...,n } P i G (4) Everymaximalsubgroupisnormal (5) Foreverypropersubgroup H of G N G ( H ) >H Proof. Seeforexample[ 5 ,Theorem1.26]. Thepenultimatecategoryofgroupsissolvablegroups.Asolvablegroupisagroup withacompositionseries,allofwhosecompositionfactorsarecyclicofprimeorder.The Jordan-HolderTheoremprovidesinformationaboutthecompositionfactorsofanontrivial nitegroup. Theorem2.1.4. (Jordan-HolderTheorem) Let G beanontrivialnitegroup.Then (1) G hasacompositionseries (2) Thecompositionfactorsinacompositionseriesareunique,uptoisomorphism. Proof. Seeforexample[ 3 ,Theorem3.22]. 12 PAGE 13 Thecompositionseriesofagroupassistsinbreakingdownthegroupintosmaller pieces.Therearemanyresultswhichdictatesituationswhenagivengrouphascomposition factorswhicharecyclicofprimeorder.Forexample,in1911,W.Burnsideprovedthe followingresult: Theorem2.1.5. (BurnsideTheorem) Forprimes p and q ,everygroupoforder p a q b is solvable. Proof. Seeforexample[ 5 ,Theorem7.8],aproofusingnitegrouptheory. ItwasW.Burnsidewhoconjecturedthateverynonabeliansimplegroupwasofeven order.Thisfactwasprovenin1963byW.FeitandJ.Thompson. Theorem2.1.6. (Feit-ThompsonTheorem) Let G beagroup.If G hasoddorder,then G issolvable. Proof. See[ 13 ]. Theremainingcategoryofgroupsincludesallgroups,orgroupsthatmayormay nothaveacompositionserieswithonlycycliccompositionfactors.Thecomposition factorsofeverynitegrouparesimplegroups.Sinceagroupcanbeconstructedfrom itscompositionsfactors,theJordan-HolderTheoremillustratestheimportanceof understandingnitesimplegroups.Thustheclassicationofnitesimplegroupswas amajorachievement.Theclassicationtookover50yearsandmanymathematiciansto complete. Theorem2.1.7. (ClassicationofFiniteSimpleGroups,2004) Let G beanitesimple group.Then G isisomorphictooneofthefollowinggroups: (1) Acyclicgroupofprimeorder (2) Analternatinggroupwithdegreeatleast5 (3) AsimplegroupofLietype (4) Oneofthe26sporadicsimplegroups. 13 PAGE 14 Proof. See[ 12 ]. Becausethenitesimplegroupsareknown,wecanstudyarbitrarygroupsthrough theircompositionfactors.Itisnaturaltonowfocusonstudyingthemaximalsubgroupsof nitegroups. 2.2MaximalSubgroups Onewaytoinvestigatethestructureofanitegroupistostudyitsmaximal subgroups.Sincebeforetheclassicationofnitesimplegroups,researchershavebeen studyingthemaximalsubgroupsofallnitegroups.Thedenitionofmaximalsubgroups isseenbelow: Denition2.2.1. Asubgroup M ofagroup G iscalledamaximalsubgroupif M ) = G andtheonlysubgroupscontaining M are M and G Innontrivialnitegroups,maximalsubgroupswillalwaysexistbecausethesubgroups formapartiallyorderedsetunderinclusion.Sincethesetofsubgroupsisnite,this partiallyorderedsetwillhaveamaximalelement.However,notallgroupswillhave maximalsubgroups.Forexample,therationalnumbersunderadditionhavenomaximal subgroups. Wewillproceedbystatingsomewell-knownresultsaboutmaximalsubgroupsofnite groups.Wemayusesomeoftheseresultsinthelaterchapterswithoutexplicitlystating so. Lemma2.2.2. Let H beapropersubgroupofanitegroup G .Thenthereisamaximal subgroupof G containing H Proof. Seeforexample[ 3 ,Exercise2.4.16]. Inadditiontoeverypropersubgroupbeingcontainedinamaximalsubgroup, maximalsubgroupsofcertainnitegroupsenjoysomeveryniceproperties.Afewofthese propertiescanbeseeninthelemmasbelow: 14 PAGE 15 Lemma2.2.3. Let G beanitegroup.If M isamaximalsubgroupof G ,then M is normalin G or M isself-normalizing. Proof. Seeforexample[ 3 ,Exercise4.3.23]. Lemma2.2.4. Everymaximalsubgroupofanitenilpotentgroupisnormalofprime index. Proof. Seeforexample[ 3 ,Theorem3.22]and[ 5 ,Theorem1.20]. Lemma2.2.5. Everymaximalsubgroupofanitesolvablegrouphasprimepowerindex. Proof. Seeforexample[ 5 ,Exercise3B.1]. Animportantsubgroupofanitegroupinvolvingitsmaximalsubgroupsiscalledthe Frattinisubgroup. Denition2.2.6. Let G beanitegroup.TheFrattinisubgroupof G ,denoted"( G ),is theintersectionofthemaximalsubgroupsof G Next,westatesomepropertiesofFrattinisubgroupsofnitegroups. Lemma2.2.7. TheFrattinisubgroupofanitegroupisnilpotent. Proof. Seeforexample[ 3 ,Exercise6.1.25]. Lemma2.2.8. Let G beanitegroup.Then ( G ) isacharacteristicsubgroupof G Proof. Seeforexample[ 3 ,Exercise6.1.21]. Lemma2.2.9. Let p beaprimenumber.If P isa p -group,then P/ ( P ) iselementary abelian. Proof. Seeforexample[ 5 ,Exercise1D.8]. Understandingthemaximalsubgroupsofanitegroupcanprovideawealthof informationintothegroupitself.Forexample,undercertainconditions,knowingonlythe numberofmaximalsubgroupsinagroupisenoughtoidentifythegroup.Themaximal subgroupsofnitesimplegroupshavebeenresearchedandtheyareunderstood.The 15 PAGE 16 maximalsubgroupsofsmallnitesimplegroupsarelistedinthe AtlasofFiniteGroups [ 16 ].Moregeneralsituationshavealsobeenconsidered.In[ 1 ],M.Aschbacherclassiesthe maximalsubgroupsofclassicalgroups.Heprovesthateverymaximalsubgroupofclassical groupsbelongstooneofeightclassesofsubgroupsorhasageneralizedFittingsubgroup whichisanonabeliansimplegroup.In[ 7 ],P.KleidmanandM.Liebeckstudytheeight classesintroducedbyAschbacher.Westatetheirresultsbelow. Set G = PGL ( n,q )andlet V betheunderlying n -dimensionalvectorspaceover GF ( q ),aniteeldwith q elements. C 1 : Stabilizersofsubspacesof V C 2 : Stabilizersofdirectsumdecompositionsof V intosubspacesofthesamedimension C 3 : Stabilizersofextensioneldsof GF ( q )ofprimedegree C 4 : Stabilizersoftensorproductdecompositions V = V 1 V 2 C 5 : Lineargroupsoversubeldsofprimeindex C 6 : Normalizersof r -groupsofsymplectictype,where r isaprimedi#erentfrom p C 7 : Stabilizersoftensorproductdecompositions V = t i =1 V i ,where V i allhavethesame dimension C 8 : Classicalgroups Theorem2.2.10. (Aschbacher'sTheorem) Let H beasubgroupof PGL ( n,q ) not containing PSL ( n,q ) .Thenoneofthefollowingholds: (1) H iscontainedinamemberofoneoftheclasses C 1 ,C 2 ,...,C 8 (2) H isalmostsimpleandisinducedbyanabsolutelyirreduciblesubgroupmodulo scalars. Proof. See[ 1 ]. Itturnsoutthatmoreinformationabout H canbestatedifitisnotamemberofone oftheclasses C 1 ,C 2 ,...,C 8 16 PAGE 17 Theorem2.2.11. Let H fallundercase (2) inAschbacher'sTheorem(Theorem 2.2.10 ). Theneither H is S m or A m ,with m = n +1 or m = n +2 ,or H hasorderatmost q 3 n Proof. See[ 7 ]. TheO'Nan-ScottTheoremgivesaclassicationofthemaximalsubgroupsofthe symmetricgroup. Theorem2.2.12. (O'Nan-ScottTheorem) If H isanypropersubgroupof S n (except A n ), then H isasubgroupofoneormoreofthefollowingsubgroups: (1) Anintransitivegroup S k S m ,where n = k + m (2) Animprimitivegroup S k + S m ,where n = km (3) Aprimitivewreathproduct, S k + S m ,where n = km (4) Ana!negroup AGL d ( p ) = p d : GL d ( p ) ,where n = p d (5) Agroupofshape T m (Out( T ) S m ) ,where T isanon-abeliansimplegroup,acting onthecosetsofthediagonalsubgroup Aut( T ) S m ,where n = | T | m 1 (6) Analmostsimplegroupactingonthecosetsofamaximalsubgroup. Proof. See[ 11 ]. Becausethemaximalsubgroupsofanitegroupcontainsuchavastamountof information,therearecurrentlyopenquestionsregardingthemaximalsubgroupsofan arbitrarynitegroup. 2.3NotationandSomeDenitions Oneopenquestionregardingthemaximalsubgroupsofanitegroupisthefollowing: Withlimitedassumptionsonthestructureofanitegroup G ,whatcanwesayaboutthe numberofmaximalsubgroupsin G ?Toaidinansweringthisquestion,wewillintroduce thefollowingdenitionsandnotation. Denition2.3.1. Let H beasubgroupofthenitegroup G .Asubgroup K of G is calledacomplementfor H in G if G = HK and H K =1. 17 PAGE 18 Denition2.3.2. AHallsubgroupofanitegroupisasubgroupwhoseorderiscoprime toitsindex. Denition2.3.3. Let beanysetofprimenumbers.Anitegroup G issaidtobe -separable,ifthereexistsanormalserieswhereeachfactorisa -groupora " -group. Throughoutthisdissertation,weusethefollowingnotation.Below,weassumethat G isanitegroupand p isaprimenumber. 1 ( G )= x p : x # G Aut( G )denotestheautomorphismgroupof G C G ( H )denotesthecentralizerofasubgroup H in G [ K,L ]= k 1 l 1 kl : k # K,l # L ,where K and L aresubgroupsof G m ( G )denotesthesetofmaximalsubgroupsof G m p ( G )denotesthesetofmaximalsubgroupsof G whoseindexin G isapowerof p N G ( H )denotesthenormalizerofasubgroup H in G n p ( G )denotesthenumberofSylow p -subgroupsof G ( G )denotestheFrattinisubgroup ( G )denotesthesetofprimeswhichdividetheorderof G Syl p ( G )denotesSylow p -subgroupsof G$ 1 ( G )= x # G : x p =1 Out( G )denotestheouterautomorphismgroupof G O p ( G )denotesthesmallestnormalsubgroupof G whoseindexisapowerof p O p ( G )denotesthelargestnormal p -subgroupof G Z ( G )denotesthecenterof G 2.4OntheNumberofMaximalSubgroups Forsomegroups G ,wecanmakeminimalassumptionsonitsstructureandstill calculatetheexactnumberofmaximalsubgroups.Wediscusssomesuchsituationsbelow. 18

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When G isanitecyclicgroupitiseasytoseethat | m ( G ) | = | ( G ) | .If p isaprime numberand G isanoncyclic p -subgroup,then | m ( G ) | canbecalculatedtoo.Forsucha G ,if G/ ( G )isanelementaryabeliangroupofrank k where k $2,then | m ( G ) | = p k 1 p 1 Becausethenumberofmaximalsubgroupscanbecalculatedforall p -groups,thenumber ofmaximalsubgroupscanalsobecalculatedforallnilpotentgroups.Weshallseethe numberofmaximalsubgroupsofanilpotentgroupisthesumofthenumberofmaximal subgroupsofeachofitsSylowsubgroups. Itisnaturaltonextconsiderthenumberofmaximalsubgroupsforbothsolvable andnonsolvablegroups.However,withtheseassumptionswecannotexpecttocalculate theexactnumberofmaximalsubgroupsforanarbitrarygroup.However,ifgivenanite group,itispossibletocalculatethenumberofmaximalsubgroups.ResourceslikeGAP [ 4 ]andthe AtlasofFiniteGroups [ 16 ]canbeusedtoaidinsuchatask.Weseekto ndthenumberofmaximalsubgroupsofageneralnitegroup,whilemakingminimal structuralassumptions.Thusweattempttonarrowtherangeforthepossiblenumberof maximalsubgroupsofanarbitrarynitegroup.Onewaytoapproachthisistoconsider bothupperandlowerboundsforthenumberofmaximalsubgroupsinanitegroup. 19 PAGE 20 CHAPTER3 UPPERBOUNDSFORTHENUMBEROFMAXIMALSUBGROUPS 3.1Introduction Therehavebeenseveralmathematicianswhoestimatedthenumberofmaximal subgroupsinnitegroups.Inparticular,theirworkfocusesontheupperboundfor thenumberofmaximalsubgroupsinnitesolvablegroups.Theupperboundforthe numberofmaximalsubgroupshasgraduallybeensharpenedbymakingfurtherand furtherassumptionsonthegroup.Therstupperbound,provenbyG.E.Wall,only consideredtheorderofthegroup.TogetherR.J.Cook,J.WiegoldandA.G.Williamson alsoconsideredthesmallestprimethatdividestheorderofthegroup.LaterM.Herzog andO.Manzstudiedthesmallestprimeandlargestprimewhichdividetheorderofthe groupaswellastheorderofthegroup.Finally,B.Newtonusedalloftheprimesthat dividetheorderofthegroup,alongwiththeirmultiplicities,tostatehisupperboundfor thenumberofmaximalsubgroupsofnitegroup.Alloftheaforementionedresultsare explicitlystatedinSection 3.2 below. Inthischapter,wemakemoreassumptionsonthestructureofanitegroupto furtherimprovetheupperboundforthenumberofmaximalsubgroupsinanitesolvable group.Foranitegroup G ,recallwedenote O p ( G )asthesmallestnormalsubgroupof G whoseindexin G isapowerof p anddenote m p ( G )asthesetofmaximalsubgroupsof G whoseindexin G isapowerof p .Ourmainresultofthischapteristhefollowing: Theorem3.1.1. Let G beanitesolvablegroupandsuppose p # ( G ) .If P # Syl p ( G ) is cyclic,then | m p ( G ) | #{ 1 ,p } .Moreprecisely, | m p ( G ) | = #$ p, if O p ( G )= G 1 if O p ( G )
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3.2UpperBoundsProvenbyOthers Forsolvablegroups,duringthelastftyyearsestimationsontheupperboundforthe numberofmaximalsubgroupshavebeengraduallysharpened.Theearliestsuchestimate wasprovenbyG.E.Wall.Hecomputedanupperboundforthenumberofmaximal subgroupsinanitesolvablegroupin[ 14 ]. Theorem3.2.1. (Wall) Let G beanitesolvablegroup.Then | m ( G ) | < | G | Proof. See[ 14 ]. Almostthirtyyearslater,R.J.Cook,J.WiegoldandA.G.Williamsonimproved Wall'sresultin[ 15 ]. Theorem3.2.2. (Cook,Wiegold,Williamson) Let G beanitesolvablegroupand suppose p thesmallestprimein ( G ) .Then | m ( G ) | / | G | ( 1 p ( 1 Thisboundisachievedifandonlyif G iselementaryabelian. Proof. See[ 15 ]. Thelargestpossiblenumberofmaximalsubgroupsinasolvablegroupwasnarrowed furtherin[ 8 ]byM.HerzogandO.Manz. Theorem3.2.3. (Herzog-Manz) Let G beanitesolvablegroupwithFrattinisubgroup ( G ) .Wedenoteby q thelargestprimein ( G ) andby p thesmallestprimein ( G ) Then | m ( G ) | / q | G/ ( G ) | ( p p ( q ( 1) Proof. See[ 8 ]. ThemostaccurateresulttodatewasprovenbyB.Newtonin[ 10 ],assumingthatthe nitegroupissolvable.Foraprime p ,herstprovesaboundforthenumberofmaximal subgroupsinanitesolvablegroupwith p -powerindex. 21

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Theorem3.2.4. (Newton) Let p beaprimenumberandlet G beanitesolvablegroupof order p k m ,where p m .Supposethat [ G : O p ( G )]= p r .Then | m p ( G ) | / p r ( 1 p ( 1 + p k r +1 ( p p ( 1 Moreover, | m p ( G ) | isnevergreaterthan p k +1 ( p p ( 1 ,andif O p ( G )
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maximaland M
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Therefore G = O p ( G ) Q .Let R # Syl p ( G )with R 2 Q andassume R ) = Q .Then Q 0 R p sothat( Q O p ( G )) / O p ( G ) 0 ( G/ O p ( G )) p ,acontradiction.Thus R = Q ,contradicting ourchoiceof M L .Therefore L = 1 and | m p ( G ) | = |N| =1,asdesired. 24

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CHAPTER4 LOWERBOUNDSFORTHENUMBEROFMAXIMALSUBGROUPS 4.1Introduction Theaforementionedresultsregardingupperboundsforthenumberofmaximal subgroupsofnitesolvablegroupsarevaluableinboundingthenumberofmaximal subgroupsinanitegroup.Tocontinuetonarrowtherangeforthepossiblenumberof maximalsubgroupsofagivennitegroup,wenowconsiderthesmallestpossiblenumber ofmaximalsubgroups.Inthischapter,weproducethreelowerboundsforthenumberof maximalsubgroupsinanynitegroup. Fixapositiveinteger m .Itisnaturaltosuspectthatamongallgroupsoforder m | m ( G ) | issmallestwhen G isacyclicgroup.If G isthecyclicgroup,thenthenumberof maximalsubgroupsisequaltothenumberofdistinctprimeswhichdividetheorderofthe group.Weconrmbothfactsbelow(Corollary 4.1.2 ). Givenasetofprimenumbers # with | # | = n ,weinvestigatelowerboundsfor | m ( G ) | when G isanitenoncyclicgroupwith ( G )= # .Weobtainthefollowingmaintheorem: Theorem4.1.1. Let n beapositiveintegerandsupposethat G isanitenoncyclicgroup suchthat | ( G ) | = n .Assumethat p # ( G ) isthesmallestprimewhichdivides | G | .Then | m ( G ) | $n + p .Furthermore,if G hasanoncyclicSylowsubgroupand q # ( G ) isthe smallestprimesuchthat Q # Syl q ( G ) isnoncyclic,then | m ( G ) |$ n + q ThelowerboundsinTheorem 4.1.1 arebestpossibleandthisfactisconrmedin Proposition 4.3.5 below.Animmediateconsequenceofthistheoremisthefollowinglower boundonthenumberofmaximalsubgroupsinanarbitrarynitegroup.Notethatitalso yieldsanewcharacterizationofcyclicgroups. Corollary4.1.2. If G isanitegroup G ,then | m ( G ) | $| ( G ) | .Furthermore,equality holdsifandonlyif G iscyclic. 4.2PreliminaryResults Here,wecollectsomeresultsneededtoproveTheorem 4.1.1 25 PAGE 26 Denition4.2.1. Agroup G issupersolvableifthereexistsnormalsubgroups N i with 1= N 0 0 N 1 0 0 N r = G, whereeachfactor N i /N i 1 iscyclicfor1 / i / r Theorem4.2.2. IfalltheSylowsubgroupsofanitegroup G arecyclic,then G is supersolvable. Proof. By[ 5 ,Theorem5.16],both G and G/G arecyclic.Thus G issupersolvable. Theorem4.2.3. Everyminimalnormalsubgroup M ofanitegroupisadirectproduct ofisomorphicsimplegroups.Furthermore,if M issolvablethenitisanelementary abelian p -groupforsomeprime p Proof. Seeforexample[ 5 ,Exercise2A.5andLemma3.11]. Theorem4.2.4. Let G beanitegroup.If G isnotsolvableand P # Syl 2 ( G ) ,then P is noncyclic. Proof. Suppose P # Syl 2 ( G )iscyclic.Then G hasanormal2-complement N (seefor example[ 5 ,Corollary5.14]).Since N isofoddorder,itissolvablebytheFeit-Thompson Theorem.Andbecause G/N iscyclic, G isalsosolvable. Theorem4.2.5. (Schur-Zassenhaus) EverynormalHallsubgroupofanitegrouphasa complement. Proof. Seeforexample[ 5 ,Theorem3.8]. Theorem4.2.6. (Maschke) Let G beanitegroupandlet F beaeldwhosecharacteristicdoesnotdivide | G | .If V isany FG -moduleand U isanysubmoduleof V ,then V has asubmodule W suchthat V = U 3 W Proof. Seeforexample[ 3 ,Sec.18.1Theorem1]. ThefollowinglemmasinthissectionareusefulintheproofofTheorem 4.1.1 26 PAGE 27 Lemma4.2.7. Supposethat G isanitegroup,andlet ( G )= { p 1 ,p 2 ,...,p n } .Foreach i # { 1 2 ,...,n } ,let P i # Syl p i ( G ) .If G = P 1 P 2 " P n ,then | m ( G ) | = | m ( P 1 ) | + | m ( P 2 ) | + + | m ( P n ) | Proof. Notethatbecause G isnilpotent,allmaximalsubgroupsof G willhaveprimeindex in G .Foreach i # { 1 2 ,...,n } ,let m p i ( G )denotethesetofmaximalsubgroupsof G with index p i .Dene$ i : m p i ( G ) 4 m ( P i )by $i ( M )= P i M ,forall M # m p i ( G ).Since G = P i M [ P i : P i M ]=[ G : M i ]= p i Then P i M P i andso P i / ( P i M )isofprimeorder.Thus P i M # m ( P i ). Nowdene % i : m ( P i ) 4 m p i ( G )by % i ( N i )= P 1 " N i " P n forall N i # m ( P i ).Since N i isamaximalsubgroupofa p i -group, N i P i and[ P i : N i ]= p i .Thus % i ( N i ) G and[ G : % i ( N i )]= p i .So % i ( N i ) # m p i ( G ). Since$ i and % i areinversesofeachother, $i isabijectionand | m p i ( G ) | = | m ( P i ) | Observethat | m ( G ) | = | m p 1 ( G ) | + | m p 2 ( G ) | + + | m p n ( G ) | becauseeverymaximalsubgroupof G hasprimeindex.Thus | m ( G ) | = | m ( P 1 ) | + | m ( P 2 ) | + + | m ( P n ) | asdesired. Lemma4.2.8. Let N beanonabelianminimalnormalsubgroupofanitegroup G .Let p bethelargestprimein ( N ) .Thenthereareatleast p maximalsubgroupsof G whichdo notcontain N 27 PAGE 28 Proof. Let P beaSylow p -subgroupof N .Set N 0 = N G ( P ).Then G = NN 0 bythe FrattiniArgument.Because P & N and N isaminimalnormalsubgroupof G N 0 ) = G andthereexistsamaximalsubgroup M of G containing N 0 .Observethat N )0 M becauseotherwise G = NM 0 M .Therefore M isamaximalsubgroupof G which doesnotcontain N .Infact,since N isthedirectproductofisomorphicsimplegroupsby Theorem 4.2.3 ,thereexistsasimplenormalsubgroup S of N whichisnotcontainedin M Set R = S M and n =[ S : R ].Since S actsbyleftmultiplicationonthesetof leftcosetof R in S ,thereexistsanontrivialhomomorphismfrom S tothesymmetric groupon n elements.Then,because S isasimplegroup, S mustbeisomorphictoa subgroupofthealternatinggroupon n elements.Thus p / n .Since S ) = R ) =1, R isnotanormalsubgroupof S .Itfollowsthat M isnotanormalsubgroupof G .Then M isself-normalizingbecauseitisamaximalsubgroupof G .Thusthereareatleast p S -conjugatesof M ,yieldingatleast p conjugatesof M in G .Therefore G hasatleast p maximalsubgroupswhichdonotcontain N ,asdesired. Lemma4.2.9. Let n$ 3 beaninteger.Suppose S isasetof n distinctprimes.Assume that p and q arethelargestandsmallestelementsof S ,respectively.Then p $n + q Proof. Observethatallelementsof S areatleast2unitsapartfromeachother,exceptfor possiblythetwosmallestelementsof S whichare1unitapart.Therefore p ( q$ 2( n ( 2)+1 whichisatleast n as n $3.So p ( q$ n and p $n + q ,asdesired. Lemma4.2.10. Let G beanitegroupwithanoncyclicSylowsubgroup.Supposethat p # ( G ) isthesmallestprimesuchthat P # Syl p ( G ) isnoncyclic.If N isasolvable minimalnormalsubgroupof G andallSylowsubgroupsof G/N arecyclic,then N isa p -group. 28 PAGE 29 Proof. Supposethat N isa q -group,where q # ( G )and p ) = q .Then P N =1and P = PN/N / G/N ,bytheSecondIsomorphismTheorem.Since PN/N # Syl p ( G/N ), PN/N iscyclicandcannotbeisomorphictothenoncyclicgroup P ,acontradiction. Therefore, N isa p -group. Lemma4.2.11. Let N beasolvablenormalsubgroupofanitegroup G .Supposethat H isaHallsubgroupof N .Then G = N G ( H ) N Proof. Let g # G .Since H g 0 N g = N and H g hasthesameorderofaHallsubgroupof N H g isaHallsubgroupof N .Therefore H and H g are N -conjugate,say H gn = H for some n # N .Then gn # N G ( H )and g # N G ( H ) n 1 0 N G ( H ) N. Therefore G = N G ( H ) N ,asdesired. Lemma4.2.12. Let G beanitesolvablegroup.If H isaHallsubgroupof G and K isa subgroupwhichcontains N G ( H ) ,then K isself-normalizing. Proof. Let g # G satisfy K g = K .Then H g isaHallsubgroupof K .SinceallHall subgroupsof K areconjugate,thereexists n # K suchthat H gn = H .So gn # N G ( H ) and g # N G ( H ) n 1 0 K. Therefore K isself-normalizing. 4.3ProofofMainTheorem WiththegoalofprovingTheorem 4.1.1 ,werstinvestigate | m ( G ) | inacyclicgroup G Theorem4.3.1. Foranitecyclicgroup G | m ( G ) | = | ( G ) | 29 PAGE 30 Proof. Let G beanitecyclicgroup.Then G isisomorphicto Z p 1 1 Z p 2 2 " Z p n n where | G | = p 1 1 p 2 2 p n n .ByLemma 4.2.7 | m ( G ) | = | m ( Z p 1 1 ) | + | m ( Z p 2 2 ) | + + | m ( Z p n n ) | Foreach p i # ( G ), Z p i i hasauniquemaximalsubgroupisomorphicto Z p i 1 i andso | m ( Z p i i ) | =1.Therefore | m ( G ) | = n = | ( G ) | asdesired. Next,weproveTheorem 4.1.1 when G isa p -groupandtheninthecasewhereall Sylowsubgroupsof G arecyclic. Lemma4.3.2. Let p beaprimenumberandsupposethat P isanoncyclic p -subgroup. Let P/ ( P ) betheelementaryabeliangroupofrank k ,where k$ 2 and ( P ) denotesthe Frattinisubgroupof P .Then | m ( P ) | = p k ( 1 p ( 1 Inparticular, | m ( P ) | $p +1 Proof. Since P/ ( P )isanelementaryabeliangroupofrank k ,wehave ' m ( P/ ( P ) ) ' = p k ( 1 p ( 1 Because" ( P )iscontainedineverymaximalsubgroupof P | m ( P ) | = ' m ( P/ ( P ) ) ' = p k ( 1 p ( 1 Observethat p k 1 p 1 issmallestwhen k =2.Inthiscase, p k 1 p 1 = p +1,sothat | m ( P ) |$ p +1. Theorem4.3.3. Suppose G isanitenoncyclicgroupsuchthat | ( G ) | = n andlet p # ( G ) bethesmallestprimedividing | G | .IfallSylowsubgroupsof G arecyclic,then | m ( G ) | $n + p 30 PAGE 31 Proof. Supposethisisfalse.Let G beacounterexampleofminimumorder.Then | m ( G ) | p .ByTheorem 4.3.1 | m ( G/N ) | = n ( 1.These n ( 1maximalsubgroupsof G/N willcorrespondtoatleast n ( 1maximalsubgroupsof G whichcontain N .Thus | m ( G ) | >n ( 1+ p ,anal contradiction. Finally,weprovetheremainingcaseofTheorem 4.1.1 31 PAGE 32 Theorem4.3.4. Let G beanitenoncyclicgroupsuchthat | ( G ) | = n ,where n$ 2 Assumethat G hasanoncyclicSylowsubgroupand p # ( G ) isthesmallestprimesuch that P # Syl p ( G ) isnoncyclic.Then | m ( G ) | $n + p Proof. Supposethisisfalse.Let G beacounterexampleofminimumorder.Then | m ( G ) | p and | m ( G/N ) |$ n ( 1+ q $n + p. Ineithercase,weobtain n + p maximalsubgroupsof G/N ,whichwillcorrespondtoat least n + p maximalsubgroupsof G whichcontain N ,acontradiction.ThereforeallSylow 32 PAGE 33 subgroupsof G/N arecyclic.So G/N issupersolvableandthus G issolvable.Then G has aHall p -subgroup,say H Foreachprime q # ( G )with q ) = p ,let H q beaHall q -subgroupcontaining P Let M q beamaximalsubgroupwhichcontains H q ,yielding n ( 1distinctmaximal subgroupsof G .If H G ,then G/H isanoncyclic p -groupandLemma 4.3.2 implies | m ( G/H ) |$ p +1.These p +1maximalsubgroupsof G/H willcorrespondtoatleast p +1maximalsubgroupsof G thatcontain H ,andarethereforedistinctfromthe n ( 1 M q 's.Thus | m ( G ) | $( n ( 1)+( p +1)= n + p, acontradiction. Therefore H isnotnormalin G .Let M p beamaximalsubgroupof G whichcontains N G ( H ).Then M p isself-normalizingbyLemma 4.2.12 .Therefore M p hasatleast p conjugatesallofwhicharemaximal.Since | m ( G ) | PAGE 34 Supposethat H actsirreduciblyon P .Then N P ( H ) # { 1 } ,P + .Since H isnot normalin G N P ( H ) ) = P .If N P ( H )istrivial,then H ismaximaland [ G : H ]= | P |$ p 2 So H hasatleast p 2 conjugates,allofwhicharemaximalanddonotcontain P .Therefore | m ( G ) | $( n ( 1)+ p 2 >n + p, acontradiction.Thus H doesnotactirreduciblyon P .Then P hasanontrivialproper H -invariantsubgroup P 1 .ByMaschke'sTheorem,thereexistsanontrivialproper H -invariantsubgroup P 2 of P suchthat P = P 1 P 2 .Now H mustactnon-trivially onatleastoneof P 1 or P 2 .Withoutlossofgenerality,assume H actsnon-triviallyon P 1 Let M beamaximalsubgroupwhichcontains N G ( H ).Thenthe p conjugatesof M will notcontain P 1 .Thereforeif M isamaximalsubgroupwhichcontains P 1 H ,itisdistinct fromtheaforementioned n ( 1+ p maximalsubgroupsof G .Hence | m ( G ) |$ ( n ( 1+ p )+1= n + p, analcontradiction. Theorem 4.1.1 followsimmediatelyfromLemma 4.3.2 ,Theorem 4.3.3 andTheorem 4.3.4 WeconcludethissectionbyprovingthelowerboundinTheorem 4.1.1 isbestpossible. Proposition4.3.5. Let n beapositiveinteger.Givenanysetofdistinctprimes { p 1 ,p 2 ,...,p n } where p 1 isthesmallest,thereexistsanitegroup G with ( G )= { p 1 ,p 2 ,...,p n } and | m ( G ) | = n + p 1 Proof. Considerthenoncyclicgroup G = Z p 1 Z p 1 Z p 2 " Z p n 34

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ByLemma 4.3.2 | m ( Z p 1 Z p 1 ) | = p 1 +1.Also, | m ( Z p i ) | =1for i # { 2 ,...,n } by Lemma 4.3.1 .Thus | m ( G ) | =( n ( 1)+( p 1 +1)= n + p 1 byLemma 4.2.7 ,andthelowerboundinTheorem 4.1.1 isbestpossible. 4.4ASpecialCase WeseektoimprovethelowerboundinTheorem 4.1.1 .Thusweconsideranite group G suchthat | ( G ) | =2. Theorem4.4.1. Let G beanitenoncyclicgroupsuchthat ( G )= { p,q } ,where p
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alsononcyclic.If P isnoncyclic,then p 2 dividestheorderof G .Then, | m ( P ) | $p +1and | m ( Q ) |$ 1sothat | m ( G ) | $p +2,acontradiction.Therefore Q mustbenoncyclicand possibly p 2 | G | .Eitherway, | m ( Q ) |$ q +1and | m ( P ) | $1,forcing | m ( G ) |$ q +2,a contradiction. Therefore G isnotnilpotent.Then G hasamaximalsubgroup M whichisnotnormal byTheorem 2.1.3 .Thus M has[ G : N G ( M )]=[ G : M ]conjugatesin G .Byatheorem ofBurnside(Theorem 2.1.5 ), G issolvableandso[ G : M ]isaprimepower.If p | [ G : M ], thenbyLemma 4.4.2 p 2 | [ G : M ]because M isnotanormalsubgroupof G .Thus p 2 dividestheorderof G and | m ( G ) | $p 2$ p +2,acontradiction.Therefore p 2 doesnot dividetheorderof G and p [ G : M ].But[ G : M ] ) =1,sothat q | [ G : M ].Hence | m ( G ) | $q .Noticethat Q isnotcontainedinanyconjugateof M .Thereforeif K isany maximalsubgroupof G containing Q ,then K isdi#erentfromalltheconjugatesof M Thus | m ( G ) |$ q +1,forcing p ( q ( 1). Let n p ( G )denotethenumberofSylow p -subgroupsof G andlet n q ( G )denotethe numberofSylow q -subgroupsof G .BySylow'sthirdtheorem, n q ( G ) | p and n p ( G ) | q b Then Q G and n p ( G ) | q b .Since G isnotnilpotent n p ( G ) ) =1.Also n p ( G ) ) = q because p ( q ( 1).Therefore n p ( G ) $q 2 .Since"( Q )isacharacteristicsubgroupof Q andevery characteristicsubgroupofanormalsubgroupisnormal"( Q ) G .Suppose"( Q ) ) =1, Then | G/ ( Q ) | < | G | and q +2 / | m ( G/ ( Q )) | /| m ( G ) | wheretherstinequalityholdsbyminimality,acontradiction.Therefore," ( Q )=1and Q iselementaryabelianbecause Q isnilpotent.Let L beamaximal P -invariantsubgroup of Q .Then L G andbyminimality L istrivial.Therefore P actsirreducibleon Q and C Q ( P )istrivial.Thus N G ( P )= P C Q ( P )= P. 36 PAGE 37 Supposethat N G ( P )isnotamaximalsubgroupof G .Let M 0 beamaximalsubgroupof G containing N G ( P ).ByDedekind'sLemma(Lemma 4.4.3 ), M 0 = M 0 N G ( P ) Q = N G ( P )( M 0 Q ) acontradiction.Therefore N G ( P )= P isamaximalsubgroupsof G ,whichhas [ G : N G ( P )]=[ G : P ]= q n$ q +2 analcontradiction. Proposition4.4.4. Givenanytwoprimes p and q suchthat p
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CHAPTER5 SOMEFURTHERLOWERBOUNDSFORTHENUMBEROFMAXIMAL SUBGROUPS 5.1Introduction Somelowerboundsforthenumberofmaximalsubgroupswereinvestigatedinthe previouschapter.Ingeneral, | m ( G ) | $| ( G ) | andequalityholdsifandonlyif G iscyclic. Underfurtherassumptionsonthestructureof G ,thislowerboundwasimproved.If G is noncyclicthen | m ( G ) |$| ( G ) | + p ,where p # ( G )isthesmallestprimethatdivides | G | Additionally,if G hasanoncyclicSylowsubgroupand q # ( G )isthesmallestprimesuch that Q # Syl q ( G )isnoncyclic,thenitisproven | m ( G ) | $| ( G ) | + q. (5-1) Observethatgivenanysetofdistinctprimes { q 1 ,q 2 ,...,q n } where q 1 isthesmallest,there existsanitegroup G with ( G )= { q 1 ,q 2 ,...,q n } and | m ( G ) | = | ( G ) | + q 1 Forexample, G = Z q 1 Z q 1 q 2 q n achievesthelowerboundinInequality 5-1 .However,the previouslystatedlowerboundscanberestrictivebecausetheyconsideratmostoneprime in ( G ).Forinstance,let G = S 10 Z 11 Z 143 ,where S 10 isthesymmetricgroupon10 letters.Then | ( G ) | =6and G hasanoncyclicSylow2-subgroup.Therefore q =2and Inequality 5-1 yields | m ( G ) |$ 8,however | m ( G ) | =4002.ThelowerboundinInequality 5-1 canbedrasticallyimprovedwithminimalassumptionsonthestructureofthegroup. Inthepresentchapter,weimprovetheexistinglowerboundfor | m ( G ) | bypartitioning ( G )intothreesets.Weconsiderthesetofprimes p forwhich G isnot p -solvable, denoted # ( G ),andsetofprimes q forwhich G is q -solvable.Wefurthersubdividethe lattersetofprimesintothesetofprimeswhichcorrespondtocyclicSylowsubgroupsof G ,denoted & ( G ),andthesetofprimeswhichcorrespondtononcyclicSylowsubgroups of G ,denoted ( G ).Undertheseconsiderationsweobtaintwonewlowerboundsfor 38

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| m ( G ) | .Inadditiontothesets & ( G ), ( G )and # ( G ),therstlowerboundincludesa furtherassumptionon G .Weassumethatthenonabeliancompositionfactorsof G satisfy atechnicalproperty,namelyPropertyB,whichisdenedinDenition 5.2.6 Theorem5.1.1. Let G beanitegroup.Ifthenonabeliancompositionfactorsof G satisfyPropertyB,then | m ( G ) | $| ( G ) | + % p # ( G ) p + % p # # ( G ) p. (5-2) PropertyBisnotaveryrestrictivepropertyandweconcludethischapterby providingsomeexamplesofsimplegroupsthatsatisfyPropertyB(seeSection 5.8 ). However,notallnitesimplegroupssatisfyPropertyBandknowingthecomposition factorsofagroupmayincreasethepossiblelowerbound.Foragroup G ,weintroduce belowtheinvariant ( G )whichtakesintoaccountmorenelytheinuenceofthe compositionfactorsintoourproblem.Thefollowingtheoremdoesnotrequireanyspecial conditiononthecompositionfactors. Theorem5.1.2. If G isanitegroup,then | m ( G ) |$| & ( G ) | + % p # ( G ) ( p +1)+ ( G ) % p # # ( G ) ( p +1) (5-3) NoticethatTheorem 5.1.2 impliesTheorem 5.1.1 when ( G ) $1.Theinvariant ( G ) iscloselyconnectedtothevalues ( C )as C runsoverthenonabeliancompositionfactors of G .Inthepenultimatesection,(seeSection 5.7 ),weanalyze ( G )andseethat ( C ) tendstoinnityformanyfamiliesofnitesimplegroups C .Although,weprovidemany examplesofnitesimplegroupswhichsatisfyboth ( C ) > 1andPropertyB,ourresults donotusetheclassicationofnitesimplegroups. BothTheorem 5.1.1 andTheorem 5.1.2 improvethelowerboundinInequality 5-1 .Reconsideringthegroup G = S 10 Z 11 Z 143 ,thelowerboundinInequality 5-3 yields | m ( G ) |$ 4001,avastimprovementoverthelowerboundinInequality 5-1 InmostcasesTheorem 5.1.2 givesastrongerresultthanTheorem 5.1.1 .However,it 39

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ispossibleforthelowerboundinInequality 5-2 tobesharperthanthelowerboundin Inequality 5-3 becausenotallnitegroups G satisfy ( G ) $1.Forinstance, ( A 6 ) < 1 where A 6 isthealternatinggroupon6letters,and ( PSL (2 7) ) < 1.Both A 6 and PSL (2 7)satisfyPropertyB,thusTheorem 5.1.1 isapplicable.Insuchacase,say G = PSL (2 7) Z 11 Z 143 ,thelowerboundinInequality 5-3 yields | m ( G ) |$ 21,butthe lowerboundinInequality 5-2 yields | m ( G ) | $28. 5.2Denitions Webeginwithabasicdenition. Denition5.2.1. Let G beanitegroupandlet p beaprimenumber.Then G is p -solvableifallofthenonabeliancompositionfactorsof G donothave p dividingtheir order. Toimprovetheexistinglowerboundsfor | m ( G ) | ,rstpartition ( G )intothreesets: & ( G ), ( G )and # ( G ). Denition5.2.2. Let G beanitegroup. Dene & ( G )asthesetof p # ( G )suchthat G is p -solvableand P # Syl p ( G )is cyclic. Dene ( G )asthesetof p # ( G )suchthat G is p -solvableand P # Syl p ( G )is noncyclic. Dene # ( G )asthesetof p # ( G )suchthat p dividestheorderofanonabelianchief factorof G Toaidincountingsomeofthemaximalsubgroupsof G withindexa # ( G )-number, wemakethefollowingdenitions: Denition5.2.3. Let S beanonabeliansimplegroupandassume M isamaximal subgroupof S .Wesaythat M satisesPropertyAif N Aut( S ) ( M ) S =Aut( S ).Let m A ( S ) denotethesetofmaximalsubgroupsof S thatsatisfyPropertyA. Denition5.2.4. Let G beanitegroupandsuppose C isanonabeliancomposition factorof G .Dene ( ( C )= p #$ ( C ) ( p +1)and ( ( G )= p # # ( G ) ( p +1). 40

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Denition5.2.5. Let G beanitegroupandsuppose C isanonabeliancomposition factorof G .Set ( C )= | m A ( C ) | % ( C ) .If G isnotsolvable,set ( G )=min { ( C ): C isanonabeliancompositionfactorof G } andif G issolvable,thenset ( G )=0. Denition5.2.6. Let S beanonabeliannitesimplegroup.Wesaythat S satises PropertyBif | m ( G ) | $( ( S ),forallsubgroups G suchthat S / G / Aut( S ). 5.3PreliminaryResults Inthissection,wecollectsomenumbertheoreticresultswhichwillaidusinbounding boththenumberofmaximalsubgroupsofagroupthatsatisfyPropertyAaswellas groupswhichsatisfyPropertyB. Lemma5.3.1. Let n$ 3 beaninteger.Suppose S isasetof n distinctprimes.Assume that p and q arethelargestandsmallestelementsof S ,respectively.Then p $n + q Proof. Observethatallelementsof S areatleast2unitsapartfromeachother,except forpossiblythetwosmallestelementsof S whichare1unitapart.Therefore p ( q$ 2( n ( 2)+1,whichisatleast n as n $3.So p ( q$ n and p $n + q ,asdesired. Lemma5.3.2. Let l$ 3 beaninteger.If p 1 ,p 2 ,...,p l areprimenumberssuchthat p 1

p i +1forall i # { 1 2 ,...,l ( 1 } .Hence p l l>p l l 1 i =1 ( p i +1) $p l + l 1 % i =1 ( p i +1) andthus p l l$ l % i =1 ( p i +1) asdesired. 41

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Lemma5.3.3. Let q =2 f $8 andsuppose n # { q ( 1 ,q +1 } .If n isnotaprimenumber, then n$ % p | n ( p +1)+2 Proof. Suppose n = p 1 p 2 p l ,where p 1 $p i forall i # { 2 3 ,...,l } .If n =9,theresult isclear.Thusweassume n> 9andproceedbyinductionon l .Since n isoddand q isa poweroftwo, p 1$ 5.Then( p 1 ( 1)( p 2 ( 1) $5andsothat p 1 p 2$ 2 % i =1 ( p i +1)+2 Thuswesuppose p 1 p 2 p l 1 $l 1 % i =1 ( p i +1)+2 Because( p 1 p 2 p l 1 ( 1)( p l ( 1)$ 2, n = p 1 p 2 p l $p 1 p 2 p l 1 + p l +1$ l 1 % i =1 ( p i +1)+2+ p 1 +1 = l % i =1 ( p i +1)+2 $% p | n ( p i +1)+2 asdesired. Lemma5.3.4. If q =2 f$ 8 ,then 2( q +1) $% p | ( q 3 q ) ( p +1) 42 PAGE 43 Proof. Toprovethisresult,wedividetheproofintothreecases.Firstsuppose q +1is prime.Then q ( 1isnotaprimenumberand q ( 1$ % p | ( q 1) ( p +1)+2 byLemma 5.3.3 .Adding( q +1)+2tobothsidesoftheinequalityaboveyields 2( q +1) $% p | ( q 3 q ) ( p +1) Thusassumethat q ( 1isprime.Then q +1isnotprimeand q +1$ % p | ( q +1) ( p +1)+2 byLemma 5.3.3 .Adding( q ( 1)+2tobothsidesoftheinequalityaboveyields 2( q +1) $% p | ( q 3 q ) ( p +1) Finally,supposethat q +1nor q ( 1isprime.ThenLemma 5.3.3 implies q +1+ q ( 1$ % p | ( q 2 1) ( p +1)+4 and 2( q +1) $% p | ( q 3 q ) ( p +1) asdesired. 5.4 | m ( G ) | for p -SolvableGroups Foranitesolvablegroup,wewillmakeuseofthewellknownfactthatevery maximalsubgrouphasprimepoweredindex.For p # & ( G ),wecountthenumberof maximalsubgroupsof G bycountingthenumberofmaximalsubgroupswhoseindexin G isapowerof p .Recalldenotes m p ( G )thesetofmaximalsubgroupsof G whoseindexin G isapowerof p .Thersttwolemmasofthissectionarestandardresultswhichwillaid usinbounding | m p ( G ) | frombelow. 43 PAGE 44 Lemma5.4.1. Let G beanite p -solvablegroup,where p # ( G ) .If H isaHall p -subgroupof G and K isasubgroupwhichcontains N G ( H ) ,then K isself-normalizing. Proof. Let g # G satisfy K g = K .Then H g isaHall p -subgroupof K .SinceallHall p -subgroupsof K areconjugate,thereexists n # K suchthat H gn = H .So gn # N G ( H ) and g # N G ( H ) n 1 0 K. Therefore K isself-normalizing. Lemma5.4.2. Let N beasolvablenormalsubgroupofanitegroup G .Supposethat H isaHallsubgroupof N .Then G = N G ( H ) N Proof. Let g # G .Since H g 0 N g = N and H g hasthesameorderofaHallsubgroupof N H g isaHallsubgroupof N .Therefore H and H g are N -conjugate,say H gn = H for some n # N .Then gn # N G ( H )and g # N G ( H ) n 1 0 N G ( H ) N. Therefore G = N G ( H ) N ,asdesired. Let p # ( G )foranitegroup G .Recallthat O p ( G )isthelargestnormal p -subgroup of G .Thefollowingtheoremgivesalowerboundon | m p ( G ) | Theorem5.4.3. Suppose G isanitegroupandassume p # & ( G ) 5 ( G ) .Then | m p ( G ) |$ 1 and | m p ( G ) | $p +1 when p # ( G ) Proof. Supposethisisfalse.Let G beacounterexampleofminimumorderandassume P # Syl p ( G ).Since G hasamaximalsubgroupwhichcontainsaHall p -subgroup, | m p ( G ) |$ 1.Therefore | m p ( G ) |

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H actsirreduciblyon P .Then N P ( H ) # { 1 } ,P + .Byminimality, O p ( G )=1andso H ) G .Then N P ( H )= { 1 } and H ismaximal.Therefore [ G : H ]= | P | $p 2 and H hasatleast p 2 conjugates,allofwhicharemaximalandliein m p ( G ),acontradiction. Thus H doesnotactirreduciblyon P .Then P hasanontrivialproper H -invariant subgroup P 1 .ByMaschke'sTheorem(Theorem 4.2.6 ),thereexistsanontrivialproper H -invariantsubgroup P 2 of P suchthat P = P 1 P 2 .Now H mustactnon-triviallyon atleastoneof P 1 or P 2 .Withoutlossofgenerality,assume H actsnon-triviallyon P 1 Let M beamaximalsubgroupwhichcontains N G ( H ) P 2 .Then M isself-normalizingby Lemma 5.4.1 ,andsothe p conjugatesof M willnotcontain P 1 andareelementsof m p ( G ). If L isamaximalsubgroupwhichcontains P 1 H ,then L # m p ( G )andisdistinctfromthe aforementioned p maximalsubgroupsof m p ( G ).Hence | m p ( G ) |$ p +1,acontradiction. Therefore, P ) G and O p ( G )isapropersubgroupof P .Let J beaHall p -subgroup of O p p ( G ).Then J ) =1and G = N G ( J ) O p p ( G ) 0 N G ( J ) O p ( G ) wheretheequalityholdsbyLemma 5.4.2 .Noticethat N G ( J )isapropersubgroupof G because O p ( G )=1byminimality.If M isamaximalsubgroupof G whichcontains N G ( J ),itisself-normalizingbyLemma 5.4.1 .So M hasatleast p conjugatesin m p ( G ), allofwhichdonotcontain O p ( G ).Byminimality, | m p ( G/ O p ( G )) | $1andsothereisat leastonemaximalsubgroupin m p ( G )whichcontains O p ( G ).Therefore | m p ( G ) |$ p +1, analcontradiction. ObservethatthetheoremaboveprovesbothTheorem 5.1.1 andTheorem 5.1.2 for allnitesolvablegroups.Suppose G isanitesolvablegroup.Then ( G )=0andthe 45

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nonabeliancompositionfactors G satisfyPropertyB.Thus,theorem 5.4.3 yieldsatleast | & ( G ) | + % p # ( G ) ( p +1)= | ( G ) | + % p # ( G ) p distinctmaximalsubgroupsof G 5.5 | m ( G ) | forNonsolvableGroups Throughoutthissectionwewillmakeuseofthefactthateveryminimalnormal subgroupofanitegroupisthedirectproductofisomorphicsimplegroups.Fora nonsolvablegroup,wewillcountthemaximalsubgroupsbyseparatingthemintothose thatcontainaminimalnormalsubgroup N andthosethatdonotcontain N Lemma5.5.1. Let N beanonabelianminimalnormalsubgroupofanitegroup G and supposethat N = S 1 S 2 " S l ,where S i isanonabeliansimplegroupand S i = S j for all i,j # { 1 2 ,...,l } .Let S = S i andxsome $i : S 4 S i tobeanisomorphism.Let M beamaximalsubgroupof S anddene Q M =$ 1 ( M ) $2 ( M ) ...$ l ( M ) .If M satises PropertyA,then G = N N G ( Q M ) Proof. Since N G N N G ( Q M )isasubgroupof G .Let g # G andobservethat gQ M g 1 isapropersubgroupof N .Denethemap ) g : N 4 N by ) g ( n )= gng 1 forall n # N Then ) g # Aut( N ).So ) g willsendminimalnormalsubgroupsof N tominimalnormal subgroupsof N ,thus ) g permutestheminimalnormalsubgroupsof N .Let S l bethe symmetricgroupon l objects.Thenthereexists % # S l suchthat gS & 1 ( i ) g 1 = S i ,forall i # { 1 2 ,...,l } Denethemap # g,i : S 4 S by # g,i ( s )= $1 i ( g$ & 1 ( i ) ( s ) g 1 ) forall s # S .Then # g,i # Aut( S )forall i # { 1 2 ,...,l } .SinceAut( S )= N Aut( S ) ( M ) S ,thereexistssome s g,& 1 ( i ) # S suchthat s g,& 1 ( i ) Ms 1 g,& 1 ( i ) = $1 i ( g$ & 1 ( i ) ( M ) g 1 ) 46

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Applying $i ,weobtaintheequality$ i ( s g,& 1 ( i ) Ms 1 g,& 1 ( i ) )= g $& 1 ( i ) ( M ) g 1 Set n g = ($ 1 ( s g,& 1 (1) ) $2 ( s g,& 1 (2) ) ,...,$ l ( s g,& 1 ( l ) ) ) # N .Then gQ M g 1 = g $& 1 (1) ( M ) g 1 " g$ & 1 ( l ) ( M ) g 1 = $1 ( s g,& 1 (1) Ms 1 g,& 1 (1) ) "$ l ( s g,& 1 ( l ) Ms 1 g,& 1 ( l ) ) = n g ( $1 ( M ) "$ l ( M ) ) n 1 g = n g Q M n 1 g So n 1 g gQ M g 1 n g = Q M andtherefore n 1 g g # N G ( Q M )and g # n g N G ( Q M ).Since n g # N g # N N G ( Q M )and G = N N G ( Q M ). Lemma5.5.2. Let N beanonabelianminimalnormalsubgroupofanitegroup G andsupposethat N = S 1 S 2 " S l ,where S i isanonabeliansimplegroupand S i = S j forall i,j # { 1 2 ,...,l } .Let S = S i andassume M isamaximalsubgroupof S whichsatisesPropertyA.Fixsome $i : S 4 S i tobeanisomorphismanddene Q M =$ 1 ( M ) $2 ( M ) ...$ m ( M ) .Then N G ( Q M ) isamaximalsubgroupof G Proof. Since N isaminimalnormalsubgroupof G and Q M & N N G ( Q M )isaproper subgroupof G .Suppose K isapropersubgroupof G whichcontains N G ( Q M ).Then N isnotcontainedin K byLemma 5.5.1 .Sothereexistssome i # { 1 2 ,...,l } such that K doesnotcontain S i ,but K S i = $i ( M ).Observethat G actstransitively on { S 1 ,S 2 ,...,S l } and N actstriviallyon { S 1 ,S 2 ,...,S l } ,so K actstransitivelyon { S 1 ,S 2 ,...,S l } .Foreach j # { 1 2 ,...,l } ,thereissome k j # K suchthat k j S i k 1 j = S j and K S j = k j ( K S i ) k 1 j .Therefore K S j ) = S j andso K S j =$ j ( M )as $j ( M ) 0 K S j .So K N = Q M and N G ( Q M ) N = Q M .So G = KN = N G ( Q M ) N 47 PAGE 48 and K = N G ( Q M )( K N )= N G ( Q M ) Therefore N G ( Q M )isamaximalsubgroupof G Thefollowingtwolemmascountthenumberofmaximalsubgroupswhichdonot contain N ,therstlemmausingPropertyAandthesecondlemmausingPropertyB. Lemma5.5.3. Suppose N isanonabelianminimalnormalsubgroupofanitegroup G andassume N = S 1 S 2 " S l ,where S i isasimplegroupand S i = S j forall i,j # { 1 2 ,...,l } .Thenthereareatleast | m A ( S 1 ) | maximalsubgroupsof G whichdonot contain N Proof. Let S = S i for i # { 1 2 ,...,l } .Fixsome$ i : S 4 S i tobeanisomorphism. Set n = | m A ( S ) | ,sothat S has n maximalsubgroups,say M 1 ,M 2 ,...,M n ,suchthat N Aut( S ) ( M j ) S =Aut( S )forall j # { 1 2 ,...,n } .Dene Q M j = $1 ( M j )$ 2 ( M j ) " $l ( M j ).Then G = N N G ( Q M j )byLemma 5.5.1 and N G ( Q M j )isamaximalsubgroupof G byLemma 5.5.2 .Therefore G hasatleast n maximalsubgroupswhichdonotcontain N ,asdesired. Lemma5.5.4. Suppose N isanonabelianminimalnormalsubgroupofanitegroup G andassume N = S 1 S 2 " S l ,where S i = S j forall i,j # { 1 2 ,...,l } .If S 1 satises PropertyB,thenthereareatleast ( ( S 1 ) maximalsubgroupsof G whichdonotcontain N Proof. Let p bethelargestprimein ( N )andsuppose P isaSylow p -subgroupof N .Then G = N N G ( P )bytheFrattiniArgument.Because P & N and N isa minimalnormalsubgroupof G N G ( P ) ) = G andthereexistsamaximalsubgroup L of G containing N G ( P ).Observethat N )0 L becauseotherwise G = NL 0 L. Therefore L ismaximalsubgroupof G whichdoesnotcontain N 48 PAGE 49 For i # { 1 2 ,...,l } ,dene R i = S i L .Observethat L actstransitivelyon { S 1 ,S 2 ,...,S l } .Thus R i and R j are L -conjugate,for i,j # { 1 2 ,...,l } andhavethesame order.Since N isthedirectproductofisomorphicsimplegroups, S i isnotcontainedin L forall i # { 1 2 ,...,l } .Also S i ) = R i ) =1and R i isnotanormalsubgroupof S i .Itfollows that L isnotanormalsubgroupof G ,andthus L isaself-normalizingmaximalsubgroup of G Because S i actsbyleftmultiplicationonthesetofleftcosetof R i in S i ,thereexists anontrivialhomomorphismfrom S i tothesymmetricgroupon[ S i : R i ]elementsand S i mustbeisomorphictoasubgroupofthealternatinggroupon[ S i : R i ]elements.Thus p / [ S i : R i ].Let s i # S i anddene g =( s 1 ,s 2 ,s 3 ,...,s l ) # N .Conjugating L by g ,we obtainatleast p l distinctmaximalsubgroupsof G whichdonotcontain N If l$ 2,then G hasatleast p | ( N ) | maximalsubgroupswhichdonotcontain N as p> | ( N ) | byLemma 5.3.1 .Since p | ( N ) | = p | ( S 1 ) | $( ( S 1 ) byLemma 5.3.2 ,thereareatleast ( ( S 1 )maximalsubgroupswhichdonotcontain N .So assumethat l =1.Then N = S 1 and N C G ( N ) G .Then G/C G ( N )isisomorphicto H ,where N / H / Aut( N ) Since N satisesPropertyB,thereareatleast ( ( H )$ ( ( S 1 )maximalsubgroupsof G whichdonotcontain N ,asdesired. Weconcludethissection,withtwotheoremsthatallowustocountsomeofthe maximalsubgroupsofanonsolvablegroupbytheirindex. Theorem5.5.5. If G beanitegroup,thenithasatleast ( ( G ) ( G ) maximalsubgroups of G withindexa # ( G ) -number. 49

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Proof. Supposethisisfalse.Let G beacounterexampleofminimumorder.Assume N isaminimalnormalsubgroupof G suchthat N = S 1 S 2 ... S l ,where S i isa simplegroupand S i = S j forall i,j # { 1 2 ,...,l } .First,supposethat N issolvable. Then # ( G )= # ( G/N )and ( G )= ( G/N ).Byminimality, G/N hasatleast ( ( G ) ( G ) maximalsubgroupswithindexa # ( G )-number.Thus G hasatleast ( ( G ) ( G )maximal subgroupswhichcontain N andhaveindexa # ( G )-number,acontradiction. Therefore N isnotsolvable.Byminimality, G/N hasatleast ( ( G/N ) ( G/N ) maximalsubgroupswithindexa # ( G/N )-number.So G hasatleast ( ( G/N ) ( G/N ) maximalsubgroupswhichcontain N andhaveindexa # ( G )-number.Inaddition,there areatleast | m A ( S 1 ) | maximalsubgroupswhichdonotcontain N byLemma 5.5.3 Observethat ( N )= ( S 1 )= | m A ( S 1 ) | ( ( S 1 ) = | m A ( S 1 ) | ( ( N ) andthus | m A ( S 1 ) | = ( ( N ) ( N ).Intotal, G hasatleast ( ( G/N ) ( G/N )+ ( ( N ) ( N ) $( ( G ) ( G ) maximalsubgroupswithindexa # ( G )-number,analcontradiction. Theorem5.5.6. Let G beanitegroup.Ifthenonabeliancompositionfactorsof G satisfyPropertyB,thenthereexistsatleast ( ( G ) maximalsubgroupsof G withindexa # ( G ) -number. Proof. Supposethisisfalse.Let G beacounterexampleofminimumorderandassume N isaminimalnormalsubgroupof G .First,supposethat N issolvable.Then # ( G )= # ( G/N )and G/N hasatleast ( ( G )maximalsubgroupswithindexa # ( G )-number,by minimality.Thus G hasatleast ( ( G )maximalsubgroupswhichcontain N andhaveindex a # ( G )-number,acontradiction. Therefore N isnotsolvable.Suppose N = S 1 S 2 " S l ,where S i isanonabelian simplegroupand S i = S j forall i,j # { 1 2 ,...,l } .Byminimality, G/N hasatleast 50 PAGE 51 ( ( G/N )maximalsubgroupswithindexa # ( G/N )-number.So G hasatleast ( ( G/N ) maximalsubgroupswhichcontain N andhaveindexa # ( G )-number.ByLemma 5.5.4 G hasatleast ( ( S 1 )maximalsubgroupswhichdonotcontain N .Intotal, G hasatleast ( ( G/N )+ ( ( S 1 )= ( ( G/N )+ ( ( N )$ ( ( G ) maximalsubgroupswithindexa # ( G )-number,analcontradiction. 5.6ProofofMainTheorems ProofofTheorem 5.1.1 Weobtainatleast | & ( G ) | + % p # ( G ) ( p +1) distinctmaximalsubgroupsof G withindexa & ( G ) 5 ( G )-numberbyTheorem 5.4.3 .By Theorem 5.5.6 G hasatleast ( ( G )maximalsubgroupsof G withindexa # ( G )-number. Therefore, | m ( G ) | $| & ( G ) | + % p # ( G ) ( p +1)+ ( ( G )= | ( G ) | + % p # ( G ) p + % p # # ( G ) p, asdesired. ProofofTheorem 5.1.2 Weobtainatleast | & ( G ) | + % p # ( G ) ( p +1) distinctmaximalsubgroupsof G withindexa & ( G ) 5 ( G )-numberbyTheorem 5.4.3 ByTheorem 5.5.5 G hasatleast ( ( G ) ( G )maximalsubgroupsof G withindexa # ( G )-number.Therefore, | m ( G ) |$| & ( G ) | + % p # ( G ) ( p +1)+ ( ( G ) ( G ) asdesired. 51

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WeconcludethissectionbyprovingtheboundinTheorem 5.1.2 isbestpossible.The sameexampleprovesthatthelowerboundinTheorem 5.1.1 isbestpossible. Proposition5.6.1. Givenanynitesetofprimes $andanysubset$ of $,thereexists anitesolvablegroup G with ( G )=$ ( G )= $and | m ( G ) | = | & ( G ) | + % p # ( G ) ( p +1)+ ( ( G ) ( G ) Proof. Supposethat$= { p 1 ,p 2 ,...,p n } and $= { p 1 ,p 2 ,...,p l } where0 / l / n Considerthenoncyclicgroup G = Z p 1 Z p 1 Z p 2 Z p 2 " Z p l Z p l Z p l +1 Z p l +2 " Z p n whichisoforder p 2 1 p 2 2 p 2 l p l +1 p l +2 p n .Then$= ( G ), $= ( G )and ( G )=0.By Lemma 4.3.2 | m ( Z p i Z p i ) | = p i +1for i # { 1 2 ,...,l } andobservethat | m ( Z p i ) | =1for i # { l +1 ,...,n } .Thus | m ( G ) | = | & ( G ) | + l % i =1 ( p i +1) byLemma 4.2.7 ,andthelowerboundinTheorem 5.1.2 isbestpossible. ThereexistsnonsolvablegroupswhichdemonstratethelowerboundinTheorem 5.1.2 isbestpossible.Forexample,suppose G isanitesimplegroupsuchthat m ( G )= m A ( G ).Then & ( G )= 1 = ( G ) and | m ( G ) | = | m A ( G ) | = ( ( G ) ( G ) andthus G achievesthelowerboundinTheorem 5.1.2 5.7Remarkson ( G ) Inthissection,wegivesomegroups G whichsatisfy ( G ) > 1. Lemma5.7.1. Let A n denotethesimplealternatinggroupon n letters.If n$ 9 ,then ( A n ) >n .Otherwise ( A 5 )= 21 13 ( A 6 )= 10 13 ( A 7 )=3 and ( A 8 )= 127 21 52

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Proof. If5 / n / 8,theresultholdsbytheAtlasofFiniteGroups[ 16 ].Soweassume that n $9.Let S n denotethesymmetricgroupon n letters.Foreachinteger k with 1 n ,wehaveatleast n ( n 2 ( 2)= n ( n 4) 2 distinct maximalsubgroupsof A n .Observethattheaforementionedmaximalsubgroupsare distinctfromthe n conjugatesofthemaximalsubgroup A n 1 in A n .Let M beoneof theaforementioned n ( n 4) 2 + n maximalsubgroupsof A n .SinceAut( A n )= S n and M is normalizedbyanelementof S n \A n ,itfollowsthat N Aut( A n ) ( M ) A n =Aut( A n ).Therefore | m A ( A n ) |$ n ( n 4) 2 + n and n 2 2 n 2 / % ( A n ) >n ,asdesired. Lemma5.7.2. Let q =2 f .If q $8 ,then ( PSL (2 ,q ))$ q 2 + q +1 2( q +1) If q =4 ,then ( PSL (2 ,q ))= 21 13 Proof. Let S = PSL (2 ,q ).If q =4,theresultholdsbytheresultholdsbythe Atlas ofFiniteGroups [ 16 ].Soassumethat q $8.Then D 2( q 1) D 2( q +1) and Z f 2 Z q 1 are maximalsubgroupsof S byDickson'sTheorem(see[ 2 ]).Since S issimple, D 2( q 1) has [ q 3 ( q :2( q ( 1)]=2 f 1 ( q +1)conjugates, D 2( q +1) has[ q 3 ( q :2( q +1)]=2 f 1 ( q ( 1) conjugatesand[ q 3 ( q : Z f 2 Z q 1 ]= q +1.Thus S hasatleast q 2 + q +1distinctmaximal subgroups.Theouterautomorphismgroupof S isgeneratedbyaeldautomorphism$ oforder f ,whichxesallthreeoftheaforementionedconjugacyclassesofmaximal subgroups.Therefore | m A ( S ) | $q 2 + q +1.Since | S | = q 3 ( q ,Lemma 5.3.4 implies ( S )$ q 2 + q +1 p # # ( S ) ( p +1) $q 2 + q +1 2( q +1) asdesired. 53 PAGE 54 Table5-2. Valuesof ( G ). G ( G ) 6 ( G ) 7 G ( G ) 6 ( G ) 7 M 11 309 25 12 Fi 22 703 131 313 47 14 960 240 M 12 2 850 25 114 Fi 23 > 10 10 > 10 10 M 22 1 948 33 59 Fi 24 > 10 10 > 10 10 M 23 44 414 57 779 M c L 495 529 33 15 016 M 24 1 507 537 57 26 448 He 6 862 629 39 175 964 J 1 13 016 53 245 Ru 367 882 130 65 5 659 725 J 2 16 964 21 807 Suz 291 965 788 47 6 212 038 J 3 68 410 33 2 073 HN 8 935 530 772 53 168 594 920 J 4 > 10 10 > 10 10 O N 83 897 408 85 987 028 Co 3 424 818 005 57 7 452 947 Th> 10 10 > 10 10 Co 2 3 581 796 533 57 62 838 535 Ly> 10 10 > 10 10 Co 1 > 10 10 > 10 10 B> 10 10 > 10 10 HS 68 410 33 2 073 M> 10 10 > 10 10 Lemma5.7.3. Let q = p f ,where p isanoddprimenumber.If q$ 13 ,then ( PSL (2 ,q )) $q 2 2( q +1) Otherwise ( PSL (2 7))= 8 15 ( PSL (2 9))= 10 13 and ( PSL (2 11))= 67 25 Proof. ThisproofisomittedduetoitssimilaritiestotheproofofLemma 5.7.2 Weconcludethissectionbygivingvaluesof ( G ),where G isasporadicsimple group.Thefollowingvaluesgiveninthetablebelowwerecalculatedfromthe Atlasof FiniteGroups [ 16 ]. 5.8RemarksonPropertyB Inthissection,wegivensomeexamplesofnitesimplegroupswhichsatisfyProperty B. Lemma5.8.1. If n$ 5 ,then A n satisesPropertyB. 54

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Proof. If n =6,theresultholdsbytheresultholdsbythe AtlasofFiniteGroups [ 16 ].Soassumethat n ) =6.SinceAut( A n )= S n ,theresultfollowsfromtheproofof Lemma 5.7.1 Lemma5.8.2. Suppose q = p f ,where p isaprimenumber.If q \$ 4 ,then PSL (2 ,q ) satisesPropertyB. Proof. If n # { 7 9 } ,thentheresultholdsbythe AtlasofFiniteGroups [ 16 ].Soassume that n )#{ 7 9 } .SinceOut( PSL (2 ,q )) = Z (2 ,q 1) Z d ,theresultfollowsfromtheproofsof Lemma 5.7.2 andLemma 5.7.3 55

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REFERENCES [1] M.Aschbacher, Onthemaximalsubgroupsoftheniteclassicalgroups ,Invent.Math. 76 (1984),469514. [2] L.E.Dickson, Lineargroups:Withanexpositionofthegaloiseldtheory ,Dover PublicationInc.,NewYork,1958. [3] D.S.Dummit&R.M.Foote, Abstractalgebra ,3rded.,JohnWhily&Sons,Inc., Hoboken,NJ,2004. [4] TheGAPGroup, Gapgroups,algorithms,andprogramming ,4.7.2,2013. [5] I.M.Isaacs, Finitegrouptheory ,GraduateStudiesinMathematics,vol.92,American MathematicalSociety,Providence,RI,2008. [6] L.K.Lauderdale, Lowerboundsonthenumberofmaximalsubgroupsinanite group ,Arch.Math. 101 (2013),915. [7] P.Kleidman&M.Liebeck, Thesubgroupstructureofniteclassicalgroups CambridgeUniversityPress,NewYork,NY,1990. [8] M.Herzog&O.Manz, Onthenumberofsubgroupsinnitesolvablegroups ,J. Austral.Math.Soc. 58 (1995),134141. [9] J.Fisher&J.McKay, Thenonabeliansimplegroups g | g | < 10 6 maximalsubgroups ,Math.Comp. 32 (1978),12931302. [10] B.Newton, Onthenumberofsubgroupsinnitesolvablegroups ,Arch.Math. 96 (2011),501506. [11] L.L.Scott, Representationsincharacteristicp ,Proc.Sympos.PureMath. 37 (1980), 318331. [12] D.Gorenstein&R.Lyons&R.Solomon, Theclassicationofnitesimplegroups vol.1-6,AmericanMathematicsSociety,2004. [13] W.Feit&J.G.Thompson, Solvabilityofgroupsofoddorder ,PacicJ.Math. 13 (1963),7751029. [14] G.E.Wall, Someapplicationsoftheeulerianfunctionsofanitegroup ,J.Austral. Math.Soc. 2 (1961/1962),3559. [15] R.J.Cook&J.Wiegold&A.G.Williamson, Enumeratingsubgroups ,J.Austral. Math.Soc. 43 (1987),220223. [16] J.H.Conway&R.T.Curtis&S.P.Norton&R.A.Parker&R.A.Wilson, Atlasof nitegroups ,OxfordUniversityPress,1985;reprintedwithcorrections,2004. 56

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BIOGRAPHICALSKETCH Lindsey-KayLauderdalewasborninChampaign,Illinois.Shegraduatedfrom CentennialHighSchoolandcompletedaBachelorofScienceinmathematicsfromthe UniversityofIllinoisatUrbana-Champaignin2009.In2011,Lindsey-Kayearneda MasterofScienceinmathematicsfromtheUniversityofFlorida.Inherfreetimeshe enjoysshopping,watchingtelevisionandnailart. 57