The theory of groups [incomplete scan]

121 Pages·1970·11.69 MB·English

by  Ian D. Macdonald

2 Someexamplesofgroups
(iv) Someidentity elementehasthe followingproperty: given
1
any element a inathere isat least onesolution x to the
=
equation ax e, each solution being called an inverse
to a and written as a-I.
Some examples of groups Itisimportant tonote that wearenot assuminguniquenessof
identities and inverses in a given group. Another fact that
requires carefulconsideration isthat weare usingwhat mightbe
calledright identities andright inverses. Supposewemakejust
=
one changein the above axioms,replacing the equation ax e
SINOEexamples areofgreat importance inthe theory ofgroups in (iv)by xa = e, sothat (in an obvious sense)we are dealing
we devote this chapter to a survey of the examples that are with right identities and left inverses; then many systems that
co~monly encountered and used. We must of courseindicate are far from being groups satisfy the new axioms. Wemay for
w~ch of the several possible definitions of 'group" wehave in instance take any set S,suchasthe set ofallteapots, anddefine
mind, and the reason for our particular choice issimplicity in products by putting xy = xfor eachpair x,y ofelements ofS.
proving that certain systems weshallpresent areinfactgroups. Axioms(i)and (ii)arereadily verified,and everyelement y ofS
Ifthe reader is~amiliarwith most ofthe facts in this chapter serves asan identity asrequired in (iii). To solvethe equation
he mayyet findIt ofvalue asasystematic revision ofhisknow xa =y forxwith agiveny, wesimplytake x = y. The axioms
ledge. aretherefore satisfiedifSnon-empty. But Sdoesnot constitute
a group if it has more than one element, for the equation
DEFINITIO..N: A group is a set and a binary operation+ in
that set satisfying certain conditiona.f that is the elements ax= y has no solution x ify =F a.
a Wepassontoexamplesofsystemsthat aregroups. Themost
a, b (in that order) in the group define an element usually
written as ab and usually called the product of a and b the familiar and readily available aregroups ofnumbers.
binary operation being (conventionally) called multiplic~tion. Example 1.01. The set Z of allintegers with addition asits
The conditions are: operation formsagroup. Tohold fast to the conventions inour
(i) aisclosedunder multiplication. definition would involve writing 3+7 as 3.7 and calling 10
(ii) The associativelawformultiplication holds, that is the 'product' of 3 and 7. We therefore have every excuse to
a(bc) = (ab)c callthe binary operation addition and to usethe familiar nota
tion in this special case. The group axioms are easily verified:
for allelements a,b,oinG.
(iii) There is at least one identity element in a, that is an (i)issimplythat the sumoftwointegers isaninteger, and (ii)is
element esuch that ae =a for all ain a. wellknown, plausible, and a consequence of any axiom system
forintegers. Wecantake the number 0asanidentity, forsurely
t For more details ~bout binary operations seethe Appendix. a+O=afor allelements aofZ. As for (iv), aninverseofais
. +
The reader whoWIshesto assumemore, for instance that identities and =
mverses havosomeofthe two-sided and uniquenees properties enumerated in -a sincewehave, just as simply, a+(-a) O.
Thoo,,?ms2.01-2.05, may do so. He willnot then have to prove the corre
Example 1.02. Theset Qofallrational numbersformsagroup
sponding theorems, but he wiUhavethe additional task ofverifying that all
the ~xampl~ presented inthisandsubsequent chapters aregroupsinthesense under addition, the proof being similar to that for Z.
ofhis definition.
858187 B
Some exa1111Jle.osf groups 3 4 Some examples of groups
Example 1.03. The set R of allreal numbers forms agroup Example 1.07. The non-zero complex numbers form agroup
under addition. Againthe reader should be ableto findaproof under the usualmultiplication.
quite like that given for Z.
Example 1.08. The complex numbers of unit modulus form
Example 1.04.Theset Cofallcomplexnumbersformsagroup a group under the usual multiplication.
under addition.
Example 1.09. For each prime number p there is asubset of
Itmay be wellto mention somesets ofnumbers that do not the complex numbers that forms a multiplicative group of
form groups. The odd integers do not form a group under someinterest. Let Z; denote the followingset:
addition (though the evenintegers do); a sufficient reason for {z:z ECand zP"= 1for someinteger n}.
this is that (i)doesnot hold; another reason is that there is no
identity element, which impliesthat (iii)and (iv)cannot hold. (Note that the positive integer n depends on the element e.)
The non-negative integers donot form agroup under addition Weverifythe axiomsindividually forthis important group. Let
for in their case (iv)doesnot hold (it is an empirical fact that Z1'Z2beelements ofZ;withzf" = 1,zf"= 1. Then (ZtZ2)P' = 1
Z;
many naturally occurringsetswithmultiplication arenotgroups whereristhe largerofmandn,so isclosedunder multiplica
because (iv) fails). The integers with the binary operation of tion. The associative lawholds simplybecauseitisalwaysvalid
Z;
subtraction donot form a group because the associative law is for multiplication of complex numbers. Clearly contains
not valid; for instance 5-(4-1)= 2and (5-4)-1= o. the complexnumber 1,which willserve as an identity element.
Since multiplication is the other familiar operation on (real IfzP' = 1then (l/z)p·= 1,so ifzis in Z; sois lIz-note that
Z;
and complex) numbers, it is reasonable to ask what sets of z =f=O-and l/z will act as an inverse of z. Therefore is a
numbers form groups when the group multiplication is taken multiplicative group.
to be the ordinary multiplication of numbers. The set of all A slight knowledge of complex numbers reveals that the
Z;
real numbers is not a candidate for group status in this sense elements of are precisely all those numbers ofthe form
for~tco~taiDBthe number 0and (since the number 1is clearl; 2k7r .. 2k7T
oos-+~sm--
an identrtyelement) werequire for axiom (iv) a solution x to p" p"
the equation Ox=1;and becausethere is no such x wedo not
have a group. forn =0,1,2, ...,where k isan integer with 0 :::;;k <p»;
Further important types of groups have residue classes
Example 1.05. The set ofpositive rational numbers forms a instead ofnumbers as elements. Aresidue classmodulo n is a
group under (ordinary) multiplication. To see this, note that certain equivalence class of integers:'] thus the residue class
(i) and (ii) are easily checked, take e= 1 in (iii), and take (aJmodulo n isthe set ofall solutions x to the congruenoe
o.
a-I = l/a in (iv); l/a exists since a =f= We remark that the
x == amodulon,
non-zero rationals form agroup for similar reasons.
where n is a fixed positive integer. More explicitly but less
Example 1.06. The set of positive (or of non-zero) real precisely (in the striot mathematioal sense)wehave
numbers forms a group under multiplication of numbers. As
(aJ ={..,.a-n, a,a+n, a+2n, ...}.
the reader should have bynowacquired somefacility inproving
such a statement there isno need to give further details. t As explained inthe Appendix.
Some examples of groups 5 6 S01T/,e examples of groups
The set ofresidue classes modulo nisdenoted by Zn;itcontains The subject ofmultiplication ofresidue classes and the forma
nelements, each being an infinite set of integers. tion of multiplicative groups requires some care. The usual
definition ofmultiplication isthe obvious one:
Example 1.10. We define a binary operation called addition
on residue classes by the following equation: [abJ== [aJ[bJ.
It isessential toverify that this isaproper definition. This may
[a+b] == [aJ+[bJ.
be carried out in a way similar to that explained in the case of
This.merely says that the sum of the residue classes containing addition, the key step being the remark that
the integers a and brespectively is to be that containing a+b. (a+kn)(b+ln) == ab+(la+kb+kln)n
yve ought first, however, to have verified that the result [a+b] and so [(a+kn)(b+ln)J == Cab].
ISthe same whateverrepresentatives we choose in [a]and [b],so The reader ought to fillin all the details for himself.
that the definition depends onlyonthe classescontaining aand b,
Unfortunately this multiplication of residue classes will not
not on these numbers themselves. This weproceed to do. Take
generally yield groups. Just as the number 0 cannot occur in
any element of[a];it willbe a+kn for someinteger k. Similarly multiplicative groups ofnumbers sothe residue class [OJcannot
an arbitrary element of [b]will be b+ln. Thus occur in multiplicative groups of residue classes when n> 1,
[a]== [a+kn], [b]== [b+ln]. forhow can w*e' find an inverse element for [O]? Remember that
[O][x] == [0] [1] for all x. There is another complication.
We require [a+knJ+[b+lnJ == [a+bJ Consider residue classes modulo 4, for which
if we are to have a proper definition ofsum. But [2][2] == [4] == [OJ;
and sothe product oftwo non-zero classes may be [0]' We must
[a+kn]+[b+ln] == [(a+kn)+(b+ln)], ~xclude [OJfrom any group, and having done that there are still
and since (a+kn)+(b+ln) == a+b modulon cases when we do not have closure.
We can, however, salvage something.
we have [(a+kn)+(b+ln)] == [a+bJ,
Example l.ll. The set of non-zero residue classes modulo p
and so all iswell. forms a group under multiplication, pbeing any prime number.
J
The group axioms are easily verified now that the delicate The associative law and the existence ofan identity element [1
matter ofdefinition has been dealt with. Closure isobvious, the are obvious facts, but axioms (i) and (iv) require care in their
associative law is well-known or alternatively follows from the consideration. Let [aJ be a non-zero residue class modulo p.
corresponding axiom for integers, [0] is an identity element This is precisely the same as supposing that p does not divide
[-a Jisan inverse for [aJ. Therefore wehave agroup ofresidu; the integer a. Anelementary and well-known result ofnumber
classes, denoted by Z,,, with addition as its binary operation. theory+ states that forsuitable integers u,vwehave the equality
Notice that it contains precisely nelements where nis an arbi au+pv == 1.
.trary positive integer. We therefore have a solution to the Thus in terms ofresidue classes modulo pwehave
problem of finding a group with a given finite number of ele [aJ[uJ== [auJ== [au+pvJ == [l].
ments.
t Seethe Appendix.
Some examples of groups 7 8 Some examples of groups
Nowconsider(i),the closurerequirement, and supposethat [a] THEOREM1.12. The set of allmappings of afixed (non-empty)
and [b] are non-zeroresidue classesfor which [a][b] = [OJ. It set into itself forms a semigroup -0
followsfromthe existence ofu and from the commutative and To be quite explicit about the multiplication of mappings
associativelawsthat used here, let .pand .pbe two mappingsofour fixedset X into
[b] =[a][u][b] =['u][a][b]=[u][O]=[0], itself; then 4,.p is defined by putting x(#) = (x.p).pfor each
and wearrive at acontradiction. Therefore (i)holds. Sincewe element x in X.
havefoundaninverse[u] to [a], wehave alsoverified(iv). This Wenowhaveameansofconstructing many semigroups. For
showsthat wehave a multiplicative group whoseelements are groups ofmappings,however,werequire inversesofmappings,
the p-I non-zeroresidue classesmodulop. andwerecallthat amapping fromasetX into X hasaninverse
if and onlyif it isboth one-one and cnto.] This explains our
Allour examples sofar have been groups with the property
interest in the concept of 'permutation'.
that everypair ofelements commutes, that isab isalwaysequal
to 00, and this, of course, has been a consequenceofthe com DEFINITION. Apermutation ofa set isa one-one mapping of
mutative laws for addition and multiplication of real and the set ontoitself.
complexnumbers. We omit detailed verification.
Example 1.13.ThesetSxofallpermutations ofa(non-empty)
set X forms a group. Itis easy to seethat Sx is closed,and
DEFINITION. The elements a, b of the group G commute if
(sinceSx is a subset of the set of all mappings of X into X)
ab =ba. Thegroup Giscommutative orabelian ifeverypair of
associativity is a consequence of Theorem 1.12. An identity
its elements commutes.
element ofSx isthe permutation which leavesevery member
L
Apair ofelements ofthe form z, xx in any group commutes of X fixed, and which therefore has the property that pL =P
because ofthe associative law. foreachelementpofSx. Finally, sincepis one-oneandontoX,
There are many groups that are non-abelian, and they are each permutation p h~san inverse mapping p-l, with pp-l = L,
oftenbestpresented asgroupsofmappingsofonesortoranother. and it iseasilyverifiedthat p-l is one-one and onto X, that is
This seemsto be their natural context. To make this remark p-l is amember of Sx'
precise, we recall that there is a natura.l multiplication of Thisgroup Sx isknown as the symmetric group onthe setX.
mappings and that this multiplication is associative; details In a sensethat willlater be made precise,the structure ofthe
are tobe foundinthe Appendix. The namesemigroup isgiven group Sx dependsonlyon the number ofelementsin X. Since
to any non-empty set-with an associative multiplication under there are n! distinct permutations ofa finiteset ofn elements,
whichit isclosed. Thus everygroup isasemigroupwithcertain weseethat Sx hasn! elements ifX hasn elements.
additional properties; note that axiom(iii)impliesthat nogroup WhenX isfiniteand smallit ispracticable towritedownthe
isempty. Onthe other hand many semigroups (forexample, elements ofSx explicitly and to carry out exploratory calcula
the even integers with the usual binary operation of multi tions. IfX ={I,2,3}then Sx is
plication) failto satisfy group axioms (iii)and (iv). {L, (23),(31),(12),(123),(132)}
These remarks, together with the facts presented in the
in the usual permutation notation. This particular Sx isnon-
Appendix, should make the following theorem and its proof
apparent. t Seethe Appendix.
Some examples of groups 9 10 Someexamplesofgroups
abelian, sincewehave, using the usual multiplication ofmap mapping; clearlyeach translation hasan inverse,whichisagain
pings, (23)(31)= (132), a translation; clearly each reflection c/> has c/> as an inverse,
because # = t.
(31)(23)=(123),
Hencewefind two more groups.
sothat (23)(31)=1= (31)(23).
Example 1.16.Thesetofalltranslations of.!l'(withtheusual
Example 1.14. Let Rx be the set of permutations p of the multiplication) formsan abelian group. Itisnot hard tovisual
setX suchthat eachpleavesallbut afinite number ofelements
izethe group properties for this set.
of X fixed. Itturns out that Rx is an interesting group. To
prove closure the reader should show that if upis the set of Example 1.17. The set ofalloongruencemappings ofthe line
elements of X that are displaced by the permutation p then .!l'formsagroup. Acongruencemapping may be presented as
a combination of suitable translations and reflections, which
u"'Pt S; up!Uup,. The other axioms are readily verified. This
group Rx isknownasthe restrictedsymmetric grouponthesetX. should enable the reader to seehowthe effectof a congruence
mappingmaybe'undone' andaninverseconstructed. Theother
Example 1.15. Let Ax bethe setofpermutations pofX such
laws are easierto grasp. Among points that the reader might
that eachpleavesallbut afinite number ofelementsofX fixed,
ponder on is the fact that the product of two reflections is a
and hasevenparity (that is,p isthe product ofanevennumber
translation.
of transpositions). Onceagain the reader is invited to prove
The congruence mappings of the plane & into itself are a
thegroupproperties ofAx undertheappropriate multiplication.
little more complicated to describe. There are:
Thegroup Ax iscalledthe alternating group onthe setX.
(i) Translations-a mapping c/>of& into & isa translation
Wenext considercertain permutations ofrather specialsets,
if the segment joining P to Pc/>isofconstant length and
namelythesetsofpoints makingupEuclidean spaceofone,two,
direction for allpoints P.
or three dimensions. Initially our account of these groups of
(ii) Rotations about a fixed point O-a mapping c/> of F/'
mappings will be geometrical and tentative, and it should not
intoF/'isarotation if P and Pc/>areat the samedistance
be regarded as rigorous. After these intuitive considerations
from 0 and ifthe angle subtended at 0by P and Pc/>is
weshallexaminegroups ofmatrices, both thosethat arisewhen
constant for allpoints P.
coordinates are introduced in Euclidean geometry and those
(iii) Reflectionsin a fixed line.!l'-a mapping c/> ofF/'into &
that result fromgeneralization.
isareflectionin.!l' if eachpoint P maps onto its mirror
Let .!l'be a line-a one-dimensional Euclidean space. Itis
imagein.!l'.
obviousinthe intuitive sensethat twosorts ofcongruence(that
We note several of the numerous groups formed by sets of
is,distance-preserving) mappingsof.!l'onto .!l'exist:
congruencemappings ofF/'.
(i) Translations-a mapping c/>of.!l' into.!l'isatranslation if
c/>movesevery point afixeddistance to the right or left. Example 1.18.Thesetofalltranslations ofF/'formsanabelian
(ii) Reflectionsin afixed point O-a mapping c/>of.!l' into .!l' group.
is a reflection if c/> maps each point P into the point P,p which
ison the other side of0 but at the same distance from O. Example 1.19.Thesetofallrotations ofF/'aboutafixedpoint
o
Clearly there is a translation t which acts as an identity formsan abelian group.
Some examples of groups 11 12 Some examples of qroup«
Example 1.20. The setofallcongruencemappings off!Jforms mappings are rotations of f!J about 0 through the angle 2kTr/n
agroup. Note that this groupincludesallrotations about every for k=0.1.2,...,n-l, and reflections in the linesjoining 0 to
point 0off!J. eachvertex and to the mid-point ofeach side. Itwillbe found
that there are 2n such mappings and that these form a group,
Inthe case of congruence mappings of three-dimensional
aso-calleddihedral group.
space 9' into 9' weindicate somefamiliar types:
Weshallnowtry tointroduce somemathematical rigour into
(i) translations;
our geometrical considerations. We take coordinates in the
(ii) rotations about afixedline;
usual fashion, though we shall not try to relate all the groups
(iii) reflections in afixedplane.
that now force themselves on our attention to Examples
It isaremarkable factthat anycongruencemapping of9' which
1.16-1.24.
isacombination oftranslations and rotations and.whichleaves .z
Coordinates for arise from the real numbers. Groups
onepoint fixed isequivalent to a rotation about a linethrough
t corresponding to the congruencegroups 1.16and 1.17are:
that fixedpoint.
The following groups of congruence mappings of 9' are of Example 1.26. Thesetofallmappings4>d'withdreal,suchthat
interest. x4>d =x+d
forallrealnumbers z, forms an abeliangroup. The verification
Example 1.21. Thesetofalltranslations of9'formsan abelian
isomitted.
group.
Example 1.27.Thesetofallmappingsofthepreviousexample
Example 1.22. The set ofallrotations of9' about afixedline
together with all mappings <Pa of the type x<pa =;= 2a-x forms
forms an abelian group.
agroup. The verification isomitted.
Example 1.23. The set ofall rotations of 9' leaving a given
In higher dimensions points are regarded as ordered sets of
point fixed forms a group.
real numbers when coordinates are introduced, and we write
Example 1.24. Theset ofallcongruencemappings of9' forms
such ordered sets as row vectors.. Now an nXn matrix is
a group.
essentially a mapping of the set of row vectors into itself, the
Many other important groupsariseassetsofmappings which row vector being multiplied by the matrix in the usual way.
leavesomegeometrical figureinvariant. Weshall giveonecase Let therefore Yen denote the set of allrow vectors such as
whichgives agroup that weshall mention several times, refer x =(x1,XZ ••••,x,,),
ring the interested reader elsewhere'[for afuller account. where XllXZ, ...,X" are real numbers; and let .At", be the set of
Example 1.25.The setofallcongruencemappings off!J'which allnxn matrices, sothat an element A of .AI" has the form
leavefixedaregular n-sidedpolygon isagroup. Wemean that
the polygon is unchanged as a whole, but not necessarily
unchanged point by point; and of coursenmust be an integer
A=
greaterthan 2. Ifthecentreofthe polygonis0then the required
t Aproofmay be foundinJ. L. Syngeand B. A. Griffith,PrillcVples oj
mechaniC8,p.280 (3rded.,McGraw·HilI. 1959).
t For instance to P. S. Alexandroff,Introduction to the theory oj groups, a"l a,,2
Chapter 6(Blackie. 1959). with the entries ailreal.
Some examples of groups 13 14 Someexamplesofgroups
Then A definesamapping of&In into gen' namely the mapping Example 1.31. Theset ofmatrices
whichtakes each x into xA where
{(~ ~):x real and non-zero}
xA =(aUXl+aZlXZ+· ..+anlxn, ...,alnxl+aznxz+ ...+annxn)'
formsan abeliangroupunder matrix multiplication. Theproof
Example 1.28. Theset ..Itnofallnon-singular nXn matrices dependsonthefactthat
withrealentriesformsagroupunder matrix multiplication. The
closurelawisobvious. Theassociativelawfollows,forexample, X x)(y y) = (XY xy)
(0000 00'
from the facts that amatrix represents amapping and that the
product oftwomatricesrepresents theproduct ofthe associated whioh proves closure, shows that an identity element is the
mappings. The unit matrix serves as an identity element, and matrix inwhichx =1,and indicates howto findinversessince
an inverse of the element A is the inversematrix A Sowe
-1.
have a group.
Our final example is of lesser interest than multiplicative
Example 1.29. The set ofall real nXn matrices with deter
groups ofmatrices.
minant 1formaagroupunder matrix multiplication. The least
obviouspoint intheproofofthis statement isthe closurelaw. Example 1.32.Theset ofallmXn matrices formsanabelian
groupunder matrix addition. Theentries mayberational num
Many variations onthese examples arepossible. In addition
bers, real numbers, complexnumbers, etc.
to having infinitely many possibilities for n, we can choose
the rational numbers or the complex numbers or other sets
Problems
of numbers for entries. We can even choose residue classes
(Harderproblemsarestarred)
modulo somepositive integer.
1. Discussthe followingsystemswith a viewto determiningwhichof
Example 1.30. The following set of matrices, in which i themaregroups.
denotes a complex number for whichiZ = -1,forms a group (i) Thepositiverealnumbers,withthebinaryoperationofdivision.
(ii) Theintegersthat aredivisibleby 10,withaddition.
under matrix multiplication:
(iii) Thenon-zerorationalnumbers,with abtaken asthe 'product'
i) -i)
ofa and b.
{(oI 0l') (-10 -10') (-10 01') (01 -10)' (i0 0' (0-i 0 ' (iv){a+b..J2:a andbrational,not bothzero}withthebinaryopera-
tion ofnumbermultiplication.
0) (i O)} (v)Theintegers,with2a+3btakenasthe' product' ofaandb.
-i
(o i!' 0 -i . (vi)Therationals,witha+b-ab taken asthe 'product' ofaandb,
(vii)The set ofall 2x2singularmatriceswith real entries, under
Theverification isomitted. Thisgroup iscalledthe quaternion matrix addition.
(viii)The set of all vectors in three-dimensionalEuclidean space,
group.
withthe binaryoperationofvectorproduct.
The multiplicative semigroupofallnXnmatrices with (say) (ix) Theset{,pk: kreal}ofallmappings,pk' where
real entries contains many subsets which form multiplicative Xc/>k =kx,
groups not containing the unit matrix. We exhibit one non of2'into 2'.
(x) {a+b..J2+c..J3: a, b, c rational, not all zero}with tho binary
trivial specimen.
operationofnumbermultiplication.
Someexamplesofqroups 15
2. Whichofthe groupsinthe previousquestionare abelian? Whichof
2
the restare semigroups?
3. Showthat thenon-zeroresidueclassesmodulo6donot formagroup
under multiplication.
Provethat ifthenon-zeroresidueclaesesmodulonformagr\)upunder
multiplication andn > 1then nisprime. Basic theorems and concepts
4. Aset ofmatricesformsagroup under matrix multiplication. Show
that oithereverymemberofthesetissingularoreverymemberisnon
singular.
5. Let p beany prime number. Showthat the followingset formsan
additive group:
{m/pn: mand nintegers}.
THE theorems inthis chapter areoftwo sorts. First, wemust
x4> x
Prove that themapping4>forwhich =p»,where isanarbitrary
deducesomefacts fromthe axioms. Second,wehave to discuss
element ofthe set,isapermutation.
the important concepts of homomorphism and isomorphism.
6. Let Gbeagroupofpermutations onasetX. Showthat thoseper
mutations whichleavefixeda givenolemontx ofX formagroup. The facts appear rather pedantic unlessthey areviewedagainst
a background of systems that are not groups, while the con
7. LetTnbothosetofn Xn uniteiangulurmatriceswithrealentries(that
ismatriceswith 1foreachentry onthe principaldiagonaland0every cepts are ofan importance that cannot be over-emphasized.
wherebelowthisdiagonal). Showthat Tnisamultiplicative group. Werecall our definition ofagroup asasemigroupwith right
S. Let T~bethe setofnXnunitriangular matricesTw:ith residueclasses identities andright inverses,inasensemadepreciseinChapter 1.
modulopforentries(pisafixedprime). Showthat isamultiplicative Sometimes a definition is given that states that each right
group,and findthe number ofelementsinit.
identity isalsoaleft identity and that eachright inverseisalso
*9. LetSbeanysetand Gbeanygroup.' Provethat ifmultiplicationof
a left inverse; but wecan deducethese facts fromour axioms.
tho.mappingsiseuitablly definedthenthe setofallmappingsofS into G
formsagroup.
THEOREM2.01. If the element x of any group G has X-I as a
10.ThesetofallorderedtriplesofintegersisdenotedbyG;thusatypical
edleefmineendtaosffGolilsow(txs,:fl,y)wheretx,fl,yare integers. MultiplicationinGis t(hreignhtx)-Iinxve=rsee,.80 thatxx-I =efor some (right)identity elemente,
(tx,fl,y)(g,,),,) = (tx+(-l)llg: {3+(-l)Y7], (-I)fyH). Proof. Let ebe a (right) identity element of Gof the kind
Showthat Gwiththismultiplication isagroup.
mentioned inaxiom (iv). We haveto usethe associativelawin
+11.Showthat there isalargest multiplicative group of2X2matrices the form
(0 0)
W.ithitr'ea enrrescontam..ing 5 _5 ' and find thl.Sgroup.
The left-hand side of this equation reduces to x-Ie and so to
+12.'AsemigroupShastheproperty that foreachelementxinS thero X-I, by axioms (iii)and (iv);therefore
iSsaangeroleump?entdenoted byX-IinS suchthat y=-1 =YforallyinS. Is X-l = (x-Ix)x-I•
**13.Let Sbeasemigroupwith thofollowingproperties: But, byaxiom (iv),X-I itself has aninverse, denoted by (X-I)-I,
(i)There isanelementesuchthat ex= x forallx inS. such that X-I(X-I)-I =e; and wehave
(ii) XFo-Irfeoarcwhhsiucchh=e-le1m=enet.eand foreachx inS there isanelement X-I(X-I)-I = {(x-IX)X-I}(X-1)-1.
Prove that the set{Be: BES},whereeisfixed,formsagroup. The left-hand side is e by (iv); and the right-hand side is
Some example.<! ofgroup8 15
2. Whichofthe groupsinthe previousquestionare abelian? Whichof
2
the restare semigroups?
3. Showthat the non-zeroresidueclassesmodulo6donotformagroup
under multiplication.
Provethat ifthenon-zeroresidueclassesmodulonformagt'(lupunder
multiplication andn > 1then nisprime. Basic theorems and concepts
4. A sotofmatricesformsagroup under matrix multiplication. Show
that either everymemberofthe setissingularoreverymemberisnon
singular.
5. Let p beany prime number. Showthat the followingRetformsan
additive group: .
{rn/pn: mand nintegers},
Prove that the mapping", forwhichxrp =px,wherex isanarbitrary THE theorems inthis chapter areoftwo sons. First, wemust
element ofthe set, isapermutation. deducesomefactsfromthe axioms. Second,wehaveto discuss
the important concepts of homomorphism and isomorphism.
6. Let Gbeagroupofpermutations onasetX. Showthat thoseper
mutations whichleavefixedagivenelementx ofX formagroup_ Thefacts appearrather pedantic unlessthey areviewedagainst
a background of systems that are not groups, whilethe con
7. LetTnbethesetofnXnunitriangular matriceswithrealentries(that
ismatriceswith 1foreachentry ontheprincipaldiagonaland0every cepts are ofan importance that cannot be over-emphasized.
wherebelowthisdiagonal). Showthat Tnisamultiplicative group. Werecallourdefinitionof agroup as asemigroupwith right
8. Let T~bethesetofnXnunitriangular matriceswith residueclasses identities andrightinverses,inasensemadepreciseinChapter 1.
modulopforentries(pisafixedprime). Showthat ~':isamultiplicative Sometimes a definition is given that states that each right
group, and findthe number ofelementsinit.
identity isalsoaleftidentity andthat eachright inverseisalso
*9. Let Sbeanysetand Gbeanygroup.'Prove that ifmultiplicationof
a leftinverse; but wecan deducethese facts fromour axioms.
tho.mappingsissuitably definedthen the setofallmappingsofS into G
formsagroup.
THEOREM2.01. If the element x of any group Ghas X-I as a
10. ThesetofallorderedtriplesofintegersisdenotedbyG;thusatypical
elementofGis(0<,p,y)where0<,p,Yareintegers. MultiplicationinGis (right) inverse, 80thatxx-l =efor some(right)identity elemente,
definedasfollows: then x-Ix = e.
(o<,p,y)(g,,,),O =(o<+(-l).Bg;p+(-l)Y"), (-I)fy+S). Proof. Let ebe a (right) identity element of Gof the kind
Showthat G withthis multiplication isagroup.
mentioned.in axiom (iv). We haveto usethe associativelawin
*1.1. ShoIw' that there i.s.alar(g0est0m)ultiplicative group of2X2matrices the form
Withrea entriescontammg 5 _ 5 ' and find thi.s group.
The left-hand side ofthis equation reduces to x-Ie and so to
*12.'AsemigroupS hastheproperty that foreachelementxin Sthere x-I,by axioms(iii)and (iv); therefore
isanelementdenoted byx-1inSsuchthat y=-l = YforallyinS. Is =
S agroup? X-I (x-1X)X-1.
"13. Let Sbeasemigroupwith the followingproperties: But, by axiom (iv),X-I itselfhasaninverse,denoted by (X-I)-I,
(i)ThereisBJlelementesuchthat ex =x forallx inS. such that X-1(X-I)-1 =e;and wehave
(ii) xF-o1rfeoarcwhhsuicchh=e-le1m=ente.eand foreachx inS there isan element X-1(X-I)-1 = {(x-IX)X-1}(x-1.)-1.
Prove that the set {se:eES}, whereeisfixed,formsagroup. The left-hand side is e by (iv); and the right-hand side is
Basic theorems and concepts 17 lR Basictheoremsandconcepts
(x-IX){X-I(X-I)-I} by the associativelaw,and clearlythis equals In allourexamples,there wasonlyoneidentity elementand
X-IX. Thus wehave proved that X-IX isthe identity element e each element had only oneinverse.
as required.0 '
The above theorem impliesthat Xand X-I alwayscommute. TIrEOREM 2.04. In any group G there is only one identity
This isperhaps not surprising to us, for it happened in all the element, and eachelement of Ghas only oneinver8e.
examples ofthe previous chapter. The reader should consider Proof. Suppose wehad two identity elements, say eland e2•
~het~er th:~e is a correspondingtheorem in a semigroupwith Sincee2isaright identity ele2 = el,andsince(byTheorem2.02)
right identities and left inverses, such as a set withxy defined el is a left identity e1e2=e2• Therefore el = e2, and Gcannot
to be x. (Wediscussed such objects in Chapter I. Here isa have more than oneidentity element.
specificinstance. Itwill be found that ifSis Supposethat theelement aofGhasboth X and y asinverses.
= =
Then ax ay, and by left cancellation X y; therefore a-I is
unique.0
then matrix multiplication in Sissuch that the product oftwo THEOREM 2.05. In any group, x = (X-I)-I.
matrices isthe left-hand factor.)
Proof. Wehavedefined(X-l)-lasasolutiony ofthe equation
Another readily acceptable fact follows.
x-Iy =e;weknow that eachofe,x-l, y isunique. But clearly
=
THEOREM2.02. If xisany elementofany group Gwithidentity y x is one solution to x-1y =e, by Theorem 2.01. Sinceit
=
element e,such that gg-l =efor all gin G,then ex =z. isthe onlysolution, wehave z (X-I)-I·O
The next result isofa different character. Itisageneraliza
Proof. The previous theorem and the axioms are used as
tion ofthe associative law that appeared as axiom (ii) and it
follows:
ex = (=-l)X =x(x-1x) = xe = x. states, roughly, that a product of group elements can be
Thus the result isproved; x and ealways commute.0 bracketed in any way onepleases.
TH~OREM2.03. If a and bareelements of any group G,and if THEOREM 2.06. Letfixed elementsaI'a2,... ,anbetaken in order
there ~8an element xfor whichxa = xb oran elementyfor which in any group G. Then thevalue oftheproduct ofall theseelement8
ay = by, then a =b. in thegiven orderis 1~naffectedby the 8equencein which product«
PrOOf· Suppose xa = xh. The existence of inverses gives areformed.
X-l(xa) =x-1(xb). Proof. Themeaningofthis theoremissubtle. Forthepurpose
ofillustration letn=3.Thestatement thenisthat theelements
Useofthe associative lawand Theorem 2.01gives
al(a2aS) and(ala2)a3arethesame,forthereareonlytwopossible
ea= eb. ways of forming products; either form a2a3 first or form ala2
Theorem 2.02nowgivesa =b. The deduction from ay =by first. The statement does not suggest that (ala2)aS = (a2aI)a3,
nee~ only the ax.iomsand being thus easier isomitted.0 which is certainly false in general. When n= 4 the possible
This theorem grves us left and right cancellation laws in an products ofaI'a2, aa,a4inthat orderareal{a2(aaa4)}, (ala2)(aaa4),
obvious sense. The reader should consider whether there are al{(a2a3)aJ, {al(a2aS)}a4' {(a1a2)aa}a4, and 110 others. The
such lawsin the set S ofmatrices mentioned above. theorem impliesthat these products are all equal.
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