Herstein 3th editon

Struktur Aljabar

2. ABSTRACTALGEBRA

3. ABSTRACTALGEBRA
ThirdEdition
I.N.Herstein
LateProfessorofMathematics
University
ofChicago
IIPRENTICE-HALL,UpperSaddleRiver,NewJersey07458
@

4. LibraryofCongressCataloginginPublicationData
Herstein,I.N.
Abstractalgebra/I.N.Herstein.-3rded.
p.em.
Includesindex.
ISBN0-13-374562-7(alk.paper)
1.Algebra,Abstract.I.Title
QA162.H471995 95-21470
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5. ToBiska

6. CONTENTS
Preface ix
1ThingsFamiliar andLessFamiliar1
1 AFewPreliminaryRemarks 1
2SetTheory3
3Mappings8
4A(S)(TheSet of1-1Mappingsof SontoItself) 16
5TheIntegers 21
6MathematicalInduction 29
7ComplexNumbers 32
2Groups 40
1DefinitionsandExamplesofGroups 40
2SomeSimpleRemarks 48
3Subgroups 51
4Lagrange'sTheorem 56
5HomomorphismsandNormalSubgroups 66
6FactorGroups 77
7TheHomomorphismTheorems 84
8Cauchy'sTheorem 88
9DirectProducts 92
10FiniteAbelianGroups(Optional) 96
11ConjugacyandSylow'sTheorem(Optional) 101
vii

7. viiiContents
3TheSymmetricGroup 108
1Preliminaries 108
2CycleDecomposition 111
3OddandEvenPermutations119
4RingTheory 125
1
2
3
4
5
6
7
5Fields
1
2
3
4
5
6
DefinitionsandExamples
125
SomeSimpleResults 137
Ideals,Homomorphisms,andQuotientRings 139
MaximalIdeals 148
PolynomialRings 151
PolynomialsovertheRationals166
Field
ofQuotientsofanIntegralDomain 172
176
ExamplesofFields176
ABriefExcursionintoVectorSpaces180
FieldExtensions
191
FiniteExtensions198
Constructibility
201
RootsofPolynomials 207
6SpecialTopics(Optional)215
1TheSimplicity ofAn 215
2FiniteFieldsI 221
3FiniteFieldsII:Existence 224
4FiniteFieldsIII:Uniqueness 227
5CyclotomicPolynomials 229
6Liouville'sCriterion236
7TheIrrationality
of
'IT239
Index 243

8. PREFACETOTHETHIRDEDITION
Whenwewereaskedtopreparethethirdeditionofthisbook,itwasourcon­
sensusthatitshouldnotbealteredinanysignificantway,andthatHerstein's
informalst~leshouldbepreserved.Wefeelthatone ofthebook'svirtues is
thefactthatitcoversabigchunk ofabstractalgebrainacondensedandin­
terestingway.
Atthesametime,withouttrivializingthesubject,itremainsac­
cessibletomostundergraduates.
Wehave,however,correctedminorerrors,straightenedoutinconsis­
tencies,clarifiedandexpandedsomeproofs,andaddedafewexamples.
Toresolvethemanytypographicalproblemsofthesecondedition,
PrenticeHallhashadthebookcompletely
retypeset-makingiteasierand
morepleasurabletoread.
Ithasbeenpointedouttousthatsomeinstructorswouldfindituseful
tohavetheSymmetricGroup
SnandthecyclenotationavailableinChapter
2,inordertoprovidemoreexamples ofgroups.Ratherthanalterthe
arrangementofthecontents,therebydisturbingtheoriginalbalance,
wesug­
gestanalternateroutethroughthematerial,whichaddressesthisconcern.
AfterSection2.5,onecouldspendanhourdiscussingpermutationsandtheir
cycledecomposition(Sections
3.1and3.2),leavingtheproofsuntillater.The
studentsmightthengooverseveralpastexamples
offinitegroupsandexplic­
itlysetupisomorphismswithsubgroups
of
Sn-This exercisewouldbemoti­
vatedbyCayley'stheorem,quotedinSection
2.5.Atthesametime,itwould
havethebeneficialresultofmakingthestudents morecomfortablewiththe
conceptofanisomorphism.Theinstructorcouldthenweaveinthevarious
subgroupsoftheSymmetricGroups
Snasexamplesthroughouttheremain-
ix

9. x PrefacetoThirdEdition Ch.6
derofChapter2.Ifdesired,onecouldevenintroduceSections 3.1and3.2
afterSection2.3
or2.4.
Twochangesin
theformathave beenmadesince thefirstedition.First,
aSymbolListhas
beenincludedtofacilitatekeepingtrack ofterminology.
Second,afewproblemshave
beenmarkedwithanasterisk(*).Theseserve
asavehicle
tointroduceconcepts andsimplearguments thatrelateinsome
importantway
tothediscussion.Assuch,theyshould bereadcarefully.
Finally,we
takethisopportunitytothankthemanyindividualswhose
collectiveefforts havehelped
toimprovethisedition.We thankthereview­
ers:Kwangil
KohfromNorthCarolinaStateUniversity, DonaldPassman
fromtheUniversity
ofWisconsin,andRobertZincfrom PurdueUniversity.
And,ofcourse,we thankGeorgeLobellandElaineWetterau,andothersat
PrenticeHallwhohave
beenmosthelpful.
BarbaraCortzen
David
J.Winter

10. PREFACETOTHEFIRSTEDITION
Inthelasthalf-century orsoabstractalgebrahasbecomeincreasinglyimpor­
tantnotonlyinmathematicsitself,
butalsoinavariety ofotherdisciplines.
Forinstance,theimportanceoftheresultsandconceptsofabstractalgebra
playan
ev;rmoreimportantroleinphysics,chemistry,andcomputerscience,
tociteafewsuchoutsidefields.
Inmathematicsitselfabstractalgebraplaysadualrole:that
ofaunify­
inglinkbetweendisparateparts
ofmathematicsandthatofaresearchsubject
withahighlyactivelife
ofitsown.Ithasbeenafertileandrewardingresearch
areabothinthelast100yearsandatthepresentmoment.Some
ofthegreat
accomplishmentsof
ourtwentieth-centurymathematicshavebeenprecisely
inthisarea.Excitingresultshavebeenprovedingrouptheory,commutative
andnoncommutativeringtheory,Liealgebras,Jordanalgebras,combina­
tories,andahostofotherpartsofwhat
isknownasabstractalgebra.Asub­
jectthatwasonceregardedasesoterichasbecome consideredasfairlydown­
to-earthforalargecrosssectionofscholars.
Thepurposeofthisbook
istwofold.Forthosereaderswhoeitherwant
to
goontodoresearchinmathematics orinsomealliedfieldsthatusealge­
braicnotionsandmethods,thisbookshouldserveasan
introduction-and,
westress,onlyasan introduction-tothisfascinatingsubject. Forinterested
readerswhowanttolearnwhat
isgoingoninanengagingpartofmodern
mathematics,thisbookcouldservethatpurpose,
aswellasprovidethemwith
somehighlyusabletoolstoapplyintheareasinwhichtheyareinterested.
Thechoiceofsubjectmatterhasbeenmadewiththeobjectiveofintro­
ducingreaderstosomeofthefundamentalalgebraicsystemsthatarebothin-
xi

11. xiiPrefacetoFirstEdition
terestingand ofwideuse.Moreover,ineachofthesesystemstheaimhas
beentoarriveatsomesignificantresults.There
islittlepurposeservedin
studyingsomeabstractobjectwithoutseeingsomenontrivialconsequencesof
thestudy.Wehope
thatwehaveachievedthegoalofpresentinginteresting,
applicable,andsignificantresultsineachofthesystemswehavechosento
discuss.
Asthereaderwillsoonsee,therearemanyexercisesinthebook.They
areoftendividedintothreecategories:easier,middle-level,andharder(with
anoccasionalveryhard).Thepurpose
oftheseproblems istoallowstudents
totesttheirassimilationofthematerial,tochallengetheirmathematicalinge­
nuity,topreparethegroundformaterial
thatisyettocome,andtobea
means
ofdevelopingmathematicalinsight,intuition,andtechniques.Readers
shouldnotbecomediscouragediftheydonotmanagetosolvealltheprob­
lems.Theintent
ofmanyoftheproblems isthattheybe tried-evenifnot
solved-forthepleasure(andfrustration)ofthereader.Someoftheprob­
lemsappearseveraltimesinthebook.Tryingtodotheproblems
isundoubt­
edlythebestway
ofgoingaboutlearningthesubject.
Wehavestrivedtopresentthematerialinthelanguageandtoneofa
classroomlecture.Thusthepresentation
issomewhatchatty;wehopethat
thiswill
putthereadersattheirease. Anattemptismadetogivemanyand
revealingexamples
ofthevariousconceptsdiscussed.Someoftheseexam­
plesarecarriedforwardtobeexamples
ofotherphenomena
that_comeup.
Theyareoftenreferredtoasthediscussionprogresses.
Wefeel
thatthebook isself-contained,exceptinone section-thesec­
ondlastone
ofthebook-wherewemakeimplicituse ofthefactthatapoly­
nomialoverthecomplexfieldhascomplexroots(that
isthecelebratedFun­
damentalTheorem ofAlgebraduetoGauss),andinthelastsectionwherewe
makeuse
ofalittleofthecalculus.
Wearegratefultomanypeoplefortheircommentsandsuggestionson
earlierdraftsofthebook.Many
ofthechangestheysuggestedhavebeenin­
corporatedandshouldimprovethereadability
ofthebook.Weshouldliketo
express
ourspecialthankstoProfessorMartinIsaacsforhishighlyuseful
comments.
Wearealsogratefulto
FredFlowersforhisusualsuperbjoboftyping
themanuscript,andtoMr.Gary
W.OstedtoftheMacmillanCompanyfor
hisenthusiasmfortheprojectandforbringingittopublication.
Withthiswewishallthereadersahappyvoyageonthemathematical
journeytheyareabouttoundertakeintothisdelightfulandbeautifulrealm
ofabstractalgebra.
I.N.H.

12. aES
a$S
SeT,T:JS
S=T ~
o
AUB
AnB
{sESissatisfiesP}
A-B
A'
(a,b)
AXB
IR
f:S~T
f(s)
i:S~S,is
f-l(t)
f-l(A)
fog,fg
A(S)
Sn
n!
7L
O(s)
SYMBOLLIST
aisanelementofthesetS,3
aisnotanelement ofthesetS,3
S
isasubsetofthesetT,3
ThesetsSandTareequal(havethesameelements),4
Theemptyset,4
TheunionofthesetsAandB,4
TheintersectionofthesetsAandB,4
ThesubsetofelementsofSsatisfyingP,4
ThedifferenceofthesetsAandB,4
ThecomplementofA,5
Orderedpairconsistingofa,b(seealsobelow),5
TheCartesianproductof AandB,5
Thesetofrealnumbers,8
Functionfrom
thesetStothesetT,8
Image
oftheelementsunderthefunctionf,8
Theidentityfunctionon S,9
Inverseimage
oftunderf,10
Inverseimage ofasubsetAofTunderf:S
~T,10
Compositionorproductoffunctionsfandg,11,18
Setof1-1mappingsfromasetStoS, 16
Symmetricgroup ofdegreen,16,109
nfactorial,17
Setofintegers, 21
Orbitofsrelativetomappingf,21
xiii

13. xivSymbolList
N
mIn
m~n
(a,b)
C
i,-i
z=a+bi
z=a-bi
lIz
Izl
r(cos9+isin9)
9
n
Q
En
IGI
C(a)
(a)
Z(G)
a---b
a==bmodn
a==ben)
[a]
cl(a)
o(a)
iG(H)
7L
n
Un
'P(n)
Hb
aH
G=G'
'P(G)
Ker'P
N<JG
GIN
AB
G
1
XG
2
X...XG
n
(
a
b c)
u v w
(a,b,,c)
An
ao+ali+a2j+a3k
detx
Setofpositiveintegers, 21
mdividesn,22
mdoesnotdivide n,22
Greatestcommondivisorof a,b(seealsoabove), 23
Setofcomplexnumbers,32
Squarerootsof
-1,32
Complexnumber zwithrealpart aandimaginary
part
b,32
Conjugateofcomplexnumber z=a+bi,32
Inverseofthecomplexnumber z,33
Absolutevalueofcomplexnumber z,34
Polarformofacomplexnumber, 35
Primitiventhrootofunity,36, 42
Setofrationalnumbers, 42
Groupofnthrootsofunity, 42
OrderofthegroupG, 42
Centralizerof ainG,53,102
Cyclicgroupgeneratedby
a,53
CenterofgroupG, 53
aisequivalentto binaspecifiedsense, 57
aiscongruentto bmodulon(longform), 57
aiscongruentto bmodulon(shortform),
;'7
Classofallbequivalentto a,58
Conjugacyclassof a,58,101
Orderofelementaofagroup, 60
Indexof HinG,59
Setofintegersmod n,61
Groupofinvertibleelementsof 7L
n
,62
Euler'Pfunction(phifunction), 62
Rightcoset ofsubgroupH,58
LeftcosetofsubgroupH, 64
GroupG isisomorphictogroup G',68
Imageofhomomorphism,70
Kernelofthehomomorphisms'P,70,140
NisanormalsubgroupofG, 72
QuotientofagroupGbyasubgroup N,78
Productofsubsets A,Bofagroup, 79
DirectproductofG},G
2
,
•••,G
n
,93
Permutationsending atou,btov,...,ctow,110
Cyclesending atob,...,ctoa,111
Alternatinggroupofdegree n,121,215
Quaternion,131
Determinateofthe 2x2matrixx,136

14. H(F)
R(f)S
(a)
F[x]
degp(x)
(g(x))
g(x)If(x)
R[x]
F(x)
vEV
av
atvt+ ·..+anVn
(VbV2'• • • ,Vn)
V(f)W
dimF(V)
U+W
[K:F]
F[a]
F(a)
E(K)
f'(x)<f>n(X)
SymbolList xv
Ringofquaternionsover F,136
Directsumofrings R,S,146
Idealgeneratedby ainacommutativering, 145
Polynomialringoverthefield F,152
Degreeofpolynomial p(x),153
Idealgeneratedby g(x)inapolynomialring, 157
Polynomialg(x)dividesf(x),157
Polynomialringoverring R,163
Fieldofrationalfunctionsin xoverF,177
VectorVinavectorspace V,180
Scalaratimesvector v,180
Linearcombinationofvectors Vt,•••,V
n
,181
SUbspacespannedby Vt,V2'•••,Vn,181
Directsum ofvectorspaces V,W,181
Dimensionof VoverF,186
SumofsubspacesU,WofV,190
Degreeof KoverF,191
Ringgeneratedby aoverF,196
Fieldextensionobtainedbyadjoiningto atoF,196
Fieldofalgebraicelementsof KoverF,198
Formalderivativeofpolynomialf(x), 227
nthcyclotomicpolynomial,230

15. ABSTRACTALGEBRA

16. 1
THINGSFAMILIAR
ANDLESSFAMILIAR
1.AFEWPRELIMINARY REMARKS
Formanyr9adersthisbookwillbetheirfirstcontactwithabstractmathe­
matics.
l'he·subjecttobediscussed isusuallycalled"abstractalgebra,"but
thedifficultiesthatthereadermayencounterarenotsomuchduetothe"al­
gebra"part
astheyaretothe"abstract"part.
Onseeingsomeareaofabstractmathematicsforthefirsttime,beitin
analysis,topology,orwhat-not,thereseemstobeacommonreactionforthe
novice.Thiscanbestbedescribedbyafeelingofbeingadrift,ofnothaving
somethingsolidtohangonto.This
isnottoosurprising,forwhilemanyofthe
ideasarefundamentallyquitesimple,theyaresubtleandseemtoeludeone's
graspthefirsttimearound.Onewaytomitigatethisfeelingoflimbo,orasking
oneself"What
isthepointofallthis?," istotaketheconceptathandandsee
what
itsaysinparticularcases.Inotherwords,thebestroadtogoodunder­
standingofthenotionsintroduced
istolookatexamples.This istrueinallof
mathematics,butit
isparticularlytrueforthesubjectmatterofabstractalgebra.
Canone,withafewstrokes,quicklydescribetheessence,purpose,and
backgroundforthematerialweshallstudy?Let'sgiveitatry.
Westartwithsomecollectionofobjects
Sandendowthiscollection
withanalgebraicstructurebyassumingthatwecancombine,inoneorsev­
eralways(usuallytwo),elementsofthisset
Stoobtain,oncemore,elements
ofthisset
S.Thesewaysofcombiningelementsof SwecalloperationsonS.

17. 2 ThingsFamiliarandLessFamiliar Ch.1
Thenwetrytoconditionorregulatethenatureof Sbyimposingcertain
rulesonhowtheseoperationsbehaveon
S.Theserulesareusuallycalledthe
axiomsdefiningtheparticularstructureon S.Theseaxiomsareforustode­
fine,butthechoicemadecomes,historicallyinmathematics,fromnoticing
thattherearemanyconcretemathematicalsystemsthatsatisfytheserulesor
axioms.Weshallstudysomeofthebasicaxiomaticalgebraicsystemsinthis
book,namely
groups,rings, andfields.
Ofcourse,onecouldtrymanysetsofaxiomstodefinenewstructures.
Whatwouldwerequireofsuchastructure?Certainlywewouldwantthat
theaxiomsbeconsistent,that
is,thatweshouldnotbeledtosomenonsensi­
calcontradictioncomputingwithintheframeworkoftheallowablethingsthe
axiomspermitustodo.Butthat
isnotenough.Wecaneasilysetupsuchal­
gebraicstructuresbyimposingasetofrulesonaset
Sthatleadtoapatho­
logicalorweirdsystem.Furthermore,theremaybevery
fewexamplesof
somethingobeyingtheruleswehavelaiddown.
Timehasshownthatcertainstructuresdefinedby"axioms"playan
im­
portantroleinmathematics(andotherareas aswell)andthatcertainothers
areofnointerest.Theoneswementionedearlier,namelygroups,rings,and
fields,havestoodthetestoftime.
Awordabouttheuseof"axioms."Ineverydaylanguage"axiom"
meansaself-evidenttruth.Butwearenotusingeveryday
languige;weare
dealingwithmathematics.
Anaxiomisnotauniversaltruth-butoneofsev­
eralrulesspellingoutagivenmathematicalstructure.Theaxiom
istruein
thesystemwearestudyingbecausewehaveforcedittobetruebyhypothe­
sis.It
isalicense,intheparticularstructure,todocertainthings.
Wereturntosomethingwesaidearlieraboutthereactionthatmany
studentshaveontheirfirstencounterwiththiskindofalgebra,namelyalack
offeelingthatthematerial
issomethingtheycangettheirteethinto.Donot
bediscouragediftheinitialexposureleavesyouinabitofa
fog.Stickwith
it,trytounderstandwhatagivenconceptsays,andmost importantly,lookat
particular,concreteexamplesoftheconceptunderdiscussion.
PROBLEMS
1.LetSbeasethavinganoperation *whichassignsanelement a*bofS
foranya,bES.Letusassumethatthefollowingtworuleshold:
1.Ifa,bareanyobjectsin S,thena*b=a.
2.Ifa,bareanyobjectsin S,thena*b=b*a.
Showthat Scanhaveatmostoneobject.

18. Sec.2 Set Theory 3
2.LetSbethesetofallintegers0,±1,±2,...,±n,....Fora,binSdefine
*bya*b=a-b.Verifythefollowing:
(a)a*b=1=b*aunlessa=b.
(b)(a*b)*c=1=a*(b*c)ingeneral. Underwhatconditionsona,b,cis
(a*b)*c=a*(b*c)?
(c)Theinteger0hasthepropertythata*0=aforeveryainS.
(d)ForainS,a*a=O.
3.LetSconsistofthetwoobjectsDand~.Wedefinetheoperation*onS
bysubjectingDand~tothefollowingconditions:
1.D*~=~=~*D.
2.D*D=D.
3.~*~=D.
Verifybyexplicitcalculation thatifa,b,careanyelementsofS(i.e.,a,b
andccanbeanyofDor~),then:
(a)a*bisinS.
(b)(a*b)*c=a*(b*c).
(c)a*b=b*a.
(d)ThereisaparticularainSsuch thata*b=b*a=bforallbinS.
(e)GivenbinS,thenb*b=a,whereaistheparticularelementinPart
(d).
2.SETTHEORY
Withthechangesin themathematicscurriculumin theschoolsin theUnited
States,manycollegestudentshavehadsomeexposuretosettheory.Thisin­
troduction
tosettheoryintheschoolsusuallyincludes theelementaryno­
tions
andoperationswithsets. Goingontheassumptionthatmanyreaders
willhave someacquaintancewithsettheory,weshallgivea rapidsurveyof
thoseparts ofsettheorythatweshallneedinwhatfollows.
First,however,we
needsomenotation.Toavoidtheendlessrepetition
ofcertainphrases,we introduceashorthandforthesephrases.LetSbea
collection
ofobjects;theobjectsofSwecall theelementsofS.Todenote
thatagivenelement,a,isanelementofS,wewrite aES-thisisread"ais
anelementofS."Todenotethecontrary,namelythatanobjectaisnotan
elementofS,wewrite a
fl.S.So,forinstance,ifS denotesthesetofallposi­
tiveintegers
1,2,3,...,n,...,then165ES,whereas-13
f£S.
Weoftenwanttoknoworprovethatgiventwosets SandT,oneof
theseisapartoftheother.WesaythatSisa subsetofT,whichwewrite
SeT(read"SiscontainedinT")ifeveryelementofSisanelementofT.

19. 4 ThingsFamiliarandLessFamiliar Ch.1
Intermsofthenotationwenowhave: SeTifsE Simpliesthat sET.We
canalsodenotethisbywriting
T
=>S,read"TcontainsS."(Thisdoesnotex­
cludethepossibilitythatS
=T,thatis,thatSandThaveexactlythesame
elements.)Thus,if
TisthesetofallpositiveintegersandS isthesetofall
positiveevenintegers,then
SeT,andS isasubsetof T.Inthedefinition
givenabove,S
=>SforanysetS;thatis,Sisalwaysasubsetofitself.
Weshallfrequentlyneedtoshowthattwosets
SandT,definedper­
hapsindistinctways,areequal,thatis,theyconsistofthesamesetofele­
ments.Theusualstrategyforprovingthis
istoshowthatboth SeTand
TCS.Forinstance,ifS isthesetofallpositiveintegershaving6asafactor
and
Tisthesetofallpositiveintegershavingboth2and3asfactors,then
S
=T.(Prove!)
Theneedalsoarisesforaverypeculiarset,namelyonehavingnoele­
ments.Thisset
iscalledthe nulloremptysetandisdenotedby 0.Ithasthe
propertythatit
isasubsetof anysetS.
LetA,Bbesubsetsofagivenset S.Wenowintroducemethodsofcon­
structing
othersubsetsofSfromAandB.Thefirstofthese istheunionofA
andB,writtenAUB,whichisdefined:AUBisthatsubsetofSconsisting
ofthoseelementsofSthatareelementsof
Aorareelementsof B.The"or"
wehavejustused
issomewhatdifferentinmeaningfromtheordinaryusage
oftheword.Herewemeanthatanelementc isinAUBifitisinA,orisin
B,orisinboth.The"or"isnotmeanttoexcludethepossibilitythatboth
thingsaretrue.Consequently,forinstance,A
UA=A.
IfA={I,2,3}andB={2,4,6,IO},thenAUB={I,2,3,4,6,IO}.
Wenowproceedtooursecondwayofconstructingnewsetsfromold.
Againlet
AandBbesubsetsofaset
S;bytheintersectionofAandB,writ­
ten
AnB,weshallmeanthesubset ofSconsistingofthoseelementsthat
arebothin
AandinB.Thus,intheexampleabove, AnB={2}.Itshould
beclearfromthedefinitionsinvolvedthat
AnBCAandAnBcB.
Particularexamplesofintersectionsthatholduniversallyare: AnA=A,
AnS=A,An0=0.
Thisisanopportunemomenttointroduceanotationaldevicethatwill
beusedtimeaftertime.Givenaset S,weshalloftenbecalledontode­
scribethesubset
AofS,whoseelementssatisfyacertainproperty P.We
shallwritethisas
A={sESissatisfiesPl.Forinstance,if A,Baresubsets
ofS,thenAUB={sESisEAorsEB}whileAnB={sESisEA
andsEB}.
Althoughthenotionsofunionandintersectionofsubsets ofShave
beendefinedfortwosubsets,it isclearhowonecandefinetheunionandin­
tersection
ofanynumberofsubsets.
Wenowintroduceathirdoperationwecanperformonsets,the
differ­
ence
oftwosets.If A,BaresubsetsofS,wedefineA-B ={aEAIa
f£B}.

20. Sec.2 Set Theory 5
SoifAisthesetofallpositiveintegersand Bisthesetofallevenintegers,
then
A-B isthesetofallpositiveoddintegers.Intheparticularcasewhen
Aisasubsetof S,thedifferenceS -A iscalledthe complementofAinS
andiswrittenA'.
Werepresentthesethreeoperationspictorially.If Ais
®andBis®,
then
1.AUB=.istheshadedarea.
2.AnB=GBistheshadedarea.
3.A-B =G1Vistheshadedarea.
4.B-A =~ istheshadedarea.
Notetherelationamongthethreeoperations,namelytheequality
AUB=(AnB)U(A-B)U(B-A).Asanillustrationofhowonegoes
aboutprovingtheequalityofsetsconstructedbysuchset-theoreticconstruc­
tions,
we
pfovethislatterallegedequality.Wefirstshowthat (AnB)U
(A-B)U(B-A)CAUB;thispartiseasyfor,bydefinition, AnBCA,
A-B CA,andB-A CB,hence
(AnB)U(A- B)U(B-A)CAUAUB=AUB.
Nowfortheotherdirection,namelythat AUBC(AnB)U(A-B)U
(B-A).GivenIIEAUB,ifuEAanduEB,thenuEAnB,soitiscer­
tainlyin
(AnB)U
(.1-1-B)U(B- A).Ontheotherhand,if uEAbut
uft.B,then,bythev~rydefinitionof A-B, uEA-B, soagainit iscer­
tainlyin
(AnB)U(A-B)U(B-A).Finally,if uEBbutu
$A,then
uEB-A,soagainit isin(AnB)U(A-B)U(B-A).Wehavethus
coveredallthepossibilitiesandhaveshownthat
AUBC(AnB)U
(A- B)U(B-A).Havingthetwooppositecontainingrelationsof AUB
and(AnB)U(A-B)U(B-A),weobtainthedesiredequalityofthese
twosets.
Weclosethisbriefreviewofsettheorywithyetanotherconstruction
wecancarryoutonsets.This istheCartesianproduct definedforthetwo
sets
A,BbyAXB={(a,b)IaEA,bEB},wherewe declaretheordered
pair
(a,b)tobeequaltotheorderedpair
(abb
t
)ifandonlyif a=atand
b=b
I
.Here,too,weneednotrestrictourselvestotwosets;forinstance,we

21. 6 ThingsFamiliarandLessFamiliar Ch.1
candefine,forsets A,B,C,theirCartesianproductasthesetofordered
triples(a,b,c),whereaEA,bEB,CECandwhereequalityoftwoor­
deredtriplesisdefinedcomponent-wise.
PROBLEMS
EasierProblems
1.Describethefollowingsetsverbally.
(a)S=={Mercury,Venus, Earth,...,Pluto}.
(b)S=={Alabama,Alaska, ...,Wyoming}.
2.Describe
thefollowingsetsverbally.
(a)S=={2,4,6,8,...}.
(b)S =={2,4,8,16,32,...}.
(c)S=={I,4,9,16,25,36,...}.
3.IfAisthesetofallresidentsoftheUnitedStates,BthesetofallCana­
diancitizens,
andCthesetofallwomenintheworld,describethesets
AnBnC,A-B,A- C,C -Averbally.
~
4.IfA=={I,4,7,a}andB=={3,4,9,II}andyouhave beentoldthat
AnB=={4,9},whatmustabe?
5.IfACBandBCC,provethatACC.
6.IfACB,provethatAUCCBUCforanyset C.
7.ShowthatAUB==BUAandAnB==BnA.
8.Provethat(A-B)U(B-A)==(AUB)-(AnB).Whatdoesthis
looklikepictorially?
9.
ProvethatAn(BUC)==(AnB)U(AnC).
10.Prove thatAU(BnC)==(AUB)n(AUC).
11.Writedownall thesubsetsofS=={I,2,3, 4}.
Middle-LevelProblems
*12.IfCisasubsetofS,letC'denotethecomplementofCinS.Provethe
DeMorganRules forsubsetsA,BofS,namely:
(a)(AnB)'==A'UB'.
(b)(AUB)'==A'nB'.
*13.LetSbeaset.Foranytwosubsets ofSwedefine
A+B==(A-B)U(B-A)andA·B==AnB.

22. Sec.2 SetTheory 7
Provethat:
(a)A+B=B+A.
(b)A+0=A.
(c)A·A=A.
(d)A+A=0.
(e)A+(B+C)=(A+B)+C.
(f)IfA+B=A+C,thenB=C.
(g)A. (B+C)=A·B+A.C.
*14.IfCisafiniteset,let m(C)denotethenumberofelementsinC.IfA,B
arefinitesets, provethat
meAUB)=meA)+m(B)-meAnB).
15.Forthreefinitesets A,B,Cfinda formulaformeAUBUC).(Hint:
Firstconsider D=BUCandusetheresultofProblem14.)
16.Takeashot
atfindingmeAlUA
2U...UAn)fornfinitesets AI,A
2
,
•••,
An·
17.UsetheresultofProblem14toshowthatif800/0ofallAmericanshave
gonetohighschool and70%ofallAmericansreadadailynewspaper,
thenatleast500/0ofAmericanshavebothgonetohighschool andreada
dailynewspaper.
18.A
pubircopinionpollshows that930/0ofthepopulationagreedwiththe
governmentonthefirstdecision, 84%onthesecond,and740/0onthe
third,for threedecisionsmadebythegovernment.Atleastwhatper­
centageofthepopulationagreedwiththegovernmentonallthreedeci­
sions?(Hint:
UsetheresultsofProblem15.)
19.Inhis
bookATangledTale, LewisCarrollproposedthefollowingriddle
aboutagroupofdisabledveterans: "Saythat700/0havelost aneye,750/0
anear,800/0anarm,85%aleg.Whatpercentage,atleast,musthavelost
all
four?"SolveLewis Carroll'sproblem.
*20.Show,forfinitesets A,B,thatmeAXB)=m(A)m(B).
21.IfSisasethavingfiveelements:
(a)Howmanysubsetsdoes Shave?
(b)Howmanysubsetshaving fourelementsdoesShave?
(c)Howmanysubsetshavingtwo elementsdoesShave?
HarderProblems
22.(a)Showthatasethavingnelementshas2
n
subsets.
(b)If0<m<n,howmanysubsetsaretherethathaveexactlymele­
ments?

23. 8 ThingsFamiliarandLessFamiliar Ch.1
3.MAPPINGS
Oneofthetrulyuniversalconcepts thatrunsthroughalmosteveryphase of
mathematicsis thatofafunctionormappingfromonesettoanother.One
couldsafelysay thatthereisnopartofmathematicswhere thenotiondoes
notariseorplayacentralrole. Thedefinitionofafunctionfrom onesetto
anothercanbegiveninaformalwayinterms ofasubsetoftheCartesian
productofthesesets.Instead,here,weshallgiveaninformal andadmittedly
nonrigorousdefinition
ofamapping(function)from onesettoanother.
LetS,Tbesets;a functionormappingffromStoTisarulethatas­
signsto
eachelementsESauniqueelementtET.Let'sexplainalittle
morethoroughlywhatthismeans. Ifsisagiven elementofS,thenthereis
onlyoneelementtinTthatisassociatedtosbythemapping.Assvaries
over
S,tvariesoverT(inamannerdependingons).Notethatbythedefini­
tiongiven,
thefollowingisnotamapping.LetSbethesetofallpeoplein
theworldandTthesetofallcountriesin theworld.Letfbetherulethatas­
signstoevery
personhisorhercountryofcitizenship.Thenfisnotamap­
pingfromS
toT.Whynot?Because therearepeoplein theworldthatenjoy
adualcitizenship;forsuchpeople
therewouldnotbeauniquecountryofcit­
izenship.Thus,if
MaryJonesisbothanEnglishandFrench
citiz~,fwould
notmakesense,asamapping,whenappliedto MaryJones.Ontheother
hand,therulef:IR~IR,whereIRisthesetofrealnumbers,definedby
f(a)=a
2
foraEIR,isaperfectlygoodfunctionfromIRtoIR.Itshouldbe
notedthatf(-2)=(-2)2=4=f(2),andf(-a)=f(a)forallaEIR.
WedenotethatfisamappingfromS toTbyf:S~Tandforthe
tETmentionedabovewewrite t=f(s);wecallttheimageofsunderf.
Theconceptishardlyanew oneforany ofus.Sincegradeschoolwe
haveconstantly
encounteredmappingsandfunctions,oftenin theformof
formulas.Butmappingsneednotberestrictedtosets ofnumbers.Aswesee
below,theycanoccurinanyarea.
Examples
1.lletS
={allmenwhohave everlived}andT={allwomenwhohaveever
lived}.Define
f:S
~Tbyf(s)=motherofs.Therefore,f(JohnF.Ken­
nedy)=RoseKennedy, andaccordingtoourdefinition,RoseKennedy is
theimageunderfofJohnF.Kennedy.
2.LetS={alllegallyemployedcitizens oftheUnitedStates}andT={posi­
tiveintegers}.Define,for
sES,f(s)byf(s)
=SocialSecurity Numberofs.
(Forthepurposeofthistext,letusassume thatalllegallyemployedcitizens
oftheUnitedStateshaveaSocialSecurityNum~er.) Thenfdefinesamap­
pingfromS
toT.

24. Sec.3 Mappings 9
3.LetSbetheset ofallobjectsforsaleinagrocerystoreandlet T={all
realnumbers}.Define f:S~Tbyf(s)=priceofs.Thisdefinesamapping
fromSto
T.
4.LetSbetheset ofallintegersandlet T
=S.Definef:S~Tbyf(m)=
2mforanyinteger m.Thustheimageof6underthismapping, f(6),isgiven
by
f(6)
=2 . 6=12,whilethatof-3,f(-3),isgivenby f(-3)=2(-3)=
-6.IfSJ,S2ESarein Sandf(Sl)=f(S2),whatcanyousayabout SlandS2?
5.LetS=Tbetheset ofallrealnumbers;define f:S~Tbyf(s)=S2.
Doeseveryelement ofTcomeupasanimage ofsomesES?Ifnot,how
wouldyoudescribetheset
ofallimages{f(s)Is
ES}?Whenisf(s1)=
f(S2)?
6.LetS=Tbetheset ofallrealnumbers;define f:S~Tbyf(s)=S3.This
isafunctionfromSto T.Whatcanyousayabout {f(s)IsES}?Whenis
f(s1)=f(S2)?
7.LetTbeanynonemptysetandletS=TXT,theCartesianproductof T
withitself.Definef:TXT~Tbyf(t
1
,t2)=t
1
•Thismappingfrom TXT
toTiscalledthe projectionofTXTontoitsfirstcomponent.
8.LetSbetheset ofallpositiveintegersandlet Tbethesetofallpositive
rationalnumbers.Define
f:SXS
~Tbyf(m,n)=min.Thisdefinesa
mappingfromS
XStoT.Notethatf(l,2)
=~whilef(3,6)=~=~=
f(l,2),although(1,2)=1=(3,6).Describethesubset ofSXSconsistingof
those(a,b)suchthatf(a,b)=~.
ThemappingstobedefinedinExamples9 and10aremappingsthat
occurforanynonemptysetsand
playaspecialrole.
9.LetS,Tbenonemptysets,andlet tobeafixedelement ofT.Define
f:S
~Tbyf(s)=t0foreverysES;fiscalleda constantfunctionfrom
StoT.
10.LetSbeanynonemptysetanddefine i:S~Sbyi(s)=sforevery
sES.WecallthisfunctionofStoitselfthe identityfunction (oridentitymap­
ping)
onS.Wemay,attimes,denoteitby is(andlaterinthebook, bye).
Nowthatwehavethenotion ofamappingweneedsomeway ofidenti­
fyingwhentwomappingsfromonesetto
anotherareequal.This isnot
Godgiven;it isforustodecidehow todeclaref
=gwheref:S~Tand
g :S~T.Whatismorenaturalthantodefinethisequalityviatheactions of
fandgontheelements ofS?Moreprecisely,wedeclare thatf=gifand
onlyif
f(s)
=g(s)foreverysES.IfSisthesetofallrealnumbersand fis
definedonSby f(s)=S2+2s+1,whilegisdefinedonSbyg(s)=
(s+1)2,ourdefinitionoftheequalityoffandgismerelyastatement ofthe
familiaridentity
(s+1)2
=S2+2s+1.

25. 10 ThingsFamiliarandLessFamiliar Ch.1
Havingmadethedefinitionofequalityoftwomappings,wenowwant
tosingleoutcertaintypes ofmappingsby thewaytheybehave.
Definition.Themapping[:S~TisontoorsurjectiveifeverytET
istheimageunder[ofsomesES;thatis,ifandonlyif,giventET,there
existsan
s
ESsuchthatt=[(s).
Intheexampleswegaveearlier,in Example1themappingisnotonto,
since
noteverywomanthateverlivedwas themotherofamalechild.Simi­
larly,inExample2
themappingisnotonto,for noteverypositiveinteger is
theSocialSecurity NumberofsomeU.S.citizen. ThemappinginExample4
failsto
beontobecausenoteveryinteger iseven;andinExample5,again,
themappingisnotonto,for thenumber-1,forinstance,isnotthesquareof
anyrealnumber.However,themappinginExample6 isontobecauseevery
real
numberhasauniquerealcuberoot. Thereadercandecidewhether or
notthegivenmappings areontointheotherexamples.
Ifwedefine[(S)={[(s)
ETisES},anotherwayofsayingthatthe
mapping
[:S
~Tisontoisbysayingthat[(S)=T.
Anotherspecifictype ofmappingplaysan importantandparticular
roleinwhatfollows.
Definition.A mapping[:S
~Tissaidto beone-to-one(written1-1)
orinjectiveiffor SI=1=S2inS,[(SI)=1=[(S2)inT.Equivalently,[is1-1if
[(SI)=[(S2)impliesthatSI=S2.
Inotherwords,amapping is1-1ifittakesdistinctobjectsintodistinct
images.Intheexamples
ofmappingswegaveearlier,themapping ofExample
1
isnot1-1,sincetwobrotherswouldhavethesamemother.HoweverinEx­
ample2themapping
is1-1becausedistinctU.S.citizenshavedistinctSocial
Securitynumbers(provided
thatthereisnogoof-upinWashington,which is
unlikely).Thereadershouldcheckifthevarious otherexamplesofmappings
are
1-1.
Givenamapping [:S
~TandasubsetACT,wemaywanttolook at
B={sES1[(s)EA};\veusethenotation[-l(A)forthisset B,andcall
P-l(A)theinverseimage o[Aunder[.Ofparticularinterest is[-l(t),thein­
verseimage
ofthesubset{t}ofTconsistingoftheelementt
ETalone.If
theinverseimage of{t}consistsofonlyoneelement,say sES,wecouldtry
todefine[-I(t)bydefining[-I(t)=s.Aswenotebelow,this neednotbea
mappingfrom
TtoS,butissoif[is1-1andonto.Weshalluse thesameno­
tation
[-Iincasesofbothsubsetsandelements.This [-Idoesnotingeneral
defineamappingfrom
TtoSforseveralreasons.First,if [isnotonto,then

26. Sec.3 Mappings 11
thereissometinTwhichisnottheimageofanyelements,soI-let)=0.
Second,if [isnot1-1,thenforsomet ETthereareatleasttwodistinct
Sl=1=S2inSsuch thatfest)=t=[(S2).SOI-let)isnotauniqueelement of
S-somethingwerequirein ourdefinitionofmapping.However,if [isboth
1-1andontoT,then[-Iindeeddefinesamapping ofTontoS.(Verify!)This
bringsus
toaveryimportantclass ofmappings.
Definition.
Themapping[:S
~Tissaidtobea1-1correspondence
orbijectionif[isboth1-1andonto.
Now
thatwehavethenotionofamappingandhavesingled outvarious
types
ofmappings,wemightverywellask: "Goodandwell,butwhatcanwe
dowiththem?"Asweshallseeina moment,wecanintroduce anoperation
ofcombiningmappingsincertaincircumstances.
Consider
thesituationg : S
~Tand[:T~u.GivenanelementsES,
thengsendsitinto theelementg(s)inT;sog(s)isripeforbeingacted on
byfThuswegetan elementleges»)EU.Weclaimthatthisprocedurepro­
videsuswithamappingfromS
toU.(Verify!)Wedefinethis moreformally
inthe
Defi~ition. Ifg:S~Tand[:T~U,thenthecomposition(orprod­
uct),denot~d by[0g,isthemapping[0g:S~Udefinedby ([0g)(s)
leges)~foreverysES.
Notethattocomposethetwomappings[ andg-thatis,for[0gto
haveany sense-theterminalset, T,forthemappinggmustbetheinitialset
forthemapping f.Onespecialtimewhenwecanalwayscomposeanytwo
mappingsiswhenS=l~=U,thatis,whenwemapSintoitselfAlthough
special,thiscase isoftheutmostimportance.
Weverifyafewproperties
ofthiscompositionofmappings.
Lemma1.3.1.Ifh:S
~T,g:T~U,and[:U~V,then[0(g0h)=
{fog)oh.
ProofHowshallwego aboutprovingthislemma? Toverifythattwo
mappings
areequal,wemerelymustcheck thattheydothesamething to
everyelement. Notefirstofallthatboth[0(g0h)and([0g)0hdefinemap­
pingsfromS
toV,soitmakessense tospeakabouttheirpossibleequality.
Ourtask,then, istoshowthatforevery sES,([0(g0h)(s)
(([0g)0h)(s).Weapplythedefinitionofcompositiontoseethat
([0(g0h))(s)=[((g0h)(s)=[(g(h(s)).

27. 12 ThingsFamiliarandLessFamiliar
Unraveling
((f0g)0h)(s)=(f0g)(h(s»=f(g(h(s»),
wedoindeedseethat
(f0(g0h»(s)=((f0g)0h)(s)
foreverys
ES.Consequently,bydefinition, fo(g0h)=(fog)0h.D
Ch.1
(Thesymbol Dwillalwaysindicatethattheproofhasbeencompleted.)
Thisequality
isdescribedbysayingthatmappings,undercomposition,
satisfytheassociativelaw.Becauseoftheequalityinvolvedthere
isreallyno
needforparentheses,so
wewritefo(g0h)asfogoh.
Lemma1.3.2.If
g:S
~Tandf:T~Uareboth1-1,thenfog:S~U
isalso1-1.
ProofLetussupposethat (fog)(Sl)=(fog)(S2);thus,by definition,
f(g(Sl»
=f(g(S2»·Sincef is1-1,wegetfromthisthatg(Sl) =g(S2);how­
ever,g
isalso1-1,thus Sl=S2follows.Since (fog)(Sl)=(fog)(S2)forces
Sl=S2,themappingfogis1-1.D
WeleavetheproofofthenextRemarktothereader.
Remark.If
g:S
~Tandf:T~Uarebothonto,thenfo g:S~Uis
alsoonto.
Animmediateconsequence ofcombiningtheRemarkandLemma
1.3.2
istoobtain
Lemma1.3.3.If
g:S
~Tandf:T~Uarebothbijections,then
fog : S~Uisalsoabijection.
If
fisa1-1correspondenceofSonto T,thenthe"object" f-
l
:
T
~S
definedearliercaneasilybeshowntobea
1-1mappingof TontoS.Inthis
caseit
iscalledtheinverseof fInthissituation wehave
Lemma1.3.4.If
f:S
~Tisabijection,then fof-
l
=iTandf-
l
0f=
is,whereisand iTaretheidentitymappingsof SandT,respectively.
ProofWeverifyoneofthese.Ift ET,then(fof-l)(t)=f(f-l(t».
Butwhatisf-l(t)?Bydefinition,f-l(t)isthatelement SoESsuchthat

28. Sec.3 Mappings 13
t=f(so).Sof(f-l(t»=f(so)=t.Inotherwords,(fof-l)(t)=tforevery
tET;hencefof-
1
=iT,theidentitymapping onT.D
Weleavethelastresultofthissectionforthe readertoprove.
Lemma1.3.5.
Iff:S
~TandiTistheidentitymapping ofTontoit­
selfand
isisthatofSontoitself,theniT0f=fandfois=f.
PROBLEMS
EasierProblems
1.Forthegivensets S,Tdetermineifamapping f:S
~Tisclearlyandun-
ambiguouslydefined;ifnot,saywhynot.
(a)
S=setofallwomen, T=setofallmen,f(s)=husbandofs.
(b)S=setofpositiveintegers, T=S,f(s)=s-1.
(c)S=setofpositiveintegers, T=setofnonnegativeintegers, f(s)=
s-1.
(d)S =set ofnonnegativeintegers, T=S,f(s)=s-1.
(e)S
~setofallintegers,T=S,f(s)=s-1.
(f)S=~setofallrealnumbers, T=S,f(s)=~.
(g)S=setofallpositiverealnumbers, T=S,f(s)=~.
2.Inthoseparts ofProblem1where fdoesdefineafunction,determineif
it
is1-1,onto,orboth.
*3.Iffisa1-1mappingofSontoT,provethatf-
1
isa1-1mappingofT
ontoS.
*4.Iffisa1-1mappingofSontoT,provethatf-
1
0f=is.
S.Givea proofoftheRemarkafterLemma1.3.2.
*6.Iff:S
~Tisontoandg:T~Uandh:T~Uaresuchthatg0f=h0f,
provethatg=h.
*7.Ifg:S~T,h:S~T,andiff:T~Uis1-1,show thatiffog=foh,
theng=h.
8.LetSbetheset ofallintegersandT={I,-I};f:S~Tisdefinedby
f(s)=1ifsiseven,f(s)=-1ifsisodd.
(a)DoesthisdefineafunctionfromS
toT?
(b)Show thatf(SI+sz)=f(st)f(sz).Whatdoesthissay abouttheinte­
gers?
(c)Is
f(Stsz)=f(SI)f(sz)alsotrue?

29. 14ThingsFamiliarand LessFamiliar Ch.1
9.LetSbetheset ofallrealnumbers.Define f:S~Sbyf(s)=S2,and
g:S~Sbyg(s)=s+1.
(a)Findfog.
(b)Findg0f.
(c)Isfog=gof?
10.Let Sbetheset ofallrealnumbers andfora,bES,wherea=1=0;define
fa,b(S)=as+b.
(a)Showthatfa,b0fc,d=fu,uforsomereal u,v.Giveexplicitvaluesfor
u,vintermsofa,b,C,andd.
(b)Isfa,b0fc,d=fc,d0fa,balways?
(c)Findallfa,bsuchthatfa,b0fl,1=fl,10fa,b'
(d)Showthatf-;,1existsandfinditsform.
11.LetSbe
thesetofallpositive integers.Define f:S
~Sbyf(l)=2,
f(2)=3,f(3)=1,andf(s)=sforanyothersES.Showthat fo fo[=
is.Whatis[-Iinthiscase?
Middle-LevelProblems
12.LetSbetheset
ofnonnegativerationalnumbers, thatis,S={minIm,n
nonnegativeintegers, n
=1=O},andletTbethesetofallintegers.
(a)Doesf:S~Tdefinedby f(mln)=2
m
3
n
definealegitimatefunction
from
StoT?
(b)Ifnot,howcouldyoumodifythedefinition of[soastogetalegiti­
matefunction?
13.LetSbethesetofallpositiveintegers
oftheform 2
m
3
n
,
wherem>0,
n>0,andletTbethesetofallrationalnumbers.Define [:S
~Tby
[(2
m
3
n
)
=min.Provethat fdefinesafunctionfrom StoT.(Onwhat
properties
oftheintegersdoesthisdepend?)
14.Let
[:S
~S,whereS isthesetofallintegers,bedefinedby [(s)=
as+b,wherea,bareintegers.Findthenecessaryandsufficientcondi­
tions
ona,binorderthatf0[=is.
15.Findall [oftheformgiveninProblem14such that[0[0f=is.
16.If[isa1-1mappingofSontoitself,showthat ([-1)-1=f.
17.IfSisafinitesethaving m>0elements,howmanymappingsarethere
ofSintoitself?
18.InProblem17,howmany
1-1mappingsarethere ofSintoitself?
19.
LetSbethesetofallrealnumbers, anddefinef:S
~Sby[(s)
S2+as+b,wherea,barefixedrealnumbers.Provethatfornovalues
ata,bcanfbeontoor1-1.

30. Sec.3 Mappings 15
20.LetSbethesetofallpositiverealnumbers.Forpositivereals a,cand
nonnegativereals b,d,isiteverpossiblethatthemappingf:S~Sde­
finedby
f(s)=(as+b)/(cs+d)satisfiesf
of=is?Findallsucha,b,c,d
thatdothetrick.
21.LetSbethesetofallrationalnumbersandletfa,b:S
~Sbedefinedby
fa,b(S)=as+b,wherea=1=0,barerationalnumbers.Findallfe,dofthis
forms~tisfying fe,d0fa,b=fa,b0fe,dforeveryfa,b·
22.LetSbethesetofallintegersanda,b,crationalnumbers.Define
f:S~Sbyf(s)=as
2
+bs+c.Findnecessaryandsufficientconditions
ona,b,c,sothatfdefinesa mappingonS[Note:a,b,cneednotbeinte­
gers;forexample,
f(s)=
~s(s+1)=~S2+~sdoesalwaysgiveus anin­
tegerforintegrals.]
HarderProblems
23.LetSbethesetofallintegersoftheform2
m
3
n
,
m
2::0,n2::0,andletT
bethesetofallpositiveintegers. Showthatthereisa1-1correspondence
ofSontoT.
24.Provethatthereisa1-1correspondenceofthesetofallpositiveintegers
ontotITe:setofallpositiverationalnumbers.
25.LetSbethesetofallrealnumbersandTthesetofallpositivereals.
Finda1-1mapping fofSontoTsuchthatf(S1+S2)=f(S1)f(S2)forall
SbS2ES.
26.ForthefinProblem25,findf-
1
explicitly.
27.Iff,garemappingsofSintoSandfogisaconstantfunction,then
(a)Whatcanyousayaboutfifgisonto?
(b)WhatcanyousayaboutgiffisI-I?
28.IfSisafinitesetandfisamappingofSontoitself,showthatfmustbe
1-1.
29.IfSisafinitesetandfisa1-1mappingofSintoitself,show thatfmust
besurjective.
30.IfSisafinitesetandfisa1-1mappingofS,showthatforsomeinteger
n>0,
fofofo...of=i
\ I s.
ntimes
31.IfShasmelementsinProblem30,findann>°(intermsofm)that
workssimultaneouslyforall1-1mappings ofSintoitself.

31. 16 ThingsFamiliarandLessFamiliar Ch.1
4.A(S)(THESETOF1-1 MAPPINGS OF5 ONTOITSELF)
Wefocusourattentioninthissection onparticularlynicemappings ofanon­
emptyset,
S,intoitself.Namely,weshallconsider theset,A(S),ofall1-1map­
pings
ofSontoitself.Althoughmost oftheconcernin thebookwillbeinthe
caseinwhich
Sisafiniteset,we donotrestrictourselves tothatsituationhere.
WhenShasafinite numberofelements,say n,thenA(S)hasaspecial
name.
Itiscalledthesymmetricgroup ofdegreen andisoftendenotedbySn'
ItselementsarecalledpermutationsofS.Ifweareinterestedinthestructure
ofSn'itreallydoes notmattermuchwhatourunderlyingsetSis.So,you
can
thinkofSasbeingtheset{I,...,n}.Chapter3willbedevotedtoa
study,in
somedepth,ofSn'Intheinvestigationoffinitegroups, Snplaysa
centralrole.
TherearemanypropertiesofthesetA(S)onwhichwecouldconcen­
trate.
Wehavechosentodevelopthoseaspectsherewhichwillmotivate the
notionofagroupandwhichwillgive thereadersomeexperience, andfeel­
ingfor,workingina
group-theoreticframework.Groupswillbediscussedin
Chapter2.
Webeginwitharesult thatisreallya compendiumofsomeofthere­
sults
obtainedinSection3.
Lemma1.4.1.A(S)satisfiesthefollowing:
(a)
f,g
EA(S)impliesthatfogEA(S).
(b)f,g,hEA(S)impliesthat(fog)0h=fo(g0h).
(c)Thereexistsanelement-theidentitymapping i-suchthatfoi
iof=fforeveryfEA(S).
(d)Given f
EA(S),thereexistsagEA(S)(g=f-l)suchthatfog=
gof=i.
ProofAllthesethingsweredoneinSection3, eitherinthetextmate­
rial
orintheproblems.Weleaveit tothereadertofindtherelevantpartof
Section3 thatwillverifyeach ofthestatements(a)through(d).D
WeshouldnowliketoknowhowmanyelementsthereareinA(S)
whenSisafinitesethaving nelements.Todoso,wefirst makeaslightdi­
gression.
Suppose
thatyoucan doacertainthingin rdifferentways andasec­
ondindependentthingin sdifferentways. Inhowmanydistinctwayscan
you
doboththingstogether?Thebestwayoffindingoutistopicturethisin

32. Sec.4 A(S)(TheSet of1-1MappingsofSOntoItself) 17
aconcretecontext.Suppose thattherearerhighwaysrunningfromChicago
toDetroitandshighwaysrunningfrom DetroittoAnnArbor.Inhowmany
wayscanyougofirst toDetroit,thentoAnnArbor?Clearly,forevery road
youtakefromChicago toDetroityouhave swaysofcontinuingontoAnn
Arbor.YoucanstartyourtripfromChicagoin rdistinctways,henceyou
cancompleteitin
IS+S+S+...+s,=rs
rtimes
differentways.
It
isfairlyclear thatwecan extendthisfromdoingtwo independent
thingstodoingmindependentones,foraninteger m>2.Ifwecandothe
firstthingsin rIdistinctways, thesecondin r2ways,, themthinrmdis-
tinctways,
thenwecandoallthesetogetherinrlr2r mdifferentways.
Let'srecallsomething
manyofushavealreadyseen:
Definition.
Ifnisapositiveinteger, thenn!(read"nfactorial")isde­
finedby
n!=1·2. 3...n.
Lemma1.4.2.IfShasnelements,thenA(S)hasn!elements.
Proof:LetfEA(S),whereS={XbX2,...,x
n
}.Howmanychoices
does
fhaveasaplacetosendxI?Clearlyn,forwecansend xIunderftoany
elementofS.ButnowfisnotfreetosendX2anywhere,forsince fis1-1,we
musthave
f(XI)
=1=f(X2).Sowecan sendX2anywhereexcept ontof(XI).
Hencefcansend X2inton-1differentimages.Continuingthisway,wesee
thatfcansend Xiinton-(i- 1)differentimages. Hencethenumberof
suchf'sisn(n-1)(n- 2)..·1=n!D
Example
Thenumbern!getsverylargequickly. Tobeabletoseethepictureinitsen­
tirety,welook
atthespecialcase n=3,wheren!isstillquitesmall.
Consider
A(S)=S3'whereSconsistsofthethreeelements Xl,X2,X3.We
listalltheelements ofS3'writingouteachmappingexplicitlybywhatitdoes
toeach
ofXl,X2,X3.
1.
i:XI~XbX2~X2,X3~X3.
2. f:XI~X2,X2~X3,X3~XI.
3.g:Xl~X2,X2~XbX3~X3·
4.gof:XI~XbX2~X3,X3~X2. (Verify!)

33. 18 ThingsFamiliarandLessFamiliar
5.fog:Xl~X3,X2~X2,X3~Xl.(Verify!)
6.fof:Xl~X3'X2~Xl,X3~X2.(Verify!)
Ch.1
Sincewehavelistedheresixdifferentelements ofS3,andS3hasonly
sixelements,wehaveacompletelist
ofalltheelements ofS3.Whatdoes
thislisttellus?
Tobeginwith,we notethatfog
=1=gof,soonefamiliarrule
ofthekindofarithmeticwehave beenusedtoisviolated.Sinceg ES3and
g
ES3 ,wemusthave gogalsoin S3.Whatisit?Ifwecalculategog,we
easilyget
gog=i.Similarly,weget
(f0g)0(f0g)=i=(g0f)0(g0f).
Notealso thatfo(fof)=i,hencef-
l
=fof.Finally,weleaveittothe
readertoshowthatg0f=f-
l
0g.
Itisalittlecumbersome towritethisproductin A(S)usingthe o.From
nowonweshalldropitandwrite fogmerelyasfg.Also,weshallstartusing
theshorthandofexponents,toavoidexpressionslike fo fofo...0f.Wede­
fine,forf
EA(S),fa=i,f2=fof=ff,andsoon.Fornegativeexponents
-nwedefinef-nbyf-n=(f-l)n,wheren isapositiveinteger.Theusual
rules
ofexponentsprevail,namely frfs=fr+sand(fr)s=frs.Weleave
theseas
exercises-somewhattediousonesat that-forthereader.
Example
Donotjumptoconclusionsthatallfamiliarproperties ofexponentsgoover.
Forinstance,in theexampleofthef,gES3definedabove,weclaimthat
(fg)2
=1=f2g2.Toseethis,we notethat
sothat(fg)2:Xl~Xl,X2~X2,X3~X3,thatis,(fg)2=i.Ontheother
hand,f2=1=iandg2=i,hencef2g2=f2=1=i,whence(fg)2=1=f2g2inthiscase.
However,some
otherfamiliarproperties dogoover.Forinstance,if
f,g,h
areinA(S)andfg=fh,theng=h.Why?Because,fromfg =fhwe
have
f-l(fg)=f-l(fh);therefore,g =ig=(f-If)g=f-l(fg)=f-l(fh)=
(f-If)h=ih=h.Similarly,gf=hfimpliesthatg=h.Sowecancancelan
elementinsuch
anequationprovided thatwedonotchangesides.In S3our
f,gsatisfygf=f-lg,butsincef
=1=/-1wecannotcanceltheghere.
PROBLEMS
Recallthatfgstandsfor fogand,also,what fmmeans.S,withoutsubscripts,
willbeanonemptyset.

34. Sec.4
EasierProblems
A(S)(TheSet of1-1MappingsofSOntoItself) 19
1.IfSl=1=S2areinS,showthatthereisanfEA(S)suchthatf(sl)=S2.
2.IfSlES,letH==={fEA(S)If(Sl)=sd.Showthat:
(a)iEH.
(b)Iff,gEH,thenfgEH.
(c)IffEH,thenf-
1
EH.
3.SupposethatSl=1=S2areinSandf(Sl)=S2,wherefEA(S).ThenifHis
asinProblem2 andK={gEA(S)Ig(S2)=S2},showthat:
(a)
IfgEK,thenf-lgfEH.
(b)IfhEH,thenthereissomegEKsuchthath=f-lgf.
4.Iff,g,hEA(S),showthat (f-lgf)(f-1hf)=f-l(gh)f.Whatcanyousay
about
(f-lgf)n?
5.Iff,gEA(S)andfg=gf,showthat:
(a)(fg)2=f2g2.
(b)(fg)-l=f-lg-l.
6.Pushtheresult ofProblem5,for thesamefandg,toshowthat(fg)m=
fmgmforallintegers m.
*7.Verify_therulesofexponents,namely frfs=fr+sand(fr)s=frsfor
fEA
(5)andpositiveintegers r,s.
8.Iff,gEA(S)and(fg)2=f2g2,provethatfg=gf.
9.IfS={x},X2,X3'X4},letf,gES4bedefinedby
and
Calculate:
(a)f2,f3,f4.
(b)g2,g3.
(c)fg.
(d)gf.
(e)(fg)3,(gf)3.
(f)f-
1
,
g-l
10.IffES3,showthatf6=i.
11.Canyoufindapositiveinteger msuchthatfm=iforallfES4?

35. 20 ThingsFamiliarandLessFamiliar
Middle-LevelProblems
Ch.1
*12.IffESn,showthatthereissomepositiveintegerk,dependingonf,such
thatfk=i.(Hint:Considerthepositivepowersoff.)
*13.Showthatthereisapositiveintegertsuchthatft=iforallfESn.
14.Ifm<n,showthatthereisa1-1mappingF:Sm~SnsuchthatF(fg)=
F(f)F(g)foralII,gESm.
15.IfShasthreeormoreelements,showthatwecanfindf,gEA(S)such
thatfg=1=gf.
16.LetSbeaninfinitesetandletMeA(S)bethesetofallelements
fEA(S)suchthatf(s)=1=sforatmostafinitenumberofsES.Prove
that:
(a)f,gEMimpliesthatfgEM.
(b)fEMimpliesthatf-
l
EM.
17.ForthesituationinProblem16,show,if fEA(S),thatf-
1
Mf=
{f-lgflgEM}mustequalM.
18.LetS~TandconsiderthesubsetVeT)={fEA(S)If(t)ETforevery
tET}.Showthat:
(a)iEVeT).
(b)f,gEVeT)impliesthatfgEVeT).
19.IftheSinProblem18hasnelementsandThasmelements,how
manyelementsarethereinVeT)?Showthatthereisamapping
F:VeT)~SmsuchthatF(fg)=F(f)F(g)forf,gEVeT)andFisonto
Sm·
20.Ifm<n,canFinProblem19everbeI-I?Ifso,when?
21.InSnshowthatthemappingfdefinedby
[i.e.,f(xJ=Xi+lifi<n,f(xn)=xtlcanbewrittenasf=gtg2gn-l
whereeachgiESninterchangesexactlytwoelements ofS={Xl,,xn},
leavingtheotherelementsfixedinS.
HarderProblems
22.IffESn,showthatf=hlh
2
•••hmforsomeh
j
ESnsuchthat
h;=i.
*23.Call anelementinSnatranspositionifitinterchangestwoelements,leav­
ing
theothersfixed.ShowthatanyelementinSnisaproductoftranspo­
sitions.(This sharpenstheresultofProblem22.)
24.
Ifnisatleast3,showthatforsomefinSn,fcannotbeexpressedinthe
formf=g3foranyginSn.

36. Sec.5 TheIntegers 21
25.IffESnissuchthatf=1=ibutf3=i,showthatwecan numberthe
elementsofSinsuchaway thatf(Xl)=X2,f(X2)=X3,f(X3)=Xl,
f(X4)=Xs,f(xs)=X6,f(X6)=X4,...,f(X3k+l)=X3k+2'f(X3k+2)=X3k+3,
f(X3k+3)=X3k+lforsome k,and,forall theotherXtES,f(xt)=Xt.
26.Viewafixedshuffle ofadeckof52cardsasa1-1mapping ofthedeck
ontoitself.Show thatrepeatingthisfixedshuffleafinite(positive)num­
beroftimeswillbring thedeckback toitsoriginalorder.
*27.IffEA(S),call,for sES,theorbitofs(relativeto f)thesetO(s)=
{fj(s)Iallintegersj}.Showthatif s,tES,theneitherO(s)nOCt)=0or
O(s)=OCt).
28.IfS={Xl'X2,...,X12}andfES12isdefinedby f(xt)=Xt+lifi=1,2,...,
11andf(x12)=Xl,findtheorbitsofalltheelementsofS(relativetof).
29.IffEA(S)satisfiesf3=i,showthattheorbitofanyelementofShas
one
orthreeelements.
*30.Recallthataprimenumber isanintegerp>1suchthatpcannotbefac­
toredasaproductofsmallerpositiveintegers.
IffEA(S)satisfiesfP=i,
whatcanyousayaboutthesize oftheorbitsoftheelementsofSrelativeto
f?
Whatpropertyoftheprimenumbersareyouusingtogetyouranswer?
31.Prove
thatifShas morethantwoelements,thentheonlyelements fain
A(S)
~chthatfof=ffoforallfEA(S)mustsatisfyfa=i.
*32.Wesay thatgEA(S)commuteswithfEA(S)iffg=gf.Findall
theelementsin A(S)thatcommutewithf:S~Sdefinedby
f(x1)=X2 ,f(X2)=Xl,andf(s)=sifs=1=XbX2 .
33.InSnshowthattheonlyelementscommutingwith fdefinedby f(Xi)=
Xi+1i"fi<n,f(x
n
)=Xbarethepowersoff,namelyi=fa,f,f2,...,f
n
-I
.
34.ForfEA(S),letC(f)={gEA(S)Ifg=gf}.Provethat:
(a)g,hEC(f)impliesthatghEC(f).
(b)gEC(f)impliesthatg-lEC(f).
(c)C(f)isnotempty.
5.THEINTEGERS
Themathematicalsetmostfamiliar toeverybodyisthatofthepositiveinte­
gers
1,2,...,whichweshalloftencall
N.Equallyfamiliar istheset,7L,ofall
integers-positive,negative,andzero.Because ofthisacquaintancewith7L,
weshallgive herearathersketchysurvey ofthepropertiesof7Lthatweshall
useoftenin
theensuingmaterial.Most oftheseproperties arewellknown to
allof
us;afewarelesswellknown.
Thebasicassumptionwe makeaboutthesetofintegersisthe

37. 22ThingsFamiliarand LessFamiliar Ch.1
Well-OrderingPrinciple. Anynonemptysetofnonnegativeintegers
hasasmallestmember.
Moreformally,whatthisprinciplestates isthatgivenanonemptyset V
ofnonnegativeintegers, thereisanelementVoEVsuchthatVo
~vforevery
vEV.Thisprinciplewillserveas thefoundationfor ourensuingdiscussion
oftheintegers.
Thefirstapplicationwe makeofitistoshowsomethingweallknow
andhavetakenforgranted,namely thatwecandivide oneintegerbyan­
othertogeta remainderthatissmaller.This isknownas Euclid'sAlgorithm.
Wegiveita moreformalstatementandaproofbasedonwell-ordering.
Theorem1.5.1(Euclid'sAlgorithm). Ifmandnareintegerswith
n>0,thenthereexistintegersqandr,with0::5r<n,suchthatm=qn+r.
Proof.LetWbethesetofm-tn, wheretrunsthroughallthe
integers,i.e.,
W={m-tnItE
Z}.NotethatWcontainssomenonnegative
integers,forif
tislargeenoughandnegative,thenm- tn>O.Let
V={vEWIv
2::O};bythewell-orderingprinciple Vhasasmallestelement,
r.SincerEV,r2::0,andr=m-qn forsome q(forthatistheformofall
elementsin
W
~V).Weclaimthatr<n.Ifnot,r=m-qn~n,hence
m-(q+l)n2::O.Butthisputsm-(q +l)ninV,yetm-(q+l)n<r,
contradictingtheminimalnatureofrinV.Withthis,Euclid'sAlgorithm is
proved.D
Euclid'sAlgorithmwillhaveahost ofconsequencesforus,especially
aboutthenotionofdivisibility.Sincewe arespeakingabouttheintegers,
beitunderstoodthatalllettersusedinthissectionwillbeintegers. Thiswill
savealot
ofrepetitionofcertainphrases.
Definition.Givenintegers
m
=1=0andnwesaythatmdividesn, writ­
tenasmin,ifn=cmforsomeinteger c.
Thus,forinstance,2114, (-7)114,41(-16).Ifmin,wecallmadivi­
sor
orfactorofn,andnamultipleofm.Toindicatethatmisnotadivisorof
n,wewritem
tn;so,forinstance,3t5.
Thebasicelementarypropertiesofdivisibilityarelaidoutin
Lemma1.5.2.Thefollowingaretrue:
(a)
11nforalln.
(b)Ifm
=1=0,thenm10.

38. Sec.5
(c)IfminandnIq,thenmIq.
(d)IfminandmIq,thenmI(un+vq)forallu,v.
(e)Ifmil,thenm=1orm=-1.
(f)IfminandnIm,thenm=±n.
TheIntegers 23
ProofTheproofsofallthesepartsareeasy,followingimmediately
fromthedefinition
ofmin.Weleaveall butPart(d)asexercises butprove
Part(d)heretogivetheflavor ofhowsuchproofsgo.
Sosuppose
thatminandmIq.Thenn=cmandq=dmfor somec
and
d.Therefore,un+vq=u(cm)+v(dm)=(uc+vd)m.Thus,from the
definition,mI(un+vq).D
Havingtheconcept ofadivisorofaninteger,wenowwant tointroduce
thatofthegreatestcommondivisoroftwo(ormore)integers.Simply
enough,thisshould
bethelargestpossibleinteger thatisadivisorofbothin­
tegersinquestion.However,wewant
toavoidusing thesizeofaninteger­
forreasonsthatmaybecomeclearmuchlaterwhenwetalk aboutrings.So
wemakethedefinitioninwhatmayseemasastrangeway.
Definition.Given
a,b(notboth0),thentheirgreatestcommondivi­
sor
cisdefinedby:
~
(a)c>o.
(b)cIaandc lb.
(c)Ifdlaanddlb,thendlc.
Wewritethiscasc =(a,b).
Inotherwords,thegreatestcommondivisor
ofaandbisthepositive
numbercwhichdivides
aandbandisdivisiblebyevery dwhichdivides a
andb.
Definingsomethingdoes notguaranteeitsexistence.Soit isincumbent
onustoprove that(a,b)exists,andis,infact,unique. Theproofactually
showsmore,namely
that(a,b)isanicecombination ofaandb.Thiscombi­
nation
isnotunique;forinstance,
(24,9) =3=3 . 9+(-1)24=(-5)9+2 .24.
Theorem1.5.3.Ifa,barenotboth0,thentheirgreatestcommondivi­
sor
c=(a,b)exists,isunique,and,moreover,c =moa+nobforsomesuit­
able
moandno.

39. 24 ThingsFamiliarandLessFamiliar Ch.1
ProofSincenotbothaandbare0,thesetA={ma+nbIm,nEE}
hasnonzeroelements. IfxEAandx<0,then-xisalsoin Aand-x>0,
forifx=mta+ntb,then-x=(-mt)a+(-nt)b,soisinA.ThusAhas
positiveelements;hence,by
thewell-orderingprinciple thereisasmallest
positiveelement,
c,inA.SinceCEA,bytheformoftheelementsof Awe
know
thatc=moa+nobforsome mo,no.
Weclaimthatcisourrequiredgreatestcommondivisor.Firstnotethat
if
dIaanddIb,thendI(moa+nob)byPart(d)ofLemma1.5.2,that is,
dIc.So,toverifythatcisourdesiredelenient,we needonlyshowthatc Ia
andclb.
ByEuclid'sAlgorithm, a=qc+r,where0
::5r<c,thatis,a=
q(moa+nob)+r.Therefore,r=-qnob+(1-qmo)a.SorisinA.But
r<candisinA,sobythechoice ofc,rcannotbepositive.Hence r=0;in
otherwords,a=qcandsocia.Similarly,cIb.
Fortheuniquenessofc,ift>0alsosatisfied tIa,tIbanddItforalld
suchthatdIaanddIb,wewouldhave ticandcit.ByPart(f)ofLemma
1.5.2weget
thatt=c(sincebotharepositive).D
Let'slookatanexplicitexample,namely a=24,b=9.Bydirectexami­
nationweknow
that(24,9) =3;notethat3 =3·9+(-1)24.What is
(-24,9)?
Howisthisdoneforpositivenumbers aandbwhichmaybequitelarge?
Ifb>a,interchangeaandbsothata>b>O.Thenwecanfind (a,b)by
1.observingthat(a,b)=(b,r)wherea=qb+rwitha
::5r<b(Why?);
2.finding(b,r),whichnow iseasiersince oneofthenumbersissmaller
thanbefore.
So,forexample,wehave
(100,28)
=(28,16)
(28,16)
=(16,12)
(16,12)
=(12,4)
Thisgivesus
(100,28)
=(12,4)=4.
since100 =3(28)+16
since28
=1(16)+12
since16
=1(12)+4
Itispossibletofindtheactualvalues ofmoandnosuchthat
4=mo100+no28

40. Sec.5 TheIntegers 25
bygoingbackwardsthroughthecalculationsmadetofind 4:
Since16= 1(12)+ 4,
Since28= 1(16)+ 12,
Since100=3(28)+ 16,
Butthen
4=16+(-1)12
12=28+(-1)16
16=100+(-3)28
4=16+(-1)12=16+(-1)(28+(-1)16)
=
(-1)28+(2)16=(-1)28+(2)(100+ (-3)28)
=(2)
100+(-7)28
sothatmo=2andno=-7.
ThisshowshowEuclid'sAlgorithmcanbeusedtocompute (a,b)for
anypositiveintegers
aandb.
Weshallincludesomeexercisesattheendofthissection onother
propertiesof(a,b).
Wecometotheveryimportant
DefilJ!tion.Wesaythat aandbarerelativelyprimeif(a,b)=1.
Sotheintegers aandbarerelativelyprimeiftheyhavenonontrivial
commonfactor.
Animmediatecorollaryto Theorem1.5.3is
Theorem1.5.4.Theintegers aandbarerelativelyprimeif andonlyif
1
=rna+nbforsuitableintegers mandn.
Theorem1.5.4hasanimmediateconsequence
Theorem1.5.5.If
aandbarerelativelyprime andaIbc,thenaIc.
ProofByTheorem 1.5.4,ma+nb= 1forsome mandn,hence
(ma+nb)c=c,thatis,mac+nbc=c.Byassumption,aIbcandbyobser­
vation
aImac,henceaI(mac+nbc)andso aIc.D
Corollary.If bandcarebothrelativelyprimeto a,thenbcisalsorel­
,ativelyprimeto
a.
ProofWepickupthe proofofTheorem1.5.5atmac+nbc=c.Ifd=
(a,bc),thendIaanddIbc,hencedI(mac+nbc)=c.SincedIaanddIc

41. 26ThingsFamiliarandLessFamiliar Ch.1
and(a,c)=1,wegetthatd=1.Since1 =d=(a,be),wehavethat beis
relativelyprimeto a.D
Wenowsingle outanultra-importantclass ofpositiveintegers,which
we
metbeforeinProblem30,Section 4.
Definition.A primenumber,oraprime,isanintegerp>1,suchthat
foranyinteger
aeitherpIaorpisrelativelyprime toa.
Thisdefinitioncoincideswiththeusualone, namelythatwecannotfac­
torpnontrivial/yo
Forifpisaprimeasdefinedaboveand p=abwhere
1
~a<p,then(a,p)=a(Why?)andpdoesnotdivide asincep>a.Itfol­
lows
thata=1,sop=b.Ontheotherhand,if pisaprimeinthesensethat
itcannotbefactorednontrivially,andif
aisanintegernotrelativelyprimeto
p,then(a,b)isnot1anditdivides aandp.Butthen(a,b)equalsp,byour
hypothesis,so pdividesa.
Anotherresultcoming outofTheorem1.5.5is
Theorem1.5.6.Ifpisaprime andpI(ala2...an),thenpIajfor
some
iwith1
$is;n.
ProofIfpIaI,thereisnothingtoprove.Supposethat p{a1;thenp
andalarerelativelyprime. Butpial(a2...an),hencebyTheorem1.5.5,
pIa2...an'Repeattheargumentjustgivenon az,andcontinue.0
Theprimesplayaveryspecialrolein thesetofintegerslargerthan1in
thateveryinteger n>1iseitheraprime oristheproductofprimes.We
shallshowthisinthenexttheorem.Inthetheoremafterthenext
weshall
show
thatthereisauniquenessabouttheway n>1factorsintoprimefac­
tors.Theproofs
ofboththeseresultsleanheavily onthewell-orderingprin­
ciple.
Theorem1.5.7.If n>1,theneithernisaprimeornistheproductof
primes.
Proof.Supposethatthetheorem isfalse.Thentheremustbean
intger
m>1forwhichthe theoremfails.Therefore,theset Mforwhichthe
theoremfailsisnonempty,so,bythewell-orderingprinciple, Mhasaleast
element
m.Clearly,since m
EM,mcannotbeaprime,thus m=ab,where
1
<a<mand1<b<m.Becausea<mandb<mandmistheleast
elementin
M,wecannothave aEMorbEM.Sincea
$.M,b$.M,bythe
definition
ofMthetheoremmustbetrueforboth aandb.Thusaandbare

42. Sec.5 TheIntegers 27
primesortheproductofprimes;from m=abwegetthatmisaproduct
ofprimes.Thisputsmoutside ofM,contradictingthatmEM.Thisproves
thetheorem.
D
Weassertedabove thatthereisacertainuniqueness aboutthedecom­
position
ofanintegerintoprimes.Wemakethisprecisenow. Toavoidtrivi­
alities
ofthekind6 =2 . 3=3 . 2(so,inasense,6hastwofactorizationsinto
theprimes2and3),weshallstate
thetheoreminaparticularway.
Theorem1.5.8.Givenn >1,thenthereisoneandonlyonewayto
writenintheform n=
p~lp~2...pC;/,wherePI<P2<...<Pkareprimes
andtheexponents
aI,a2,...,akareallpositive.
ProofWestartaswedidabovebyassuming thatthetheoremis
false,so thereisaleastintegerm >1forwhichit isfalse.Thism musthave
twodistinctfactorizations asm=
p~lp~2...p%k= q~lq~2...q~ewhere
PI<P2<...<Pk'ql<q2<...<qeareprimesandwheretheexponents
ab...,akandbb...,beareallpositive.Since PIIp~l..·p%k=q~l...q~e,
byTheorem1.5.6PIIq7
1
forsomei;hence,againby Theorem1.5.6,PIIqi,
hencePI=qi'Bythesametokenql=Pjforsome j;thusPI~Pj=
ql~qi=Pl'Thisgivesus thatPI=qi'Nowsince m/PI<m,m/PI
hastheuniquefactorizationproperty. Butm/PI=p~1-1 p~2...p%k=
p~1-1 q~2>:.q~eandsincem/PIcanbefactoredin oneandonlyonewayin
thisform,weeasilyget
k=
e,P2=q2,...,Pk=qk,al-1=b
i
-1,
a2=b2,...,ak=bk.Sowesee thattheprimesandtheirexponentsarising
inthefactorization
ofmareunique.Thiscontradicts thelackofsuchunique­
nessfor
m,andsoprovesthetheorem.D
Whattheselasttwotheoremstellus isthatwecanbuildup theintegers
fromtheprimesinaverypreciseandwell-definedmanner.
Onewouldex­
pectfromthis
thatthereshouldbemany-thatis,aninfinity-ofprimes.
Thisoldresultgoesback
toEuclid;infact, theargumentweshallgive isdue
toEuclid.
Theorem1.5.9.Thereisaninfinitenumberofprimes.
ProofIftheresultwerefalse,wecould enumeratealltheprimesin
PI'P2'...,Pk'Considertheinteger q=1+PIP2...Pk'Sinceq>Pifor
everyi
=1,2,...,k,qcannotbeaprime.Since Pi
~q,forwegetaremain­
derof1ondividingqbyPi,qisnotdivisiblebyany ofPb'..,Pk'Soqis
notaprime norisitdivisiblebyanyprime.Thisviolates Theorem1.5.7,
therebyprovingthetheorem.
D

43. 28ThingsFamiliarandLessFamiliar Ch.1
Resultsmuchsharperthan Theorem1.5.9existabouthowmanyprimes
thereareuptoagivenpoint.Thefamousprimenumbertheoremstatesthat
forlarge
nthenumberofprimeslessthan orequalto nis"moreorless"
n/log
en,wherethis "moreorless"ispreciselydescribed.Therearemany
openquestions
abouttheprimenumbers.
PROBLEMS
EasierProblems
1.Find(a,b)andexpress(a,b)asma+nbfor:
(a)(116,-84).
(b)(85,65).
(c)(72,26).
(d)(72,25).
2.
Provealltheparts ofLemma1.5.2,except part(d).
3.Show
that(ma,mb)=mea,b)ifm >o.
4.ShowthatifaImandb Imand(a,b)=1,then(ab)1m.
5.Factorthefollowingintoprimes.
(a)36.
(b)120.
(c)720.
(d)5040.
6.Ifm=
p~l...p%kandn =pr
1
•••pZ\wherep t ,••.,Pkaredistinct
primesand
at,...,akarenonnegativeand b
t
,
...,b
karenonnegative,
express(m,
n)as
p~l...p~kbydescribingthec'sinterms ofthea'sand
b's.
*7.Definetheleast commonmultipleofpositiveintegersmandntobethe
smallestpositiveinteger
vsuchthatbothmIvandnIv.
(a)Showthatv=mn/(m,n).
(b)IntermsofthefactorizationofmandngiveninProblem 6,whatis
v?
8.Findtheleastcommonmultiple
ofthepairsgiveninProblem 1.
9.Ifm,n>0aretwointegers,showthatwecanfindintegers u,vwith
-n/2
~v~n/2suchthatm=un+v.
10.Tocheck
thatagiveninteger n>1isaprime,prove thatitisenoughto
show
thatnisnotdivisiblebyanyprimepwithp
~yj,.

44. Sec.6 MathematicalInduction 29
11.Checkif thefollowingareprime.
(a)301.
(b)1001.
(c)473.
12.Startingwith 2,3,5,7,,constructthepositiveintegers1 +2·3,
1
+2·3·5,1+2 . 3·5.7, Doyoualwaysgetaprime numberthis
way?
Middle-LevelProblems
13.
IfPisanoddprime,show thatpisoftheform:
(a)4n+1or4n+3forsome n.
(b)6n+1or6n+5forsome n.
14.AdapttheproofofTheorem1.5.9toprove:
(a)Thereisaninfinitenumberofprimesoftheform4n+3.
(b)Thereisaninfinitenumberofprimesoftheform6n+5.
15.Show thatnointegeru=4n+3canbewrittenas u=a
2
+b
2
,
where
a,bareintegers.
16.IfTisaninfinitesubset of
N,thesetofallpositiveintegers,show that
thereisa1-1mappingofTontoN.
17.Ifpis_aprime,prove thatonecannotfindnonzerointegers aandbsuch
thata2~pb
2
.
(Thisshows thatvIPisirrational.)
6.MATHEMATICAL INDUCTION
Ifwelookback atSection5,weseethatatseveralplaces-forinstance,in
the
proofofTheorem1.5.6-wesay"argueasaboveandcontinue."This is
notverysatisfactoryasameans ofnailingdownanargument. Whatisclear
isthatweneedsometechnique ofavoidingsuchphraseswhenwewant to
proveaproposition aboutallthepositiveintegers.Thisisprovidedforusby
the
PrincipleofMathematicalInduction;infact,thiswill betheusualmethod
thatweshalluseforproving theoremsaboutallthepositiveintegers.
Theorem1.6.1.Letpen)beastatementaboutthepositiveintegers
suchthat:
(a)
P(l)istrue.
(b)
IfP(k)happenstobetrueforsomeinteger k
2::1,thenP(k+1)isalso
true.
ThenP(n)istrueforalln
2::1.

45. 30 ThingsFamiliarandLessFamiliar Ch.1
ProofActually,theargumentsgiveninprovingTheorems1.5.7and
1.5.8area
prototypeoftheargumentwegivehere.
Suppose
thatthetheoremisfalse;then,bywell-ordering, thereisa
leastinteger
m
2::1forwhich P(m)isnottrue.Since P(l)istrue,m=I=-1,
hencem>1.Now1~m-1<m,sobythechoiceofm,P(m- 1)mustbe
valid.
Butthenbytheinductivehypothesis [Part(b)]wemusthave thatP(m)
istrue.Thiscontradicts thatP(m)isnottrue.Thus therecanbe nointeger
forwhich
Pisnottrue,andsothetheoremisproved.D
Weillustratehow touseinductionwithsome ratherdiverseexamples.
Examples
1.Supposethatntennisballs areputinastraightline,touchingeachother.
Thenweclaimthattheseballs maken-1contacts.
ProofIfn=2,thematterisclear.Ifforkballswehave k-1con­
tacts,
thenaddingoneball(on aline)addsonecontact.So k+1ballswould
have
kcontacts.Soif pen)iswhatisstatedaboveaboutthetennisballs,we
see
thatifP(k)happenstobetrue,thensoisP(k+1).Thus,bythetheo­
rem,
P(n)istrueforalln
2::1.D
2.IfpisaprimeandpIala2...an'thenpIaiforsome1~i~n.
ProofLetpen)bethestatementinExample2.ThenP(l)istrue,for
if
pial,itcertainlydivides aiforsome 1
::::;i~1.
Supposeweknow thatP(k)istrue,andthatpIala2...akak+l.Thus,
by
Theorem1.5.6,since pI(ala2...ak)ak+leitherpIak+l(adesiredcon­
clusion)
orpial·..ak.Inthissecondpossibility,since P(k)istruewehave
thatpIaiforsome 1
:5i::::;k.Combiningbothpossibilities,weget thatpIaj
forsome1~j~k+1.SoPart(b)ofTheorem1.6.1holds;hence pen)is
trueforalln2::1.0
3.Forn2::1,1+2+...+n=~n(n+1).
ProofIfpen)isthepropositionthat1+2+...+n=~n(n+1),then
P(l)iscertainlytrue,for1 =~(1+1).IfP(k)shouldbetrue,thismeans that
1+2+...+k=~k(k+1).
Thequestionis:IsP(k+1)thenalsotrue, thatis,is1+2+...+k+
(k+1)=~(k+l)((k+1)+I)?Now1 +2+...+k+(k+1)=
(1+2+...k)+(k+1)=~k(k+1)+(k+1),sinceP(k)isvalid.But
~k(k+1)+(k+1)=~(k(k+1)+2(k+1»=~(k+l)(k+2),which
assuresus
thatP(k+1)istrue.Thustheproposition1 +2+...+n=~n(n+1)istrueforalln2::1.0

46. Sec.6 MathematicalInduction 31
Wemustemphasizeonepointhere:Mathematicalinductionisnota
methodforfindingresults aboutintegers;itisa meansofverifyingaresult.
Wecould,byothermeans,findtheformulagivenabovefor1+2+...+n.
Part(b)ofTheorem1.6.1isusuallycalled theinductionstep.
Intheproblemsweshallgive someotherversionsoftheprincipleofin­
duction.
PROBLEMS
EasierProblems
1.Provethat1
2
+2
2
+3
2
+...+n
2
=~n(n+1)(2n+1)byinduction.
2.
Provethat1
3
+2
3
+...+n
3
=
~n2(n+1)2byinduction.
3.Provethatasethavingn~2elementshas~n(n-1)subsetshavingex­
actly
twoelements.
4.Provethatasethavingn
~3elementshasn(n-l)(n-2)/3!subsets
havingexactly
threeelements.
5.Ifn
;:::4andSisasethavingnelements,guess(fromProblems3and4)
howmanysubsetshavingexactly4 elementsthereareinS.Thenverify
yourguessusing mathematicalinduction.
*6.
CompletetheproofofTheorem1.5.6,replacing thelastsentencebyan
inductionargument.
7.Ifa
=1=1,provethat1+a+a
2
+...+an=(a
n
+
1
-
l)/(a- 1)byinduc­
tion.
8.Byinduction,showthat
1+1+...+ 1 n
N 2 . 3 n(n+1)n+1 .
*9.SupposethatP(n)isapropositionaboutpositiveintegersnsuchthat
P(no)isvalid,andifP(k)istrue,somustP(k+1)be.Whatcanyousay
aboutpen)?Proveyourstatement.
*10.Letpen)beapropositionaboutintegersnsuchthatP(l)istrueand
suchthatifP(j)istrueforallpositiveintegersj<k,thenP(k)istrue.
ProvethatP(n)istrueforallpositiveintegersn.
Middle-LevelProblems
11.Giveanexampleofapropositionthatisnottrueforanypositiveinteger,
yetforwhichtheinductionstep[Part(b)ofTheorem1.6.1]holds.
12.
Provebyinductionthatasethavingnelementshasexactly2 nsubsets.

47. 32ThingsFamiliarandLessFamiliar Ch.1
13.Provebyinduction onnthatn
3
-
nisalwaysdivisibleby 3.
14.Usinginduction onn,generalizetheresultinProblem 13to:Ifpisaprime
number,then
n
P
-nisalwaysdivisibleby p.(Hint:Thebinomialtheorem.)
15.Provebyinductionthatforasethaving nelementsthe numberof1-1
mappingsofthissetontoitselfisn!.
7.COMPLEX NUMBERS
Weallknowsomething abouttheintegers,rationalnumbers, andrealnum­
bers-indeed,thisassumptionhas beenmadeforsome ofthetextmaterial
andmanyoftheproblemshavereferred tothese numbers.Unfortunately,
thecomplexnumbers
andtheirproperties aremuchlessknown topresent­
daycollegestudents.
Atonetimethecomplexnumberswerea partofthe
highschoolcurriculum
andtheearlycollegeone.This isnolongerthecase.
Soweshall
doarapiddevelopment ofthisveryimportantmathematicalset.
Thesetofcomplexnumbers,
C,isthesetofalla+bi,wherea,bare
realandwherewe declare:
1.a+bi=c+di,fora,b,c,dreal,if andonlyif a=candb=d.
2.(a+bi)±(c+di)=(a±c)+(b±d)i.
3.(a+bi)(c+di)=(ac- bd)+(ad+bc)i.
Thislastproperty-multiplication-canbestberememberedbyusing
i
2
=-1andmultiplyingoutformallywiththisrelationinmind.
Forthecomplexnumberz=a+bi,aiscalledthe realpartofzandb
theimaginarypartofz.Ifais0,wecallzpurelyimaginary.
Weshallwrite0 +Oias0anda+Oiasa.Notethatz+0=z,z1=z
foranycomplex numberz.
Givenz=a+bi,thereisacomplexnumberrelatedtoz,whichwe
writeasZ,definedbyZ=a-bi. Thiscomplex number,Z,iscalledthe
complexconjugateofz.Takingthecomplexconjugategivesusamapping
ofContoitself.Weclaim
Lemma1.7.1.Ifz,wEC,then:
(a)(Z)=z.
(b)(z+w)=Z+w.
(c)(zw)=zw
(d)zzisrealandnonnegativeandis,infact,positiveif z=1=O.

48. Sec.7 ComplexNumbers 33
(e)z+zistwicetherealpartofz.
(f)z-zistwicetheimaginarypartofztimesi.
ProofMostofthepartsofthislemmaarestraightforwardandmerelyin­
volveusingthedefinition
ofcomplexconjugate. WedoverifyParts(c)and(d).
Suppose
thatz=a+bi,W=e+di,wherea,b,e,darereal.So zw=
(ae-bd)+(ad+be)i,hence
(zw)=(ae-bd)+(ad+be)i=(ae-bd)-(ad +be)i.
Ontheotherhand,
z=a-biandW=e -di,hence,bythedefinition ofthe
productinC,zW=(ae-bd)-(ad +be)i.Comparingthiswith theresult
thatweobtainedfor
(zw),weseethatindeed(zw)=
Zw.ThisverifiesPart(c).
WegonexttotheproofofPart(d).Suppose thatz=a+bi=1=0;then
z=a-bi andzz=a
2
+b
2
•
Sincea,barerealandnotboth0,a
2
+b
2
is
realandpositive,asassertedin Part(d).D
TheproofofPart(d)ofLemma1.7.1shows thatifz=a+bi
=1=0,then
zz=a
2
+b
2
¥Oandz(zl(a
2
+b
2
»=1,so
actsliketheinverselizofz.Thisallowsus tocarryoutdivisioninC,staying
inCwhiledoingso.
Wenowlistafew
propertiesof
C.
Lemma1.7.2.Cbehavesunderitssumandproductaccordingtothe
following:Ifu,v,wEC,then
(a)u+v=v+u.
(b)(u+v)+w=u+(v+w).
(c)uv=VUe
(d)(uv)w=u(vw).
(e)u=1=0impliesthatu-
1
=11uexistsinCsuchthatuu-
1
=1.
ProofWeleavetheproofsofthesevariouspartstothereader.0
ThesepropertiesofCmakeofCwhatweshallcalla field,whichwe
shallstudyinmuch
greaterdepthlaterinthebook.Whatthelemmasaysis
thatweareallowedtocalculatein
Cmoreorlessaswedidwithreal num­
bers.However,Chasamuchricher structurethanthesetofrealnumbers.

49. 34 ThingsFamiliarandLessFamiliar
Wenowintroducea"size"functiononC.
Ch.1
Definition.Ifz=a+biEC,thenthe absolutevalue ofz,writtenas
Izl,isdefinedby Izi=-v.a=Ya
2
+b
2
•
Weshallsee,inafewmoments,whatthislastdefinitionmeansgeomet­
rically.Inthemeantimeweprove
Lemma1.7.3.If
u,vE
C,thenluvl=lullvl.
ProofBydefinition,IuI=~andIvI=~.Now
luvl=Y(uv)(uv)=Y(uv)(uv)(byPart(c)ofLemma1.7.1)
=Y(uu)(vv) (byLemma 1.7.2)
=~~ =lullvl·D
Anotherwayofverifyingthislemma istowriteu=a+bi,v=e+di,
uv=(ae-bd) +(ad+bc)iandtonotetheidentity
(ae-bd)2 +(ad+be)2=(a
2
+b
2
)(e
2 +d
2).
Noteseveralsmallpointsaboutconjugates.Ifz E
C,thenz isrealif
andonlyifz=z,andzispurelyimaginaryifandonlyifz= -z.If
z,wEC,then
-
(zW+zw)=zW+zW=zw+zW,
sozW+ZWisreal.Wewanttogetan upperboundforIzw+zwl;thiswill
comeupinthe
proofofTheorem1.7.5below.
Butfirstwemustdigressforamomenttoobtainastatementabout
quadraticexpressions.
Lemma1.7.4.
Leta,b,ebereal,with a>O.Ifaa
2
+ba+e
2::0for
everyreal
a,thenb
2
-4ae
:5O.
ProofConsiderthequadraticexpressionfora=-bI2a.Weget
a(-bI2a)2+b(-bI2a)+e2::O.Simplifyingthis,weobtainthat (4ae-b
2
)/4a
~0,andsincea>0,weendupwith 4ae-b
2
~0,andso b
2
-4ae:5O.D
Weusethisresultimmediatelytoprovetheimportant
Theorem1.7.5(TriangleInequality).
Forz,wE
C,Iz+wI:5IzI+IwI·
ProofIfz=0,thereisnothingtoprove,sowemayassumethatz=I=-0;
thuszz>O.Now,forareal,

50. Sec.7 ComplexNumbers
0::::;laz+wl
2
=(az+w)(az+w)=(az+w)(az+W)
=a
2
zz+a(zW+ZW)+wW.
35
Ifa=ZZ>0,b=zW+zw,c=wW,thenLemma1.7.4tells usthat
b
2
-
4ac=
(zw+ZW)2-4(zz)(ww)::::;0,hence(zw+ZW)2::::;4(zz)(ww)=
41z1
2
1w1
2
•
Therefore,
zW+zw::::;2lzllwl.
Fora=1above,
Iz+wl
2
=zz+wW+zW+zw=Izl
2
+Iwl
2
+zW+zw
::::;Izl
2
+Iwl
2
+21z1Iwl
fromtheresultabove.Inotherwords, Iz+W1
2
::::;(IzI+IwI)2;takingsquare
roots
wegetthedesiredresult, Iz+
wi::::;IzI+Iwi.0
Whyisthisresultcalledthetriangleinequality?Thereasonwillbeclear
once
weviewthecomplexnumbersgeometrically.Representthecomplex
number
z=a+biasthepointhavingcoordinates (a,b)inthex-yplane.
Thedistance
rofthispointfromtheorigin is
Va
2
+b
2
,
inotherwords,Izl.
Theangle ()iscalledthe argumentofzand,aswesee,tan ()=b/a.Also,
a=rcos(),b=rsin
();therefore,z=a+bi=r(cos()+isin().Thisrep­
resentationof
ziscalledits polarform.
Givenz=a+bi,w=c+di,thentheirsum isz+w=(a+c)+
(b+d)i.Geometrically,wehavethepicture:
(a+c,b+d)

51. 36 ThingsFamiliarandLessFamiliar Ch.1
wheren~1isaninteger.
Thestatement
Iz+
wi::::;Izi+IwImerelyreflectsthefactthatinatriangle
oneside
isofsmallerlengththanthesum ofthelengthsoftheothertwo
sides;thustheterm
triangleinequality.
Thecomplexnumbersthatcomeupinthepolarformcos ()+isin()
areveryinterestingnumbersindeed.Specifically,
Icos()+isin()j=
Vcos
2
()+sin
2
()
=
Vi=1,
sotheygiveusmanycomplexnumbers ofabsolutevalue 1.Intruththeygive
us
allthecomplexnumbersofabsolutevalue
1;toseethisjustgobackand
lookatthepolarform
ofsuchanumber.
Let'srecalltwobasicidentitiesfromtrigonometry,cos(
()+
t/J)=
cos()cost/J-sin()sint/Jandsine ()+t/J)=sin()cost/J+cos()sint/J.There­
fore,if
z=r(cos()+isin()andw=s(cos
t/J+isint/J),then
zw=r(cos()+isin().s(cost/J+isint/J)
=rs(cos()cost/J-sin()sint/J)+irs(sin()cost/J+cos()sint/J)
=rs[cos(()+t/J)+isine()+t/J)].
Thus,inmultiplyingtwocomplexnumbers,theargumentoftheproduct is
thesumoftheargumentsofthefactors.
Thishasanotherveryinterestingconsequence.
Theorem1.7.6 (DeMoivre'sTheorem). Foranyinteger n
;:::1,
[r(cos()+isin()]n=rn[cos(n()+isin(n()].
ProofWeproceedbyinductionon n.Ifn=1,thestatement is
obviouslytrue.Assumethenthatforsome k,[r(cos()+isin()]k
r
k
[cosk()+isink()].Thus
[r(cos()+isin()]k+l=[r(cos()+isin()]k.r(cos()+isin()
=r
k
(
cosk()+isink().r(cos()+isin()
=rk+1 [cos(k+1)()+isin(k+1)()]
bytheresult oftheparagraphabove.Thiscompletestheinductionstep;
hencetheresult
istrueforallintegers n
;:::1.0
Intheproblemsweshallseethat DeMoivre'sTheorem istrueforall
integersm;infact,it istrueevenif misrational.
Considerthefollowingspecialcase:
()
21T..21T
n=cos- +lsIn-,
n n

52. Sec.7
ByDeMoivre'sTheorem,
ComplexNumbers 37
((
21T).'(21T))n
cosn+lsInn
=cos(nen
7T
))+isin(n(2n
7T))
=cos21T+isin21T=1.
SoO~=1;youcanverifythat0'::=1=1if0<m<n.ThispropertyofOnmakes
itoneoftheprimitiventhrootsofunity,whichwill beencounteredinProb­
lem26.
PROBLEMS
EasierProblems
1.Multiply.
(a)(6-7i)(8+i).
(b)(~+~i)(~-~i).
(c)(6+7i)(8-i).
2.
ExpressZ-1intheformZ-1=a+bifor:
(a)z=6+8i.
(b)z=
6
1
8i.1 .
(c)z=v2+v2
l
.
*3.Showthat(Z)-l=(Z-l).
4.Find(cos0+isin0)-1.
5.Verifypartsa,b,e,fofLemma1.7.1.
*6.showthatzisrealifandonlyifz=z,andispurelyimaginaryifand
onlyifz= -z.
7.VerIfythecommutativelawofmultiplicationzw=wzinC.
8.Showthatforz=1=0,Iz-
1
1=1/1zl.
9.Find:
(a)16-4il·
(b)I~+~il·
(c)l~+~il·
10.Show thatIzi=Izi.

53. 38 ThingsFamiliarandLessFamiliar Ch.1
11.Find thepolarformfor
)
V21.
(az=="2"-V2l.
(b)z==4i.
(
66.
c)z==V2+V2l.
(d)z==-13+39_i.
22V3
12.Prove that(cos(~0)+isin(~0»2==cos0+isino.
13.BydirectmUltiplicationshow that(~+~V3i)3==-1.
Middle-LevelProblems
14.Show that(cos0+isino)'n==cos(mO)+isin(mO)forallintegersm.
15.Show that(cos0+isino)r==cos(rO)+isin(rO)forallrationalnum­
bers
r.
16.Ifz
ECandn;:::1isanypositiveinteger,show thattherearendistinct
complex
numberswsuchthatz==w
n
•
17.Findthenecessaryandsufficientconditionon ksuchthat:
(
(
27Tk)..(27Tk))n1
cosn+1SInn ==
((
27Tk)..(27Tk))rn1
cosn+1SInn ¥
and
if0<m<n.
18.Viewing thex-yplaneas thesetofallcomplexnumbersx+iy,showthat
multiplicationby iinducesa 90°rotationofthex-yplaneinacounter­
clockwisedirection.
19.
InProblem18,interpretgeometricallywhatmultiplicationbythecom­
plex
numbera+bidoesto thex-yplane.
*20.Prove
thatIz+w1
2
+Iz-wl
2
==2(lz/
2
+IwI
2
).
21.Considertheset A=={a+biIa,b
EZ}.Provethatthereisa1-1corre­
spondence
ofAonto
N.(AiscalledthesetofGaussianintegers.)
22.Ifaisa(complex)rootofthepolynomial
where
the
(Xiarereal,show thatamustalso bearoot.[risarootofa
polynomialp(x)ifp(r)==0.]

54. Sec.7
HarderProblems
ComplexNumbers 39
23.Findthenecessaryandsufficientconditionson zandwinorderthat
Iz+wi=Izl+Iwl·
24.Findthenecessaryandsufficientconditionson Z1,...,Zkinorderthat
IZI+...+zkl=IZII+...+IZkl·
*25.Thecomplexnumber ()issaidtohaveordern;:::1if()n=1and()m=1=1
for0
<m<n.Showthatif ()hasordernand()k=1,wherek>0,then
nIk.
*26.Findallcomplexnumbers ()havingordern.(Thesearetheprimitive nth
rootsofunity.)

55. 2
GROUPS
1.DEFINITIONSANDEXAMPLES OFGROUPS
WehaveseeninSection4 ofChapter1thatgivenany nonemptyset,theset
A(S)ofall1-1mappings ofSontoitselfisnotjustasetalone, buthasafar
richertexture.
ThepossibilityofcombiningtwoelementsofA(S)toget
yetanotherelementofA(S)endowsA(S)withanalgebraicstructure.We
recallhowthiswasdone:If
f,gEA(S),thenwecombine themtoformthe
mapping
fgdefinedby (fg)(s)=f(g(s»forevery sES.Wecalledfgthe
productoffandg,andverifiedthatfgEA(S),andthatthisproductobeyed
certainrules.
Fromthemyriadofpossibilitieswesomehowselectedfourpar­
ticularrules
thatgovernthe behaviorofA(S)relativetothisproduct.
Thesefourruleswere
1.Closure,namelyif f,gEA(S),thenfgEA(S).WesaythatA(S)is
closedunderthisproduct.
2.Associativity,thatis,givenf,g,hEA(S),thenf(gh)=(fg)h.
3.Existenceofaunitelement,namely,thereexistsaparticularelement
iEA(S)(theidentitymapping)such thatfi=if=fforallfEA(S).
4.Existenceofinverses,thatis,givenfEA(S)thereexistsanelement,
denotedby
f-I,inA(S)suchthatff-
1
=f-lf=i.
Tojustifyormotivatewhythesefourspecificattributes ofA(S)were
singledout,incontradistinction
tosomeothersetofproperties,isnoteasyto
40

56. Sec.1 DefinitionsandExamples ofGroups 41
do.Infact,in thehistoryofthesubjectit tookquitesometimetorecognize
thatthesefourpropertiesplayed thekeyrole.Wehavetheadvantageofhis­
toricalhindsight,
andwiththishindsightwechoose themnotonlytostudy
A(S),butalsoas thechiefguidelinesforabstracting toamuchwidercontext.
Althoughwesaw
thatthefourpropertiesaboveenabledustocalculate
concretelyin
A(S),thereweresomedifferenceswith thekindofcalculations
weareusedto.
IfShasthreeormoreelements,wesawin Problem15,
Chapter1,Section4 thatitispossiblefor f,gEA(S)tohavefg
=1=gf.How­
ever,thisdid
notpresentuswithinsurmountabledifficulties.
Withoutany
furtherpolemicswego tothe
Definition.A nonemptysetG issaidtobeagroupifinG thereisde­
finedan
operation*suchthat:
(a)
a,bEGimpliesthata*bEG.(Wedescribethisbysaying thatGis
closedunder*.)
(b)Given
a,b,c EG,thena*(b*c)=(a*b)*c.(Thisisdescribedby
saying
thatthe
a$sociativelawholdsin G.)
(c)Thereexistsaspecial elementeEGsuchthata*e=e*a=aforall
aEG(eiscalledtheidentityorunitelementofG).
(d)ForeveryaEGthereexistsan elementbEGsuchthata*b=
b*a=e.(Wewritethis elementbasa-Iandcallittheinverseof
ainG.)
Thesefourdefiningpostulates(called thegroupaxioms)fora group
were,afterall, patternedafterthose thatholdin A(S).Soitisnotsurprising
thatA(S)isagrouprelative totheoperation"compositionofmappings."
Theoperation*inG isusuallycalled theproduct,butkeepinmind
thatthishasnothing todowithproductasweknowitfor theintegers,ratio­
nals,reals,
orcomplexes.Infact,asweshallseebelow,in manyfamiliarex­
amples
ofgroupsthatcomefromnumbers,whatwecall theproductinthese
groups
isactuallytheaddition ofnumbers.However,ageneralgroup need
havenorelationwhatsoevertoaset ofnumbers.Wereiterate:Agroup isno
more,noless, thananonemptysetwithanoperation *satisfyingthefour
groupaxioms.
Beforestarting
tolookinto thenatureofgroups,welook atsomeex­
amples.
ExamplesofGroups
1.Let7Lbethesetofallintegersandlet*betheordinaryaddition, +,in7L.
That7Lisclosedandassociativeunder*arebasicpropertiesoftheintegers.
Whatservesas theunitelement, e,of7Lunder*?Clearly,since a=a*e=

57. 42 Groups Ch.2
a+e,wehave e=0,and0istherequiredidentityelementunderaddition.
Whatabouta-I?Heretoo,sincee=0=a*a-I=a+a-I,thea-Iinthisin­
stanceis
-a,andclearlya*(-a)=a+(-a)=o.
2.Let
0bethesetofallrationalnumbersandlettheoperation*on0be
theordinaryaddition ofrationalnumbers. Asabove,0iseasilyshownto be
agroupunder*.Notethat7LC0andboth7Land0aregroupsunderthe
same
operation*.
3.Let
0'bethesetofallnonzerorationalnumbers andlettheopera­
tion
*on
0'betheordinarymultiplication ofrationalnumbers.Bythefa­
miliar
propertiesoftherationalnumbersweseethat
0'formsagroupt:ela­
tiveto *.
4.LetIR+bethesetofallpositivereal numbersandlettheoperation*onIR+
betheordinaryproductofrealnumbers.Againit iseasytocheckthatIR+is
agroupunder*.
5.LetEnbetheset ofO~,i=0,1,2,...,n-1,whereOnisthecomplexnum­
berOn=Cos(27T/n)+isin(27T/n).LetO~*0h=O~+j,theordinaryproductof
thepowers
of
Onascomplexnumbers.By DeMoivre'sTheoremwesawthat
O~=1.Weleaveittothereadertoverifythat Enisagroupunder *.Theele­
ments
ofEnarecalledthe n'throotsofunity.Thepicturebelowillustratesthe
group£6,whoseelementsarerepresentedbythedots
ontheunitcircleinthe
complexplane.
Noteonestrikingdifference betweentheExamples1to4andExample
5;thefirstfourhave aninfinitenumberofelements,whereas Enhasafinite
number,n,ofelements.
Definition.A groupG issaidtobeafinitegroup ifithasafinite
numberofelements.ThenumberofelementsinG iscalledthe orderofGand
isdenotedbyIGI.

58. Sec.1 DefinitionsandExamples ofGroups 43
ThusEnaboveisafinitegroup, andIEnI=n.
Alltheexamples
presentedabovesatisfytheadditionalproperty that
a*b=b*aforanypair ofelements.This neednotbetrueinagroup.Just
witnessthecase
ofA(S),where Shasthreeormoreelements;therewesaw
thatwecouldfind f,gEA(S)suchthatfg
=1=gf.
Thispromptsustosingleoutasspecialthosegroups ofGinwhich
a*b=b*aforalla,bEG.
Definition.A groupGissaidtobeabelianifa*b=b*aforall
a,bEG.
Thewordabelianderivesfromthename ofthegreatNorwegianmathematician
NielsHenrik
Abel(1802-1829),one ofthegreatestscientistsNorwayhasever
produced.
Agroupthatisnotabelianiscallednonabelian,anottoosurprising
choice
ofname.
Wenowgiveexamples
ofsomenonabeliangroups. Ofcourse,the
A(S)affordusaninfinitefamily ofsuch.Butwepresentafewotherexam­
plesinwhichwecancomputequitereadily.
6.Let
IRbethesetofallrealnumbers, andletG bethesetofallmappings
Ta,b:IR~IRdefinedby Ta,b(r)=ar+bforanyreal numberr,wherea,b
arerealnumbers anda=1=o.Thus,forinstance, T
S,-6issuchthatT
S,-6(r)=
5r-6;Ts,-6(14)=5 .14- 6 =64,TS,-6(1T)=571"-6.TheTa,bare1-1map­
pingsofIRontoitself,andweletT
a
, b*T
e
, dbetheproductoftwoofthese
mappings.So
(Ta,b*Te,d)(r)=Ta,b(Te,d(r»=aTe,d(r)+b=a(er+d)+b
=(ae)r+(ad+b)=Tae,ad+b(r).
Sowehavetheformula
Ta,b*Te,d=Tae,ad+b· (1)
Thisresultshowsus thatTa,b*Te,disinG-foritsatisfiesthemembership
requirementforbelongingto
G-soGisclosedunder*.Sincewe aretalking
abouttheproduct
ofmappings(i.e., thecompositionofmappings),*is
associative.TheelementT1,o=iistheidentitymapping of
IRontoitself.
Finally,what
is
T~,lb?Canwefindrealnumbers x=1=0andy,suchthat

59. 44 Groups Ch.2
Gobackto(1)above; wethuswant Tax,ay+b=TI,o,thatis,ax=1,ay+b=0.
Remembernowthata=I=-0,soifweputx =a-Iandy =-a-1b,therequired
relationsaresatisfied.Oneverifiesimmediatelythat
SoG
isindeedagroup.
What
isT
c
,
d*T
a
,
b?Accordingtotheformulagivenin(1),where were­
place
abye,ebya,bbyd,dbyb,weget
(2)
ThusTc,d*Ta,b=ifTa,b*Tc,dandonlyifbe +d=ad+b.Thisfailstobe
true,forinstance,if
a=1,b=1,e=2,d=3.SoG isnonabelian.
7.LetHCG,whereG isthegroupinExample 6,andHisdefinedby
H
={T
a
,bEGIaisrational,banyreal}.Weleaveittothereadertoverify
that
Hisagroupundertheoperation *definedon G.Hisnonabelian.
8.LetKCHeG,whereH,Gareasaboveand K={TI,bEGIbanyreal}.
Thereadershouldcheckthat
Kisagrouprelativetotheoperation *ofG,
andthat
Kis,however,abelian.
9.LetSbetheplane,that is,S={(x,y)Ix,yreal}andconsider f,gEA(S)
definedbyf(x,y) =(-x,y)andg(x,y) =(-y,
x);fisthereflectionabout
they-axisand
gistherotationthrough 90
0
inacounterclockwisedirection
abouttheorigin.WethendefineG
={figjIi=0,
1;j=0,1,2,3},andlet*
inGbetheproductofelements inA(S).Clearly,f2 =g4=identitymapping;
(f*g)(x,y) =(fg)(x,y) =f(g(x,y»=f(-y,x)=(y,x)
and
(g*f)(x,y)=g(f(x,y»=g(-x,y)=(-y,-x).
Sog*f
=1=f*g.Weleaveittothereadertoverifythatg *f=f*g-landG
isanonabeliangroupoforder 8.Thisgroup iscalledthedihedralgroupof
order
8.[Trytofindaformulafor (fig
j
)
*(fSgt)=fagbthatexpressesa,b
intermsofi,j,s,andt.]
10.LetSbeasinExample9and fthemapping inExample9.Letn>2and
let
hbetherotationoftheplaneabouttheoriginthroughanangleof
27T/n
inthecounterclockwisedirection.WethendefineG ={fkh
j
Ik=0,1;
j=0,1,2,...,n-I}anddefinetheproduct *inGviatheusualproductof
mappings.
Onecanverifythat f2=h
n
=identitymapping,and fh=h-Ir

60. Sec.1 DefinitionsandExamples ofGroups 45
Theserelationsallowus toshow(withsomeeffort) thatGisanonabelian
group
oforder2n.Giscalledthedihedralgroup oforder2n.
11.LetG={fEA(S)If(s)
=1=sforonlyafinite numberofsES},wherewe
supposethat
Sisaninfiniteset.Weclaim thatGisagroupundertheproduct
*inA(S).TheassociativityholdsautomaticallyinG,sinceitalreadyholdsin
A(S).Also,iEG,sincei(s)=sforallsES.SowemustshowthatG is
closedundertheproductandif fEG,thenf-
1
EG.
Wefirstdispose
oftheclosure.Supposethat f,gE
G;thenf(s)=s
except,say,for
SI,S2,•••'Snandg(s)=sexceptfors{,
s~, ,s~.Then
(fg)(s)=f(g(s»=sforalls otherthanSI,S2,•••,Sn,s{,,s/n(and
possiblyevenforsome
ofthese).So fgmovesonlyafinite numberofele­
ments
ofS,sofgEG.
Finally,if
f(s)=sforalls otherthan
SbS2'•••,Sn,thenf-l(f(s»=
f-
1
(s),
butf-l(S)=f-l(f(s»=(f-lf)(s)=i(s)=s.Soweobtain that
f-
1
(s)=sforallsexcept SI,•••'Sn.Thusf-
1
EGandGsatisfiesallthegroup
axioms,henceG
isagroup.
12.LetGbetheset ofallmappings
T(j,whereToistherotationofagivencir­
cleaboutitscenterthroughanangle
()intheclockwisedirection.InGdefine
*bythecomposition ofmappings.Since, asisreadilyverified, To*
T",=T0+'"'
Gisclosedunder *.Theothergroupaxiomscheck outeasily.Note that
121T=To=theidentitymapping,andr;l=T-o=121T-o.Gisanabeliangroup.
Aswedidfor
A(S)weintroducetheshorthandnotation anfor
,a*a*a···*a
ntimes
anddefine
a-
n
=(a-
1
)n,fornapositiveinteger,and aO=e.Theusualrules
ofexponentsthenhold,thatis, (am)n=a
mn
andam*an=a
m+
n
foranyinte­
gers
mandn.
Notethatwiththisnotation,ifG isthegroupofintegersunder+,then
anisreallyna.
Havingseenthe 12examplesofgroupsabove,the readermightgetthe
impressionthatall,
oralmostall,setswithsomeoperation *formgroups.
This
isfarfromtrue.Wenowgivesomeexamples ofnongroups.Ineachcase
wecheckthefourgroupaxioms
andseewhichofthesefail tohold.
Nonexamples
1.LetGbetheset ofallintegers,andlet*betheordinaryproductofinte­
gersin
G.Sincea *b=ab,fora,bEGweclearlyhave thatGisclosedand
associativerelativeto *.Furthermore,the number1servesastheunitele-

61. 46 Groups Ch.2
ment,since a*1=a1=a=1a=1*aforeveryaEG.Soweare three­
fourthsofthewaytoprovingthatG
isagroup.Allweneed isinversesfor
theelementsofG,relativeto
*,toliein G.Butthisjustisn'tso.Clearly, we
cannotfindaninteger bsuchthat0 *b=Ob=1,sinceOb=0forall b.But
evenotherintegersfailtohaveinversesin
G.Forinstance,we cannotfindan
integerb
suchthat3 *b=1(forthiswouldrequirethat b=
~,and~isnot
aninteger).
2.LetGbethesetofallnonzerorealnumbersanddefine,for a,bEG,a*b
=
a
2
b;thus4*5=4
2
(5)=80.Whichofthegroupaxiomsholdin Gunder
thisoperation*andwhichfailtohold?Certainly,G isclosedunder *.Is*
associative?Ifso, (a*b)*c=a*(b*c),thatis,(a*b)2C=a
2
(b*c),and
so
(a
2
b)2c=a
2
(b
2
c), whichboilsdownto a
2=1,whichholdsonlyfor
a=±1.So,ingeneral,theassociativelawdoes notholdinGrelativeto *.
WesimilarlycanverifythatGdoesnothaveaunitelement.Thusevento
discussinversesrelativeto
*wouldnotmakesense.
3.LetGbethesetofall positiveintegers,under *wherea*b=ab,the
ordinaryproductofintegers.ThenonecaneasilyverifythatGfailstobea
group
onlybecauseitfails tohaveinversesforsome(infact,most)ofitsele­
mentsrelativeto
*.
Weshallfindsomeothernonexamplesofgroupsintheexercises.
PROBLEMS
EasierProblems
1.DetermineifthefollowingsetsGwiththeoperationindicatedforma
group.Ifnot,pointoutwhichofthegroupaxiomsfail.
(a)G=setofallintegers, a*b=a-b.
(b)G=setofallintegers, a*b=a+b+abo
(c)G=setofnonnegativeintegers, a*b=a+b.
(d)G=setofallrationalnumbers
=1=-1,a*b=a+b+abo
(e)G=setofallrationalnumberswithdenominatordivisibleby5(writ-
tensothatnumeratoranddenominatorarerelativelyprime),
a*b=
a+b.
(f)G asethavingmorethanoneelement, a*b=aforalla,bEG.
2.InthegroupGdefined inExample6,showthattheset H={Ta,bIa=±1,
banyreal}formsagroupunderthe *ofG.
3.VerifythatExample7 isindeedanexampleofagroup.

62. Sec.1 DefinitionsandExamples ofGroups 47
4.ProvethatKdefinedin Example8isanabeliangroup.
5.InExample9,provethatg*f=f*g-t,andthatGisagroup,isnon­
abelian,andisoforder8.
6.LetGandHbeasinExamples6and7,respectively.Show thatif
Ta,bEG,thenTa,b*V*T;;:lEHifVEH.
7.DoProblem6withH replacedbythegroupKofExample8.
8.IfGisanabeliangroup, provethat(a*b)n=an*b
nforallintegersn.
9.IfGisa groupinwhicha
2
=eforalla
EG,showthatGisabelian.
10.
IfGisthegroupinExample6,findall Ta,b
EGsuchthatTa,b*TI,x=
Tl,x*Ta,bforallrealx.
11.
InExample10,forn=3finda formulathatexpresses(fih
j
)
*(fSh
t
)
as
fah
b
•
ShowthatGisanonabeliangroupoforder6.
12.DoProblem11forn =4.
13.Showthatanygroupoforder4orlessisabelian.
14.IfGisanygroupanda,b,c
EG,showthatifa*b=a*c,thenb=c,
andifb*a=c*a,thenb=c.
15.Express(a*b)-Iintermsofa-Iandb-
1
.
16.Using theresultofProblem15,provethatagroupGinwhich a=a-I
foreverya
EGmustbeabelian.
17.InanygroupG,provethat(a-I)-I=aforallaEG.
*18.
IfGisafinite groupofevenorder,showthattheremustbeanelement
a
=1=esuchthata=a-I.(Hint:TrytousetheresultofProblem17.)
19.InS3,showthattherearefourelementsxsatisfyingx
2
=eandthreeele­
mentsysatisfyingy3=e.
20.FindalltheelementsinS4suchthatx
4
=e.
Middle-LevelProblems
21.Show thatagroupoforder5mustbeabelian.
22.Show
thatthesetdefinedin Example10isagroup, isnonabelian,and
hasorder2n.Dothisbyfinding theformulafor(fih
j
)
*(fsh
t
)
inthe
formfah
b
•
23.InthegroupGofExample6,findall elementsU
EGsuchthat
U*Ta,b=Ta,b*Uforevery Ta,bEG.
24.
IfGisthedihedralgroup oforder2nasdefinedin Example10,provethat:
(a)IfnisoddandaEGissuchthata*b=b*aforallbEG,thena=e.
(b)Ifniseven,show thatthereisana
EG,a=1=e,suchthata*b=b*a
forallbEG.

63. 48 Groups Ch.2
(c)Ifniseven,findalltheelementsaE Gsuchthata*b=b*a
forallbEG.
25.IfGisanygroup,showthat:
(a)eisunique(i.e.,iffE Galsoactsasa unitelementforG,thenf=e).
(b)GivenaEG,thena-IEGisunique.
*26.
IfGisafinitegroup,provethat,givenaEG, thereisapositiveinteger
n,dependingona,suchthatan=e.
*27.InProblem26,showthatthereisanintegerm>0suchthatam=e
foralla EG.
HarderProblems
28.LetGbeasetwithanoperation*suchthat:
1.Gisclosed under*.
2.*isassociative.
3.
ThereexistsanelementeE Gsuch thate*x=xforallxEG.
4.GivenxEG, thereexistsayEGsuchthaty*x=e.
ProvethatGisagroup.(Thusyoumustshowthatx*e=xandx*y=e
fore,yasabove.)
29.
LetGbeafinitenonemptysetwithanoperation*suchthat:
1.Gisclosed under*.
2.*isassociative.
3.
Givena,b,cEGwitha*b=a*c,thenb=c.
4.Givena,b,c,E Gwith b*a=C*a,thenb=c.
ProvethatGmustbeagroupunder*.
30.Give anexampletoshowthattheresultofProblem29canbefalseifG
is
aninfiniteset.
31.
LetGbethegroupofallnonzerorealnumbersundertheoperation*
whichistheordinarymultiplicationofrealnumbers,andletHbethe
groupofallrealnumbersundertheoperation#,whichis theadditionof
realnumbers.
(a)ShowthatthereisamappingF:G
~HofGontoHwhichsatisfies
F(a*b)=F(a)#F(b)foralla,bEG[i.e.,F(ab)=F(a)+F(b)].
(b)ShowthatnosuchmappingFcanbe1-1.
2.SOMESIMPLEREMARKS
Inthisshortsectionweshow thatcertainformal propertieswhichfollowfrom
thegroupaxiomsholdinanygroup. Asamatteroffact,mostoftheseresults
havealready
occurredasproblemsattheendoftheprecedingsection.

64. Sec.2 Some SimpleRemarks 49
Itisalittleclumsy tokeepwritingthe *fortheproductinG,and from
nowonweshallwritethe producta*bsimplyasabforalla,bEG.
Thefirstsuchformalresultsweprovearecontainedin
Lemma2.2.1.IfG
isagroup,then:
(a)Itsidentityelement
isunique.
(b)Every aEGhasa uniqueinversea-IEG.
(c)
IfaEG,(a-l)-l==a.
(d)Fora,bEG,(ab)-l==b-Ia-
l
.
ProofWestartwith Part(a).Whatisexpectedofustocarryoutthe
proof?Wemustshow
thatife,fEGandaf==fa==aforallaEGand
ae==ea==aforallaEG,thene==fThisisveryeasy,forthen e==efand
f==ef;hencee==ef==f,asrequired.
Instead
ofprovingPart(b),weshallproveastrongerresult(listed
belowas
Lemma2.2.2),whichwillhave Part(b)asanimmediateconse­
quence.Weclaim
thatinagroupGif ab==ac,thenb==c;thatis,wecan
cancelagivenelement fromthesamesideofanequation.Toseethis,we
have,for
aEG,anelement uEGsuchthatua==e.Thusfrom ab==acwe
have
u(ab)==u(ac),
so,bytheassociativelaw, (ua)b==(ua)c,thatis,eb==ec.Henceb==eb==
ec==c,andourresultisestablished.Asimilarargumentshows thatif
ba==ca,thenb==c.However,we cannotconcludefromab==cathatb==c;
inanyabeliangroup,yes,butingeneral,no.
NowtogetPart(b)asanimplication
ofthecancellationresult.Suppose
that
b,cEGactasinversesfor
a-;thenab==e==ac,sobycancellation b==c
andweseethattheinverseof
aisunique.Weshallalwayswriteitas a-I.
ToseePart(c),notethatbydefinition a-I(a-l)-l==e;buta-Ia==e,so
bycancellationin
a-l(a-l)-l==e==a-Iawegetthat (a-l)-l==a.
Finally,for Part(d)wecalculate
(ab)(b-1a-
1
)
==((ab)b-1)a-
1
(associativelaw)
==(a(bb-
1
)a-
1
(againtheassociativelaw)
==(ae)a-
1
==aa-
1
==e.
Similarly,(b-Ia-l)(ab)==e.Hence,bydefinition, (ab)-l==b-Ia-
l
.
D

65. 50 Groups Ch.2
Wepromised tolistapiece oftheargumentgivenaboveasaseparate
lemma.We
keepthispromise andwrite
Lemma2.2.2.InanygroupG anda,b,cEG,wehave:
(a)If
ab=ac,thenb=c.
(b)Ifba=ca,thenb=c.
Beforeleavingtheseresults,
notethatifG isthegroupofreal
numbers
under+,thenPart(c)ofLemma2.2.1translatesinto thefamiliar
-(-a)=a.
Thereisonlyascantbit ofmathematicsinthissection;accordingly,
wegiveonlyafewproblems.Noindication
isgivenas tothedifficultyofthese.
PROBLEMS
1.SupposethatGisasetclosed underanassociativeoperationsuchthat
1.givena,yEG,thereisanxEGsuchthatax=y,and
2.givena,wEG,thereisauEGsuchthatua=w.
ShowthatGisagroup.
*2.IfGisafinitesetclosedunderanassociativeoperationsuch thatax=ay
forcesx=yandua=waforcesu=w,foreverya,x,y,u,wEG,prove
thatGisagroup.(This isarepeatofaproblemgivenearlier. Itwillbe
usedin thebodyofthetextlater.)
3.
IfGisagroupinwhich (ab)i=aib
iforthreeconsecutiveintegers i,prove
thatGisabelian.
4.Show
thattheresultofProblem3wouldnotalwaysbetrueiftheword
"three"werereplacedby"two." Inotherwords,show thatthereisa
groupG
andconsecutivenumbersi,i+1suchthatGisnotabelianbut
doeshave thepropertythat(ab)i=aib
iand(ab)i+l=ai+1b
i
+
1
forall
a,binG.
5.LetGbeagroupinwhich (ab)3=a
3
b
3and(ab)5=a
5
b
5foralla,bEG.
ShowthatGisabelian.
6.
LetGbeagroupinwhich(ab)n=anb
nforsomefixed integern>1for
all
a,bEG.Foralla,bEG,provethat:
(a)(ab)n-l=bn-1an-1.
(b)anb
n
-1=bn-1a
n
.
(c)(aba-1b-1)n(n-l)=e.
[Hintfor Part(c):Notethat(aba-1)r=abra-
1 forallintegers r.]

66. Sec.3
3.SUBGROUPS
Subgroups 51
Inorderforustofind outmoreaboutthemakeup ofagivengroupG,itmay
betoomuch
ofatasktotackleall ofGhead-on.Itmightbedesirableto
focusourattention
onappropriatepieces ofG,whicharesmaller,overwhich
wehavesomecontrol,andaresuch
thattheinformationgatheredabout
themcanbeusedtogetrelevantinformationandinsight
aboutGitself.The
questionthenbecomes:Whatshouldserveassuitablepiecesforthiskindof
dissection
ofG?Clearly,whateverwechooseassuchpieces,wewant them
toreflectthefact thatGisagroup,notmerelyanyoldset.
Agroup
isdistinguishedfromanordinarysetbythefact thatitisen­
dowedwithawell-behavedoperation.It
isthusnaturalto demandthatsuch
piecesabovebehavereasonablywithrespect
totheoperationofG.Once
this
isgranted,weareledalmostimmediatelytotheconcept ofasubgroup
ofagroup.
Definition.A nonemptysubset, H,ofagroupG iscalleda subgroup
ofGif,relativetotheproductinG, Hitselfformsagroup.
Westressthephrase"relative
totheproductin G."Take,forinstance,
thesubsetA
={I,-I}in7L,thesetofintegers.Underthemultiplicationof
integers,Aisagroup.ButAisnotasubgroupof
7Lviewedasagroupwith
respectto
+.
EverygroupGautomaticallyhastwoobvioussubgroups,namelyGit­
selfandthesubgroupconsisting
oftheidentityelement, e,alone.Thesetwo
subgroupswecall
trivialsubgroups. Ourinterestwillbeintheremaining
ones,the
propersubgroupsofG.
Beforeproceedingtoacloserlookatthegeneralcharacterofsub­
groups,wewanttolookatsomespecificsubgroups
ofsomeparticular,ex­
plicit
~roups.Someofthegroupsweconsiderarethoseweintroducedasex­
ample~inSection1;wemaintainthenumberinggiventhereforthem.In
someoftheseexamplesweshallverifythatcertainspecifiedsubsetsareindeed
subgroups.Wewouldstronglyrecommendthatthereadercarryoutsucha
verificationinlots
oftheothersandtrytofindotherexamplesforthemselves.
Intryingtoverifywhether
ornotagivensubset ofagroupisasub­
group,wearesparedcheckingone
oftheaxiomsdefiningagroup,namely
theassociativelaw.SincetheassociativelawholdsuniversallyinagroupG,
givenanysubset
AofGandanythreeelementsof A,thentheassociative
lawcertainlyholdsforthem.Sowemustcheck,foragivensubset
AofG,
whether
AisclosedundertheoperationofG,whether eisinA,andfinally,
given
aEA,whethera-Iisalsoin A.

67. 52 Groups Ch.2
Notethatwecansaveonemorecalculation.Supposethat AeGis
nonemptyand thatgivena,bEA,thenabEA.Supposefurtherthatgiven
aEA,thena-IEA.Thenweassertthat eEA.ForpickaEA;thena-IEA
bysupposition,hence aa-
lEA,againbysupposition.Since aa-
l=e,we
havethat
eEA.ThusaisasubgroupofG.Inotherwords,
Lemma
2.3.1.Anonemptysubset AeGisasubgroupofGifand
onlyif
AisclosedwithrespecttotheoperationofGand,given aEA,then
a-IEA.
Wenowconsidersomeexamples.
Examples
1.LetGbethegroup 7Lofintegersunder+andletHbethesetofevenin­
tegers.Weclaim
thatHisasubgroupof 7L.Why?Is Hclosed,that is,given
a,bEH,isa+bEH?Inotherwords,if a,bareevenintegers, isa+ban
eveninteger?
Theanswerisyes,so Hiscertainlyclosedunder +.Nowtothe
inverse.Sincetheoperationin
7Lis+,theinverseofaE7Lrelativetothisop­
eration
is-a.IfaEH,thatis,ifaiseven,then -aisalsoeven,hence
-aEH.Inshort,Hisasubgroupof7Lunder+.
2.LetGonceagain bethegroup 7Lofintegersunder +.InExample 1,H,
thesetofevenintegers,can bedescribedin anotherway:namely Hconsists
ofallmultiplesof2.ThereisnothingparticularinExample1thatmakesuse
of2itself.Letm>1beanyinteger andletHmconsistofallmultiplesof m
in7L.Weleaveittothe readertoverifythatHmisasubgroupof7Lunder+.
3.LetSbeanynonemptysetandletG =A(S).IfaES,letH(a)=
{fEA(S)If(a)=a}.Weclaimthat H(a)isasubgroupofG.Forif
f,gEH(a),then(fg)(a)=f(g(a»=f(a)=a,sincef(a)=g(a)=a.Thus
fgEH(a).Also,iffEH(a),thenf(a)=a,sothatf-l(f(a»=f-l(a).But
f-l(f(a»=f-l(a)=i(a)=a.Thus,since a=f-l(f(a»=f-l(a),wehave
thatf-
l
EH(a).Moreover,Hisnonempty.(Why?)Consequently, H(a)isa
subgroup
ofG.
4.LetGbeasinExample6 ofSection1,andHasinExample7.ThenHisa
subgroup
ofG(seeProblem3inSection1).
5.LetGbeasinExample 6,HasinExample 7,andKasinExample8in
Section
1.ThenKCHeGandKisasubgroupofbothHandofG.
6.Let
C'bethenonzerocomplexnumbersasagroupunderthemultiplica­
tion
ofcomplexnumbers. LetV={aE
C'IlaIisrational}.Then Visasub­
group
of
C'.ForiflalandIblarerational,then labl=lallblisrational,so

68. Sec.3 Subgroups 53
abEV;also,Ia-II=lilaIisrational,hence a-IEV.Therefore,Visasub­
group
of
C'.
7.LetC'andVbeasaboveandlet
U={aEC'Ia=cos()+isin(),()anyreal}.
Ifa=cos()+isin()andb=cosl/J+isinl/J,wesawin Chapter1thatab=
cos(()+l/J)+isin(()+l/J),sothatabEU,andthata-I=cos()-isin()=
cos(-()+isin(-()EU.Also,lal=1,sincelal=Vcos
2
()
+sin
2
()
=1.
Therefore,Ueve
C'andUisasubgroupbothofVandofC'.
8.LetC',U,Vbeasabove,andletn>1beaninteger.Let()n=
cos(27Tln)+isin(27T/n),andletB={I,()n,(J~,•••,(J~-l}.Since(J~=1(as
wesawby
DeMoivre'sTheorem),itiseasilychecked thatBisasubgroupof
U,V,and
C',andisofordern.
9.LetGbeanygroup andletaEG.ThesetA={a
i
Iianyinteger}isasub­
group
ofG.For,by therulesofexponents,if a
iEAanda
j
EA,thenaia
j
=
a
i+
j
,
soisinA.Also,(ai)-I=a-
i
,
so(ai)-IEA.Thismakes Aintoasub-
group
ofG.
Aisthecyclicsubgroup ofGgeneratedbyainthefollowingsense.
Definition.
Thecyclicsubgroup ofGgeneratedbyaisaset{a
i
Iiany
integer}.
Itisdenoted(a).
NotethatifeistheidentityelementofG,then(e)=tel.InExample8,
thegroup Bisthecyclicgroup (()n)of
Cgeneratedby()n.
10.LetGbeanygroup;for aEGletC(a)={gEGIga=agIoWeclaim
thatC(a)isasubgroupofG.First,theclosureofC(a).Ifg,hEC(a),then
ga=agandha=ah,thus(gh)a=g(ha)=g(ah)=(ga)h=(ag)h=
a(gh)~by therepeateduseoftheassociativelaw), henceghEC(a).Also,if
gEtea),thenfromga=agwehave g-I(ga)g-l=g-I(ag)g-I,which
simplifies
toag-
I =g-Ia;whenceg-lEC(a).So,C(a)istherebyasub­
group
ofG.
Theseparticularsubgroups
C(a)willcome uplaterforusandtheyare
givenaspecialname. WecallC(a)thecentralizerofainG.Ifinagroup
ab=ba,wesaythataandbcommute.ThusC(a)isthesetofallelementsin
G
thatcommutewith a.
11.LetGbeanygroup andletZ(G)={zEGIzx=xzforallxEG}.We
leaveit tothereadertoverifythatZ(G)isasubgroupofG.Itiscalledthe
centerofG.

69. 54 Groups Ch.2
12.LetGbeanygroupand Hasubgroupof G.ForaEG,leta-1Ha=
{a-1haIhEH}.Weassertthat a-1Haisasubgroupof G.Ifx=a-1h1aand
y=a-
1
h
2awherehI,h
2
EH,thenxy=(a-
1
h1
a)(a-
1
h
2
a)=a-
l
(h1h
2
)a
(associativelaw),andsince HisasubgroupofG, h
1h
2EH.Therefore,
a-
1
(h
1
h
2
)aEa-1Ha,whichsaysthat xyEa-1Ha.Thusa-1Haisclosed.
Also,if
x=a-1haEa-1Ha,then,as iseasilyverified,X-I=(a-1ha)-1=
a-1h-1aEa-1Ha.Therefore,a-1Haisasubgroupof G.
Anevendozenseemstobeabouttherightnumberofexamples,so we
goontootherthings.Lemma2.3.1pointsoutforuswhatweneed inorder
thatagivensubsetofagroupbeasubgroup.Inanimportantspecialcase
we
canmakeaconsiderablesavingincheckingwhetheragivensubset Hisa
subgroupof
G.Thisisthecaseinwhich Hisfinite.
Lemma2.3.2.SupposethatG isagroupand Hanonemptyfinitesub­
setofGclosedundertheproductin
G.ThenHisasubgroupof G.
Proof.ByLemma2.3.1wemustshow thataEHimpliesa-IEH.If
a=e,thena-1=eandwearedone.Supposethenthata
=1=e;consider
theelementsa,a
2
,
•••,a
n
+
1
,
wheren=IHI,theorderofH.Herewe
havewrittendownn+1elements,all oftheminHsinceHisclosed, andH
hasonlyndistinctelements. Howcanthisbe?Onlyifsometwo oftheele­
mentslistedareequal;
putanotherway,onlyif a
i
=a
j
forsome1
~i<
j~n+1.Butthen,bythecancellation propertyingroups,a
j
-
i=e.Since
j-i
2::1,a
j
-
iEH,henceeEH.However,j-i-I
2::0,soa
j
-
i-1EH
andaa
j
-
i-1=a
j
-
i=e,whencea-1=a
j
-
i-1EH.Thisprovesthe
lemma.D
Animmediate,butneverthelessimportant,corollarytoLemma 2.3.2isthe
Corollary.IfG
isafinitegroupand HanonemptysubsetofGclosed
undermultiplication,then
Hisasubgroupof G.
PROBLEMS
EasierProblems
1.IfA,BaresubgroupsofG,showthat AnBisasubgroupof G.
2.Whatisthecyclicsubgroupof 7Lgeneratedby -1under+?
3.LetS3bethesymmetricgroupofdegree 3.Findallthesubgroupsof S3'
4.Verifythat Z(G),thecenterofG, isasubgroupof G.(SeeExample11.)

70. Sec.3 Subgroups 55
5.IfC(a)isthecentralizerofainG(Example10),prove thatZ(G)
naEGC(a).
6.ShowthataEZ(G)ifandonlyif C(a)=G.
7.InS3,findC(a)foreachaES3.
8.IfGisanabeliangroup andifH={aE GIa
2
=e},showthatHisa
subgroup
ofG.
9.Giveanexample
ofanonabeliangroupforwhichtheHinProblem8is
notasubgroup.
10.
IfGisanabeliangroup andn>1aninteger,let An={anIaEG}.
ProvethatAnisasubgroupofG.
*11.
IfGisanabeliangroup andH={aE GIan(a)=eforsome n(a)>1de­
pending
ona},provethatHisasubgroupofG.
Wesay
thatagroupGiscyclicifthereexistsanaEGsuchthat
everyxEGisapowerofa,thatis,x=a
j
forsomej.Inotherwords,G
iscyclicifG =(a)forsome aEG,inwhichcasewesay thataisagen­
eratorforG.
*12.Prove
thatacyclicgroup isabelian.
13.
IfGiscyclic,show thateverysubgroup ofGiscyclic.
14.
IfGhasnopropersubgroups,prove thatGiscyclic.
15.
IfGisagroupandHanonemptysubsetofGsuchthat,given a,bEH,
thenab-1EH,prove thatHisasubgroup ofG.
Middle-Level
Problems
*16.IfGhasnopropersubgroups,prove thatGiscyclicoforderp,wherep
isaprime number.(Thissharpens theresultofProblem14.)
17.
IfGisagroupanda,xEG,prove thatC(x-
1
ax)=x-
1
C(a)x.[SeeEx­
amples10
and12forthedefinitionsofC(b)andofx-1C(a)x.]
18.IfSisanonemptysetandxeS,showthatT(X)={fEA(S)If(X)c
X}isasubgroupofA(S)ifXisfinite.
19.IfA,Baresubgroupsofanabeliangroup
G,letAB=tabIaEA,bEB}.
ProvethatABisasubgroupofG.
20.Giveanexample
ofagroupG andtwosubgroups A,BofGsuchthat
ABisnotasubgroupofG.
21.
IfA,BaresubgroupsofGsuchthatb-
1
AbCAforallbEB,showthat
ABisasubgroupofG.
*22.IfAandBarefinitesubgroups, ofordersm
a~dn,respectively,ofthe
abeliangroupG,prove thatABisasubgroupofordermnifmandnare
relatively
prime.

71. 56 Groups Ch.2
23.WhatistheorderofABinProblem22ifmandnarenotrelativelyprime?
24.
IfHisasubgroupofG,letN=n
xEGx-
1
Hx.ProvethatNisasubgroup
ofGsuchthaty-
1
Ny=NforeveryyEG.
HarderProblems
25.LetS,X,T(X)beasinProblem18(butXnolongerfinite).Give anex­
ample
ofasetSandaninfinitesubset XsuchthatT(X)isnotasub­
group
ofA(S).
*26.LetGbeagroup,HasubgroupofG.LetHx={hxIhEH}.Showthat,
given
a,bEG,thenHa=HborHanHb=0.
*27.IfinProblem26HisafinitesubgroupofG,provethatHaandHbhave
thesamenumberofelements.Whatisthisnumber?
28.LetM,N besubgroupsofGsuchthatX-IMxCMandX-INxCNfor
all
xEG.ProvethatMNisasubgroupofGandthatx-
1
(MN)xCMN
forallxEG.
*29.
IfMisasubgroupofGsuchthatx-
1
MxCMforallxEG,provethat
actuallyX-IMx=M.
30.IfM,Naresuchthatx-
1
Mx=Mandx-
1
Nx=NforallxEG,andif
MnN=(e),provethatmn=nmforany mEM,n EN.(Hint:Con­
sider
theelementm-
1
n-
1
mn.)
4.LAGRANGE'S THEOREM
Weareabouttoderivethefirstrealgroup-theoreticresultofimportance.
Althoughitsproofisrelativelyeasy,this theoremisliketheA-B-C'sforfi­
nitegroups andhasinterestingimplicationsin numbertheory.
Asamatteroffact,those ofyouwhosolvedProblems26and27ofSec­
tion3haveall
thenecessaryingredients toeffecta proofoftheresult.The
theoremsimplystates thatinafinitegrouptheorderofasubgroupdivides
theorderofthegroup.
Tosmooththeargumentofthistheorem-whichisduetoLagrange­
andforusemanytimeslater,we makeashortdetourintotherealmofset
theory.
Justastheconceptof"function"runsthroughoutmostphasesofmath­
ematics,soalso
doestheconceptof"relation."ArelationisastatementaRb
abouttheelementsa,bES.IfSisthesetofintegers,a=bisarelation
onS.Similarly,a<bisarelationonS,asisa
~b.

72. Sec.4 Lagrange'sTheorem 57
Definition.Arelation'"'-'onasetS iscalledan equivalencerelation if,
forall
a,b,cES,itsatisfies:
(a)
a
'"'-'a(reflexivity).
(b)a'"'-'bimpliesthatb'"'-'a(symmetry).
(c)a'"'-'b,b'"'-'cimpliesthata'"'-'c(transitivity).
Ofcourse,equality, =,isanequivalencerelation,so thegeneralnotion
ofequivalencerelation isageneralizationofthatofequality.Inasense,an
equivalencerelationmeasuresequalitywithregard
tosomeattribute.This
vague
remarkmaybecomeclearerafterweseesomeexamples.
Examples
1.LetSbealltheitemsforsaleinagrocerystore;wedeclare a
'"'-'b,for
a,bES,ifthepriceofaequalsthatofb.Clearly,thedefiningrules ofan
equivalencerelationholdforthis'"'-'.Notethatinmeasuringthis"generalized
equality"
onSweignoreall propertiesoftheelementsofSotherthantheir
price.So a
'"'-'biftheyareequalasfaras theattributeofpriceisconcerned.
2.LetSbetheintegersandn>1afixedinteger. Wedefinea'"'-'bfora,
bESifnI(a- b).Weverifythatthisisanequivalencerelation.Since nI0
and0=a-a,wehavea'"'-'a.BecausenI(a-b)impliesthatnI(b-a), we
have
thata
'"'-'bimpliesthatb'"'-'a.Finally,if a'"'-'bandb'"'-'c,then
nI(a-b)andnI(b-c);hencenI((a- b)+(b-c»,thatis,nI(a-c).
Therefore,a'"'-'c.
Thisrelationontheintegersisofgreatimportancein numbertheory
andiscalledcongruencemodulon;whena'"'-'b,wewritethisas a==bmodn
[or,sometimes,as a==ben)],whichisread"acongruenttobmodn."We'll
berunningintoitveryoftenfromnowon. Asweshallsee,this isaspecial
case
ofamuchwider phenomenoningroups.
3.WegeneralizeExample 2.LetGbeagroupandHasubgroupofG.For
a,bEG,definea
'"'-'bifab-
1
EH.SinceeEHande=aa-
1
,
wehavethat
a
'"'-'a.Also,if ab-
1
EH,thensince HisasubgroupofG,(ab-1)-1EH.But
(ab-
1
)-1=(b-
1
)-la-1=ba-t,soba-
1
EH,henceb'"'-'a.Thistellsus that
a'"'-'bimpliesthatb'"'-'a.Finally,if a'"'-'bandb'"'-'c,thenab-1EHand
bc-
1
EH.But(ab-
1
)(bc-
1
)
=ac-
1
,
whenceac-
1
EHandtherefore
a
'"'-'c.Wehaveshown thetransitivityof'"'-',thus---isanequivalencerelation
onG.
Note
thatifG =7L,thegroupofintegersunder+,andHisthesub­
groupconsisting
ofallmultiplesofn,forn>1 afixedinteger, thenab-
1
EH

73. 58 Groups Ch.2
translatesinto a==ben).Socongruencemodnisaveryspecialcase ofthe
equivalencewehavedefinedinExample
3.
Itisthisequivalencerelation thatweshalluseinprovingLagrange's
theorem.
4.LetGbeanygroup. Fora,bEGwedeclarethata
---bifthereexistsan
xEGsuchthatb=x-lax.Weclaimthatthisdefinesanequivalencerela­
tiononG.First,
a
---afora=e-lae.Second,if a---b,thenb=x-lax,hence
a=(X-l)-lb(x-
l
),
sothatb
---a.Finally,if a---b,b---c,thenb=x-lax,c=
y-lbyforsome x,yEG.Thusc=y-l(x-lax)y=(xy)-la(xy),andso a---c.
Wehaveestablished thatthisdefinesanequivalencerelation onG.
Thisrelation,too,playsanimportantroleingrouptheoryand
isgiven
thespecialname
conjugacy.Whena
---bwesaythat"aandbareconjugate
in
G."NotethatifGisabelian,then a
---bifandonlyif a=b.
Wecouldgoonandontogivenumerousinterestingexamples ofequiva­
lencerelations,butthiswouldsidetrackusfromourmaingoalinthissection.
Therewillbenolack
ofexamplesintheproblemsattheendofthissection.
Wego
onwithourdiscussionandmakethe
Definition.If
---isanequivalencerelationon S,then[a],theclassof
a,isdefinedby [a]={bESIb---a}.
Letusseewhattheclass ofaisinthetwoexamples,Examples3and 4,
justgiven.
InExample
3,a
---bifab-
l
EH,thatis,ifab-
l
=h,forsome hEH.
Thusa---bimpliesthata=hb.Ontheotherhand,if a=kbwherekEH,
thenab-
l
=(kb)b-
l
=kEH,soa---bifandonlyif aEHb={hbIhEH}.
Therefore,[b]=Hb.
ThesetHbiscalleda rightcosetofHinG.WeranintosuchinProb­
lem
26ofSection3.Notethat bEHb,sinceb=ebandeEH(alsobecause
bE[b]
=Hb).Rightcosets, andlefthandedcounterpartsofthemcalled left
cosets,
playimportantrolesinwhatfollows.
InExample
4,wedefineda
---bifb=x-laxforsome xEG.Thus[a]=
{x-laxIxEG}.Weshalldenote [a]inthiscaseas cl(a)andcallitthe conju­
gacyclass
ofainG.IfG isabelian,then cl(a)consistsof aalone.Infact,if
aEZ(G),thecenterofG,then cl(a)consistsmerely ofa.
Thenotionofconjugacyanditspropertieswillcropupagainoften,es­
peciallyinSection
11.
Weshallexaminetheclass ofanelementainExample2laterinthis
chapter.
Theimportantinfluencethatanequivalencerelationhasonaset isto
breakitupandpartitionitintonicedisjointpieces.

74. Sec.4 Lagrange'sTheorem 59
Theorem2.4.1.If'"'"'isanequivalencerelationon S,thenS =Ural,
wherethisunionrunsoveroneelementfromeachclass,andwhere
[a]
=1=[b]
impliesthat [a]n[b]=0.Thatis,'"'"'partitionsSintoequivalenceclasses.
ProofSincea E[a],wehaveUaES[a]=S.Theproofofthesecondas­
sertion
isalsoquiteeasy.Weshowthatif [a]
=1=[b],then[a]n[b]=0,or,
what
isequivalenttothis,if [a]n[b]
=1=0,then[a]=[b].
Suppose,then,that [a]n[b]=1=0;letcE[a]n[b].Bydefinitionof
class,
C
'"'"'asincec E[a]andc~bsincec E[b].Therefore,a~cbysym­
metryof'"'"',andso,sincea'"'"'candc'"'"'b,wehavea~b.Thusa E[b];if
xE[a],thenx'"'"'a,a~bgivesusthat x'"'"'b,hencexE[b].Thus[a]C[b].
Twargumentisobviouslysymmetricin aandb,sowehave [b]C[a],
whence[a]=[b],andourassertionabove isproved.
Thetheorem
isnowcompletelyproved. D
WenowcanproveafamousresultofLagrange.
Theorem2.4.2(Lagrange'sTheorem).IfG
isafinitegroupand His
asubgroupofG,thentheorderof Hdividestheorderof G.
ProofLetuslookbackatExample 3,whereweestablishedthatthe
relation
a
'"'"'bifab-1EHisanequivalencerelationandthat
[a]=Ha={halhEH}.
Let
kbethenumberofdistinct classes-eallthemHa 1,...,Hak.ByTheo­
rem2.4.1,G
=HalUHa2U...UHakandweknowthatHa
jnHai=0if
i
=1=j.
WeassertthatanyHaihas IHI=orderofHnumberofelements.Map
H~Haibysendingh~hai.We claimthatthismap is1-1,forifhai =
h'ai,thenbycancellationinG wewouldhaveh =h';thusthemap is1-1.It
isdefinitelyontobytheverydefinitionofHai.So HandHaihavethesame
number,
IHI,ofelements.
SinceG
=HalU...UHakandtheHaiaredisjointandeachHaihas
IHIelements,wehavethat IGIkIHI.ThusIHIdividesIGIandLa-
grange'sTheorem
isproved.D
AlthoughLagrangesoundslikeaFrenchname, J.L.Lagrange(1736-1813)was
actuallyItalian,havingbeenbornandbroughtupinTurin.
Hespentmostof
hislife,however,inFrance.Lagrangewasagreatmathematicianwhomade
fundamentalcontributionstoalltheareas
ofmathematicsofhisday.
IfGisfinite,thenumberofrightcosetsof HinG,namely IGI/IHI,is
calledthe indexofHinGandiswrittenasi
G(H).

75. 60 Groups Ch.2
RecallthatagroupG issaidtobe cyclicifthereisanelementaEG
suchthateveryelementinG
isapowerofa.
Theorem2.4.3.A groupGofprimeorder iscyclic.
ProofIfHisasubgroupofGthen,byinvokingLagrange'sTheorem,
IHIdividesIGI=p,paprime,so IHI=1orp.SoifH
=1=(e),thenH=G.If
aEG,a=1=e,thenthepowers ofaformasubgroup (a)ofGdifferentfrom
(e).Sothissubgroup isallofG.Thissaysthatany xEGisoftheform x=a
i
•
Hence,Giscyclicbythedefinitionofcyclicgroup. D
IfGisfiniteand aEG,wesawearlierintheproof ofLemma2.3.2that
an(a)=eforsomen(a)
2::1,dependingon a.Wemakethe
Definition.IfGisfinite,thenthe orderofa,written0(a),istheleast
positiveinteger
msuchthat am=e.
Supposethata
EGhasorderm.Considertheset A={e,a,a
2
,
•••,a
m
-
1~;
weclaimthat Aisasubgroupof G(sinceam=e)andthatthe melements
listedin
Aaredistinct.Weleavetheverificationoftheseclaimstothe
reader.Thus
IAI=m=o(a).SinceIAIIIG/,wehave
Theorem2.4.4. IfGisfiniteand a
EG,then 0(a)IIGI.
IfaEG,whereG isfinite,wehave,byTheorem 2.4.4,IGI
Thus
Wehaveprovedthe
k.o(a).
Theorem2.4.5.IfGisafinitegroup ofordern,thenan=eforall
aEG.
Whenweapplythislastresulttocertainspecialgroupsarising innum­
bertheory,weshallobtainsomeclassicalnumber-theoreticresultsdueto
FermatandEuler.
Let
7Lbetheintegersandlet n>1beafixedinteger.Wegobackto
Example2
ofequivalencerelations,where wedefineda
==bmodn(a
congruentto bmodn)ifnI(a-b).Theclass ofa,raj,consistsofall
a+nk,wherekrunsthroughalltheintegers.Wecallitthe congruenceclass
ofa.

76. Sec.4 Lagrange'sTheorem 61
ByEuclid'sAlgorithm,givenanyinteger b,b=qn+r,where
o~r<n,thus[b]=[r].Sothenclasses[0], [1],...,[n-1]giveusall
thecongruenceclasses.Weleaveittothereadertoverifythattheyare
distinct.
LetEn={[OJ,[1],...,[n-I]}.Weshallintroducetwooperations, +
and·inEn.Under+Enwillformanabeliangroup; under·Enwillnotform
agroup,butacertainpieceofitwillbecomeagroup.
Howtodefine
[a]+[b]?Whatismorenaturalthantodefine
[a]+[b]=[a+b].
Butthere isaflyintheointment.Isthisoperation +in
Enwell-defined?
Whatdoesthatmean?Wecanrepresent
[a]bymanya's-forinstance,if
n=3,[1]=[4]=[-2]=...,yetweareusinga particulara todefinethe
addition.What
wemustshow isthatif[a]=[a']and[b]=[b'],then[a+b]
=[a'+b'],forthenwewillhave [a]+[b]=[a+b]=[a'+b']=
[a']+[b'].
Supposethat [a]=
[a'];thennI(a-a').Alsofrom [b]=[b'],
n
I(b-b'),hencenI«a-a')+(b-b'»=«a+b)-(a'+b'».There­
fore,
a+b
==(a'+b')modn,andso [a+b]=[a'+b'].
Sowenowhaveawell-definedadditioninEn.Theelement [0]actsas
theidentityelementand [-a]actsas- [a],theinverseof [a].Weleaveitto
thereadertocheckoutthatEnisagroupunder +.Itisacyclicgroupof
orderngeneratedby
[1].
Wesummarizethisall as
Theorem2.4.6.
Enformsacyclicgroupundertheaddition [a]+[b]=
[a+b].
HavingdisposedoftheadditioninEn,weturntotheintroductionofa
multiplication.Again,what
ismorenaturalthandefining
[a]·[b]=[ab]?
So,forinstance,if n=9,[2][7]=[14]=[5],and[3][6]=[18]=[0].Under
thismultiplication-weleavethefactthatit iswell-definedtothe reader­Endoesnotform agroup.Since [0][a]=[0]foralla,andtheunitelement
undermultiplication
is[1],[0]cannothaveamultiplicativeinverse.Okay,
whynottrythenonzeroelements
[a]
=1=[0]asacandidateforagroupunder
thisproduct?
Hereagainit isnogoifnisnotaprime. Forinstance,if n=6,
then[2]
=1=[0],[3]=1=[0],yet[2][3]=[6]=[0],sothenonzeroelementsdo
not,ingeneral,giveusagroup.
Soweask:CanwefindanappropriatepieceofEnthatwillforma

77. 62 Groups Ch.2
groupundermultiplication?Yes!Let Un={[a]E7LnI(a,n)=I},noting
that
(a,n)=1ifandonlyif (b,n)=1for[a]=[b].BytheCorollaryto
Theorem
1.5.5,if(a,n)=1and(b,n)=1,then(ab,n)=1.So[a][b]=[ab]
yieldsthatif [a],[b] EUn,then[ab]EUnandUnisclosed.Associativity is
easilychecked,followingfromtheassociativityoftheintegersundermulti­
plication.Theidentityelement
iseasytofind,namely [1].Multiplicationis
commutativein Un.
Notethatif [a][b]=[a][c]where[a]EUn'thenwehave[ab]=[ac],
andso [ab-ac]=[0].Thissaysthat nIa(b- c)=ab-ac;butaisrela­
tivelyprimeto
n.ByTheorem 1.5.5onemusthavethat nI(b-c),andso
[b]=[c].Inotherwords, wehavethecancellationpropertyin Un.ByProb­
lem2ofSection
2,Unisagroup.
What
istheorderof Un?Bythedefinitionof Un,IUnI=number ofin­
tegers1
~m<nsuchthat (m,n)=1.Thisnumbercomesupoftenand we
giveitaname.
Definition.The
Eulercp-function,
cp(n),isdefinedbycp(1)=1and,
for
n>1,
cp(n)=thenumberofpositiveintegers mwith1::::;m<nsuch
that
(m,n)=1.
ThusIUnl=
cp(n).Ifn=p,aprime,wehavecp(p)=p-1.Wesee
thatcp(8)=4foronly 1,3,5,7arelessthan8andpositiveandrelatively
primeto
8.Wetryanotherone,
cp(15).Thenumbers 1~m<15relatively
primeto
15are1,2,4,7,8,11,13,14,so
cp(15)=8.
Letuslookatsomeexamplesof Un.
1.Us={[I],[3],[5],[7]}.Notethat [3][5]=[15]=[7],[5]2=[25]=[1].In
fact,
Usisagroupoforder4inwhich a
2
=eforeveryaEUs.
2.V
15={[I],[2], [4], [7],[8],[11],[13],[14]}. Notethat [11][13]=[143]=
[8],[2]4=[1],andsoon.
Thereadershouldverifythat
a
4
=e=[1]foreveryaEU
15
•
3.Ug={[I],[2],[4],[5], [7],[8]}.Notethat [2]1=[2],[2]2=[4],[2]3=[8],
[2
4
]
=[16]=[7],[2]5=[32]=[5];also[2]6=[2][2]5=[2][5]=[10]=[1].
Sothepowersof [2]giveuseveryelementin Ug•ThusUgisacyclicgroupof
order
6.Whatotherelementsin UggenerateUg?
InparalleltoTheorem 2.4.6wehave
Theorem2.4.7.
Unformsanabeliangroup,undertheproduct
[a][b] =[ab],oforder
cp(n),wherecp(n)istheEulercp-function.

78. Sec.4 Lagrange'sTheorem 63
Animmediateconsequence ofTheorems2.4.7and2.4.5 isafamousre­
sultinnumbertheory.
Theorem2.4.8(Euler). Ifaisanintegerrelativelyprimeto n,thenacp(n)==1modn.
ProofUnformsagroup ofordercp(n),sobyTheorem2.4.5,acp(n)=e
foralla
EUn'Thistranslatesinto [aCP(n)]=[a]cp(n)=[1],whichin turntrans­
latesinto
nI
(aCP(n)-1)foreveryinteger arelativelyprime top.Inother
words,acp(n)==1modn.D
Aspecialcase,where n=pisaprime,isduetoFermat.
Corollary(Fermat). Ifpisaprimeandp~a,then
a
P
-
1
==1modp.
Foranyintegerb,b
P==bmodp.
ProofSincecp(p)=p-1,if(a,p)=1,wehave,by Theorem2.4.8,
thata
P
-1==l(p),hencea
1
•
a
P
-
1==a(p),sothata
P==a(p).Ifp Ib,then
b==O(p)andb
P==O(p),so thatb
P==b(p).D
LeonardEuler(1707-1785)wasprobably thegreatestscientist thatSwitzerland
hasproduced.
Hewasthemostprolific ofallmathematiciansever.
PierreFermat(1601-1665)wasagreatnumbertheorist.Fermat'sLastThe­
orem-whichwasinfactfirstprovedin1994byAndrew Wiles-statesthatthe
equation
an+b
n
=c
n
(a,b,c,nbeingintegers)hasonlythetrivialsolutionwhere
a=0orb=0orC=0ifn>2.
OnefinalcautionarywordaboutLagrange'sTheorem.Itsconversein
general
isnottrue.Thatis,ifG isafinitegroupof ordern,thenitneed not
betruethatforeverydivisor mofnthereisasubgroupofGoforderm.A
groupwiththisproperty
isveryspecialindeed, anditsstructurecan be
spelledoutquitewellandprecisely.
PROBLEMS
EasierProblems
1.Verifythattherelation
'"'-'isanequivalencerelationontheset Sgiven.
(a)S=IRreals,a'"'-'bifa-bisrational.
(b)S=C,thecomplexnumbers, a'"'-'biflaI=Ib\.
(c)S=straightlinesintheplane, a'"'-'bifa,bareparallel.
(d)S=setofallpeople,a'"'-'biftheyhavethesamecoloreyes.

79. 64 Groups Ch.2
2.Therelation'"-ontherealnumbersIRdefinedby a'"-bifbotha>band
b>aisnotanequivalencerelation.Whynot? Whatpropertiesofan
equivalencerelationdoesitsatisfy?
3.Let
'"-bearelationonasetSthatsatisfies(1) a'"-bimpliesthatb'"-a
and(2)a~bandb--cimpliesthata~c.Theseseem toimplythat
a'"-a.Forifa'"-b,thenby(1),b'"-a,soa--b,b'"-a,soby(2), a'"-a.If
thisargumentiscorrect,thentherelation'"-mustbeanequivalencerela­
tion.
Problem2showsthatthisisnotso.Whatiswrongwiththeargu­
mentwehavegiven?
4.LetSbeaset,{Sa}nonemptysubsetssuch thatS=UaSa andSan
Sf3=
oifa=1={3.Defineanequivalencerelation onSinsuchaway thattheSa
arepreciselyall theequivalenceclasses.
*5.LetGbeagroupandHasubgroupofG.Define,for a,bEG,a--bif
a-IbEH.Provethatthisdefinesanequivalencerelation onG,andshow
that[a]=aH={ahIhEH}.ThesetsaHarecalledleftcosetsofH
inG.
6.
IfGisS3andH={i,f},wheref:S
~Sisdefinedby f(XI)=
X2,f(X2)=Xl,f(X3)=X3'listalltherightcosets ofHinGandlistall
theleftcosetsofHinG.
7.
InProblem6,iseveryrightcoset ofHinGalsoaleftcoset ofHinG?
8.Ifeveryrightcoset ofHinG isaleftcoset ofHinG,provethat
aHa-
1
=HforallaEG.
9.In
E
16
,writedownall thecosetsofthesubgroupH={[OJ,[4],[8],[12]}.
(Since
theoperationin
Enis+,writeyourcosetas [a]+H.Wedon't
needtodistinguishbetweenrightcosets andleftcosets,sinceEnis
abelianunder+.)
10.InProblem9,whatistheindexofHinE
16?(Recallthatwedefinedthe
indexiG(H)asthenumberofrightcosetsinG.)
11.
ForanyfinitegroupG,showthatthereareasmanydistinctleftcosets of
HinGas therearerightcosets ofHinG.
12.IfaHandbHaredistinctleftcosets ofHinG,areHaandHbdistinct
rightcosets
ofHinG?Provethatthisistrueorgiveacounterexample.
13.FindtheordersofalltheelementsofU
l8
•IsU
l8cyclic?
14.Find
theordersofalltheelementsofU
20
•IsU
20cyclic?
*15.
Ifpisaprime,show thattheonlysolutions ofx
2
==1modpareX==
1modporx==-1modp.
*16.IfGisafiniteabelian groupandaI,...,anareallitselements,show
thatX=ala2...anmustsatisfyx
2=e.
17.IfGisofoddorder,whatcanyousay aboutthexinProblem16?

80. Sec.4 Lagrange'sTheorem 65
18.Usingtheresults ofProblems15and16,provethatifpisanoddprime
number,
then(p-I)!
==-1modp.(Thisisknownas Wilson'sTheo­
rem.)Itis,ofcourse,also trueifp=2.
19.Findallthedistinctconjugacyclasses ofS3 •
20.InthegroupG ofExample6 ofSection1,find theconjugacyclass ofthe
elementT
a
, b•Describeitinterms ofaandb.
21.LetG bethedihedralgroup oforder8(seeExample 9,Section1).Find
theconjugacyclassesinG.
22.VerifyEuler'sTheoremforn=14anda=3,andforn=14anda=5.
23.InU41,showthat thereisanelementasuchthat[a]2=[-1],thatis,an
integer
asuchthata
2
==-1mod41.
24.Ifpisaprimenumberoftheform4n+3,showthatwecannotsolve
x
2
==-1modp
[Hint:UseF'ermat's Theoremthata
P
-
1
==1modpifp{a.]
25.Showthatthenonzeroelementsin 7L
nformagroup undertheproduct
[a][b]=[ab]ifandonlyifnisaprime.
Middle-LevelProblems
26.LetGbeagroup,
HasubgroupofG,andletSbetheset ofalldistinct
rightcosets
ofHinG,Tthesetofallleftcosets ofHinG.Provethatthere
isa1-1mappingofSontoT.(Note:Theobviousmapthatcomestomind,
whichsends
HaintoaH,isnottherightone.SeeProblems5and12.)
27.IfaH=bHforcesHa=HbinG,show thataHa-
1
=HforeveryaEG.
28.IfGisacyclicgroup ofordern,showthatthereare
'P(n)generatorsfor
G.Givetheirformexplicitly.
29.IfinagroupG, aba-
1
=bi,showthatarba-
r =b
ir
forallpositiveinte­
gers
r.
30.IfinGa
5
=eandaba-
1
=b
2
,
findo(b)ifb
=1=e.
*31.Ifo(a)=mandas=e,provethatmIs.
32.LetG
beafinitegroup, HasubgroupofG.Letf(a)betheleastpositive
msuchthatamEH.Provethatf(a)lo(a).
33.If i
=1=fEA(S)issuchthatfP=i,paprime,andifforsome
sES,fj(s)=sforsome1~j<p,showthatf(s)=s.
34.IffEA(S)hasorderp,paprime,show thatforeverysEStheorbitof
sunderfhasoneorpelements.[Recall: Theorbitofsunderfis
{fj(s)Ijanyinteger}.]

81. 66 Groups Ch.2
35.If1EA(S)hasorderp,paprime,andSisafinitesethavingnelements,
where(n,p)==1,showthatforsomesES,I(s)==s.
HarderProblems
36.Ifa>1isaninteger,showthatnIq;(a
n
-1),whereq;istheEuler
q;-function.[Hint:Considertheintegersmod(a
n
-
1).]
37.
Inacyclicgroupofordern,showthatforeachpositiveintegermthatdi­
vides
n(includingm==1andm==n)thereare
q;(m)elementsoforderm.
38.Using theresultofProblem37,showthatn==~mlnq;(m).
39.LetGbeafiniteabeliangroupofordernforwhichthenumberofsolu­
tions
ofx
m
==eisatmostmforanymdividingn.ProvethatGmustbe
cyclic.[Hint:Let
l/J(m)bethenumberofelementsinGoforderm. Show
thatl/J(m)::::;q;(m)anduseProblem38.]
40.
UsingtheresultofProblem39,showthatUp,ifpisaprime,iscyclic.
(This
isafamousresultinnumbertheory;itasserts theexistenceofa
primitiveroot modp.)
41.UsingtheresultofProblem40,showthatifpisaprimeoftheform
p==4n+1,thenwecansolvex
2
==-1modp(withxaninteger).
42.
UsingWilson'sTheorem(seeProblem28),showthatifpisaprimeof
theformp==4n+1andif
p-1(p-1)
y==1·2·3···-2-==-2-!,
theny2
==-1modp.(Thisgives anotherproofoftheresultin Problem
41.)
43.LetGbeanabeliangroupofordern,andal,...,anitselements.Let
x==ala2···an.Showthat:
(a)IfGhasexactlyoneelementb=1=esuchthatb
2
==e,thenx==b.
(b)IfGhasmorethanoneelementb=1=esuchthatb
2
==e,thenx==e.
(c)Ifnisodd,thenx==e(seeProblem16).
5.HOMOMORPHISMS ANDNORMAL SUBGROUPS
Inacertainsensethesubjectofgrouptheoryisbuiltupoutofthreebasic
concepts:
thatofahomomorphism,thatofanormalsubgroup,andthatof
thefactororquotientgroupofagroupbyanormalsubgroup.Wediscussthe
firsttwooftheseinthissection, andthethirdinSection6.
Withoutfurtheradoweintroducethefirstofthese.

82. Sec.5 Homomorphisms andNormalSubgroups 67
Definition.LetG,G'betwogroups;thenthemappingq;:G~G'is
ahomomorphismifq;(ab)==q;(a)q;(b)foralla,bEG.
(Note:Thisq;hasnothingtodowiththeEulerq;-function.)
Inthisdefinitiontheproductontheleft side-inq;(ab)-isthatofG,
whiletheproductq;(a)q;(b)isthatofG'.Ashortdescription ofahomomor­
phism
isthatitpreservestheoperationof G.Wedonotinsistthat
q;beonto;if
it
is,we'llsaythatit is.Beforeworkingoutsomefactsabouthomomorphisms,
wepresentsomeexamples.
Examples
1.LetGbethegroupofallpositiverealsunderthemultiplicationofreals,
and
G'thegroupofallreals underaddition.Let
q;:G~G'bedefinedby
q;(x)==IOgloXforxEG.SinceIOglO(XY)==IOgloX+IOgloY,wehave
q;(xy)==q;(x)+q;(y),soq;isahomomorphism.Italsohappenstobeonto
and
1-1.
2.LetGbean abeliangroupandlet
q;:G~Gbedefinedbyq;(a)==a
2
.
Sinceq;(ab)==(ab)2==a
2
b
2 ==
q;(a)q;(b),q;isahomomorphismofGintoit­
self.
Itneednotbeonto;thereadershouldcheckthatin Us(seeSection4)
a
2==eforallaEUs,so
q;(G)==(e).
3.Theexampleof Usabovesuggeststheso-called trivialhomomorphism. Let
Gbeanygroupand G'anyother;defineq;(x)==e',theunitelementof G',
forallxEG.Trivially,q;isahomomorphismofGinto G'.Itcertainlyisnot
averyinterestingone.
Anotherhomomorphismalwayspresent
istheidentitymapping, i,ofany
groupGintoitself.Since
i(x)==xforallxEG,clearlyi(xy)==xy==i(x)i(y).
Themap iis1-1andonto,but,again, isnottoointerestingasahomomor­
phism.
4.LetGbethegroup ofintegersunder +andG'=={I,-I},thesubgroupof
therealsundermultiplication.Define
q;(m)==1ifmiseven,q;(m)==-1ifm
isodd.Thestatementthatq;isahomomorphismismerelyarestatementof:
even
+even==even,even +odd==odd,and odd+odd==even.
5.LetGbethegroupofallnonzerocomplexnumbersundermultiplication
andlet
G'bethegroupofpositiverealsundermultiplication.Let
q;:G~G'
bedefinedbyq;(a)==IaI;thenq;(ab)==IabI==IaIIbI==q;(a)q;(b),soq;isa
homomorphism
ofGintoG'.Infact,
q;isonto.
6.LetGbethegroupinExample6 ofSection1,andG'thegroup of
nonzerorealsundermultiplication.Defineq;:G~G'byq;(Ta,b)==a.That

83. 68 Groups Ch.2
q;isahomomorphismfollowsfrom theproductruleinG,namely, Ta,bTc,d=
Tac,ad+b'
7.LetG=7Lbethegroupofintegersunder+andletG'=7L
n
•Define
q;:G~7L
nbyq;(m)=[m].Sincetheadditionin 7L
nisdefinedby [m]+[r]=
[m+r],weseethatq;(m+r)=q;(m)+q;(r),soq;isindeedahomomor­
phism
of7Lonto7L
n
•
8.Thefollowinggeneralconstructiongivesrisetoawell-knowntheorem.
LetGbeanygroup,andlet A(G)bethesetofall 1-1mappingsofGontoit­
self-hereweareviewingGmerelyasaset,forgettingaboutitsmultiplication.
Define
Ta:G
~GbyTa(x)=axforeveryxEG.Whatistheproduct,TaTb, of
TaandTb asmappingson G?Well,
(weused
theassociativelaw).Sowesee thatTaTb =Tab'
Definethemapping
q;:G~A(G)byq;(a)=Ta,foraEG.Theprod­
uctrulefor
theT'stranslatesinto
q;(ab)=Tab=TaTb =q;(a)q;(b),soq;isa
homomorphismofGintoA(G).Weclaimthatq;is1-1.Suppose thatq;(a)=
q;(b),thatis,Ta=Tb.Therefore,a=Ta(e)=Tb(e)=b,soq;isindeed1-1.
Itisnotontoingeneral-forinstance,ifGhas ordern>2,thenA(G)has
ordern!,andsincen!>n,q;doesn'thaveaghost ofachanceofbeingonto.
Itiseasytoverifythattheimage of'fJ,q;(G)={T
aIaEG},isasubgroup
ofA(G).
Thefactthatq;is1-1suggeststhatperhaps1-1homomorphismsshould
playaspecialrole. Wesinglethemoutinthefollowingdefinition.
Definition.
Thehomomorphism
q;:G~G'iscalleda monomor­
phism
if
q;is1-1.A monomorphismthatisontoiscalledan isomorphism.An
isomorphismfromG toGitselfiscalledanautomorphism.
Onemoredefinition.
Definition.TwogroupsG
andG'aresaidto beisomorphicifthereis
anisomorphismofGontoG'.
WeshalldenotethatGandG'areisomorphicbywritingG
===G'.
Thisdefinitionseemstobeasymmetric,but,in pointoffact,it isnot.
..ForifthereisanisomorphismofGontoG',thereisoneofG'ontoG(see
Problem2).
Weshalldiscuss morethoroughlylaterwhatitmeansfortwogroupsto
beisomorphic.Butnowwesummarizewhatwedidin Example8.

84. Sec.5 Homomorphisms andNormalSubgroups 69
Theorem2.5.1(Cayley'sTheorem).EverygroupG isisomorphicto
somesubgroupof
A(S),foranappropriate S.
Theappropriate SweusedwasGitself.Buttheremaybebetter
choices.Weshallseesome
intheproblemstofollow.
WhenG
isfinite,wecantaketheset SinTheorem2.5.1tobefinite,in
whichcase
A(S)isSnanditselementsarepermutations.Inthiscase,Cay­
ley'sTheorem
isusuallystatedas: Afinitegroupcanberepresented asa
group
ofpermutations.
(ArthurCayley(1821-1895)wasanEnglishmathematicianwhoworkedinma­
trixtheory,invarianttheory,
andmanyotherpartsofalgebra.)
Thisisagoodplacetodiscusstheimportanceof"isomorphism."Let
q;
beanisomorphism ofGontoG'.Wecanview G'asarelabelingofG,
usingthelabelq;(x)fortheelement x.Isthislabelingconsistentwiththe
structureofG
asagroup?That is,ifxislabeled
q;(x),ylabeledq;(y),what
isxylabeledas?Sinceq;(x)q;(y)=q;(xy),weseethatxyislabeledas
q;(x)q;(y),sothisrenamingoftheelements isconsistentwiththeproductin
G.Sotwogroupsthatare isomorphic-althoughtheyneednotbe equal-in
acertainsense,asdescribedabove,areequal.Often,it isdesirabletobeable
toidentifyagivengroup
asisomorphictosomeconcretegroupthatwe
know.
We
goonwithmoreexamples.
9.LetGbeanygroup, aEGfixedinthediscussion.Define
q;:G~Gby
q;(x)=a-IxaforallxEG.Weclaimthatq;isanisomorphismofGonto
itself.First,
soq;isatleastahomomorphismofGintoitself.It is1-1forifq;(x)=q;(y),
thena-Ixa=a-1ya,sobycancellationinGweget x=y.Finally,q;isonto,
for
x=a-I(axa-I)a=
q;(axa-
I
)foranyxEG.
Hereq;iscalledthe innerautomorphism ofGinducedbya.Thenotion
of
automorphismandsomeofitspropertieswillcomeupintheproblems.
Onefinalexample:
10.LetGbethegroupofrealsunder +andletG'bethegroupofall
nonzerocomplexnumbersundermultiplication.Define
q;:G~G'by
q;(x)=cosx+isinx.
Wesawthat(cos x+isinx)(cosy+isiny)=cos(x+y)+isin(x+y),

85. 70 Groups Ch.2
hence'P(x)'P(Y)='P(x+y)and'PisahomomorphismofGintoG'.'Pisnot
1-1because,forinstance,'P(D)='P(27T)=1,noris'Ponto.
Now
thatwehaveafewexamplesinhand,westartalittleinvestigation
ofhomomorphisms.Webeginwith
Lemma2.5.2.If
'PisahomomorphismofGintoG',then:
(a)'P(e)=e',theunitelement ofG'.
(b)'P(a-1)='P(a)-1forallaEG.
ProofSincex=xe,'P(x)='P(xe)='P(x)'P(e);bycancellationin G'
weget'P(e)=e'.Also,'P(aa-
1
)='P(e)=e',hencee'='P(aa-
1
)=
'P(a)'P(a-
1
),whichprovesthat'P(a-
I
)='P(a)-I.D
Definition.Theimageof'P,'P(G),is'P(G)={'P(a)IaEG}.
Weleavetothereadertheproofof
Lemma2.5.3.If'Pisahomomorphism ofGintoG',thentheimage
of'PisasubgroupofG'.
Wesingledoutcertainhomomorphismsandcalled"themmonomor­
phisms.
Theirpropertywasthattheywere1-1.Wewant
t6measurehowfar
agivenhomomorphism
isfrombeingamonomorphism.Thispromptsthe
Definition.
If
'Pisahomomorphism ofGintoG',thenthe kernelof
'P,Ker'P,isdefinedby Ker'P={aEGI'P(a)=e'}.
Ker'Pmeasuresthelack of1-11nessat'onepointe'.Weclaimthat
thislack
isratheruniform.Whatis,W={x
EGI'P(x)=w'}foragiven
w'EG'?Weshowthatif'P(x)=w'forsome xEG,then W=
{kxIkEKer'P}=(Ker'P)x.Clearly,if kEKer'Pand'P(x)=w',then
'P(kx)='P(k)'P(x)=e''P(x)=w',sokxEW.Also,if'P(x)='P(Y)=w',
then'P(x)='P(Y),hence'P(Y)'P(X)-I=e';but'P(X)-l='P(x-
1
)by
Lemma2.5.2,so e'='P(Y)jP(X)-1='P(Y)'P(x-
l
)='P(yx-
1
),whence
yx-1EKer'PandsoyE(Ker'P)x.Thustheinverse image ofanyelement
w'in'P(G)EG'istheset(Ker'P)x,wherexisanyelementinGsuchthat
'P(x)=w'.
Westatethisas
Lemma2.5.4.Ifw'EG'isoftheform'P(x)
{yEGI'P(Y)=w'}=(Ker'P)x.
w',then

86. Sec.5 Homomorphisms andNormalSubgroups 71
Wenowshallstudysomebasicpropertiesofthekernels ofhomomor­
phisms.
Theorem2.5.5.If'PisahomomorphismofGinto G',then
(a)Ker'PisasubgroupofG.
(b)GivenaEG,a-I(Ker'P)aCKer'P.
ProofAlthoughthis issoimportant,its proofiseasy.If a,bEKer'P,
then'P(a)=='P(b)==e',hence'P(ab)=='P(a)'P(b)==e',whenceabEKer'P,
soKer'Pisclosedunderproduct.Also'P(a)==e'impliesthat'P(a-I)==
'P(a)-I==e',andso a-IEKer'P.Therefore,Ker'Pisasubgroupof G.If
kEKer'PandaEG,then'P(k)==e'.Consequently,'P(a-Ika)==
'P(a-I)'P(k)'P(a)=='P(a-I)e''P(a)=='P(a-I)'P(a)=='P(a-Ia)=='P(e)==e'.This
tells
usthata-Ika
EKer'P,hencea-I(Ker'P)aEKer'P.Thetheorem is
nowcompletelyproved. D
Corollary.If'PisahomomorphismofGinto G',then'Pisa
monomorphismifandonlyifKer'P==(e).
ProofThisresult isreallyacorollarytoLemma2.5.4.Weleavethe
fewdetailstothereader.
D
Property(b)ofKer
'PinTheorem2.5.5 isaninterestingandbasicone
forasubgrouptoenjoy.Weranintothispropertyinthetextmaterialand
problemsearlieronseveraloccasions.Weuseittodefinetheultra-important
classofsubgroupsofagroup.
Definition.ThesubgroupNofG
issaidtobeanormalsubgroup of
Gifa-INaCNforeverya
EG.
Ofcourse,Ker'P,foranyhomomorphism, isanormalsubgroupofG.
Asweshallseeinthenextsection,everynormalsubgroup
ofGisthekernel
ofsomeappropriatehomomorphismofGintoanappropriategroup
G'.Soin
acertainsensethenotionsofhomomorphismandnormalsubgroupswillbe
showntobeequivalent.
Althoughwedefinedanormalsubgroupvia
a-INaCN,weactually
have
a-INa==N.Forifa-INaCNforalla
EG,thenN ==a(a-INa)a-
1
C
aNa-
1
==(a-
I)-INa-1CN.SoN ==aNa-
1
foreverya
EG.Transposing,
wehaveNa
==aN;thatis,everyleftcoset ofNinG isarightcoset of
NinG.
Ontheotherhand,ifeveryleftcosetof NinGisarightcoset,thenthe
leftcosetaN,whichcontains
a,mustbeequaltotherightcosetcontaining a,

87. 72 Groups Ch.2
namelyNa.Thus,aN=NaandN=a-INaforallaEG,whichistosaythat
NisnormalinG.
Wewrite
"Nisanormalsubgroup ofG"bytheabbreviatedsymbol
N<JG.
Notethata-INa=Ndoesnotmeanthat a-Ina=nforeverynEN.
No-merelythatthesetofalla-Inaisthesameastheset ofalln.
Wehaveproved
Theorem2.5.6.N<JGifandonlyifeveryleftcoset ofNinGisa
rightcoset
ofNinG.
Beforegoinganyfurther,wepausetolook
atsomeexamples ofkernels
ofhomomorphismsandnormalsubgroups.
IfG
isabelian,theneverysubgroup ofGisnormal,for a-Ixa=xfor
every
a,xEG.Theconverseofthisisnottrue.Nonabeliangroupsexistin
which
everysubgroup isnormal.Seeifyoucanfindsuchanexampleoforder
8.Suchnonabeliangroupsarecalled Hamiltonian,aftertheIrishmathemati­
cian
W.R.Hamilton(1805-1865). Thedesiredgroupof order8canbefound
inthe
quaternionsofHamilton,whichweintroducein Chapter4,Section1.
InExample 1,
cp(x)=10gloX,andKercp={xIlogloX=O}={I}.In
Example
2,whereG isabelian,and
cp(x)=X
2
,
Ker
cp={xEGIx
2
=e}
Thekernelofthetrivialhomomorphism ofExample3 isallofG.InExam­
ple
4,Ker
cpisthesetofallevenintegers.InExample 5,Kercp=
{aEC'IiaI=I},whichcan beidentified,fromthepolarform ofacomplex
number,as
Ker
cp={cosx+isinxIxreal}.InExample 6,Kercp=
{Tt,bEGIbreal}.InExample 7,Kercpisthesetofallmultiplesofn.InEx­
amples8
and9,thekernelsconsists ofealone,forthemapsaremonomor­
phisms.InExample10,wesee
thatKer
cp={27TmImanyi~ger}.
Ofcourse,allthekernelsabovearenormalsubgroups oftheirrespec­
tivegroups.Weshouldlook
atsomenormalsubgroups,intrinsicallyinGit­
self,withoutrecourse
tothekernelsofhomomorphism.Wegobacktothe
examples
ofSection1.
1.InExample7,H={Ta,bEGIarational}.If Tx,yEG,weleaveittothe
readertocheckthat
T;'~HTx,yCHandsoH<JG.
2.InExample9thesubgroup {i,g,i,g3}<JG.Heretooweleavethe
checkingtothereader.
3.InExample10thesubgroup H=Ii,h,h
2
,
•••,h
n
-I
}
isnormalin G.This
wealsoleave
tothereader.

88. Sec.5 Homomorphisms andNormalSubgroups 73
4.IfGisanygroup,Z(G),thecenterofG,isa normalsubgroupofG(see
,Example
11ofSection3).
5.
IfG=S3'Ghastheelementsi,f,g,
i,fg,andgf,wheref(XI)=X2,
f(X2)=Xt,f(X3)=X3andg(Xt)=X2'g(X2)=X3,g(X3)=Xt.Weclaim
thatthesubgroupN=Ii,g,i}<lS3'Aswesawearlier(orcancompute
now),fgf-t=g-t=i,fg2f-
t
=g.(fg)g(fg)-l=fggg-If-l=fgf-t=
g2,andsoon.SoN<lS3follows.
Thematerialinthissection hasbeenaratherrichdiet.Itmaynotseem
so,buttheideaspresented,althoughsimple,arequitesubtle.Werecom­
mendthatthereaderdigesttheconceptsandresultsthoroughlybeforegoing
on.
Onewayofseeinghowcompletethisdigestionis,is totakeastabat
manyofthealmostinfinitelist ofproblemsthatfollow.Thematerialofthe
nextsectionisevenaricherdiet,andevenhardertodigest.Avoidamathe­
maticalstomachachelaterbyassimilatingthissectionwell.
PROBLEMS
EasierProblems
1.Determineineachofthepartsifthegivenmappingisahomomorphism.
Ifso,identifyits kernelandwhetherornotthemappingis1-1oronto.
(a)G=
Zunder+,G'=Zn,cp(a)=[a]foraEZ.
(b)Ggroup,cp:G~Gdefinedbycp(a)=a-IforaEG.
(c)Gabeliangroup,cp:G~Gdefinedbycp(a)=a-IforaEG.
(d)Ggroupofallnonzerorealnumbersundermultiplication,G'=
{I,-I},cp(r)=1ifrispositive,cp(r)=-1ifrisnegative.
(e)Ganabeliangroup,n>1 afixedinteger, andcp:G~Gdefinedby
cp(a)=anforaEG.
2.Recall
thatG
~G'meansthatGisisomorphictoG'.Provethatforall
groupsG
bG
2
,G
3
:
(a)G
t~G
t
.
(b)G
I~G
2impliesthatG
2
~G
I
•
(c)G
t~G
2
,G
2~G
3impliesthatG
I
~G
3
•
3.LetGbeanygroupandA(G)thesetofall1-1mappingsofG,asaset,
ontoitself.DefineLa:G
~GbyLa(x)=xa-
t
.
Provethat:
(a)LaEA(G).
(b)LaLb=Lab'
(C)Themapping
t/J:G~A(G)definedbyt/J(a)=Laisamonomor­
phismofGintoA(G).

89. 74 Groups Ch.2
4.InProblem3provethatforalla,bEG,TaLb =LbTa,whereTaisde­
finedasin
Example8.
5.InProblem4,showthatifVEA(G)issuchthatT
aV=VT
aforall
aEG,thenV=L
bforsomebEG.(Hint:Acting oneEG,findout
whatbshouldbe.)
6.
Provethatif
cp:G~G'isahomomorphism,thencp(G),theimageofG,
isasubgroupofG'.
7.Show
that
cp:G~G',wherecpisahomomorphism,isamonomorphism
if
andonlyif Ker
cp=(e).
8.FindanisomorphismofG,thegroupofallrealnumbersunder+,onto
G',thegroupofallpositivereal numbersundermultiplication.
9.VerifythatifG isthegroupinExample6ofSection1, andH=
{T
a
,bEGIarational},thenH<lG,thedihedralgroupoforder8.
10.Verify
thatinExample9ofSection1, thesetH={i,g,g2,g3}isanor­
mal
subgroupofG,thedihedralgroupoforder8.
11.VerifythatinExample10ofSection1,thesubgroup
H={i,h,h
2
,
•••,h
n
-
I
}
isnormalinG.
12.ProvethatifZ(G)isthecenterofG,thenZ(G)<lG.
13.IfGisafiniteabelian groupofordernand
cp:G~Gisdefinedby
cp(a)=amforallaEG,findthenecessaryandsufficientcondition that
cpbeanisomorphismofGontoitself.
14.IfGisabelianandcp:G~G'isahomomorphismofGontoG',prove
thatG'isabelian.
15.
IfGisanygroup, N<JG,and
cp:G~G'ahomomorphismofGonto
G',provethattheimage,cp(N),ofNisanormalsubgroupofG'.
16.IfN<lGandM<lGandMN={mn1m EM,nEN},provethatMNis
asubgroupofGandthatMN<lG.
17.IfM<lG,N<lG,provethatMnN<lG.
18.IfHisanysubgroupofGandN=n
aEGa-IHa,provethatN<lG.
19.
IfHisasubgroupofG,letN(H)bedefinedbytherelationN(H)=
{aEGIa-1Ha=H}.Provethat:
(a)N(H)isasubgroupofGandN(H)
~H.
(b)H<lN(H).
(c)IfKisasubgroupofGsuchthatH<lK,thenKCN(H).[SoN(H)
isthelargestsubgroupofGinwhich Hisnormal.]
20.
IfM<lG,N<lG,andMnN=(e),showthatfor mEM,nEN,mn=nm.

90. Sec.5 Homomorphisms andNormalSubgroups 75
21.LetSbeanysethaving morethantwoelements andA(S)thesetof
all1-1mappingsofSontoitself.IfsES,wedefine H(s)=
{fEA(S)If(s)=s}.ProvethatH(s)cannot beanormalsubgroup ofA(S).
22.LetG=S3,thesymmetricgroupofdegree3andletH={i,f},where
f(Xl)=x2,f(X2)=xl,f(X3)=X3·
(a)Writedownall theleftcosetsofHinG.
(b)Writedownall therightcosets ofHinG.
(c)Iseveryleftcoset ofHarightcoset ofH?
23.LetGbeagroupsuch thatallsubgroups ofGarenormalinG.If
a,bEG,provethatba=ajbforsomej.
24.IfG
bG
2aretwogroups,letG=G
lXG
2
,theCartesianproductofG
b
G2[i.e.,Gisthesetofallorderedpairs(a,b)whereaE Gl, bE G2].
Definea productinGby (al'bl)(a2'b2)
=(ala2'blb2 ).
(a)ProvethatGisagroup.
(b)Showthatthereisamonomorphism'PlofGlintoGsuchthat
'Pl(Gl)<lG,givenby'Pl(al)=(abe2),wheree2istheidentityele­
mentofG
2
•
(c)Findthesimilarmonomorphism'P2ofG
2intoG.
(d)Usingthemappings'Pb'P2ofParts(b)and(c),prove that
'Pl(G
1)'P2(G2) =Gand'Pl(Gl)n'P2(G2)istheidentityelementofG.
(e)ProvethatG
l
XG
2
~G
2XG
l
•
25.LetGbeagroupandletW=G X GasdefinedinProblem24.Provethat:
(a)Themapping'P:G~Wdefinedby'P(a)=(a,a)isamonomorphism
ofGintoW.
(b)Theimage'P(G)inW[i.e.,{(a,a)IaEG}]isnormalinWifandonly
ifGisabelian.
Middle-Level
Problems
*26.IfGisagroupandaEG,define
O"a:G~GbyO"a(g)=aga-
l
. Wesaw
inExample9
ofthissectionthat
O"aisanisomorphismofGontoitself,so
O"aEA(G),thegroupofall1-1mappings ofG(asaset) ontoitself.De­
finet/J:G~A(G)byt/J(a)=O"aforallaEG.Provethat:
(a)t/JisahomomorphismofGintoA(G).
(b)Kert/J=Z(G),thecenterofG.
27.
If
0isanautomorphismofGandN<lG,provethatO(N)<lG.
28.Let0,t/JbeautomorphismsofG,andletOt/Jbetheproductof0andt/Jas
mappings
onG.Provethat
Ot/JisanautomorphismofG,andthatO-lis
anautomorphismofG,sothatthesetofallautomorphismsofGisitself
agroup.

91. 76 Groups Ch.2
*29.A subgroupTofagroupWiscalledcharacteristicif'P(T)CTforallau­
tomorphisms,'P,ofW.Provethat:
(a)McharacteristicinGimplies thatM<lG.
(b)M,NcharacteristicinGimplies thatMNischaracteristicinG.
(c)Anormalsubgroupofagroupneednotbecharacteristic.(This is
quitehard; youmustfindanexampleofagroupGandanoncharac­
teristic
normalsubgroup.)
30.Suppose
thatIGI=pm,wherep
(mandpisaprime.IfHisanormal
subgroup
oforderpinG,provethatHischaracteristic.
31.Suppose
thatGisanabeliangroupoforderpnmwherep
(misaprime.
If
HisasubgroupofGoforderpn,provethatHisacharacteristicsub­
groupofG.
32.
DoProblem31evenifGisnotabelianifyou happentoknowthatfor
somereasonorotherH<lG.
33.Suppose
thatN<lGandMeNisacharacteristicsubgroup ofN.Prove
thatM<JG.(ItisnottruethatifM<lNandN<lG,thenMmustbe
normalinG.See Problem50.)
34.
LetGbeagroup,
.:fI(G)thegroupofallautomorphismsofG.(SeeProb­
lem28.)
LetI(G)=
{UaIaEG},whereU
aisasdefinedin Problem26.
ProvethatI(G)<l.:fI(G).
35.Show thatZ(G),thecenterofG,isacharacteristicsubgroup ofG.
36.
IfN<lGandHisasubgroupofG,showthatHnN<lH.
HarderProblems
37.IfGisa nonabeliangroupoforder6,provethatG
~S3 .
38.LetGbeagroupandHasubgroupofG.LetS={HaIaEG}bethe
setofallrightcosets ofHinG.Define, forbEG,Tb:S~SbyTb(Ha)
=Hab-
1
•
(a)ProvethatTbTe=Tbeforall b,cEG[thereforethemapping
l/J:G~A(S)definedbyl/J(b)=T
bisahomomorphism].
(b)DescribeKerl/J,thekernelofl/J:G~A(S).
(c)ShowthatKerl/JisthelargestnormalsubgroupofGlyinginH
[largestin
thesensethatifN<lGandNCH,thenNCKer
l/J].
39.UsetheresultofProblem38toredoProblem37.
Recall
thatifHisasubgroupofG,thentheindexofHinG,iG(H),is
thenumberofdistinctrightcosets ofHandG(ifthis numberisfinite).
40.
IfGisafinitegroup, HasubgroupofGsuchthatn
(iG(H)!wheren=
IGI,provethatthereisanormalsubgroupN =I=-(e)ofGcontainedin H.

92. Sec.6 FactorGroups 77
41.Suppose thatyouknow thatagroupG oforder21containsan elementa
oforder7.ProvethatA=(a),thesubgroupgeneratedbya,isnormalin
G.(Hint:Use
theresultofProblem40.)
42.Supposethatyouknow thatagroupG oforder36hasasubgroup Hor
order9.ProvethateitherH<JGorthereexistsasubgroup N<JG,
NcH,andINI=3.
43.Provethatagroupoforder9mustbeabelian.
44.Provethatagroupoforderp2,Paprime,hasanormalsubgroup of
orderp.
45.UsingtheresultofProblem44,provethatagroupoforderp2,Pa
prime,must
beabelian.
46.LetGbeagroupoforder15;show thatthereisanelementa
=1=einG
such
thata
3
=eandanelementb
=1=esuchthatb
5
=e.
47.InProblem46,show thatbothsubgroupsA={e,a,a
2
}
andB=
fe,b,b
2
,
b
3
,
b
4
}
arenormalinG.
48.FromtheresultofProblem47,show thatanygroup oforder15iscyclic.
Very
HardProblems
49.LetGbeagroup,HasubgroupofGsuchthati
G(H)isfinite.Prove that
thereisasubgroupNcH,N <JGsuchthati
G(N)isfinite.
50.ConstructagroupGsuch thatGhasa normalsubgroupN,andNhasa
normalsubgroup
M(i.e.,N<JG,M<JN),yetMisnotnormalinG.
51.
LetGbeafinitegroup,
q;anautomorphismofGsuchthatq;2istheiden­
tityautomorphism
ofG.Supposethat
'P(x)=ximpliesthatx=e.Prove
thatGisabelianandq;(a)=a-IforallaEG.
52.LetGbeafinitegroup andq;anautomorphism ofGsuchthatq;(x)=
X-Iformorethanthree-fourths oftheelementsofG.Provethat'P(Y)=
y-
1
forallyEG,andsoGisabelian.
6.FACTORGROUPS
LetGbeagroup andNanormalsubgroup ofG.InprovingLagrange's The­
oremweused,foranarbitrarysubgroup H,theequivalencerelation a'"'-'bif
ab-
1
EH.Let'strythisoutwhenNisnormalandseeifwecansayalittle
morethan
onecouldsayforjustanyoldsubgroup.
So,let
a
'"'-'bifab-1ENandlet[a]={xEGIx'"'-'a}.Aswesaw
earlier,
[a]=Na,therightcoset ofNinGcontaining a.Recallthatin
looking
at7L
nwedefinedforit anoperation+via[a]+[b]=[a+b].Why

93. 78 Groups Ch.2
nottrysomethingsimilarforanarbitrarygroupGandanormalsubgroup
NofG?
SoletM ={[a]IaEG},where [a]={xEGIxa-
1
EN}=Na.Wede­
fineaproductinMvia
[a][b]=[ab].WeshallsoonshowthatM isagroup
underthisproduct.
Butfirstandforemost wemustshowthatthisproduct in
Miswell-defined.Inotherwords,wemustshowthatif [a]=[a']and [b]=
[b'],then [ab]=[a'b'],forthiswouldshowthat [a][b]=[ab]=[a'b']=
[a'][b'];equivalently,thatthisproduct ofclassesdoesnotdependonthepar­
ticularrepresentatives
weusefortheclasses.
Thereforeletussupposethat
[a]=[a']and[b]=[b'].Fromthedefi­
nitionofourequivalence
wehavethat a'=na,wheren EN.Similarly,
b'=mb,wheremEN.Thusa'b'=namb=n(ama-1)ab;sinceN <lG,
ama-
1
isinN,so n(ama-
1
)
isalsoinN.Soif weletnl=n(ama-
1
),then
nlENanda'b'=nlab.Butthistellsusthat a'b'ENab,sothat
a'b'
~ab,fromwhichwehavethat[a' b']=[ab],theexactthing were­
quiredtoensurethatourproductin
Mwaswell-defined.
ThusM
isnowendowedwithawell-definedproduct[a][b] =[able
Wenowverifythegroupaxiomsfor M.Closurewehavefromtheverydef­
initionofthisproduct.If
[a],[b],and[c]arein M,then[a]([b][c]) =
[a][bc]=[a(be)]=[(ab)c](sincetheproductinG isassociative)=
[ab][c]=([a][b])[ c].Therefore,theassociativelawhasbeenestablished
fortheproductin
M.Whataboutaunitelement?Why nottrytheobvious
choice,namely
[e]?Weimmediatelyseethat [a][e]=rae]=[a]and
[e][a]=rea]=[a],so[e]doesactastheunitelementfor M.Finally,what
aboutinverses?Here,too,theobviouschoice
isthecorrectone.If aEG,
then[a][a-
1
]
=[aa-
1
] =[e],hence[a-I]actsastheinverseof [a]relative
totheproductwehavedefinedin
M.
Wewanttogive Maname,andbetterstill,asymbolthatindicatesits
dependenceonGand
N.Thesymbol weuseforM isGIN(read"GoverNor
GmodN")and GINiscalledthefactorgroup orquotientgroupofGby N.
Whatwehaveshown istheveryimportant
Theorem
2.6.1.IfN<lGand
GIN={[a]/aEG}={NalaEG},
thenGIN isagrouprelativetotheoperation [a][b]=[ab].
Oneobservationmustimmediatelybemade,namely
Theorem
2.6.2.IfN<lG,thenthere isahomomorphism
f/JofGonto
GINsuchthatKerf/J,thekerneloff/J,isN.

94. Sec.6 FactorGroups 79
ProofThemostnaturalmappingfromG toGINistheonethat
doesthetrick.Definet/J:G~GINbyt/J(a)=[a].Ourproductasdefined
in
GINmakesof
t/Jahomomorphism,fort/J(ab)=[ab]=[a][b]=
t/J(a)t/J(b).Sinceevery elementXEGINisoftheformX=[b]=t/J(b)for
some
bEG,
t/Jisonto.Finally, whatisthekernel,Kert/J,oft/J?Bydefini­
tion,
Ker
t/J={aE GIt/J(a)=E},whereEistheunitelementofGIN.But
whatisE?NothingotherthanE=[e]=Ne=N,andaEKert/Jifand
onlyif E=N=t/J(a)=Na.ButNa=Ntellsusthata=eaENa=N,so
wesee
thatKer
t/JCN.ThatNCKert/J-whichiseasy-weleavetothe
reader.So Kert/J=N.D
Theorem2.6.2substantiatestheremarkwemadeintheprecedingsec­
tionthateverynormalsubgroup
NofGisthekernelofsomehomomor­
phismofGontosomegroup.
The"somehomomorphism" isthe
t/Jdefined
aboveandthe"some group"
isGIN.
ThisconstructionofthefactorgroupGby Nispossiblythesinglemost
importantconstructioningrouptheory.In
otheralgebraicsystemsweshall
haveanalogousconstructions,asweshallseelater.
Onemightask:Whereinthiswholeaffairdid thenormalityofNinG
enter?Why
notdothesamethingforanysubgroup HofG?Solet'stry and
seewhathappens.Asbefore,wedefine
W={[a]laEG}={HalaEG}
wheretheequivalence a
'"'-'bisdefinedby ab-
1
EH.Wetrytointroducea
productin
Waswedidfor GINbydefining[a][b]=[ab].Isthisproductwell
defined?If
hEH,then[hb]=[b],sofortheproductto bewelldefined,we
wouldneedthat
[a][b]=[a][hb],thatis,[ab]=[ahb].Thisgivesusthat Hab
=Hahb,andsoHa=Hah;thisimpliesthat H=
Haha-t,whenceaha-
1
EH.
Thatis,forallaEGandallhEH,aha-
1
mustbeinH;inotherwords,H
mustbenormalinG. Soweseethatinorderfortheproductdefined inWto
bewell-defined,Hmustbeanormalsubgroup ofG.
Weviewthis matterofthe quotientgroupinaslightlydifferentway. If
A,BaresubsetsofG,letAB=tabIaEA,bEB}.IfHisasubgroupofG,
then
HHCHisanotherway ofsayingthatHisclosedundertheproduct
ofG.
LetGIN={NaIaEG}bethesetofallrightcosets ofthenormalsub­
group
NinG.Usingtheproduct ofsubsetsofGasdefinedabove,what is
(Na)(Nb)?Bydefinition,(Na)(Nb)consistsofallelementsoftheform
(na)(mb),wheren,mEN,andso
(na)(mb)=(nama-1)(ab)=ntab,

95. 80 Groups Ch.2
wherenl=nama-
1
isinN,sinceN isnormal.Thus(Na)(Nb) CNab.Onthe
otherhand,if
nEN,then
n(ab)
=(na)(eb)E(Na)(Nb),
sothatNab
C(Na)(Nb).Inshort,wehaveshownthatthe product-assub­
setsof
G-ofNaandNb isgivenbytheformula(Na)(Nb) =Nab.Allthe
othergroupaxiomsforGIN,asdefinedhere,arenowreadilyverifiedfrom
thisproductformula.
Anotherwayofseeingthat(Na)(Nb)
=Nabistonotethatbythenor­
malityofN,
aN=Na,hence(Na)(Nb) =N(aN)b=N(Na)b=NNab=
Nab,sinceNN=N(becauseN isasubgroupofG).
Howeverweview
GIN-asequivalenceclasses orasasetofcertain
subsetsof
G-wedogetagroupwhosestructure isintimatelytiedtothatof
G,viathenaturalhomomorphism
l/JofGontoGIN.
Weshallseeverysoonhow
wecombineinductionandthestructureof
GINtogetinformationabout
G.
WhenG isafinitegroupand N<JG,thenthenumberofrightcosetsof
NinG,i
G(N),isgiven-astheproofofLagrange'sTheorem showed-by
iG(n)=IGI/INI.ButthisistheorderofGIN,which isthesetofalltheright
cosetsofNin
G.ThusIGINI=IGIIINI.Westatethismoreformally as
Theorem2.6.3.IfG isafinitegroupand N<JG,thenIGINI
IGI/INI·
Asanapplicationofwhatwehavebeentalkingabouthere,
weshall
proveaspecialcaseofatheoremthat
weshallproveinitsfullgenerality
later.Theproofwe
give-fortheabeliancase-isnotaparticularlygood
one,butitillustratesquiteclearlyageneraltechnique,thatofpullingback
informationabout
GINtogetinformationaboutGitself.
Thetheorem weareabouttoprove isduetothegreatFrenchmathematician
A.L.Cauchy(1789-1857),whosemostbasiccontributionswereincomplex
variabletheory.
Theorem2.6.4(Cauchy).IfG isafiniteabeliangroupoforder IGI
andpisaprimethatdividesIG I,thenGhas anelementoforder p.
ProofBeforegettinginvolvedwiththeproof,wepointouttothe
readerthatthetheorem
istrueforanyfinitegroup.Weshallproveitinthe
generalcaselater,withaproofthatwillbemuchmorebeautifulthantheone
weareabouttogiveforthespecial,abeliancase.
Weproceedbyinductionon
IGI.Whatdoesthismeanprecisely?Weshall

96. Sec.6 FactorGroups 81
assumethetheoremtobetrueforallabeliangroupsoforderlessthan IGIand
showthatthisforcesthetheoremtobetrueforG.If
IGI
=1,thereisnosuchp
andthetheorem
isvacuouslytrue.So wehaveastartingpointforourinduction.
Supposethatthere
isasubgroup(e)
=1=N=1=G.SinceINI<IGI,ifpdi­
vides
INI,byourinductionhypothesistherewouldbeanelement oforderp
in
N,henceinG, andwewould bedone.Sowemaysuppose thatp
(INI.
SinceG isabelian,everysubgroup isnormal,sowecanform GIN.Becausep
dividesIGIandp(INI,andbecauseIGINI=IGI/INI,wehavethat p
dividesIGINI.Thegroup GINisabelian,sinceG is(Prove!)andsince
N=1=(e),INI>1,soIGINI=IGI/INI<IGI.Thus,againbyinduction,there
existsanelementin
GINoforderp.Inotherwords,thereexistsan aEG
suchthat
[a]P=[e],but[a]
=1=[e].Thistranslatestoa
PEN,af!:.N.Soif
m=INI,then(aP)m=e.So(am)P=e.Ifwecouldshowthat b=am =1=e,
thenbwouldbetherequiredelement oforderpinG. Butifam=e,then
[a]m=[e],andsince[a]hasorderp,pIm(seeProblem 31ofSection4).But,
byassumption,
p
(m=INI.SOwearedoneifGhasa nontrivialsubgroup.
ButifGhasnonontrivialsubgroups,itmustbecyclic ofprimeorder.
(SeeProblem
16ofSection3,whichyoushouldbeabletohandlemoreeas­
ilynow.)What
isthis"prime order"?Becausepdivides IGI,wemusthave
IGI=p.Butthenanyelement a
=1=eEGsatisfiesa
P
=eandisoforderp.
Thiscompletestheinduction,andsoprovesthetheorem. D
Weshallhave otherapplicationsofthiskind ofgroup-theoreticargu­
ment
intheproblems.
Thenotion
ofafactorgroup isaverysubtleone, andofthegreatest
importanceinthesubject.Theformation
ofanewsetfromanoldoneby
using
aselementsofthisnewsetsubsets oftheoldone isstrangetotheneo­
phyteseeingthiskind
ofconstructionfor thefirsttime.Soit isworthwhile
lookingatthiswhole
matterfromavariety ofpointsofview.Weconsider
GINfromanotheranglenow.
Whatarewedoingwhenweform
GIN?Sure,wearelookingatequiva­
lencesclassesdefinedvia
N.Let'slookatit anotherway.Whatwearedoing
isidentifyingtwoelementsinGiftheysatisfytherelation ab-1EN.Ina
sense
weareblottingoutN.SoalthoughGINisnotasubgroupofG,wecan
lookatitasG,with
Nblottedout, andtwoelementsasequaliftheyare
equal
"uptoN."
Forinstance,informing TLIN,where7Listhegroupofintegersand Nis
thesetofallmultiplesof5in7L,whatwearedoing isidentifying1with6,11,
16,-4,-9,andsoon,andweareidentifyingallmultiplesof5with O.The
nicethingaboutallthis isthatthisidentificationjibeswithadditionin 7L
whenwegooverto 7LIN.
Let'slookatafewexamplesfromthispoint ofview.

97. 82 Groups Ch.2
1.LetG ={T
a
,
bIa
=1=0,breal}(Example6 ofSection1).Let N=
{T1,bibreal}CG;wesawthatN<JG,soitmakessensetotalkabout GIN.
NowT
a
,bandT
a
,°areinthesameleftcoset ofNinG,soin GINweareget­
tinganelementbyidentifying
T
a
,
bwithT
a
,
0.Thelatterelementjustdepends
ona.Moreover,theTa,bmultiplyaccordingto Ta,bTc,d=Tac,ad+bandifwe
identify
Ta,bwithTa,o,Tc,dwithTc,o,thentheirproduct,which isTac,ad+b'is
identifiedwith T
ac
,°.SoinGINmultiplicationislikethatofthegroup of
nonzerorealnumbers undermultiplication,andinsomesense(whichwillbe
mademoreprecisein
thenextsection)GINcanbeidentifiedwiththisgroup
ofrealnumbers.
2.LetGbethegroupofrealnumbers under+andlet
7Lbethegroupofin­
tegers
under+.SinceG isabelian,
7L<JG,andsowecantalkabout GI7L.
Whatdoes GI7Lreallylooklike? InformingGI7L,weareidentifyinganytwo
real
numbersthatdifferbyaninteger.So0isidentifiedwith -1,-2,-3,...
and1,2,3,...
;~isidentifiedwith~, ~, -~, -~,....Everyrealnumber a
thushasamate,ii,where0::::;ii<1.So,in GI7L,thewholereallinehasbeen
compressedintotheunitinterval[0,
1].Butalittlemore istrue,forwehave
alsoidentifiedthe
endpointsofthisunitinterval.Sowearebendingtheunit
intervalaroundso
thatitstwoendpointstouchandbecomeone.Whatdo
wegetthisway?Acircle,
ofcourse!So GI7Lislikeacircle,inasensethat
canbemadeprecise,andthiscircle
isagroupwithanappropriateproduct.
3.LetGbethegroup ofnonzerocomplexnumbersandlet N=
{aEGilaI=I}whichistheunitcircleinthecomplexplane.Then Nisa
subgroup
ofGandisnormalsinceG isabelian.Ingoingto GINwearede­
claring
thatanycomplexnumber ofabsolutevalue1willbeidentifiedwiththe
realnumber
1.Nowany aEG,initspolarform,canbewritten asa=
r(cos()+isin(),where r=1aI,andIcos()+isin()I=1.Inidentifying
cos
()+isin()with1,weareidentifying awithr.Soinpassingto GINevery
element
isbeingidentifiedwithapositiverealnumber,andthisidentification
jibeswiththeproductsinGandinthegroup
ofpositiverealnumbers,since
lab1=1aIIbl·SoGINisinaveryrealsense(nopunintended)thegroupof
positiverealnumbersundermultiplication.
PROBLEMS
1.IfGisthegroupofallnonzerorealnumbers undermultiplicationand N
isthesubgroupofallpositiverealnumbers,write outGINbyexhibiting
thecosets
ofNinG,andconstructthemultiplicationin GIN.
2.IfGisthegroup ofnonzerorealnumbers undermultiplicationand

98. Sec.6 FactorGroups 83
N={I,-I},showhowyoucan"identify" GINasthegroupofall
positiverealnumbers
undermultiplication.Whatarethecosetsof
NinG?
3.IfGisagroupandN<lG,showthatifMisasubgroupofGINandM=
{aEGINaEM},thenMisasubgroupofG,andM
~N.
4.IfMinProblem3 isnormalin GIN,showthatthe Mdefinedisnormal
in
G.
5.InProblem3,show thatMINmustequalM.
6.Arguingasin theExample2,whereweidentifiedGIlLasacircle,where
Gisthegroupofrealsunder+andlLintegers,consider thefollowing:
let
G={(a,b)Ia,breal},where+inGisdefinedby (a,b)+(c,d)=
(a+c,b+d)(soGistheplane),andletN={(a,b)EGIa,bareinte­
gers}.Show
thatGINcanbeidentifiedasatorus(donut), andsowecan
definea
productonthedonutsothatitbecomesagroup. Here,youmay
thinkofatorusas theCartesianproductoftwocircles.
7.
IfGisacyclicgroup andNisasubgroupofG,showthatGINisacyclic
group.
8.
IfGisanabeliangroup andNisasubgroupofG,showthatGINisan
abeliangroup.
9.
DoProblems7 and8byobserving thatGINisahomomorphicimage
ofG.
10.LetGbeanabeliangroup oforder
p~lP22...p%k,wherePhP2,,Pk
aredistinctprimenumbers.Show thatGhassubgroups S1,S2, ,Skof
ordersp~l,P2
2
,...,p%k,respectively.(Hint:UseCauchy's Theoremand
passtoafactorgroup.)Thisresult,whichactuallyholdsforallfinite
groups,
isafamousresultingroup theoryknownas Sylow'sTheorem.
WeproveitinSection11.
11.
IfGisagroupandZ(G)thecenterofG,showthatifGIZ(G)iscyclic,
thenG
isabelian.
12.
IfGisagroupandN<lGissuchthatGINisabelian,prove that
aba-1b-
1 ENforalla,bEG.
13.IfGisagroupandN<lGissuchthat
foralla,bEG,provethatGINisabelian.
14.
IfGisanabeliangroup oforderPIP2...Pk,where
PhP2,...,Pkare
distinctprimes,prove thatGiscyclic.(See Problem15.)

99. 84 Groups Ch.2
15.IfG isanabeliangroupandifGhasanelement ofordermandone of
ordern,wheremandnarerelativelyprime,provethatGhasanelement
of
ordermn.
16.LetG beanabeliangroup oforderpnm,wherepisaprimeand p
(m.
LetP={aEGIa
pk
=eforsome kdependingona}.Provethat:
(a)PisasubgroupofG.
(b)GIPhasnoelements oforderp.
(c)Ipi=pn.
17.LetGbeanabeliangroup ofordermn,wheremandnarerelatively
prime.
LetM={aEGIam=e}.Provethat:
(a)MisasubgroupofG.
(b)GIMhasnoelement, x,otherthantheidentityelement,suchthat
x
m
=unitelement ofGIM.
18.LetGbeanabeliangroup(possiblyinfinite)andlettheset T=
{aEGIam=e,m>1dependingon a}.Provethat:
(a)TisasubgroupofG.
(b)GIThasnoelement-otherthanitsidentity element-offiniteorder.
7.THEHOMOMORPHISM THEOREMS
LetGbeagroupand
cpahomomorphismofGontoG'.IfKisthekernelof
cp,thenKisanormalsubgroup ofG,hencewecanform GIK.Itisfairlynat­
uraltoexpect
thatthereshould beaverycloserelationshipbetween G'and
GIK.TheFirstHomomorphismTheorem, whichweareabouttoprove,
spells
outthisrelationshipinexactdetail.
Butfirstlet'slookbackatsome oftheexamplesoffactorgroupsin
Section6toseeexplicitlywhattherelationshipmentionedabovemightbe.
1.LetG={T
a
,
bIa
=1=0,breal}andletG'bethegroup ofnonzeroreals
undermultiplication.
Fromtheproductrule oftheseT's,namely
Ta,bTc,d=Tac,ad+b'wedeterminedthatthemapping
cp:G~G'defined
byCP(Ta,b)=aisahomomorphism ofGontoG'withkernel K=
{T1,bibreal}.Ontheotherhand,inExample1 ofSection6wesawthat
GIK={KTa,°Ia=1=0real}.Since
(KTa,o)(KTx,o)=KTax,o
themapping ofGIKontoG',whichsendseach KTa,oontoa,isreadily
seento
beanisomorphismofGIKontoG'.Therefore,
GIK::::=G'.
2.InExample3,Gwasthegroupofnonzerocomplexnumbersundermultipli­
cationand
G'thegroupofallpositiverealnumbersundermultiplication.

100. Sec.7 TheHomomorphism Theorems 85
Letcp:G~G'definedbycp(a)=laIforaEG.Then,since Iabl=
IaIIbI,cpisahomomorphismofGonto G'(canyouseewhyit isonto?).
Thusthekernel
Kof
cpispreciselyK={aEGilaI=I}.Butwehaveal­
readyseenthatif
IaI=1,thenaisoftheformcos ()+isin().Sotheset
K={cos()+isin()I0
~()<27T}.Ifaisanycomplexnumber,then
a=r(cos()+isin(),where r=IaI,isthepolarformof a.ThusKa=
Kr(cos()+isin()=K(cos()+isin()r=Kr,sinceK(cos()+isin()=K
becausecos ()+isin()EK.SoG/K,whoseelementsarethecosets Ka,
fromthisdiscussion,hasallitselementsoftheform Kr,wherer>O.The
mappingof
G/KontoG'definedbysending Krontorthendefinesan iso­
morphism
ofG/KontoG'.So,here,too, G/K
~G'.
Withthislittleexperiencebehinduswearereadytomakethejumpthe
wholeway,namely,to
Theorem2.7.1(FirstHomomorphismTheorem).Letcpbeahomo­
morphismofG
ontoG'withkernel K.ThenG'
~G/K,theisomorphismbe­
tweenthesebeingeffectedbythemap
I/J:G/K~G'
definedbyI/J(Ka)=cp(a).
ProofThebestwaytoshowthat G/KandG'areisomorphicistoex­
hibitexplicitlyanisomorphismofG/
KontoG'.Thestatementofthetheo­
remsuggestswhatsuchanisomorphismmightbe.
Sodefine
I/J:G/K~G'byI/J(Ka)=cp(a)foraEG.Asusual,ourfirst
task
istoshowthat
I/Jiswelldefined,that is,toshowthatif Ka=Kb,then
I/J(Ka)=I/J(Kb).Thisboilsdowntoshowingthatif Ka=Kb,thencp(a)=
cp(b).ButifKa=Kb,thena=kbforsome kEK,hencecp(a)=cp(kb)=
cp(k)cp(b).SincekEK,thekernelofcp,thencp(k)=e',theidentityelement
of
G',soweget
cp(a)=cp(b).ThisshowsthatthemappingI/Jiswelldefined.
BecausecpisontoG',givenxEG',thenx=cp(a)forsome aEG,
thus
x=
cp(a)=I/J(Ka).ThisshowsthatI/JmapsG/KontoG'.
IsI/JI-I?SupposethatI/J(Ka)=I/J(Kb);thencp(a)=I/J(Ka)=
I/J(Kb)=cp(b).Therefore,e'=cp(a)cp(b)-l=cp(a)cp(b-
1
)=cp(ab-
1
).Be­
cause
ab-
1
isthusinthekernelof
cp-whichisK-wehaveab-
1
EK.This
impliesthat
Ka=Kb.Inthisway
I/Jisseentobe 1-1.
Finally,isI/Jahomomorphism?Wecheck:1/J((Ka)(Kb»=I/J(Kab)=
cp(ab)=cp(a)cp(b)=I/J(Ka)I/J(Kb),usingthatcpisahomomorphismand
that
(Ka)(Kb)=Kab.Consequently,
I/Jisahomomorphismof G/KontoG',
andTheorem 2.7.1isproved.D

101. 86 Groups Ch.2
HavingtalkedabouttheFirstHomomorphismTheoremsuggeststhat
thereareothers.Thenextresult,however,
isanextensionoftheFirst
HomomorphismTheorem,and
istraditionallycalledtheCorrespondence
Theorem.Inthecontextofthetheoremabove,itexhibitsa
1-1correspon­
dencebetweensubgroupsofG
I
andthosesubgroupsofGthatcontain K.
Theorem2.7.2(CorrespondenceTheorem).Letthemap
q;:G~G'be
ahomomorphismofGonto
G'withkernel K.IfH'isasubgroupof G'andif
H
={aEGI
q;(a)EH'},
thenH
isasubgroupofG,H
:>K,andHIK==H'.Finally,if H'<JG',then
H<JG.
ProofWefirstverifythattheHabove isasubgroupof G.Itisnot
empty,sincee
EH.Ifa,bEH,then
q;(a),q;(b)EH',henceq;(ab)=
q;(a)q;(b)EH',sinceH'isasubgroupof G';thisputsabin H,soH is
closed.Further,ifa EH,thenq;(a)EH',henceq;(a-
I
)=q;(a)-IisinH',
againsince H'isasubgroupof G',whencea-IEH.Therefore,H isasub­
groupof
G.
Because
q;(K)={e'}CH',wheree'istheunitelementof G',wehave
thatK
CH.SinceK <JGandK CH,itfollowsthatK <JH.Themapping
q;
restrictedtoHdefinesahomomorphismofHonto H'withkernelK.Bythe
FirstHomomorphismTheoremweget
HIK
==H'.
Finally,if H'<JG'andifa EG,thenq;(a)-IH'q;(a)CH',so
q;(a-I)H'q;(a)CH'.Thistellsusthatq;(a-IHa)CH',soa-IHaCH.This
provesthenormalityof
HinG.D
ItisworthnotingthatifK isanynormalsubgroupof G,and
q;isthe
naturalhomomorphismofGonto
G/K,thenthetheoremgivesusa 1-1cor
..
respondencebetweenallsubgroups H'ofG/KandthosesubgroupsofG
thatcontain
K.Moreover,thiscorrespondencepreservesnormalityinthe
sensethat
H'isnormalin G/Kifandonlyif Hisnormalin G.(SeeProblem
7,aswellasthelastconclusionofthetheorem.)
WenowstatetheSecondHomomorphismTheorem,leavingitsproof
tothereaderinProblem
5.
Theorem2.7.3(SecondHomomorphismTheorem).Let Hbeasub­
groupofagroupGandNanormalsubgroupof
G.ThenHN=
{hnIhEH,nEN}isasubgroupofG,H nNisanormalsubgroupofH,
andHI(H
nN)
=:(HN)/N.

102. Sec.7 TheHomomorphism Theorems 87
Finally,wego ontotheThirdHomomorphism Theorem,whichtellsus
alittlemore
abouttherelationshipbetween NandN'whenN'<JG'.
Theorem2.7.4(ThirdHomomorphismTheorem). Ifthemapcp:G~G'isahomomorphismofGontoG'withkernelKthen,if N'<JG'
andN ={aEGIcp(a)EN'},weconcludethatGIN~G'IN'.Equivalently,
GIN~(GIK)/(NIK).
ProofDefine themappingl/J:G~G'IN'byl/J(a)=N'cp(a)forevery
a
EG.Since
cpisontoG'andevery elementofG'IN'isacosetoftheform
N'x',andx'=cp(x)forsomex EG,wesee thatl/JmapsGontoG'IN'.
Furthermore,l/Jisahomomorphism ofGontoG'IN',forl/J(ab)=
N'cp(ab)=N'cp(a)cp(b)=(N'cp(a»(N'cp(b»=l/J(a)l/J(b),sinceN'<JG'.
Whatisthekernel,M, ofl/J?IfaEM,thenl/J(a)istheunitelement of
G'IN',thatis,l/J(a)=N'.Ontheotherhand,by thedefinitionofl/J,l/J(a)=
N'cp(a).BecauseN'cp(a)=N'wemusthavecp(a)EN';butthisputsainN,
bytheverydefinition
ofN.ThusMeN.ThatNCMiseasyandisleftto
thereader.Therefore,M =N,so
l/Jisahomomorphism ofGontoG'IN'
withkernel N,whence,bytheFirstHomomorphismTheorem, GIN~G'IN'.
Finally,againbyTheorems2.7.1 and2.7.2,G'~GIK,N'~NIK,which
leads
ustoGIN
~G'IN'~(GIK)/(NIK).D
Thislastequality ishighlysuggestive;wearesort of"cancelingout"the
Kinthenumeratoranddenominator.
PROBLEMS
1.ShowthatM~NintheproofofTheorem2.7.3.
2.LetGbethegroup ofallreal-valuedfunctions ontheunitinterval[0,1],
wherewedefine,for
I,gEG,additionby (I+g)(x)=I(x)+g(x)for
everyx
E[0,1].IfN={IEGI
I(~)=O},provethatGIN~realnum­
bers
under+.
3.LetGbethegroup ofnonzerorealnumbers undermultiplicationandlet
N
={1,-1}.ProvethatGIN
~positiverealnumbers undermultiplication.
4.IfG
bG
2aretwogroups andG=G
lXG
2={(a,b)IaEG
l
,bEG
2
},
wherewedefine (a,b)(c,d) =(ac,bd),showthat:
(a)
N={(a,e2)IaEG
l
},wheree2istheunitelementofG2,isanormal
subgroup
ofG.
(b)
N~G
l
.
(c)GIN
~G
2
•

103. 88 Groups Ch.2
5.LetGbeagroup,HasubgroupofG,andN<JG.LetthesetHN=
{hnIhEH,nEN}.Provethat:
(a)HnN<JH.
(b)HNisasubgroupofG.
(c)NCHNandN<JHN.
(d)(HN)IN~HI(HnN).
*6.IfGisagroupandN<JG,showthatifaEGhasfinite ordero(a),then
NainGINhasfiniteorderm,wheremlo(a).(Provethis byusingthe
homomorphismofGontoGIN.)
7.Ifq;isahomomorphismofGontoG'andN<JG,showthatq;(N)<JG'.
8.CAUCHY'STHEOREM
InTheorem2.6.4-Cauchy'sTheorem-weprovedthatifaprime pdivides
theorderofafiniteabeliangroupG,thenGcontainsanelementoforderp.
WedidpointouttherethatCauchy'sTheoremistrueevenif thegroupisnot
abelian.Weshallgiveavery neatproofofthishere;this proofisdueto
McKay.
Wereturnforamomenttosettheory,doingsomething thatwemen­
tionedintheproblemsinSection4.
LetSbeaset,fEA(S),anddefinearelation onSasfollows:s
'"'-'tif
t=fi(S)forsomeintegeri(icanbepositive,negative, orzero).Weleaveit
tothereaderasaproblemthatthisdoes indeeddefineanequivalencerela­
tiononS.Theequivalenceclass ofs,[s],iscalledtheorbitofsunderfSoS
isthedisjointunionoftheorbitsofitselements.
Whenfisoforderp,paprime,wecansaysomething aboutthesizeof
theorbitsunderf;thoseofthereaderswhosolvedProblem34ofSection4
already
knowtheresult.Weproveitheretoputitontherecordofficially.
[Iffk(s)=s,ofcourseftk(s)=sfor everyintegert.(Prove!)]
Lemma2.8.1.IffEA(S)isoforderp,paprime,thentheorbitof
anyelementofSunderfhas1orpelements.
ProofLetsE
S;iff(s)=s,thentheorbitofsunderfconsistsmerely
ofsitself,sohas oneelement.Suppose thenthatf(s)=1=s.Considertheele­
mentss,f(s),f
2
(s),...,f
p
-1
(s);weclaimthatthesepelementsaredistinct
andconstitutetheorbitofsunderfIfnot,thenfi(S)=fj(s)forsome
o
~i<j~p-1,whichgivesus thatfj-i(S)=s.Letm=j-i;then
0<m~p-1andfm(s)=s.ButfP(s)=sandsincep~m, ap+bm=1for
someintegersaandb.Thusfl(S)=fap+bm(s)=fap(fbm(s»=fap(s)=S,

104. Sec.8 Cauchy'sTheorem 89
sincefm(s)=fP(s)=s.Thiscontradictsthatf(s)=1=s.Thustheorbitofs
underfconsistsofs,f(s),f
2
(s),...,fP
-1(s),soaspelements.D
WenowgiveMcKay's proofofCauchy'sTheorem.
Theorem2.8.2(Cauchy). IfpisaprimeandpdividestheorderofG,
thenGcontainsan
elementoforderp.
ProofIfp=2,theresultamountstoProblem 18inSection1.Assume
that
p
=1=2.LetSbethesetofallorderedp-tuples(aba2,...,ap-bap),where
aba2'...' apareinGandwhere ala2...ap-lap =e.WeclaimthatShasnP-
l
elementswhere n=IGI.Why?Wecanchooseab ,ap-larbitrarilyin G,
andbyputting ap=(ala2...ap-l)-t,thep-tuple(aba2,,ap-bap)thensatisfies
so
isinS.ThusShas n
P
-1elements.
Note
thatifala2...ap-1ap=e,thenapala2...ap-l=e(forifxy=ein
agroup,then
yx=e).Sothemapping f:S
~Sdefinedbyf(ab...,ap) =
(ap,aI,a2,.··,ap-l)isinA(S).Notethatf=1=i,theidentitymaponS,and
thatfP=i,sofisoforderp.
Iftheorbitofsunderfhasoneelement,thenf(s)=s.Ontheother
hand,if f(s)=1=s,weknow thattheorbitofsunderfconsistsprecisely ofp
distinctelements;thiswehaveby Lemma2.8.1.Nowwhen isf(s)=1=s?We
claimthatf(s)=1=sifandonlyif whens=(aI'a2,...,ap),thenforsome
i=1=j,ai=1=aj.(Weleavethisto thereader.)So f(s)=sifandonlyif s=
(a,a,...,a)forsome aEG.
LetmbethenumberofsESsuchthatf(s)=s;sincefor s=
(e,e,...,e),f(s)=s,weknow thatm2::1.Ontheotherhand,if f(s)=1=s,
theorbitofsconsistsofpelements,andtheseorbits aredisjoint,forthey are
equivalenceclasses. Ifthereareksuchorbitswhere f(s)=1=s,wegetthat
n
P
-
1=m+kp,forwehaveaccountedthiswayforevery elementofS.
ButpinbyassumptionandpI(kp).Sowemusthave pIm,sincem=
n
P
-
l
-
kp.Becausem=1=0andpIm,wegetthatm>1.Butthissaysthat
thereisans=(a,a,...,a)=1=(e,e,...,e)inS;fromthedefinitionofSthis
implies
thata
P=e.Sincea
=1=e,aistherequiredelementoforderp.D
Notethattheprooftellsus thatthenumberofsolutionsinG ofx
P
=e
isapositivemultiple ofp.
Westronglyurge thereaderwhofeelsuncomfortablewith theproof
justgiventocarry outitsdetailsfor p=3.Inthiscase theactionoffonS
becomesclear andourassertionsaboutthisactioncan becheckedexplicitly.

105. 90 Groups Ch.2
Cauchy'sTheoremhasmanyconsequences.Weshallpresentoneof
these,inwhichwedeterminecompletelythenatureofcertaingroupsof
orderpq,wherepandqaredistinctprimes. Otherconsequenceswillbe
foundintheproblemsettofollow,andinlatermaterialongroups.
Lemma
2.8.3.LetGbeagroupoforder pq,wherep,qareprimes
and
p>q.IfaEGisoforderpandAisthesubgroupofGgeneratedby a,
thenA<JG.
ProofWeclaimthat AistheonlysubgroupofG oforderp.Suppose
that
Bisanothersubgroupof orderp.Considertheset AB ==
{xyIxEA,yE
B};weclaimthat ABhasp2distinctelements.Forsuppose
that
xy==uvwherex,uEA,y,vE
B;thenu-lx==vy-l.Butu-lxEA,
vy-lEB,andsinceu-lx==vy-t,wehaveu-lxEAnB.SinceB=1=Aand
AnBisasubgroupof AandAisofprimeorder,weareforcedtoconclude
thatAnB==(e)andsou-lx==e,thatis,u==x.Similarly,v==y.Thusthe
number
ofdistinctelementsin ABisp2.ButalltheseelementsareinG,
whichhasonly
pq<p2elements(since p>q).Withthiscontradictionwe
seethat
B==AandAistheonlysubgroup oforderpinG.Butif xEG,B==
x-lAxisasubgroupofGoforderp,inconsequenceofwhichweconclude
thatx-lAx==
A;henceA<JG.D
Corollary.IfG, aareasinLemma2.8.3andif xEG,then
x-lax==ai,where0 <i<p,forsome i(dependingon x).
ProofSincee=1=aEAandx-lAx==A,x-laxEA.Buteveryelement
ofAisoftheform ai,0::;i<p,andx-lax=1=e.Inconsquence,x-lax==ai,
where0 <i<p.D
Wenowprovearesult ofadifferentflavor.
Lemma
2.8.4.IfaEGisofordermandbEGisofordern,where
mandnarerelativelyprimeand ab==ba,thenc ==abisofordermn.
ProofSupposethat Aisthesubgroupgeneratedby aandBthatgener­
atedby
b.BecauseIAI==mandIBI==nand(m,n)==1,wegetAnB==(e),
whichfollowsfromLagrange'sTheorem,for IAnBIInandIAnBIIm.
Supposethatc
i==e,wherei>
0;thus(ab)i==e.Sinceab==ba,e ==
(ab)i==aib
i
;thistellsusthat a
i==b-
i
EAnB==(e).Soa
i==e,whencemIi,
andb
i==e,whencenIi.Because(m,n)==1andmandnbothdivide i,mn
dividesi.Soi
2::mn.Since(ab)mn==a
mn
b
mn
==e,weseethatmnisthesmall­
estpositiveinteger
isuchthat (ab)i==e.Thissaysthat abisofordermn,as
claimedinthelemma. D

106. Sec.8 Cauchy'sTheorem 91
Beforeconsideringthe moregeneralcase ofgroupsoforderpq,let's
lookataspecialcase,namely,agroupG
oforder15.ByCauchy'sTheorem,
Ghaselements boforder3andaoforder5.BytheCorollarytoLemma
2.8.3,b-1ab=ai,where0 <i<5.Thus
b-
2
ab
2
=b-1(b-1ab)b=b-1aib=(b-1ab)i=(ai)i=a
i2
andsimilarly,b-
3
ab
3
=a
i3
.Butb
3
=e,soweget a
i3
=a,whencea
i3
-
1
=e.
Sinceaisoforder5,5mustdivide i
3
-
1,thatis,i
3
==1(5).However,byFer­
mat's
Theorem(Corollaryto Theorem2.4.8),i
4
==1(5).Thesetwoequations
for
itellusthati
==1(5),so,since0 <i<5,i=1.Inshort,b-1ab=a
i=a,
whichmeans thatab=ba.Sinceaisoforder5andboforder3,byLemma
2.8.4,c =abisoforder15.Thismeansthatthe15powerse=co,c,c
2
,
•••,
C
14
aredistinct,somustsweep outallofG.Inaword,G mustbecyclic.
Theargumentgivenfor 15couldhave beenmadeshorter,buttheform
inwhichwedidit
istheexactprototypefortheproofofthemoregeneral
Theorem2.8.5.LetGbeagroupoforderpq,wherep,qareprimes
and
p>q.Ifq
(p-1,thenGmustbecyclic.
ProofByCauchy'sTheorem,Ghasanelementaoforderpandan
elementboforderq.BytheCorollarytoLemma2.8.3,b-1ab=a
i
forsome i
with0 <i<p.Thusb-rab
r
=airforallr2::0(Prove!),andsob-qab
q
=a
iq
.
Butb
q=e;therefore,a
iq
=aandsoa
iq
-
1=e.Becauseaisoforderp,we
conclude
thatpIi
q
-
1,whichistosay,i
q
==l(p).However,by Fermat's
Theorem,i
P
-
1
==l(p).Sinceq(p-1,weconcludethati==l(p),andsince
o<i<p,i=1follows.Therefore, b-1ab=a
i=a,henceab=ba.By
Lemma2.8.4,c =abhasorderpq,sothepowersofcsweepoutallofG.
ThusG
iscyclic,andthetheoremisproved.D
PROBLEMS
Middle-LevelProblems
1.IntheproofofTheorem2.8.2,show thatifsometwoentriesin s
(aba2,...,ap)aredifferent,thenI(s)=1=s,andtheorbitofsunder1
haspelements.
2.Prove
thatagroupoforder35iscyclic.
3.Using
theresultofProblem40ofSection5,give anotherproofof
Lemma2.8.3.(Hint:Usefor Hasubgroupoforderp.)
4.Constructanonabeliangroupof order21.(Hint:Assume thata
3
=e,

107. 92 Groups Ch.2
b
7
=eandfindsome isuchthata-1ba=a
i
=1=a,whichisconsistentwith
therelationsa
3
=b
7
=e.)
5.LetGbeagroupoforderpnm,wherepisprimeandp~m.Supposethat
Ghasanormalsubgroup
Poforderpn.ProvethatO(P)=Pforevery
automorphism
0ofG.
6.LetG beafinitegroupwithsubgroups A,BsuchthatIAI>
VTGIand
IBI>VTGI.ProvethatAnB=1=(e).
7.IfGisagroupwithsubgroups A,Bofordersm,n,respectively,where
mandnarerelativelyprime,prove thatthesubset ofG,
AB={abIaEA,bEB},hasmndistinctelements.
8.Provethatagroupoforder99hasanontrivialnormalsubgroup.
9.Provethatagroupoforder42hasanontrivialnormalsubgroup.
10.FromtheresultofProblem9,provethatagroupoforder42hasanor­
malsubgroup
oforder21.
HarderProblems
11.
IfGisagroupandA,Bfinitesubgroups ofG,provethatthesetAB=
tabIaEA,bEB}has(IAliB1)/IAnBIdistinctelements.
12.Provethatanytwononabeliangroups oforder21areisomorphic.(See
Problem4.)
VeryHardProblems
13.
Usingthefactthatanygroup oforder9isabelian,provethatanygroup
oforder99isabelian.
14.Letp>qbetwoprimessuch thatqIp-1.Provethatthereexistsa
nonabeliangroup
oforderpq.(Hint:Usetheresult ofProblem40of
Section4,namelythatUpiscyclicif pisaprime,andtheideaneededto
do
Problem4above.)
15.Provethatifp>qaretwoprimessuch thatqIp-1,thenanytwonon­
abeliangroups
oforderpqareisomorphic.
9.DIRECTPRODUCTS
Inseveraloftheproblemsandexamplesthatappearedearlier,wewent
throughthefollowingconstruction:
IfG1,G2aretwogroups,thenG =
G
1
XG
2isthesetofallorderedpairs(a,b),whereaEG
1andbEG
2and

108. Sec.9 DirectProducts 93
wheretheproductwasdefined component-wisevia(aI,b1)(a2'b2)
(a1a2'b1b2),theproductsineachcomponentbeingcarried outintherespec­
tivegroupsG
1andG
2
•Weshouldliketoformalizethisprocedurehere.
Definition.If
G1,G2,...,Gnarengroups,thentheir (external)di­
rectproduct
G
1
XG
2
XG
3
X...XGnisthesetofallorderedn-tuples
(aba2'...'an)whereaiEGi,fori=1,2,...,n,andwheretheproductin
G
1
XG
2
X...XG
nisdefinedcomponent-wise,that is,
ThatG =G
1
XG
2
X...XG
nisagroupisimmediate,with
(e1'e2'...,en)asitsunitelement,where eiistheunitelement ofGi,and
h ( )
-1-(-1-1 -1)
were
aba2,...,an-a1,a2,...,an.
GismerelytheCartesianproduct ofthegroupsG 1,G2,...,Gnwitha
productdefinedinGbycomponent-wisemultiplication.Wecallit
external,
sincethegroupsG
1
,G
2
,
•••,G
nareanygroups,withnorelatiohnecessarily
holdingamongthem.
Considerthesubsets
G
jCG
1
XG
2x··.XG
n
=G,where
inotherwords,
Gjconsistsofalln-tupleswhereinthe ithcomponentanyel­
ement
ofG
icanoccurandwhereevery othercomponentistheidentityele­
ment.Clearly,
G
iisagroupand isisomorphicto G
i
.bytheisomorphism
1Ti:Gj
~Gidefinedby 1Ti(e1'e2'...,ai'...,en)=ai.Furthermore,not
onlyisG
jasubgroupofGbutG
j<JG.(Prove!)
Givenanyelement
a=
(aba2'...,an)EG,then
that
is,everyaEGcanbewrittenas a=
ala2·..an,whereeacha
iEG
i
•
Moreover,acanbewritteninthiswayinauniquemanner, thatis,ifa=
ala2·.·an=blb2·· ·bn,wheretheaiEGiandbiEG
i
,then
a
l
=b
i
,...,an=bn.SoG isbuiltupfromcertainnormalsubgroups, the
G
j
,asG=G
IG
2
• • •G
ninsuchaway thateveryelement aEGhasa
uniquerepresentationintheform a=
ala2. · ·anwithajEG
j
•
Thismotivatesthefollowing
Definition.
ThegroupG issaidtobe the(internal)directproduct of
itsnormalsubgroups
NbN2, ...,NnifeveryaEGhasa uniquerepresenta­
tionintheform
a=ala2...an'whereeach aiENifori=1,2,...,n.
Fromwhatwehavediscussedabovewehavethe

109. 94 Groups Ch.2
Lemma2.9.1.IfG=G
I
XG
2
X...XG
nistheexternaldirectprod­
uctof
G
I
,G
2
,
•••,G
n
,thenGistheinternaldirectproductofthenormal
subgroups
G
bG
2
,
•••,G
ndefinedabove.
Wewanttogointheotherdirection,namelytoprovethatifG
isthein­
ternaldirectproductofitsnormalsubgroups
N
bN
2
,.••,N
n
,thenG isiso­
morphicto
N
I
XN
2
X...XN
n
•Todoso, wefirstgetsomepreliminaryresults.
TheresultweareabouttoprovehasalreadyoccurredasProblem
20,
Section5.Forthesakeofcompletenessweproveithere.
Lemma
2.9.2.LetGbeagroup, M,NnormalsubgroupsofGsuch
that
MnN=(e).Then,given mEMandnEN,mn=nm.
ProofConsidertheelement a=mnm-In-
I
.
Viewingaasbracketed
oneway,
a=(mnm-I)n-
I
;
then,since N<JGandnEN,mnm-
I
EN,so
a=(mnm-I)n-
I
isalsoin N.Nowbracket aintheotherway, a=m(nm-In-
I
).
SinceM<JGandm-
I
EM,wehave nm-In-
I
EMandso a=
m(nm-In-
I
)
EM.ThusaEMnN=(e),whichistosay,mnm-In-
I
=e.
Thisgivesusthat mn=nm,asrequired.D
IfGistheinternaldirectproduct ofthenormalsubgroups NI,
N
2
,
•••,N
n
,weclaimthat N
inN
j=(e)fori
=1=j.Forsupposethat
aEN
inNj;thena=e.e. · .eae...e,wherethe aoccursintheithplace.
Thisgives
usonerepresentationof ainG=N
IN
2
•••N
n
•Ontheotherhand,
a=e.e...e.a.e ...e,wherethe aoccursinthe jthplace,so ahasthesecond
representation
asanelementof N
IN
2
•••N
n
•Bytheuniquenessoftherepre­
sentation,weget
a=e,andso N
inN
j
=(e).
Perhapsthingswouldbeclearerifwedoitfor n=2.Sosupposethat
N
I<JG,N
2<JG,andeveryelement aEGhasauniquerepresentation as
a=al.a2,wherealENI,a2EN2.Supposethat aENIn
N2;thena=
a.eisarepresentationofa=al.a2withal=aEN
l
,
a2=eEN2.How­
ever
a=e.a,soa=b
l
.b
2
,whereb
l=eE
Nbb
2=aEN2.Bythe
uniqueness
oftherepresentationwemusthave aI=bl,thatis,a=e.
SoN
InN
2=(e).
Theargumentgivenabovefor
N
b
...,N
nisthesameargument asthat
givenfor
n=2,butperhapsislesstransparent.Atanyratewehaveproved
Lemma
2.9.3.IfGistheinternaldirectproductofitsnormalsub­
groups
N
bN
2
,
.•.,N
n
,then,for i
=1=j,N
inN
j=(e).
Corollary.IfG isasinLemma2.9.3,thenif i=1=jandaiENiand
ajENj,wehaveaiaj=ajai'

110. Sec.9 DirectProducts 95
ProofByLemma2.9.3, N
inN
j=(e)fori=1=j.Sincethe N'sarenor­
malinG,byLemma2.9.2wehave
thatanyelementin N
icommuteswithany
elementin
Nj,thatis,aiaj=ajaiforaiENi,ajENj.0
Withthesepreliminaries outofthewaywecannowprove
Theorem2.9.4.LetGbeagroupwithnormalsubgroups
N
I
,N
2
,
•••,
Nn.Thenthemapping
t/J(aba2,...,an)=aIa2...anisanisomorphism
from
N
I
XN
2
X...XN
n(externaldirectproduct) ontoGifandonlyifG is
theinternaldirectproductof NI,N2 ,•••,Nn.
ProofSupposeG isaninternaldirectproductof N
b...,N
n
•Since
everyelement
ainGhasarepresentation a=ala2...an,withthe aiENi,
wehave thatthemapping
t/Jisonto.Weassert thatitisalso1-1. Forif
t/J((aba2'···'an»=t/J((b1,b2,···,bn»,thenbythedefinition oft/J,
aIa2...an=bIb2...bn.Bytheuniqueness oftherepresentationofanele-
mentinthisformwededucethat
al=
b
ha2=b2,...,an=bn.Hencet/Jis1-1.
Allthatremains istoshowthatt/Jisahomomorphism.So,consider
t/J((a
ba2,...,an)(b
1
,b
2
,
•••,bn»=
t/J((a1b
ba
2b2,...,anbn»
=(a1b1)(a2b
2
)• • •(anbn)
Sinceb
iE
Nbitcommuteswith ai'bifori>1bytheCorollaryto Lemma
2.9.3.Sowecanpullthe b
iacrossalltheelements totherightofittoget
aIbIa2b2·..anbn=aIa2b2a3b3anbnbl.Nowrepeatthisprocedurewith b2,
andsoon,togetthat aIbIa2b2 anbn=(aIa2..·an)(bIb2 ...bn).Thus
t/J((a1,a2,·· .,an)(b1,b2,·..,bn»=a1b1a2b2. · ·anbn
=(aa...a)(bb ...b )
1 2 n 1 2 n
=t/J((aba2,...,an»t/J((b1b2,·..,bn»·
Inotherwords,t/Jisahomomorphism.
Ontheotherhand,suppose thatt/Jisanisomorphism.Thentheconclu­
sionthatG
istheinternaldirectproduct ofNI,N2,...,N
neasilyfollows
fromthefactthat
t/Jisontoand1-1.
Withthistheproof
ofTheorem2.9.4iscomplete.D
Corollary.LetGbeagroupwithnormalsubgroups N
I
,N
2
•ThenG
istheinternaldirectproduct ofNIandN2ifandonlyifG =N1N2and
N
InN
2=(e).

111. 96 Groups Ch.2
ProofThisfollowseasilyfrom thefactthatl/J:N
lXN2
~G,whichis
givenbyl/J(a},a2)=aIa2,isanisomorphismif andonlyif NIN2=Gand
N
InN
2=(e).0
InviewoftheresultofTheorem2.9.4anditscorollary,we dropthead­
jectives
"internal"and"external"andmerelyspeakaboutthe"directprod­
uct."
WhennotationG=N
I
XN
2isuseditshould beclearfromcontext
whetheritstandsfor theinternalorexternaldirectproduct.
Theobjectiveisoftentoshowthatagivengroupisthedirectproduct
ofcertainnormalsubgroups.Ifonecandothis,thestructureofthegroup
can
becompletelydeterminedifwehappentoknowthoseofthenormal
subgroups.
PROBLEMS
1.IfG
IandG
2aregroups,provethatG
IXG
2
=G
2XG
I
•
2.IfG
landG
2arecyclicgroups ofordersmandn,respectively,prove that
GIXG2iscyclicif andonlyifmandnarerelativelyprime.
3.
LetGbeagroup,A=GXG.InAletT={(g,g)IgEG}.
(a)Prove thatT
=G.
(b)
ProvethatT<JAifandonlyifG isabelian.
4.
LetGbeanabeliangroup oforder
p'{'lp'{'2.. ·p'!:k,wherePI,P2'···'Pk
aredistinctprimes andmi>0,m2>0,...,mk>O.ByProblem10 of
Section6,foreach i,Ghasasubgroup Pioforderpr'.Showthat
G=PIXP
2
X...XP
k
•
5.LetGbeafinitegroup, N
I
,N
2
,
•••,N
knormalsubgroups ofGsuchthat
G=NIN 2•••NkandIGI=INIIIN21···INkl.ProvethatGisthedirect
productof
N},N
2
,
•••,N
k
•
6.LetGbeagroup,NI,N2,...,N
knormalsubgroupsofGsuchthat:
1.G=NIN
2
•
••Nk.
2.Foreachi,Nin(NIN2
•••Ni-INi
+
1
•••Nk)=(e).
ProvethatGisthedirectproductof
N},N
2
,
•••,Nk.
10.FINITEABELIANGROUPS(OPTIONAL)
Wehavejustfinisheddiscussing theideaofthedirectproductofgroups.If
weweretoleavethattopicatthepointwherewe ended,itmightseemlike
anicelittleconstruction,
butsowhat?Togivesomemoresubstancetoit,

112. Sec.10 FiniteAbelianGroups(Optional) 97
weshouldproveatleastonetheoremwhichsays thatagroupsatisfyinga
certaincondition
isthedirectproductofsomeparticularlyeasygroups.For­
tunately,suchaclass
ofgroupsexists,thefiniteabeliangroups.Whatwe
shallprove
isthatanyfiniteabeliangroup isthedirectproductofcyclic
groups.Thisreducesmostquestionsaboutfiniteabeliangroupstoquestions
aboutcyclicgroups,areduction
thatoftenallowsustogetcompletean­
swerstothesequestions.
Theresultsonthestructureoffiniteabeliangroupsarereallyspecial
casesofsomewideranddeepertheorems.
Toconsiderthesewouldbegoing
toofarafield,especiallysincethestoryforfiniteabeliangroups
issoimpor­
tantinitsownright.Thetheoremweshallprove
iscalledthe Fundamental
TheoremonFiniteAbelianGroups,andrightfullyso.
Beforegettingdowntotheactualdetails
oftheproof,weshouldliketo
giveaquicksketchofhowweshallgoaboutprovingthetheorem.
Ourfirststepwillbetoreducetheproblemfromanyfiniteabelian
grouptoonewhoseorder
ispn,wherepisaprime.Thisstepwillbefairly
easytocarryout,andsincethegroupwillhave
orderinvolvingjustone
prime,thedetails
oftheproofwillnotbeclutteredwithelementswhoseor­
dersaresomewhatcomplicated.
Soweshallfocusongroups
oforderpn.LetGbeanabeliangroupof
order
pn.WewanttoshowthatthereexistcyclicsubgroupsofG,namely
AI,A2,...,Ak,suchthateveryelement xEGcanbewritten asx=
bIb2·..b
k
,whereeach biEAi'inauniqueway.Otherwiseput,since
each
Aiiscyclicandgeneratedby ai,say,wewanttoshowthat x=
a';la'{'2...a7:\wheretheelementsa'r
iareunique.
Adifficultyappearsrightaway,forthere
isnotjustonechoiceforthese
elements
aI,...,ak.Forinstance,ifG istheabeliangroup oforder4with
elements
e,a,b,ab, wherea
2=b
2
=eandab=ba,thenwecanseethatif
A,B,Carethecyclicsubgroupsgeneratedby a,b,andab,respectively,then
G
=AXB=AXC=BXC.Sothereisalackofuniquenessinthechoiceof
the
ai.Howtogetaroundthis?
Whatweneed
isamechanismforpicking alandwhich,whenapplied
after
wehavepicked aI,willallowustopicka2,andsoon.Whatshouldthis
mechanismbe?
OurcontrolontheelementsofGliesonlyinspecifyingtheir
orders.It
istheorderofthe element-whenproperlyused-thatwillgiveus
themeanstoprovethetheorem.
Supposethat
G=AlXA
2
X...XA
k
,whereIGI=pnandtheA's
havebeennumbered,sothat IAil=pn
1andnl
;:::n2;:::...;:::nk,andeachAi
iscyclicgeneratedby ai.Ifthisweresoand x=a';l...a7:\then

113. 98 Groups Ch.2
becausenl2::ni,pnilpnt,sosinceeverya'r
lpfl
,=e,thusx
pfl1
=e.Inother
words,alshouldbeanelement
ofGwhoseorder isaslargeasitcanpossibly
be.
Fine,wecannowpick al.Whatdowedofor a2?IfG =
G/A
bthento
getthefirstelementneededtorepresentG
asadirectproductofcyclic
groups,weshouldpickanelementinGwhoseorder
ismaximal.Whatdoes
thistranslateintoinGitself?Wewantanelement
a2suchthat a2requiresas
highapower aspossibletofallinto A1.Sothatwillbetheroadtotheselec­
tionofthesecondelement.However,ifwepickanelement
a2withthis
property,itmaynotdothetrick;wemayhavetoadaptitsothatitwill.The
doingofallthis
isthetechnicalpartoftheargumentanddoesgothrough.
Thenonerepeatsitappropriatelytofindanelement
a3,andsoon.
This
istheprocedureweshallbegoingthroughtoprovethetheorem.
Buttosmoothoutthesesuccessivechoices
ofaI,a2,.'"weshalluseanin­
ductionargumentandsomesubsidiarypreliminaryresults.
Withthissketchasguidewehopetheproofofthetheoremwillmake
sensetothereader.
Oneshouldnotconfusethebasicideainthe proof­
whichisquitereasonable-withthetechnicaldetails,whichmaycloudthe
issue.Sowenowbeginto
fillinthedetails ofthesketchoftheproofthatwe
outlinedabove.
Lemma
2.10.1.LetGbeafiniteabeliangroupoforder mn,wherem
andnarerelativelyprime.If M={xEGIx
m
=e}andN={xEGIx
n
=e},
thenG =MXN.Moreover,ifneither mnornis1,thenM
=1=(e)andN=1=(e).
ProofThesets MandNdefinedintheassertionabovearequickly
seentobesubgroupsofG.Moreover,if
m
=1=1,thenbyCauchy'sTheorem
(Theorem2.6.4)wereadilyobtain
M
=1=(e),andsimilarlyif n=1=1,that
N=1=(e).Furthermore,since MnNisasubgroupofbothMandN,byLa­
grange'sTheorem,
1MnNIdividesIMI=mandINI=n.Becausemandn
arerelativelyprime,weobtain
1MnNI=1,henceMnN=(e).
Tofinishtheproof,weneedtoshowthatG =MNandG =MXN.
Sincemandnarerelativelyprime,thereexistintegers randssuchthat
rm+sn=1.IfaEG,thena=a
l=a
sn
+
rm
=asna
rm
;since(asn)m=
a
snm
=e,wehavethat a
sn
EM.Similarly,armEN.Thusa=asna
rm
isin
MN.InthiswayG =MN.ItnowfollowsfromCorollarytoTheorem2.9.4
thatG
=MXN.D
Animmediateconsequence isthe
Corollary.
LetGbeafiniteabeliangroupandlet pbeaprimesuch
that
pdividesIGI.ThenG=PXTforsomesubgroups PandT,where
/PI=pm,m>0,andITIisnotdivisibleby p.

114. Sec.10 FiniteAbelianGroups(Optional) 99
ProofLetP={xEGIx
Ps
=eforsome s}andletthesubset
T={xEGIXl=efortrelativelyprimetop}.ByLemma2.10.1,G=PXT
andP=1=(e).Sinceevery elementinPhasorderapowerofp,Ipiisnotdivis­
iblebyany
otherprime(byCauchy's Theorem),soIpi
=pmforsome m.
Itiseasytosee thatptITIbymakinguseofLagrange's Theorem.Thus
wereallyhave
thatPisnotmerelysomesubgroup ofGbutiswhatiscalled
ap-Sylowsubgroup
ofG.(SeeSection 11).0
Wenowcome tothekeystepin theproofofthetheoremweseek.
Theproofisalittledifficult, butoncewehavethisresult therestwillbe
easy.
Theorem2.10.2.LetGbeanabeliangroup oforderpn,p aprime,
andlet
aEGhavemaximal orderofalltheelementsinG. ThenG
=
AXQ,whereAisthecyclicsubgroup generatedbya.
ProofWeproceedbyinductiononn.Ifn=1,thenIGI=pandGis
alreadyacyclicgroup
generatedbyanya
=1=einG.
Wesuppose
thetheoremtobetrueforallm<n.Wefirstshow that
thetheoremiscorrectif thereexistsanelementbEGsuchthatb
tt.A=(a)
andb
P
=e.LetB=(b),thesubgroupofGgeneratedbyb;thus
AnB=(e)(seeProblem1).
LetG=G/B;byassumptionB=1=(e),henceIGI<IGI.InG,whatis
theorderofa=Ba?Weclaimthato(a)=o(a).Tobeginwith,we know
thato(a)Io(a)(seeProblem6ofSection2.7). Ontheotherhand,ao(a)=e,
soao(a)EB.Sinceao(a)EA,weseethatao(a)EAnB=(e),whence
ao(a)=e.Thistellsus thato(a)Io(a).Henceo(a)=o(a).
SinceaisanelementofmaximalorderinG,by theinductionweknow
thatG=(a)XTforsomesubgroup TofG.BytheCorrespondence
Theoremwealsoknow thatT=Q/Bforsomesubgroup QofG.Weclaim
thatGistheinternaldirect productAXQ.ThatG=AQisleftto
thereader.ItremainstoshowthatAnQ=(e).Leta
i
EAnQ.Then
a
i
EQ/B=T,andsince(a)nT=(e),wehavethata
i
=e.Butsinceo(a)=
o(a),thisimplies a
i
=e.Therefore,AnQ=(e)andweobtain that
G=AXQ.
Suppose,then, thatthereisnoelementbinG,bnotinA,suchthat
b
P=e.WeclaimthatthisforcesG=A=(a),inwhichcaseG isacyclic
group.Suppose
thatG
=1=AandletxEG,xf!:.Ahavesmallestpossible
order.Because
o(x
P
)<o(x),wehave,by ourchoiceofx,thatx
P
EA,hence
x
P
=a
i
forsome i.
Weclaim thatpIi.Leto(a)
=pS,andnotethatthemaximality

115. 100 Groups Ch.2
oftheorderofaimpliesthat x
Ps
=e.Butx
Ps
=(XP)pS-l=(ai)pS-l=e.Since
o(a)
=pS,wehavep Ii.
ThusxP=ai,wherep Ii.Lety=a-uP.x.ThenyP=a-ix
p
=a-ia
i
=e.
Moreover,y
f£(a)=A,becausexf£A.Butthisputsusbackinthesituation
discussedabove,wherethereexists
abEG,b
f£Asuchthat b
P
=e;inthat
casewesaw
thatthetheoremwascorrect.SowemusthaveG =(a),andG
isacyclicgroup.Thisfinishestheinductionandprovesthetheorem. D
Wearenowabletoprovetheverybasicandimportant
Theorem2.10.3(FundamentalTheoremonFiniteAbelianGroups).
Afiniteabeliangroup isthedirectproduct ofcyclicgroups.
ProofLetGbeafiniteabeliangroupandpaprimethatdivides IGI.
BytheCorollaryto Lemma2.10.1,G =PXT,whereIPI=pn.ByTheorem
2.10.2,P =AlXA
2X...XA
k
,wherethe
Aiarecyclicsubgroups ofP.Ar­
guingbyinductiononIGI,wemaythusassumethat
T=T
I
XT
2
X...XT
q
,
wherethe
T
iarecyclicsubgroups ofT.Thus
G=(AIXA
2
X X A
k
)
X(T
1
XT
2
X X T
q
)
=AlXA
2
X X A
k
XT
I
XT
2
X X T
q
•
Thisveryimportant theoremisnowproved.D
WereturntoabeliangroupsG oforderpn.Wenowhaveathandthat
G
=AlXA2
X...XAk,wherethe
Aiarecyclicgroups oforderpn,.We
canarrangethenumberingsothatn
1
2::n22::...2::nk·Also,IGI=
IA
IXA
2X...XA
kI=IA
IIIA
2
/·•·IA
kI,whichgivesus that
hencen =ni+n2+...+nk.Thustheintegersni2::0giveusapartition of
n.Itcanbeshownthattheseintegersn 1,n2,...,nk-whicharecalledthe
invariants
ofG-areunique.In otherwords,twoabeliangroups oforderpn
areisomorphicif
andonlyiftheyhavethesameinvariants. Grantedthis,it
followsthatthe
numberofnonisomorphicabeliangroups oforderpnisequal
tothe
numberofpartitionsofn.
Forexample,if n=3,ithasthefollowingthreepartitions:3 =3,3=
2+1,3=1+1+1,sotherearethreenonisomorphicabeliangroups of
orderp3(independentofp).Thegroupscorrespondingtothesepartitions
areacyclicgroup
oforderp3,thedirectproductofacyclicgroupoforder p2
byoneoforderp,andthedirectproduct ofthreecyclicgroups oforderp,
respectively.

116. Sec.11 Conjugacyand Sylow'sTheorem(Optional) 101
Forn=4weseethepartitionsare4 =4,4=3+1,4=2+2,4=
2+1+1,4=1+1+1+1,whicharefiveinnumber.Thus therearefive
nonisomorphicgroups
oforderp
4
•
Canyoudescribe themviathepartitions
of4?
Givenanabeliangroup
ofordern=
p~lp~2...p%\wherethe Piare
distinctprimesand
theaiareallpositive, thenGisthedirectproduct ofits
so-called
Pi-Sylowsubgroups(see,e.g., theCorollarytoLemma2.10.1).For
eachprime Pithereareasmanygroups oforder
pfiastherearepartitionsof
ai.Sothenumberofnonisomorphicabeliangroups ofordern=p~l...p%k
isf(al)f(a2)..·f(ak),wherefern)denotesthenumberofpartitionsofm.
Thusweknowhowmanynonisomorphicfiniteabeliangroups therearefor
anygivenorder.
Forinstance,howmanynonisomorphicabeliangroups arethereof
order144?Since
144=2
4
3
2
,
andtherearefivepartitions of4,twopartitions
of2,thereare10nonisomorphicabeliangroups oforder144.
Thematerial
treatedinthissectionhas beenhard,thepathsomewhat
tortuous,
andtheefforttounderstandquiteintense. Tosparethereadertoo
muchfurtheragony,weassignonly threeproblemstothissection.
PROBLEMS
1.LetA beanormalsubgroup ofagroupG, andsupposethatbEGisan
element
ofprimeorderp,andthatb
f£A.ShowthatAn(b)=(e).
2.LetG beanabeliangroup oforderpn,paprime,andletaEGhavemax­
imalorder.Show
thatxo(a)=eforall xEG.
3.LetG
beafinitegroup,with N<JGandaEG.Provethat:
(a)TheorderofaNinGINdividestheorderofainG, thatis,
o(aN)lo(a).
(b)If(a)nN=(e),theno(aN)=o(a).
11.CONJUGACY ANDSYLOW'STHEOREM (OPTIONAL)
IndiscussingequivalencerelationsinSection4wementioned,asanexample
ofsucharelationinagroupG,
thenotionofconjugacy.Recallthattheele­
ment
binGissaidtobeconjugatetoaEG(ormerely,aconjugate ofa)if
thereexistsanx
EGsuchthatb=x-lax.WeshowedinSection4 thatthis
definesanequivalencerelation
onG.Theequivalenceclass ofa,whichwe
denoteby
cl(a),iscalledthe conjugacyclass ofa.

117. 102 Groups Ch.2
Forafinitegroupanimmediatequestionpresentsitself: Howlargeis
cl(a)?Ofcourse,this dependsstronglyontheelementa.Forinstance,if
aEZ(G),thecenterofG,thenax==xaforallxEG,hencex-lax=a;in
otherwords,theconjugacyclass ofainthis caseconsistsmerely oftheele­
mentaitself.Ontheotherhand,if cl(a)consistsonly oftheelementa,then
x-lax=aforallxEG.Thisgivesus thatxa=axforallxEG,hence
aEZ(G).SoZ(G)ischaracterizedas thesetofthoseelements ainG
whoseconjugacyclasshasonly
oneelement,aitself.
ForanabeliangroupG,sinceG
==Z(G),twoelements areconjugateif
andonlyif theyareequal.Soconjugacy isnotaninterestingrelationfor
abeliangroups;however,for
nonabeliangroupsit isahighlyinterestingno­
tion.
Given
aEG,cl(a)consistsofallx-laxasxrunsoverG.Sotodeter­
minewhich
arethedistinctconjugates ofa,weneedtoknowwhentwocon­
jugates
ofacoincide,which isthesameasasking:Whenisx-lax=y-Iay?In
thiscase,transposing,we
obtaina(xy-l)=(xy-l)a;inotherwords,xy-l
mustcommutewitha.Thisbringsus toaconceptintroducedasExample10
inSection3,
thatofthecentralizerofainG.We repeatsomethingwedid
there.
Definition.
IfaEG,thenC(a),thecentralizerofainG,isdefined
by
C(a)
=={xEGIxa==ax}.
WhenC(a)aroseinSection3we showedthatitwasasubgroup ofG.
Werecordthisnow moreofficiallyas
Lemma2.11.1.ForaEG,C(a)isasubgroupofG.
Aswesawabove, thetwoconjugates x-laxandy-1ayofaareequal
onlyif
xy-lEC(a),thatis,onlyif xandyareinthesamerightcoset ofC(a)
inG.Ontheotherhand,if xandyareinthesamerightcoset ofC(a)inG,
thenxy-lEC(a),hencexy-la
==axy-l.Thisyields thatx-lax=y-1ay.Sox
andygiverise tothesameconjugateofaifandonlyif xandyareinthe
samerightcoset
ofC(a)inG.Thusthereareas manyconjugatesofainGas
therearerightcosets
ofC(a)inG.Thisis mostinterestingwhenG isafinite
group,forin
thatcasethenumberofrightcosets ofC(a)inGiswhatwe
calledtheindex,ic(C(a)),ofC(a)inG,andisequaltoIGIIIC(a)l.
Wehaveproved
Theorem2.11.2.LetGbeafinitegroupandaEG;thenthenumber
ofdistinctconjugates ofainGequals theindexofC(a)inG.

118. Sec.11 ConjugacyandSylow'sTheorem(Optional) 103
Inotherwords,thenumberofelementsin cl(a)equalsiG(C(a»=
IGI/IC(a)l·
Thistheorem,althoughitwasrelativelyeasytoprove, isveryimpor-
tantandhasmanyconsequences.Weshallseeafewofthesehere.
Onesuchconsequence
isakindofbookkeepingresult.Sinceconjugacy
isanequivalencerelationonG,G istheunionofthedisjointconjugacy
classes.Moreover,byTheorem2.11.2,weknowhowmanyelementsthere
areineachclass.Puttingallthisinformationtogether,weget
Theorem2.11.3(TheClassEquation).IfG
isafinitegroup,then
IGI=
~ic(C(a»=~ I~~)I'
wherethesumrunsoverone afromeachconjugacyclass.
It
isalmostasacredtraditionamongmathematicianstogive,asthefirst
applicationoftheclassequation,aparticulartheoremaboutgroupsoforder
pn,wherepisaprime.Notwantingtobeaccusedofheresy,wefollowthis
traditionandprovethepretty andimportant
Theorem2.11.4.IfG
isagroupof orderpn,wherepisaprime,then
Z(G),thecenterofG, isnottrivial(i.e.,thereexistsanelement a
=1=einG
suchthat
ax=xaforallxEG).
ProofWeshallexploittheclassequationtocarryouttheproof. Let
z=IZ(G)I;aswepointedoutpreviously, zisthenthenumber ofelements
inGwhoseconjugacyclasshasonlyoneelement.Since
eEZ(G),z
~1.
Foranyelement boutsideZ(G),itsconjugacyclasscontainsmorethanone
elementand
IC(b)1<IGI.Moreover,since IC(b)1dividesIGIbyLagrange's
theorem,
IC(b)1=pn(b),where1
::::;neb)<n.Wedividethepiecesofthe
classequationintotwoparts:thatcomingfromthecenter,andtherest.We
get,thisway,
pn=IGI=Z+
L---lQL=Z+L ~;h)=Z+L pn~n(h).
b~Z(G)IC(b)1 n(b)<nP n(b)<n
Clearly,pdividestheleft-handside, pn,anddividesLn(b)<npn-n(b).Thenet
resultofthis
isthatpIz,andsincez
~1,wehavethatzisatleastp.Sosince
z=IZ(G)I,theremustbeanelement a=1=einZ(G),whichprovesthetheo­
rem.D
Thislasttheoremhasaninterestingapplication,whichsomereaders
mayhaveseeninsolvingProblem
45ofSection5.Thisis

119. 104 Groups Ch.2
Theorem2.11.5.IfG isagroupoforderp2,wherep isaprime,then
G
isabelian.
ProofByTheorem2.11.4,Z(G)
=1=(e),so thatthereisanelement,a,
oforderpinZ(G).IfA=(a),thesubgroupgeneratedby a,thenACZ(G),
henceACC(x)forallxEG.GivenxEG,x$.A,thenC(x):JAand
xEC(x);soIC(x)1>p,yetIC(x)1mustdivide p2.Thenetresultofthisis
thatIC(x)1=p2,soC(x)=G,whencexEZ(G}.Sinceeveryelement ofG
isinthecenter ofG,Gmustbeabelian. D
Intheproblemstocomeweshallgivemanyapplications ofthenature
ofgroupsoforderpn,wherep isaprime.Thel1aturalattackonvirtuallyall
theseproblemsfollowsthelines
ofthe
argumelltweareabouttogive.We
chooseone
ofawidepossibleset ofchoicestoillustratethistechnique.
Theorem2.11.6.IfG isagroupoforderpn,paprime,thenGcon­
tainsanormalsubgroup
oforderpn-l.
ProofWeproceedbyinduction onn.Ifn =1,thenG isoforderp
and
(e)istherequirednormalsubgroup oforderpl-l=po=1.
Supposethatweknow thatforsome keverygroup oforderpkhasa
normalsubgroup
oforderpk-l.LetGbeoforderpk+l;byTheorem2.11.4
thereexistsanelement
aoforderpinZ(G),thecenterofG.Thusthesub­
group
A=(a)generatedby aisoforderpandisnormalinG.Consider r=
G/A;risagroupoforderIGI/IAI=pk+l/p =pkbyTheorem2.6.3.Sincer
hasorder
pk,weknow thatrhasanormalsubgroup Moforderpk-l.Since
risahomomorphicimage ofG,bythe Correspondence Theorem(Theorem
2.7.2)there
isanormalsubgroupNinG,N
:JA,suchthatN/A=M.But
then
wehave
pk-l=IMI=IN/AI=
l~i'
thatis,pk-l=INI/p,leadingusto INI=pk.ThusN isourrequirednormal
subgroupinG
oforderpk.Thiscompletestheinductionandsoprovesthe
theorem.
D
Byfarthemostimportantapplicationwemake oftheclassequation is
theproofofafar-reachingtheoremduetoSylow,aNorwegianmathemati­
cian,whoproveditin1871.Wealreadyshowedthistheoremtobetruefor
abeliangroups.Weshallnowproveitforanyfinitegroup.
Itisimpossibleto
overstatetheimportance
ofSylow'sTheoreminthestudyoffinitegroups.
Withoutitthesubjectwould
notgetofftheground.

120. Sec.11 Conjugacyand Sylow'sTheorem(Optional) 105
Theorem2.11.7(Sylow'sTheorem). SupposethatGisagroupof
orderpnm,wherepisaprimeandptm.ThenGhasasubgroup oforderpn.
ProofIfn=0,thisistrivial.Wethereforeassume thatn2::1.Here,
again,weproceedbyinductionon
IGI,assumingtheresultto betrueforall
groups
HsuchthatIHI<IGI.
SupposethattheresultisfalseforG.Then,by ourinductionhypothe­
sis,
pncannotdivide IHIforanysubgroup HofGifH
=1=G.Inparticular,if
a$.Z(G),thenC(a)=1=G,hencepntIC(a)l.ThuspdividesIGI/IC(a)1=
ic(C(a»fora$.Z(G).
WritedowntheclassequationforGfollowingthelines oftheargu­
mentinTheorem2.11.4.
Ifz=IZ(G)I,thenz
2::1and
ButPIic(C(a»ifa$.Z(G),sopILaEtZ(C)ic(C(a».SincepIpnm,weget
pIz.ByCauchy'sTheoremthereisanelementaoforderpinZ(G).IfAis
thesubgroupgeneratedby a,thenIAI=pandA<lG,sinceaEZ(G).Con­
sider
f=GIA;If1=IGI/IAI=pnmlp=pn-lm.SinceIf1<IGI,byourin­
ductionhypothesis
rhasasubgroup Moforderpn-l.However,bytheCor­
respondenceTheoremthere
isasubgroupPofGsuchthatP
:JAand
PIA=M.Therefore,Ipi=IMIIAI=pn-lp=pnandPisthesought-after
subgroupofGoforder
pn,contradictingourassumptionthatGhadnosuch
subgroup.Thiscompletestheinduction,andSylow'sTheorem
isestablished.D
Actually,Sylow's Theoremconsistsofthreeparts, ofwhichweonly
provedthefirst.The
othertwoare(assuming pnm=IGI,wherep
tm):
1.Anytwosubgroupsof orderpninGareconjugate; thatis,ifIpi
IQI=pnforsubgroupsP,QofG,thenforsome xEG,Q=x-1px.
2.Thenumber ofsubgroupsoforderpninGisoftheform1 +kpanddi­
vides
IGI.
Sincethesesubgroups oforderpnpopupallovertheplace,theyare
called
p-Sylowsubgroups ofG.Anabeliangrouphasone p-Sylowsubgroup
foreveryprimepdividingitsorder.This
isfarfromtrueinthegeneralcase.
Forinstance,ifG =S3'thesymmetricgroup ofdegree3,whichhas order6
=2 .3,therearethree2-Sylowsubgroups(of order2)andone3-Sylowsub­
group(or
order3).
Forthosewhowanttoseeseveralproofsof thatpartofSylow'sTheo­
remwhichweprovedabove,and
oftheothertwoparts,theymightlook at
theappropriatesection ofourbookTopicsinAlgebra.

121. 106 Groups
PROBLEMS
Ch.2
EasierProblems
1.InS3'thesymmetricgroup ofdegree3,findall theconjugacyclasses,and
check
thevalidityoftheclassequationbydeterminingtheorders ofthe
centralizers
oftheelementsofS3 .
2.DoProblem1forG thedihedralgroup oforder8.
3.IfaEG,show thatC(x-1ax)=x-1C(a)x.
4.If
cpisanautomorphismofG,showthatC(cp(a»=cp(C(a»foraEG.
5.
IfIGI=p3andIZ(G)I
;:::p2,provethatGisabelian.
6.
IfPisap-Sylowsubgroup ofGandP<lG,provethatPistheonly
p-Sylowsubgroup
ofG.
7.IfP<lG,Pap-SylowsubgroupofG,provethat
cp(P)=Pforevery
automorphismcpofG.
8.Usetheclassequationtogivea proofofCauchy'sTheorem.
IfHisasubgroupofG,letN(H)={xEG Ix-1Hx=H}.Thisdoes
notmean
thatxa=axwheneverxEN(H),aEH.Forinstance,if
H<lG,thenN(H)=G,yetHneednotbeinthecenterofG.
9.Prove
thatN(H)isasubgroupofG,HCN(H)andinfactH<lN(H).
10.Prove thatN(x-1Hx)=x-1N(H)x.
11.IfPisap-Sylowsubgroup ofG,provethatPisap-Sylowsubgroup of
N(P)andistheonlyp-Sylowsubgroup ofN(P).
12.IfPisap-Sylowsubgroup andaEGisoforderpmforsome m,show
thatifa-1pa=PthenaEP.
13.Prove thatifGisafinitegroup andHisasubgroupofG,thenthenum­
berofdistinctsubgroups x-1HxofGequalsiG(N(H».
14.IfPisap-Sylowsubgroup ofG,showthatthenumberofdistinctX-IPx
cannot
beamultipleofp.
15.IfN<lG,letB(N)={xEGIxa=axforallaEN}.Provethat
B(N)<lG.
Middle-Level
Problems
16.Show thatagroupoforder36hasa normalsubgroupoforder3or9.
(Hint:See Problem40ofSection5.)
17.Show
thatagroupoforder108hasa normalsubgroupoforder9or27.
18.
IfPisap-SylowsubgroupofG,showthatN(N(P»=N(P).
19.IfIGI=pn,showthatGhasasubgroup oforderpmfor
alII::::;m::::;n.

122. Sec.11 Conjugacyand Sylow'sTheorem(Optional) 107
20.IfpmdividesIGI,showthatGhasasubgroupoforderpm.
21.IfIGI=pnandH=1=GisasubgroupofG,showthatN(H)~H.
22.Show thatanysubgroupoforderpn-linagroupGoforderpnisnormal
inG.
HarderProblems
23.LetGbeagroup,HasubgroupofG.Definefora,bEG,a---bifb=
h-1ahforsomehEH.Provethat
(a)thisdefinesanequivalencerelationonG.
(b)If[a]istheequivalenceclassofa,showthatifGisafinite group,
then[a]hasmelementswheremistheindexofHnC(a)inH.
24.IfGisa group,HasubgroupofG,definea relationB---Aforsub­
groupsA,BofGbytheconditionthatB=h-1AhforsomehEH.
(a)Provethatthisdefines anequivalencerelationonthesetofsub­
groupsofG.
(b)IfGisfinite,showthatthenumberofdistinctsubgroupsequivalent
toAequalstheindexofN(A)nHinH.
25.IfPisap-SylowsubgroupofG,letSbethesetofallp-Sylowsubgroups
ofG.ForQbQ2ESdefineQl---Q2ifQ2=a-1QlawithaEP.Prove,
usingthis relation,thatifQ=1=P,thenthenumberofdistincta-1Qa,with
aEP,isamultipleofp.
26.Using theresultofProblem25,showthatthenumberofp-Sylowsub­
groupsofGisoftheform1+kp.(Thisis thethirdpartofSylow'sTheo­
rem.)
27.LetPbeap-SylowsubgroupofG,andQanotherone.Supposethat
Q=1=x-1pxforanyxEG.LetSbethesetofally-lQy,asyrunsoverG.
ForQbQ2ESdefineQl---Q2ifQ2 =a-1Qla,whereaEP.
(a)Showthatthisimpliesthatthenumberofdistincty-lQyisamultiple
ofp.
(b)UsingtheresultofProblem14,showthattheresultofPart(a)can­
nothold.
(c)Provefromthisthatgivenanytwop-SylowsubgroupsPandQofG,
thenQ=x-1pxforsomexEG.
(Thisis
thesecondpartofSylow'sTheorem.)
28.IfHisasubgroupofGoforderpmshowthatHiscontainedinsome
p-SylowsubgroupofG.
29.
IfPisap-SylowsubgroupofGanda,b
EZ(P)areconjugateinG,
provethattheyarealreadyconjugateinN(P).

123. 3
THESYMMETRICGROUP
1.PRELIMINARIES
Letusrecalla theoremprovedin Chapter2forabstractgroups.Thisresult,
knownas
Cayley'sTheorem (Theorem2.5.1),asserts thatany groupG isiso­
morphictoasubgroup
ofA(S),thesetof1-1mappingsofthesetSontoit­
self,forsomesuitable
S.Infact,in theproofwegaveweusedforS the
groupGitselfviewedmerelyasaset.
Historically,groupsarosethiswayfirst,longbefore
thenotionofan
abstractgroupwasdefined.
Wefindin theworkofLagrange,Abel,Galois,
andothers,results ongroupsofpermutationsprovedinthelateeighteenth
andearlynineteenthcenturies.Yetitwasnotuntilthemid-nineteenthcen­
tury
thatCayleymoreorlessintroducedtheabstractconcept ofagroup.
Since
thestructureofisomorphicgroups isthesame,Cayley's Theorem
pointsoutacertainuniversal characterforthegroupsA(S).Ifweknew the
structureofallsubgroupsofA(S)foranyset S,wewouldknow thestructure
ofallgroups.This ismuchtoomuchtoexpect.Nevertheless, onecouldtryto
exploitthis
embeddingofanarbitrarygroupGisomorphicallyintosome
A(S).Thishas theadvantageoftransformingGasanabstractsysteminto
something
moreconcrete,namelyaset ofnicemappings ofsomesetonto
itself.
Weshallnotbeconcernedwith thesubgroupsofA(S)foranarbitrary
set
S.IfSisinfinite,A(S)isaverywild andcomplicatedobject.Evenif Sis
finite,thecompletenatureofA(S)isvirtuallyimpossibletodetermine.
108

124. Sec.1 Preliminaries 109
Inthischapterweconsideronly A(S)forSafiniteset.Recallthatif S
hasnelements,then wecallA(S)thesymmetricgroup ofdegreen, andde­
noteitbySn-Theelementsof Snarecalledpermutations;weshalldenote
tI:tembylowercaseGreekletters.
Since
wemultipliedtwoelements
u,rEA(S)bytherule (ur)(s)=
u(res»~thiswillhavetheeffectthatwhen weintroducetheappropri­
atesymbolstorepresenttheelementsof
Sn,thesesymbols, orpermutations,
willmultiplyfrom
righttoleft. Ifthereaderslookatsomeotherbookonal­
gebra,theyshouldmakesurewhichwaythepermutationsarebeingmulti­
plied:righttoleft
orlefttoright.Veryoften,algebraistsmultiplypermuta­
tionsfrom
lefttoright.Tobeconsistentwith ourdefinitionofthe
compositionofelementsin
Sn,wedoitfromrighttoleft.
ByCayley'sTheoremweknowthatifG
isafinitegroupoforder n,
thenG isisomorphictoasubgroupof SnandSnhasn!elements.Speaking
loosely,
weusuallysaythatG isasubgroupof Sn'Sincenissomuchsmaller
than
n!fornevenmodestlylarge,ourgroupoccupiesonlyatinylittlecorner
in
Sn.ItwouldbedesirabletoembedGinan Snfornassmallaspossible.For
certainclassesoffinitegroupsthis
isachievableinaparticularlyniceway.
Let
Sbeafinitesethaving nelements;wemight aswellsupposethat
S=
{x},X2,...,xn}.GiventhepermutationuESn=A(S),then
U(Xk)ESfork=1,2,...,n,soU(Xk)=Xi
k
forsome ik,1~ik
~n.Because
Uis1-1,ifj=1=k,thenXi)=u(Xj)=1=U(Xk)=Xi
k
•
Therefore,thenumbers
iI,i2,...,inaremerelythenumbers1,2, ...,nshuffledaboutinsomeorder.
Clearly,theactionof
UonSisdeterminedbywhatUdoestothesub­
script
jofXj'sothesymbol "x"isreallyexcessbaggageand,assuch,canbe
discarded.Inshort,wemayassumethat
S={I,2,...,n}.
Let'srecallwhat ismeantbythe productoftwoelementsof A(S).If
u,rEA(S),thenwedefinedurby(ur)(s)=u(r(s»foreverysES.We
showed
inSection4ofChapter1that A(S)satisfiedfourpropertiesthatwe
usedlater
asthemodeltodefinethenotionofanabstractgroup.Thus Sn,in
particular,
isagrouprelativetotheproductofmappings.
Ourfirstneed issomehandywayofdenotingapermutation,that is,an
element
uinSn.Oneclearway istomakeatableofwhatudoestoeachele­
mentof
S.Thismightbecalledthe graphof
u.Wedidthisearlier,writing
outu,sayuES3'inthefashion:u:Xl~X2,X2~X3'X3~Xl'Butthisis
cumbersomeandspaceconsuming.Wecertainlycanmakeitmorecompact
bydroppingthe
x'sandwriting
u=(~; ~).Inthissymbolthenumberin
thesecondrow
istheimageunder
(J'ofthenumberinthefirstrowdirectly
aboveit.There
isnothingholyabout3inallthis;itworksequallywellfor
any
n.

125. 110 TheSymmetricGroup Ch.3
If0'ESnand0'(1)=i},0'(2)=i
2
,
•••,u(n)=in,weusethesymbol
(
1~...~)torepresent0'andwewrite0'=(~ ~ ~)oNote
itl2•••In Itl2 In
thatitisnotnecessarytowritethefirstrowintheusual order1 2...n;
anywaywewritethefirstrow,aslongaswecarrythe i/salongaccord­
ingly,westillhaveu.Forinstance,intheexamplein S3cited,
U=(~ ~ ~)=(~ ~ ~)=(~ ~ ~).
Ifweknow0'=(~~ ~),whatisu-
1
?Itiseasy,just flipthe
Itl2•••In
symbolforuovertogetu-
1
=(i~ ~).(Prove!)Inourexample
u=G~ ~}u-
1
=(i~ ~)=Gi~).Theidentity element-
whichweshallwrite ase-ismerelye=(~;::::).
Howdoestheproductin Sntranslateinterms ofthesesymbols?Since
UTmeans:"FirstapplyTandtotheresult ofthisapply0',"informingthe
product
ofthesymbolsfor
0'andTwelookatthenumber kinthefirstrow
ofTandseewhat numberi
kisdirectlybelow kinthesecondrowofT.We
thenlook
atthespoti
kinthefirstrow of
0'andseewhat isdirectlybelowit
inthesecondrow
of
u.ThisistheimageofkunderUTeWethenrunthrough
k=1,2,...,nandgetthesymbolforUTeWejustdothisvisually.
Weillustratethiswithtwopermutations
u=(~ ~i~ ~)andT=(~ ~ ;1;)
inS5'ThenUT=G; ~ ~;).
Eventheeconomyachievedthisway isnotenough.Afterall,thefirst
row
isalways1 2 ...n,sowecoulddispensewithit,andwrite
u=
(
1 2
...n)("0 0)Tho"fib0h "
o
0 "asll,l2,•••,In'ISISne,utInt enextsectIonwe
Itl2•••In
shallfinda betterandbrieferway ofrepresentingpermutations.
PROBLEMS
1.Findtheproducts:
(
1 2 3 4 5
63)(212 3 4 5 6)
(a)6 4 5 2 1 3 4 5 6 1"

126. Sec.2 CycleDecomposition 111
(b)G
234
5)C2
34
;).
1 3 4 5 3 2 1 4
(c)(~
2 3 45r
1
C2 3 45)C2 3 45)
13252134541325·
2.Evaluateallthepowersofeachpermutationa(i.e.,finda
k
forallk).
(a)G
2 3 4 5
~}345 6
(b)G
2 3 4 567)
1 3 4657·
(c)(~
23 45
~}4 5 2 1
3.Prove
that
(~
2
Zr
1
=(1
i
2
~n).
IIi
2 2
4.FindtheorderofeachelementinProblem 2.
5.FindtheorderoftheproductsyouobtainedinProblem 1.
2.CYCLEDECOMPOSITION
Wecontinuetheprocessofsimplifyingthenotationusedtorepresentagiven
permutation.Indoingso,wegetsomethingmorethanjustanewsymbol;we
getadevicetodecomposeanypermutationasaproduct
ofparticularlynice
permutations.
Definition.Let
i
bi
2
,
•••,i
kbekdistinctintegersinS ={I,2,...,n}.
Thesymbol (i
li
2
•••i
k
)willrepresentthepermutation
aESn,where
a(i
l
)=i
2
,a(i
2
)=i
3
,
•••,a(i
j
)=i
j+
lforj<k,
u(i
k
)=i
band
a(s)=sforanysESifsisdifferentfrom i
l
,i
2
,
•••,i
k
•
Thus,inS 7,thepermutation (13 54) isthepermutation
(
1 2 3 4 5 6
7)
3 2 5 1 4 6 7·Wecallapermutationoftheform (ili2•••ik)
ak-cycle.Forthespecialcase k=2,thepermutation(i
li
2
)iscalleda trans­
position.
Notethatif
a=(i
li
2
•••i
k
),thenaisalso(i
k
i
li
2
•••i
k
-
l
),
(i
k
-
1i
ki
1i
2
•••i
k
-
2
),andsoon.(Prove!) Forexample,
(13 54) =(41 35) =(54 13) =(35 41).
Twocycles,saya
k-cycleandan m-cycle,aresaidtobe disjointcycles if

127. 112 TheSymmetricGroup Ch.3
theyhavenointegerincommon.Whence (135)and(42 67)in S7
aredisjointcycles.
Giventwodisjointcyclesin
Sn,weclaimthattheycommute.Weleave
theproof
ofthistothereader,withthesuggestion thatif
u,Taredisjoint
cycles,thereadershouldverifythat(uT)(i)=(Tu)(i)forevery iES=
{I,2,...,n}.Westatethisresultas
Lemma3.2.1. Ifu,TESnaredisjointcycles,thenUT=TU.
Let'sconsideraparticular k-cycleU=(12...k)inSn.Clearly,
u(l)=2bythedefinitiongivenabove;how is3relatedto I?Sinceu(2)=3,
wehaveal(l)=u(2)=3.Continuing,weseethatu
j
(l)=j+1for
j~k-1,whileUk(1)=1.Infact,wesee thatUk=e,whereeistheidentity
elementin
Sn.
Therearetwothingstobeconcludedfromtheparagraphabove.
1.Theorderofak-cycle,asanelementof Sn,isk.(Prove!)
2.If
U=(i
li
2
•••ik)isak-cycle,thenthe orbitofi
lunderU(see
Problem
27inSection4 ofChapter1) is{il, i2
,
•••,ik}.Sowecansee
thatthe
k-cycle
U=(i1i
2
•••ik)is
U=(ilu(il)u
2
(i
l
)...Uk-l(i
l ».
GivenanypermutationTin Sn,foriE{I,2,...,n},considerthe orbit
ofiunderT.Wehavethatthisorbit is{i,T(i),T
2
(i),...,T
s
-l
(i)},
where
TS(i)=iandsisthesmallestpositiveintegerwiththisproperty.Considerthe
s-cycle
(iT(i)T
2
(i)...TS-l(i»;wecallitthe cycleofTdeterminedby i.
Wetakeaspecificexampleandfindallitscycles.Let
(
1 2 3 4 5 6 7 8 9)
T=394156278;
whatisthecycleofTdeterminedby I?Weclaimthatit is(134).Why?
Ttakes1into3,3into 4and4into1,andsinceT(l)=3,T
2
(1)=T(3)=4,
T
3
(1)=T(4)=1.Wecangetthisvisuallybyweavingthrough
(
~5 6 7 89)
39415627 8
withthethinpath.What isthecycle ofTdeterminedby2?Weaving
through
(
12~)
3941 56278

128. Sec.2 CycleDecomposition 113
withthethinpath,weseethatthecycleofTdeterminedby2 is
(29 87).ThecyclesofTdeterminedby5and6are(5)and(6),respec­
tively,since5and6areleftfixedby
T.SOthecyclesof Tare(134),
(29 87),(5),and(6).Therefore wehavethat
T=(134)(29 87)
(5)(6),where
weviewthese cycles-asdefinedabove-aspermutationsin Sg
becauseeveryintegerin S={I,2,...,9}appearsinoneandonlyonecycle,
andtheimageofany
iunder
Tisreadofffromthecycleinwhichitappears.
There
isnothingspecialaboutthepermutation
Tabovethatmadethe
argument
wegavegothrough.Thesameargumentwouldholdfor anyper­
mutationin
Snforanyn.Weleavetheformalwritingdownoftheproofto
thereader.
Theorem3.2.2.Everypermutation inSnistheproductofdisjoint
cycles.
Inwritingapermutation
uasaproductofdisjointcycles,weomitall
I-cycles;that
is,weignorethei'ssuch that
u(i)=i.Thuswewrite
u=(123)(45)(6)(7)simplyasu=(123)(45).In otherwords,writing
uasaproductofk-cycles,with k>1,weassumethatuleavesfixedanyin­
tegernotpresentinanyofthecycles.Thusinthegroup
S11thepermuta­
tion
T=(156)(23 9 87)leavesfixed 4,10,and11.
Lemma3.2.3.IfTinSnisak-cycle,thentheorderofTisk;thatis,
T
k=eandT
j
=I=-efor0<j<k.
ConsiderthepermutationT=(12)(34 56)(78 9)inSg.
Whatisitsorder?Sincethedisjointcycles (12),(34 56), (789)
commute,
T
m
=(12)m(34 56)m(78
9)m;inorderthat T
m
=ewe
need(12)m=e,(34 56)m=e,(789)m=e.(Prove!)Tohave
(789)m=e,wemusthave3 Im,since(789)isoforder3;tohave
(34 56)m=e,wemusthave4 1m,because(34 56) isoforder4,
andtohave (12)m=e,wemusthave 21m,because(12)isoforder2.
Thistellsusthatmmustbedivisibleby 12.
Ontheotherhand,
SoTisoforder12.
Here,again,thespecialpropertiesofTdonotenterthepicture.What
wedidforTworksforanypermutation.Toformulatethisproperly,recall
thattheleastcommonmultipleof
mandnisthesmallestpositiveinteger v
whichisdivisibleby mandby n.(SeeProblem 7,Chapter1,Section5.)Then
wehave

129. 114 TheSymmetricGroup Ch.3
Theorem3.2.4.LetUESnhaveitscycledecompositioninto disjoint
cyclesoflengthmbm2,.•.,mk'ThentheorderofUistheleastcommon
multipleof
ml,m2,•••,mk'
ProofLet
U=TlT2.••Tk,wherethe Tiaredisjointcyclesoflength mi'
Sincethe Tiaredisjointcycles, TiTj=TjTi;thereforeif Mistheleastcommon
multipleof
ml,m2,""mk'then
u
M
=(TIT2.•.Tk)M=T~T~...Tk
M
=e
(sinceTt;t=ebecauseTiisofordermiandmiIM).Therefore,theorderof
uisatmostM.Ontheotherhand,ifUN=e,thenTfT!j•••TZ=e.This
forceseachTf=e,(prove!)because Tiaredisjointpermutations,so m
iIN,
sinceTiisofordermi'ThusNisdivisiblebytheleastcommonmultipleof
mbm2'...,mk'soMIN.Consequently,weseethatuisoforderMas
claimedinthetheorem. D
Notethatthedisjointnessofthecycles inthetheorem isimperative.
Forinstance,(12)and(13),whicharenotdisjoint,areeachoforder 2,
buttheirproduct (12)(13) =(132)isoforder3.
Let'sconsiderTheorem3.2.4inthecontextofacardshuffle.Suppose
that
weshuffleadeck of13cardsinsuchawaythatthetopcard isputinto
thepositionofthe3rdcard,thesecondinthatofthe4th,
...,theithintothe
i+2position,workingmod 13.Asapermutation,
u,of1,2,...,13,the
shufflebecomes
(
1 2 3 4 5 6 7 8 9
10111213)
u=3 45 6 7 8 9 1011121312'
anduismerelythe13-cycle (135 7 9 11132 46 81012),
souisoforder13.Howmanytimesmustwerepeatthisshuffletogetthe
cardsbacktotheiroriginalorder?Theanswer
ismerelytheorderof
u,that
is,13.Soittakes 13repeatsoftheshuffletogetthecardsbacktotheirorigi­
nalorder.
Let'sgiveatwisttotheshuffleabove.Supposethatweshufflethe
cardsasfollows.Firsttakethetopcardandputitintothesecond-to-Iast
place,andthenfollowitbytheshufflegivenabove.Howmanyrepeatsare
nowneededtogetthecardsbacktotheiroriginalorder?Thefirstoperation
istheshufflegivenbythepermutation
T=(11211109 8 7 6 5
432)followedbyuabove.SowemustcomputeUTandfinditsorder.
But
UT=(13 5 7 9 11132 4 6 8 1012)
X(11211109 8 7 6 5 4 3 2)
=(1)(23 4 5 6 7 8 9 10111213),

130. Sec.2 CycleDecomposition 115
soisoforder12.Soitwould take12repeatsoftheshuffletogetbacktothe
originalorder.
Canyoufindashuffle ofthe13cardsthatwouldrequire42repeats?Or
20repeats?Whatshufflewouldrequirethegreatestnumberofrepeats,and
whatwouldthis numberbe?
Wereturntothegeneraldiscussion.Considerthepermutation
(123);wesee that(123) =(13)(12). Wecanalsosee that
(123) =(23)(13).Sotwothings areevident.First,we canwrite
(123)as theproductoftwotranspositions,andinatleasttwodistinct
ways.Given
thek-cycle(i
1i2...ik),then(i
1i2...ik)
(i
1ik)(i1ik-1 )···(i
1i2),soeveryk-cycleisa productofk- 1transposi­
tions(if
k>1)andthiscan bedoneinseveralways, sonotinauniqueway.
Because
everypermutationistheproductofdisjointcycles andeverycycleis
a
productoftranspositionswe have
Theorem3.2.5.EverypermutationinSnistheproductoftransposi­
tions.
This
theoremisreallynotsurprisingforitsays, afterall,nothingmore
orlessthanthatanypermutationcanbeeffectedbycarryingoutaseriesof
interchangesoftwoobjectsatatime.
Wesawthatthereisalackofuniquenessinrepresentingagivenper­
mutationasaproductoftranspositions.But,asweshall seeinSection3,
someaspects
ofthisdecompositionareindeedunique.
Asafinalwordofthissection,we wouldliketopointouttheconve­
nience
ofcyclenotation.Whenwerepresentelementsofapermutation
groupasproductsofdisjointcycles, manythingsbecometransparent-for
example,theorderofthepermutationisvisibleataglance.Toillustratethis
point,we
nowgiveafewexamples ofcertaingeometricgroups,which arein
fact
permutationgroupsthathavealreadyappearedinChapter2underdif­
ferentguises.
Examples
1.Informally,a motionofageometricfigureisapermutationofitsvertices
thatcanberealizedbyarigidmotioninspace.Forexample,thereareeight
motionsofasquare,whosevertices arenumbered1,2,3,4asbelow:
4 3
2

131. 116 TheSymmetricGroup Ch.3
a=(13)isthereflectionabouttheaxisofsymmetryjoiningvertices2and
4intheoriginalposition,
~=(1234)isthecounterclockwiserotationby 90°,
~2=(13)(24)isthecounterclockwiserotationby 180°,
~3=(1432)isthecounterclockwiserotationby 270°,
a~=(12)(34)isthereflectionin theverticalaxis ofsymmetry,
a~2=(24)isthereflectionin theotherdiagonalaxis,
a~3=(14)(23)isthereflectionin thehorizontalaxis,and, ofcourse
a
2
=~4=(1)isthe"motion"thatleavestheverticesunchanged.
Wealsohave
therelation
~a=a~3.
Thesemotions, orsymmetriesofasquare,formasubgroup ofS4whichis
calledtheoetiegroup, orthedihedralgroup oforder8.Thisgroup(or,
strictlyspeaking,agroupisomorphic
toit)wasintroducedinExample9 of
Section2.1withoutmention ofpermutations.
2.Thereareonlyfoursymmetries ofanon-squarerectangle:
4
1
---u _i---_---1
3
2
thereflectionsin thetwoaxesofsymmetry,rotationby 180°andtheidentity.
Thesemotionscan
beidentifiedwithpermutations(1),(14)(23),(12)(34),
(13)(24),
andformasubgroup ofthegroupobtainedinExample 1.Thissub­
group
isoftencalled Klein's4-group.
3.Weleaveit tothereadertoverifythatthegroupofallmotionsofan
equilateraltriangle
isthefullsymmetricgroup S3.
3
2
4.Themotionsofaregularhexagonform thedihedralgroup oforder12,
generatedbythepermutations
a=(15)(24),corresponding toareflection

132. Sec.2 CycleDecomposition 117
aboutoneoftheaxesofsymmetry,andJ3=(123456),correspondingtothe
counterclockwiserotationby60°.
5 4
6 3
2
Ingeneral,thedihedralgroup oforder2n,whichwasfirst introducedinEx­
ample
10ofSection2.1,canbeinterpretedasthegroupofsymmetriesofa
regularn-gon
(apolygonwith nedgesofequallength).
PROBLEMS
EasierProblems
2 3 4 5 6 7)
3 15674'
3.Expressas theproductofdisjointcycles andfindtheorder.
(a)(12 3 5 7)(24 76).
(b)(12)(13)(14).
(c)(12 345)(12 3 46)(12 3 47).
(d)(123)(132).
(e)(123)(35 79)(123)-1.
(f)(12 3 45)3.
4.Giveacomplete proofofTheorem3.2.2.
5.Showthatak-cyclehasorderk.
6.Findashuffle ofadeckof13cardsthatrequires42repeatstoreturnthe
cardstotheiroriginalorder.
1.Showthatif
u,Taretwodisjointcycles, thenUT=TU.
2.Findthecycledecomposition andorder.
(
1 2 3 4 5 6 7 8 9)(a)3 1 4 2 7 6 9 8 5'
(
1 2 3 4 5 6
7)
(b)7 6 5 4 3 2 l'
(c)
G~;~ ~ ~ i)G

133. 118 TheSymmetricGroup Ch.3
7.DoProblem6forashufflerequiring 20repeats.
8.ExpressthepermutationsinProblem3astheproductoftranspositions.
9.Giventhetwotranspositions (12)and (13),findapermutation0­
suchthatu(12)u-
1
=(13).
10.Provethatthere
isnopermutation
Usuchthatu(12)u-
1
=(123).
11.Provethatthere
isapermutation
usuchthatu(123)u-
1
(456).
12.Provethatthere
isnopermutation
usuchthatu(123)u-
1
(124)(567).
Middle-LevelProblems
13.Provethat
(12)cannotbewritten astheproductofdisjoint 3­
cycles.
14.Provethatforanypermutation
u,UTU-
I
isatranspositionifTisatrans­
position.
15.ShowthatifTisak-cycle,thenUTU-
I
isalsoa k-cycle,foranypermuta­
tionu.
16.Let<I>beanautomorphismof S3'Showthatthere isanelementUES3
suchthat<1>(T)=U-ITUforeveryTES3.
17.Let (12)and(12 3...n)beinSn'Showthatanysubgroupof Sn
thatcontainsbothofthesemustbeallofSn(sothesetwopermutations
generate
Sn)'
18.If TIandT2aretwotranspositions,showthat TIT2canbeexpressedasthe
productof3-cycles(notnecessarilydisjoint).
19.Provethatif
TI'T2,andT3aretranspositions,then TIT2T3
=1=e,theidentity
elementof
Sn.
20.If
T},T2aredistincttranspositions,showthat TIT2isoforder2 or3.
21.Ifu,TaretwopermutationsthatdisturbnocommonelementandUT=e,
provethatU=T=e.
22.FindanalgorithmforfindingUTU-
I
foranypermutationsu,TofSn.
23.Letu,Tbetwopermutationssuchthattheybothhavedecompositions
intodisjointcyclesofcyclesoflengths
ml,m2,...,mk'(Wesaythat
theyhavesimilardecompositionsintodisjointcycles.)Provethatfor
somepermutation
p,
T=pUp-I.
24.Findtheconjugacyclassin Snof(12...n).Whatistheorderofthe
centralizerof
(12...n)inSn?
25.DoProblem24for
U=(12)(34).

134. Sec.3
3.ODDANDEVENPERMUTATIONS
OddandEvenPermutations 119
WenoticedinSection2thatalthougheverypermutation istheproductoftrans­
positions,thisdecomposition
isnotunique.Wedidcomment,however,thatcer­
tainaspectsofthiskindofdecompositionareunique.Wegointothisnow.
Let'sconsiderthespecialcase
ofS3'forherewecanseeeverythingex­
plicitly.
Let
f(XbX2'X3)=(Xl-X2)(XI-X3)(X2-X3)beanexpressionin
thethreevariablesXbX2,X3.WeletS3actonf(x)=f(XbX2,X3)asfollows.
If0'ES3'then
Weconsiderwhat0'*doesto f(x)forafew oftheo"SinS3.
Consider0'=(12).Then0'(1)=2,0'(2)=1,and0'(3)=3,sothat
O'*(f(x»=(Xu(l)-Xu(2»(Xu(l)-Xu(3»(Xu(2)-Xu(3»
=(x2- Xl)(X2-X3) (Xl-X3)
-
(Xl-X2) (Xl-X3) (X2-X3)
-
f(x).
So
u*comingfrom0'=(12)changesthesign off(x).Let'slookattheac­
tion
ofanotherelement,
,.,=(123),ofS3onf(x).Then
T*(f(x»=(XT(I)-XT (2»(XT(I)-XT(3»(X7{2)-X7{3»
=(x2-X3 )(X2-XI )(X3-Xl)
=(Xl-X2) (Xl-X3)(X2-X3)
=f(x),
soT*comingfrom,., =(123)leavesf(x)unchanged.Whataboutthe
otherpermutationsinS3;howdothey affectf(x)?Ofcourse,theidentityel­
ement
einducesamap e*onf(x)whichdoesnotchange f(x)atall.What
does
,.,2,for,.,above,doto f(x)?Since,.,*f(x)=f(x),weimmediatelysee that
(,.,2)*(f(x»=(XT2(1)-XT 2(2»(XT2(1)- XT2(3»(XT2(2)-XT 2(3»
=f(x).(Prove!)
Nowconsider
U'T=(12)(123) =(23);since,.,leavesf(x)aloneand0'
changesthe signoff(x),U7'mustchangethesign off(x).Similarly,(13)changes
the
signoff(x).Wehaveaccountedfortheactionofeveryelementof S3onf(x).

135. 120 TheSymmetricGroup Ch.3
Supposethat pES3isaproductp=T(T2···Tkoftranspositions
Tb...,Tk;thenpactingon f(x)willchangethesignof f(x)ktimes,since
each
Tichangesthesign off(x).Sop*(f(x»=(-1)kf(x).Ifp=
UIU2···Ut,
whereUb...,U
taretranspositions,bythesamereasoning, p*(f(x»=
(-1)tf(x).Therefore,(-1)kf(x)=(-1)tf(x),whence(-1)t=(-1)k.This
tells
usthattandkhavethe sameparity; thatis,iftisodd,then kmustbe
odd,andif
tiseven,then kmustbeeven.
Thissuggeststhatalthoughthedecompositionofagivenpermutation
U
asaproductoftransposition isnotunique,theparityofthenumberoftrans­
positionsinsuchadecomposition
of
Umightbeunique.
Westriveforthisgoalnow,suggestingtoreadersthattheycarryout
theargumentthatwedoforarbitrary
nforthespecialcase n=4.
Aswedidabove,define f(x)=
f(x},...,x
n
)tobe
whereinthisproduct
itakesonallvaluesfrom1to n-1inclusive,and jall
thosefrom2to
ninclusive.If
UESn,defineu*onf(x)by
u*(f(x»=f1(xu(i)-xU(j»·
1<]
Ifu,TESn'then
=u*(T*(11(Xi-Xj»))=U*(T*(f(X)))=(U*T*)(f(X»
SO(UT)*=U*T*whenappliedto f(x).
WhatdoesatranspositionTdotof(x)?Weclaimthat T*(f(x»=
-f(x).Toprovethis,assumingthatT=(ij)wherei<j,wecountupthe
numberof
(xu-xv),withu<v,whichgettransformedintoan (x
a
-Xb)with
a>b.Thishappensfor (xu-Xj)ifi<u<j,for(Xi-Xv)ifi<v<j,andfi­
nally,for (Xi-Xj).Eachoftheseleadstoachangeofsignon f(x)andsince
thereare
2(j-i-1)+1such,that is,anoddnumberofthem, wegetan
oddnumberofchangesofsignon
f(x)whenactedonby T*.Thus
T*(f(x»=-f(x).Therefore,ourclaimthat T*(f(x»=-f(x)forevery
transposition
Tissubstantiated.

136. Sec.3 OddandEvenPermutations 121
If0'isanypermutationinSnand0'=T1T2...TbwhereTbT2'...,Tk
aretranspositions,then0'*=(T1T2...Tk)*=TiT~...Tkasactingonf(x),
andsinceeachT~(f(x» =-f(x),weseethatO'*(f(x»=(-l)kf(x).Simi-
larly,if0'='1'2···'t,where'b'2'...''taretranspositions,then
O'*(f(x»=(-l)tf(x).ComparingthesetwoevaluationsofO'*(f(x»,we
conclude
that(-l)k=(-l)t.Sothesetwodecompositionsof
O'astheprod­
uctoftranspositionsareofthesameparity.Thusanypermutation iseither
theproduct
ofanoddnumberoftranspositionsortheproductofaneven
number
oftranspositions,andnoproduct ofanevennumberoftranspositions
canequalaproduct
ofanoddnumberoftranspositions.
Thissuggests thefollowing
Definition.Thepermutation
0'ESnisanoddpermutationif0'isthe
productofanoddnumberoftranspositions,andisanevenpermutation if0'
istheproductofanevennumberoftranspositions.
Whatwehaveprovedaboveis
Theorem3.3.1.A permutationinSniseitheranoddoranevenper­
mutation,butcannotbeboth.
WithTheorem3.3.1behinduswe candeduceanumberofitsconse­
quences.
LetAnbethesetofallevenpermutations;if
0',TEAn,thenweimme­
diatelyhave
that
O'TEAn.SinceAnisthusafiniteclosed subsetofthe(fi­
nite)
groupSn,AnisasubgroupofSn,byLemma2.3.2.Aniscalledthealter­
natinggroup
ofdegreen.
Wecanshow thatAnisasubgroupofSninanotherway.Wealready
sawthatAnisclosedundertheproductofSn,sotoknowthatAnisasub­
groupofSnwemerelyneedshowthat
0'ESnimpliesthat0'-1ESn.Forany
permutation0'weclaimthat0'and0'-1areofthesameparity.Why?Well,if
0'=Tt'T2·..Tk'wheretheTiaretranspositions,then
sinceTil=Ti.Therefore,weseethattheparityof0'and0'-1is(-l)k,so
theyareofequalparity.Thiscertainlyshows that0'EAnforces0"-1EAn,
whenceAnisasubgroupofSn.
Butitshowsalittlemore, namelythatAnisanormalsubgroup ofSn.
Forsupposethat0"EAnandpESn.Whatistheparityofp-
1
0"p?Bythe

137. 122 TheSymmetricGroup Ch.3
above,pandp-lareofthesameparityandITisanevenpermutationsop-1lTp
isanevenpermutation,hence isinAn.ThusAn isanormalsubgroupof Sn.
Wesummarizewhatwehavedonein
Theorem3.3.2.
An'thealternatinggroup ofdegreen,isanormal
subgroupof
Sn.
Welook atthisinyet anotherway.Fromtheverydefinitionsinvolved
wehavethefollowingsimplerulesfortheproduct
ofpermutations:
1.Theproduct oftwoevenpermutations iseven.
2.Theproduct oftwooddpermutationsiseven.
3.Theproduct ofanevenpermutationbyan oddone(orofanoddone
byanevenone)
isodd.
If
ITisanevenpermutation,let ()(IT)=1,andifITisanoddpermuta­
tion,let
()(
IT)=-1.Theforegoingrulesaboutproductstranslateinto
()(lTr)=()(IT)()(r),so()isahomomorphismofSnontothegroup E={I,-I}
oforder2undermultiplication.Whatisthekernel,N,of()?Bytheverydefi­
nition
ofAnweseethatN=An.SobytheFirstHomomorphismTheorem,
E
~Sn/An.Thus2 =lEI=ISn/Anl=ISnl/IAnl,if n>1.Thisgivesusthat
IAnl=!ISnl=!n!.
Therefore,
Theorem3.3.3.Forn>1,Anisanormalsubgroup ofSnoforder!n!.
Corollary.Forn>1,Sncontains!n!evenpermutationsand!n!odd
permutations.
Afinalfewwordsaboutthe
proofofTheorem3.3.1beforeweclose
thissection.Manydifferentproofs
ofTheorem3.3.1areknown.Quite
frankly,wedo
notparticularlylikeany ofthem.Someinvolvewhatmightbe
calleda"collectionprocess,"whereonetries
toshowthatecannotbewrit­
tenastheproduct ofanoddnumberoftranspositionsbyassumingthatit is
suchashortestproduct,andbytheappropriatefinaglingwiththisproduct,
shorteningittogetacontradiction.
Otherproofsuse otherdevices.The
proofwegaveexploitsthegimmickofthefunction [(x),which,insome
sense,
isextraneoustothewholeaffair.However,the proofgivenisprobably
themosttransparent
ofthemall,whichiswhyweusedit.
Finally,thegroup
An'forn
2::5,isanextremelyinterestinggroup.We

138. Sec.3 OddandEvenPermutations 123
shallshowin Chapter6thattheonlynormalsubgroupsofAn'forn2::5,are
(e)andAnitself.A groupwiththis propertyiscalleda simplegroup(notto
beconfusedwith aneasygroup).Theabelianfinitesimple groupsaremerely
thegroupsofprimeorder.TheAnforn2::5provideuswithaninfinitefam­
ily
ofnonabelianfinitesimplegroups. Thereareotherinfinitefamilies offinite
simplegroups.
Inthelast20years orsotheheroicefforts ofalgebraistshave
determinedallfinitesimplegroups. Thedeterminationofthesesimplegroups
runs
about10,000printedpages.Interestinglyenough, anynonabelianfinite
simple
groupmusthaveevenorder.
PROBLEMS
EasierProblems
1.Findtheparityofeachpermutation.
(a)(12 3 4 5 6 7 8 9)
2451 3 7 896·
(b)(12 3 4 5 6)(789).
(c)(12 3 4 5 6)(12 3 4 5 7).
(d)(12)(123)(45)(568)(179).
2.
If
(]"isak-cycle,showthat(]"isanoddpermutationifkiseven,andisan
evenpermutationifkisodd.
3.Provethat(]"andT-
1
(]"T,foranyu,TESn,areofthesameparity.
4.
Ifm<n,wecanconsiderSmCSnbyviewing
uESnasactingon
1,2,...,m,...,nasitdidon1,2,...,manduleavesj>mfixed.
ProvethattheparityofapermutationinSm,whenviewedthiswayas an
elementofSn,doesnotchange.
5.
Supposeyouaretoldthatthepermutation
(
1 2 3 4 5 6 7 8 9)312 7896
inSg,wheretheimagesof5and4havebeenlost,is anevenpermuta­
tion.Whatmusttheimagesof5and4be?
Middle-LevelProblems
6.Ifn
2::3,showthateveryelementinAnisaproductof3-cycles.
7.Show
thateveryelementinAnisaproductofn-cycles.
8.FindanormalsubgroupinA
4oforder4.

139. 124 TheSymmetricGroup Ch.3
HarderProblems(Infact,veryhard)
9.Ifn2::5and(e)=1=NCAnisanormalsubgroupof An'showthat Nmust
containa3-cycle.
10.Usingtheresult ofProblem9,showthatifn
2::5,theonlynormalsub­
groups
ofAnare(e)andAnitself.(Thus thegroupsAnforn
2::5giveus
aninfinitefamily
ofnonabelianfinitesimplegroups.)

140. 4
RINGTHEORY
1.DEFINITIONS ANDEXAMPLES
Sofarinourstudyofabstractalgebra,wehavebeenintroducedtoonekind
ofabstractsystem,whichplaysacentralroleinthealgebraoftoday.That
wasthenotionofagroup.Becauseagroup
isanalgebraicsystemwithonly
oneoperation,andbecauseagroupneednotsatisfytherule
ab=ba,itran
somewhatcounterto
ourpriorexperienceinalgebra.Wewereusedto sys­
temswhereyoucouldbothaddandmultiplyelementsandwheretheele­
mentsdidsatisfythecommutativelawofmultiplication
ab=ba.Further­
more,thesesystemsofouracquaintanceusuallycamefromsetsof
numbers-integers,rational,real,andforsome,complex.
Thenextalgebraicobjectweshallconsider
isaring.Inmanyways
thissystemwillbemorereminiscent
ofwhatwe hadpreviouslyknown
thanweregroups.
Foronethingringswillbeendowedwithaddition and
multiplication,andthesewillbesubjected tomanyofthefamiliarrules
weallknowfromarithmetic.
Ontheotherhand,rings neednotcome
from
ourusuall1umbersystems,and,infact,usuallyhavelittle todowith
thesefamiliarones.Althoughmany
oftheformalrules ofarithmetic
hold,many
strange-orwhatmayseemas strange-phenomenadotake
place.Asweproceedandseeexamples ofrings,weshallseesome of
thesethingsoccur.
Withthispreambleoverwearereadytobegin.Naturallyenough,the
firstthing
weshoulddo istodefinethatwhichwe'llbetalkingabout.
125

141. 126 RingTheory Ch.4
Definition.A nonemptyset Rissaidtobea ringifinRtherearetwo
operations
+and·suchthat:
(a)
a,bERimpliesthata+bER.
(b)a+b=b+afora,bER.
(c)(a+b)+c=a+(b+c)fora,b,c ER.
(d)Thereexistsanelement 0ERsuchthata+0=aforeveryaER.
(e)Given aER,thereexistsabERsuchthat a+b=O.(Weshallwrite
bas-a.)
Notethatsofarallwehavesaid isthatRisanabeliangroupunder +.We
nowspell
outtherulesforthemultiplicationin R.
(f)a,bERimpliesthata.bER.
(g)a·(b.c)=(a·b).cfora,b,c ER.
Thisisallthatweinsistonasfarasthemultiplicationbyitself isconcerned.
Butthe+and·arenotallowedtoliveinsolitarysplendor.Weinterweave
thembythetwo
distributivelaws
(h)a·(b+c)=a·b+a·cand
(b+c).a=b.a+c .a,fora,b,c ER.
Theseaxiomsforaringlookfamiliar.Theyshouldbe,fortheconcept
ofringwasintroducedasageneralizationofwhathappensintheintegers.
Because
ofAxiom(g),theassociativelaw ofmultiplication,theringswede­
finedareusuallycalled
associativerings. Nonassociativeringsdoexist,and
someoftheseplayanimportantroleinmathematics.
Buttheyshallnotbe
ourconcernhere.Sowheneverweusetheword"ring"weshallalwaysmean
"associativering."
AlthoughAxioms
(a)to(h)arefamiliar,therearecertainthingsthey
donotsay.Welook atsomeofthefamiliarrules thatarenotinsistedupon
forageneralring.
First,we
donotpostulatetheexistenceofanelement1 E Rsuchthat
a.1=1 .a=aforeveryaER.Manyoftheexamplesweshallencounter
willhavesuchanelement,andin
thatcasewesay thatRisaringwithunit.
Inallfairnessweshouldpoint outthatmanyalgebraistsdodemandthata
ringhaveaunitelement.Wedoinsistthat1
=1=0;thatis,theringconsisting
of0aloneisnotaringwithunit.

142. Sec.1 DefinitionsandExamples 127
Second,in ourpreviousexperiencewiththings ofthissort,whenever
a.b=0weconcluded thata=0orb=O.Thisneednotbetrue,ingeneral,
inaring.Whenitdoeshold,thering
iskindofniceand isgivenaspecial
name;it
iscalleda domain.
Third,nothing issaidintheaxiomsforaringthatwillimplythe com­
mutativelaw
ofmultiplicationa.b =b.a.Therearenoncommutativerings
wherethislawdoesnothold;weshallseesomesoon.
Ourmainconcernin
thischapterwill
bewithcommutativerings, butformany oftheearlyresults
thecommutativity
oftheringstudiedwill notbeassumed.
Aswementionedabove,somethingsmakecertainringsnicerthanoth­
ers,andsobecomeworthy
ofhavingaspecialname.Wequicklygivealist of
definitionsforsome ofthesenicerrings.
Definition.A commutativering Risanintegraldomain ifa.b=0in
Rimpliesthat a=0orb=O.
Itshouldbepointedoutthatsomealgebrabooksinsist thatanintegral
domaincontainaunitelement.Inreading
anotherbook,thereadershould
checkifthis
isthecasethere. Theintegers,7L,giveusanobviousexampleof
anintegraldomain.Weshallseeother,somewhatlessobviousones.
Definition.A ringRwithunit issaidtobea divisionring ifforevery
a
*0inRthereisanelementbER(usuallywrittenas a-I)suchthat
a.a-I=a-I.a=1.
Thereasonforcallingsucharingadivisionringisquiteclear,forwe
candivide(atleastkeepingleftandrightsidesinmind).Althoughnoncom­
mutativedivisionringsexistwithfairfrequency
anddoplayanimportant
roleinnoncommutativealgebra,theyarefairlycomplicatedandweshall
giveonlyoneexampleofthese.Thisdivisionring
isthegreatclassiconein­
troducedbyHamiltonin1843and
isknownasthering ofquaternions.(See
Example
13below.)
Finally,wecometoperhapsthenicestexampleofaclassofrings,the
field.
Definition.A ringRissaidtobeafieldifRisacommutativedivision
ring.
Inotherwords,afield isacommutativeringinwhichwecandivide
freelybynonzeroelements.Otherwiseput,
Risafieldifthenonzeroele­
ments
ofRformanabeliangroup under·,theproductin R.

143. 128 RingTheory Ch.4
Forfieldswedohavesomereadyexamples:therationalnumbers,the
realnumbers,thecomplexnumbers.
Butweshallseemanymore,perhaps
lessfamiliar,examples.Chapter5willbedevotedtothestudyoffields.
Wespendtherestofthetimeinthissectionlookingatsomeexamples
ofrings.Weshalldrop
the·fortheproductandshallwritea·bsimply asabo
Examples
1.Itisobviouswhichring weshouldpick asourfirstexample,namely 7L,the
ringofintegersundertheusualadditionandmultiplicationofintegers.Natu­
rallyenough,
7Lisanexampleofanintegraldomain.
2.Thesecondexample isequallyobviousasachoice.Let
Qbethesetofall
rationalnumbers.Asweallknow,Qsatisfiesalltherulesneededforafield,
soQisafield.
3.Therealnumbers,IR,alsogiveusanexampleofafield.
4.Thecomplexnumbers,C,formafield.
NotethatQeIReC;wedescribethisbysayingthatQisasubfieldof
IR(andofC)andIRisasubfieldofC.
5.LetR=7L
6
,theintegersmod 6,withtheadditionandthemultiplication
definedby
[a]+[b]=[a+b]and[a][b]=[ab].
Notethat [0]isthe0requiredbyouraxiomsforaring,and [1]istheunit
elementofR.Note,however,that
7L
6isnotanintegraldomain,for
[2][3]=[6]=[0],yet[2]
=1=[0]and[3]=1=[0].Risacommutativeringwithunit.
Thisexamplesuggeststhe
Definition.
Anelementa
=1=0inaringR isazero-divisorinRifab=0
forsome
b
=1=0inR.
Weshouldreallycallwhatwedefinedaleftzero-divisor;however,
since
weshallmainlytalkaboutcommutativerings,weshallnotneedany
left-rightdistinctionforzero-divisors.
Notethatboth
[2]and[3]in7L
6arezero-divisors.Anintegraldomain
is,ofcourse,acommutativeringwithoutzero-divisors.
6.LetR=7Ls,theringofintegersmod 5.Ris,ofcourse,acommutativering
withunit.Butit
ismore;infact,it isafield.Itsnonzeroelementsare [1], [2],
[3],[4]andwenotethat [2][3]=[6]=[1],and[1]and[4]aretheirownin­
verses.Soeverynonzeroelementin
7Lshasaninversein 7Ls.
Wegeneralizethistoanyprime p.

144. Sec.1 DefinitionsandExamples 129
7.Let7L
pbetheintegersmod p,wherepisaprime.Again 7L
pisclearlya
commutativeringwith
1.Weclaimthat 7L
pisafield.Toseethis,notethatif
[a]
=1=[0],thenp(a.Therefore,byFermat'sTheorem(CorollarytoTheorem
2.4.8),aP-
1
==l(p).Fortheclasses[.]thissaysthat [aP-1
]
=[1].But[aP-1
]
=[alP-I,so[a]p-l=[1];therefore,[a]p-2istherequiredinversefor [a]in7Lp,
hence7Lpisafield.
Because
7L
phasonlyafinitenumberofelements,it iscalleda finite
field.
Laterweshallconstructfinitefieldsdifferentfromthe 7L
p's.
8.Let
Qbetherationalnumbers;if aEQ,wecanwrite a=mIn,wherem
andnarerelativelyprimeintegers.Callthisthe reducedform fora.LetRbe
thesetofall
aE
Qinwhosereducedformthedenominator isodd.Under
theusualadditionandmultiplicationinQthesetRformsaring. Itisaninte­
graldomainwithunitbut
isnotafield,for
~,theneededinverseof 2,isnot
inR.ExactlywhichelementsinRdohavetheirinversesinR?
9.LetRbethesetofall aE
Qinwhosereducedformthedenominator is
notdivisiblebyafixedprime p.Asin(8),R isaringundertheusualaddi­
tionandmultiplicationinQ,isanintegraldomain butisnotafield.What
elementsofRhavetheirinversesinR?
BothExamples8and9aresubrings
of
Qinthefollowingsense.
Definition.IfRisaring,thena subringofRisasubsetS ofRwhich
isaringiftheoperations abanda+barejusttheoperations ofRappliedto
theelements
a,bES.
ForStobeasubring,it isnecessaryandsufficientthatSbenonempty
andthat
ab,a±b ESforall a,bES.(Prove!)
Wegiveonefurthercommutativeexample.Thisonecomesfromthe
calculus.
10.LetRbethesetofallreal-valuedcontinuousfunctionsontheclosedunit
interval
[0,1].Forf,gERandxE[0,1]define(f+g)(x)=f(x)+g(x),
and(f·g)(x)=f(x)g(x). Fromtheresultsinthecalculus, f+gandf·gare
againcontinuousfunctionson
[0,1].WiththeseoperationsR isacommuta­
tivering.It
isnotanintegraldomain. Forinstance,if f(x)=-x+
~for
o::::;x::::;~andf(x)=0for~<x::::;1,andifg(x)=0for0::::;x::::;~and
g(x)=2x-1for~<x::::;1,thenf,gERand,as iseasytoverifY,f' g=O.
Itdoeshaveaunitelement,namelythefunction edefinedby e(x)=1forall
xE[0,1].WhatelementsofRhavetheirinversesinR?

145. 130 RingTheory Ch.4
Weshouldnowliketoseesomenoncommutativeexamples.Theseare
notsoeasytocomeby,althoughnoncommutativeringsexistinabundance,
becausewe
arenotassuminganyknowledge oflinearalgebra onthereader's
part.
Theeasiestandmostnaturalfirstsource ofsuchexamples isthesetof
matricesoverafield.So,in
ourfirstnoncommutativeexample,weshall
reallycreate
the2X2matriceswithrealentries.
11.
LetFbethefieldofrealnumbersandletRbethesetofallformalsquare
arrays
where
a,b,c,dareanyrealnumbers. Forsuchsquarearrayswedefineaddi­
tionina
naturalwaybydefining
ItiseasytoseethatRformsanabeliangroupunderthis+with
(~ ~)act-
(
-a-b) (ab)
ingasthezeroelementand_C _dthenegativeofcd.Tomakeof
Raring,we needamultiplication.Wedefineoneinwhatmayseemahighly
unnaturalwayvia
(
a
b)(rs)=(ar+btas+bU).
cd t U cr+dtcs+du
Itmaybealittlelaborious, butonecancheckthatwiththeseoperations Risa
noncommutativeringwith
(~ ~)actingasitsmultiplicativeunitelement.
Notethat
(~ ~)(~ ~)=(~ ~)
while
so

146. Sec.1 DefinitionsandExamples 131
Notethat(~ ~)and(~ ~)arezero-divisors;infact,
(~ ~r=(~ ~)
so
isanonzeroelementwhosesquare isthe0element ofR.ThisRisknownas
the
ringofall2X2matricesover F,therealfield.
Forthoseunfamiliarwiththesematrices,andwhoseenosenseinthe
productdefinedforthem, let'slookathowwedocomputetheproduct.
To
getthetopleftentryintheproduct AB,we"multiply"thefirstrowof Aby
thefirstcolumn
ofB,whereA,BER.Forthetoprightentry,it isthefirst
rowof
Aversusthesecondcolumn ofB.Thebottomleftentrycomesfrom
thesecondrow
ofAversusthefirstcolumnof B,andfinally,thebottom
rightentry
isthesecondcolumnof Aversusthesecondcolumnof B.
Weillustratewithanexample:Let
A= (1
~)
-32
and
Thenthefirstrowof
Ais1,
~andthefirstcolumnof Bis~,71";we"multiply"
thesevia1 .~+~.71"=71"/2+~,andsoon.Soweseethat
AB= (~+71"/2
-1+271"
~-71"/2)
-~-271".
Intheproblemsweshallhavemanymatrixmultiplications,sothatthe
readercanacquiresomefamiliaritywiththisstrangebutimportantexample.
12.LetRbeanyringandlet
s=
{(:~) Ia,b,C,dER}
with+and·asdefinedinExample 11.Onecanverifythat Sisaring,also,
undertheseoperations.It
iscalledthe ringof2X2matricesover R.
Ourfinalexample isoneofthegreatclassicalexamples,the realqua­
ternions,
introducedbyHamilton(asanoncommutativeparalleltothecom­
plexnumbers).
13.Thequaternions.LetFbethefieldofrealnumbersandconsidertheset
ofallformalsymbols
ao+ali+a2j+a3k,whereao,aba2'a3EF.
Equalityandadditionofthesesymbolsareeasy,viatheobviousroute

147. 132 RingTheory Ch.4
ao+ali+a2j+a3k=f30+f31i+f32j+f33k
ifandonlyifao=f3o,a1=f3l'a2=f32anda3=f33 ,and
(ao+ali+a2j+a3k)+(f30+f31i+f32j+f33k)
=(ao+f30)+(a1+f31)i+(a2+f32)j+(a3+f33)k.
Wenowcome tothetrickypart,themultiplication.WhenHamiltondiscov­
ereditonOctober6,1843,hecutthebasicrulesofthisproduct outwithhis
penknifeonBroughamBridgeinDublin.
Theproductisbasedon i
2=j2=
k
2=-1,ij=k,jk=i,ki=jandji= -k,kj= -i,ik= -j.Ifwegoaround
thecircleclockwise
theproduct
ofanytwosuccessiveones isthenextone,andgoingaround
counterclockwisewegetthenegatives.
Wecanwrite
outtheproductnow ofanytwoquaternions,accordingto
therulesabove,
declaringbydefinitionthat
(ao+ali+a2j+a3k)(f30+f31i+f32j+f33k)
=1'0+1'1i+1'2j+1'3k,
where
1'1=aof31+a1f30+a2f33-a3f32
1'2=aof32-a1f33+a2f30+a3f31
1'3=aof33+a1f32-a2f31+a3f30
(I)
Itlookshorrendous, doesn'tit?Butit'snotas badasallthat.Weare
multiplying
outformallyusingthedistributivelaws andusingtheproduct
rulesforthe
i,j,kabove.
Ifsome
aiis0inx=ao+ali+a2j+a3k,weshallomititinexpress­
ingx;thus0+Oi+OJ+Okwillbewrittensimplyas 0,1+Oi+OJ+Okas1,
o+3i+4j+Okas3i+4j,andsoon.
Acalculationrevealsthat
(II)
=
a6+ai+a~+a~.

148. Sec.1 DefinitionsandExamples 133
Thishasaveryimportantconsequence;forsuppose thatx=ao+ali+
a2j+a3k =1=0(sosomeai=1=0).Then,since thea'sarereal,f3=a~+ai+
a~+a~ =1=o.Thenfrom(II)weeasilyget
(
. .
k)(ao
al·a2·a3
k
)1
ao+all+a2]+a3Ii-lil-Ii]-Ii=.
So,ifx=1=0,thenxhasaninversein thequaternions.Thusthequaternions
formanoncommutativedivisionring.
Although,aswementionedearlier, thereisnolackofnoncommutative
divisionrings,
thequaternionsabove (orsomepiece ofthem)areoftenthe
onlynoncommutativedivisionrings
thatevenmanyprofessionalmathemati­
cianshaveeVerseen.
Weshallhave
manyproblems-someeasyandsomequiteabit
harder-aboutthetwoexamples: the2X2matricesandthequaternions.
Thisway thereaderwillbeabletoacquiresomeskillwithplayingwith
noncommutativerings.
Onefinalcommentinthissection:If
1'0,1'1,1'2,1'3areasin(I), then
(a~+ai+a~+a~)(f3~+f3i+f3~+f3~)
= I'~+I'i+ I'~+I'~.
(III)
This
isknownas Lagrange'sIdentity; itexpressestheproductoftwosums of
foursquaresagainasasum offoursquares.Itsverificationwill beoneofthe
exercises.
PROBLEMS
EasierProblems
*1.Findalltheelementsin
E
24thatareinvertible(i.e.,haveamultiplicative
inverse)inE
24
•
2.Showthatanyfieldisanintegraldomain.
3.Show
that
Enisafieldif andonlyifnisaprime.
4.VerifythatExample8 isaring.Findallitsinvertibleelements.
5.
DoProblem4forExample 9.
6.InExample11, the2X2matricesover thereals,check theassociative
law
ofmultiplication.

149. 134 RingTheory Ch.4
7.Workoutthefollowing:
(
12)(1~)
(a)4-701·
(
1
1)2
(b)1 1 .
(
1
1)3
(c)
gg.
(d)(:~)(~ ~)-(~ ~)(: ~).
8.Findallmatrices(:~)suchthat(:~)(~ ~)=(~ ~)(: ~)-
9.Findall2 X2matrices(:~)thatcommutewithall2 X2matrices.
10.
LetRbeanyringwithunit,Sthering of2X2matricesover R.(SeeEx­
ample12.)
(a)Checktheassociativelaw ofmultiplicationin S.(Remember:Rneed
notbecommutative.)
(b)Show
that{(
~:)Ia,b,C,ER}isasubringofS.
(c)Show that(~:)hasaninversein Sifandonlyif aandChave
inversesin
R.Inthatcasewritedown
(~ ~r
l
explicitly.
11.Let
F:
C~Cbedefinedby F(a+bi)=a-bi.Showthat:
(a)F(xy)=F(x)F(y)forx,yEC.
(b)F(xx)=Ix1
2
•
(c)UsingParts(a) and(b),show that
(a
2
+b
2
)(e
2 +d
2
)=(ae-bd)2 +(ad+be)2.
[Note:F(x)
ismerely
x.]
12.Verifytheidentityin Part(e)ofProblem11directly.
13.Find
thefollowingproducts ofquaternions.
(a)(i+j)(i-j).
(b)(1-i+2j-2k)(1 +2i-4j+6k).
(c)(2i-3j +4k)2.
(d)
i(ao+ali+a2j+a3k)-(aD+ali+a2j+a3k)i.

150. Sec.1 DefinitionsandExamples 135
14.Show thattheonlyquaternionscommutingwith iareoftheforma+f3i.
15.Findthequaternionsthatcommutewithbothiandj.
16.Verifythat
(ao+ali+a2j+a3k)(aO-ali-a2j-a3k)
==a~+ai+a~+a~.
17.VerifyLagrange'sIdentitybyadirectcalculation.
Middle-LevelProblems
18.Inthequaternions,define
Show
thatIxyI==IxIIyIforanytwoquaternions xandy.
19.Show thatthereisaninfinitenumberofsolutionstox
2
==-1inthe
quaternions.
20.Inthequaternions,consider thefollowingsetGhavingeightelements:
G
-{+l+.+.+k}
- -,-l,-j,- .
(a)ProvethatGisagroup(undermultiplication).
(b)Listallsubgroups ofG.
(c)WhatisthecenterofG?
(d)ShowthatGisanonabeliangroupall ofwhosesubgroupsare
normal.
21.Showthatadivisionring isadomain.
22.Giveanexample,in thequaternions,ofanoncommutativedomain that
isnotadivisionring.
23.Define
themap*inthequaternionsby
Showthat:
(a)x**==(x*)*==x.
(b)(x+y)*==x*+y*.
(c)xx*==x*xisrealandnonnegative.
(d)(xy)*==y*x*.
[NotethereversaloforderinPart(d).]
24.Using*,defineIxl==
Vxx*.ShowthatIxyl==Ixllylforanytwoqua­
ternions
xandy,byusingParts(c) and(d)ofProblem23.

151. 136 RingTheory Ch.4
25.UsetheresultofProblem24toproveLagrange'sIdentity.
InProblems26to30,letRbethe2X2matricesoverthereals.
26.
If(
~ ~)ER,showthat(~ ~)isinvertibleinR ifandonlyifad-be*-O.
Inthatcasefind(~ ~)-1.
27.Definedet(:~)=ad-be.Forx,yERshowthatdet(xy)
(detx)(dety).
28.Show that{xERIdetx=1=o}formsagroup,G, undermatrixmultiplica­
tion
andthatN={xERIdetx=I}isanormalsubgroupofG.
29.
IfxERisazero-divisor,show thatdetx=0,and,conversely,if x
=1=°is
suchthatdetx=0,thenxisazero-divisorin R.
30.InR,showthat{ ( _~:)Ia,breal}isafield.
HarderProblems
31.LetRbetheringofall2 X2matricesover7L
p
,paprime.Show thatif
det(:~)=ad-be*-0,then(~ ~)isinvertibleinR.
32.
LetRbeasinProblem31.Show thatforx,yER,det(xy)=det(x)det(y).
33.LetGbethesetofelementsxintheringRofProblem31suchthat
det(x)
=1=0.
(a)ProvethatGisagroup.
(b)FindtheorderofG.(Quitehard)
(c)FindthecenterofG.
(d)Findap-SylowsubgroupofG.
34.
LetTbethegroupofmatricesAwithentriesinthefield 7L
2suchthatdet A
isnotequalto0.ProvethatTisisomorphictoS3'thesymmetricgroup
ofdegree3.
35.ForRasinExample10,show thatS={fERIfisdifferentiableon(0,I)}
isasubringofRwhichisnotanintegraldomain.
IfFisafield,let H(F)betheringofquaternionsover F,thatis,the
set
ofall
aa+ali+a2j+a3k,whereaa,aba2,a3EFandwhereequal­
ity,addition,andmultiplication
aredefinedasfor therealquaternions.

152. Sec.2 Some SimpleResults 137
36.IfF=C,thecomplexnumbers,showthatH(C)isnotadivisionring.
37.
In
H(C),findanelementx=1=0suchthatx
2
=O.
38.ShowthatH(F)isadivisionringif andonlyifa6+ai+a~+a~=0
foraba2,a3'a4inFforcesao=al=a2=a3=O.
39.IfQisthefieldofrationalnumbers,showthatH(Q)isadivisionring.
40.
Provethatafinitedomainisadivisionring.
41.
UseProblem40toshowthat7L
p
isafieldif pisaprime.
2.SOMESIMPLERESULTS
Nowthatwehaveseensomeexamplesofringsandhavehadsomeexperi­
enceplayingaroundwiththem,itwouldseemwisetodevelopsomecompu­
tationalrules.Thesewillallowus toavoidannoyingtrivialitiesthatcould
besetacalculationwemightbemaking.
Theresultsweshallproveinthissectionarenotverysurprising,not
toointeresting,andcertainlynotatallexciting.Neitherwaslearningtheal­
phabet,butitwassomethingwehadtodobeforegoingontobiggerandbet­
terthings.Thesameholdsfortheresultsweareabouttoprove.
Sincearing Risatleastanabeliangroupunder+,thereare
certainthingsweknowfromourgrouptheorybackground,forinstance,
-(-a)=a,-(a+b)=(-a)+
(-b);ifa+b=a+c,thenb=c,and
soon.
Webeginwith
Lemma4.2.1.LetRbeanyringandleta,bER.Then
(a)aO=Oa=o.
(b)a(-b)=(-a)b=-(ab).
(c)(-a)(-b)=abo
(d)If1ER,then(-l)a=-a.
ProofWedotheseinturn.
(a)Since0 =0+0,aO=a(O+0)=aO+aO,henceaO=O.Wehave
usedtheleftdistributivelawinthis proof.Therightdistributivelawgives
Oa=O.
(b)ab+a(-b)=a(b+(-b»=aO=0fromPart(a).Therefore,
a(-b)=-(ab).Similarly,(-a)b=-(ab).
(c)ByPart(b),(-a)(-b)=-«-a)b)=-(-(ab»=ab,sinceweare
inanabeliangroup.

153. 138 RingTheory Ch.4
(d)If1ER,then(-l)a+a==(-l)a+(l)a==(-1+l)a==Oa==O.So
(-l)a==-abythedefinitionof-a.D
Anothercomputationalresult.
Lemma4.2.2.InanyringR,(a+b)2==a
2+b
2+ab+bafora,bER.
ProofThisisclearlytheanalogof
(a+f3)2==Q?+2af3+f32inthein­
tegers,say,
butkeepinginmindthatRmaybenoncommutative.So,toit.By
therightdistributivelaw (a+b)2==(a+b)(a+b)==(a+b)a+(a+b)b==
a
2+ba+ab+b
2
,exactlywhatwasclaimed.D
Canyouseethenoncommutativeversionofthebinomialtheorem?Try
itfor(a+b)3.
Onecuriosityfollows fromthetwodistributivelaws whenRhasa unit
element.Thecommutativelawofadditionfollowsfromtherest.
Lemma4.2.3.IfRisasystemwith1satisfyingall theaxiomsofa
ring,
exceptpossiblya+b==b+afora,bER,thenRisaring.
ProofWemustshowthata+b==b+afora,bER.Bytherightdis­
tributivelaw
(a+b)(l+1)==(a+b)l+(a+b)l==a+b+a+b.Onthe
otherhand,bytheleftdistributivelaw (a+b)(l+1)==a(l+1)+b(l+1)
==a+a+b+b.Butthena+b+a+b==a+a+b+b;sincewe areina
groupunder+,wecancancelaontheleftandbontherighttoobtainb+a
==a+b,asrequired.Risthereforearing.D
Weclosethis briefsectionwitharesultthatisalittlenicer. Wesaythat
aringRisaBooleanring [aftertheEnglishmathematicianGeorgeBoole
(1815-1864)]if x
2==xforeveryxER.
Weproveaniceresult onBooleanrings.
Lemma4.2.4.A Booleanringiscommutative.
ProofLetx,yER,aBooleanring.Thusx
2==x,y2==y,(x+y)2==
X+y.But(x+y)2==x
2+xy+yx+y2==X+xy+yx+y,byLemma4.2.2,
so
wehave(x+y)==(x+y)2==X+xy+yx+y,fromwhichwe have
xy+yx==O.Thus0==x(xy+yx)==x
2
y+xyx==xy+xyx,while0 ==
(xy+yx)x==xyx+yx
2
==xyx+yx.Thisgivesus xy+xyx==xyx+yx,and
soxy==yx.Therefore,Riscommutative.D

154. Sec.3 Ideals,Homomorphisms,andQuotientRings
PROBLEMS
139
1.LetRbearing:since Risanabeliangroupunder+,nahasmeaningfor
usfor
nE7L,aER.Showthat(na)(mb)=(nm)(ab)ifn,mareintegers
anda,bER.
2.IfRisanintegraldomain andab=acfora
=1=0,b,cER,showthatb=c.
3.IfRisafiniteintegraldomain,show thatRisafield.
4.IfRisaringandeERissuchthate
2=e,showthat(xe- exe)2=
(ex- exe)2=0forevery xER.
5.LetRbearinginwhich x
3
=xforeveryxER.ProvethatRiscommu­
tative.
6.
Ifa
2
=0inR,showthatax+xacommuteswitha.
7.LetRbearinginwhich x
4
=xfor everyxER.ProvethatRiscommu­
tative.
8.IfFisafinitefield,showthat:
(a)Thereexistsaprime psuchthatpa=0forallaEF.
(b)IfFhasqelements,thenq=pnforsomeintegern.(Hint:Cauchy's
Theorem)
9.Letpbeanoddprimeandlet1+
~+...+1/(p-1)=alb,wherea,b
areintegers.Show thatpIa.(Hint:AsarunsthroughUp,sodoesa-I.)
10.Ifpisaprimeandp>3,showthatif1+~+...+1/(p- 1)=alb,where
a,bareintegers,thenp2Ia.(Hint:Consider1Ia
2
asarunsthrough Up.)
3.IDEALS,HOMOMORPHISMS, ANDQUOTIENTRINGS
Instudyinggroups,it turnedoutthathomomorphisms,andtheirkernels­
thenormalsubgroups-playedacentralrole.Thereisnoreasontoexpect
thatthesamethingshould notbetrueforrings.Asamatteroffact,the
analogs,in thesettingofrings,ofhomomorphismandnormalsubgroupdo
playakeyrole.
With
thebackgroundwehaveacquiredaboutsuchthingsingroupthe­
ory,
theparalleldevelopmentforringsshould beeasyandquick.Anditwill
be!Withoutany
furtherfusswe makethe
Definition.Themapping
'P:R~R'oftheringRintotheringR'isa
homomorphismif
(a)'P(a+b)='P(a)+'P(b)and
(b)'P(ab)='P(a)'P(b)foralla,bER.

155. 140 RingTheory Ch.4
Sincearinghastwo operations,itisonly naturalandjustthatwede­
mandthatboththeseoperationsbepreservedunderwhatwewouldcalla
ring
homomorphism.Furthermore,Property(a)intheDefinitiontellsusthatq;isahomomorphismofRviewedmerelyasanabeliangroupunder+into
R'(alsoviewedasa groupunderitsaddition).Sowe cancallon,andexpect,
certainresultsfromthisfactalone.
Justaswesawin Chapter2,Section5forgroups, theimageofRunder
ahomomorphismfromRtoR',isasubringofR',asdefinedin Chapter4,
Section1(Prove!).
Letq;:R~R'bearinghomomorphismandletKerq;=
{xERIq;(x)=O},the0beingthatofR'.WhatpropertiesdoesKerq;enjoy?
Clearly,
fromgrouptheoryKer
q;isanadditivesubgroupofR.Butmuch
moreistrue.IfkEKerq;andrER,thenq;(k)=0,soq;(kr)=q;(k)q;(r)=
Oq;(r)=0,andsimilarly,q;(rk)=O.SoKerq;swallowsupmultiplication
fromtheleftandtherightby arbitraryringelements.
This
propertyofKer
q;isnowabstractedtodefinetheimportantanalog
inring
theoryofthenotionofnormalsubgroupingrouptheory.
Definition.
LetRbearing.A nonemptysubsetIofRiscalledan
idealofRif:
(a)IisanadditivesubgroupofR.
(b)GivenrER,aEI,thenraEIandarEI.
Weshallsoonseesomeexamplesofhomomorphismsandideals.But
firstwe notethatPart(b)inthedefinitionofidealreallyhasaleft andaright
part.Wecouldsplitit anddefinea setLofRtobealeftidealofRifLisan
additive
subgroupofRandgivenrER,aEL,thenraEL.Sowerequire
onlyleftswallowing-up
foraleftideal. Wecansimilarlydefine rightideals.
Anidealaswedefinedit isbothaleftandarightideal ofR.Byallrightswe
shouldthencallwhatwecalledanideala two-sidedidealofR.Indeed,in
workingin
noncommutativeringtheoryoneusesthisname; here,by"ideal"
weshall
alwaysmeanatwo-sidedideal.Exceptforsomeoftheproblems,we
shall
notusethenotionofone-sidedidealsinthis chapter.
Beforegoingon,we recordwhatwasdoneaboveforKer
q;as
Lemma4.3.1.Ifq;:R~R'isahomomorphism,thenKerq;isan
ideal
ofR.
Weshallsoonseethateveryidealcanbemadethekernelofahomo­
morphism.
Shadesofwhathappensfornormalsubgroupsofgroups!

156. Sec.3 Ideals,Homomorphisms,andQuotientRings 141
Finally,let Kbeanidealof R.SinceKisanadditivesubgroupof R,the
quotientgroup
R/Kexists;it ismerelythesetofallcosets a+Kasaruns
over
R.ButRisnotjustagroup;it is,afterall,aring.Nor isKmerelyanad­
ditivesubgroupof
R;itismorethanthat,namelyanidealof R.Weshould
beabletoputallthistogethertomakeof
R/Karing.
Howshouldwedefineaproductin
R/Kinanaturalway?Whatdowe
wanttodeclare
(a+K)(b+K)tobe?Thereasonablething istodefine
(a+K)(b+K)=ab+K,whichwedo.Asalways,thefirstthingthat
comesup
istoshowthatthisproduct iswell-defined.Isit?Wemustshow
that
ifa+K=a'+Kandb+K=b'+K,then(a+K)(b+K)=ab+K=
a'b'+K=(a'+K)(b'+K).However,if a+K=a'+K,thena-a'EK,
so(a-a')bEK,sinceKisanidealof R(infact,sofar,since Kisaright
idealof
R).Becauseb+K=b'+K,wehaveb-b'EK,soa'(b-b')EK,
sinceKisanidealof R(infact,since Kisaleftidealof R).Soboth(a-a')b=
ab-a'b anda'(b-b') =a'b-a'b' areinK.Thus(ab-a'b) +(a'b-a'b') =
ab-a'b'EK.
Butthistellsus(justfromgrouptheory)that ab+K=a'b'+K,ex­
actlywhat
weneededtohavetheproductwell-defined.
So
R/Kisnowendowedwithasumandaproduct.Furthermore,the
mapping
q;:R~R/Kdefinedbyq;(a)=a+KforaERisahomomor­
phismof
RontoR/Kwithkernel K.(Prove!)Thistellsusrightawaythat
R/Kisaring,beingthehomomorphicimage oftheringR.
Wesummarizeallthisin
Theorem
4.3.2.LetKbeanidealof R.Thenthequotientgroup R/K
asanadditivegroup isaringunderthemultiplication (a+K)(b+K)=
ab+K.Furthermore,themap
q;:R~R/Kdefinedbyq;(a)=a+KforaER
isahomomorphismof RontoR/KhavingKasitskernel.So R/Kisahomo­
morphicimageof
R.
Justfromgroup-theoreticconsiderationof Rasanadditivegroup,we
havethatif
q;isahomomorphismof RintoR',thenit is1-1ifandonlyif
Kerq;=(0).Asingroups,wedefineahomomorphismtobea monomor­
phismifitis1-1.Amonomorphismwhich isalsoonto iscalledan isomor­
phism.
WedefineRandR'tobeisomorphicifthereisanisomorphismofR
onto
R'.
Anisomorphismfromaring Rontoitselfiscalledan automorphismof
R.Forexample,suppose Risthefield
Cofcomplexnumbers.Thenthemap­
pingfromCtoCsendingeachelementofCtoitscomplexconjugate isan
automorphismofC.(Prove!)

157. 142 RingTheory Ch.4
Onewouldhave tobeanawfulpessimisttoexpectthatthehomomor­
phismtheoremsprovedinSection7
ofChapter2failinthissetting.Infact
theydohold,withtheslightlyobviousadaptationneeded
tomaketheproofs
gothrough.Westatethehomomorphismtheoremswithoutanyfurtherado,
leavingthefewdetails
neededtocompletetheproofs tothereader.
Theorem4.3.3(FirstHomomorphismTheorem). Letthemapping
cp:R~R'beahomomorphism ofRontoR'withkernel K.ThenR'===RIK;
infact,themappingl/J:RIK~R'definedbyl/J(a+K)=cp(a)definesan
isomorphism
ofRIKontoR'.
Theorem4.3.4(CorrespondenceTheorem). Letthemapping
cp:R~R'
beahomomorphism ofRontoR'withkernel K.IfI'isanidealofR',let
I={aERIcp(a)EI'}.ThenIisanidealofR,I:>Kand11K===I'.Thissets
upa
1-1correspondencebetweenalltheideals ofRIandthoseideals ofR
thatcontainK.
Theorem4.3.5(SecondHomomorphismTheorem). LetAbeasub­
ringofaring
RandIanidealofR.ThenA+I={a+iIaEA,iEI}isa
subring
ofR,Iisanidealof A+I,and(A+1)11
===AI(AnI).
Theorem4.3.6(ThirdHomomorphismTheorem). Letthemapping
cp:R~RIbeahomomorphism ofRontoRIwithkernel K.IfI'isanideal
ofR'andI={aERIcp(a)EI'},thenRII=R'II'.Equivalently,if Kisan
ideal
ofRand
I:>KisanidealofR,thenRII=(RIK)/(IIK).
Weclosethissectionwithaninspection ofsomeofthethingswehave
discussedinsomeexamples.
Examples
1.Asusualweuse
7L,theringofintegers,for ourfirstexample.Let n>1be
afixedinteger
andletInbethesetofallmultiplesofn;thenInisanidealofI.If7Lnistheintegersmod n,definecp:7L~7Lnbycp(a)=[aJ.Asiseasily
seen,cpisahomomorphism of7Lonto7Lnwithkernel In.SobyTheorem
4.3.3,7L
n=7Lll
n
•(Thisshouldcomeasnosurprise,forthat ishowweorigi­
nallyintroduced
7L
n
.)
2.LetFbeafield;whatcantheideals ofFbe?Supposethat I
=1=(0)isan
idealof
F;leta
=1=0EI.Then,since Iisanidealof F,1=a-1aEI;but
now,since1 EI,rl=rEIforeveryrEF.Inshort,I=F.SoFhasonly
thetrivialideals
(0)andFitself.
3.LetRbethering ofallrationalnumbershaving odddenominatorsintheir
reducedform.
LetIbethoseelements ofRwhichinreducedformhavean

158. Sec.3 Ideals,Homomorphisms,andQuotientRings 143
evennumerator;it iseasytoseethatIisanidealofR.Defineq;:R~7L
2
,the
integersmod
2,by
q;(alb)=0ifaiseven(a,bhavenocommonfactor)and
q;(alb)=1ifaisodd.Weleaveittothe readertoverifythatq;isahomo­
morphism
ofRonto7L
2withkernel I.Thus7L
2
:::::::RII.Givetheexplicitiso­
morphism
ofRIIonto7L
2
•
4.LetRbethering ofallrationalnumberswhosedenominators(wheninre­
ducedform)arenotdivisibleby
p,pafixedprime. LetIbethoseelementsin
Rwhosenumerator isdivisibleby p;IisanidealofRand
RII:::::::7L
p
,theinte­
gers
modp.(Prove!)
5.LetRbethering ofallreal-valuedcontinuousfunctionsontheclosedunit
intervalwhere
(f+g)(x)=f(x)+g(x)and(fg)(x)=f(x)g(x)forf,gER,
xE[0,1].LetI={fERI
f(~)=OJ.Weclaim thatIisanidealofR.
Clearly,it isanadditivesubgroup.Furthermoreif fEIandg ER,then
f(~)=0,so(fg)(~)=f(~)g (~)=Og(~)=O.ThusfgEI;sinceIiscommuta­
tive,
gfisalsoin I.SoIisanidealofR.
WhatisRII?Givenany fER,then
f(x)=(f(x)-
f(~»+f(~)=g(x)+f(~),
whereg(x)=f(x)-f(~).Becauseg(~)=f(~)-f(~)=0,gisinI.So
g
+I=I.Thusf+I=
(f(~)+g)+I=f(~)+I.Becausef(~)isjustareal
number,
RIIconsistsofthecosets
a+Iforareal.Weclaim thateveryreal
acomesup. Foriff(~)=f3=1=0,then
af3-1f+I=(af3-
1
+I)(f+I)=(af3-1+I)(f(~)+I)
=(af3-1+1)(/3+I)=af3-1f3+I=a+I.
SoRIIconsistsofalla+I,areal.Thuswehaveshowndirectly thatRIIis
isomorphictotherealfield.
Wenowuse
Theorem4.3.3toshow thatRII
:::::::realfield.Letq;:R~~
bedefinedbyq;(f)=f(~).Thenitiseasytoverify thatq;isahomomor­
phism,q;isontoand Kerq;={fERIf(~)=O};inotherwords,Kerq;=I.
SoRII:::::::imageofq;=IR.
6.LetRbetheintegralquaternions,in otherwords,R={aa+ali+a2j+
a3kIaa,al,a2,a3E7L}.Forafixedprime p,let
Thereadershouldverifythat
I
pisanidealofRandthatRlI
p
:::::::H(7L
p
)(see
Problem36
ofSection1andtheparagraphbeforeit).

159. 144 RingTheory Ch.4
7.LetR={(~ ~) Ia,bE~};Risasubringofthe2 X2matricesoverthe
reals.
LetI= {
(~ ~)IbE~};Iiseasilyseen tobeanadditivesubgroup ofR.Isit
anideal
ofR?Consider
(
X
Y)(Ob)=(0Xb).
OxOO00'
soitisinI.Similarly,
(
0
b)(XY)=(0bX)
OOOx00'
soit,too, isinI.SoIisanidealofR.WhatisR/I?Weapproachitfromtwo
points
ofview.
Given(
~:)ER ,
then(~:)=(~ ~)+(~ ~)sothat
(~:)+I=((~ ~)+(~ ~))+I=(~ ~)+L
since(~ ~)isin1.Thusallthecosets ofIinRlooklike(~ ~)+1.Ifwemap
thisonto
a,thatis,
1\1( (~ ~)+I)=a,wecancheckthat'"isanisomorphism
ontotherealfield.So R/I::::::IR.
Weseethat R/I"'"~anotherway.Definecp:R~~bYCP(~ ~)=a.We
claim
that
cPisahomomorphism.For,given(~ ~).(~ ~).then
cP(~ ~)=a,cP(~ ~)=c,
(
a
b)+(cd)=(a+Cb+d)
oa 0C 0a+c'

160. Sec.3 Ideals,Homomorphisms,andQuotientRings 145
and(~ ~)(~ ~)=(~ad~be}hence
~((~ ~)+(~ ~))=~(a~e ~:~)
=a+e=~(~:)+~(~ ~)
and~((~:)(~ ~))=~(~ad~be)
=ae=~(~ ~)~(~ ~).
So~isindeedahomomorphism ofRontoRWhatisKer~?If(~ ~)EKer~,
(
a
b) (ab).(ab)
then'P0a=a,butalso'P0a=0,SInce0aEKer'P.
Thusa=O.Fromthiswesee thatI=Ker'P.SoR/I~imageof'P=IRby
Theorem4.3.3.
8.LetR={(_
~:)Ia,bE~}andletCbethefieldofcomplexnumbers.
Define"':R~Cby'"( _~ ~)=a+bi.Weleaveit tothereadertoverifythat
l/JisanisomorphismofRontoC.SoRisisomorphictothefieldofcomplex
numbers.
9.LetRbeanycommutativeringwith 1.IfaER,let(a)={xaIxER}.We
claimthat
(a)isanidealofR.Toseethis,suppose thatu,vE(a);thus
u=xa,v=yaforsomex,yER,whence
u±v=xa±ya=(x±y)aE(a).
Also,if uE(a)andrER,thenu=xa,henceru=r(xa)=(rx)a,soisin(a).
Thus(a)isanidealofR.
NotethatifRisnotcommutative,then(a)neednotbeanideal;butit
iscertainlyaleftideal ofR.

161. 146 RingTheory
PROBLEMS
Ch.4
EasierProblems
1.
IfRisacommutativering andaER,letL(a)={xERIxa=OJ.Prove
thatL(a)isanidealofR.
2.IfRisacommutativeringwith1 andRhasnoidealsotherthan(0)and
itself,provethatRisafield.(Hint: LookatExample9.)
*3.If
q;:R~R'isahomomorphismofRontoR'andRhasa unitelement,
1,show
that
q;(1)istheunitelementofR'.
4.IfI,IareidealsofR,defineI+IbyI+I={i+jliEI,jEI}.Prove
thatI+IisanidealofR.
5.IfIisanidealofRandAisasubring ofR,showthatInAisanidealofA.
6.IfI,IareidealsofR,showthatInIisanidealofR.
7.Givea completeproofofTheorem4.3.2.
8.Givea completeproofofTheorem4.3.4.
9.Letq;:R~R'beahomomorphismofRontoR'withkernelK.IfA'is
asubringofR',letA={aERIq;(a)EA'}.Showthat:
(a)AisasubringofR,A:>K.
(b)A/K~A'.
(c)IfA'isaleftideal ofR',thenAisaleftideal ofR.
10.ProveTheorem4.3.6.
11.InExample3,givetheexplicitisomorphism ofR/Ionto7L
2
•
12.InExample4,showthatR/I
~7L
p
•
13.InExample6,showthatR/I
p~H(7L
p
)'
14.InExample8,verifythatthemappingt/JgivenisanisomorphismofRontoC.
15.IfI,IareidealsofR,letIIbethesetofallsumsofelementsoftheform
ij,whereiEI,jEI.ProvethatIIisanidealofR.
16.Showthattheringof2X2matricesovertherealshasnontrivialleft
ideals
(andalsonontrivialrightideals).
17.ProveTheorem4.3.5.
IfR,Sarerings,define thedirectsumofRandS,R
EBS,by
REBS={(r,s)IrER,sES}
where(r,s)=(rbSl)ifandonlyif r=rbs=Sbandwhere
(r,s)+(t,u)=(r+t,s+u),(r,s)(t, u)=(rt,su).

162. Sec.3 Ideals,Homomorphisms,andQuotientRings 147
18.ShowthatREBSisaringandthatthesubrings{(r,0)IrER}and
{CO,s)IsES}areidealsofREBSisomorphictoRandS,respectively.
19.IfR={(~ :)Ia,b,creal}andI={(~ ~)lbreal},Showthat:
(a)Risaring.
(b)IisanidealofR.
(c)R/I~FEBF,whereFisthefieldofrealnumbers.
20.IfI,fareidealsofR,letR
1=RIIandR
2=Rlf.Showthat'P:R~R
1EBR
2
definedby'P(r)=(r+I,r+f)isahomomorphismofRintoR
1EBR
2
suchthatKer'P=Inf.
21.LetZ15betheringofintegersmod15.Show thatZ15~Z3EBZ5.
Middle-LevelProblems
22.LetZbetheringofintegersandm,ntworelativelyprimeintegers,1m
themultiplesofminZ,andInthemultiplesofninZ.
(a)Whatis1mnIn?
(b)UsetheresultofProblem20toshowthatthereisaone-to-one
homomorphismfromZIImntoZIImEBZII
n
•
23.Ifm,narerelativelyprime,prove thatZmn~ZmEBZn.(Hint:Useacount­
ingargument
toshowthatthehomomorphismofProblem22(b)isonto.)
*24.UsetheresultofProblem23toprovetheChineseRemainderTheorem,
whichasserts thatifmandnarerelativelyprimeintegersanda,banyin­
tegers,we
canfindanintegerxsuchthatx
==amodmandx==bmodn
simultaneously.
25.
LetRbetheringof2X2matricesovertherealnumbers;supposethatI
isanideal ofR.ShowthatI=(0)orI=R.(Contrastthiswith theresultof
Problem16.)
HarderProblems
26.LetRbearingwith1 andletSbetheringof2X2matricesoverR.IfI
isanideal ofSshowthatthereisanidealfofRsuchthatIconsistsofall
the2X2matricesoverf.
27.If
PbP2'...,Pnaredistinctoddprimes,show thatthereareexactly2
n
solutionsofx
2
==xmod(p1•••Pn),where0::;x<Pl·..Pn·
28.Suppose thatRisaringwhoseonlyleftideals are(0)andRitself.Prove
that eitherRisadivisionring orRhasPelements,paprime,andab=0
forevery
a,bER.

163. 148 RingTheory Ch.4
29.LetRbearingwith 1.AnelementaERissaidtohavealeftinverseif
ba
=1forsomebER.Showthatiftheleftinverseb ofaisunique,then
ab=1(sobisalsoarightinverse ofa).
4.MAXIMALIDEALS
Thiswill beasectionwith onemajortheorem.Theimportanceofthisresult
willonly
becomefullytransparentwhenwediscussfieldsin Chapter5.How­
ever,it
isaresultthatstandsonitsowntwofeet.Itisn'tdifficulttoprove,
butinmathematicsthecorrelationbetweendifficultandimportantisn'tal­
ways
thathigh.Therearemanydifficultresults thatareofverylittleinterest
andofevenlessimportance,andsomeeasy resultsthatarecrucial.Of
course,therearesomeresults-many,many-whichareofincrediblediffi­
culty
andimportance.
Lemma4.4.1.LetRbeacommutativeringwith unitwhoseonly
ideals
are(0)anditself.ThenRisafield.
ProofLeta
=1=0beinR.Then(a)={xaIxER}isanidealofR,as
weverifiedinExample9intheprecedingsection.Sincea
=laE(a),(a)
=1=(0).
Thus,
byourhypothesisonR,(a)=R.Butthen,bythedefinitionof(a),
everyelementiERisamultiplexaofaforsomexER.Inparticular,be­
cause1
ER,1=baforsomebER.Thisshows thatahastheinversebin R.
SoRisafield.0
InTheorem4.3.4-theCorrespondenceTheorem-wesawthatif
'P:R~R'isahomomorphismofRontoR'withkernelK,thenthereisa
1-1
correspondencebetweenidealsofR'andidealsofRthatcontainK.Sup­
posethattherearenoidealsotherthanKitselfandRwhichcontainK.
Whatdoesthisimply aboutR'?Since(0)in R'correspondstoKinR,and
R'correspondstoallofR,wemustconcludethatinthiscase R'hasno
idealsotherthan(0)anditself.Soif R'iscommutativeandhasaunitele­
ment,then,byLemma4.4.1,R'mustbeafield.
This
promptsthefollowingdefinition.
Definition.AproperidealM ofRisa maximalidealofRiftheonly
ideals
ofRthatcontainMareMitselfandR.
Thediscussionprecedingthisdefinitionhas alreadyalmostproved
forus

164. Sec.4 MaximalIdeals149
Theorem4.4.2. LetRbeacommutativeringwith 1,andletMbea
maximalideal
ofR.ThenRIMisafield.
ProofThereisahomomorphismofRontoR'=RIM,andsince1 ER
wehavethat R'has1+Masitsunitelement.(SeeProblem 3,Section3).
Because
Misamaximalideal ofR,wesawinthediscussionabovethat R'
hasnonontrivialideals.Thus,by Lemma4.4.1,R'=RIMisafield.0
Thistheoremwillbe ourentryintothediscussion offields,foritwill
enableustoconstructparticularlydesirablefieldswheneverweshall
need
them.
Theorem4.4.2hasaconverse.This
is
Theorem4.4.3.If Risacommutativeringwith1and ManidealofR
suchthatRIMisafield,then Misamaximalideal ofR.
Proof
WesawinExample2 ofSection3 thattheonlyidealsinafield
Fare(0)andFitself.Since RIMisafield,ithasonly(0) anditselfasideals.
Butthen,bythecorrespondencegivenusby
Theorem4.3.4,therecanbeno
idealof
Rpropertybetween MandR.ThusMisamaximalideal ofR.D
Wegiveafewexamples ofmaximalidealsincommutativerings.
Examples
1.Let7Lbetheintegersand Manidealof 7L.Asanideal of7Lwecertainly
havethat
Misanadditivesubgroup of7L,somustconsist ofallmultiplesof
somefixedinteger n.Thussince RIM
==7L
nandsince7L
nisafieldifandonly
if
nisaprime,weseethat Misamaximalidealof 7Lifandonlyif Mconsists
ofallthemultiples
ofsomeprime p.Thusthesetofmaximalidealsin 7Lcor­
respondstotheset
ofprimenumbers.
2.Let7Lbetheintegers,andlet R={a+biIa,bE7L},asubringof
C(i
2
=-1).LetMbethesetofalla+biinR,where31aand31b.We
leaveittothereadertoverify
thatMisanidealofR.
WeclaimthatMisamaximalideal ofR.ForsupposethatN
:JMand
N=1=Misanidealof R.Sothereisanelementr+siEN,where3does not
divideror3doesnotdivides.Therefore,3(r
2
+S2).(Proveusingcongru­
encesmod3
!)Butt=r
2
+S2=(r+si)(r-si), soisinN,sincer+siEN
andNisanidealof R.SoNhasaninteger t=r
2
+S2notdivisibleby 3.Thus
ut+3v=1forsomeintegers u,
v;buttEN,henceutENand3EMeN,
so3vEN.Therefore,1 =ut+3vEN.Therefore,(a+bi)1EN,sinceNis
anidealof R,foralla+biER.Thistellsus thatN=R.Sotheonlyideal of
RaboveMisRitself.Consequently, Misamaximalidealof R.

165. 150 RingTheory Ch.4
ByTheorem4.4.2we knowthatRIMisafield.Itcanbeshown(see
Problem2)thatRIMisafieldhaving nineelements.
3.LetRbeasinExample2andletI={a+biISlaandSIb}.Weassertthat
Iisnotamaximalideal ofR.
InRwecanfactorS=(2+i)(2-i). LetM={x(2+i)IxER}.Mis
anidealofR,andsinceS=(2+i)(2- i)isinM,wesee thatICM.Clearly,
I=1=Mfor2+iEMandisnotinIbecauseS{2.SoI=1=M.CanM=R?
Ifso,then(2+i)(a+bi)=1forsomea,b.Thisgives 2a-b=1and
2b+a=0;thesetwoequationsimplythatSa=2,soa=~,b=-t.But
~fl.7L,-t$.7L;theelementa+bi=~-tiisnotinR.SoM =1=R.
Onecanshow,however, thatMisamaximalideal ofR.(SeeProblem 3.)
4.LetR={a+bv'21a,bintegers},whichisasubring oftherealfieldunder
thesumandproductofrealnumbers.ThatRisaringfollowsfrom
(a+bv'2)+(c+dv'2)=(a+c)+(b+d)v'2
and
(a+bv'2)(c+dv'2)=(ac+2bd)+(ad+bc)v'2.
LetM={a+bv'2ERISlaand5Ib}.MiseasilyseentobeanidealofR.
Weleaveit tothereadertoshowthatMisamaximalideal ofRandthat
RIMisafieldhaving 2Selements.
5.
LetRbetheringofallreal-valuedcontinuousfunctionsontheclosed
unitinterval[0,1]. WeshowedinExampleSofSection3 thatifM=
{fERIf
(~)=O},thenMisanidealofRandRIMisisomorphictothereal
field.Thus,
byTheorem4.4.3,Misamaximalideal ofR.
Ofcourse,ifwelet
My={fERIf(y)=O},whereyE[0,1],thenMyis
alsoamaximalideal. Itcanbeshownthateverymaximalidealin Risofthe
formMyforsomeyE[0,1],buttoproveitwewouldrequiresomeresultsfrom
realvariabletheory.
Whatthisexampleshowsis thatthemaximalidealsin Rcorrespondto
thepointsof[0,1].
PROBLEMS
1.Ifa,bareintegersand3
{aor3{b,showthat3{(a
2
+b
2
).
2.Show thatinExample2,RIMisafieldhaving nineelements.
3.
InExample3,showthatM={x(2+i)IxER}isamaximalideal ofR.

166. Sec.5 PolynomialRings 151
4.InExample3,show thatRIM~7Ls.
5.InExample3,show thatRII
~7LsEB7Ls.
6.InExample4,showthatMisamaximalideal ofR.
7.InExample4,show thatRIMisafieldhaving 25elements.
8.UsingExample2asamodel,constructafieldhaving49elements.
Wemakeashortexcursionbacktocongruences
modp,wherepisanodd
prime.If aisanintegersuch thatp
~aandr==amodphasasolution xin7L,
wesaythat aisaquadraticresidue modp.Otherwise,aissaidtobea quadratic
nonresidue
modp.
9.Showthat(p-1)/2ofthenumbers1,2,...,p-1arequadratic
residues
and(p-1)/2arequadraticnonresiduesmodp.[Hint:Show
that{x
2
1x
=1=0E7L
p
}formsagroup oforder(p-1)/2.]
10.Letm>0bein7L,andsupposethatmisnotasquarein7L.LetR=
{a+Y;;;bIa,bE7L}.Provethatundertheoperationsofsumand
productofrealnumbersRisaring.
11.
Ifpisanoddprime,letusset I
p
={a+
Y;;;bIpiaandpib},where
a+Y;;;bER,theringin Problem10.Show thatI
pisanidealofR.
12.Ifmisaquadraticnonresidue modp,showthattheidealI
pinProblem
11isamaximalideal ofR.
13.InProblem12show thatRll
pisafieldhaving p2elements.
5.POLYNOMIAL RINGS
Thematerialthatweconsiderinthissectioninvolves thenotionofpolyno­
mialandtheset
ofallpolynomialsoveragivenfield. Wehopethatmost
readerswillhavesomefamiliaritywith
thenotionofpolynomialfrom their
highschooldays andwillhaveseensome ofthethingsonedoeswithpolyno­
mials:factoringthem,lookingfor
theirroots,dividing onebyanothertoget
aremainder,
andsoon.Theemphasisweshallgive totheconceptandalge­
braicobjectknownasapolynomialringwill
beinaquitedifferentdirection
from
thatgiveninhighschool.
Bethatasitmay,whatweshallstrive todohereistointroducethe
ringofpolynomials overafieldandshowthatthisringisamenabletoacare­
fuldissection
thatrevealsitsinnermoststructure. Asweshallsee,thisring is
verywell-behaved.Thedevelopmentshouldremindus ofwhatwasdonefor
theringofintegersinSection5 ofChapter1.Thusweshallrunintotheana­
log
ofEuclid'salgorithm,greatest commondivisor,divisibility, andpossibly
mostimportant,
theappropriateanalogofprimenumber.Thiswilllead to

167. 152 RingTheory Ch.4
uniquefactorizationofageneralpolynomialintothese"primepolyno­
mials,"
andtothenatureoftheidealsandthemaximalidealsinthisnew
setting.
Butthepolynomialringenjoys onefeaturethattheringofintegersdid
not:
thenotionofarootofapolynomial.Thestudyofthenatureofsuch
roots-whichwillbedone,forthemostpart,inthenextchapter-consti­
tutesalarge
andimportantpartofthealgebraichistory ofthepast.Itgoes
underthetitleTheoryofEquations,andinitshonorablepast,alargevariety
ofmagnificentresultshave beenobtainedinthisarea.Hopefully,weshall
see
someoftheseasourdevelopmentprogresses.
Withthissketchyoutlineofwhatweintendtodooutoftheway,we
nowgetdowntothenitty-grittyofdoingit.
LetFbeafield.BytheringofpolynomialsinxoverF,whichweshall
alwayswriteas
F[x],wemeanthesetofallformalexpressions p(x)=
ao+alx+...+an_lX
n
-
1
+anx
n
, n
2::0,wheretheai'thecoefficientsof
thepolynomialp(x), areinF.Wesometimesemploythealternativenota­
tion:
p(x)=aox
n+alx
n
-
1
+...+an.InF[x]wedefineequality,sum, and
productoftwopolynomialssoas tomakeofF[x]acommutativeringasfol­
lows:
1.Equality.Wedeclarep(x)=ao+alx+...+anx
nandq(x)=
bo+b1x+...+bnx
ntobeequalifandonlyif theircorrespondingcoeffi­
cients
areequal,thatis,ifandonlyif ai=biforalli
2::o.
Wecombinethisdefinitionofequalityofpolynomialsp(x)andq(x)
withtheconventionthatif
q(x)=bo+b1x+ · · . +bnx
n
andifb
m
+
1=...=b
n=0,thenwecandropthelastn-m termsandwrite
q(x)as
q(x)=bo+b1x+...+bmX
m
•
Thisconventionisobservedinthedefinitionofadditionthatfollows,
wheresisthelargerofmandnandweaddcoefficientsan+l=...=as=0
if
n<sorb
m
+
1=...=b
s=0ifm<s.
2.Addition.Ifp(x)=ao+alx+...+anx
nandq(x)=bo+b1x
+...+bmx
m
,
wedeclarep(x)+q(x)=Co+CIX+...+csx
s
,wherefor
eachi,Ci=ai+bi.
Soweaddpolynomialsbyaddingtheircorrespondingcoefficients.
Thedefinitionofmultiplicationisalittle morecomplicated.Wedefine
itloosely
atfirstandthenmoreprecisely.

168. Sec.5 PolynomialRings 153
3.Multiplication.Ifp(x)=ao+alx+...+anx
n andq(x)=
bo+b1x+...+bmx
m
, wedeclarep(x)q(x)=Co+CIX+...+ctX
t
,
where
theCiaredeterminedbymultiplyingtheexpressionoutformally,using the
distributivelaws andtherulesofexponentsXUXV=x
u
+
v
,
andcollecting
terms.
Moreformally,
Weillustratethese operationswithasimpleexample, butfirsta nota­
tionaldevice: Ifsomecoefficientis0,wejustomitthatterm.Thuswewrite
9
+Ox+7x
2
+Ox
3
-
14x
4
as9+7x
2
-
14x
4
•
Letp(x)=1+3x
2
,
q(x)=4 -5x+7x
2
-
x
3
•
Thenp(x)+q(x)=
5 -5x+10x
2
-
x
3
while
p(x)q(x)=(1+3x
2
)(4-5x +7x
2
-
x
3
)
=4 -5x+7x
2
-
x
3
+3x
2
(4-5x+7x
2
-
x
3
)
=4 -5x+7x
2
-
x
3
+12x
2
-
15x
3
+21x
4
-
3x
5
=4 -5x+19x
2
-
16x
3
+21x
4
-
3x
5
•
TrythisproductusingtheCiasgivenabove.
Insomesensethisdefinition ofF[x]isnotadefinitionatall.Wehave
indulgedin
somehandwavinginit. Butitwilldo. Wecouldemployse­
quences
toformallydefine F[x]moreprecisely,butitwouldmerelycloud
whattomost
readersiswellknown.
Thefirstremarkthatwemake-anddonotverify-isthatF[x]isa
commutativering.
Togothroughthedetailsofcheckingtheaxiomsfora
commutativeringisastraightforward
butlaborioustask. However,itisim­
portanttonote
Lemma4.5.1.F[x]isacommutativeringwithunit.
Definition.
Ifp(x)=ao+alx+...+anx
nandan
=1=0,thenthe
degreeofp(x),denotedbydegp(x),isn.
Sothedegree ofapolynomialp(x)isthehighestpowerofxthatoccursin
theexpressionfor
p(x)withanonzerocoefficient.Thus deg(x-x
2
+x
4
)
=4,
deg(7x)=1,deg7 =O.(Notethatthisdefinitiondoesnotassigna degreeto
O.Itis,however,sometimesconvenient toadopttheconventionthatthede­
greeof0
be-00,inwhichcase manydegreerelatedresultswill holdinthis
extendedcontext.)Thepolynomialsofdegree0andthepolynomial0 are
calledtheconstants;thusthesetofconstantscan beidentifiedwith F.
Thedegreefunction onF[x]willplayasimilarrole tothatplayedby

169. 154 RingTheory Ch.4
thesizeoftheintegersin7L,inthatitwillprovideuswithaversionofEu­
clid'sAlgorithmfor
F[x].
Oneimmediateandimportantproperty ofthedegreefunction isthatit
behaveswellforproducts.
Lemma
4.5.2.Ifp(x),q(x)arenonzeroelements ofF[x],then
deg(p(x)q(x»=degp(x)+degq(x).
ProofLetm=degp(x)andn=degq(x);thusthepolynomialp(x) =
ao+alx+...+amx
m
, wheream
=1=0,andthepolynomial q(x)=bo+
b1x+...+bnx
n
, wherebn
=1=0.Thehighestpowerof xthatcanoccurin
p(x)q(x)isx
m
+
n
,
fromourdefinitionoftheproduct.What isthecoefficient
of
x
m+
n
?
Theonlywaythat x
m+
n
canoccur isfrom(amxm)(bnx
n
) =
ambnx
m
+
n
.
Sothecoefficient ofx
m+
n
inp(x)q(x)isambn,whichisnot0,since
am
=1=0,b
n
=1=0.Thusdeg(p(x)q(x»=m+n=degp(x)+degq(x),as
claimedinthelemma.
D
Onealsohassomeinformationabout deg(p(x)+q(x».Thisis
Lemma4.5.3.Ifp(x),q(x)EF[x]andp(x)+q(x)
=1=0,then
deg(p(x)+q(x»::;max(degp(x),degq(x».
Weleavethe proofofLemma4.5.3tothereader. Itwillplaynorolein
what
istocome,whereasLemma4.5.2willbeimportant.Weputitinsothat
the
"+"shouldnotfeelslightedvis-a-vistheproduct.
Animmediateconsequence ofLemma4.5.2 is
Lemma4.5.4.F[x]isanintegraldomain.
ProofIfp(x)
=1=°andq(x)=1=0,thendegp(x);:::0,degq(x);:::0,so
deg(p(x)q(x»=degp(x)+degq(x);:::0.Therefore,p(x)q(x)hasadegree,
socannotbe
°(whichhasnodegreeassignedtoit).Thus F[x]isanintegral
domain.
D
Oneofthethingsthatwewereonceforced tolearnwastodivide
onepolynomialbyanother. Howdidwedothis? Theprocesswas
called
longdivision. Weillustratewithanexamplehowthiswasdone,
forwhatwe
dointheexampleisthemodelofwhatweshalldointhegen­
eralcase.
Wewanttodivide
2x
2
+1intox
4
-
7x+1.Wedoitschematicallyas
follows:

170. Sec.5
andweinterpretthis as:
PolynomialRings
-7x+1
-7x+1
-7x+1~
155
x
4
-
7x+1=(2x
2+1)(~x2-~)+(-7x+~)
and-7x+~iscalledthe remainderinthisdivision.
Whatexactlydidwedo?First,wheredidthe~X2comefrom?Itcame
fromthefactthatwhenwemultiply
2x
2
+1by
~x
2wegetx
4
,
thehighest
poweroccurringin
x
4
-
7x+1.Sosubtracting
~x2(2x2+1)fromx
4
-
7x+1
getsridofthex
4
termandwegoontowhat isleftandrepeattheprocedure.
This"repeattheprocedure"suggestsinduction,and
thatishowwe
shallcarryouttheproof.Butkeepinmind
thatallweshallbedoing iswhat
wedidintheexampleabove.
Whatthisgivesus
issomethinglikeEuclid'sAlgorithm,intheintegers.
However,herewecallitthe
DivisionAlgorithm.
Theorem4.5.5(DivisionAlgorithm). Giventhepolynomials f(x),
g(x)EF[x],whereg(x)
=1=0,then
f(x)=q(x)g(x)+rex),
whereq(x),rex) EF[x]andrex)=0ordegrex)<degg(x).
Proof
Wegobyinductionon degf(x).Ifeitherf(x)=0ordegf(x)<
degg(x),thenf(x)=Og(x)+f(x),whichsatisfiestheconclusionofthe
theorem.
Sosuppose
thatdegf(x)
2::degg(x);thusthepolynomial f(x)=
ao+atX+...+a mxm,wheream=1=0andthepolynomial g(x)=bo+btx
+...+bnx
n
, wherebn
=1=0andwhere m2::n.
Consider
:mxm-ng(x)=~mxm-n(b
o+b1x+ · · · +bnx
n
)
n n

171. 156 RingTheory Ch.4
Thus(amlbn)xm-ng(x)hasthesamedegreeandsameleadingcoefficient asdoes
f(x),sof(x)-(amibn)xm-ng(x) =h(x)issuchthattherelationdeg h(x)<
degf(x)holds.Thus,byinduction,
h(x)=qt(x)g(x)+r(x),whereqt(x),r(x)EF[x]
andr(x)=0ordegr(x)<degg(x).Rememberingwhat h(x)is,weget
h(x)=f(x)-~mxm-ng(x)=qJ(x)g(x)+r(x)
n
so
If
q(x)=(amlbn)x
m
-
n +ql(X),wehaveachievedtheformclaimedinthe
theorem.D
TheDivisionAlgorithmhasoneimmediateapplication: Itallowsusto
determinethe
natureofalltheideals ofF[x].Asweseeinthenexttheorem,
anideal
ofF[x]mustmerelyconsist ofallmultiples,byelementsof F[x],of
some
fixedpolynomial.
Theorem
4.5.6.IfI
=1=(0)isanidealof F[x],thenI={f(x)g(x)If(x)E
F[x]};thatis,Iconsistsofallmultiplesofthefixedpolynomial g(x)bytheel­
ements
ofF[x].
ProofToprovethetheorem,weneedtoproducethatfixedpolyno­
mial
g(x).Wherearewegoingtodigitup? Theonecontrolwehavenumeri­
callyonagivenpolynomial
isitsdegree.Sowhynotusethedegreefunction
asthemechanismforfinding
g(x)?
SinceI
=1=(0)thereareelementsin Ihavingnonnegativedegree.So
there
isapolynomialg(x)
=1=0inIofminimaldegree;that is,g(x)=1=0isinI
andif0=1=t(x)EI,thendeg t(x)2::degg(x).Thus,bythedivisionalgorithm,
t(x)=q(x)g(x)+r(x),wherer(x)=0ordegr(x)<degg(x).Butsince
g(x)EIandIisanidealofF[x],wehavethat q(x)g(x)EI.Byassumption,
t(x)EI,thust(x)-q(x)g(x) isinI,sor(x)=t(x)-q(x)g(x) isinI.Since
g(x)hasminimaldegreefortheelements ofIandr(x)EI,degr(x)cannot
belessthandeg g(x).Soweareleftwith r(x)=O.Butthissaysthat t(x)=
q(x)g(x).Soeveryelementin Iisamultipleof g(x).Ontheotherhand,

172. Sec.5 PolynomialRings 157
sinceg(x)EIandIisanidealof F[x],f(x)g(x)EIforallf(x)EF[x].The
netresultofallthis
isthatI={f(x)g(x)If(x)EF[x]}.D
Definition.Anintegraldomain Riscalleda principalidealdomain if
everyideal
IinRisoftheform I={xaIxER}forsomeaEI.
Theorem4.5.6canbestated as:F[x]isaprincipalidealdomain.
Weshallwritethe idealgeneratedbyagivenpolynomial,g(x), namely
{f(x)g(x)If(x)EF[x]},as(g(x».
Theproofshowedthatif Iisanidealof F[x],thenI=(g(x»,where
g(x)isapolynomialoflowestdegreecontainedin I.Butg(x)isnotunique,for
if
a
=1=0EF,thenag(x)isinIandhasthesamedegree asg(x),soI=(ag(x».
Togetsomesortofuniquenessinallthis,wesingleoutaclassofpoly­
nomials.
Definition.f(x)EF[x]isamonicpolynomial ifthecoefficientofits
highestpower
is1.
Thusf(x)ismonicmeansthat
Weleavetothereadertoshowthatif
Iisanidealof F[x],thenthere is
onlyone monicpolynomialoflowestdegreein I.Singlingthisoutasthegen­
eratorof
Idoesgiveusa"monic"uniquenessforthegenerationof I.
Ournextstepinthisparalleldevelopmentwithwhathappensinthein­
tegers
istohavethenotionofonepolynomialdividinganother.
Definition.Supposef(x),g(x)EF[x],withg(x)
=1=O.Wesaythat
g(x)dividesf(x),writtenasg(x)If(x),iff(x)=a(x)g(x)forsome a(x)E
F[x].
Notethatif f(x)=1=0andg(x)If(x),thendeg g(x)~degf(x)byLemma
4.5.2.Moreover,theideals (f(x»and(g(x»ofF[x],generatedby f(x)and
g(x),respectively,satisfythecontainingrelation (f(x»c(g(x».(Prove!)
Weagainemphasizetheparallelismbetween
7L,theintegers,and F[x]
byturningtothenotionof greatestcommondivisor. Inordertogetsomesort
ofuniqueness,weshallinsistthatthegreatestcommondivisoralwaysbea
monicpolynomial.

173. 158 RingTheory Ch.4
Definition.Foranytwopolynomials f(x)andg(x)EF[x](notboth0),
thepolynomial
d(x)EF[x]isthegreatestcommondivisor off(x),g(x) if
d(x)isamonicpolynomialsuchthat:
(a)
d(x)If(x)andd(x)Ig(x).
(b)Ifhex)If(x)andhex)Ig(x),thenhex)Id(x).
Althoughwedefinedthegreatestcommondivisor oftwopolynomials,
weneitherknow,asyet,thatitexists,
norwhatitsformmaybe.Wecould
defineitinanother,andequivalent,wayasthe
monicpolynomial ofhighest
degreethatdividesboth
f(x)andg(x).Ifwedidthat,itsexistencewouldbe
automatic,
butwewouldnotknowitsform.
Theorem4.5.7.Given
f(x)andg(x)
=1=0inF[x],thentheirgreatest
commondivisor
d(x)EF[x]exists;moreover, d(x)=a(x)f(x)+b(x)g(x)
forsomea(x),b(x)EF[x].
ProofLetIbetheset ofallr(x)f(x)+s(x)g(x)asrex),sex) run
freelyover
F[x].WeclaimthatI isanidealofR.For,
(rl(x)f(x)+Sl(X)g(X»+(r2(x)f(x)+S2(X)g(X»
=(rl(x)+r2(x»f(x)+(Sl(X)+S2(X»g(x),
soisagainin I,andfort(x)EF[x],
t(x)(r(x)f(x)+s(x)g(x»=(t(x)r(x»f(x)+(t(x)s(x»g(x),
soit,too, isagainin I.ThusIisanidealofF[x].Sinceg(x)
=1=0,weknow
thatI=1=0,sincebothf(x)andg(x)arein[.
Since[=1=0isanidealofF[x],itisgeneratedbya uniquemonic polyno­
mial
d(x)(Theorem4.5.6).Since f(x),g(x)areinI,theymustthenbemulti­
ples
ofd(x)byelements ofF[x].Thisassuresus thatd(x)If(x)and
d(x)Ig(x).
Becaused(x)EIand[ isthesetofallr(x)f(x)+
s(~)g(x), wehave
thatd(x)=a(x)f(x)+b(x)g(x)forsomeappropriate a(x),b(x) EF[x].
Thusif hex)If(x)andhex)Ig(x),thenhex)I(a(x)f(x)+b(x)g(x»=d(x).
Sod(x)isthegreatestcommondivisor off(x)andg(x).
Thisprovesthetheorem;theuniquenessof d(x)isguaranteedbythe
demand
thatwehavemade thatthegreatestcommondivisorbemonic. 0
Anotherwaytoseetheuniqueness ofd(x)isfrom
Lemma4.5.8.If
f(x)
=1=0,g(x)=1=0arein F[x]andf(x)Ig(x)and
g(x)If(x),thenf(x)=ag(x),whereaEF.

174. Sec.5 PolynomialRings 159
ProofBythemutualdivisibilityconditionon f(x)andg(x)wehave,
byLemma4.5.2,
degf(x)
:::;degg(x):::;degf(x),sodegf(x)=degg(x).But
f(x)=a(x)g(x),so
degf(x)=dega(x)+degg(x)=dega(x)+degf(x),
inconsequenceofwhichdeg a(x)=0,soa(x)=a,anelementofF.D
Weleavetheproof oftheuniquenessofthegreatestcommondivisor
viaLemma4.5.8tothereader.
Definition.Thepolynomialsf(x),g(x)inF[x]aresaidtoberelatively
prime
iftheirgreatestcommondivisor is1.
Althoughit ismerelyaveryspecialcaseof Theorem4.5.7,toempha­
sizeitandtohaveittoreferto,westate:
Theorem4.5.9. Iff(x),g(x)EF[x]arerelativelyprime,then
a(x)f(x)+b(x)g(x)=1forsome a(x),b(x) EF[x].Conversely,if
a(x)f(x)+b(x)g(x)=1forsome a(x),b(x)EF[x],thenf(x)andg(x)are
relativelyprime.
ProofWeleavethis"conversely" parttothereaderasexercise.D
Aswiththeintegers,wehave
Theorem4.5.10. Ifq(x)andf(x)arerelativelyprimeandif
q(x)If(x)g(x),thenq(x)Ig(x).
Proof
ByTheorem4.5.9a(x)f(x)+b(x)q(x)=1forsome a(x),
b(x)
EF[x].Therefore,
a(x)f(x)g(x)+b(x)q(x)g(x)=g(x). (1)
Sinceq(x)Ib(x)g(x)q(x)andq(x)If(x)g(x)byhypothesis,q(x)dividesthe
left-handside
oftherelationin(1).Thus q(x)dividestheright-handside
of(1),thatis, q(x)Ig(x),thedesiredconclusion. 0
Wearenowreadytosingle outtheimportantclass ofpolynomialsthat
willplaythesameroleasprimeobjectsin F[x]asdidtheprimenumbersin 7L.
Definition.Thepolynomialp(x)EF[x]isirreducibleifp(x)isofpos­
itivedegreeandgivenany
polynomialf(x)inF[x],theneitherp(x)If(x)or
p(x)isrelativelyprimeto f(x).

175. 160 RingTheory Ch.4
Weshouldnoteherethatthedefinitionimpliesthat apolynomialp(x)
ofpositivedegree isirreducibleinF[x] ifandonlyifp(x)cannotbefactored
asaproductoftwopolynomialsofpositivedegree.Inotherwords, if p(x)=
a(x)b(x),wherea(x)andb(x)areinF[x],theneithera(x)isaconstant(that
is,anelementof F),orb(x)isconstant.Theproofofthisfact isverysimilar
totheproofofananalogousobservationconcerningtwoequivalentdefini­
tionsofaprimenumber.
Firstsupposethat
p(x)(ofpositivedegree)cannotbefactored asa
productoftwonon-eonstantpolynomials.Then,givenany
f(x)inF[x],we
haveonlytwopossibilitiesfor (p(x),f(x»,namely1 oramonicpolynomialof
theform
c·p(x),wherec isanelementof F.Thus(p(x),f(x»=1orp(x)If(x),
whichshowsthat p(x)isirreducible.
Nowlet
p(x)beirreduciblein F[x]andsuppose p(x)=a(x)b(x)for
some
a(x),b(x) inF[x].Accordingtothedefinition,wemusthave
p(x)Ia(x)or(P(x),a(x» =1.Ifp(x)Ia(x),thenb(x)mustbeaconstant.If,
ontheotherhand,
p(x)anda(x)arerelativelyprime,thenbyTheorem
4.5.10,p(x)Ib(x),andinthiscase a(x)mustbeaconstant.Thisshowsthatan
irreduciblepolynomialcannotbefactored
asaproductoftwonon-constant
polynomials.
Notethattheirreducibilityofapolynomialdependsonthefield
F.For
instance,thepolynomial
x
2
-
2isirreduciblein
Q[x],whereQisthefieldof
rationalnumbers,but
x
2
-2isnotirreduciblein
lR[x],whereIRisthefieldof
realnumbers,forinlR[x]
x
2
-
2=(x-
v2)(x+v2).
CorollarytoTheorem 4.5.10.Ifp(x)isirreduciblein F[x]and
p(x)Iat(x)a2(x)...ak(x),whereat(x),...,ak(x)areinF[x],thenp(x)Iai(x)
forsome i.
ProofWeleavetheprooftothereader.(SeeTheorem1.5.6.) D
Asidefromitsotherproperties,anirreduciblepolynomial p(x)inF[x]
enjoysthepropertythat (p(x»,theidealgeneratedby p(x)inF[x],isamax­
imalidealof
F[x].Weprovethisnow.
Theorem
4.5.11.Ifp(x)EF[x],thentheideal (p(x»generatedby
p(x)inF[x]isamaximalidealof F[x]ifandonlyif p(x)isirreducibleinF[x].
ProofWefirstprovethatif p(x)isirreduciblein F[x],thentheideal
M=(p(x»isamaximalidealof F[x].For,supposethat Nisanidealof
F[x],andN
~M.ByTheorem4.5.6,

176. Sec.5 PolynomialRings 161
N=(f(x»forsome f(x)EF[x].
Becausep(x)EMeN,p(x)==a(x)f(x),sinceeveryelementin Nisofthis
form.
Butp(x)isirreduciblein F[x],hencea(x)isaconstantorf(x)isacon­
stant.If
a(x)==aEF,thenp(x)==af(x),sof(x)
=a-Ip(x).Thusf(x)EM,
whichsaysthat NCM,henceN==M.Ontheotherhand,if f(x)==bEF,
then1 ==b-1bEN,sinceNisanidealof F[x],thusg(x)1ENforall
g(x)EF[x].Thissaysthat N==F[x].Therefore,wehaveshown Mtobea
maximalidealof
F[x].
Intheotherdirection,supposethat M==(p(x»isamaximalideal of
F[x].Ifp(x)isnotirreducible,then p(x)==a(x)b(x),wheredeg a(x)
;:::1,
degb(x);:::1.LetN==(a(x»;then,since p(x)==a(x)b(x),p(x)EN.There­
fore,
MeN.Sincedeg a(x)
;:::1,N==(a(x»=1=F[x],sinceeveryelementin
(a(x»hasdegreeatleastthatof a(x).Bythemaximalityof Mweconclude
that
M==N.Butthen a(x)EN==M,whichtellsus thata(x)==f(x)p(x);
combinedwith p(x)==a(x)b(x)==b(x)f(x)p(x),wegetthat b(x)f(x)==1.
Sincedeg1 ==°<degb(x)
~deg(b(x)f(x»==deg1 ==0,wehavereacheda
contradiction.Thus
p(x)isirreducible.D
Thistheorem isimportantbecauseittells usexactlywhatthemaximal
idealsof
F[x]are,namelytheidealsgeneratedbytheirreduciblepolynomi­
als.If
Misamaximalideal ofF[x],F[x]/Misafield,andthisfieldcontains
F(ormoreprecisely,thefield {a+MIaEF},whichisisomorphicto F).
Thisallows ustoconstructdecentfields K
:)F,thedecencyofwhichliesin
that
p(x)hasa rootinK.Theexactstatementandexplanationofthiswe
postponeuntilChapter
5.
Thelasttopicinthisdirectionthatwewanttodiscuss isthefactoriza­
tionofagivenpolynomialasaproduct
ofirreducibleones.Notethatif
p(x)==aox
n+atXn-1+...+an-IX+an,ao
=1=0,isirreduciblein F[x],then
so
isao1p(x)irreduciblein F[x];however,ao1p(x)hastheadvantage of
beingmonic.Sowehavethismonicirreduciblepolynomialtriviallyobtain­
ablefrom
p(x)itself.Thiswillallowustomakemoreprecisetheuniqueness
partofthenexttheorem.
Theorem
4.5.12.Letf(x)EF[x]beofpositivedegree.Theneither
f(x)isirreduciblein F[x]orf(x)istheproductofirreduciblepolynomialsin
F[x].Infact,then,
f(x)==apl(X)m
1
p2(X)m
2
•••
Pk(X)mk,
whereaistheleadingcoefficient off(x),Pt(x),...,Pk(X)aremonicandir­
reduciblein
F[x],mi>0,...,mk>°andthisfactorizationinthisform is
uniqueuptotheorderofthe p;(x).

177. 162 RingTheory Ch.4
Pr~ofWefirstshowthefirsthalfofthetheorem,namelythat f(x)is
irreducibleortheproductofirreducibles.Theproof isexactlythesame as
thatofTheorem1.5.7,withaslight,obviousadjustment.
Wegobyinductionon
degf(x).Ifdegf(x)=1,thenf(x)=ax+b
witha
=1=0andisclearlyirreduciblein F[x].Sotheresult istrueinthiscase.
Suppose,then,thatthetheorem
iscorrectforall a(x)EF[x]suchthat
deg
a(x)<degf(x).Iff(x)isirreducible,thenwehavenothingtoprove.
Otherwise,
f(x)=a(x)b(x),a(x) andb(x)EF[x]anddeg a(x)<degf(x)
anddeg b(x)<degf(x).Bytheinduction, a(x)[andb(x)]isirreducibleor is
theproductofirreducibles.Butthen f(x)istheproductofirreduciblepoly­
nomialsin
F[x].Thiscompletestheinduction,andsoprovestheopeninghalf
ofthetheorem.
Nowtotheuniquenesshalf.Againwegobyinductionon
degf(x).If
degf(x)=1,thenf(x)isirreducibleandtheuniqueness isclear.
Supposetheresulttrueforpolynomialsofdegreelessthan
degf(x).
Supposethat
f(x)=apl(X)m
1
p2(x)m
2
•••Pk(X)m
k=aql(X)n
1
•••qr(x)n
r
,
wherethe Pi(X),qi(X) aremonicirreduciblepolynomialsandthe mi'niare
allpositiveand
aistheleadingcoefficientof f(x),thatis,thecoefficient
ofthehighestpowertermof
f(x).SincePl(X)If(x),wehavethat
Pl(X)Iql(X)n
1
•••qr(x)n
r
,sobythecorollarytoTheorem 4.5.10,Pl(X)Iqi(X)
forsome i.Sinceqi(X)ismonicandirreducible, asisPl(X),wegetPl(X)=
qi(X).Wecansuppose(onrenumbering) thatpl(x)=ql(X).Thus
;1«:))=ap1(X)m,-1p2(x)mz· · ·Pk(X)m
k
=apl(X)n
1-lq2(x)n
2
•
••qr(x)n
r
•
Byinductionwehaveuniquefactorizationintherequiredformfor
f(X)/Pl(X),whosedegree islessthan degf(x).Henceweobtainthat
ml-1=nl-1(soml=nl),m2=n2,...,mk=nk,r=kandP2(X)=
q2(X),...,Pk(X)=qk(X),onrenumberingtheq'sappropriately.Thiscom­
pletestheinductionandprovesthetheorem.
0
Wehavepointedouthowsimilarthesituation isfortheintegers 7Land
thepolynomialring
F[x].Thissuggeststhatthereshouldbeawiderclassof
rings,ofwhichthetwoexamples
7LandF[x]arespecialcases,forwhich
muchoftheargumentationworks.Itworkedfor
7LandF[x]becausewehad
ameasureofsizeinthem,eitherbythesizeofaninteger
orthedegreeofa
polynomial.ThismeasureofsizewassuchthatitallowedaEuclid-typealgo­
rithmtohold.

178. Sec.5 PolynomialsRings 163
Thisleadsus todefineaclass ofrings,theEuclideanrings.
Definition.AnintegraldomainRisaEuclideanring ifthereisa
function
dfromthenonzeroelementsofRtothenonnegativeintegers that
satisfies:
(a)
Fora
=1=0,b=1=0ER,d(a)::;d(ab).
(b)Given a=1=0,b=1=0,thereexistqandrERsuchthatb==qa+r,where
r==0ord(r)<d(a).
Theinterestedstudentshouldtry toseewhichoftheresultsprovedfor
polynomialrings
(andtheintegers)holdinageneralEuclideanring.Aside
fromafewproblemsinvolvingEuclideanrings,weshall
notgoanyfurther
withthisinterestingclass ofrings.
Thefinalcommentwemakehereisthatwhatwedidforpolynomials
overafieldwecould
trytodoforpolynomialsover anarbitraryring. Thatis,
givenanyring
R(commutativeornoncommutative),wecoulddefine the
polynomialring R[x]inxoverRbydefiningequality,addition, andmultipli­
cationexactlyaswedidin
F[x],forFafield.Theringsoconstructed, R[x],is
averyinterestingring,whosestructure istightlyinterwovenwith thatofRit­
self.
Itwouldbetoomuchtoexpect thatall,orevenany, ofthetheorems
provedinthissectionwould carryovertoR[x]forageneralring R.
PROBLEMS
Inthefollowingproblems, Fwillalways denoteafield.
EasierProblems
1.IfFisafield,show thattheonlyinvertibleelementsin F[x]arethe
nonzeroelements ofF.
2.IfRisaring,weintroducethering R[x]ofpolynomialsin xoverR,justas
wedid
F[x].Definingdegf(x)forf(x)ER[x]aswedidin F[x],showthat:
(a)deg(f(x)g(x»
::;degf(x)+degg(x)iff(x)g(x)=1=O.
(b)Thereisacommutativering Rsuchthatwecanfind f(x),g(x)in
R[x]withdeg(f(x)g(x»<degf(x)+degg(x).
3.Findthegreatestcommondivisor ofthefollowingpolynomialsoverQ,
thefieldofrationalnumbers.
(a)x
3
-
6x+7andx+4.
(b)x
2
-1and2x
7
-
4x
5
+2.

179. 164 RingTheory Ch.4
(c)3x
2+1andx
6
+x
4
+X+1.
(d)x
3
-1andx
7
-
x
4
+x
3
-1.
4.ProveLemma4.5.3.
5.
InProblem3,letI={f(x)a(x)+g(x)b(x)},wheref(x),g(x)runoverQ[x]anda(x)isthefirstpolynomial andb(x)thesecondoneineachpart
oftheproblem.Findd(x),sothatI=(d(x»forParts(a),(b),(c),
and(d).
6.Ifg(x),f(x)EF[x]andg(x)If(x),showthat(f(x»c(g(x».
7.Provetheuniquenessofthegreatestcommondivisoroftwopolynomials
in
F[x]byusingLemma4.5.8.
8.Iff(x),g(x)EF[x]arerelativelyprimeandf(x)Ih(x)andg(x)Ih(x),
showthatf(x)g(x)Ih(x).
9.ProvetheCorollarytoTheorem4.5.10.
10.Showthatthefollowingpolynomialsareirreducibleover thefieldFindi­
cated.
(a)x
2
+7overF=realfield=
IR.
(b)x
3
-3x +3overF=rationalfield =Q.
(c)x
2
+X+loverF=7L
2
•
(d)x
2
+loverF=7L
19
•
(e)x
3
-90verF=7L
13
•
(f)x
4
+2x
2
+2overF=
Q.
11.Ifp(x)EF[x]isofdegree3 andp(x)=aox
3+aIx
2+a2x+a3'show
thatp(x)ifirreducibleoverFifthereisnoelementrEFsuchthat
p(r)=aor
3
+aIr
2+a2r+a3=O.
12.IfFeKaretwofieldsandf(x),g(x)EF[x]arerelativelyprimein F[x],
showthattheyarerelativelyprimeinK[x].
Middle-LevelProblems
13.Let
IRbethefieldofrealnumbersandCthatofcomplexnumbers.Show
thatlR[x]/(x
2
+1)~C.[Hint:IfA=lR[x]/(x
2
+1),letubetheimageof
xinA;showthateveryelementinAisoftheforma+bu,where
a,bElRandu
2
=-1.]
14.LetF=7L
II
,theintegersmod11.
(a)Letp(x)=x
2+1;showthatp(x)isirreduciblein F[x]andthat
F[x]/(p(x»isafieldhaving121elements.
(b)Letp(x)=x
3+X+4EF[x];showthatp(x)isirreduciblein F[x]
andthatF[x]/(p(x»isafieldhaving 11
3
elements.

180. Sec.5 PolynomialsRings 165
15.LetF=7L
pbethefield ofintegersmodP,wherePisaprime,andlet
q(x)EF[x]beirreducibleofdegreen.ShowthatF[x]/(q(x»isafield
having
atmostpnelements.(SeeProblem 16foramoreexactstate­
ment.)
16.Let
F,q(x)beasinProblem
15;showthatF[x]/(q(x»hasexactlypnele­
ments.
17.Let
Pl(X),P2(X),...,Pk(X)EF[x]bedistinctirreduciblepolynomials
andlet
q(x)=PI(X)P2(X)...Pk(X).Showthat
~=--.fuLEBF[x]EB'"EBF[x]
(q(x»(PI(X» (P2(X» (Pk(X»"
18.LetFbeafinitefield.Show thatF[x]containsirreduciblepolynomials of
arbitrarilyhighdegree.(Hint:TrytoimitateEuclid's proofthatthereis
aninfinityofprimenumbers.)
19.Constructafieldhaving
P2elements,for Panoddprime.
20.IfRisaEuclideanring,show thateveryideal ofRisprincipal.
21.IfRisaEuclideanring,show thatRhasaunitelement.
22.IfRistheringofevenintegers,show thatEuclid'salgorithm isfalsein R
byexhibitingtwoevenintegersforwhich thealgorithmdoes nothold.
HarderProblems
23.LetF=7L
7andletp(x)=x
3
-
2andq(x)=x
3
+2bein F[x].Show
that
p(x)andq(x)areirreduciblein F[x]andthatthefieldsF[x]/(p(x»
andF[x]/(q(x»areisomorphic.
24.Let
Qbethefieldofrationalnumbers, andletq(x)=x
2+X+1bein
Q(x).Ifaisacomplexnumbersuchthata
2
+a+1=0,showthatthe
set
{a+
baIa,bEQ}isafieldintwoways;thefirstbyshowingittobe
isomorphictosomethingyouknow
isafield,thesecondbyshowing that
ifa+ba
=1=0,thenitsinverse isofthesameform.
25.IfPisaprime,showthat q(x)=1+x+x
2+...x
p
-l
isirreduciblein
Q[x].
26.LetRbeacommutativeringinwhich a
2
=0onlyif a=O.Showthatif
q(x)ER[x]isazero-divisorin R[x],then,if
q(x)=a x
n
+a x
n
-I
+...+a
o 1 n'
thereisanelementb
=1=0inRsuchthatbaa=bal=...=ban=O.

181. 166 RingTheory Ch.4
27.LetRbearingand Ianidealof R.IfR[x]andI[x]arethepolynomial
ringsin
xoverRandI,respectively,showthat:
(a)I[x]isanidealof R[x].
(b)R[x]/I[x]
~(R/I)[x].
VeryHardProblems
*28.
DoProblem26evenifthecondition "a
2
=0onlyif a=0"doesnothold
inR.
29.LetR={a+biIa,bintegers}C
C.Letd(a+bi)=a
2
+b
2
•
Showthat
RisaEuclideanringwhere disitsrequiredEuclideanfunction. (Ris
knownasthering ofGaussianintegers andplaysanimportantrole in
numbertheory.)
6.POLYNOMIALS OVERTHE RATIONALS
Inourconsiderationofthepolynomialring F[x]overafield F,theparticular
natureof
Fneverenteredthepicture.Alltheresultsholdforarbitraryfields.
However,thereareresultsthatexploittheexplicitcharacterofcertainfields.
Onesuchfield
isthatoftherationalnumbers.
Weshallpresenttwoimportanttheoremsfor
Q[x],thepolynomialring
overtherationalfieldQ.Theseresultsdependheavilyonthefactthat weare
dealingwithrationalnumbers.Thefirstofthese,
Gauss'Lemma,relatesthe
factorizationovertherationalswithfactorizationovertheintegers.Thesec­
ondone,known
astheEisensteinCriterion, givesusamethodofconstructing
irreduciblepolynomialsofarbitrarydegree,at
will,in
Q[x].InthisthefieldQ
ishighlyparticular.Forinstance,there isnoeasyalgorithmforobtainingirre­
duciblepolynomialsofarbitrarydegree
noverthefield 7L
poftheintegers
mod
p,paprime.Evenover 7L
2suchanalgorithm isnonexistent;itwouldbe
highlyusefultohave,especiallyforcodingtheory.Butitjustdoesn't
exist­
sofar.
Webeginourconsiderationwithtwoeasyresults.
Lemma
4.6.1.Letf(x)E
Q[x];then
where
u,m,ao,...,anareintegersandthe ao,
ab...,anhavenocommon
factorgreaterthan1(i.e.,arerelativelyprime)and
(u,m)=1.

182. Sec.6 PolynomialsOver theRationals 167
ProofSincef(x)EQ[x],f(x)=qox
n+qlX
n
-
l +...+qn,wherethe
qiarerationalnumbers.So fori=0,1,2,...,n,qi=bilci'wherebi'ciare
integers.Thus
clearingofdenominatorsgivesus
f(x)=COC
l
.1..C
n
(uox
n
+UlX
n
-l
+ · · · +un),
wheretheUiareintegers.Ifwisthegreatestcommondivisorofun,Ul,...,Un'
theneachUi=wai,whereao,
ab...,anarerelativelyprimeintegers.Then
cancelingoutthegreatestcommonfactorofwandCOCl•••C
ngivesus
whereU,marerelativelyprimeintegers,asisclaimedin thelemma.D
Thenextresultisaresultaboutaparticularhomomorphicimageof
R[x]foranyringR.
Lemma4.6.2.IfRisanyringandIanidealofR,thenI[x],thepoly­
nomialringin
xoverI,isanidealofR[x].Furthermore,R[x]II[x]
==(RII)[x],
thepolynomialringin xoverRII.
ProofLetR=RII;thenthereisahomomorphismq;:R~R,defined
byq;(a)=a+I,whosekernelisI.Define<I>:R[x]~R[x]by:If
f(x)=aox
n
+a1x
n
-
1
+...+an,
then
<I>(f(x»=q;(ao)x
n
+q;(a1)x
n
-
1
+...+q;(an).
Weleaveit tothereadertoprovethat<I>isahomomorphismofR[x]onto
R[x].Whatisthekernel,Ker<1>,of<I>?Iff(x)=aox
n
+...+anisinKer<1>,
then<I>(f(x»=0,the°elementofR[x].Since
<I>(f(x»=q;(ao)x
n
+q;(a1)x
n
-
1
+ · · . +q;(an)=0,
weconcludeq;(ao)=0,q;(al)=0,...,q;(an)=0,bytheverydefinitionof
whatwemeanbytheO-polynomialinapolynomialring. Thuseachaiisin

183. 168 RingTheory Ch.4
thekernelof'P,whichhappens tobeI.Becauseao,al,...,anareinI,[(x)=
aox
n
+alX
n
-
l
+...+anisinI[x].SoKer<1>CI[x].ThatI[x]CKer<1>isim­
mediatefromthedefinition
ofthemapping
<1>.HenceI[x]=Ker<1>.Bythe
FirstHomomorphism
Theorem(Theorem4.3.3),thering I[x]isthenan
ideal
ofR[x]and
R[x]~R[x]/Ker<1>=R[x]ll[x].Thisprovesthelemma,re­
membering
that
R=RII.0
Asaveryspecialcase ofthelemmawehavethe
Corollary.Let7Lbetheringofintegers,paprimenumberin 7L,and
I=(p),theidealof7Lgeneratedby p.Then7L[x]ll[x]~7L
p[x].
ProofSince7L
p
~TLII,thecorollaryfollowsbyapplyingthelemmato
R=7L.0
Weareready toprovethefirst ofthetwomajorresultsweseekinthis
section.
Theorem4.6.3(Gauss'Lemma). If[(x)E7L[x]isamonicpolynomial
and
[(x)=a(x)b(x),wherea(x)andb(x)arein
Q[x].Then[(x)=
al(x)bl(x), whereal(x),b
l(x)aremonicpolynomialsin 7L[x]anddeg al(x)=
dega(x),degb
l(x)=degb(x).
ProofSupposethat[(x)E7L[x]andthat[(x)=a(x)b(x),wherea(x),
b(x)
E
Q[x],anddeg a(x)=S,degb(x)=r.ByLemma4.6.1,wecanex­
presseach
ofa(x),b(x) asaproduct ofarationalnumberandapolynomial
withintegercoefficients.Moreprecisely,
a(x)=
.!!l(a'x
S
+a'x
s
-1
+...+a')=
.!!la(x)
m
1
0 1 smIl'
wherea~,aI,...,a;arerelativelyprimeintegers and
b(x)=!!1-(b,x
r
+b'x
r
-1
+...+b')=
!!1-b(x)
m
2
0 1
rm
2
1 ,
where
b~,bl,...,b;arerelativelyprime.Thus
where
vandwarerelativelyprime,bycanceling outthecommonfactorof
UlU2andmlm2.Therefore,w[(x)=val(x)bl(x), and[(x),al(x),b
l(x)are
allin7L[x].Ofcourse,wemayassumewithnoloss ofgeneralitythatthelead­
ingcoefficients
ofal(x)andb
l(x)arepositive.
If
w=1,then,since [(x)ismonic,weget that
vabb~=1andthisleads

184. Sec.6 PolynomialsOver theRationals 169
easilyto v=1,ab=bb=1andso f(x)=al(x)bl(x), wherebothal(x)and
b
l(x)aremonicpolynomialswithintegercoefficients.This ispreciselythe
claimofthetheorem,sincedeg
al(x)=dega(x)anddegb
l(x)=degb(x).
Supposethen thatw
=1=1;thusthereisaprimepsuchthatpIwand,
since(v,w)=1,p~v.Also,sincethecoefficientsab,at,...,a;ofal(x)
arerelativelyprime,there isanisuchthatp~ai;similarly,there isaj
suchthat p~bj.LetI=(p)betheidealgeneratedby pinZ;then
Z/I:::=7Lpand,bytheCorollaryto Lemma4.6.2,Z[x]/I[x]:::=7Lp[x],soisanin­
tegraldomain.However,since
pIw,
lV,theimage ofwinZ[x]/I[x],is0,
andsincep~v,vtheimage ofvinZ[x]/I[x]isnotO.Thus01(x)=
val(x)bl(x),wherev¥0andaleX)¥0,bl(x)¥0becausep~a;andp~bjfor
thegiven
i,jabove.Thiscontradicts that
Z[x]/I[x]isanintegraldomain.So
w=1=1isnotpossible,andthe theoremisproved.D
Itmightbeinstructiveforthe readertotrytoshowdirectlythatif
x
3
+6x-7istheproductoftwopolynomialshavingrationalcoefficients,
thenit
isalreadytheproductoftwomonicpolynomialswithintegercoeffi­
cients.
Oneshouldsaysomethingabout C.F.Gauss(1777-1855),consideredbymany
tobethegreatestmathematicianever.Hiscontributionsinnumbertheory,al­
gebra,geometry,andsoon,are
ofgiganticproportions.Hiscontributionsin
physicsandastronomyarealsoofsuchagreatproportionthathe
isconsidered
byphysicistsasoneoftheirgreats,andbytheastronomersasoneoftheimpor­
tantastronomersofthepast.
Asweindicatedatthebeginning ofthissection,irreduciblepolynomi­
als
ofdegreenoveragivenfield Fmaybevery hardtocomeby.However,
overtherationals,duetothenexttheorem,theseexistinabundanceandare
veryeasytoconstruct.
Theorem4.6.4(TheEisensteinCriterion). Letf(x)=x
n+alX
n
-1
+...+anbeanonconstantpolynomialwithintegercoefficients.Suppose
thatthere
issomeprime psuchthatpial,pIa2,...,pIan,butp2
~an.Then
f(x)isirreducibleinQ[x].
ProofSupposethatf(x)=u(x)v(x),whereu(x),vex) areofpositive
degree
andarepolynomialsin
Q[x].ByGauss'Lemmawemayassume that
bothu(x)andvex)aremonicpolynomialswithintegercoefficients. Let
I=(p)betheidealgeneratedby pinZ,andconsiderZ[x]/I[x],whichisan
integraldomain,sinceweknowbytheCorollaryto
Lemma4.6.2that
7L[x]/I[x]
:::=(Z/I)[x]:::=Zp[x].Theimageof f(x)=x
n+alX
n
-1
+...+anin
Z[x]/I[x]isx
n
,sinceplab...,pIan.Soifu(x)istheimage ofu(x)andvex)

185. 170 RingTheory Ch.4
thatof vex)inZ[x]/I[x],thenx
n=
u(x)v(x).Sinceu(x)lx
n
,
v(x)lx
n
in
Z[x]/I[x],
wemusthavethatu(x)=x
r
,vex)=x
n
-
r
forsome1::;r<n.Butthen
u(x)=x
r+pg(x)andv(x)=x
n
-
r+ph(x),whereg(x)andhex)arepolyno­
mialswithintegercoefficients.Since
u(x)v(x)=x
n+pxrh(x)+pxn-rg(x)+
p2g(x)h(x),andsince1 ::;r<n,theconstanttermof u(x)v(x)isp2st,where
sistheconstantterm ofg(x)andttheconstanttermof hex).Becausef(x)=
u(x)v(x),theirconstanttermsareequal,hence an=p2st.Sincesandtarein­
tegers,wegetthat
p
2
1an,acontradiction.Inthiswayweseethat f(x)isirre­
ducible.
D
WegivesomeexamplesoftheusetowhichtheEisensteinCriterion
canbeput.
1.Letf(x)=x
n
-p,p anyprime.Thenoneseesataglancethat f(x)is
irreduciblein
Q[x],fortheEisensteinCriterionapplies.
2.Letf(x)=x
5
-
4x+22.Since2122,2
2t22and2dividestheotherrel­
evantcoefficients
off(x),theEisensteinCriteriontellsusthat f(x)isir­
reduciblein
Q[x].
3.Letf(x)=XlI-6x
4
+12x
3
+36x-6.Weseethat f(x)isirreducible
inQ[x]byusingeither2 or3tochecktheconditions oftheEisenstein
Criterion.
4.Letf(x)=5x
4
-
7x+7;f(x)isnotmonic,butwecanmodify f(x)
slightlytobeinapositionwherewecanapplytheEisensteinCriterion.
Let
g(x)=5
3
f(x)=5
4
x
4
- 7·5
3
x+7 .53=(5X)4-175(5x)+875.
Ifwelet y=5x,theng(x)=hey)=y4-175y+875.Thepolynomial
hey)isirreduciblein
Z[y]byusingtheprime7andapplyingtheEisen­
steinCriterion.Theirreducibility
ofh(y)impliesthat ofg(x),andso
that
off(x),in
Q[x].
ThissuggestsaslightgeneralizationoftheEisensteinCriterionto
nonmonicpolynomials.(SeeProblem4.)
5.Letf(x)=x
4+x
3
+x
2+X+1;asitstandswecannot, ofcourse,
applytheEisensteinCriterionto
f(x).Wepasstoapolynomial g(x)
closelyrelatedto f(x)whoseirreducibilityin
Q[x]willensurethatof
f(x).Letg(x)=f(x+1)=(x+1)4+(x+1)3+(x+1)2+(x+1)+
1=x
4+5x
3
+10x
2
+lOx+5.TheEisensteinCriterionappliesto
g(x),usingtheprime 5;thusg(x)isirreduciblein
Q[x].Thisimplies
that
f(x)isirreduciblein
Q[x].(SeeProblem1.)

186. Sec.6 PolynomialsOvertheRationals 171
GottholdEisenstein(1823-1852)inhisshortlife madefundamentalcontribu­
tionsinalgebra
andanalysis.
PROBLEMS
1.InExample 5,showthatbecauseg(x)isirreduciblein
Q[x],thensois
f(x).
2.Provethatf(x)=x
3
+3x+2isirreducibleinQ[x].
3.Show thatthereisaninfinitenumberofintegersasuchthatf(x)
x
7
+15x
2
-
30x+aisirreduciblein
Q[x].Whata'sdoyousuggest?
4.Prove
thefollowinggeneralization oftheEisensteinCriterion. Let
f(x)=aox
n+alX
n
-
1
+...+anhaveintegercoefficients andsuppose
thatthereisaprimepsuchthat
p
(ao,pial'pia2,•••,pIan-I,PIan,
butp2(an;thenf(x)isirreducibleinQ[x].
5.Ifpisaprime,show thatf(x)=x
p
-1
+x
p
-2
+...+x+1isirreducible
in
Q[x].
6.LetFbethefieldand'PanautomorphismofF[x]suchthat'P(a)=afor
every
aEF.Iff(x)EF[x],provethatf(x)isirreduciblein F[x]ifand
onlyifg(x)=
'P(f(x»is.
7.LetFbeafield.Define themapping
cp:F[x]~F[x]by'P(f(x»=f(x+1)
forevery
f(x)EF[x].Provethat
'PisanautomorphismofF[x]suchthat
'P(a)=aforeveryaEF.
8.LetFbeafield andb=1=0anelementofF.Definethemapping
'P:F[x]~F[x]by'P(f(x»=f(bx)foreveryf(x)EF[x].Provethat'Pis
anautomorphismofF[x]suchthat'P(a)=aforeveryaEF.
9.LetFbeafield,b=1=0,celementsofF.Definethemapping
'P:F[x]~F[x]by'P(f(x»=f(bx+c)forevery f(x)EF[x].Provethat
'PisanautomorphismofF[x]suchthat'P(a)=aforeveryaEF.
10.Let'PbeanautomorphismofF[x],whereFisafield,such that'P(a)=a
foreveryaEF.Provethatiff(x)EF[x],thendeg'P(f(x»=degf(x).
11.Let'PbeanautomorphismofF[x],whereFisafield,such that'P(a)=a
foreveryaEF.Provethereexistb=1=0,cin Fsuchthat'P(f(x»=
f(bx+c)forevery f(x)EF[x].

187. 172 RingTheory Ch.4
12.Findanonidentityautomorphism'PofQ[x]suchthat'P
2
istheidentity
automorphism
of
Q[x].
13.ShowthatinProblem 12youdonotneedtheassumption'P(a)=afor
every
aE
QbecauseanyautomorphismofQ[x]automaticallysatisfies
'P(a)=aforeveryaEQ.
14.LetCbethefield ofcomplexnumbers.Givenaninteger n>0,exhibit
anautomorphism'PofC[x]ofordern.
7.FIELDOF QUOTIENTS OFANINTEGRALDOMAIN
Giventheintegraldomain7L,theringofintegers,thenintimatelyrelatedto7L
isthefieldQofrationalnumbersthatconsistsofallfractionsofintegers;that
is,allquotientsmin,wherem,n=1=0arein7L.Notethatthere isnounique
wayofrepresenting~,say,inQbecause~= ~=(-7)/(-14)=....
Inotherwords,weareidentifying~with~,(-7)/(-14),andsoon.Thissug­
geststhatwhat
isreallygoingoninconstructingtherationalsfromtheinte­
gers
issomeequivalencerelationonsomesetbasedontheintegers.
Therelationof
Qto7Lcarriesovertoanyintegraldomain D.Givenan
integraldomain
D,weshallconstructafield F
~Dwhoseelementswillbe
quotients
albwitha,bED,b
=1=O.Wegothroughthisconstructionformally.
Let
Dbeanintegraldomain,andletS ={(a,b)Ia,bED,b
=1=O};Sis
thusthesubset ofDXD-theCartesianproductof Dwithitself-inwhich
thesecondcomponent
isnotallowedtobe o.Thinkof (a,b)asalbforamo­
ment;ifso,whenwouldwewanttodeclarethat
(a,b)=(e,d)?Clearly,we
wouldwantthisif alb=eld,whichin Ditselfwouldbecome ad=be.With
thisas
ourmotivatingguidewedefinearelation
---onSbydeclaring:
(a,b)---(e,d)for(a,b),(e,d)inS,ifandonlyif ad=be.
Wefirstassertthatthisdefinesanequivalencerelationon S.Wego
throughthethreerequirementsforanequivalencerelationtermbyterm.
1.(a,b)
---(a,b),forclearlyab=ba(sinceDiscommutative).So---is
reflexive.
2.(a,b)---(e,d)impliesthat(e,d)---(a,b),for(a,b)---(e,d)means
ad=be;for(e,d)---(a,b)weneedeb=da,butthisistrue,since
eb=be=ad=da.So---issymmetric.
3.(a,b)---(e,d),(e,d)---(e,f)impliesthat ad=be,ef=de,soadf=
bef=bde;butd=1=0andweareinanintegraldomain,hence af=
befollows.Thissaysthat (a,b)---(e,f).Sotherelation istransitive.

188. Sec.7 FieldofQuotientsofanIntegralDomain 173
Wehaveshown that'"'-'definesanequivalencerelationonS.LetFbe
thesetofalltheequivalenceclasses [a,b]oftheelements(a,b)ES.Fisour
requiredfield.
ToshowthatFisafield,we mustendowitwithanadditionandmulti­
plication.First
theaddition;whatshoulditbe?Forgettingallthefancytalk
aboutequivalencerelationandthelike,wereally want[a,b]tobealb.Ifso,
whatshould[a,b]+[e,d]beotherthantheformallycalculated
~+£.=ad+be?
b d bd.
Thismotivatesus todefine
[a,b]+[e,d]=[ad+be,bdl. (1)
Notethatsinceb
=1=0,d=1=°andDisadomain,thenbd=1=0,hence
[ad+be,bd]isalegitimateelementofF.
Asusualwe areplaguedwithhaving toshowthattheadditionsode­
finedin
Fiswell-defined.Inotherwords,we mustshowthatif[a,b]=
[a',b']and[e,d]=[e',d'],then[a,b]+[e,d]=[a',b']+[e',d'].From(1)
wemustthusshow that[ad+be,bd]=[a'd'+b'e',b'd'],whichistosay,
(ad+be)b'd'=bd(a'd'+b'e').Since[a,b]=[a',b']and[e,d]=[e',d'],
ab'=ba'andcd'=de'.Therefore,(ad+be)b'd'=ab'dd'+bb'ed'=
ba'dd'+bb'dc' =(a'd'+b'e')bd,asrequired.Thus"+"iswell-definedin F.
Theclass[0, b],b
=1=0,actsas the°under"+,"wedenoteitsimplyas0,
andtheclass[-a,b]isthenegativeof[a,b].ToseethatthismakesofFan
abeliangroupiseasy,butlaborious,forallthatreallyneedsverificationis
theassociativelaw.
Now
tothemultiplicationin F.Againmotivatedbythinkingof[a,b]as
alb,wedefine
[a,b][e,d]=[ae,bdl. (2)
Againsinceb
=1=0,d=1=0,wehavebd=1=0,sotheelement[ae,bd]is
alsoalegitimate
elementofF.
Asforthe"+"wemustshowthattheproductsointroducediswell-de­
fined;thatis,if[a,b]=[a',b'],[e,d]=[e',d'],then
[~,b][e,d]=[ae,bd]=[a'e',b'd']=[a',b/][e', d'].
Weknowthatab'=ba'andcd'=de',soaeb'd'=ab'cd'=ba'de'=bda'e',
whichisexactlywhatweneedfor[ae,bd]=[a'e',b'd'].Thustheproductis
well-definedin
F.
Whatactsas1in F?Weclaimthatforanya
=1=0,b=1=°inD,[a,a]=
[b,b](sinceab=ba)and[e,d][a,a]=[ea,da]=[e,d],since(ea)d=(da)e..

189. 174 RingTheory Ch.4
So[a,a]actsas 1,andwewriteitsimplyas1 =[a,a](foralla=1=°inD).
Given[a,b]=1=0,thena=1=0,so[b,a]isinF;hence,because [a,b][b,a]=
[ab,ba]=[ab,ab]=1,[a,b]hasaninversein F.Allthatremainstoshow
thatthenonzeroelementsof Fformanabeliangroupunderthisproduct is
theassociativelawandcommutativelaw.Weleavethesetothereader.
Toclinch
thatFisafield,weneedonlynowshowthedistributivelaw.
But[ad+be,bd][e,f]=[(ad+be)e,bdf],so
([a,b]+[e,d])[e,f]=[ade+bee,bdf],
while
[a,b][e,f]+[e,d][e,f]
=rae,bf]+[ee,df]=[aedf+bfee,bdf2]
=[(ade+bee)f,bdf2]=[ade+bee,
bdf][f:f]
=[ade+bee,bdf],
whichwehaveseen is([a,b]+[e,d])[e,fl.Thedistributivelaw isnowes­
tablished,so
Fisafield.
Let
a
=1=°beafixedelementin Dandconsider[da,a]foranydED.
Themapq;:d~[da,a]isamonomorphismofDintoF.Itiscertainly1-1,
forifq;(d)=[da,a]=0,thenda=0,sod=0,sinceDisanintegraldo­
main.Also,q;(dld2
)=[dld2a, a]whileq;(dl)q;(d2
)=[dla,a][d
2a, a]
[dld2a
2
,
a
2
]
=[dld2a,a][a,a]=[dld2a,a]=q;(dld2
).Furthermore,
[d
1a,a]+[d2a,a]=[dta
2+a
2
d2, a
2
]
=[dta+d2a,a][a,a]
=[edt+d2)a,a]
so
q;(dl+d
2
)=[(dl+dl)a,a]=[dla,a]+[d
2a,a]=q;(dl)+q;(d
2
).Thus
q;mapsDmonomorphicallyinto F.So,DisisomorphictoasubringofF,and
wecanthusconsider
Das"embedded"inF.Weconsidereveryelement
[a,b]ofFasthefraction alb.
Theorem4.7.1.LetDbeanintegraldomain.Thenthereexistsafield
F
:JDwhichconsistsofallfractions alb,asdefinedabove,ofelementsin D.
ThefieldFiscalledthe fieldofquotientsofD.WhenD=7L,thenFis
isomorphictothefieldQofrationalnumbers.Also,if Disthedomainof
evenintegers,then
Fisalsotheentirefield
Q.
Whatwedidaboveinconstructingthefield ofquotientsof Dwasa
long,formal,wordy,andprobablydullway
ofdoingsomethingthat isbyits

190. Sec.7 FieldofQuotientsofanIntegralDomain 175
naturesomethingverysimple. Wereallyaredoingnothingmorethanform­
ingallformalfractions
alb,a,b
=1=0inD,whereweaddandmultiplyfrac­
tionsasusual.However,it
issometimesnecessarytoseesomethingdoneto
itslastdetail, painfulthoughitmaybe.Mostofushadneverseenareally
formal
andpreciseconstructionoftherationalsfromtheintegers.Now
thatwehaveconstructedFfromDinthisformalmanner,forgetthede­
tails
andthinkofFasthesetofallfractionsofelementsofD.
PROBLEMS
1.Provetheassociativelaw ofadditionin F.
2.Provethecommutativelaw ofadditionin F.
3.ProvethattheproductinFiscommutativeandassociative.
4.
IfKisanyfieldthatcontainsD,showthatK
~F.(SoFisthesmallest
field
thatcontainsD.)

191. 5
FIELDS
Thenotionofaringwasunfamiliarterritoryformostof us;thatofafield
touchesmorecloselyonourexperience.Whiletheonlyring,otherthana
field,thatwemighthaveseeninourearlytrainingwastheringofintegers,
wehadabitmoreexperienceworkingwithrationalnumbers,realnumbers,
and,someofus,complexnumbers,insolvinglinearandquadraticequations.
Theabilitytodividebynonzeroelementsgave
usabitofleeway,which we
mightnothavehadwiththeintegers,insolvingavarietyofproblems.
Soatfirstglance,whenwestartworkingwithfieldswefeelthatweare
onhomeground.Aswepenetratedeeperintothesubject,westartrunning
acrossnewideasandnewareasofresults.Onceagainwe'llbeinunfamiliar
territory,buthopefully,aftersomeexposuretothetopic,thenotions
willbe­
comenaturaltous.
Fieldsplayanimportantroleingeometry,
inthetheoryofequations,
and
incertainveryimportantpartsofnumbertheory.Weshalltouchupon
eachoftheseaspects
asweprogress.Unfortunately,becauseofthetechnical
machinery
wewouldneedtodevelop, wedonotgointoGaloistheory,avery
beautifulpartofthesubject.Wehopethatmanyofthereaders
willmakecon­
tactwithGaloistheory,andbeyond,intheirsubsequentmathematicaltraining.
1.EXAMPLES OFFIELDS
Let'srecallthatafield Fisacommutativeringwithunitelement1suchthatfor
everynonzero
aEFthereisanelementa-IEFsuchthat aa-
1
=1.Inother
words,fieldsare"somethinglike"therationals
Q.Butaretheyreally?The in-
176

192. Sec.1 ExamplesofFields177
(verify!),
tegersmod
p,7L
p
,wherepisaprime,formafield;in 7L
pwehavetherelation
pI=,I+1+...+I,=0
(ptimes)
Nothingakin
tothishappensin
Q.Thereareevensharperdifferences
among
fields-howpolynomialsfactor overthem,special propertiesofwhich
we'llseeinsomeexamples,
andsoon.
Webeginwithsomefamiliarexamples.
Examples
1.
Q,thefieldofrationalnumbers.
2.IR,thefieldofrealnumbers.
3.C,thefieldofcomplexnumbers.
4.LetF={a+biIa,bEQ}cC.ToseethatFisafieldisrelativelyeasy.
Weonlyverifythatifa+bi=1=0isinF,then(a+bi)-lisalsoin F.But
whatis(a+bi)-l?Itismerely
a bi
(a
2
+b
2
)
(a
2
+b
2
)
andsincea
2
+b
2
=1=0andisrational,thereforeal(a
2
+b
2
)
andalso
bl(a
2
+b
2
)
arerational,hence(a+bi)-lisindeedinF.
5.LetF={a+
bV21a,bEQ}CIR.AgaintheverificationthatFisafield
isnottoohard.Here,too,weonlyshow theexistenceofinversesin Ffor
thenonzeroelementsofF.Supposethata+bVi=1=0isinF;then,since
V2isirrational,a
2
-
2b
2=1=O.Because
(a+bV2)(a-bVi)=a
2
-
2b
2
,
weseethat(a+
bV2)(ale-Vible)=1,wheree=a
2
-
2b
2
•
There­
quiredinversefor
a+
bV2isale-Vible,whichiscertainlyanele­
mentofF,sincealeandblearerational.
6.LetFbeanyfieldandletF[x]betheringofpolynomialsin xoverF.
SinceF[x]isanintegraldomain,ithasafield ofquotientsaccordingto
Theorem4.7.1,whichconsists ofallquotientsf(x)lg(x),wheref(x)and
g(x)areinF[x]andg(x)=1=O.ThisfieldofquotientsofF[x]isdenotedby
F(x)andiscalledthefieldofrationalfunctionsinxoverF.
7.7L
p
,
theintegersmodulotheprimep,isa(finite)field.
8.InExample2inSection4 ofChapter4wesawhow toconstructafield
havingnineelements.
Theseeightexamples
arespecificones.Using thetheoremswe
have
provedearlier,wehave somegeneralconstructions offields.

193. 178 Fields Ch.5
9.IfDisanyintegeraldomain,thenithasafieldofquotients,byTheorem
4.7.1,whichconsistsofallthefractions
alb,whereaandbareinDand
b
=1=O.
10.IfRisacommutativeringwithunitelement1and Misamaximalideal
of
R,thenTheorem4.4.2tells usthatRIMisafield.
Thislastexample,forparticular
R's,willplayanimportantrole inwhat
istofollowinthischapter.
Wecouldgoon,especiallywithspecialcasesofExamples9and
10,to
seemoreexamples.
Butthe10thatwedidseeaboveshowusacertainvari­
etyoffields,andweseethatit
isnottoo hardtorunacrossfields.
InExamples7and8thefieldsarefinite.IfF
isafinitefieldhaving qel­
ements,viewing
Fmerelyasanabelian groupunderitsaddition, "+,"we
have,byTheorem2.4.5,that qx=0forevery xEF.Thisisabehaviorquite
distinctfromthatwhichhappensinthefieldsthatweareusedto,such
asthe
rationalsandreals.
Wesingleoutthiskindofbehaviorinthe
Definition.Afield
Fissaidtohave(or,tobeof) characteristicp
=1=0
ifforsomepositiveinteger
p,px=0forall xEF,andnopositiveinteger
smallerthan
penjoysthisproperty.
Ifafield
Fisnotofcharacteristicp
=1=0foranypositiveinteger p,we
callitafieldof
characteristicO.So
Q,IR,Carefieldsofcharacteristic 0,while
7L
3isofcharacteristic3.
Inthedefinitiongivenabovetheuseoftheletter pforthecharacteris­
tic
ishighlysuggestive,for wehavealwaysusedptodenoteaprime.Infact,
asweseeinthenexttheorem,thisusageof
pisconsistent.
Theorem5.1.1.Thecharacteristicofafield
iseither0 oraprime
number.
ProofIfthefield Fhascharacteristic0,thereisnothingmoretosay.
Supposethenthat
mx=0forall xEF,wheremisapositiveinteger.Let p
bethesmallestpositiveintegersuchthat px=0forall xEF.We claimthat
pisaprime.Forifp=UV,whereU>1andv>1areintegers,thenin F,
(u1)(v1)=(uv)l=0,where1 istheunitelementof F.ButF,beingafield, is
anintegraldomain(Problem
1);therefore,u1=0orvI=O.Ineithercase
wegetthat0
=(u1)(x)=ux[or,similarly,0 =(v1)x=vx]foranyx inF.
Butthiscontradictsourchoiceof pasthesmallestintegerwiththisproperty.
Hence
pisaprime.D

194. Sec.1 ExamplesofFields179
NotethatwedidnotusethefullforceoftheassumptionthatFwasa
field.
WeonlyneededthatFwasanintegraldomain(with1).Soifwedefine
thecharacteristicofanintegraldomaintobe0orthesmallestpositiveinte­
ger
psuchthatpx=0forallxEF,weobtainthesameresult.Thusthe
Corollary.IfDisanintegraldomain, thenitscharacteristiciseither0
oraprimenumber.
PROBLEMS
1.Showthatafieldisanintegraldomain.
2.Prove
theCorollaryevenif Ddoesnothavea unitelement.
3.Givenaring R,letS=R[x]betheringofpolynomialsin xoverR,and
letT=S[y]betheringofpolynomialsin yoverS.Showthat:
(a)Anyelementf(x,y)inThastheform
'L'LaijXiyj,wheretheaijare
inR.
(b)Intermsoftheformoff(x,y)inTgivenin Part(a),givethecondi­
tionfor
theequalityoftwoelementsf(x,y)andg(x,y)inT.
(c)Intermsoftheformfor f(x,y)inPart(a),give theformulafor
f(x,y)+g(x,y),forf(x,y),g(x,y)inT.
(d)Givetheformfor theproductoff(x,y)andg(x,y)iff(x,y)and
g(x,y) areinT.(Tiscalledtheringofpolynomialsintwovariables
overR,
andisdenotedbyR[x,y].)
4.IfDisanintegraldomain,show thatD[x,y]isanintegraldomain.
5.
IfFisafieldandD=F[x,y],thefieldofquotientsofDiscalledthe
fieldofrationalfunctionsintwovariablesover F,andisusuallydenoted
byF(x,y). GivetheformofthetypicalelementofF(x,y).
6.ProvethatF(x,y)isisomorphictoF(y,x).
7.IfFisafieldofcharacteristicp
=1=0,showthat(a+b)P=a
P+b
Pforall
a,bEF.(Hint:Usethebinomialtheoremandthefactthatpisaprime.)
8.IfFisafieldofcharacteristicp
=1=0,showthat(a+b)m=am+b
m
,
wherem=pn,foralla,binFandanypositiveintegern.
9.LetFbeafieldofcharacteristicp=1=0andletcp:F~Fbedefinedby
cp(a)=a
P
forallaEF.
(a)Showthatcpdefinesa monomorphismofFintoitself.
(b)Giveanexample ofafieldFwherecpisnotonto.(Veryhard.)
10.IfFisafinitefield ofcharacteristicp,showthatthemappingcpdefined
above
isonto,hence isanautomorphismofF.

195. 180 Fields Ch.5
2.ABRIEF EXCURSION INTOVECTORSPACES
Togetintothethingsweshouldliketodoinfieldtheory, weneedsome
technicalequipmentthat
wedonothave asyet.Thisconcernstherelationof
twofields
K
~Fandwhatwewouldliketoconsider assomemeasureofthe
sizeof
Kcomparedtothatof F.Thissize iswhatweshallcallthe dimension
ordegreeofKoverF.
However,intheseconsiderations,muchless isneededof Kthanthatit
beafield.Wewouldberemissifweprovedtheseresultsonlyforthespecial
contextoftwofields
K
~Fbecausethesameideas,proofs,andspirithold in
afarwidersituation.Weneedthenotionofa vectorspace overafield F.
Asidefromthefactthatwhatwedoinvectorspaceswillbeimportant inour
situationoffields,theideasdevelopedappear
inallpartsofmathematics.
Studentsofalgebramustseethesethingsatsomestageoftheirtraining.
An
appropriateplace isrighthere.
Definition.A
vectorspaceV overafield Fisanabeliangroupunder
"+"suchthatforevery
aEFandevery vEVthereisanelementavEV,
andsuchthat:
1.a(vl+V2)=aVl+aV2,foraEF,VbV2EV.
2.(a+(3)v=av+f3v,fora,f3EF,vEV.
3.a(f3v)=(af3)v,fora,f3EF,vEV.
4.Iv=vforallvEV,where1 istheunitelementof F.
Indiscussingvector spaces-whichwewilldovery briefly-weshall
uselowercaseLatinlettersforelementsof
VandlowercaseGreeklettersfor
elementsof
F.
Ourbasicconcernherewillbewithonlyoneaspectofthetheoryofvec­
torspaces:thenotionofthedimensionof
VoverF.Weshalldevelopthisno­
tion
asexpeditiouslyaspossible,notnecessarilyinthebest ormostelegant
way.Wewouldstronglyadvisethereaderstoseetheothersidesofwhat
is
doneinvectorspaces inotherbooksonalgebraorlinearalgebra(forinstance,
ourbooks
APrimeronLinearAlgebra andMatrixTheoryandLinearAlgebra).
Beforegettingdowntosomeresults,welookatsomeexamples.We
leavetothereaderthedetails
ofverifying,ineachcase,thattheexample
really
isanexampleofavectorspace.
Examples
1.LetFbeanyfieldandlet V={(
al,a2,...,an)IaiEFforalli}betheset
ofn-tuplesover
F,withequalityandadditiondefinedcomponent-wise.For

196. Sec.2 ABriefExcursion intoVectorSpaces 181
v=(aba2'...'an)andf3EF,definef3vbyf3v=(f3abf3a2'···'f3an).
Visavectorspaceover F.
2.LetFbeanyfieldandlet V=F[x]betheringofpolynomialsin xoverF.
Forgettingtheproductofanyarbitraryelementsof F[x]butusingonlythat
ofapolynomialbyaconstant,forexample,wefindthat
f3(ao+alx+...+anx
n
) =f3ao+f3a
1x+ . · .+f3anx
n
.
Inthisway Vbecomesavectorspaceover F.
3.LetVbeasinExample2andlet W={f(x)EVIdeg(f(x»::;n}.ThenW
isavectorspaceover F,andWCVisasubspaceof Vinthefollowingsense.
Definition.A
subspaceofavectorspace Visanonemptysubset Wof
Vsuchthat
awEWandWI+W2EWforallainFandW,WbW2EW.
Notethatthedefinitionofsubspace WofVimpliesthat Wisavector
spacewhoseoperationsarejustthoseof
Vrestrictedtotheelementsof W.
4.LetVbethesetofallreal-valueddifferentiablefunctionson [0,1],the
closedunitinterval,withtheusualadditionandmultiplicationofafunction
byarealnumber.Then
Visavectorspaceover
IR.
5.LetWbeallthereal-valuedcontinuousfunctionson [0,1],againwiththe
usualadditionandmultiplicationofafunctionbyarealnumber.
W,too,isa
vectorspaceover
IR,andtheVinExample4 isasubspaceof W.
6.LetFbeanyfield, F[x]theringofpolynomialsin xoverF.Let[(x)
beinF[x]andJ=(f(x»theidealof F[x]generatedby f(x).LetV=
F[x]/J,wherewedefine a(g(x)+J)=ag(x)+J.Visthenavectorspace
over
F.
7.Let
IRbetherealfieldandlet Vbethesetofallsolutionstothedifferen­
tialequation
d
2y/dx
2
+y=0.Visavectorspaceover
IR.
8.LetVbeanyvectorspaceoverafield F,VI,V2,...' Vnelementsof V.Let
(VbV2'···,Vn)={aIVI+a2v2+...+anVnIaI,a2,··.,anEF}.Then
(VbV2,,Vn)isavectorspaceover Fandisasubspaceof V.Thissubspace
(VbV2',Vn)iscalledthe subspaceofVgeneratedorspanned byVb...,Vn
overF;itselementsarecalled linearcombinations ofVI,...,Vn.Weshall
soonhaveagreatdealtosayabout
(VI,V2'...,Vn).
9.LetVandWbevectorspacesoverthefield Fandlet V
EBW=
{(v,w)IVEV,wEW},withequalityandadditiondefinedcomponentwise,

197. 182 Fields Ch.5
andwherea(v,w) =(av,aw). ThenVEf)Wiseasilyseen tobeavector
space
overF;itiscalledthedirectsum ofVandW.
10.LetK
~Fbetwofields,wheretheaddition"+"isthatofKandwhere
av,foraEFandvEKistheproduct,aselements ofthefieldK.ThenCon­
ditions1
and2definingavectorspace aremerelyspecialcases ofthedistrib­
utivelaws
thatholdin K,andCondition3ismerelyaconsequence ofthe
associativityoftheproductinK.Finally,Condition 4isjusttherestate­
mentofthefactthat1istheunitelementofK.SoKisavectorspace
over
F.
Inonerespectthereisasharpdifferenceamongtheseexamples.We
specifythisdifferencebyexaminingsome
oftheseexamplesinturn.
1.InExample1,if
VI=(1,0,...,0),V2=(0,1,0,...,0),...'V
n=(0,0,...,1),
theneveryelementvinVhasa uniquerepresentationin theformv=
aivi+ ·..+ anvn,where
ab...,anareinF.
2.InExample3,if VI=1,V2=X,•••,Vi=xi-t,...,Vn+I=x
n
,thenevery
vEVhasa uniquerepresentationasv=aivi+...+anvn, withthe(ti
inF.
3.InExample7,everysolution ofd
2y/dx
2
+y=°isoftheuniqueform
y=
acosx+{3sinx,withaand{3real.
4.InExample8,everyvE(VI'...,vn)hasarepresentation-albeitnotnec­
essarily
unique-asv=aIVt+...+anVnfromtheverydefinitionof
(Vt,...,v
n
).Uniquenessofthisrepresentation dependsheavily ontheele­
ments
VI,...,Vn.
5.Inthespecialcase ofExample10,where K=
C,thefieldofcomplex
numbers,
andF=
IRthatoftherealnumbers, theneveryVECisofthe
uniqueformv=a+f3i,a,(3EIR.
6.ConsiderK=F(x)~F,thefieldofrationalfunctionsin xoverF.We
claim-andleavetothereader-thatwecannotfindanyfiniteset ofele­
mentsin
Kwhichspans KoverF.Thisphenomenonwasalsotrueinsome
oftheotherexampleswegave ofvectorspaces.
Thewholefocus ofourattentionherewillbeonthisnotionofavector
spacehavingsomefinitesubset
thatspansit overthebasefield.
Beforestartingthisdiscussion,wemustfirstdispose
ofalistofformal
propertiesthatholdinavectorspace.You, dearreader,arebynowso

198. Sec.2 ABriefExcursion intoVectorSpaces 183
sophisticatedindealingwiththeseformal,abstractthings thatweleavethe
proofofthenextlemma toyou.
Lemma
5.2.1.IfVisavectorspace overthefieldF,then,forevery
aEFandeveryuEV:
(a)aO=0,where0 isthezero-elementofV.
(b)Ou=0,wherethefirst0 isthezeroin F.
(c)au=0impliesthata=0oru=o.
(d)(-a)u=-(au).
Inviewofthislemmaweshall notrunintoanyconfusionifweuse the
symbol0 bothforthezero ofFandthatofV.
Weforgetvectorspacesfora momentandlookatsolutionsofcertain
systems
oflinearequationsinfields.Take,forexample, thetwolinearhomo­
geneousequationswithrealcoefficients,
xI+X2+X3=0and3xI -X2+X 3
=O.Weeasilysee thatforany Xl,X3suchthat4Xl+2X3=0andX2=
-(x1+x3)'wegetasolution tothissystem.Infact, thereexistsaninfinity of
solutionstothissystemotherthanthetrivialoneXl=0,X2=0,X3=O.Ifwe
lookatthisexample
andaskourselves:"Whyisthereaninfinityofsolutions
tothissystemoflinearequations?",wequicklycome totheconclusionthat,
becausetherearemorevariables
thanequations,wehave roomtomaneuver
toproducesolutions.This isexactlythesituation thatholdsmoregenerally,
asweseebelow.
Definition.
LetFbeafield;thenthen-tuple
(f3b•••,f3n),wherethe
f3iareinF,andnotallofthemare 0,issaidtobeanontrivialsolutioninF to
thesystemofhomogeneouslinearequations
al1x1+a12x2+...+alnxn=0
a21xl+a22x2+ · . · +a2nxn=0
(*) =0
wheretheaijareallin F,ifsubstitutingXl=f31,•••,xn=f3nsatisfiesall
theequations
of(*).
Forsuchasystem(*)wehave thefollowing

199. 184 Fields Ch.5
Theorem5.2.2. Ifn>r,thatis,ifthenumberofvariables(un­
knowns)exceeds
thenumberofequationsin(*), then(*)hasanontrivialso­
lutionin
F.
ProofThemethodisthat,whichsome ofuslearnedinhighschool, of
solvingsimultaneousequationsbyeliminating oneoftheunknownsandat
thesametimecutting thenumberofequationsdownbyone.
We
proceedbyinductiononr,thenumberofequations.Ifr=1,the
system(*)reduces
toallxl+...+alnXn=0,andn>1.Ifallthe
ali=0,
thenXl=X2=...=Xn=1isanontrivialsolution to(*).So,onrenumber­
ing,wemayassume
that
all=1=0;wethenhavethesolutionto(*),which is
nontrivial:X2=...=Xn=1andXl=-(I/all)(a12+..· +aln).
Supposethattheresultiscorrectfor r=kforsome kandsupposethat
(*)isasystemofk+1linearhomogeneousequationsin n>k+1vari­
ables.
Asabove,we mayassumethatsome
aij=1=0,and,withoutlossofgen­
erality,
that
all=1=0.
Weconstructa relatedsystem,(**), ofklinearhomogeneousequations
in
n-1variables;since n>k+1,wehavethatn-1>k,sowecanapply
induction
tothisnewsystem(**). Howdowegetthisnewsystem?Wewant
toeliminateXlamongtheequations.Wedosobysubtracting
ail/alltimes
thefirstequationfromtheithoneforeachofi=2,3,...,k+1.Indoing
so,we
endupwiththenewsystem ofklinearhomogeneousequationsin
n-1variables:
(**)
where
f3ij=aij-ail/allfori=2,3,...,k+1andj=2,3,...,n.
Since(**) isasystemofklinearhomogeneousequationsin n-1vari­
ablesand
n-1>k,byourinduction(**)hasanontrivialsolution
(1'2,...,I'n)
inF.Let1'1=-(a121'2+...+alnl'n)/all;weleaveit tothereadertoverify
thatthe(1'1,1'2'...,I'n)soobtainedisarequirednontrivialsolutionto(*).
Thiscompletes
theinductionandsoprovesthetheorem.D
Withthisresultestablished,we arefreetouseitin ourstudyofvector
spaces.
Wenowreturntothesespaces. Werepeat,foremphasis,something
wedefinedearlierinExample
8.

200. Sec.2 ABriefExcursion intoVectorSpaces 185
Definition.LetVbeavectorspaceover FandletVbVZ,...,V
nbein
V.Theelement VEVissaidtobealinearcombination ofVI,Vz,...,V
nif
V=aivi+...+anV
nforsome
aI,...aninF.
AsweindicatedinExample 8,theset (VI,Vz,...,V
n
)ofalllinearcom­
binations
of
Vbvz,...,V
nisavectorspaceover F,andbeingcontainedin V,
isasubspaceof V.Whyisitavectorspace? Ifaivi+...+anV
nand
f3IVI+...+f3nv naretwolinearcombinations ofVb...,Vn'then
(aiV
I+...+anvn)+(f3IVl+ . · . +f3nvn)
=(a
l+f31)VI+...+(an+f3n)v
n
bytheaxiomsdefiningavectorspace,andso isin(vI,...,Vn).IfyEFand
aIvl+...+anVnE(Vb...'Vn),then
and
isalsoin
(Vb...,v
n).Thus
(Vb...,v
n
)isavectorspace.Aswecalledit
earlier,it
isthesubspaceofVspannedoverFby
Vb...,Vn.
Thisleadsustotheultra-importantdefinition.
Definition.ThevectorspaceVoverF isfinitedimensionaloverFif
V=(Vb...,v
n
)forsomeVb...,V
ninV,thatis,ifVisspanned overFbya
finiteset
ofelements.
Otherwise,wesaythatV
isinfinitedimensionaloverFifit isnotfinite
dimensionalover
F.Notethatalthoughwehavedefinedwhat ismeantbya
finite-dimensionalvectorspace,westillhave
notdefinedwhat ismeantbyits
dimension.
Thatwillcomeinduecourse.
Supposethat
Visavectorspaceover FandVI,...,V
ninVaresuch
thateveryelement
Vin
(Vb...,v
n
)hasauniquerepresentationintheform
V=aIvl+...+anv
n
,where
ab...,anEF.Since
oE(VI,...,v
n
)and0=OV
I+...+OVn'
bytheuniquenesswehaveassumedweobtainthatif aivi+...+anV
n=0,
then
al=az=..· =an=O.Thispromptsasecondultra-importantdefini­
tion.
Definition.LetVbeavectorspaceover
F;thentheelementsVb.••,V
n
inVaresaidtobelinearlyindependentoverFif aivi+...+anV
n=0,
whereab...,anareinF,impliesthatal=az=...=an=O.

201. 186 Fields Ch.5
IftheelementsVb...,VninVarenotlinearlyindependentover F,
thenwesaythattheyare linearlydependentover F.Forexample,if
~isthe
fieldofrealnumbersand
Visthesetof3-tuplesover
~asdefinedinExam­
ple
1,then(0,0,1),(0,1,0),and(1,0,0)arelinearlyindependentover
~
(Prove!)while(1, -2,7),(0,1,0),and(1,-3,7)arelinearlydependentover
IR,since1(1,-2,7)+(-1)(0,1,0)+(-1)(1,- 3,7)=(0,0,0)isanontrivial
linearcombinationoftheseelementsoverIR,whichistheO-vector.
Notethatlinearindependencedependsonthefield
F.If
C~~arethe
complexandrealfields,respectively,thenCisavectorspaceover~,butitis
alsoavectorspaceoverCitself.Theelements 1,iinCarelinearlyindepen­
dentover~butarenotsooverC,sincei1+(-l)i=°isanontriviallinear
combinationof
1,iover
C.
Weprove
Lemma
5.2.3.IfVisavectorspaceover FandVI,...,UninVarelin­
earlyindependentover
F,theneveryelement VE
(Vb'..,vn)hasaunique
representation
as
with
ab...,aninF.
ProofSupposethat VE(VI,""vn)hasthetworepresentations as
U=alv+...+anvn=f3IUI+...+f3nvnwiththea'sandf3'sinF.This
gives
usthat
(al-f31)VI+...+(an-f3n)vn=0;sinceVb'.',Vnarelin­
earlyindependentover
F,weconcludethat
al-f31=0,...,an-f3n=0,
yieldingforustheuniquenessoftherepresentation. D
Howfiniteisafinite-dimensionalvectorspace?Tomeasurethis,calla
subset
VI,...,VnofVaminimalgeneratingset forVoverFifV=(VI,...,vn)
andnosetoffewerthan nelementsspans VoverF.
Wenowcometothethirdvitallyimportantdefinition.
Definition.If
Visafinite-dimensionalvectorspaceover F,thenthe
dimensionofVoverF,writtendimp(V), isn,thenumberofelements ina
minimalgeneratingsetfor
VoverF.
Intheexamplesgiven,
dim~(C) =2,since1,iisaminimalgenerating
setforCoverIR.However,dimc(C)=1.InExample1,dimp(V)=nandin
Example
3,dimp(V)=n+1.InExample7thedimensionof VoverFis2.
Finally,if (VI,...,Vn)CV,thendimp(vI,...,Vn)isatmostn.
Wenowprove

202. Sec.2 ABriefExcursion intoVectorSpaces 187
Lemma5.2.4. IfVisfinitedimensionalover Fofdimensionn andif
theelementsVb...,V
nofVgenerateVoverF,thenVI,...,V
narelinearly
independentover
F.
ProofSupposethatVI,••.,V
narelinearlydependentover
F;thusthere
isalinearcombinationaivi+...+anV
n=0,wherenotalltheaiareO.We
maysuppose,withoutloss
ofgenerality,that
al =1=0;thenVI=
(-1IaI)(a2v2+...+anv
n
).GivenVEV,becauseVb"',v
nisagenerating
setfor
VoverF,
V=f31Vl+ . · · +f3nvn
= (
-~)(azuz+ · · · +anun)+f3zuz+ · · · +f3nun;
thusV2'..•,V
nspanVoverF,contradictingthatthesubset
VbV2,.••,Vnisa
minimalgeneratingset
ofVoverF.D
Wenowcome toyetanotherimportantdefinition.
Definition.
LetVbeafinite-dimensionalvectorspace over
F;then
Vb..•,V
nisabasisofVoverFiftheelementsVb...,V
nspanVoverFand
arelinearlyindependentover F.
ByLemma5.2.4anyminimalgeneratingset ofVoverFisabasisofV
overF.Thus,finite-dimensionalvectorspaceshavebases.
Theorem5.2.5.Supposethat
Visfinitedimensionalover F.Thenanytwo
basesof
VoverFhavethesame numberofelements,andthisnumberisex­
actlydimp
(V).
ProofLet
Vb'••,V
nandWb•••,W
mbetwobases ofVoverF.We
wanttoshow
thatm=n.Supposethatm>n.BecauseVI,••.,V
nisabasis
ofVoverF,weknow thateveryelementin Visalinearcombination ofthe
VioverF.Inparticular,WI,...,W
mareeachalinearcombination of
Vb..•,V
n
overF.Thuswehave

203. 188 Fields Ch.5
wheretheaijareinF.
Consider
f31Wl+
Thesystemoflinearhomogeneousequations
alif31+a2if32+...+amif3m=0,i=1,2,...,n,
hasanontrivialsolutionin FbyTheorem5.2.2,since thenumberofvari­
ables,
m,exceedsthenumberofequations,n. Iff3l'...,13missuchasolution
in
F,then,bytheabove,f3lWl+ . · . +f3mwm=0,yetnotallthef3iare0.This
contradicts
thelinearindependenceof
Wb•••,W
moverF.Therefore,m::;n.
Similarly,n::;m;hencem=n.Thetheoremisnowproved,sinceaminimal
generatingset
ofVoverFisabasisofVoverFandthenumberofelements
inthisminimalgenerating
setisdimp(V),bydefinition.Therefore,bythe
above,n=dimp(V),completingtheproof.D
Afurtherresult,whichweshalluseinfieldtheory, ofasimilarnature
tothethingswehave beendoingis
Theorem5.2.6.LetVbeavectorspaceoverFsuchthatdimp(V)=
n.Ifm>n,thenanymelementsofVarelinearlydependentoverF.
ProofLet
Wb•.•,W
mEVandletVb...,V
nbeabasisofVoverF;
heren=dimp(V)byTheorem5.2.5.Therefore,
Theproofgivenin Theorem5.2.5,thatifm>nwecanfind f3l'...,13minF,
andnotall0,suchthatf3lWl+...+f3mwm=0,goesoverwordforword.
ButthisestablishesthatWl,...,W
marelinearlydependentoverF.D
Weclosethissectionwithafinal theoremofthesameflavorasthepre­
cedingones.
Theorem5.2.7.LetVbeavectorspaceoverFwithdimp(V) =n.
ThenanynlinearlyindependentelementsofVformabasis ofVoverF.

204. Sec.2 ABriefExcursion intoVectorSpaces 189
Proof.Wewant toshowthatifVb...,V
nEVarelinearlyindependent
over
F,thentheyspan VoverF.LetVE
V;thenv,VI,...,V
naren+1ele­
ments,hence,by
Theorem5.2.6,theyarelinearly dependentoverF.Thus
thereexistelements
a,ab...,aninF,notall0,such thatav+aivi+ ·..+
anV
n=0.Theelement
acannotbe0,otherwiseaivi+...+anV
n =0,and
notallthe
aiare0.Thiswouldcontradict thelinearindependence oftheele-
mentsVb , VnoverF.Thusa=1=0,andsoV=(-I/a)(alvl+...+anv
n
)=
f3IVI+ + f3nvn,wheref3i=-ai/aleTherefore,VI,...,VnspanVoverF,
andthusmustformabasis ofVoverF.D
PROBLEMS
EasierProblems
1Determineifthefollowingelementsin
V,thevectorspace of3-tuples
over
IR,arelinearlyindependent overIR.
(a)(1,2,3), (4,5,6), (7,8,9).
(b)(1,0,1),(0,1,2),(0,0,1).
(c)(1,2,3),(0,4,5),(~,3,2
41).
2.Findanontrivialsolutionin7Lsofthesystemoflinear homogeneous
equations:
Xl+2x2+3x3=°
3xI+4x2+2x3=°
3.IfVisavectorspace ofdimensionn over7L
p
,paprime,show thatVhas
pnelements.
4.Proveall ofLemma5.2.1.
5.Let
Fbeafieldand V=F[x],thepolynomialringin XoverF.Considering
Vasavectorspaceover F,provethat Visnotfinitedimensionalover F.
6.IfVisafinite-dimensionalvectorspace overFandifWisasubspaceof
V,provethat:
(a)Wisfinitedimensional overFanddimp(W)
:sdimp(V).
(b)Ifdimp(W)=dimF(V),thenV=W.
*7.Definewhatyoufeelshould beavectorspacehomomorphismfjJofV
intoW,whereVandWarevectorspacesover F.Whatcanyousay about
thekernel,K,offjJwhereK={vEVIfjJ(v)=o}?Whatshouldavector
spaceisomorphismbe?

205. 190 Fields Ch.5
8.IfVisavectorspaceover FandWisasubspaceof V,definetherequi­
siteoperationsin
V/WsothatV/Wbecomesavectorspaceover F.
9.IfVisafinite-dimensionalvectorspaceover FandVI,•••,V
rninVare
linearlyindependentover
F,showwecanfind WI,•••,W
rinV,where
m+r=dimF(V),suchthatVI,'••,V
rn
,
Wb•..,W
rformabasis ofV
overF.
10.Ift/J:V~V'isahomomorphismofVontoV'withkernel K,showthat
V'=V/K(asvectorspacesover F).(SeeProblem7).
11.Show
thatifdimF(V)=nandWisasubspaceofVwithdimF(W)=m,
thendimp(V/W)=n-m.
12.IfVisavectorspaceover Fofdimensionn,provethatVisisomorphic
tothevectorspace
ofn-tuplesover F(Example1).(SeeProblem7).
Middle-LevelProblems
13.
LetK
:JFbetwofields;suppose thatK,asavectorspaceover F,hasfi­
nitedimension n.ShowthatifaEK,thenthereexistaa,ab...,aninF,
notall0,suchthat
14.LetFbeafield,F[x]thepolynomialringin xoverF,andf(x)=1=0in
F[x].ConsiderV=F[x]/Jasavectorspaceover F,whereJistheideal
ofF[x]generatedby f(x).Provethat
dimp(V)=degf(x).
15.IfVandWaretwofinite-dimensionalvectorspacesover F,provethat
VEBWisfinitedimensionalover Fandthatdimp(VEBW)=
dimF(V)+dimp(W).
16.LetVbeavectorspaceover FandsupposethatUandWaresubspaces
ofV.DefineU+W={u+WIuEU,WEW}.Provethat:
(a)U+WisasubspaceofV.
(b)U+Wisfinitedimensionalover FifbothUandWare.
(c)UnWisasubspaceofV.
(d)U+Wisahomomorphicimage ofUEBW.
(e)IfUandWarefinitedimensionalover F,then
dimp(U+W)=dimp(U)+dimp(W)- dimp(UnW).

206. Sec.3
HarderProblems
FieldExtensions 191
17.Let K~Fbetwofieldssuchthatdimp(K) =m.SupposethatVisavec­
torspaceover K.Provethat:
(a)Visavectorspaceover F.
(b)IfVisfinitedimensionalover K,thenit isfinitedimensionalover F.
(c)IfdimK(V)=n,thendimp(V) =mn[i.e.,dimp(V) =
dimK(V)dimp(K)].
18.Let
K
~Fbefieldsandsuppose thatVisavectorspaceover Ksuchthat
dimp(V)isfinite.Ifdimp (K)isfinite,showthat dimK(V)iffiniteandde­
termineitsvalueinterms
ofdimp(V)and dimp(K).
19.Let Dbeanintegraldomainwith 1,whichhappensto beafinite-dimen­
sionalvectorspaceoverafield
F.ProvethatDisafield.(Note:Since Fl,
whichwecanidentifywith F,isinD,theringstructure ofDandthevec­
torspacestructure
ofDoverFareinharmonywitheachother.)
20.LetVbeavectorspaceoveran infinitefieldF.Showthat Vcannotbethe
set-theoreticunionofa
finitenumberof propersubspacesof V.(Veryhard)
3.FIELDEXTENSIONS
Ourattentionnowturnstoarelationshipbetweentwofields KandF,where
K
~F.WecallKanextension(orextensionfield) ofF,andcallFasubfield
ofK.Theoperationsin Farejustthose ofKrestrictedtotheelementsofF.
Inallthat followsinthissectionitwill beunderstoodthatFCK.
Wesaythat Kisafiniteextension ofFif,viewedasavectorspaceover
F,dimp(K)isfinite.Weshallwrite dimp(K)as[K:F]andcallitthe degree
ofKoverF.
Webeginourdiscussionwithwhat isusuallythefirstresultoneproves
intalkingaboutfiniteextensions.
Theorem
5.3.1.LetL
~K~Fbethreefieldssuchthatboth [L:K]
and[K:F]arefinite.Then Lisafiniteextension ofFand[L:F]=
[L:K][K:F].
Proof.Weshallprove thatLisafiniteextension ofFbyexplicitlyex­
hibitingafinitebasis
ofLoverF.Indoingso,weshallobtainthestrongerre­
sultassertedinthetheorem,namelythat
[L:F] =[L:K][K:F].
Supposethat [L:K]=mand[K:F]=
n;thenLhasabasis VI,V2'"•'V
m
overK,andKhasabasisWbW2,•.•,W
noverF.Weshallprove thatthemn

207. 192 Fields Ch.5
elementsViWj,wherei=1,2,...,mandj=1,2,...,n,constituteabasisof
LoverF.
Webeginbyshowingthat,atleast,theseelementsspan LoverF;this
will,ofcourse,showthat
Lisafiniteextensionof F.LetaE
L;sincetheele­
mentsVb••.,V
mformabasisof LoverK,wehavea=klVI+ .·. +kmv
m
,
wherek
l
, k
2
, •••, k
mareinK.SinceWI,.••, w
nisabasisof KoverF,we
canexpresseach k
ias
wherethe
hjareinF.Substitutingtheseexpressionsforthe k
iintheforego­
ingexpressionof
a,weobtain
a=(!IIWI+!12W2+ . · . +!lnWn)VI
+..· +(!mIWI+!m2W2+...+!mnWn)Vm·
Therefore,onunscramblingthissumexplicitly, weobtain
Thusthe
mnelementsViWjinLspanLover
F;therefore,[L:F]isfiniteand,
infact,
[L:F]
::::;mn.
Toshowthat [L:F] =mn,weneedonlyshowthatthe mnelements
ViWjabovearelinearlyindependentover F,forthen-togetherwiththefact
thattheyspan
LoverF-wewouldhavethattheyformabasisof LoverF.
ByTheorem5.2.5wewouldhavethedesiredresult [L:F]=mn=
[L:K][K:F].
Supposethenthatforsome b
ijinFwehavetherelation
Reassemblingthissum,weobtain
CIVI+C2V2+...+CmV
m=0,whereCl=
bI1WI+...+bInw
n
,
•..,C
m=bmlWI+...+bmnw
n
•Sincethe Ciareele­
mentsof
Kandtheelements
Vb•.•,V
ninLarelinearlyindependentover K,
weobtain CI=C2=...=C
m=O.
Recallingthat Ci=bnWI+...+binw
n
,wherethe b
ijarein Fand
where
WI,•.•,WninKarelinearlyindependentover F,wededucefromthe
factthat
CI=C2=...=C
m=0thatevery b
ij=o.Thusonlythetriviallinear
combination,witheachcoefficient
0,oftheelements ViWjoverFcanbe o.
Hencethe ViWjarelinearlyindependentover F.Wesawabovethatthiswas
enoughtoprovethetheorem.
D

208. Sec.3 FieldExtensions 193
ThereadershouldcompareTheorem 5.3.1withtheslightlymoregen­
eralresult
inProblem17ofSection2.Thereadershouldnowbeable to
solveProblem 17.
Asaconsequenceofthetheoremwehavethe
Corollary.If
L
~K~Farethreefieldssuchthat [L:F]isfinite,
then
[K:F] isfiniteanddivides [L:F].
ProofSinceL
~K,K cannothavemorelinearlyindependentele­
mentsover
Fthandoes L.Because,byTheorem5.2.6, [L:F]isthesizeof
thelargestsetoflinearlyindependentelementsin
LoverF,wethereforeget
that
[K:F]
::::;[L:F],somustbefinite.SinceL isfinitedimensionalover F
andsinceKcontainsF,Lmustbefinitedimensionalover K.Thusallthe
conditionsofTheorem
5.3.1arefulfilled,whence [L:F]=[L:K][K:F].
Consequently,[K:F]divides[L:F],asisassertedintheCorollary. 0
IfKisafiniteextensionof F,wecansayquiteabitaboutthebehavior
oftheelementsof
Kvis-a-visF.
Theorem5.3.2.Supposethat Kisafiniteextensionof Fofdegreen.
Then,givenanyelement uinKthereexistelements
ao,at,...,aninF,not
allzero,suchthat
ProofSince[K:F]=dimF(K)=nandtheelements 1,u,u
2
,...,un
aren+1innumber,byTheorem5.2.6theymustbelinearlydependentover
F.Thuswecanfind
ao,at,...,aninF,notall0,suchthatao+atU+
a2u2+...+anu
n =0,provingthetheorem. D
Theconclusionofthetheoremsuggeststhatwesingleoutelementsin
anextensionfieldthatsatisfyanontrivialpolynomial.
Definition.If
K
~Farefields,then aEKissaidtobe algebraicover
Fifthereexistsapolynomial p(x)=1=0inF[x]suchthat pea)=o~
Bypea)weshallmeanthe element aoa
n+atan-t+...+
aninK,
wherep(x)=aox
n
+atXn-
t+...+
an.
IfKisanextensionof Fsuchthateveryelementof Kisalgebraicover
F,wecallKanalgebraicextension ofF.InthesetermsTheorem5.3.2canbe
restated
as:IfKisafiniteextension ofF,thenKisanalgebraicextensionofF.
Theconverseofthis isnottrue;analgebraicextensionof Fneednotbe
offinitedegreeover
F.Canyoucomeupwithanexampleofthissituation?

209. 194 Fields Ch.5
Anelementof Kthatisnotalgebraicover Fissaidtobe transcendental
overF.
Let'sseesomeexamplesofalgebraicelementsinaconcretecontext.
ConsiderC:JQ,thecomplexfield asanextensionoftherationalone.Thecom­
plexnumbera
=1+iisalgebraicover
Q,sinceitsatisfiesa
2
-
2a+2=0
over
Q.Similarly,therealnumber b=J1+\0/1+v'2isalgebraicoverQ,
sinceb
2
=1+~1+0,so(b
2
-
1)3=1+
\/2,andtherefore
((b
2
-
1)3-1)2=2.Expandingthisout, wegetanontrivialpolynomialex­
pressionin
bwithrationalcoefficients,which isO.Thusbisalgebraicover
Q.
ItispossibletoconstructrealnumbersthataretranscendentaloverQ
fairlyeasily(seeSection6 ofChapter6).However,ittakessomerealeffort
toestablishthetranscendence
ofcertainfamiliarnumbers.Thetwofamil­
iarnumbers
eand1Tcan beshown tobetranscendentalover
Q.Thateis
suchwasprovedbyHermitein1873;theproof that1Tistranscendental
overQismuchharderandwasfirstcarriedoutbyLindemannin1882.We
shallnotgointothe
proofherethatanyparticularnumber istranscenden­
talover
Q.However,inSection7 ofChapter6weshallatleastshowthat
1Tisirrational.Thismakesitapossiblecandidateforatranscendentalnum­
berofQ,forclearlyanyrational numberbisalgebraicoverQbecauseit
satisfiesthepolynomialp
(x)=x-b,whichhasrationalcoefficients.
Definition.A complexnumber issaidtobean algebraicnumberifit
isalgebraicover
Q.
Asweshallsoonsee,thealgebraicnumbersformafield,which isa
subfieldofC.
Wereturntothegeneraldevelopmentofthetheoryoffields.Wehave
seen
inTheorem5.3.2thatif Kisafiniteextensionof F,theneveryelement
of
Kisalgebraicover F.Weturnthismatteraroundbyasking:If Kisanex­
tension
ofFandaEKisalgebraicover F,canwesomehowproduceafinite
extensionof
Fusinga?Theanswer isyes.Thiswillcome asaconsequenceof
thenext
theorem-whichweproveinacontextalittlemoregeneralthan
what
wereallyneed.
Theorem5.3.3.
:-LetDbeanintegraldomainwith1which isafinite­
dimensionalvectorspaceoverafield
F.ThenDisafield.
ProofToprovethetheorem,wemustproducefor a
=1=0inDanin­
verse,
a-I,inD
suchthataa-
I
=1.

210. Sec.3 FieldExtensions 195
Asinthe proofofTheorem5.3.2,ifdimp(D) =n,then1,a,a
2
,, an
inDarelinearlydependentover F.Thusforsomeappropriateao,aI,,an
inF,notallofwhichare 0,
Letp(x)=f3ox'+f3IX,-1+...+f3,=1=°beapolynomialin F[x]oflowest
degreesuchthat
pea)=0.Weassertthat
f3,=1=0.Foriff3,=0,then
°=f30a
r+f31a
r
-
1+ . · · +f3r-la
=(f3oa
r
-
1 +f31a
r
-2
+ ·..+f3r-l)a.
SinceDisanintegraldomainand a
=1=0,weconclude thatf3oa,-1+
f3la,-2+ +f3,-1=0,henceq(a)=0,whereq(x)=f3ox,-1+
(31X,-2+ + f3,-linF[x]isoflowerdegreethan p(x),acontradiction.
Thusf3,=1=0,hencef3;1isinFand
a(f3oa
r
-
1+...+f3r-l)=-1
f3r '
givingusthat -(f3oa,-l+...+f3,-I)If3"whichisinD,isthea-IinDthat
werequired.Thisprovesthetheorem. D
HavingTheorem5.3.3inhand,wewanttomakeuse ofit.So,howdo
weproducesubrings
ofafieldKthatcontainFandarefinitedimensional
over
F?Suchsubrings,assubrings ofafield,areautomaticallyintegraldo­
mains,andwouldsatisfythehypothesis
ofTheorem5.3.3.Themeanstothis
endwillbetheelementsin
Kthatarealgebraicover F.
Butfirstadefinition.
Definition.
TheelementaintheextensionK ofFissaidtobe alge­
braicofdegreenifthereisapolynomialp(x)inF[x]ofdegreensuchthat
pea)=0,andnononzeropolynomial oflowerdegreein F[x]hasthisprop­
erty.
Wemayassume
thatthepolynomialp (x)inthisdefinition ismonic,for
wecoulddivide thispolynomialbyitsleadingcoefficient
toobtainamonic
polynomial
q(x)inF[x],ofthesamedegreeas p(x),andsuchthatq(a)=0.
Wehenceforthassume thatthispolynomialp(x)ismonic;wecallitthe min­
imalpolynomial
foraoverF.
Lemma5.3.4.LetaEKbealgebraicover Fwithminimalpolynomial
p(x)inF[x].Thenp(x)isirreduciblein F[x].

211. 196 Fields Ch.5
ProofSupposethatp(x)isnotirreduciblein F[x];thenp(x)=
f(x)g(x)wheref(x)andg(x)areinF[x]andeachhas positivedegree.Since
o=pea)=f(a)g(a),andsincef(a)andg(a)areinthefield K,weconclude
that
f(a)=0org(a)=0,bothofwhichareimpossible,sinceboth f(x)and
g(x)areoflowerdegreethan f(x).Therefore,p(x)isirreduciblein F[x].D
LetaEKbealgebraicofdegreenoverFandletp(x)EF[x]beits
minimalpolynomialover
F.Givenf(x)EF[x],thenf(x)=q(x)p(x)+rex),
whereq(x)andrex)areinF[x]andrex)=0ordegrex)<degp(x)follows
fromthedivisionalgorithm.Therefore,
f(a)=q(a)p(a)+rea)=rea),since
p
(a)=O.Inshort,anypolynomialexpressionin aoverFcanbeexpressedas
apolynomialexpressionin
aofdegreeatmostn- 1.
LetF[a]={f(a)If(x)EF[x]}.Weclaimthat F[a]isasubfieldof Kthat
containsboth
Fanda,andthat [F[a]:F]=n.Bytheremarkmadeabove,
F[a]isspannedover Fby1,a,a
2
, •
••,an-I,SOisfinitedimensionalover F.
Moreover,as iseasilyverified,F[a]isasubringof
K;asasubringofK,F[a]is
anintegraldomain.Thus,by Theorem5.3.3,F[a]isafield.Sinceit isspanned
over
Fby1,a,a
2
,
•••,an-I,wehavethat [F[a]:F]
~n.Toshowthat
[F[a]:F]=nwemustmerelyshowthat 1,a,a
2
,
•••,a
n
-
larelinearlyinde-
pendentover
F.Butif
aa+ala+ +an_Ia
n
-
1
=0,withtheaiinF,then
q(a)=0,whereq(x)=aa+alx+ +an_Ix
n
-
1
isinF[x].Sinceq(x)isof
lowerdegreethan
p(x),whichistheminimalpolynomialfor ainF[x],weare
forcedtoconclude
thatq(x)=O.Thisimpliesthat
aa=al=...=an-l=O.
Therefore,1,a,a
2
,
•••,a
n
-
larelinearlyindependentover Fandformabasis
ofF[a]overF.Thus[F[a]:F]=n.SinceF[a]isafield,notmerelyjustaset
ofpolynomialexpressions ina,weshalldenote F[a]byF(a).Notealsothatif
Misanyfieldthatcontainsboth Fanda,thenMcontainsallpolynomialex­
pressionsin
aoverF,henceM
:>F(a).SoF(a) isthesmallestsubfield ofK
containingbothFand
a.
Definition.F(a)iscalledthefieldorextensionobtainedby adjoining
atoF.
Wenowsummarize.
Theorem5.3.5.Let K
:>Fandsupposethat ainKisalgebraicover F
ofdegreen.ThenF(a),thefieldobtainedbyadjoining atoF,isafiniteex­
tension
ofF,and
[F(a):F]=n.
Beforeleaving Theorem5.3.5,let'slook atitinaslightlydifferentway.
LetF[x]bethepolynomialringin xoverF,andletM=(p(x»betheideal

212. Sec.3 FieldExtensions 197
ofF[x]generatedby p(x),theminimalpolynomialfor ainKoverF.
ByLemma5.3.4, p(x)isirreduciblein F[x];hence,by Theorem4.5.11,
Misamaximalideal ofF[x].Therefore,F[x]/(p(x))isafieldby Theorem
4.4.2.
Definethemappingf/!:F[x]~Kbyf/!(f(x))=f(a).Themappingf/!is
ahomomorphismofF[x]intoK,andtheimageofF[x]inKismerelyF(a)
bythedefinition ofF(a).Whatisthekerneloff/!?ItisbydefinitionJ=
{f(x)EF[x]If/!(f(x))=O},andsinceweknowf/!(f(x))=f(a),J={f(x)E
F[x]If(a)=Ol.Sincep(x)isinJandp(x)istheminimalpolynomialfor a
overF,p(x)isofthelowestpossibledegreeamong theelementsofJ.Thus
J=(p(x))bytheproofofTheorem4.5.6,andsoJ=M.BytheFirstHomo­
morphismTheoremforrings,F[x]/M==imageofF[x]underf/!=F(a),and
sinceF[x]/M isafield,wehave thatF(a)isafield.Weleave theproof,from
thispoint
ofview,of[F(a): F]=degp(x)tothereader.
PROBLEMS
1.Showthatthefollowingnumbersin
Carealgebraicnumbers.
(a)V2+V3.
(b)v7+VUe
(c)2+iV3.
(d)cos(21T/k)+isin(21T/k),k apositiveinteger.
2.DeterminethedegreesoverQofthenumbersgiveninParts(a) and(c)
ofProblem1.
3.WhatisthedegreeofCOS(21T/3)+isin(21T/3)overQ?
4.WhatisthedegreeofCOS(21T/8)+isin(21T/8)overQ?
5.Ifpisaprimenumber,prove thatthedegreeofCOS(21T/p)+isin(21T/p)
overQisp-1andthat
f(x)=1+x+x
2
+...+x
p
-1
isitsminimalpolynomialover
Q.
6.(Forthosewhohave hadcalculus)Show that
1 1 1
e=I+-+-+···+-+···
I!2! n!
isirrational.
7.IfainKissuchthata
2
isalgebraicoverthesubfieldFofK,showthata
isalgebraicover F.

213. 198 Fields Ch.5
8.IfFcKandf(a)isalgebraicover F,wheref(x)isofpositivedegreein
F[x]anda EK,provethata isalgebraicover F.
9.InthediscussionfollowingTheorem5.3.5,showthatF[x]/M isofdegree
n
=degp(x)overF,andso[F(a): F]=n=degp(x).
10.Provethatcos1
0
isalgebraicover
Q.(1
0
=onedegree.)
11.IfaEKistranscendentalover F,letF(a)={f(a)/g(a)If(x),g(x)=1=0E
F[x]}.ShowthatF(a) isafieldand isthesmallestsubfieldofKcontain­
ingboth
Fanda.
12.IfaisasinProblem11,showthatF(a)
=F(x),whereF(x) isthefieldof
rationalfunctionsin
xoverF.
13.LetKbeafinitefieldandFasubfieldofK.If [K:F]=nandFhasq
elements,showthat Khasqnelements.
14.UsingtheresultofProblem
13,showthatafinitefieldhaspnelements
forsomeprime
pandsomepositiveinteger n.
15.Constructtwofields KandFsuchthat Kisanalgebraicextensionof F
butisnotafiniteextensionof F.
4.FINITEEXTENSIONS
Wecontinueintheveinoftheprecedingsection.Again K
:JFwillalways
denotetwofields.
Let
E(K)bethesetofallelementsinKthatarealgebraicover F.Cer­
tainly,F
CE(K).Ourobjectiveistoprovethat E(K)isafield.Oncethis is
done,we'llseealittleofhow E(K)sitsinK.
Withoutfurtherado
weproceedto
Theorem5.4.1.
E(K)isasubfieldofK.
ProofWhatwemustshow isthatifa,bEKarealgebraicover F,then
a
±b,ab,andalb(ifb=1=0)areallalgebraicover F.Thiswillassure usthat
E(K)isasubfieldofK.We'lldoallofa±b,ab,andalbinoneshot.
LetKo
=F(a)bethesubfieldofKobtainedbyadjoiningato F.Since
aisalgebraicover F,sayofdegree m,then,byTheorem5.3.5, [Ko:F]=
m.
Sincebisalgebraicover FandsinceKocontainsF,wecertainlyhavethat b
isalgebraicover Ko.Ifbisalgebraicover Fofdegreen,thenit isalgebraic
overKoofdegreeatmost
n.ThusK
1
=Ko(b),thesubfieldofKobtainedby
adjoiningbtoKo,
isafiniteextensionofKoand [K
1
:Ko]
~n.
Thus,byTheorem5.3.1,
[K
1
:F]=[K
1
:Ko][Ko: F]
:smn;thatis,K
1is
afiniteextensionof F.Assuch,byTheorem5.3.2, K1isanalgebraicexten-

214. Sec.4 FiniteExtensions 199
sionofF,soallitselementsarealgebraic overF.SinceaEK0cK1andbE
K1,thenalloftheelementsa±b,ab,alb areinK1,hencearealgebraic over
F.Thisisexactlywhatwewanted. Thetheoremisproved.D
Ifwelook attheproofalittlemorecarefully,wesee thatwehaveactu­
allyprovedalittlemore,namely
the
Corollary.IfaandbinKarealgebraicover Fofdegreesmandn,
respectively,then a±b,ab,andalb(ifb
=1=0)arealgebraic overFofdegree
atmost
mn.
Aspecialcase, butoneworthnoting andrecording,isthecaseK=
C
andF=Q.Inthatcasewecalled thealgebraicelementsin CoverQtheal­
gebraicnumbers. SoTheorem5.4.1inthiscasebecomes
Theorem5.4.2.Thealgebraicnumbersformasubfield ofC.
Forallweknow atthemoment, thesetofalgebraicnumbersmayvery
wellbeall
of
C.Thisisnotthecase,fortranscendentalnumbers doexist;we
showthis
tobetrueinSection6 ofChapter6.
Wereturntoageneralfield K.ItssubfieldE(K)hasaveryparticular
quality,whichweprovenext.This
propertyisthatanyelementinKwhichis
algebraicover E(K)mustalready beinE(K).
Inordernottodigressin thecourseoftheproofweareabouttogive,
weintroducethefollowingnotation.
If
aba2,""anareinK,then
F(ab".,an)willbethefieldobtainedasfollows:K
1=F(al),K2 =
K1(a2)=F(aba2),K3 =K2(a3)=F(al'a2,a3),...,Kn=Kn-1(an) =
F(aba2,· · . ,an)'
Wenowprove
Theorem5.4.3.IfuinKisalgebraicover E(K),thenuisinE(K).
ProofToprovethetheorem,allwemustdo isshowthatuisalgebraic
over
F;thiswillputuinE(K),andwewillbedone.
Since
uisalgebraicover E(K),thereisanontrivialpolynomial f(x)=
x
n+alx
n
-1
+a2Xn-2+ +an,where
aba2,""anareinE(K),suchthat
f(u)=O.Sinceaba2, , anareinE(K),theyarealgebraicover Flofde-
grees,say,mbm2"'"mn,respectively.Weclaim that[F(al"'"an):F]is
atmostmlm2...mn.Toseethis,merelycarry outnsuccessiveapplications
ofTheorem5.3.1tothesequenceKbK2,...,Knoffieldsdefinedabove.We
leaveits
prooftothereader.Thus,since uisalgebraicoverthefieldK
n=F(aba2,""an)[afterall, thepolynomialsatisfiedby uisf(x)=

215. 200 Fields Ch.5
x
n+alx
n
-1
+...+an,whichhasallitscoefficientsinF(a},a2,...,an)],the
field
Kn(u)isafiniteextension ofKn,andsinceKnisafiniteextension ofF,
wehave,againby Theorem5.3.1,thatKn(u)isafiniteextension ofF.Be­
cause
uEKn(u),weobtainfrom Theorem5.3.2thatuisalgebraicover F.
Thisputs uinE(K)bytheverydefinitionofE(K),therebyprovingthetheo­
rem.D
ThereisafamoustheoremduetoGauss,oftenreferred toastheFun­
damentalTheoremofAlgebra,whichasserts(interms ofextension)thatthe
onlyfiniteextension
of
C,'thefieldofcomplexnumbers, isCitself.Inreality
thisresult
isnotapurelyalgebraicone,itsvaliditydependingheavily on
topologicalpropertiesofthefieldofrealnumbers. Bethatasitmay,it isan
extremely
importanttheoreminalgebraandinmanyotherpartsofmathe­
matics.
TheformulationoftheFundamentalTheoremofAlgebrainterms of
thenonexistenceoffiniteextensions of
Cisalittledifferentfrom thatwhich
isusuallygiven. Themostfrequentforminwhichthisfamousresult isstated
involves
theconceptofarootofapolynomial,aconceptweshalldiscuss at
somelengthlater. Intheseterms theFundamentalTheoremofAlgebrabe­
comes:Apolynomial
ofpositivedegreehavingcoefficientsin
Chasatleast
onerootinC.Theexactmeaning ofthisstatementanditsequivalencewith
theotherformofthetheoremstatedabovewillbecomeclearerlater,after
thedevelopmentofthematerialonroots.
Afield
Lwiththepropertyof
Cdescribedintheparagraphsabove is
saidtobealgebraicallyclosed. IfwegrantthatCisalgebraicallyclosed
(Gauss'Theorem),then,by
Theorem5.4.3,wealsohave
Thefieldofalgebraicnumbersisalgebraicallyclosed.
PROBLEMS
1.Showthata=
v2-V3isalgebraicoverQofdegreeatmost4byex­
hibitinga
polynomialf(x)ofdegree4 over
Q.suchthatf(a)=O.
2.IfaandbinKarealgebraicoverFofdegreesmandn,respectively,and
if
mandnarerelativelyprime,show that[F(a,b):F]=mn.
3.IfaE
Cissuchthatp(a)=0,where
p(x)=x
5
+v2x
3
+V5x
2
+v7X+VU,
showthataisalgebraicoverQofdegreeatmost80.

216. Sec.5 ConstructibiI ity201
4.IfK~Fissuchthat[K:F]=p,paprime,show thatK=F(a)forevery
ainKthatisnotinF.
5.If[K:F]=2
n
andTisasubfield ofKcontainingF,showthat[T:F]=2
m
forsome m~n.
6.Giveanexampleoftwoalgebraicnumbersaandbofdegrees2 and3,re­
spectively,such
thatabisofdegreelessthan6over
Q.
7.IfK~Farefieldsandab...,anareinK,showthatF(ab...,an)equals
F(au (1)'•••,au(n»foranypermutation
(Tof1,2,...,n.
5.CONSTRUCTIBILITY
InancientGreece,unlikein theotherculturesofthetime,theGreekmathe­
maticianswereinterestedinmathematicsasanabstractdiscipline
rather
thanasapragmaticbag oftrickstodoaccountsortocarryoutmeasure­
ments.
Theydevelopedstronginterests andresultsin numbertheoryand,
mostespecially,ingeometry.
Intheseareasthey posedpenetratingques­
tions.
Thequestionstheyaskedin geometry-twoofwhichwill makeupthe
topictreatedhere-arestillofinterestandsubstance.TheEnglishmathe­
maticianG.H.Hardy,inhissad
butcharminglittle bookAMathematician's
Apology,
describestheancientGreekmathematiciansas"colleaguesfrom
anothercollege."
Two
oftheseGreekquestionswill beourconcerninthissection.But,
asa
matteroffact,theanswertobothwillemergeasaconsequence ofthe
criterionforconstructibility,whichwewillobtain. Westatethesequestions
nowandwillexplainalittle
laterwhatisentailedinthem.
QUESTIONl
Canoneduplicateacubeusingjuststraight-edge andcompass?(Byduplicat­
ingacube,we
meandoublingitsvolume.)
QUESTION2
Canonetrisectanarbitraryangleusing juststraight-edgeandcompass?
Despite
theseeminglyinfinite numberofangle-trisectorsthatcropup
everyyear, theanswertobothquestionsis"no."Asweshallsee,it isimpos­
sibletotrisect
60°usingjuststraight-edge andcompass.Ofcourse,some
angles
aretrisectable,forinstance, 0°,90°,145°,180°, ...,butmostangles
(inaveryprecisemeaningof
"most")arenot.

217. 202 Fields Ch.5
Beforegettingtotheexactmeaning ofthequestionsthemselves,we
want
tospelloutinexplicittermsexactlywhat therulesofthegameare.
Byastraight-edgewedo notmeanaruler-thatis,aninstrumentformea­
suringarbitrarylengths.
No!Astraight-edgeismerelyastraightline,
withnoquantitative
ormetricpropertiesattributedtoit.Wearegivena
line
segment-towhichweassignlength I-andallotherlengthsthatwe
getfromthismust
beobtainablemerelyemployingastraight-edgeand
compass.
Letuscallanonnegativereal number,b,aconstructiblelength if,by
a
finitenumberofapplicationsofthestraight-edgeandcompassand the
pointsofintersectionobtainedbetweenlinesandcirclessoconstructed,
wecanconstructalinesegment
oflengthb,startingoutfromthelineseg­
mentwehaveassignedlength 1.
Fromourhighschoolgeometrywerecallsomethingswecandointhis
framework.
1.Whateverlengthweconstruct ononelinecanbeconstructedonany
otherlinebyuse ofthecompassactingasatransferagent.
2.Wecandrawalineparalleltoagivenlinethatgoesthroughagiven
point.
3.Wecanconstructalength nforanynonnegativeinteger n.
Fromtheseandbyusingresults aboutthesimilarityoftriangles,wecan
constructanynonnegativerationallength.We
don'tdothatatthismoment
foritwillcome
outasaspecialcase ofwhatweareabout todo.
Weclaimthefollowingproperties:
1.Ifaandbareconstructiblelengths,thenso isa+b.ForifABisa
lengthsegment
oflengthaandCDisoneoflength b,wecantransferthis
linesegment
CD,bymeansofacompass,toobtaintheline ABE,whereAB
isoflengthaandBEisoflengthb.Thusthelinesegment AEisoflength
a+b.Ifb>a,howwouldyouconstruct b-a?
2.Ifaandbareconstructiblelengths,thenso isaboWeInayassume
that
a
=1=0andb=1=0,otherwise,thestatementistrivial.Considerthefollow­
ingdiagram
P L)
J B

218. Sec.5 Constructibility203
WhereL
1andL
2aretwodistinctlinesintersectingat P,andsuchthat PA
haslengtha,PBhaslength b,andPIhaslength1.LetL
3bethestraightline
through
IandAandL4thelineparallelto L3passingthrough B.IfCisthe
pointofintersectionofL
1andL
4
,wehavethediagram
p-----+---#--------L
2
Alloftheseconstructionscanbecarried outbystraight-edgeandcompass.
Fromelementarygeometrythelengthof
PCisaboTherefore,abiscon­
structible.
3.Ifaandbareconstructibleand b
=1=0,thenalbisconstructible.Con­
siderthediagram
--+----....---------L
2
whereP,A,B,I,L
bandL
2areasinProperty2above.LetL sbetheline
throughAandBandletL
6bethelinethroughIparalleltoL s.IfDisthe
pointofintersectionofL
1andL
6
,then,againbyelementarygeometry,the
lengthof
PDisalb.Westressagainthatalltheconstructionsmadecanbe
carriedoutbystraight-edgeandcompass.
Ofcourse,thisshowsthatthenonnegativerationalnumbersarecon­
structiblelengths,sincetheyarequotientsofnonnegativeintegers,whichwe
knowtobeconstructiblelengths.Butthereareotherconstructiblelengths,
forinstance,theirrationalnumber
v2.Becausewecanconstructbystraight­
edgeandcompasstheright-angletriangle

219. 204 Fields Ch.5
withsidesABandBCoflength1,weknow,bythePythagoreanTheorem,
that
ACisoflength
v2.Sov2isaconstructiblelength.
InProperties1to3
weshowedthattheconstructiblelengthsalmost
formafield.What
islackingisthenegatives.Togetaroundthis, wemake
the
Definition.Therealnumber
aissaidtobea constructiblenumberif
IaI,the absolutevalueof a,isaconstructiblelength.
Asfar
aswecansayatthemoment,anyrealnumbermightbeacon­
structibleone.Weshallsoonhaveacriterionwhichwilltellusthatcertain
realnumbersarenotconstructible.
Forinstance,weshallbeabletodeduce
fromthiscriterionthatboth
~andcos 20
0
arenotconstructible.This in
turnwillallowustoshowthattheanswertobothQuestions1and2 is"no."
Butfirstwestate
Theorem5.5.1.Theconstructiblenumbersformasubfieldofthefield
ofrealnumbers.
ProofProperties1to3almostdothetrick;wemustadaptProperty1
slightlytoallowfornegatives.Weleavethefewdetailstothereader.
D
Ournextgoal istoshowthataconstructiblenumbermustbeanalge­
braic
number-notanyoldalgebraicnumber,butonesatisfyingarather
stringentcondition.
Note,first,thatif
a
2::0isaconstructiblenumber,thenso is~.Con­
siderthediagram
B
Itisofasemicircleofradius (a+1)/2,centerat C,ADisoflengtha,DBis
oflength1,andDEistheperpendicularto ABatD,intersectingthecircleat
E.Allthisisconstructiblebystraight-edgeandcompass.Fromelementary
geometrywehavethat
DEisoflength
~,hence~isconstructible.
Wenowheadforthenecessaryconditionthatarealnumberbecon­
structible.Let
Kbethefieldofconstructiblenumbers,andlet Kobeasub­
fieldof
K.BytheplaneofKoweshallmeanthesetofallpoints (a,b)inthe
realEuclideanplanewhosecoordinatesaand
bareinKo.If(a,b)and(c,d)

220. Sec.5 ConstructibiI ity205
areintheplaneof Ko,thenthestraightlinejoiningthemhastheequation
(y-b)/(x-a)=(b-d)/(a- c),soisoftheform ux+vy+w=0,where
u,v,andwareinKo.Giventwosuchlines U1X+V1Y+W1=0and
U2X+V2Y+W2=0,where
UbVbW1andU2,V2,W2areallin Ko,eitherthey
areparallelortheirpointofintersection
isapointin Ko.(Prove!)
Givenacirclewhoseradius
risinK0andwhosecenter (a,b)isinthe
planeof
Ko,thenitsequation is(x-a)2+(y-b)2=r
2
,whichwesee,on
expanding,
isoftheform
XL-+y2+dx+ey+f=0,whered,e,andfareinKo.
Toseewherethiscircleintersectsalineintheplaneof Ko,ux+vy+w=0,
wesolvesimultaneouslytheequationsofthelineandofthecircle.Forin­
stance,if
V
=1=0,theny=-(ux+w)/v;substitutingthisfor yintheequation
ofthecircle
x
2+y2+dx+ey+f=0leadsustoaquadraticequationfor
thex-coordinate,
c,ofthisintersectionpoint, oftheformc
2
+S1C+S2=0,
withS1andS2inKo.Bythequadraticformula,c =(-S1±
\lsi-4s
2)/2,
andifthelineandcircleintersectintherealplane,thensi-4s22:O.If
s=si-4s
2
2:0andif K1=Ko(~), thenweseethatthex-coordinate, C,
liesinK
1
•If~EKo,thenK
1
=Ko;otherwise,[K
1
:Ko]=2.Sincethe
y-coordinate
d=(-uc+w)/v,wehavethat disalsoin K
1
•Thustheinter­
sectionpoint(c,
d)liesintheplaneof K1where[K1 :K0]=1or2.Thestory
issimilarif v=0andU
=1=o.
Finally,togettheintersectionoftwocircles x
2+y2+dx+ey+f=0
and
x
2+y2+gx+hy+k=0intheplaneof Ko,subtractingoneofthese
equationsfromtheothergivesustheequationofthelineintheplaneof
Ko,
(d-g)x+(e-h)y+(f-k)=O.Sotofindthepointsofintersectionof
twocirclesintheplaneof
Koisthesameasfindingthepointsofintersection
ofalineintheplaneof
Kowithacircleinthatplane.This ispreciselythesitua­
tion
wedisposedofabove. Soifthetwocirclesintersect intherealplane,their
pointsofintersectionlieintheplaneofanextensionof
Koof degree1or 2.
Toconstructaconstructiblelength, a,westartintheplaneof
Q,thera­
tionals;thestraight-edgegives
uslinesintheplaneof
Q,andthecompass
circlesintheplaneofQ.Bytheabove,theseintersectatapointintheplane
ofanextensionofdegree1
or2of
Q.Togetto a,wegobythisprocedure
fromtheplaneofQtothatof L
1
,say,where [L
1
:Q]=1or2,thentothat
of
L
2
,where[L
2
:
Ltl=1or2,andcontinueafinitenumberoftimes.We
get,thisway,afinitesequenceQ=LoeLlc...cLnoffields,whereeach
[L
i
:L
i
-1
]=1or2andwhere aisinLn.
ByTheorem5.3.1, [Ln:
Q]=[Ln: Ln-1 ][Ln-1:Ln-2] ..·[L1:Q]and
sinceeachof
[L
i
:L
i
-1]=1or2,weseethat [Ln:
Q]isapowerof2.Since
aEL
n
,wehavethatQ(a)isasubfieldof L
n
,hencebytheCorollarytoTheo­
rem
5.3.1,
[Q(a):Q]mustdivideapower of2,hence[Q(a):Q]=2
m
for
somenonnegativeintegerm.
Equivalently,byTheorem5.3.5,theminimal

221. 206 Fields Ch.5
polynomialfor aover0musthavedegree apowerof 2.Thisisanecessary
conditionthat abeconstructible.Wehaveprovedtheimportantcriterionfor
constructibility,namely
Theorem
5.5.2.Inorderthattherealnumber abeconstructible,it is
necessarythat
[O(a):0]beapowerof 2.Equivalently,theminimalpolyno­
mialof
aover
0musthavedegreeapowerof 2.
Toduplicateacubeofsides 1,soofvolume 1,bystraight-edgeand
compasswouldrequire
ustoconstructacubeofsidesoflength bwhosevol­
umewouldbe
2.Butthevolumeofthiscubewouldbe b
3
,
sowewouldhave
tobeabletofindaconstructiblenumber
bsuchthat b
3
=2.
Givenarealnumber bsuchthat b
3
=2,thenitsminimal polynomial
over
0isp(x)=x
3
-
2,forthispolynomial ismonicandirreducibleover
0(if
youwant,bytheEisensteinCriterion),and
p(b)=O.Also,asiscleartothe
eye,
p(x)isofdegree3.Since3 isnotapowerof 2,byTheorem5.5.2,there is
nosuchconstructible b.Therefore,thequestionoftheduplicationofthecube
bystraight-edgeandcompasshas anegativeanswer.Wesummarizethis
in
Theorem5.5.3.Itisimpossibletoduplicateacubeofvolume1by
straight-edgeandcompass.
WenowhavedisposedoftheclassicalQuestion
1,soweturnouratten­
tiontoQuestion
2,thetrisectionofanarbitraryanglebystraight-edgeand
compass.
Ifwecouldtrisecttheparticularangle
60°,wewouldbeabletocon­
structthetriangle
ABCinthediagram
where
()=20°andACisoflength1,bystraight-edgeandcompass.Since AB
isoflengthcos 20°,wewouldhavethat b=cos20°isaconstructiblenumber.
Wewanttoshowthat
b=cos20°isnotaconstructiblenumberbypro­
ducingitsminimalpolynomialover
0,andshowingthatthispolynomial isof
degree
3.Tothisendwerecallthetriple-angleformulafromtrigonometry,
namelythatcos
3cP=4cos
3
cP-3coscPoIfb=cos20°,then,since
cos(3.
20°)=cos60°=
~,thistrigonometricformula becomes 4b
3
-
3b=
~,
andso 8b
3
-
6b-1=O.Ifc=2b,thisbecomesc
3
-
3c- 1=O.Ifbiscon­
structible,thenso
isc.Butp(c)=0,wherep(x)=x
3
-
3x- 1,andthis

222. Sec.6 RootsofPolynomials 207
polynomialisirreducible overQ.(Prove!)So p(x)istheminimalpolynomial
for
cover
Q.Becausep(x)isofdegree3,and3isnotapowerof2,byTheo­
rem5.5.2wehave thatcisnotconstructible.Sowe cannottrisect60
0
by
straight-edgeandcompass.Thisanswers Question2inthenegative.
Theorem5.5.4.Itisimpossibletotrisect60
0
bystraight-edgeand
compass.
Wehopethatthistheoremwilldissuade anyreaderfromjoiningthe
hordesofangle-trisectors.Therearemoreprofitableandpleasanterwaysof
wastingone'stime.
Thereisyetanotherclassicalproblemofthiskindtowhichtheanswer
is"no."Thisisthequestionofsquaringthecircle.This questionasks:Can
weconstructa squarewhoseareaisthatofacircleofradius1 bystraight­
edge
andcompass?This isequivalenttoaskingwhether
V;isacon­
structible
number.Ifthiswerethecase,thensince
1T=(V;)2,thenumber1T
wouldbeconstructible.ButLindemannprovedin1882that1Tisinfacttran­
scendental,socertainly
isnotalgebraic,andsocannotbeconstructible.There­
fore,thecircle ofradius1 cannotbesquaredbystraight-edgeandcompass.
Ofcourse,whatwedidabovedoesnotconstitutea proofoftheimpos­
sibility
ofsquaringthecircle,since wehavepresupposedLindemann'sresult
withoutprovingit.
Toprovethat
1Tistranscendentalwouldtakeustoofar
afield.
Onemightexpect thatitwouldbeeasiertoprovethat
1Tisnotcon­
structible
thantoprovethatitisnotalgebraic.This doesnotseemtobethe
case.Untilnowallproofs that
1Tisnotconstructiblegovia therouteofex­
ploiting
thetranscendenceof
1T.
PROBLEMS
1.CompletetheproofofTheorem5.5.1.
2.Prove
thatx
3
-
3x-1isirreducibleover
Q.
3.Showthattheconstructiongivenfor~, a2::0doesindeedgiveus~.
4.Provethattheregularheptagon(seven-sidedpolygonwithsides ofequal
length)isnotconstructiblebystraight-edgeandcompass.
6.ROOTSOF POLYNOMIALS
LetF[x],asusual, bethepolynomialringinx overthefieldF andletKbean
extensionfield ofF.IfaEKand

223. 208 Fields Ch.5
thenbyf(a)weunderstandtheelement
f(a)=aa+ala+...+ana
n
inK.Thisis theusagewehave madeofthisnotationthroughoutthischap­
ter.
Wewillnow beinterestedinthosea'sinKsuchthatf(a)=O.
Definition.TheelementaEKisarootofthepolynomialf(x)EF[x]
iff(a)=O.
Inwhatwehave doneupuntilnowwehavealways hadanextension
field
KofFgiventousandweconsideredtheelementsin Kalgebraicover
F,thatis,thoseelementsofKthatarerootsofnonzeropolynomialsin F[x].
WesawthatifaEKisalgebraicoverFofdegreen-thatis,iftheminimal
polynomial
foraoverFisofdegreen-then[F(a): F]=n,whereF(a)is
thesubfieldofKobtainedbyadjoiningatoF.
Whatwedonowisturntheproblemaround.Wenolongerwillhave
theextensionKofFatourdisposal.Infact,ourprincipaltaskwill betopro­
duceitalmost
fromscratch.Westartwithsomepolynomial f(x)ofpositive
degreein
F[x]asouronlybit of
data;ourgoalistoconstructanextension
field
KofFinwhichf(x)willhavearoot.Oncewehavethisconstruction of
Kundercontrol,weshall elaborateonthegeneraltheme,therebyobtaining
aseries
ofinterestingconsequences.
Beforesettingoffonthissearchfor theappropriateK,wemustget
someinformationabouttherelationbetweentherootsofagivenpolynomial
andthefactorizationofthatpolynomial.
Lemma5.6.1.IfaELisarootofthepolynomialf(x)EF[x]ofde­
gree
n,whereLisanextensionfield ofF,thenf(x)factorsin L[x]asf(x)=
(x-a)q(x), whereq(x)isofdegreen-1inL[x].Conversely,if f(x)=
(x-a)q(x),withf(x),q(x),andaasabove,thenaisarootoff(x)inL.
ProofSinceFCL,F[x]iscontainedinL[x].BecauseaEL,x-a is
in
L[x];bytheDivisionAlgorithmforpolynomials,wehave f(x)=
(x-a)q(x) +r(x),whereq(x)andr(x)areinL[x]andwherer(x)=0or
degr(x)<deg(x-a) =1.Thisyields thatr(x)=b,someelementofL.
Substitutingaforxintherelationabove, andusingthefactthatf(a)=0,we
obtain0=(a-a)q(a)+b=0+b=b;thusb=o.Sincer(x)=b=0,we
have
whatwewanted,namely f(x)=(x-a)q(x).
Forthestatementthatdegq(x)=n-1wenotethatsincef(x)=
(x-a)q(x), then,byLemma4.5.2,n=degf(x)=deg(x- a)+degq(x)=
1+degq(x).Thisgivesus therequiredresult,deg q(x)=n-1.
Theconverseiscompletelytrivial. D

224. Sec.6 RootsofPolynomials 209
Oneimmediateconsequence ofLemma5.6.1 is
Theorem5.6.2.Let f(x)inF[x]havedegree n;thenf(x)canhaveat
most
nrootsinanyextension, K,ofF.
ProofWegobyinductionon n.Ifn==1,thenf(x)==ax+b,wherea
andbareinFandwhere a
=1=O.Thustheonlyrootof f(x)is-bfa,anele­
mentof
F.
Supposethatthetheorem iscorrectforallpolynomialsofdegree k-1
over
anyfield.Supposethat f(x)inF[x]isofdegreek.Iff(x)hasnorootsin
K,thenthetheorem iscertainlycorrect.Suppose,then,that aEKisarootof
f(x).ByLemma5.6.1, f(x)==(x-a)q(x), whereq(x)isofdegreek-1in
K[x].Anyroot binKoff(x)iseitheraorisarootof q(x),since0 ==f(b)==
(b-a)q(b).Byinduction,q(x)hasat mostk-1rootsin K,hencef(x)hasat
most
krootsin K.Thiscompletestheinductionandprovesthetheorem. D
Actually,theproofyieldsalittlemore.Toexplainthis"littlemore,"we
needthenotionofthemultiplicityofaroot.
Definition.
IfKisanextensionof F,thentheelement ainKisaroot
ofmultiplicityk>0off(x),wheref(x)isinF[x],iff(x)==(x-a)kq(x) for
some
q(x)inK[x]andx-a doesnotdivideq(x)(or,equivalently,where
q(a)
=1=0).
ThesameproofasthatgivenforTheorem5.6.2yieldsthesharpened
version:
Letf(x)beapolynomial ofdegreen inF[x];thenf(x)canhaveatmostn
roots
inanyextensionfieldK ofF,countingaroot ofmultiplicitykaskroots.
Theorem5.6.3.Let f(x)inF[x]bemonicofdegree nandsuppose
that
Kisanextensionof Finwhichf(x)hasnroots,countingaroot ofmulti­
plicity
kaskroots.Iftheserootsin Karea
ba2,...,am'eachhaving
multiplicitykbk2,...,kmrespectively,then f(x)factorsin K[x]asf(x)==
(x-al)k
1(x-a2)k
2
•
••(x-am)k
m
•
ProofTheproof iseasybymakinguse ofLemma5.6.1andofinduc­
tionon
n.Weleavethecarryingoutofthe prooftothereader. D
Definition.Wesaythat f(x)inF[x]splitsintolinearfactorsover(or
in)K
iff(x)hasthefactorizationin K[x]giveninTheorem5.6.3.
There
isaniceapplication ofTheorem5.6.3tofinitefields. LetFbea
finitefieldhaving
qelements,andlet
aba2'...,a
q
-lbethenonzeroele-

225. 210 Fields Ch.5
mentsof F.Sincetheseformagroupoforder q-1underthemultiplication
in
F,byTheorem2.4.5(provedeversolongago), a
q
-1
==1forany a*0in
F.Thusthepolynomial x
q
-
1
-
1inF[x]hasq-1distinctroots inF.By
Theorem5.6.3,thepolynomial
x
q
-
1
-
1
==(x-al)(x-a2)...(x-aq-l)'If
wealsoconsider
0,theneveryelement ainFsatisfiesa
q
==a,sothatthe
polynomial
x
q
-
xhastheqelementsof Fasitsdistinctroots.ByTheorem
5.6.3
wehave
Theorem5.6.4.Let
Fbeafinitefieldhaving qelements.Then x
q
-
x
factorsin F[x]as
x
q
-
x
==x(x-a1)(x-a
2
)
•••(x-a
q
-1
),
where
aba2,""aq-larethenonzeroelementsof F,and
x
q
-1
-
1
==(x-a1)(x-a
2
)
•••(x-a
q
-
1
).
Averyspecialcaseofthistheorem isthatinwhichF
==7L
p
,theintegers
modulotheprime
p.Hereq
==pandaba2,...,ap-larejust1,2,...,p-1
insomeorder.
Corollary.In
7L
p[x],thepolynomialx
p
-1
-
1factorsas
x
p
-
1
- 1
==(x-l)(x-2)...(x- (p-1».
Trythisoutfor p==5,7,and11.
Asacorollarytothecorollary, wehavearesult innumbertheory,
knownas
Wilson'sTheorem, whichweassigned asProblem18inSection4of
Chapter
2.
Corollary.lfpisaprime,then (p-I)!
==-1modp.
ProofBytheCorollaryabove,
Xp~l-1==(x-l)(x-2)...(x-(p-1»;
substitutingx==0inthisgivesus
-1==(-1)(-2)'"(-(p-1»==(-1)P-
1
1'2···(p-1)
==(-l)P-l(p-I)!
in
7L
p
•Intheintegersthistranslatesinto"congruentmod p."Thus
(-l)P-l(p-I)!
==-1modp,
andso (p-I)!==(-l)Pmodp.But(-l)P==-1modp;hencewehave
provedWilson'sTheorem.
D

226. Sec.6 RootsofPolynomials 211
Wechangedirectiontoconsidertheproblemmentionedatthebegin­
ningofthissection:given
f(x)EF[x],toconstructafiniteextension KofF
inwhichf(x)hasaroot.Asweshallseeinamoment,thisconstructionof K
willbequiteeasywhenwebringtheresultsaboutpolynomialringsprovedin
Chapter4intoplay.However,toverifythatthisconstructionworkswilltake
abitofwork.
Theorem
5.6.5.LetFbeafieldand f(x)apolynomialofpositivede­
gree
ninF[x].Thenthereexistsafiniteextension KofF,with[K:F]
~n,in
which
[(x)hasaroot.
ProofByTheorem4.5.12, [(x)isdivisiblein F[x]bysomeirreducible
polynomial
p(x)inF[x].Sincep(x)dividesf(x),degp(x)
~degf(x)=n,
andf(x)=p(x)q(x)forsomepolynomial q(x)inF[x].Ifbisarootof p(x)
insomeextensionfield,then bisautomaticallyaroot of[(x),sincef(b)=
p(b)q(b)=Oq(b)=O.Sotoprovethetheoremit isenoughtofindanexten­
sionof
Finwhichp(x)hasaroot.
Because
p(x)isirreduciblein F[x],theidealM=(p(x»ofF[x]gener­
atedby
p(x)isamaximalidealof F[x]byTheorem4.5.11.ThusbyTheorem
4.4.2,K =F[x]/Misafield.Weclaimthatthis isthefieldthatweareseeking.
Strictlyspeaking,
Kdoesnotcontain
F;aswenowshow,however, K
doescontainafieldisomorphicto F.Sinceeveryelementin Misamultiplein
F[x]ofp(x),everysuchnonzeroelementmusthavedegreeatleastthatof
p(x).Therefore,MnF=(0).Thusthehomomorphismt/J:F[x]~Kdefined
byt/J(g(x»=g(x)+Mforeveryg(x)inF[x],whenrestrictedto F,is1 - 1on
F.Therefore,theimageFofFinKisafieldisomorphicto F.Wecanidentify
F,viat/J,withFandso,inthisway,wecanconsider Kanextensionof F.
Denotex+MEKbya,sothatt/J(x)=a,aEK.Weleaveittothe
readertoshow,fromthefactthatt/JisahomomorphismofF[x]ontoKwith
kernel
M,that
t/J(g(x»=g(a)foreveryg(x)inF[x].Whatist/J(p(x»?On
theonehand,since p(x)isinF[x],t/J(p(x»=pea).Ontheotherhand,since
p(x)isinM,thekerneloft/J,t/J(p(x»=O.Equatingthesetwoevaluations of
t/J(p(x»,wegetthat pea)=O.Inotherwords,theelementa=t/J(x)inKisa
root
o[p(x).
Tofinishtheproof,allweneed istoshowthat [K:F]=degp(x)
~n.
Thiscameupearlier,inthealternativeproofwegave ofTheorem5.3.5.
Thereweleftthispointtobeprovedbythereader.Weshallbealittlemore
generoushereandcarryouttheproofindetail.
Given
hex)inF[x],then,bytheDivisionAlgorithm, hex)=p(x)q(x)+
rex)whereq(x)andrex)areinF[x],andrex)=0ordegrex)<degp(x).
Goingmodulo M,weobtainthat

227. 212 Fields
tfJ(h(x»==tfJ(p(x)q(x)+rex»~ ==tfJ(p(x)q(x»+tfJ(r(x»
==tfJ(p(x»tfJ(q(x»+tfJ(r(x»
==tfJ(r(x»==rea)
Ch.5
[sincetfJ(p(x»==pea)==0].
So,sinceeveryelementin K==F[x]/MistfJ(h(x»forsomehex)inF[x]
andtfJ(h(x»==rea),weseethateveryelementof Kisoftheform rea),where
rex)isinF[x]anddeg rex)<degp(x).Ifdegp(x)==m,thediscussionjust
madetellsusthat
1,a,a
2
,
•••,a
m
-
1
spanKoverF.Moreover,theseelements
arelinearlyindependentover
F,sincearelationofthetype
aa+ala+...
+am_la
m
-
1 ==0wouldimplythat g(a)==0whereg(x)==
aa+alx+...+
am_lX
m
-
1 isinF[x].Thisputs g(x)inM,whichisimpossiblesince g(x)isof
lowerdegreethan
p(x),unlessg(x)==O.Inotherwords,wegetacontradic­
tionunless
aa==al==...=exm-l==o.Sotheelements 1,a,a
2
,
•••,a
m
-
l
are
linearlyindependentover
F.Sincetheyalsospan KoverF,theyforma basis
ofKoverF.Consequently,
dimFK==[K:F]==m==degp(x)
::;n==deg[(x).
Thetheorem isproved.D
Wecarryoutaniterationoftheargumentusedinthelastproofto
provetheimportant
Theorem
5.6.6.Let[(x)EF[x]beofdegreen.Thenthereexistsan
extension
KofFofdegreeat most n!overFsuchthat[(x)hasnroots,count­
ingmultiplicities,in
K.Equivalently,[(x)splitsintolinearfactorsover K.
ProofWegobyinductionon n.Ifn==1,then[(x)=
a+{3x,wherea,
{3EFandwhere{3=1=O.Theonlyrootof [(x)is-exl{3,whichisinF.Thus
K==Fand[K:F]==1.
Supposethattheresultistrueforallfieldsforpolynomialsofdegree k,
andsupposethat [(x)EF[x]isofdegreek+1.ByTheorem5.6.5thereex­
istsanextension
K
1ofFwith[K
1
:F]
::;k+1inwhich [(x)hasaroot al.
Thusin K1[x],[(x)factorsas [(x)=(x-al)q(x),whereq(x)EK
1[x]isof
degree
k.Byinductionthereexistsanextension KofK1ofdegreeatmost k!
overKloverwhichq(x)splitsintolinearfactors.Butthen [(x)splitsintolin­
earfactorsover K.Since[K:F]==[K:
Ktl[Kl:F]::;(k+l)k!==(k+I)!,
theinduction
iscompletedandthetheorem isproved.D
Weleavethesubjectoffieldextensionsatthispoint.Weareexactlyat
whatmightbedescribedasthebeginningofGaloistheory.Havinganexten-

228. Sec.6 RootsofPolynomials 213
sionK ofFoffinitedegreeoverwhichagiven polynomiall(x)splitsintolin­
earfactors,thereexistsanextensionofleastdegreeenjoyingthis property.
Suchanextensioniscalleda splittingfield ofI(x)overF.Onethenproceeds
toprovethatsuchasplittingfieldis uniqueuptoisomorphism.Oncethisis
inhandtheGaloistheorygoesintofullswing,studying therelationshipbe­
tweenthegroupofautomorphismsofthissplittingfield anditssubfieldstruc­
ture.Eventually,it
leadstoshowing,amongmanyotherthings,thatthere
existpolynomials overtherationalsofalldegrees5orhigherwhoseroots
cannotbeexpressednicelyin termsofthecoefficientsofthesepolynomials.
This
isabriefandverysketchydescriptionofwherewecangofrom
hereinfieldtheory. Butthereisnohurry.Thereadersshouldassimilatethe
materialwehavepresented;thiswillputtheminagoodpositiontolearn
Galoistheoryiftheyaresoinclined.
PROBLEMS
1.ProveTheorem5.6.3.
2.
IfFisafinitefieldhaving theq-1nonzeroelementsaI,a2'...,aq-I,
provethataIa2...aq-I=(-l)q.
3.Let
Qbetherationalfieldandletp(x)=x
4
+x
3
+x
2+X+1.Show
thatthereisanextensionKof
Qwith[K:0]=4overwhichp(x)splits
into
linearfactors.[Hint: Findtherootsofp(x).]
4.Ifq(x)=x
n+aIx
n
-
I+...+an,an
=1=0,isapolynomialwithinteger
coefficientsandiftherationalnumberrisarootofq(x),provethatris
anintegerandrIan.
5.Show thatq(x)=x
3
-
7x+11isirreducibleover
Q.
6.IfFisafieldofcharacteristicp=1=0,showthat(a+b)P=a
P+bPforall
aandbinF.
7.ExtendtheresultofProblem6byshowingthat(a+b)m=am+b
m
,
wherem=pn.
8.LetF=7L
p
,paprime,andconsiderthepolynomialx
m
-
xin7L
p[x],
wherem=pn.LetKbeafiniteextensionof7L
p
overwhichx
m
-
xsplits
into
linearfactors.InKletKobethesetofallrootsofx
m
-
x.Showthat
Koisafieldhaving atmostpnelements.
9.InProblem8showthatKohasexactlypn elements.(Hint:SeeProblem
14.)
10.ConstructanextensionfieldK
nof
Qsuchthat[K
n
:0]=n,forany
n2::1.

229. 214 Fields
11.DefinethemappingB:F[x]~F[x]by
B(ao+atx+a2x2+ ...+anx
n
)
=at+2a2x+...+iaix
i
-
t +...+nanx
n
-
1
•
Ch.5
Provethat:
(a)
B(f(x)+g(x»=B(f(x»)+B(g(x».
(b)B(f(x)g(x»=f(x)B(g(x»+B(f(x»g(x)forallf(x)andg(x)inF[x].
12.IfFisofcharacteristicp=1=0,characterizeall f(x)inF[x]suchthat
B(f(x»=0.
13.Show thatiff(x)inF[x]hasa rootofmultiplicitygreater than1insome
extensionfield
ofF,thenf(x)and
B(f(x»arenotrelativelyprimein
F[x].
14.IfFisofcharacteristicp=1=0,showthatalltheroots ofx
m
-
x,where
m=pn,aredistinct.
15.
Iff(x)inF[x]isirreducibleandhasa rootofmultiplicitygreaterthan 1
insomeextension ofF,showthat:
(a)
Fmustbeofcharacteristicpforsomeprime p.
(b)f(x)=g(x
P
)
for somepolynomial g(x)inF[x].

230. 6
SPECIALTOPICS(OPTIONAL)
Inthisfinalchapter wetreatseveralunrelatedtopics. Oneofthesecomes
fromgrouptheory,andalltherestfromthetheoryoffields.Inhandling
thesespecialtopics,wedrawfrommany
oftheresultsandideasdeveloped
earlierinthebook.Althoughthesetopicsaresomewhatspecial,each
of
themhasresultsthataretrulyimportantintheirrespectiveareas.
Thereaderswhohavemanagedtosurvivesofarshouldhavepickedup
acertainsetoftechniques,experience,andalgebraicknow-howtobeableto
followthematerialwithacertaindegree
ofease.Wenowfeelfreeto treat
thevariousmattersathandinasomewhatsketchierfashionthanwehave
heretofore,leavingafewmoredetailstothereaderto
fillin.
Thematerialweshallhandledoes
notlenditselfreadilytoproblems,at
leastnottoproblemsofareasonabledegreeofdifficulty.Accordingly,we
willassignrelativelyfewexercises.Thisshouldcomeasarelieftothose
wantingtoassimilatethematerialinthischapter.
1.THE SIMPLICITYOFAn
InChapter3,wherewediscussed Sn,thesymmetricgroup ofdegreen,we
showedthatifn
;:::2,thenSnhasanormalsubgroup An,whichwecalledthe
alternatinggroupofdegreen, whichisagroupof ordern!/2.Infact,Anwas
merelythesetofallevenpermutationsinS
n.
215

231. 216 SpecialTopics(Optional) Ch.6
IndiscussingAn,wesaidthatAn,forn2::5,wasasimplegroup,that is,
thatAnhasnonormalsubgroups otherthan(e)anditself.Wepromised
there
thatwewouldprovethisfactin Chapter6.Wenowmakegoodonthis
promise.
Tomakeclearwhatit isthatweareabouttoprove,weshouldperhaps
repeatwhatwesaidabove
andformallydefinewhat ismeantbyasimple
group.
Definition.Anonabeliangroup
issaidtobe simpleifitsonlynormal
subgroupsare
(e)anditself.
Weimposetheproviso
thatGbenonabeliantoexcludethetrivialex­
amples
ofcyclicgroups ofprimeorderfromthedesignation"simple."These
cyclicgroups
ofprimeorderhavenonontrivialsubgroupsatall,so,perforce,
theyhaveno
propernormalsubgroups. Anabeliangroupwithnoproper
subgroups
iseasilyseentobecyclic ofprimeorder.
Webeginwiththeveryeasy
Lemma6.1.1.If
n
2::3andTbT2aretwotranspositionsin Sn'thenT1T2
iseithera3-cycle ortheproductoftwo3-cycles.
ProofIfT1==T2'thenT1T2==Ti==eandeiscertainlytheproduct of
two3-cycles,forinstance ase==(123)(132).
If
T1
=1=T2'thentheyeitherhaveoneletterincommon ornone.Ifthey
haveoneletterincommon,wemaysuppose,onasuitablerenumbering,that
T1==(12)and
T2==(13).ButthenT1T2 ==(12)(13)==(132),which isalreadya
3-cycle.
Finally,if
T1andT2havenoletterincommon,wemaysuppose,without
lossofgenerality,that
T1==(12)and T2==(34),inwhichcase TlT2==(12)(34)==
(142)(143),which isindeedtheproduct oftwo3-cycles.Thelemma isnow
proved.
D
Animmediateconsequence ofLemma6.1.1isthatforn
2::3the
3-cyclesgenerate
An,thealternatinggroup ofdegreen.
Theorem6.1.2.If
(Jisanevenpermutationin Sn,wheren2::3,then(J
isaproductof3-cycles.In otherwords,the3-cyclesin SngenerateAn.
ProofLet(JESnbeanevenpermutation.Bythedefinitionofthepar­
ity
ofapermutation,
(Jisaproductofanevennumberoftranspositions.
Thus(J==T1T2•.•T2i-1T2i...T2m-1T2misaproductof2mtranspositions
TbT2,...,T2m·ByLemma6.1.1,each T2i-1T2iiseithera3-cycle oraproduct

232. Sec.1 The SimplicityofAn 217
oftwo3-cycles.Sowe getthat(J"iseithera3-cycleortheproductofatmost
2m3-cycles.Thisproves thetheorem.D
Wenowgive analgorithmforcomputing theconjugateofanypermu­
tationin Sn.Let(J"ESn'andsupposethat(J"(i)==j.WhatdoesT(J"T-1look
likeifTESn?SupposethatT(i)==sandT(j)==t;thenT(J"T-
1
(S)==
T(J"(T-1(S»==T(J"(i)==T(j)==t.Inotherwords,tocomputeT(J"T-
1
replace
everysymbolin(J"byitsimageunderT.
Forinstance,if(J"==(123)andT==(143),then,since T(l)==4,T(2)==2,
T(3)==1,andT(4)==3,wesee thatT(J"T-
1
==(421)==(142).
Giventwok-cycles,say(12
...k)and(i
1i
2
•••i
k
),thentheyareconju­
gatein
Snbecauseif
Tisapermutationthatsends1intoi
b2intoi
2
,
•••,k
intoi
k
,thenT(12...k)T-
1
==(i
1i
2
, •
••,i
k
).Sinceevery permutationisthe
productofdisjointcycles andconjugationisanautomorphism,weget,from
theresultfork-cycles, thattocompute
T(J"T-1foranypermutation(J",replace
everysymbolin(J"byitsimageunderT.Inthiswaywesee thatitisextremely
easyto
computetheconjugateofanypermutation.
Giventwo permutations
(J"1and(J"2inSn,thentheyareconjugatein Sn,
usingtheobservationabove,ifin theirdecompositionsintoproductsofdis­
jointcyclestheyhave
thesamecyclelengths andeachcyclelengthwith the
samemultiplicity.Thus,forinstance,(12)(34)(567) and(37)(24)(568)are
conjugatein Sg,but(12)(34)(567)and(37)(568)arenot.
Recall
thatbyapartitionofthepositiveinteger n,wemeanadecompo­
sition
ofnasn==n1+n2+...+nk,where0
::;n1::;n2::;...::;nk.If(J"in
Snisthedisjointproductofann1-cycle,ann2-cycle, ...,annk-cycle,then
n1+n2+...+nk==n,andapermutationTisconjugateto(J"ifandonlyifT
isthedisjointproductofcyclesin thesameway.Therefore,thenumberof
conjugacyclassesin Snisequaltothenumberofpartitionsofn.
Forinstance,if n=4,thenthepartitionsof4are4=4,4=1+3,
4
==1+1+2,4=1+1+1+1,and4==2+2,whicharefive innum­
ber.Thus
S4hasfiveconjugacyclasses, namelytheclassesof(1234),(123),
(12),e,and(12)(34),respectively.
Wesummarizeeverythingwesaidabovein threedistinctstatements.
Lemma6.1.3.Tofind
T(J"T-1inSn,replaceeverysymbolin thecycle
structure
of
(J"byitsimage underT.
Lemma6.1.4.Twoelementsin Snareconjugateif theyhavesimilar
decompositionsas
theproductofdisjointcycles.
Lemma6.1.5.Thenumberofconjugacyclassesin Snisequaltothe
numberofpartitionsofn.

233. 218 SpecialTopics(Optional) Ch.6
Clearly,from theresultsabove, anytwo3-cyclesin Snareconjugatein
Sn.A3-cycleisanevenpermutation,soisinAn.Onemightwonderifany
two3-cycles
areactuallyconjugatein thesmallergroupAn.Forn
;:::5the
answeris"yes,"andisquiteeasy toprove.
Lemma6.1.6.Ifn;:::5,thenanytwo3-cyclesin Snarealreadyconju­
gateinAn.
ProofLet0"1and0"2betwo3-cyclesinSn;byLemma6.1.4they are
conjugatein Sn.Byrenumbering,we mayassumethat0"1==(123)and0"2==
T(123)T-
1
forsomeTESn.IfTiseven,thenwearedone.IfTisodd,then
p==T(45)isevenandp(123)p-l==T(45)(123)(45)-lT-1==T(123)T-
1
==0"2.
Therefore,0"1and0"2areconjugateinAn"Wethussee thatthelemmaiscor­
rect.
D
InS3thetwo3-cycles(123) and(132)areconjugatein S3butarenot
conjugatein A
3
,whichisacyclicgroupoforder3.
Wenowprovearesultthatisnotonlyimportantingrouptheory,but
alsoplaysa keyroleinfield theoryandthetheoryofequations.
Theorem6.1.7.Ifn
;:::5,thentheonlynontrivialpropernormalsub­
groupofSnisAn.
ProofSupposethatNisanormalsubgroupofSnandNisneither(e)
norSn.Let0"=1=ebeinN.SincethecenterofSnisjust(e)(SeeProblem1)
andthetranspositionsgenerateSn,thereisatranspositionTsuchthat
O"T=1=TO".ByLemma6.1.4,T1==O"TO"-lisatransposition,soTTl==TO"TO"-l =I=-e
isinN,since0"ENandTO"T==TO"T-
1
ENbecauseNisnormalinSn.SoN
containsanelementthatistheproductoftwotranspositions,namely TTl.
IfTandT1havea letterincommon,then,aswesawin theproofof
Lemma6.1.1,TTlisa3-cycle,henceNcontainsa3-cycle.By Lemma6.1.4all
3-cyclesin
Snareconjugateto
TTlsomustfallinN,bythenormalityofNin
Sn'ThusthesubgroupofSngeneratedbythe3-cycles,which,accordingto
Theorem6.1.2,isallofAn,liesin N.Notethatuptothispointwehave not
usedthatn;:::5.
Wemaythusassume thatTandT1havenoletterincommon.Without
loss
ofgeneralitywe mayassumethat
T==(12)andT1==(34);therefore,
(12)(34)
isinN.Sincen
~5,(15)isinSnhence(15)(12)(34)(15)-1 ==
(25)(34)isalsoinN;thus(12)(34)(25)(34) ==(125)isinN.Thusinthiscase
also,
Nmustcontaina3-cycle.TheargumentabovethenshowsthatN
~An.
Wehaveshown thatinbothcasesNmustcontainAn"Sincethereare
nosubgroupsstrictly betweenAnandSnandN=1=Sn,weobtainthedesired
result
thatN==
An"D

234. Sec.1
Theresultisfalse forn==4;thesubgroup
TheSimplicityofAn 219
N==fe,(12)(34),(13)(24),(14)(23)}
isapropernormalsubgroupofS4andisnotA
4
•
WenowknowallthenormalsubgroupsofSnwhenn
2::5.Canwede­
terminefromthisall thenormalsubgroupsofAnforn2::5?Theansweris
"yes";asweshall
soonsee,Anisasimplegroupifn
2::5.Theproofwegive
maystrikemanyasstrange,forithingesonthefactthat60,theorderofAs,
isnotaperfectsquare.
Theorem6.1.8.ThegroupAsisasimplegroupoforder60.
ProofSupposethatAsisnotsimple;thenithasa propernormal
subgroupNwhoseorderisas smallaspossible.LetthesubsetT==
{o'ESslNO'-
1
eN},thenormalizerofNinSs.SinceNisnormalinAs,we
knowthatT:JAs.TisasubgroupofSs,soifT=1=As,wewouldhavethat
T==Ss.Butthiswouldtellus thatNisnormalinSs,which,byTheorem
6.1.7,wouldimply thatN:JAs,givingus thatN==As,contrarytooursuppo­
sition
thatNisapropersubgroupofAs.SowemusthaveT ==As.Since(12)
isodd,it isnotinAs,henceisnotinT.Therefore,M==(12)N(12)-1
=1=N.
SinceN<lAs,wealsohavethatM<lAs(Prove!),thusbothMnN
andMN=={mn1mEM,nEN}arenormalinAs.(SeeProblem9.)Because
M=1=NwehavethatMnN=1=N,andsinceNisaminimalpropernormal
subgroupofAs,itfollowsthatMnN==(e).Ontheotherhand,
(12)MN(12)-1==(12)M(12)-1(12)N(12)-1==NM(since(12)N(12)-1==M
and(12)M(12)-1==N)==MNbythenormalityofMandNinAs.There­
fore,theelement(12)isin thenormalizerofMNinSs ,andsinceMNisnor­
malinAs,weget,aswedidabove, theMNisnormalinSs,andsoMN==As
byTheorem6.1.7.
Considerwhatwenowhave. BothMandNarenormalsubgroupsof
As,eachoforderINI,andMN==AsandMnN==(e).Weclaim,andleave
tothereader,thatMNmustthenhaveorderINI
2
•
SinceMN==As,weob­
tain
that60==IAsl=IMNI==INI
2
•
Butthisissheernonsense,since60is not
thesquareofanyinteger.Thisestablishes Theorem6.1.8.D
TogofromthesimplicityofAstothatofAnforn
2::5isnottoohard.
Notethattheargumentwegavefor Asdidnotdependon5untilthepunch
line"60isnotthesquareofanyinteger."Infact,thereasoningisvalidas
longaswe
knowthatn!/2isnotaperfectsquare.Thus,forexample,if n==6,
then6!/2==360isnotasquare,henceA
6isasimplegroup.Sinceweshall
needthisfactin thesubsequentdiscussion,werecorditbeforegoingon.

235. 220 SpecialTopics(Optional)
Corollarytothe ProofofTheorem6.1.8.A
6isasimplegroup.
Ch.6
Wereturntothequestionofwhether ornotn!/2isasquare.Asamat­
teroffact,it isnotifn>2.Thiscanbeshownasaconsequenceofthebeau­
tifultheoreminnumbertheory(theso-calledBertrandPostulate),whichas­
sertsthatfor
m>1thereisalwaysaprimebetween mand2m.Sincewedo
nothavethisresultatourdisposal,wefollowanotherroadtoshowthesim­
plicityof
Anforalln
~5.
Wenowprovethisimportanttheorem.
Theorem6.1.9.
Foralln
~5thegroup Anissimple.
ProofByTheorem6.1.8wemayassumethat n~6.Thecenterof An
forn>3ismerely(e).(Prove!)Since Anisgeneratedbythe3-cycles,if
U=1=eisinAn,then,forsome3-cycle T,UT=1=TU.
Supposethat N=1=(e)isanormalsubgroupof AnandthatU=1=eisinN.
Thus,forsome3-cycle T,UT=1=TU,whichistosay,UTU-1T-
1
=1=e.Because
Nisnormalin An,theelementTU-
1
T-
1
isinN,henceUTU-
1
T-
1
isalsoin N.
SinceTisa3-cycle,somustUTU-1alsobea3-cycle.Thus Ncontainsthe
productoftwo3-cycles,andthisproduct
isnote.Thesetwo3-cyclesinvolve
atmostsixletters,socanbeconsideredassittingin
A
6which,since n
~6,
canbeconsideredembeddedisomorphicallyin An.(Prove!)Butthen
NnA6=1=(e)isanormalsubgroupof A6,sobytheCorollaryabove,
NnA
6
=A
6
•Therefore,Nmustcontaina3-cycle,andsinceall3-cyclesare
conjugatein
An(Lemma6.1.6), Nmustcontainallthe3-cyclesin Sn'Since
these3-cyclesgenerate
An,weobtainthat NisallofAn,therebyprovingthe
theorem.
D
TherearemanydifferentproofsofTheorem 6.1.9-theyusuallyin­
volveshowingthatanormalsubgroup
ofAnmustcontaina 3-cycle-which
areshorterandpossiblyeasierthantheonewegave.However,welikethe
bizarretwistintheproofgiveninthatthewholeaffairboilsdowntothefact
that
60isnotasquare.Werecommendtothereadertolookatsomeother
proofsofthisveryimportanttheorem,especiallyinabookongrouptheory.
The
Anprovideuswithaninfinitefamilyoffinitesimplegroups.There
areseveral
otherinfinitefamilies offinitesimplegroupsand 26particular
onesthatdonotbelongtoanyinfinitefamily.Thisdeterminationofall
finitesimplegroups,carriedoutinthe1960sand1970sbyalargenumberof
grouptheorists,
isoneofthemajorachievementsoftwentieth-centurymath­
ematics.

236. Sec.2
PROBLEMS
FiniteFieldsI 221
*1.Provethatifn>2,thecenterofSnis(e).
*2.Provethatifn>3,thecenterofAnis(e).
3.Whatcanyousay aboutthecyclestructure oftheproductoftwo
3-cycles?
4.
Ifm<n,showthatthereisasubgroupofSnisomorphictoSm.
5.Show thatanabeliangrouphaving nopropersubgroupsiscyclicof
primeorder.
6.Howmanyconjugacyclasses arethereinS6?
7.IftheelementsaI,a2,...,angeneratethegroupG andbisanoncentral
element
ofG,provethatbai
=1=aibforsome i.
8.IfM<lNandN<JG,showthataMa-
1
isnormalinNforeveryaEG.
9.IfM<lGandN<lG,showthatMNisanormalsubgroupofG.
10.Ifn2::5isodd,show thatthen-cyclesgenerateAn.
11.Showthatthecentralizerof(12...k)inSnhasorderk(n-k)!andthat
(12···k)hasn!/(k(n-k)!)conjugatesin Sn.
12.IntheproofofTheorem6.1.8,showthatIMNI=INI
2
•
2.FINITEFIELDSI
Ourgoalinthissection andthenexttwo istogetacompletedescription of
allfinitefields. Whatweshallshow isthatthemultiplicativegroup of
nonzeroelements ofafinitefield isacyclicgroup.Thiswe dointhissection.
Inthenexttwo, theobjectiveswill betoestablishtheexistenceandunique­
nessoffinitefieldshaving
pnelementsforanyprime pandanypositiveinte­
ger
n.
Someofthethingswe areabouttodoalreadycame upintheproblem
setsingrouptheory andfieldtheoryashardproblems.Thetechniquesthat
weusecomefromgrouptheory andfieldtheory,withalittle numbertheory
thrownin.
Werecallwhat
theEuler
q;-functionis.WedefinetheEulerq;-function
by:q;(1)=1and,for n>1,q;(n)isthenumberofpositiveintegersless than
nandrelativelyprime ton.
Webeginwitharesultin numbertheorywhoseproof,however,willex­
ploitgrouptheory.Beforedoing
thegeneralcase,we doanexample.
Letn=12;then
q;(12)=4,foronly1,5,7,and11arelessthan12and
relativelyprime to12.Wecomputeq;(d)forallthedivisors of12.Wehave:

237. 222 SpecialTopics(Optional) Ch.6
cp(1)=1,cp(2)=1,cp(3)=2,cp(4)=2,cp(6)=2,andcp(12)=4.Notethat
thesum
ofall
cp(d)overallthedivisors of12is12.Thisisnoflukebut isa
specialcase
of
Theorem6.2.1. Ifn
2::1,then'Lcp(d)=n,wherethissumrunsoverall
divisors
dofn.
ProofLetGbeacyclicgroup oforderngeneratedbytheelement a.
Ifdin,howmanyelementsofGhaveorderd?Ifb =a
n
/
d
,
thenallthesolu­
tionsinGofx
d
=earethepowers e,b,b
2
,
•••,b
d
-
1
ofb.Howmanyof
thesehave
orderd?Weclaim,andleavetothereader,that b'hasorderdif
andonlyif
risrelativelyprimeto d.Sothenumber ofelementsoforder din
G,foreverydivisord
ofn,is
cp(d).EveryelementinGhasordersomedivi­
sor
ofn,soifwesumupthenumberofelementsof orderd-namely
cp(d)­
overall ddividingn,weaccountforeachelementof Gonceandonlyonce.
Hence'Lcp(d)=nifwerunoverallthedivisorsdofn.Thetheorem isnow
proved.
D
Inafinitecyclicgroup ofordernthenumber ofsolutionsofx
d
=e,the
unitelement
ofG,isexactlydforeverydthatdividesn.Weusedthisfactin
theproof
ofTheorem6.2.1.Wenowproveaconversetothis,gettingthereby
acriterionforcyclicity
ofafinitegroup.
Theorem6.2.2.LetGbeafinitegroupoforder
nwiththeproperty
thatforeverydthatdividesnthereareatmost dsolutionsof x
d
=einG.
ThenG isacyclicgroup.
ProofLet
J(d)bethenumberofelementsof Goforderd.Byhy­
pothesis,if
aEGisoforderd,thenallthesolutionsof x
d
=earethedis­
tinctpowers
e,a,a
2
,
•••,a
d
-
1,ofwhichnumber,cp(d)areoforder d.Soif
there
isanelementoforderdinG,then
tfJ(d)=cp(d).Ontheotherhand,if
there.
isnoelementinGof orderd,then
tfJ(d)=O.Soforall dinwehave
thattfJ(d)::::;cp(d).However,sinceeveryelementofGhassomeorderdthat
dividesnwehave
that
'LtfJ(d)=n,wherethissumrunsoveralldivisorsdof
n.But
n='LtfJ(d)::::;'Lcp(d)=n
sinceeachtfJ(d)::::;cp(d).Thisgivesusthat'LtfJ(d)="Lcp(d),which,together
withtfJ(d)::::;cp(d),forcestfJ(d)=cp(d)foreverydthatdividesn.Thus,in
particular,tfJ(n)=cp(n)2::1.Whatdoesthistellus? Afterall,tfJ(n)isthe
number
ofelementsin Gofordern,andsince
tfJ(n)2::1theremustbean ele­
mentainG ofordern.Therefore,theelements e,a,a
2
,
•••,a
n
-1
arealldis-

238. Sec.2 FiniteFieldsI 223
tinctandare ninnumber,sotheymustgiveall ofG.ThusG iscyclicwith a
asgenerator,proving thetheorem.D
Isthereanysituationwherewecan besurethattheequationx
d
=e
hasatmost dsolutionsinagivengroup?Certainly. IfK*isthegroupof
nonzeroelements ofafieldundermultiplication,thenthepolynomialx
n
-
1
hasatmostnrootsin
K*byTheorem5.6.2.So,ifG CK*isafinitemulti­
plicativesubgroup
ofK*,thenthenumberofsolutionsofx
d
=1inG isat
mostdforanypositiveinteger d,socertainlyforall dthatdividetheorder
ofG.ByTheorem6.2.2Gmust beacyclicgroup.Wehaveproved
Theorem
6.2.3.IfKisafieldandK*isthegroupofnonzeroele­
ments
ofKundermultiplication,thenanyfinitesubgroup ofK*iscyclic.
Averyspecialcase
ofTheorem6.2.3,butatthemomentthemostim­
portantcaseforus,
is
Theorem6.2.4.IfKisafinitefield, thenK*isacyclicgroup.
Proof.K*isafinitesubgroup ofitself,so,by Theorem6.2.3,K*is
cyclic.D
Aparticularinstance ofTheorem6.2.4isofgreatimportanceinnum­
bertheory,whereit isknownas theexistenceofprimitiverootsmodpforpa
prime.
Theorem6.2.5.Ifpisaprime,then
Z;isacyclicgroup.
PROBLEMS
1.IfaEGhasorderd,provethataralsohas orderdifandonlyifrandd
arerelativelyprime.
2.Findacyclicgenerator(primitiveroot)forZ~l.
3.DoProblem2forZi7.
4.Constructafield Khavingnineelements andfindacyclicgeneratorfor
thegroup
K*.
5.Ifpisaprimeand m=p2,then
Zmisnotafieldbuttheelements
{[a]I(a,p)=I}formagroup underthemultiplicationinZm.Provethat
thisgroup iscyclicoforderp(p-1).

239. 224 SpecialTopics(Optional) Ch.6
6.Determineallthefinitesubgroups ofC*,whereCisthefieldofcomplex
numbers.
Intherest
oftheproblemshere
q;willbetheEulerq;-function.
7.Ifpisaprime,show thatq;(pn)=pn-I(p-1).
8.If
mandnarerelativelyprimepositiveintegers,prove that
q;(mn)=q;(m)q;(n).
9.Usingtheresult ofProblems7 and8,findq;(n)intermsofthefactoriza­
tion
ofnintoprimepowerfactors.
10.Prove
thatlim
q;(n)=00.
n---..oo
3.FINITEFIELDSII:EXISTENCE
LetKbeafinitefield.Then Kmustbe ofcharacteristicp,paprime,and Kcon­
tains
0,1,2,...,p-1,thepmultiplesoftheunitelement1 ofK.SoK
:J7L
p
,
or,moreprecisely, Kcontainsafieldisomorphicto 7L
p
•SinceKisavector
spaceover
7L
pandclearlyisoffinitedimensionover 7L
p
,if[K:7L
p
]=n,then
Khaspnelements.Toseethis,let
VbV2,...,VnbeabasisofKover7L
p
•Then
foreverydistinctchoiceof(a},a2,'..,an),wheretheaiarein 7L
p
,theelements
aredistinct.Thus,sincewecanpick(aI,a2,...,an)inpnways,Khaspn
elements.
ThemultiplicativegroupK* ofnonzeroelements ofKisagroupof
orderpn- 1.So,wehave thata
m
-I
=1,wherem=pn,forevery ainK,
henceam=a.Sincethis isalsoobviouslytruefor a=0,wehavethat am=a
foreveryainK.Therefore,thepolynomial x
m
-
xin7L
p[x]hasm=pndis­
tinctrootsin
K,namelyalltheelements ofK.Thusx
m
-
xfactorsin K[x]as
where
aba2,'..'amaretheelementsof K.
Everythingwejustsaidwealreadysaid,inmore orlessthesameway,
inSection6
ofChapter5.Sincewewantedtheseresultstobefreshinthe
reader'smind,we
repeatedthismaterialhere.
Wesummarizewhatwejustdidin
Theorem6.3.1.LetKbeafinitefield ofcharacteristicp,paprime.
ThenKcontainsm=pnelementswhere n=[K:7L
p
],andthepolynomial
x
m
-
xin7L
p[x]splitsintolinearfactorsin K[x]as

240. Sec.3 FiniteFields II:Existence225
whereaI,a2,...,amaretheelementsofK.
Twonaturalquestionspresentthemselves:
1.Forwhatprimespandwhatintegersndoesthereexistafieldhaving pn
elements?
2.Howmanynonisomorphicfieldsaretherehavingpnelements?
Weshallanswerbothquestionsinthissection andthenext.Thean­
swerswill
be
1.Foranyprimepandanypositiveintegernthereexistsafinitefield
having
pnelements.
2.Twofinitefieldshaving thesamenumberofelementsareisomorphic.
Itistothesetworesultsthatwenowaddressourselves.First,wesettle
thequestionoftheexistenceoffinitefields.Webeginwitha generalremark
aboutirreduciblepolynomials.
Lemma6.3.2.LetFbeanyfieldandsupposethatp(x)isanirre­
duciblepolynomialin
F[x].Supposethatq(x)inF[x]issuchthatinsomeex­
tensionfield
ofF,p(x)andq(x)haveacommonroot.Thenp(x)dividesq(x)
inF[x].
Proof.
Supposethatp(x)doesnotdivideq(x);sincep(x)isirreducible
in
F[x],p(x)andq(x)mustthereforeberelativelyprimeinF[x].Thusthere
arepolynomialsu(x)andv(x)inF[x]suchthat
u(x)p(x)+v(x)q(x)=1.
SupposethattheelementainsomeextensionKofFisarootofbothp(x)
andq(x);thuspea)=q(a)=O.Butthen1=u(a)p(a)+v(a)q(a)=0,a
contradiction.Sowe
getthatp(x)dividesq(x)inF[x].0
Notethatwecanactuallyprovealittlemore,namely
Corollary.Iff(x)andg(x)inF(x)arenotrelativelyprimeinK[x],
whereKisanextensionofF,thentheyarenotrelativelyprimeinF[x].
LetFbeafieldofcharacteristicp
=1=O.Weclaimthatthepolynomial
f(x)=x
m
-
x,wherem=pn,cannothaveamultiplerootinanyextension

241. 226 SpecialTopics(Optional) Ch.6
fieldKofF.Doyourememberwhatamultipleroot ofapolynomialis?We
refreshyourmemory.
Ifg(x)isinF[x]andifKisanextensionfield ofF,then
ainKisamultiplerootofg(x)ifg(x)=(x-a)2q(x)forsomeq(x)inK[x].
Wereturntothepolynomialf(x)=x
m
-x above.Since f(x)=
x(x
m
-
t
-
1)and0isnotarootofx
m
-
t
- 1,itisclearlytruethat0isasimple
(Le.,
notmultiple)rootoff(x).Supposethat
aEK,K:::>F,isarootoff(x);
thusam=a.Ify=x-a, then
f(y)==ym-y=(x-a)m-(x-a) =x
m
-~-(x-a)
(sincewe areincharacteristicp=1=0andm=pn)
=x
m
-x (becauseam=a)=f(x).
So
f(x)=f(y)=ym-y=(x-a)m-(x-a)
=(x-a)«x-a)m-t- 1),
andclearlythis isdivisibleby x-aonlytothefirstpower,since x-adoes
notdivide(x-a)m-l- 1.Soaisnotamultiplerootoff(x).
Wehaveproved
Theorem6.3.3.Ifn>0,thenf(x)=x
m
-
x,wherem=pn,hasno
multiplerootsinanyfield
ofcharacteristicp.
Weshouldaddawordtotheproofabovetonaildown thestatementof
Theorem6.3.3aswegaveit. Anyfieldofcharacteristicp
=1=0isanextension
of7L.
p
,andthepolynomialf(x)isin7L.
p[x].Sotheargumentabove,with K
anyfieldofcharacteristicpandF=7L.
p
,provesthe theoreminitsgivenform.
Wehaveexactly
whatweneedtoprovetheimportant
Theorem6.3.4.Foranyprime pandanypositiveinteger nthereex­
istsafinitefieldhaving
pnelements.
ProofConsiderthepolynomialx
m
-x in7L.
p[x],wherem=pn.By
Theorem5.6.6thereexistsafiniteextension Kof7L
psuchthatinK[x]the
polynomial
xm-xfactorsas
where
at,a2'...,amareinK.ByTheorem6.3.3,x
m
-x doesnothaveany
multiplerootsin
K,hencetheelements at,a2'...,amarem=pndistinctel­
ements.Wealsoknow
that
aba2,.'.'amarealltherootsofx
m
-x inK,
sincex
m
-
xisofdegreem.

242. Sec.4 FiniteFieldsIII:Uniqueness 227
LetA={aEKIam=a};aswejustsaw, Ahasmdistinctelements.We
claim
thatAisafield.Ifa,bEA,thenam=aandb
m
=b,hence(ab)m=
amb
m
=ab,thusabEA.Becausewe areincharacteristicp
=1=0andm=pn,
(a+b)m=am+b
m
=a+b,hencea+bisinA.
SinceAisafinitesubset ofafieldandisclosedwithrespect tosumand
product,AmustbeasubfieldofK.SinceAhasm=pnelements,Aisthus
thefieldwhoseexistencewasassertedin
thestatementofthetheorem.With
thisthe
theoremisproved.0
PROBLEMS
*1.GivethedetailsoftheproofoftheCorollarytoLemma6.3.2.
Thenexttwoproblemsarea repeatofonesgivenearlierin thebook.
2.
Iff(x)=aox
n+atXn-t+...+anisinF[x],letf'(x)betheformal
derivative
off(x)definedby thefollowingequation: f'(x)=naoxn-t
+(n-l)atxn-2+...+(n-i)aixn-i-t+ ·.· +an-t.Provethat:
(a)([(x)+g(x»'=['(x)+g'(x)
(b)([(x)g(x»'=f'(x)g(x)+[(x)g'(x)forallf(x)andg(x)inF[x].
*3.Provethat [(x)inF[x]hasamultiple rootinsomeextension ofFifand
onlyif[(x)and['(x)arenotrelativelyprime.
4.Iff(x)=x
n
-x isinF[x],provethatf(x)doesnothaveamultiple rootin
anyextension
ofFifFiseitherofcharacteristic0 orofcharacteristic
p
=1=0,wherepdoesnotdividen-1.
s.UsetheresultofProblem4 togiveanotherproofofTheorem6.3.3.
6.
IfFisafieldofcharacteristicp
=1=0,constructapolynomialwithmultiple
roots
oftheformx
n
-
x,wherepI(n-1).
7.
IfKisafieldhaving pnelements,show thatforeverymthatdividesn
thereisasubfieldofKhavingpmelements.
4.FINITEFIELDSIII: UNIQUENESS
Nowthatweknow thatfinitefieldsexisthaving pnelements,foranyprime p
andanypositiveinteger n,wemightask: Howmanyfinitefieldsare there
withpnelements?Forthistomakeanysenseatall,whatwe arereallyasking
is:Howmanydistinctnonisomorphicfields aretherewithpnelements?The
answertothisisshortandsweet:one. Weshallshow herethatanytwofinite
fieldshavingthesame
numberofelementsareisomorphic.

243. 228 SpecialTopics(Optional) Ch.6
LetKandLbetwofinitefieldshaving pnelements.ThusKandLare
bothvectorspaces ofdimensionnover7L
p
•Assuch,KandLareisomorphic
as
vectorspaces. Ontheotherhand,K*andL*arebothcyclicgroups of
orderpn-1byTheorem6.2.4;hence K*andL*areisomorphicas multi­
plicativegroups.
Onewouldimagine thatonecouldputthesetwoisomor­
phisms
togethertoprovethatKandLareisomorphicas fields.Butitjust
isn'tso.
Theproofdoesnottakethisdirectionatall.ButthefinitenessofK
andLtogetherwiththesetwoisomorphisms(oftwostructurescarriedby K
andL)dosuggestthat,perhaps, KandLareisomorphicasfields.This isin­
deedthecase,aswenow proceedtoshow.
Webeginwith
Lemma6.4.1.Ifq(x)in7L
p[x]isirreducibleofdegreen,then
q(x)I(x
m
-x), wherem=pn.
Proof.
ByTheorem4.5.11theideal(q(x»of
Zp[x]generatedbyq(x)is
amaximalideal of7L
p[x]sinceq(x)isirreduciblein7L
p[x].LetA=
7L
p[x]/(q(x»;byTheorem4.4.3,Aisafieldofdegreenover7Lp,hencehas pn
elements.Therefore, u
m=uforeveryelementuinA.
Leta=x+(q(x»bethecosetofxinA=7L
p[x]/(q(x»;thusq(a)=0
andq(x)istheminimalpolynomialfor aover
7L
p
•SinceaisinA,am=a,soa
isseenasa rootofthepolynomialx
m
-x, wherem=pn.Thusx
m
-x and
q(x)haveacommonrootin A.ByLemma6.3.2wehavethat q(x)I(x
m
-x).
D
Wearenowinaposition toprovethemainresult ofthissection.
Theorem6.4.2.IfKandLarefinitefieldshaving thesamenumberof
elements,thenKandLareisomorphicfields.
ProofSupposethatKandLhavepnelements.By Theorem6.2.4,L*
isacyclicgroupgenerated,say,by theelementbinL.Then,certainly,
7L
p(b)-thefieldobtainedbyadjoining bto7L
p-isallofL.Since
[L:7L
p
]=n,byTheorem5.3.2bisalgebraicover7L
pofdegreen,with
n=deg(q(x»,whereq(x)istheminimalpolynomialin7L
p[x]forb,andisir­
reduciblein7Lp[x].
Themappingt/J:7L
p[x]~L=7L
p(b)definedbyt/J(f(x»=f(b)isa
homomorphismof7L
p[x]ontoLwithkernel(q(x»,theidealof7Lp[x]gener­
atedbyq(x).SoL=7L
p[x]/(q(x».
Becauseq(x)isirreduciblein7Lp[x]ofdegreen,byLemma6.4.1q(x)
mustdivide x
m
-
x,wherem=pn.However,by Lemma6.3.1,thepolyno­
mial
x
m
-
xfactorsin K[x]as
x
m
- X=(x-at)(x-a
2
)
•••(x-am),

244. Sec.5 CyclotomicPolynomials 229
whereaba2,""amarealltheelementsof K.Therefore,q(x)divides
(x-al)(x- a2)...(x-am)'BytheCorollarytoTheorem4.5.10, q(x)can­
notberelativelyprimetoallthe
x-ai inK[x],henceforsome j,q(x)and
x-aj haveacommonfactorofdegreeatleast 1.Inshort,x-aj must
divide
q(x)inK[x],soq(x)=(x-aj)h(x)forsome hex)inK[x].There­
fore,
q(aj)=O.
Sinceq(x)isirreduciblein
Zp[x]andajisarootof q(x),q(x) mustbe
theminimalpolynomialfor
ajin
Zp[x].ThusZp(aj)==Zp[x]/(q(x»==L.This
tellsus,amongotherthings,thatwehave[Zp(aj):Zp]=n,andsince
Zp(aj)CKand[K:Zp]=nweconcludethatZp(aj)=K.Therefore,
K=Zp(aj)==L.Thuswegettheresultthatweareafter,namely,that Kand
Lareisomorphicfields.Thisprovesthetheorem. 0
CombiningTheorems6.3.4and6.4.2,wehave
Theorem
6.4.3.Foranyprime pandanypositiveintegernthereex­
ists,uptoisomorphism,oneandonlyonefieldhaving
pnelements.
5.CYCLOTOMIC POLYNOMIALS
Let
Cbethefieldofcomplexnumbers.Asaconsequenceof DeMoivre's
Theoremthecomplexnumber
On=cos
27T/n+isin27T/nsatisfiesO~=1
and0'::=1=1if0<m<n.Wecalled Onaprimitiventhrootofunity.Theother
primitiventhrootsofunityare
O
k
(27Tk)+ . .(27Tk)
=cos- lSIn-
n n n '
where
(k,n)=1and1
:sk<n.
Clearly,Onsatisfiesthepolynomialx
n
- 1inQ[x],whereQisthefield
ofrationalnumbers.Wewanttofindtheminimal(monic)polynomialfor
On
over
Q.
Wedefineasequenceofpolynomialsinductively. Atfirstglancethey
mightnotseemrelevanttothequestionoffindingtheminimalpolynomial
for
Onover
Q.Itwillturnoutthatthesepolynomialsarehighlyrelevantto
thatquestionfor,
asweshallprovelater,thepolynomial
cPn(x)thatweare
abouttointroduce
isamonicpolynomialwithintegercoefficients, isirre­
ducibleover
Q,and,moreover,cPn(8n)=o.ThiswilltellusthatcPn(x)isthe
desiredmonicminimalpolynomialfor
Onover
Q.
Wenowgoaboutthebusinessofdefiningthesepolynomials.

245. 230 SpecialTopics(Optional) Ch.6
Definition.Thepolynomials4>n(x)aredefinedinductively by:
(a)4>1(x)=x-I.
(b)Ifn>1,then4>n(x)=(x
n
-1)/TI4>d(x),whereintheproductinthe
denominator
drunsoverallthedivisorsof nexceptfor
nitself.
Thesepolynomialsarecalledthe
cyclotomicpolynomials and
cPn(x)is
calledthenth cyclotomicpolynomial.
Atthemomentit isnotobviousthatthe4>n(x)sodefinedareeven
polynomials,nordowe,asyet,haveaclueastothenatureofthecoefficients
ofthesepolynomials4>n(x).Allthiswillcomeinduetime.Butfirst wewant
tolookatsomeearlyexamples.
Examples
1.
cP2(X)=(x
2
-1)/cP1(x)=(x
2
-1)/(x- 1)=x+1.
2.cP3(X)=(x
3
-1)/4>1(x)=(x
3
-1)/(x-1) =x
2+X+1.
3.
4>4(X)=(x
4
-1)/(cPt(X)cP2(X»=(x
4
--1)/«x-1)(x+1»
(x
4
-1)/(x
2
- 1)=x
2+1.
4.
cPs(x)=(X
s
-1)/4>l(X)=(X
s
-1)/(x-
1)=x
4+x
3+x
2+X+1.
x
6
-
1
5.
4>6(X)=4>1(X)cP2(X)4>3(X)
X
6
- 1
-(x-1)(x +1)(x
2
+X+1)
x
3
+1
=X+1=x
2
-
X+1.
Wenoticeafewthingsaboutthepolynomialsabove:
1.Theyareallmonicpolynomialswithintegercoefficients.
2.Thedegreeof
cPn(x)iscp(n),wherecpistheEulercp-function,for
1::5n::56.(Checkthisout.)
3.Each4>n(x),for1::5n::56,isirreducibleinQ(x).(Verify!)
4.For1::5n::56,Onisarootof4>n(x).(Verify!)
Thesefewcasesgiveusahintastowhatthegeneralstorymightbefor
all4>n(x).Ahint,yes,butonlyahint.Toestablishthesedesiredproperties
forcPn(x)willtakesomework.
Togainsomefurtherinsightintothesepolynomials,weconsiderapar-

246. Sec.5 CyclotomicPolynomials 231
ticularcase,oneinwhich n=pm,wherepisaprime.Toavoidcumbersome
subscripts,weshalldenote4>n(x)byt/J(m)(x),wheren=pm.Theprime pwill
bekeptfixedinthediscussion.Weshallobtainexplicitformulasforthe
t/J(m)(x)'sanddeterminetheirbasicproperties.However,themethod weuse
willnotbeapplicabletothegeneralcaseof4>n(x).Tostudythegeneralsit­
uationwillrequireawideranddeepersetoftechniquesthanthoseneeded
fort/J(m)(x).
Wenoteonesimpleexample.Ifp isaprime,theonlydivisorofpthat
isnotpitselfis1.Fromthedefinitionof4>p(x)=t/J(I)(X)wehavethat
x
p
-
1
t/J(l)(X)=4>p(x)=x-I=x
p
-
1
+ · · · +x+1.
NotethatinstudyingtheEisensteinCriterionweshowedthatthispolyno­
mial
isirreduciblein
Q(x).
Whatcanwesayforthehigherf/J(m)(x)'s?
Lemma6.5.1.Forallm2:1,
Proof.Wegobyinductionon m.
Ifm=1,weshowedabovethatt/J(I)(X)=(xP-1)/(x- 1)=1+x+
x
2
+ .··+x
p
-I
,
sothelemma istrueinthiscase.
Supposethat
t/J(r)=(x
pr
-1)/(x
pr
-
1
-
1)forallr <m.Consider
t/J<m)(x).Sincetheonlyproperdivisorsof pmare1,p,p2,...,pm-I,fromthe
definitionoft/J(m)(x)wehavethat
(m) _ x
pm
- 1 .
l/J(x)-(x-l)l/J(l)(x)· · .l/J(m-l)(x)
Byinduction,t/J(r)(x)=(x
pr
-1)/(x
pr
-
1
-
1)forr <m,hence
(x-
1)t/J(1)(x). · .t/J(m-l)(x)
p1p21 pm-l1
=(x-1)x-x- ...Xm-2-=x
pm
-
1
-
1.
x-Ix
P
-
1 x
P
- 1
Butthen
pm1l/J(m)(x)=X -
x
pm
-
1
-
1
completingtheinductionandprovingthelemma. D
Noteherethat
pm_1
1/,(m)(x)=X =1+x
pm
-
1+...+x(p-l)pm-l
0/ x
pm
-
1
-
1

247. 232 SpecialTopics(Optional) Ch.6
isamonicpolynomialwithintegercoefficients.Itsdegree isclearlypm-l(p- 1),
which
isindeed
q;(pm).Finally,iff)isaprimitivepmthrootofunity,then
f)pm=1,but()pm-l=1=1,henceI/J(m)(()=0;so()isarootofI/J(m)(x).Thefinal
thing
wewanttoknow is:Is
I/J(m)(x)irreducibleoverto?
Notethat
I/J(m)(x)=1+x
pm
-
1
+...+x(p-l}pm-l=1/J(1)(X
pm
-
1
)
andweknowthatl/J(l)(X)isirreducibleinQ[x].WeshallusetheEisenstein
CriteriontoprovethatI/J(m)(x)isirreducibleinto[x].
Wedigressforamoment.If f(x)andg(x)aretwopolynomialswithin­
tegercoefficients,wedefine
f(x)
==g(x)modpiff(x)=g(x)+pr(x),where
r(x)isapolynomialwithintegercoefficients.This isequivalenttosayingthat
thecorrespondingcoefficientsof
f(x)andg(x)arecongruentmod p.Ex­
panding
(f(x)+g(x»Pbythebinomialtheorem,andusingthatallthebino­
mialcoefficientsaredivisibleby
p,sincepisaprime,wearriveat
(f(x)+g(x»P
==f(x)P+g(x)Pmodp.
Givenf(x)=aox
n
+alx
n
-
1 +...+an,wherethe aiareintegers,then,
bytheabove,
f(x\P=(ax
n+a x
n
-
1 + . · ·+a
\P==aPx
np
+aPx(n-l)p+...+aP
) 0 1 n) 0 1 n
thelattercongruencebeingaconsequenceofFermat'sTheorem(theCorol­
larytoTheorem2.4.8).Since
f(xP)
=aox
np
+al(n-l)p+...+an,weobain
that
f(xP)
==f(x)Pmodp.
Iteratingwhatwejustdid, wearriveat
forallnonnegative
k.
Wereturnto our
I/J(m)(x).SinceI/J(m)(x)=1/J(1)(X
pm
-
1
)wehave,fromthe
discussionabove,thatIfJm)(x)==IfJl)(X
pm
-
1
)modp.Therefore,
(l)(X+1)pm-l=(X+1)P-1)P
m
-
1
=(X+1)P-1)P
m
-1
I/J (x+1)- 1 x
=(xr1+pxp-2+p(p;1)xp-3+...+P(P2-
1
) x +pym-l
==1/J(1)(X)pm-1(p-l)modp==I/J(m)(x+1)modp.

248. Sec.5
Thistellsus that
CyclotomicPolynomials 233
I/J(m)(x+1)=X
pm
-
1
(P-1)+pr(x),
wherer(x)isapolynomialwithintegercoefficients.Soall thecoefficientsof
I/J(m)(x+1),withtheexceptionofitsleadingcoefficient1, aredivisiblebyp.
Ifweknewforsome reasonthattheconstanttermofhex)=I/J(m)(x+1)was
notdivisibleby p2,wecouldapply theEisensteinCriteriontoshowthathex)
isirreducible.Butwhatistheconstanttermofhex)=I/J(m)(x+I)?Itis
merelyh(O)=l/J(m)(l),which,from theexplicitform ofI/J(m)(x+1)thatwe
foundfour
paragraphsearlier,isexactlyp.Thushex)isirreduciblein
Q[x],
thatis,I/J(m)(x+1)isirreducibleinQ[x].Butthisimmediatelyimplies that
I/J(m)(x)isirreducibleinQ[x].
Summarizing,wehave proved
Theorem6.5.2.Forn=pm,wherepisanyprimeandmanynonneg­
ativeinteger,
thepolynomial
4Jn(x)isirreducibleinQ[x].
Aswepointedoutearlier,this isaveryspecialcase ofthetheoremwe
shallsoonprove;namely,
that
4Jn(x)isirreducibleforallpositiveintegers n.
Moreover,theresultandproofofTheorem6.5.2play norolein theproofof
thegeneralproposition that4Jn(x)isirreducibleinQ[x].Butbecauseofthe
resultin Theorem6.5.2andtheexplicitformof4Jn(x)whenn=pm,wedo
getaprettygoodidea ofwhatshouldbetrueingeneral.Wenowproceedto
thediscussionoftheirreducibilityof4Jn(x)forgeneraln.
Theorem6.5.3.Foreveryintegern2::1,
4Jn(x)=(x-0(1»..·(x- o(cp(n»),
where0(1),0(2),•••,o(t/J(n»arethecp(n)distinctprimitive nthrootsofunity.
ProofWeproceedbyinductiononn.
Ifn=1,then4Jl(x)=x-I,andsince1is theonlyfirstrootofunity,
theresultiscertainlycorrectinthiscase.
Suppose
thatresultis trueforallm<n.Thus,if dinandd
=1=n,then,
by
theinduction,
4Jd(X)=(x-o~~...(x-O(~(d»),wherethe0<:)arethe
primitivedthrootsofunity.Now
x
n
-1=(x-~l)(X-~2)..·(x-~n),
wherethe~irunoverallnthrootsofunity.Separatingouttheprimitiventh
rootsofunityinthisproduct,we obtain
x
n
-1=(x-0(1»...(x-o(cp(n»)v(x),

249. 234 SpecialTopics(Optional) Ch.6
wherev(x)istheproductofalltheother x-li;thusby ourinductionhy­
pothesis
v(x)istheproductofthe
<Pd(X)overalldivisors dofnwiththeex­
ceptionof
d=n.Thus,since
x-
(x
n
-1)_(x-0(1})· · ·(x-o(q>(n»)v(x)
c/>n()-v(x)- v(x)
=(x- O(l})(X-ff2})• • ·(x-ffcp(n»),
wehaveprovedtheresultclaimedinthetheorem. D
Fromtheformof<Pn(x)inTheorem6.5.3weimmediatelyseethat
<Pn(x)isamonicpolynomialinC[x]ofdegreeq;(n).Knowingthis, weprove
that,infact,thecoefficientsofcPn(x)areintegers.Why isthistrue?Proceed­
ingbyinductionon
n,wemayassumethistobethecaseif dinandd
=1=n.
Therefore,if v(x)denotesthepolynomialusedintheproofofTheorem
6.5.3,then
(x
n
- l)/v(x)=
<Pn(x)EC[x],hencev(x)Ix
n
-1inC[x].But,by
thelong-divisionprocess,dividingthemonicpolynomial
v(x)withinteger
coefficientsinto
x
n
-
1leadsustoamonicpolynomialwithintegercoeffi­
cients(andno remainder,since
v(x)I(x
n
- 1)inC[x]).Thus(x
n
- l)/v(x)=
<Pn(x)isamonicpolynomialwithintegercoefficients.Aswesaw,itsdegree is
q;(n).Thus
Theorem
6.5.4.Foreverypositiveinteger nthepolynomial
<Pn(x)isa
monicpolynomialwithintegercoefficientsofdegreeq;(n),wherecpisthe
Eulerq;-function.
KnowingthatcPn(x)isapolynomial,wecanseethatitsdegree iscp(n)
inyetanotherway.From<Pn(x)=(x
n
- l)/v(x),usinginductionon n,
deg(<pn(x))=n-deg(v(x))=n-Lq;(d),thesumoveralldivisors dofn
otherthan d=n,fromtheformof v(x).InvokingtheresultofTheorem
6.2.1,n-Lcp(d)=cp(n),whereagainthissum isoveralldin,d=1=n.We
thusobtainthatdeg(<Pn(x))=q;(n).
Theresultweareabouttoprove iswithoutquestiononeofthemost
basiconesaboutcyclotomicpolynomials.
Theorem
6.5.5.Foreverypositiveinteger nthepolynomial
<Pn(x)is
irreducibleinQ[x].
Proof.Letf(x)inQ[x]beanirreduciblepolynomialsuchthat
f(x)I<Pn(x).Thus<Pn(x)=f(x)g(x)forsomeg(x)inQ[x].ByGauss'Lemma
wemayassumethatboth
f(x)andg(x)aremonicpolynomialswithinteger
coefficients,thusarein
Z[x].Ourobjectiveistoshowthat<Pn(x)=f(x);if

250. Sec.5 CyclotomicPolynomials 235
thiswerethecase,then,since f(x)isirreducibleinQ[x],wewouldhavethat
cPn(x)isirreducibleinQ[x].
SincecPn(x)hasnomultipleroots, f(x)andg(x)mustberelatively
prime.Let
pbeaprimenumbersuchthat pdoesnotdividen.If
0isaroot
of
f(x),itisthenaroot of
cPn(x),hencebyTheorem6.5.30isaprimitiventh
rootofunity.Because pisrelativelyprimeto n,OPisalsoaprimitive nth
rootofunity,thus,byTheorem6.5.3,(}PisarootofcPn(x).Wetherefore
havethat0
=
cPn(OP)=f(OP)g(OP),fromwhich wededucethateither f(OP)=
Oarg(OP)=o.
Ouraimistoshowthat f(OP)=O.Supposenot;then g(OP)=0,hence
oisarootof g(x
P
).Because0isalsoarootoftheirreduciblepolynomial
f(x),byLemma6.3.2weobtainthat f(x)Ig(x
P).Aswesawinthecourseof
theproofofTheorem6.5.2,
g(xP)
==g(x)Pmodp.
LetJbetheidealin 7Lgeneratedby p;bytheCorollarytoTheorem
4.6.2,7L[x]/J[x]==7L
p[x],whichmeansthatreducingthecoefficientsofany
polynomial
modpisahomomorphismof 7L[x]onto7L
p[x].
Sinceallthepolynomials
cPn(x),v(x),f(x),andg(x)areinil[x],if
~n(x),v(x),!(x),andg(x)aretheirimagesin 7L
p[X],alltherelationsamong
themarepreservedgoingmod
p.Thuswehavetherelations x
n
-
1=
~n(x)v(x), ~n(x)=j(x)g(x)andl(x)Ig(x
P
)=g(x)P.
Therefore,f(x)andg(x)haveacommonroot, a,insomeextension K
of7L
p
•Nowx
n
-1=(f>n(x)v(x)=l(x)g(x),hencea,asarootofboth f(x)
andg(x),isamultiplerootof x
n
- 1.Buttheformalderivative (x
n
- 1)'of
x
n
-1isnx
n
-1=1=0,sincepdoesnotdivide n;therefore,(x
n
- 1)'isrela­
tivelyprimeto
x
n
-
1.BytheresultofProblem3ofSection3thepolyno­
mial
x
n
-1cannothaveamultipleroot.Withthiscontradictionarrivedat
fromtheassumptionthat
OPwasnotarootof f(x),weconcludethatwhen­
ever0isarootof f(x),thensomustOPbeone,foranyprime pthatdoesnot
dividen.
Repeatingthisargument,wearriveat:0'isarootoff(x)foreveryin­
teger
rthatisrelativelyprimeto n.But
0,asarootof f(x),isarootofcPn(x),
soisaprimitiventhrootofunity.Thus0'isalsoaprimitiventhrootofunity
forevery
rrelativelyprimeto n.Byrunningoverall rthatarerelatively
primeto
n,wepickupeveryprimitive nthrootofunityassomesuch
0'.
Thusalltheprimitiventhroots ofunityarerootsoff(x).ByTheorem6.5.3
weseethatcPn(x)=f(x),hencecPn(x)isirreduciblein Q[x].D
Itmaystrikethereaderasartificialandunnaturaltohaveresortedto
thepassagemod
ptocarryouttheproofoftheirreducibilityofapolynomial
withrationalcoefficients.Infact,itmayverywellbeartificialandunnatural.
Asfarasweknow,noproofoftheirreducibilityof
cPn(x)haseverbeengiven

251. 236 SpecialTopics(Optional) Ch.6
stayingcompletelyinQ[x]andnotgoingmodp.Itwouldbeestheticallysat­
isfying
tohavesuchaproof. Ontheotherhand,this isnottheonlyinstance
wherearesultisprovedbypassingtoarelatedsubsidiarysystem.Manythe­
oremsinnumbertheory-abouttheordinaryintegers-haveproofsthatex­
ploit
theintegersmodp.
Because
4>n(x)isamonicpolynomialwithintegercoefficientswhich is
irreducibleinQ[x],andsinceOn,theprimitiventhrootofunity,isarootof
4>n(x),wehave
Theorem6.5.6.4>n(x)istheminimalpolynomialinQ[x]fortheprimi­
tive
nthrootsofunity.
PROBLEMS
1.Verifythatthefirstsixcyclotomicpolynomials areirreduciblein
Q[x]bya
directfrontalattack.
2.Write
downtheexplicitformsof:
(a)
4>10(X).
(b)4>15(x).
(c)4>20(X).
3.If(x
m
-
1)I(x
n
-
1),provethatmin.
4.Ifa>1isanintegerand(am-1)I(an-1),provethatmin.
5.IfKisafiniteextension of
Q,thefieldofrationalnumbers,prove that
thereisonlyafinite numberofrootsofunityin K.(Hint:Usetheresultof
Problem10ofSection2, togetherwithTheorem6.5.6.)
6.LIOUVILLE'SCRITERION
Recallthatacomplexnumberissaidtobealgebraicofdegreenifit isthe
rootofapolynomialofdegreenoverQ,thefieldofrationalnumbers,and is
nottherootofanysuchpolynomial ofdegreelower thann.Inthetermsused
in
Chapter5,analgebraicnumberisacomplexnumberalgebraicover
Q.
Acomplexnumberthatisnotalgebraiciscalledtranscendental.Some
familiarnumbers,suchas
e,
1T,e'Tr,andmanyothers,areknowntobetran­
scendental.
Others,equallyfamiliar,suchas e+
1T,e1T,and1T
e
,
aresuspected
ofbeingtranscendentalbut,todate,thisaspectoftheirnatureisstillopen.
TheFrenchmathematicianJosephLiouville(1809-1882)gaveacrite­
rion
thatanyalgebraicnumberofdegreenmustsatisfy.Thiscriteriongives
usacondition
thatlimitstheextenttowhicharealalgebraic numberofde­
greencan
beapproximatedbyrationalnumbers.Thiscriterion isofsucha

252. Sec.6 Liouville'sCriterion 237
naturethat wecaneasilyconstructrealnumbersthatviolateitforevery
n
>1.Anysuchnumberwillthenhavetobetranscendental.Inthiswaywe
shallbeabletoproducetranscendentalnumbersatwill.However,noneof
thefamiliarnumbers
issuchthatitstranscendencecanbeprovedusingLiou­
ville'sCriterion.
InthissectionofthebookwepresentthisresultofLiouville.It
isasur­
prisinglysimpleand elementaryresulttoprove.Thistakesnothingaway
fromtheresult;inouropinionitgreatlyenhancesit.
Theorem6.6.1(Lionville).Let
abeanalgebraicnumberofdegree
n
2::2(i.e.,aISalgebraicbutnotrational).Thenthereexistsapositivecon­
stantc(whichdependsonlyon
a)suchthatforallintegers u,vwith
v>0,la-u/vl>c/v
n
.
P,oof.Let abearootofthepolynomial f(x)ofdegreen in
Q[x],
whereaisthefieldofrationalnumbers.Byclearingofdenominatorsinthe
coefficientsof
f(x),wemayassumethat f(x)=
'oX
n
+'lX
n
-
l
+...+'n'
whereallthe 'iareintegersandwhere '0>o.
Sincethepolynomial f(x)isirreducibleofdegree nithasndistinct
roots
a=al,a2,...,anin
C,thefieldofcomplexnumbers.Therefore, f(x)
factorsoverCasf(x)=ro(x-a)(x- a2)· . ·(x- an).Letu,vbeintegers
with
v>0;then
( )
n,u
n
-I f
!!:.='oU+~+...+'n-I
U
+ ,
V vn unIun'
hence
vnf(!!:')=run+run-Iv+ . · . +rUV
n
-1
+rUn
V 0 I n-I n
isaninteger.Moreover,since f(x)isirreduciblein
Q[x]ofdegreen2::2,f(x)
hasnorationalroots,so vnf(u/v)isanonzerointeger,whence Iunf(u/u)12::1.
Usingthefactoredformof f(x),wehavethat
If(u/v)1
'ol(u/v)-
azl· · ·I(u/v)-
a"I
unlf(u/u)\
hence
t(;)=ro((;)-a)((;)-az)· · · ( (;)-an)'
1(;)-al
roifl(u/v)-azl· . ·I(u/v)-~I
::> 1
-rovnl(u/u)- azl· · ·I(u/u)-ani·

253. 238 SpecialTopics(Optional) Ch.6
Letsbethelargestof lal,la21,...,lanl.Wedividetheargumentaccord­
ingas
Iu/ul>15orIUlul
::515.Iflu/ul>15,then,bythetriangleinequality,
la-(u/u)1;:::lu/ul- lal>15-s=s,and,sinceu;:::1,la-(ulu)1>slu
n
.
Ontheotherhand,ifIu/uI::52s,then,againbythetriangleinequality,
lai-(ulu)1::::;lail+lu/ul::::;s+2s=3s.Therefore,
sothat
l/t
;:::1/(3s)n-l=1/(3
n
-
l
s
n
-
l ).Goingbacktotheinequalityabove
thatla-(ulu)1
;:::1/[rov
n la2-(u/v)1...Ian-(u/u)/],wehavethat
la-(u/u)1
;:::1/(r03n-lsn-lun).Thesenumbers ro,3
n
-
1
,
sn-laredetermined
onceandforallby
aanditsminimalpolynomialf(x)anddonotdependonu
oru.Ifwelet b=1/(r03n-lsn-l),thenb>0andla-(u/u)1>blv
n
.This
coversthesecondcase,where
IUlul
::52s.
Ifcisapositivenumbersmallerthanboth bands,wehavefromthe
discussion
thatla-u/ul>c/u
nforallintegers u,u,whereu>0,thereby
provingthetheorem.
D
Let'ssee theparticularsoftheprooffortheparticularcase a=
v2.
Theminimalpolynomialfor ainQ[x]isf(x)=(x-a)(x+a),soa=aland
-a=a2'Soifuanduareintegers,and v>0,then
aninteger.So
lu
2
f(u/u)1
;:::1;:::1/u
2
.Thesaboveisthelargerof1v2/
and1-v21;thatis,s=v2.Also,the baboveis1/(3
2
-
1 (v2)2-1)=
1/(3v2),
soifcisanypositive numberlessthan1/(3v2),then1v2-uluI>clv
2
•
Whatthe theoremsaysisthefollowing:Anyalgebraicrealnumberhas
rationalnumbersasclose
asweliketoit(this istrueforallnumbers),butif
thisalgebraicreal
numberaisofdegreen
;:::2,therearerestrictionsonhow
finelywecanapproximate
abyrationalnumbers.Theserestrictionsarethe
onesimposedbyLiouville'sTheorem.
Howdoweusethisresult
toproducetranscendentalnumbers?Allwe
needdoistoproduceareal
numberT,say,such thatwhateverpositiveinte­
ger
nmaybe, andwhateverpositivecwechoose,wecanfindapairofinte­
gers
u,u,withv>0suchthatIT-u/uI<clu
n
.Wecanfindsucha
Teasilyby
writingdownaninfinitedecimalinvolving
O'sand1's,wherewemakethe O's
spreadoutbetweenthe l'sveryrapidly. Forinstance,if
T=
0.10100100000010...010...,wherethe O'sbetweensuccessive 1'sgolike
m!,thenTisanumberthatviolatesLiouville'sCriterionforevery n>O.
Thusthis numberTistranscendental.

254. Sec.7 TheIrrationalityof'IT239
Wecould,ofcourse,use otherwidespreads ofO'sbetweenthe l's­
mm,(m!)2,andsoon-toproducehordes oftranscendentalnumbers.Also,
instead
ofusingjust1's,wecoulduseany oftheninenonzerodigits toobtain
moretranscendentalnumbers.Weleave
tothereadertheverificationthat
thenumbers ofthesortwedescribed donotsatisfyLiouville'sCriterionfor
anypositiveintegern
andanypositivec.
Wecanuse thetranscendentalnumber
'Tandthevariantsofitwede­
scribed
toproveafamousresult duetoCantor.Thisresultsays thatthereis
a1-1correspondencebetweenall therealnumbers anditssubsetofreal
transcendentalnumbers.In
otherwords,insomesense, thereareasmany
transcendentalrealsas
therearereals.Wegiveabriefsketch ofhowwe
carryitout,leavingthedetails
tothereader.
First,it
iseasytoconstructa 1-1mappingoftherealsontothosereals
strictlybetween0and1(try
tofindsuchamapping).This isalsotrueforthe
realtranscendentalnumbers andthoseofthemstrictlybetween0 and1.Let
thefirstset beAandthesecondoneB,soB CA.Then,bya theoreminset
theory,itsuffices
toconstructa 1-1mappingofAintoB.
Givenany
numberinA,wecanrepresentitasaninfinitedecimal
0.ala2...an...,wheretheaifallbetween0 and9.(Wenowwave ourhands
alittle,beingalittlebitinaccurate.
Thereadershouldtry totightenup
theargument.)Definethemapping
[fromAtoBby[(0.ala2...an...)=
0.alOaZOOa3000000a4...;bytheLiouvilleCriterion,exceptforasmallset of
aba2'·..'an'··.'thenumbers 0.alOaZOOa3000000a4...aretranscendental.
Thefwewrotedown thenprovidesuswith therequiredmapping.
Onefinalword aboutthekindofapproximationofalgebraicnumbers
byrationalsexpressedin
Theorem6.6.1.Therewehavethatifaisrealalge­
braicofdegree
n
2::2,thenla-u/uI>c/u
n
forsomeappropriatepositive c.
Ifwecoulddecrease thentola-u/uI>c/u
m
form<nandsomesuitablec
(depending
onaandm),wewouldget anevensharperresult.In1955the
(then)youngEnglishmathematician K.F.Rothprovedthepowerfulresult
thateffectivelywecouldcut
thendownto 2.Hisexactresult is:Ifaisalge­
braic
ofdegreen
2::2,thenforeveryreal numberr>2thereexistsapositive
constant
c,dependingonaandr,suchthatla-u/ul>c/u'forallbutafinite
number
offractionsu/u.
7.THEIRRATIONALITY OF
'IT'
Asweindicatedearlier,Lindemannin1882provedthat1Tisatranscendental
number.Inparticular,fromthisresult
ofLindemannitfollows that
1Tisirra­
tional.Weshall
notprovethetranscendence of
1There-itwouldrequirea

255. 240 SpecialTopics(Optional) Ch.6
ratherlong detour-butwewill,atleast,provethat'TT'isirrational.Thevery
niceproofthatwegive
ofthisfactisduetoI.Niven;itappearedinhispaper
"ASimpleProofThat
'TT'IsIrrational,"whichwaspublishedintheBulletin
oftheAmericanMathematicalSociety,vol. 53(1947),p.509.Tofollow
Niven'sproofonlyrequiressomematerialfromastandardfirst-yearcalculus
course.
Webeginwith
Lemma
6.7.1.Ifuisarealnumber,thenlim unln!=O.
n
-+00
Proof.Ifu isanyrealnumber,thene
U
isawell-definedrealnumber
ande
U
=1+u+u
2
/2!+u
3
/3!+ .·. +unln!+ ··..Theseries1 +u+u
2
/2!+
·..+ unln!+...convergestoe
U
;
sincethisseriesconverges,itsnthterm
must
gotoO.Thuslimunln! =O.D
n
-+00
WenowpresentNiven'sproofoftheirrationalityof 7T.
Theorem6.7.2.'TT'isanirrationalnumber.
Proof.Supposethat
7Tisrational;then 7T=alb,whereaandbarepos­
itiveintegers.
Foreveryinteger n>0,weintroduceapolynomial,whoseproperties
willleadustothedesiredconclusion.Thebasicproperties
ofthispolynomial
willholdforallpositive
n.Thestrategyoftheproof istomakeajudicious
choiceof
nattheappropriatestage oftheproof.
Let
f(x)=xn(a-bx)nln!,where 7T=alb.This isapolynomialofde­
gree
2nwithrationalcoefficients.Expandingitout,weobtain
a x
n
+a x
n+
l
+ ·..+anx
2n
f(x)= Q 1n! '
where
_
n-lb _(-1)'n!n-ibi - (_)nbn
a
Q=an,al=na, . . .,ai- 0'( _0)'a, . . .,an- 1
l.nl.
areintegers.
Wedenotetheithderivativeof
f(x)withrespecttoxbytheusualno­
tationf(i)(x),understanding
f(O)(x)tomean f(x)itself.
Wefirstnoteasymmetrypropertyof
f(x),namely,that f(x)=
f(7T-x).Toseethis,notethat f(x)=(bnln!)x
n
(
'TT'-x)n,fromwhoseformit
isclearthat f(x)=f('TT'-x).Sincethisholdsfor f(x),itiseasytosee,from
thechainrulefordifferentiation,thatf(i)(x)
=
(-1Yf(i)(7T-x).

256. Sec.7 TheIrrationalityof1t'241
Thisstatementabout f(x)andallitsderivativesallowsustoconclude
thatforthestatementsthat
wemakeaboutthenature ofallthef(i)(O),thereare
appropriatestatementsaboutallthef(i)(
11').
Weshallbeinterestedin thevalueoff(i)(O),andf(i)(11'), forallnonneg­
ative
i.Notethatfromtheexpandedformoff(x)givenaboveweeasily ob­
tainthatf(i)(O)ismerelyi!timesthecoefficientofXiofthepolynomialf(x).
Thisimmediatelyimplies,since thelowestpowerofxappearingin f(x)is
thenth,thatf(i)(O)=0ifi<n.Fori
2::nweobtainthatf(i)(0)=i!ai-n/n!;
sincei2::n,i!/n!isaninteger,andaswepointedoutabove,ai-nisalsoan
integer;therefore f(i)(0)isanintegerforallnonnegativeintegers i.Since
f(i)(11')=(-lYf(O),wehave thatf(i)(11')isanintegerforallnonnegative
integers
i.
Weintroduceanauxiliaryfunction
Since
f(m)(x)=0ifm>2n,weseethat
= -F(x)+f(x).
Therefore,
:x(F'(x)sinx-F(x)cosx)=F"(x)sinx+F'(x)cosx
-F'(x)cosx+F(x)sinx
=(FI/(x)+F(x»sinx=f(x)sinx.
Fromthisweconclude that
i
1T
f(x)sinxdx=[F'(x)sinx-F(x)cosx]~
=(F'(11')sin11'-F(11')cos11')-(F'(O)sin0 - F(O)cos0)
=F(11')+F(O).
Butfrom"theform ofF(x)aboveandthefactthatallf(i)(O)andf(i)(17-)are
integers,weconclude thatF(11')+F(O)isaninteger.Thusfof(x)sinxdxis
aninteger.Thisstatementaboutfaf(x)sinxdxistrueforanyintegern >0
whatsoever.Wenowwanttochoose ncleverlyenoughtomakesurethatthe
statement"faf(x)sinxdxisaninteger"cannotpossiblybetrue.

257. 242 SpecialTopics(Optional) Ch.6
Wecarryoutanestimateon fof(x)sinxdx.For0<x<11"thefunc­
tion
f(x)=xn(a-bx)nln!
:s;~anln!(sincea>0),andalso0 <sinx:s;1.
Thus0<fof(x)sinxdx<fo1I"na
n
ln!dx=1I"n+la
n
lnL
Letu=1I"a;then,byLemma6.7.1,lim unln!=0,soifwepickn large
enough,wecanmakesurethatunln!<ii;,oohence1I"n+la
n
ln!='lTUnln!<1.
Butthenfof(x)sinxdxistrappedstrictlybetween0and 1.But,bywhat we
haveshown,fKf(x)sinxdxis~ninteger.Sincethere isnointegerstrictlybe­
tween0and
1,wehavereachedacontradiction.Thusthepremisethat
11"is
rationalwasfalse.Therefore,11"isirrational.Thiscompletesthe proofof
thetheorem.0

258. INDEX

260. A(S),16
Abel,43
Abeliangroup, 43
Algebraic
element,
193
extension,193
number,194
Algebraicallyclosedfield,200
Alternatinggroup,121,215
Associativelaw,12,41
Associativering,126
Automorphismofagroup,
68
inner,69
Automorphismofaring, 141
Basis,187
BertrandPostulate,220
Bijection,
11
Boole,138
Booleanring, 138
Cantor,239
Carroll,7
Cartesianproduct,5
Cauchy,
80
Cauchy'sTheorem,80, 89
Cayley,69
Cayley'sTheorem, 69
Center,53
Centralizer,53,102
Characteristicsubgroup,
76
Characteristicofafield,178
ChineseRemainderTheorem,147
ClassEquation,
103
Commutativering, 127
Commutingelements, 53
Commutingmappings, 21
Complementofaset,5
Complexnumber,
32
absolutevalueof,34
argumentof,
35
complexconjugateof, 32
imaginarypartof, 32
polarformof, 35
purelyimaginary, 32
realpartof, 32
Index 245
Compositionofmappings, 11
Congruence,57
class,60
Conjugacy,58
class,58
Conjugate
elements,
58
ofacomplexnumber,32
Constant
function,9
polynomial,153
Constructible
length,202
number,204
CorrespondenceTheorem
forgroups,
86
forrings,142
Coset
left,
64
right,58
Cycleofapermutation, 111
Cyclicgroup,53,55,60
generatorof,
53
Cyclotomicpolynomial,230
Degree
ofanalgebraicelement,
195
ofafieldextension, 191
ofapolynomial,153
DeMoivre'sTheorem, 36
DeMorganRules,6
Dihedralgroup,45,116,117
Dimension
ofavectorspace,186
Directproductofgroups,
93
Directsum
ofrings,146
ofvectorspaces,181,182
Divides,22,157
DivisionAlgorithm,
155
Divisionring, 127
Divisor,22
Domain
integral,127
principalideal,157
Duplicationofacube,
201

261. 246 Index
EisensteinCriterion, 169
Element,3
algebraic,
193
identity,41
invertible,133
orbitof, 21,65
transcendental,194
unit,
41
Emptyset,4
Equality
ofmappings,9
ofsets,4
Equivalenceclass,
58
Equivalencerelation, 57
Euclid,27
Euclideanring, 163
Euclid'sAlgorithm, 22
Euler,63
Euler
'P-function,62
Euler'sTheorem, 63
Extensionfield, 191
Factor,22
Factorgroup, 78
Factorial,17
Fermat,63
Fermat'sTheorem, 63
Field,127
algebraicallyclosed, 200
characteristicof, 178
extension,191
ofalgebraicnumbers,199,200
ofquotients,172-175
ofrationalfunctions,177,179
splitting,
213
Fieldextension, 191
algebraic,193
degreeof, 191
finite,191
Finiteabeliangroups, 96
Finitedimensionalvectorspace, 185
FirstHomomorphismTheorem
forgroups,
85
forrings,142
Formalderivative, 227
Function,8
constant,9
Euler,
62
identity,9
injective,
10
one-to-one,10
onto,10
surjective,10
FundamentalTheorem
ofAlgebra,200
onFiniteAbelianGroups, 100
Galoistheory, 212
Gauss,169
Gauss'Lemma,168
Gaussianintegers,38,
166
Greatestcommondivisor
ofintegers,
23
ofpolynomials,158
Group,41
abelian,43
alternating,121, 215
axioms,41
centerof, 53
cyclic,53, 55,60
dihedral,
45,116,117
factor,78
finite,42
Hamiltonian,72
homomorphism,67
Klein's,116
nonabelian,43
octic,116
orderof, 42
quotient,78
simple,123,216
Hamilton,
72,
134
Hamiltoniangroup, 72
Hardy,201
Hermite,194
Homomorphismofgroups,
67
imageof, 70
kernelof, 70
trivial,67
Homomorphismofrings,
139
kernelof,140
HomomorphismTheorems
forgroups,84-87
forrings,
142

262. Ideal,140
left,140
maximal,
148
right,140
trivial,
142
two-sided,140
Identityelement,
41
Identityfunction,9
Image,8
inverse,
10
Indexofasubgroup, 59
Induction,29
Inductionstep, 31
Inductivehypothesis,30
Injectivemapping,
10
Integers,21
Gaussian,38,166
relativelyprime,
25
Integraldomain, 127
Intersectionofsets,4
Invariants
ofabeliangroups,100
Inverse
ofagroupelement,
41
ofamapping,12
Invertibleelement, 133
Irreduciblepolynomial,159
Isomorphicgroups,
68
Isomorphicrings, 141
Isomorphism
ofgroups,
68
ofrings,141
Kernelofahomomorphism
forgroups,70
forrings,140
Klein's4-group,116
Lagrange,
59
Lagrange'sIdentity, 133
Lagrange'sTheorem, 59
Leadingcoefficient, 162
Leastcommonmultiple, 28
Lindemann,194
Linearcombination,
185
Lineardependence,186
Linearindependence,
185
Liouville,236
Liouville'sCriterion,236
Index 247
Mappings,8
commuting,
21
compositionof, 11
identity,9
injective,10
one-to-one,10
onto,
10
productof, 11
surjective,10
Mathematicalinduction,
29
Matrices
real2
x2,130,131
2
X2overaring, 131
Maximalideal,148
McKay,
89
Minimalgeneratingset,186
Minimalpolynomial,
195
Monomorphism
ofgroups,
68
ofrings,141
Motionofafigure, 115
Multiple,22
Multiplicity,209
Niven,240
Normalsubgroup,
71
indexof, 59
Nullset,4
Number
algebraic,194
complex,32
prime,21,
26
purelyimaginary, 32
transcendental,194, 236
Octicgroup,116
One-to-one
correspondence,
11
mapping,10
Orbit,21,65
Order
ofanelement, 60
ofagroup,42
Partition
ofaset,
58
ofapositiveinteger,100

263. 248 Index
Permutation,16,109
even,
121
odd,121
Polynomial,152,179
coefficientsof,152
constant,
153
cyclotomic,230
degreeof,
153
irreducible,159
leadingcoefficientof,162
minimal,
195
monic,157
relativelyprime,159
rootof,
208
Polynomialring,152
Primenumber,
21,26
Primitiverootofunity,37,39,229
Primitiverootmod
p,66
Principalidealdomain, 157
PrincipleofMathematicalInduction, 29
Productofmappings,11
Projection,9
Quadraticnonresidue,
151
Quadraticresidue, 151
Quaternions,131,136
Quotientgroup,
78
Rationalfunctions,177,179
Reflexivity,
57
Relation,56
congruencerelation, 57
conjugacyrelation, 58
equivalencerelation, 57
Relativelyprime
integers,
25
polynomials,159
Ring,126
associative,126
Boolean,
138
commutative,127
division,127
Euclidean,166
homomorphism,139
noncommutative,
127
ofpolynomials,152,179
withunit,126
Rootsofunity,
42
primitive,37,39,229
Rootofapolynomial,208
multiplicityof,209
Roth,239
SecondHomomorphismTheorem
forgroups,
86
forrings,142
Sets,3
Cartesianproductof,5
differenceof,4
equalityof,4
intersectionof,4
unionof,4
Simplegroup,123,216
Splittingfield,
213
Squaringofacircle, 207
Subfield,128, 191
Subgroup,51
characteristic,76
cyclic,53
indexof, 59
normal,71
proper,51
Sylow,101
trivial,51
Subring,129
Subset,3
Subspaceofavectorspace,
181
spannedbyelements, 181
Surjectivemapping, 10
Sylow,104
Sylowsubgroup,101,
105
Sylow'sTheorem, 105
Symmetricgroup,16,109
Symmetry,
57
ThirdHomomorphismTheorem
forgroups,
87
forrings,142
Transcendental
element,194
number,194,236
Transitivity,
57
Transposition,20,111
TriangleInequality, 34

264. Trisectionofanangle,201
Unionofsets,4
Unitelement
ofagroup,41
inaring,126
Vectorspace,180
basisof,
187
dimensionof,186
Index 249
finitedimensional, 185
infinitedimensional, 185
minimalgeneratingsetfor,186
Well-OrderingPrinciple,22
Wiles,
63
Wilson'sTheorem,65,210
Zerodivisor,128

265. PRENTltEBAll
IJpperSaddleRiver,NI07548
http://www.prenhall.tOm