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Transcript
```A group may be defined as follows.
3. G has an identity element e. There is an e in G such that x * e 5 e * x 5 x for all
x [ G.
Groups
!
Week 4
ition 3.1 ■4. G
Group
contains
inverses. For each a [ G, there exists b [ G such that a * b 5 b * a 5 e.
Definition
3.1 Group
Suppose the binary operation p is defined for elements of the set G. Then G is a group with
Thetophrase
“with the
respect
to p” should
be noted.hold:
For example, the set Z of all integers
respect
p provided
following
four conditions
is a group with respect to addition but not with respect to multiplication (it has no inverses
1.
G is closed
under
p.61).
That Similarly,
is, x [ G and
[G
G5
imply
in G. with respect to
elements
other
than
the yset
{1, that
21}x is
* yaisgroup
Chapter 3 for
Groups
multiplication
but notFor
withallrespect
2. p is associative.
x, y, z to
G, x * (y *Inz)most
5 (xinstances,
* y) * z. however, only one binary
operation is under consideration, and we say simply that “G is a group.” If the binary
3. G has isanunspecified,
identity element
e. There
is an e in Gnotation
such that
5 ejuxtaposition
all
* e the
* x 5 x for137
operation
the multiplicative
andxuse
xy
x
[
G.
to indicate the result of combining x and y. Keep in mind, though, that the binary operation
4. not
G necessarily
contains inverses.
For each a [ G, there exists b [ G such that a * b 5 b * a 5 e.
is
multiplication.
ition 3.2
tion 3.2
phraseGroup
“with respect to p” should be noted. For example, the set Z of all integers
■ The
Abelian
Definition
3.2
Abelian
is a group with respect toGroup
addition but not with respect to multiplication (it has no inverses
Letelements
G be a group
to p. Then the
G isset
called
commutative
group,with
or anrespect
abelianto†
for
otherwith
thanrespect
61). Similarly,
G 5a {1,
21} is a group
group, if p is but
commutative.
That is,
for all
x, y in G.however, only one binary
* y 5 y *Inxmost
multiplication
not with respect
instances,
operation is under consideration, and we say simply that “G is a group.” If the binary
Remark is unspecified, we adopt the multiplicative notation and use the juxtaposition xy
operation
1 result
We can
obtain some
simple
examples
of groups
bythat
considering
toExample
indicate the
of combining
x and
y. Keep
in mind,
though,
the binaryappropriate
operation
subsets
of
the
familiar
number
systems.
is not necessarily multiplication.
1) The set of integers is a group under the OPERATION of addition:
a. The set C of all complex numbers is an abelian group with respect to addition.
■We Abelian
Group
integers
undernumbers
the OPERATION
are respect
CLOSED,
b. have
The set
Q ! seen
{0} ofthat
all the
nonzero
rational
is an abelian
group with
to
ASSOCIATIVE, have IDENTITY 0, and that any integer x has the INVERSE −x. Because
multiplication.
Let
group with
respect
to p.satisfies
Then Gall
is called
a commutative
group,
an abelian†
the G
setbeofa integers
under
four group
PROPERTIES,
it is or
a group!
+
c. The
of all positive
real
abelian
with respect to multiplicagroup,
if psetisRcommutative.
That
is,numbers
x * y 5 yis*an
x for
all x,group
y in G.
tion,
but
it is not
a group
with isrespect
(it hasitno
identityalland
no
2) The
set
{0,1,2}
under
not a to
group,
because
does
not satisfy
of the
group
PROPERTIES:
it
does
not
have
the
CLOSURE
PROPERTY
(see
the
previous
■
lectures to see
Therefore,
the simple
set {0,1,2}
under of
notconsidering
a group! appropriate
Example
1 why).
We can
obtain some
examples
groupsisby
The
examples
some indication of the great variety there is in groups.
subsets
offollowing
the familiar
numbergive
systems.
(Notice also that this set is ASSOCIATIVE, and has an IDENTITY which is 0, but does not
have
theset
INVERSE
because
and −2group
are on
not
set!)
a. The
C
all PROPERTY
complex
numbers
is an−1
with
respect
Example
2 ofRecall
from Chapter
1 that
a abelian
permutation
ainsetthe
A is
a one-to-one
fromThe
A onto
A and
thatofS(A)
set of all
permutations
on A. We
havewith
seenrespect
that S(A)
b.
set Q
! {0}
all denotes
nonzerothe
rational
numbers
is an abelian
group
to
3)
The setwith
of integers
under
subtraction
is not
a group,
because
it doesand
notthat
satisfy
is closed
respect to
the binary
operation
+ of
mapping
composition
the
multiplication.
all
of
the
group
PROPERTIES:
it
does
not
have
the
ASSOCIATIVE
PROPERTY
(see
the
operation + is +associative. The identity mapping IA is an identity element:
c. The set
R of all
real
numbers is
abelian
group with
to multiplicaprevious
lectures
to positive
see why).
Therefore,
theanset
of integers
underrespect
subtraction
is not a
f + IA 5tof 5
IA + f (it has no additive identity and no
group!
tion, but it is not a group with respect
inverses).
■
all f [ S(A),
and each f [ S(A) has an inverse in S(A). Thus we may conclude from
(Notice also that this set is CLOSED, but does not have an IDENTITY and therefore also
results
infollowing
Chapter
that S(A) PROPERTY.)
is a group
respectoftothe
composition
of mappings.
The
examples
give
some with
indication
great variety
there is inHowever
groups.
does
not
have
the 1INVERSE
S(A) is not abelian since mapping composition is not a commutative operation.
■
Example
from
Chapterunder
a permutation
onaa set
A is abecause
one-to-one
mapping
4) The set2 of Recall
natural
numbers
is not
group,
it does
not
satisfy
all of
theWe
group
PROPERTIES:
does
notobtain
have an
theexplicit
IDENTITY
PROPERTY
(see
Example
shall
take
A 5the
{1,set
2,itof
3}
ofthat
S(A)
. In
from
A onto
A3and
that
S(A)
denotes
alland
permutations
on
A. Weexample
have
seen
S(A)
previous
totosee
Therefore,
off (1),
natural
numbers
isthe
closed
withlectures
respect
binary
+ set
of mapping
composition
thatthree
theis
order
to define
an
element
fthe
ofwhy).
S(A)
, weoperation
need tothe
specify
f (2),
and f (3).under
There
are
not
a group!
operation
+ is associative.
identity
mapping
IA is anthere
identity
possible
choices
for f (1).The
Since
f is to
be bijective,
areelement:
two choices for f (2) after
f + IAASSOCIATIVE,
5 f 5 IA + f but does not have the INVERSE
(Notice also that this set is CLOSED,
†
The term abelian
is used innone
honor of Nielsnegative
Henrik Abel (1802–1829).
A biographical
of Abel appears on
PROPERTY
because
are
in
the
set.)wesketch
for
all f [ S(A),
and each fof[the
S(A) has annumbers
inverse in
S(A).
Thus
may conclude from
the last page of this chapter.
results in Chapter 1 that S(A) is a group with respect to composition of mappings. However
S(A) is not abelian since mapping composition is not a commutative operation.
■1
Example 3
We shall take A 5 {1, 2, 3} and obtain an explicit example of S(A). In
Groups
!
Week 4
5) The set of rational numbers with the element 0 removed is a group under the
OPERATION of multiplication:
We have already seen that the set rational numbers with the element 0 removed under the
OPERATION of multiplication is CLOSED, ASSOCIATIVE, have IDENTITY 1, and that any
1
integer x has the INVERSE . Because the set of rational numbers with the element 0
x
removed under multiplication satisfies all four group PROPERTIES, it is a group!
6) The set of rational numbers (which contains 0) under multiplication is not a
group, because it does not satisfy all of the group PROPERTIES: it does not have the
INVERSE PROPERTY (see the previous lectures to see why). Therefore, the set rational
numbers under multiplication is not a group!
(Notice also that this set is CLOSED, ASSOCIATIVE, and has an IDENTITY which is 1.)
7) The set of rational numbers under division is not a group, because it does not
satisfy all of the group PROPERTIES: it does not have the ASSOCIATIVE PROPERTY
(see the previous lectures to see why). Therefore, the set of rational numbers under
division is not a group!
(Notice also that this set is not CLOSED because anything divided by 0 is not in the set,
does not have an IDENTITY and therefore also does not have the INVERSE PROPERTY.)
8) The set of natural numbers under division is not a group, because it does not
satisfy all of the group PROPERTIES: it does not have the IDENTITY PROPERTY (see
the previous lectures to see why). Therefore, the set of natural numbers under division is
not a group!
(Notice that this set does not have the CLOSURE, ASSOCIATIVE or INVERSE
PROPERTIES.)
9) The set of integers under multiplication is not a group, because it does not satisfy
all of the group PROPERTIES: it does not have the INVERSE PROPERTY (see the
previous lectures to see why). Therefore, the set of integers under multiplication is not a
group!
(Notice also that this set is CLOSED, ASSOCIATIVE, and has the IDENTITY ELEMENT
1.)
Definition A permutation of a set is a bijection (one to one and onto) of this set onto itself,
i.e.,
S ( A ) = { f f : A → A and f is bijective } .
Example*
2
7 Permutations
Permutations
Groups
7
!
Week 4
The aim of this section is to discuss a large class of groups, called permutation groups.
The aim ofThe
thiscentral
sectionresult
is toofdiscuss
a largeis class
of groups,
called
permutation
this section
Cayley’s
theorem,
which
states thatgroups.
actually any group
The centralis result
of thisinsection
is Cayley’s
theorem,
states that
actually
group example of a
contained
this class
of group.
Thus, which
permutation
groups
give any
a generic
is contained
in this class of group. Thus, permutation groups give a generic example of a
group.
group.
7.1
7.1 Permutations
Homework*. Take A = {1,2, 3} and obtain an explicit example of S ( A ) .
Permutations
Definition 7.1. A permutation of a set is a bijection of this set onto itself.
Definition 7.1. A permutation of a set is a bijection of this set onto itself.
Note that
Note that (i) if α : X → X is a permutation, than the inverse function α−1 : X → X is also a
(i) if αpermutation;
: X → X is a permutation, than the inverse function α−1 : X → X is also a
permutation; (ii) if α : X → X and β : X → X are permutations, one can form a composition
(ii) if αα :◦ X
X X,
andwhich
β : Xis →
permutations, one can form a composition
β :→
X→
alsoXa are
permutation.
α ◦ β : X → X, which is also a permutation.
Theorem 7.2. Let A be a nonempty set, and let S be the collection of all permutations of
TheoremTheorem
7.2.
Let A
be
nonempty
set,the
and
let
S let
be the
collection
of all
permutations
of of A. Then G
a nonempty
set,
and
G ◦.
be
collection
of all
permutations
A.
Then
SLet
isAaabe
group
under
composition
A. Then Sis isa agroup
groupunder
underthe
thecomposition
composition ◦.
Proof. Clearly, ◦ defines a binary operation on S. We need to check the three axioms G1,
Proof. Clearly, ◦ defines a binary operation on S. We need to check the three axioms G1,
G2, G3 of the group.
G2, G3 of the group.
(G1) We have seen in section 1 that composition is associative.
(G1) We have seen in section 1 that composition is associative.
(G2) Consider the mapping ι : A → A such that for all x ∈ A, ι(x) = x. Then for any
(G2) Consider the mapping ι : A → A such that for all x ∈ A, ι(x) = x. Then for any
α ∈ S one has α ◦ ι = ι ◦ α = α. Thus, ι is the identity element on S.
α ∈ S one has α ◦ ι = ι ◦ α = α. Thus, ι is the identity element
on S.
−1
−1
−1
(G3)
For
any
permutation
α
∈
S,
its
−1 inverse α −1 satisfies α −1◦ α = α ◦ α
(G3) For−1any permutation α ∈ S, its inverse α satisfies α ◦ α = α ◦ α = ι. Thus, = ι. Thus,
is the
of α with
respect
the group operation.
α−1 is the α
inverse
of αinverse
with respect
to the
grouptooperation.
Example**. Construct the Cayley (multiplication) table for S(A) in Example*.
Below
we will
denote composition
of permutations
α and
β by αβ rather
Notation Notation
Below we will
denote
composition
of permutations
α and β by αβ
rather
than α ◦ β.than α ◦ β.
7.3.A be1.any
Letset.
A be
anygroup
set. of
The
of all permutations
DefinitionDefinition
7.3.
1. Let
The
all group
permutations
of A is calledofa A is called a
on A and
symmetric symmetric
group on Agroup
and denoted
by denoted
Sym(A). by Sym(A).
2. Anyofsubgroup
Sym(A)Sym(A)
(including
Sym(A)
itself)
is called a group
permutation
group on
2. Any subgroup
Sym(A) of
(including
itself)
is called
a permutation
on
the set A. the set A.
3. Iffinite
A isset
the{1,
finite
. . , n} of nthen
elements,
Sym(A) is
commonly denoted
3. If A is the
2, . . .set
, n}{1,
of 2,n . elements,
Sym(A)then
is commonly
denoted
We have
dealt
with Sthe
Sn . We have
dealt
with the
groups
n . groups Sn .
2 ···n.
· · · (n − 1) · n.
Proposition
7.4. of
The
is group
n! = 1 S· n2 ·is· ·n!
··=
(n 1−· 1)
Proposition
7.4. The order
theorder
groupofSnthe
Proof. TheProof.
problem
of problem
computing
number the
of elements
theproblem
same as the problem
The
of the
computing
number in
of Selements
in Snasisthe
n is the same
of computing
the numberthe
of number
different of
ways
the integers
1, 2,integers
. . . , n can
be. placed
in the
n
of computing
different
ways the
1, 2,
. . , n can
be placed
in the n
Homework**.
Construct
the
Cayley
(multiplication)
table
for
S(A)
in
Homework*.
blanks indicated
each(using
integereach
just integer
once): just once):
blanks(using
indicated
!
!"
"
Example 4 (Homework) 1 2 . . . n1 .2 . . . n
.
...
...
25
25
3
igure 3.3
igure 3.3
igure 3.4
If the elements in the row headings are listed in the same order from top to bottom as the
in all positions that are symmetrically placed relative to the diagonal from upper left to lower
elements in the column headings are listed from left to right, it is also possible to use the table
and inverses
elements.
element
e is athe
lefttable
identity
if and 3.2
onlyisifnot
thesymmetric.
by e
right. In Example
3, theofgroup
is notAnabelian
since
in Figure
to check foratcommutativity.
The
operation
is
commutative
if
and
only
if
equal
elements
appear
2
2
leftdend
the! same
as r
the+column
table. Similarly,
e is a4
Groups
Week
example, g + the
r 5
is in
rowexactly
5, column
3, and
is in rowin3,the
column
5.
in all positions
are symmetrically
from
upperthe
leftsame
to lower
rightthat
identity
if and only if theplaced
columnrelative
bythe
e atdiagonal
exactly
as the
right. In Example
3, the
group
is not
since
table
3.2
is not
For
row
in the
the set
table.
it exists,
the the
inverse
ofina Figure
certain
element
asymmetric.
can
be found
by
Example
Let
G be
of Ifabelian
complex
numbers
given
by G 5
{1,
21,
i, 2i},
where
2
example, gsearching
+ r 5 dfor
is in
5, column
r + gby5asand
is in
rowin3,the
column
therow
identity
e in the3,row
again
i 5 !21 , and consider the operation of multiplication of complex numbers in G. The
If the elements in the row headings are listed in the same order from top to bottom as the
table
in Figure
thatset
is
respect
totomultiplication.
elements
inGthe
arewith
listed
from left
right,
also
use the
table
Example
4 3.3
Letshows
becolumn
the
ofclosed
complex
numbers
given
by itGis5
{1,possible
21, i,to2i},
where
Multiplication
in
G
is
associative
and
commutative,
since
multiplication
has
these
propcheck
for commutativity.
The operation
is commutative
and only ifnumbers
equal elements
i 5 !21 , toand
consider
the operation
of multiplication
of ifcomplex
in G.appear
The
erties in theinset
of
all complex
numbers.
We placed
can observe
from
Figure
3.3
that
1 isleft
thetoidenall
positions
that
are
symmetrically
relative
to
the
diagonal
from
upper
lower
table in Figure 3.3 shows that G is closed with respect to multiplication.
tity elementright.
andInthat
all elements
haveis not
inverses.
of 1table
andin21
is its
andFor
i
Example
3, the group
abelianEach
since the
Figure
3.2own
is notinverse,
symmetric.
Multiplication gin+Gr 2is5associative
and
commutative,
sincesmultiplication
has these
propd is in Thus
row 5,G
column
3, and r2group
+ g 5 with
is inrespect
row 3, column
5.
and 2i are example,
inverses of each other.
is an abelian
to multiplication.
erties in the set of all complex numbers. We can observe from Figure 3.3 that 1 is the identity3element
and21
that all
have
Each
of 1 and
its {1,
own
inverse,
i
4 elements
Let
the
setinverses.
of complex
numbers
given21
by is
G5
21,
i, 2i}, and
where
1Example
i G be2i
i
5
!21
,
and
consider
the
operation
of
multiplication
of
complex
numbers
in
G.
The
and 2i are inverses of each other. Thus G is an abelian group with respect to multiplication.
in Figure 3.3
1
1table 21
i shows2ithat G is closed with respect to multiplication.
Multiplication
3
1
21
i in G is
2iassociative and commutative, since multiplication has these prop21 21erties in1the set2i
i numbers. We can observe from Figure 3.3 that 1 is the idenof all complex
1
1tity element
21 and that
i
2i
have inverses. Each of 1 and 21 is its own inverse, and i
i
i
2i
21 all elements
1
and 2i are inverses of each other. Thus G is an abelian group with respect to multiplication.
21
1i
2i
2i 21
2i
1
21i
21
i 3 2i 1 21
i
1i
2i
21
i
2i
2i
2i
1
21corollary of Theorem 2.28 that the set
Example
51 It isi 1an immediate
Example
5 (Homework)
21 21
1
2i
i
i
i
2i
2i
2i
i
Zn 5 5 304, 314, 324, c, 3n 2 14 6
21
1
1
21
Example
5 classes
It is anmodulo
immediate
corollary
of Theorem
2.28 respect
of
congruence
n forms
an abelian
group with
■
■
■
■
Z 5 5 304, 314, 324, c, 3n 2 14 6
Example 6 Let G 5 {e, a, b,n c} with multiplication as defined by the table in Figure 3.4.
of congruence classes modulo n forms an abelian group with respect to addition.
■
■ Figure 3.3
be
ca
e
Example 5
It is an immediate corollary of Theorem 2.28 that the set
Example 6
Let G 5 {e, a, b, c} with multiplication as defined by the table in Figure 3.4.
b
c
Zn 5 5 304, 314, 324, c, 3n 2 14 6
Example
6
Let
G
5
Example 6 (Homework) {e, a, b, c} with multiplication as defined by the table in Figure 3.4.
■
e
e of congruence
a
b
cclasses modulo n forms an abelian group with respect to addition.
a?
ae
ba
cb
ec
?
be
ca
a
ca
eb
eb
ac
ac
b
e
b
?
e
a
e
e
a
b
c
c
b
b
c
e
a
a
b
c
e
From the table,
wea observe
the following:
c
c be ba c b e a
igure 3.4
1. G is closed under this multiplication.
c
c
e
a
b
■ Figure
From
the
table,
we observe
2. e 3.4
is the
identity
element.the following:
From the
table,
wemultiplication.
observe the following:
1. G is closed
under
this
2. e is the 1.identity
element.
G is closed
under this multiplication.
2. e is the identity element.
4
Groups
3. Each
3.1 Definition of a Group
141
3.1 Definition of a Group
141
Week 4
of e and b is its own inverse,! and c and a are inverses of each other.
3.1 Definition of a Group
141
This of
multiplication
commutative.
3.4. Each
e and b is itsisown
inverse, and c and a are inverses of each other.
This
multiplication
is also
associative,
but we shall not verify it here because it is a labori4. This
multiplication
3. Each of e is
andcommutative.
b is its own inverse, and c and a are inverses of each other.
ous task. It follows that G is an abelian group.
■
3.2 Properties of Group Elements
145
4. This is
multiplication
is commutative.
This multiplication
also associative,
but
we shall not verify it here because
it is a laborious
thattable
G is in
an
abelian
group.
■
This
is also
associative,
but we shall
not verify
it here because
it isset
a laboriExample
7 multiplication
The
Figure
3.5
defines
a binary
operation
p on the
S5
33. Let G be the set of all matrices in M3(R) that have the form
ous task. It follows that G is an abelian group.
■
{A, B, C, D}.
1
a
b
Example 7 The table in Figure 3.5 defines a binary operation p on the set S 5
Example 7 The table
C 0 in 1Figure
c S 3.5 defines a binary operation p on the set S 5
{A, B, C, D}.
*
A {A,
B B, C,
C D}.
D
0 0 1
realAnumbers
B a, b, and c. Prove or disprove that G is a group with respect
*A for arbitrary
AB BC
* C A DB C D
to multiplication.
AB ProveBCor disprove
CD
BACthe set
A ABBthat
A GBin Exercise 32 is a group with respect to addition.
34.
B BCCthat
B GAin Exercise 33 is a group with respect to addition.
BC ProveCAor disprove
DB
35.
C C
A D
B theCpower
D set p(A) was defined in Section 1.1 by p(A) 5
36.
A,
Dset
D
CDFor an
AA arbitrary
BB
{X 0 X 8 A}, and addition in p(A) was defined by
D
A B D D
D 3.5A B D D X 1 Y 5 (X c Y) 2 (X d Y)
■ Figure
gure 3.5
5 (X 2 Y)
c (Y 2 X).
From the table, we make the following
observations:
From the table, we make the following observations:
Prove
that we
p(A)
is p.
a group
with respectobservations:
1. a.Sthe
is closed
under
From
table,
make
the following
1.
S
is
closed
under
p.
b. If A has n distinct elements, state the order of p(A).
igure 3.5
c. 1.1, #7c @
c. 1.1, #7c @
2. C is an identity element.
1. S is closed
2. under
C is an p.
identity element.
37.
outnot
the have
elements
of p(A) since
for theDX
set 5
A 5C{a,
b,no
c}, solution.
3. Write
D does
an
inverse
has
3.
D
does
not
have
an
inverse
since
DX
5
C
has
no solution.
2. C
is
an
identity
element.
table for p(A) using addition as defined in Exercise 36.
Thus
S Ais5not
a group
respect
to
p.p(A)
Thus
Sc}.
is Prove
not
ainverse
group
with
respect
to is
p.Ca has
3. D
not
anwith
since
5
nowith
solution.
38.
Letdoes
{a, b,have
or disprove
thatDX
group
respect to the operation
of union.
Thus S is not a group with respect to p.
Finite
Group,
Infinite
Group,
Order
Group
c.
1.1,
#7c
@
39.
A 53.3
{a,
b,Finite
c}. Prove
or disprove
that
p(A) isOrder
aof
group
with
respect to the operation
Definition
3.3
■
Group,
Infinite
Group,
ofof
aaGroup
ition 3.3 Definition
■ Let
Finite
Group,
Infinite
Group,
Order
a Group
tion 3.3
3.2
■
■
■
of intersection.
If a group G has a finite number of elements, G is called a finite group, or a group of finite
group GGroup,
has a finite
number
of elements,
a finite group, or a group of finite
■If a Finite
Infinite
Group,
OrderGofis acalled
Group
order. The number of elements in G is called the order of G and is denoted by either
order. Theo(G)
number
of elements in G is called the order of G and is denoted by either
or 0 G 0 . If G does not have a finite number of elements, G is called an infinite group.
IfProperties
a group
a does
finite
number
of
elements,
G isofcalled
a finite
group,
group ofgroup.
finite
o(G)
or 0 GG0 .has
If G
not
have
a finite
number
elements,
G is
calledorana infinite
of
Group
Elements
order. The number of elements in G is called the order of G and is denoted by either
8notdefinition
Inhave
Example
the
grouprecorded
o(G)
orconsequences
0 G 0 .Example
If G does
a finite
number
of elements,
G is called an infinite group.
Several
of the
a3,group
in Theorem
Example
8 In Example*,
the ofgroup
G are
= {e, σ } has
order 3.4.
Example 8 In Example 3, the group G 5 5e, r, r2, s, g, d6 o (G ) = 2 . In Example 5,
o ( Z n ) = n. The set Z of all integers is a group2 under addition, and this is an exaple of an
G 5 5e,
r, r
s,n.g,The
d6set Z of all integers is a group under adhas
o(G) 5 6.3,Inthe
Example
5, o(Z
n) ,5
8oforder
In Example
rategy ■ Example
Parts a group.
and bdition,
the
next
are group
statements
uniqueness,
andinfinite
they can
infinite
and
thistheorem
is an example
of an infinite
If A is an
set,be
then S(A) furnishes
)
5
n.
The
set
Z
of
all
integers
is
a
o(G)
5
6.
In
Example
5,
o(Z
proved
by the
standard
type
of
uniqueness
proof:
Assume
that
two
such
quantities
exist,
2
n
an example of an infinite group.
■
G
5
5e,
r,
r
,
s,
g,
d6
and
thenand
prove
be equal.of an infinite group. If A is an infinite set, then S(A) furnishes
dition,
this
istwo
antoexample
Properties
ofthe
Group
Elements
an example
of 5
an 6.
infinite
group. 5, o(Zn) 5 n. The set Z of all integers is a group under ad■
has
order o(G)
In Example
dition, and3.4
this
is an example
of an infinite
group. If A is an infinite set, then S(A) furnishes
Exercises
Properties
of Group
Elements
orem 3.4 Theorem
■ Properties of Group Elements
an example of an infinite group.
■
3.1
True or False
Exercises
3.1
Exercises 3.1
Let G be a group with respect to a binary operation that is written as multiplication.
Label each of the following statements as either true or false.
a. The
e in Gelement
is unique.
1. element
The identity
in a group G is its own inverse.
True
oridentity
False
21
!1
b. For each x2.[ If
G,Gthe
inverse
x
in
G
unique.
is an abelian
group,isthen
" x true
for all
in G.
Label each of the following
statements
asxeither
orxfalse.
21 21
c.
For
each
x
[
G,
(x
)
5
x.
True or False
1. The identity element in a group G is its own inverse.
Label each of the following statements
as either true or false.
2. If G is an abelian group, then x!1 " x for all x in G.
1. The identity element in a group G is its own inverse.
2. If G is an abelian group, then x!1 " x for all x in G.
5
Then
Chapter 3 Groups
Groups
y 5 ey
!
since e is an identity
5 (zx)y since zx 5 e
d. Reverse order law. For any
and y inbyG,
(xy)21 5 y21x21.
5xz(xy)
associativity
Week 4
e. Cancellation laws. If a, x,5and
in G,
z(e)y aresince
xy then
5 e either of the equations ax 5 ay or
xa 5 ya implies that x 5 y.
5z
since e is an identity.
Thus y 5
and this
justifies
x21the
as others
the unique
inverse ofTo
x in
G. part b, let
Wez, prove
parts
b andthed notation
and leave
as exercises.
prove
queness Proof
¿ q) ⇒ r x [ G,
Weand
shall
use part
b in
theof
proof
ofzpart
Specifically,
we shall
suppose
that
each
y and
is and.inverse
of x. That
is, use the fact that the in21
21 21
verse (xy) is unique. This means that in order to show that y x 5 (xy) 21, we need only
xy 5
yxx21and
5 e 5calculations
zx.
to verify that (xy) (y21x21 ) 5
e 5e 5
(y21
) (xy).xzThese
are straightforward:
Then
(y21x21)(xy) 5 y21(x21x)y 5 y21ey 5 y21y 5 e
y 5 ey
and
5 (zx)y since zx 5 e
(xy)(y21x21) 5 x(yy21)x21 5 xex21 5 xx21 5 e.
5 z(xy) by associativity
z(e)x21 insince
xy 5 e order law (xy) 21 5 y21x21 is crucial
The order of the factors y215and
the reverse
5 z where since
e is2anx21
identity.
in a nonabelian group. An example
(xy) 21
y21 is requested in Exercise 5 at the
end ofy this
Thus
5 z,section.
and this justifies the notation x21 as the unique inverse of x in G.
Part
e
of
Theorem
3.4the
implies
thatpart
in d.
theSpecifically,
table for a finite
group
element
G
We shall use
part b in
proof of
we shall
useG,
theno
fact
that theofinappears
in the same
row, and
of G that
appears
twice
in the
21
21 same column.
verse
(xy)twice
is unique.
This means
thatno
in element
order to show
y21x21
5 (xy)
, we need only
These results can be21
extended
to the statement
in the following strategy box. The proof of
21
21 21
to verify that (xy) (y x ) 5 e 5 (y x ) (xy). These calculations are straightforward:
this fact is requested in Exercise 10.
(y21x21)(xy) 5 y21(x21x)y 5 y21ey 5 y21y 5 e
q) ⇒ r
■ and
In the multiplication table for a group G, each element of G appears exactly once in each
ategy
21 once
21column.
row and also appears (xy)(y
exactly
each
x21) 5inx(yy
)x21 5 xex21 5 xx21 5 e.
Remark.
em 3.5
tegy
since e is an identity
Although
our
of and
a group
a standard
one, alternative
becrucial
The
order of
thedefinition
factors y21
x21 inisthe
reverse order
law (xy) 21 forms
5 y21can
x21 is
One
of these isgroup.
given in
next theorem.
in
a nonabelian
Anthe
example
where (xy) 21 2 x21y21 is requested in Exercise 5 at the
end of this section.
e 3.5
of Theorem
3.4Conditions
implies
in athe
table for a finite group G, no element of G
Theorem
Equivalent
for
Group
■ Part
Equivalent
Conditions
for athat
Group
appears twice in the same row, and no element of G appears twice in the same column.
Let G results
be a nonempty
set that is
under aninassociative
binary
operation
called
multiThese
can be extended
to closed
the statement
the following
strategy
box. The
proof
of
plication.
Then
G
is
a
group
if
and
only
if
the
equations
ax
5
b
and
ya
5
b
have
solutions
this fact is requested in Exercise 10.
x and y in G for all choices of a and b in G.
■ In the multiplication table for a group G, each element of G appears exactly once in each
em 3.5
row and also appears exactly once in each column.
Although our definition of a group is a standard one, alternative forms can be made.
One of these is given in the next theorem.
■
Equivalent Conditions for a Group
Let G be a nonempty set that is closed under an associative binary operation called multiplication. Then G is a group if and only if the equations ax 5 b and ya 5 b have solutions
x and y in G for all choices of a and b in G.
6
tion ax 5 a has a solution x 5 u in G. We shall show that u is a right identity for every
element in G. To do this, let b be arbitrary in G. With z a solution to ya 5 b, we have
za 5 b and
Groups
!
Week 4
bu 5 (za)u 5 z(au) 5 za 5 b.
Thus u is a right identity for every element in G. In a similar fashion, there exists an
element v in G such that vb 5 b for all b in G. Then vu 5 v, since u is a right identity,
and vu 5 u, since v is a left identity. That is, the element e 5 u 5 v is an identity element for G.
Now for any a in G, let x be a solution to ax 5 e, and let y be a solution to ya 5 e.
Combining these equations, we have
x 5 ex
5 yax
5 ye
5 y,
and x 5 y is an inverse for a. This proves that G is a group.
Remark. In a group G, the associative property can be extended to products involving
moreInthan
three
factors.
For example,
if a1be
,a2extended
,a3 , and ato
are elements
G, than
then
4 products
a group
G, the
associative
property can
involvingofmore
applications
condition
2 inifDefinition
three factors.ofFor
example,
a1, a2, a3,3.1
andyield
a4 are elements of G, then applications of con-
dition 2 in Definition 3.1 yield
3a1(a2 a3)4 a4 5 3(a1 a2)a34 a4
and
(a1 a2)(a3 a4) 5 3(a1 a2)a34 a4.
These equalities suggest (but do not completely prove) that regardless of how symbols
of grouping are introduced in a product a1a2a3a4, the resulting expression can be
Chapter 3reduced
Groupsto
Chapter 3 Groups
3.6 Product
ition 3.6 Definition
■ Product
NotationNotation
3(a1 a2)a34 a4.
ition 3.6 ■
Product
Notation
Let n be a positive integer, n \$ 2. For elements a1, a2, c, an in a group G, the expression
rem 3.7
em 3.7
omplete
nduction
omplete
nduction
Withc
these observations in mind, we make the following definition.
a1a2
a is defined recursively by
Let n be a npositive integer, n \$ 2. For elements a1, a2, c, an in a group G, the expression
c ak ak11
a1a2 can is definedarecursively
by 5 (a1 a2 c ak)ak11 for k \$ 1.
1 a2
a1 a2 c ak ak11 5 (a1 a2 c ak)ak11 for k \$ 1.
We can now prove the following generalization of the associative property.
We can now prove the following generalization of the associative property.
Generalized Associative Law
Theorem
3.7 Generalized
Associative
Law
■
Associative
Law
Let nGeneralized
\$ 2 be a positive
integer, and
let a1, a2, c, an denote elements of a group G. For
■
any positive integer m such that 1 # m , n,
Let n \$ 2 be a positive integer, and let a1, a2, c, an denote elements of a group G. For
c
(a1 athat
am)(a
any positive integer m such
1#
m m11
, n,c an) 5 a1 a2 c an.
2
(a1 a2 c am)(am11 c an) 5 a1 a2 c an.
Proof For n \$ 2, let Pn denote
the statement of the theorem. With n 5 2, the only possible value for m is m 5 1, and P2 asserts the trivial equality
Proof
For n \$ 2, let Pn denote the statement of the theorem. With n 5 2, the only possi(a1)(a
5 a1 aequality
ble value for m is m 5 1, and P2 asserts
the2)trivial
2.
7
Assume now that Pk is true: For any positive
(a1)(a2) integer
5 a1 a2.m such that 1 # m , k,
c
c
(a1 a2For
ampositive
)(am11 c
ak) 5masuch
Assume now that Pk is true:
any
integer
1 a2 thata1k. # m , k,
Groups
!
Week 4
Example 1 M m×n ( R ) is an abelian group with respect to addidtion. This is an example of
another infinite group.
Example M m×n ( ) , M m×n (  ) , M m×n (  k ) , M m×n (  ) is a group with respect to addition.
Example The nonzero elements of M n ( R ) do not form a group with respect to
multiplication.
Example 2 The invertible elements of M n ( ) , M n (  ) , M n (  k ) , M n (  ) , M n ( R ) form a group
G with respect to matrix multiplication.
8
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