課
本
• 書名： Introduction to Abstract Algebra
• 作者： W. Keith Nicholson
Sets of numbers
•
•
•
•
•
N---the set of natural numbers
Z---the set of integers
Q---the set of rational numbers
R---the set of real numbers
C---the set of complex numbers
Cartesian Product
• The Cartesian product AB of two sets A
and B is defined to be the set
AB = { (a,b) | aA, bB }.
• Example.
 If A={1, 2, 3}, B={a, b}, then
 AB = {(1, a), (2, a), (3, a), (1, b), (2, b), (3, b)}
Cartesian Product
• We may write A2=AA.
• R2=RR is the Euclidean plane(歐基理德
平面).
Cartesian Product
• If A1,A2,…,An are sets, their Cartesian
product A1  A2  …  An is defined to be
the set
A1A2…An={(a1,a2,…,an ) | aiAi for each i}.
(ordered) n-tuple
• (a1,a2,…,an )=(b1,b2,…,bn )  ai=bi for each i.
Relation
• If A is a set, a subset * of A×A is called a
relation on A.
• We write a*b to mean (a,b) is an element
of *.
Mapping
• If A and B are two sets, a mapping  from
A to B, written : AB, is a subset of AB
such that
for each aA,  exactly one element
bB, (a,b) 
={(1,b),(2,a),(3,a)}
A={1,2,3}
1
2
3
a
B={a,b}
b
Mapping
• If A and B are two sets, a mapping  from
A to B, written : AB, is a subset of AB
such that
for each aA,  exactly one element
bB, (a,b) 
• The unique element bB such that (a,b)
 is denoted by b=(a).
• We refer to A and B as domain and
codomain, respectively, of the mapping .
Mapping
• Suppose  is a mapping from A to B.
 For each aA, the unique element bB such
that (a,b)  is denoted by b=(a).
 (a) is called the image of a under .
We refer to A and B as domain and codomain,
respectively, of the mapping .
Example
• Let A={1,2,3} and B={a,b}.
 1={(1,b),(2,a),(3,a)} is a mapping.
 2={(1,a),(1,b)} is not a mapping.
 3={(1,b),(2,a),(2,b),(3,a)} is not a mapping.
 4={(1,a),(2,b))} is not a mapping.
 5={(1,a),(2,a),(3,a)} is a mapping.
Theorem
• If : AB and : AB are mappings, then
=  (a)=(a) for all aA.
• Proof: ()
For each aA, we have (a,(a)).
Since =, we have (a,(a)).
By the definition of , (a)=(a).
Theorem
• If : AB and : AB are mappings, then
=  (a)=(a) for all aA.
• Proof: ()
Let (a,b), Then b=(a).
Since (a)=(a), we have b=(a)=(a).
This implies (a,b).
So, .
Similarly, we have .
Thus, =.
One-to-one mapping
• Let : AB be a mapping. We say that 
is one-to-one if for any a,bA,
ab  (a)(b)
• That is,  is one-to-one if for any a,bA,
(a) = (b)  a = b
Example
• Let A={1,2} and B={a,b,c}.
 1={(1,b),(2,a)} is 1-1.
 2={(1,a),(2,a)} is not 1-1.
Example
• If : NN is defined by (n)=2n+1 for all
n N, show that  is 1-1.
• Proof:
 Suppose (m)=(n).
 Then 2m+1 = 2n+1.
 This implies m = n.
 Thus,  is 1-1.
Claim: (a) = (b)  a = b
Example
• If : RR is defined by (x)=x2 for all x
R, show that  is not 1-1.
• Proof:
 Since (1)=1=(-1) and 1-1,  is not 1-1.
Example
• If : NN is defined by (x)=x2 for all x
N, show that  is 1-1.
• Proof:
 Suppose x,yN and (x)=(y).
 Then x2 = y2 . This implies (x+y)(x-y) = 0.
 Since x+y>0, we have x - y = 0.
 So, x = y. Thus,  is 1-1.
Claim: (a) = (b)  a = b
Onto mapping
• Let : AB be a mapping. We say that 
is onto if for any bB, there exists aA
such that b=(a).
• One-to-one and onto mappings are called
injective and surjective, respectively.
• A mapping that is both one-to-one and
onto is called a bijection.
Example
• If : NN is defined by (n)=2n+1 for all
n N, show that  is not onto.
• Proof:
 Suppose mN and (m)=2.
 Then 2m+1 = 2.
 This implies m = 1/2 , a contradiction to mN.
 Thus,  is not onto.
Example
• If : RR is defined by (x)=2x+1 for all
xR, show that  is a bijection.
• Proof:
 (x)=(y)  2x+1 = 2y+1  x = y.
 Thus,  is 1-1.
 For any yR, there exists x=(y-1)/2 R such
that (x)=2((y-1)/2)+1=(y-1)+1=y.
 So,  is onto.
 Thus,  is a bijection.
Image
• Let  be a mapping from A to B.
If SA, the set (S)={(a)| aS}.
The image of  is the set im()=(A) of all
images of element of A.
  is onto  (A)=B.
Let : AB be a mapping where A and B are
nonempty finite sets. Then any two following
conditions hold will imply the other holds.
(a) |A|=|B|
(b)  is 1-1 (c)  is onto.
Composite
• Let f: BC and g:
AB are mappings.
The mapping given by
(fg)(x)=f(g(x)) is the
composite of f and g.
• Noted that fg be a
mapping from A to C .
Example

• Let f: RR and g: RR be define by f(x) =
2x - 3 and g(x) = x2 + 1. Find the action of fg
and gf and conclude that fggf .
• Solution:
f(g(x)) = 2( g(x) ) – 3
= 2( x2 + 1 ) – 3 = 2x2 – 1
g(f(x)) = ( f(x) )2 + 1
= ( 2x – 3 )2 + 1 = 4x2 – 12x + 10
Identity mappping
• For a set A, the identity map 1A: AA is
defined by
1A(a) = a for all aA.
Theorem
• Let : AB, : BC, and : CD be
mappings. Then
1) 1A =  and 1B = .
2) ()=().
3) If  and  are both 1-1 (onto) the same is
true of .
• In (2), we may write =()=()
Composite
• If : AA is a mapping, we may write
2 = 
3 = 
4 = 
Inverse mapping
• Let : AB be a mapping. If there is a
mapping : BA such that
=1B and =1A ,
then the mapping  is said to be the
inverse of  and we say  invertible.
• If  have an inverse, the inverse is unique
and is denoted by -1.
Example
• If A={1,2,3} , define : AA by (1)=2,
(2)=3, and (3)=1. compute 2 and 3, and
also find -1.
• Solution:
 2(1)=((1))=(2)=3, 2(2)=((2))=(3)=1,
and 2(3)=((3))=(1)=2.
 3(1)=2((1))=2(2)=1, 3(2)=2((2))=2(3)=2,
and 3(3)=2((3))=2(1)=3.
Since 3=1A, we have -1=2.
Theorem
• Let : AB, : BC be mappings.
1) 1A is invertible and 1A-1 = 1A.
2) If  is invertible, then -1 is invertible and
(-1)-1 = .
3) If  and  are both invertible, the same  is
invertible and ()-1 = -1-1.
4)  is invertible   is 1-1 and onto.
Division Algorithm
• Given integers n and d, with d >0, there
exist unique integers q and r satisfying
n = qd + r, 0  r < d
• The integer q and r are called, respectively,
the quotient and remainder in the division
of a and b.
Divisor
• An integer a  0 is said to be a divisor of
an integer b, in symbol a|b, if there exists
some integer c such that b= ac.
• We write ab to indicate that b is not
divisible by a.
Theorem
•
For integers a,b,c, the following hold:
a)
b)
c)
d)
e)
f)
a|0, 1|a, a|a.
a|1 if and only if a = 1.
If a|b and b|c, then a|c.
a|b and b|a if and only if a =  b.
If a|b and b0, then |a|  |b|.
If a|b and a|c, then a|(bx+cy) for arbitrary
integers x and y.
Linear
combination of
b and c
Greatest Common Divisor
•
•
If a and b are arbitrary integers, then an
integer d is said to be a common divisor
of a and b if both d|a and d|b.
If a and b are integers, not both zero, the
greatest common divisor of a and b,
denoted by gcd(a,b), is the common
divisor d of a and b such that cd for each
common divisor c of a and b.
Theorem
• Given integer a and b, not both of which
are zero, there exists integers x and y such
that gcd(a,b) = ax + by.
• Corollary:
If a = qb + r, then gcd(a,b) = gcd(b,r).
Euclidean Algorithm
• Given integers a and b, where a  b  0. By
Division Algorithm, we have
a  q1b  r1 ,
0  r1  b
• if r1  0,
b  q2 r1  r2 ,
0  r2  r1
• if r2  0,
r1  q3 r2  r3 ,
0  r3  r2
rn  2  q n rn 1  rn ,
0  rn  rn 1
rn 1  qn 1rn  0
• We argue that rn = gcd(a,b).
Example
• Find x and y such that
gcd(12378,3054) = 12378x + 3054y
• Solution:
12378  4  3054  162
3054  18 162  138
162  1 138  24
138  5  24  18
24  1 18  6
18  3  6  0
gcd(12378,3054)=6
6  24  18
 24  (138  5  24)
 6  24  138
 6  (162  138)  138
 6 162  7 138
 6 162  7  (3054  18 162)
 132 162  7  3054
 132  (12378  4  3054 )  7  3054
 132 12378  ( 535)  3054
Relatively Prime
• Two integers a and b, not both of which
are zero, are said to be relatively prime
whenever gcd(a,b) = 1.
• Theorem
Given two integers a and b, gcd(a,b)=1 if
and only if there exist integers x and y
such that 1= ax + by.
Definition
• An integer p > 1 is called a prime if its only
positive divisors are 1 and p.
Theorem
• If p is a prime and p|ab, then p|a or p|b.
Corollary
• Any positive integer n>1 can be written
k1 k 2
kr
uniquely in canonical form n  p1 p2  pr,
where for i= 1,2…r, each ki is a positive
integer and each pi is a prime with p1<p2
< … <pr.
ki
4725  3  5  7
3
2
17460  2  3  5  7
3
2
2