Abstract Algebra1018

Chapter 1
Set Theory
1.1 Sets
A set is a well-defined collection of objects. The objects that belong to a

set are called elements or members.
We will denote sets by capital letters such as A, B, C. We write a ∈ A to

mean that a is an element of the set A. When a does not belong to A, we
write a ∈
/ A. The number of elements in a set X is called the order of X,
denote as |X|.
A set X is usually given by listing all elements in it, such as
X = {x1 , x2 , · · · , xn }.
A set whose elements are precisely those objects for which the statement p(x)
hold, we write
X = {x|p(x)}.
For example, if X is a set of nature numbers, we can write
X = {0, 1, 2, · · · }
or
X = {x|x is a natural number}.
1
CHAPTER 1. SET THEORY 2
If there are infinite elements in X, we write |X| = ∞
The followings are some important sets which will be use in the following
chapters.
N is the set of all natural numbers {0, 1, 2, 3, · · · }.
Z is the set of all integers {· · · , −2, −1, 0, 1, 2, · · · }.
Z+ is the set of all integer {1, 2, 3, · · · }.

a
Q is the set of all rational numbers: fraction of the form b, for a, b ∈ Z
and b 6= 0.
R is the set of all real numbers.
C is the set of all complex numbers.
Ø is an empty set.
Let A and B be two sets, we say A is equal to B if A and B have the same
elements, denoted by A = B. A is different from B if A is not equal to B, i.e.
A 6= B.
A is contained in B or that B contains A if every element of A is also an

element of B, write A ⊆ B or B ⊇ A. If A ⊆ B, we say that A is a subset of
A.
We have A = B if and only if A ⊆ B and B ⊆ A. If B is a subset of A

and B is different from A, we write B ⊂ A. We say the inclusion of B in A is
strict.
Remark 1.1.1. (1) A and Ø are subsets of a set A.
(2) Ø is a proper subset of a set A if and only if A 6= Ø.
The union A ∪ B of two sets A and B is the set whose elements are all
elements of A and of B:
A ∪ B = {x|x ∈ A or x ∈ B}.
The intersection of A and B is the set whose elements are elements of A and
of B:
A ∩ B = {x|x ∈ A and x ∈ B}.
T
There exist no element that belong to A and also to B, i.e., A B = Ø.
S
Moreover, in this case if X = A B, we say X is the disjoint union of A and
B. The difference of A and B is A\B = {x|x ∈ A and x ∈ / B}. Let A be a
subset of a given set X. The difference X\A is called the complement set of
A in X. We write C(A) or A0 .
Proposition 1.1.2. Let A and B be sets. Then
(1) A ∪ B = A if and only if B ⊆ A.
(2) A ∪ B = Ø if and only if A = Ø and B = Ø.
(3) A ⊆ A ∪ B and B ⊆ A ∪ B.
(4) A ∩ B ⊆ A and A ∩ B ⊆ B.
Proposition 1.1.3. Let A and B be sets. Then
(1) Commutative law of union:
A ∪ B = B ∪ A.
(2) Commutative law of intersection:
A ∩ B = B ∩ A.
(3) Associative law of union:
(A ∪ B) ∪ C = A ∪ (B ∪ C).
(4) Associative law of intersection:
(A ∩ B) ∩ C = A ∩ (B ∩ C).
(5) Distributive law of intersection with respect to union:
A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C).
(6) Distributive law of union with respect to intersection:
A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C).
Proposition 1.1.4. Let A be a subset of set X. Then
(1) (A0 )0 = A.
(2) A0 ∪ A = X.
(3) A0 ∩ A = Ø.
Proposition 1.1.5. Let A, B and C be sets. Then
(1) A\(B ∪ C) = (A\B) ∩ (A\C).
(2) A\(B ∩ C) = (A\B) ∪ (A\C).
(3) (B ∪ C)\A = (B\A) ∪ (C\A).
(4) (B ∩ C)\A = (B\A) ∩ (C\A).
Proof. We will prove (1) and leave (2)-(4) to readers.
(1) Let x ∈ A\(B ∪ C). That means x ∈ A and x ∈ / (B ∪ C). That is x ∈ A

and neither x ∈ B nor x ∈ C, i.e., x ∈ (A\B) and x ∈ (A\C). Thus, we have
x ∈ (A\B) ∩ (A\C), therefore A\(B ∪ C) ⊆ (A\B) ∩ (A\C).
On the other hand, let x ∈ (A\B)∩(A\C), i.e., x ∈ (A\B) and x ∈ (A\C).

That is x ∈ A, x ∈
/ B and x ∈ A, x ∈/ C. We have x ∈ A and A ∈ / (B ∪ C).
That is x ∈ A\(B ∪ C). Therefore (A\B) ∩ (A\C) ⊆ A\(B ∪ C). In a word,
A\(B ∪ C) = (A\B) ∩ (A\C).
(2) and (3) in next proposition are called De Morgan’s laws.
Proposition 1.1.6. Let A and B be subsets of a set X. Then
(1) A ⊆ B if and only if B 0 ⊆ A0 .
(2) (A ∪ B)0 = A0 ∩ B 0 .
(3) (A ∩ B)0 = A0 ∪ B 0 .
Definition 1.1.7. Let X be a set. The power set of X is the set whose
elements are the subsets of X. We denote it by P (X) = {A|A ⊆ X}.
Example 1.1.8. Let X = {1, 2}, then P (X) = {Ø, X, {1}, {2}}.
Remarks 1.1.9. Given any set X, we always have Ø ∈ P (X) and P (X) 6= Ø.
If there are n elements in X, then |P (X)| = 2n .
Example 1.1.10. If X = Ø, then P (X) = {Ø}. If X = {Ø}, then P (X) =

{Ø, {Ø}}.
Example 1.1.11. If X = {1, 2, 3}, then
P (X) = {Ø, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}}.
1.2 Maps
Definition 1.2.1. A map f consists of a nonempty set X, of a nonempty set

Y and of a law that assigns to each x ∈ X, exactly one element, denoted by
f (x) of Y . Denote
f : X → Y : x 7→ y.
X is called the domain of map f , Y is called the codomain of map f , f (x) is

called the image or value of x, x is called the preimage of f (x). Denote the
set of images as Im(f ).
It is obvious that the set of images is contained in Y . The set of preimages

of f is X.
Example 1.2.2. We have f : N → N : x 7→ x2 is a map. f : N → Z : x 7→ −x

is a map.
Let us consider f : Q → Z, where Q = {x|x = pq |p, q ∈ Z}, define f ( pq ) = p,

then f is not a map since f ( 21 ) = 1, but f ( 24 ) = 2.
Two maps f and g are equal if and only if they have the same domain,
codomain and the same law, that is, for every x ∈ X, we have f (x) = g(x).
There are some particular maps here.
Let X be a nonempty set. The identity map on X, denoted by 1x (or idx , or

Ix ) with X both as a domain and a codomain denoted by 1x : X → X : x 7→ x.
If A is a nonempty subset of X, the conclusion map of A in X is the function

denoted by iA with A as a domain, X as a codomain, iA : A → X given by
iA (a) = a.
Note that if A is a nonempty proper subset of X. 1a and ia have different

domain, thus 1A 6= iA . 1A = iA if and only if A = X.
Let f : X → Y be a map and A be a nonempty subset of X. The map

f |A : A → Y : a →
7 f (a) is called the restriction of f to A. In particular
iA = 1X |A .
Let X be a nonempty set and let y be a fixed element of Y . The map

fy : X → Y : x 7→ y is called the constant map. Note that fy (X) = {y} and
Im(f ) = {y} ⊆ Y .
Example 1.2.3. Let f1 : N → N : x 7→ 2x. Im(f1 ) = {y|∃ x ∈ N , such that y =

2x} is the set of even natural numbers.
f2 : N → N : x 7→ 2x + 1. Im(f2 ) = {y|∃ x ∈ N , such that y = 2x + 1} is

the set of odd natural numbers.
Example 1.2.4. Let f3 : Z → Z : x 7→ 3x, f4 : Z → Z : x 7→ −3x. Then

Im(f3 ) is the set of integers of multiple of 3, Im(f4 ) is the set of integers of
multiple of −3. We have Im(f4 ) = Im(f3 ), but f4 6= f3 .
If f5 : Q → Q : x 7→ 3x, then Im(f5 ) = Q .
Definition 1.2.5. Let f : X → Y be a map, f is said to be a surjective map

or onto if Im(f ) = Y .
It is equivalent that f is surjective if and only if for any y ∈ Y , ∃x ∈ X,

such that f (x) = y.
Definition 1.2.6. Let f : X → Y be a map, f is said to be injective or

one-to-one if given any x, x0 ∈ X, x 6= x0 implies that f (x) =
6 f (x0 ).
Definition 1.2.7. Let f : X → Y is called bijective if f is surjective and

injective.
We have that f is bijective if and only if for any y ∈ Y , there exists an

unique x ∈ X, such that f (x) = y.
If there exists a bijective map f : X → Y , we call X is isomorphic to Y ,

denoted by X ∼ = Y . Let f : X → Y and g : Y → Z be maps, such that the
codomain of f coincides with the domain of g. Then the composition of f and
g is given by g ◦ f : X → Z : x → g(f (x)).
Example 1.2.8. Let
f : Z −→ Z : x 7−→ x + 1, g : Z −→ Z : x 7−→ x2 ,
then
g ◦ f : Z −→ Z : x 7−→ (x + 1)2 .
Example 1.2.9. Let f : X −→ Y and A be a nonempty subset of X, then
f |A = f ◦ iA : a 7−→ f (a).
Note that f : Z → Z, g : Z → Z, then g ◦ f 6= f ◦ g.
Proposition 1.2.10. Let f : X → Y, g : Y → Z, h : Z → T be maps. Then
(1) f ◦ 1x = f = 1y ◦ f .
(2) h ◦ (g ◦ f ) = (h ◦ g) ◦ f .
Proof. (1) is obvious. ∀x ∈ X, we have h◦(g ◦f )(x) = h(g ◦f (x)) = h(g(f (x)).
And ((h ◦ g) ◦ f )(x) = (h ◦ g)(f (x)) = h(g(f (x)).
Proposition 1.2.11. : Let f : X 7→ Y and g : Y 7→ Z be maps
(1) If f and g are both injective, then g ◦ f is injective.
(2) If f and g are both surjective, then g ◦ f is surjective.
(3) If g ◦ f is injective, then f is injective.
(4) If g ◦ f is surjective, then g is surjective.

Proof. (1) f is injective means that x 6= x0 , then f (x) 6= f (x0 ). Since g is

injective, f (x) 6= f (x0 ), then g(f (x)) 6= g(f (x0 )). Then (g ◦ f )(x) 6= (g ◦ f )(x0 ).
(2) Note that g ◦f is surjective ⇔ ∀z ∈ Z, ∃x ∈ X, such that (g ◦f )(x) = z.

g is surjective means that ∀z ∈ Z, ∃y ∈ Y , such that g(y) = z. f is surjective,
∀y ∈ Y , ∃x ∈ X, such that f (x) = y. Thus (g ◦ f )(x) = g(f (x)) = z.
(3) Let x 6= x0 . g ◦ f is injective, then (g ◦ f )(x) 6= (g ◦ f )(x0 ). Then

g(f (x)) 6= g(f (x0 )), and f (x) 6= f (x0 ).
(4) Let z ∈ Z. Since g ◦ f is surjective, there exist a x ∈ X such that

(g ◦ f )(x) = g(f (x)) = z. That means there exist y = f (x) ∈ Y such that
g(y) = z. Then g is surjective.
Example 1.2.12. We consider the following maps
f : N −→ Z : x 7−→ −x, g : Z −→ N : x 7−→ x2 .
Then g ◦ f is injective but not surjective. So f is injective.

Example 1.2.13. Let
f : R −→ R : x 7−→ x2 , and g : R −→ R+ ∪ {0} : x 7−→ x2 .
We have g ◦ f is surjective. So g is surjective, f is not surjective.

Corollary 1.2.14. Let f : X → Y and g : Y → Z be maps,
(1) If f and g are both bijective, then also g ◦ f is bijective.
(2) If g ◦ f is bijective, then f is injective and g is surjective.
For example, let f : N → Z : x 7→ −x, and g : Z → N : x 7→ |x|. Then

g ◦ f = 1N . We have f is injective and g is surjective.
Definition 1.2.15. Let f : X → Y be a map. The map g : Y → X is called
a left inverse of f if g ◦ f = 1X . The map h : Y → X is called a right inverse
of f if f ◦ h = 1Y .
A map g : Y → X is called a two-sided inverse, if g is a left inverse and

a right inverse, i.e.
g ◦ f = 1X , f ◦ g = 1Y .
Theorem 1.2.16. Let f : X → Y be a map , then
(1) f is injective if and only if f has (at least) a left inverse.
(2) f is surjective if and only if f has (at least) a right inverse.
Proof. (1) Assume that f is a injective, we should construct a map g : Y → X

such that g ◦ f = 1X . Fix an element x0 ∈ X. We define the map g : Y → X
in the following way. Let y ∈ Y . If y ∈ Im(f ), that means there exists a
x ∈ X since f is injective. Define g(y) = x. If y ∈ / Im(f ) , define g(y) = x0 .
So g ◦ f (x) = g(y) = x = 1X . So g is a left inverse of f .
Conversely, if f has a left inverse g : Y → X, such that g ◦ f = 1X . 1X is

bijective, g ◦ f is bijective. So f is injective.
(2) Assume that f is surjective, we want to construct a map h : Y → X,

such that f ◦ h = 1Y . Given y ∈ Y , then there exist a x ∈ X such that
f (x) = y by f is surjective. We choose one element in all those elements of X
whose image under f is y. Denote it as xy . Hence, xy ∈ X, f (xy ) = y. Define
h : Y → X by h(y) = xy . For every y ∈ Y , we have f ◦ h(y) = f (xy ) = y =
1Y (y).
Conversely, assume that f has a right inverse h. Then f ◦ h = 1Y is

bijective, f is surjective by (2) 1.2.14.
Example 1.2.17. Let f : N → N be the map defined by f (x) = x + 2. Then

f is injective and Im(f ) = {x ∈ N|x ≥ 2}. The map g0 : N → N defined by
g0 (x) = x − 2 if x ≥ 2 and g0 (x) = 0 if x < 2 is a left inverse of f .
Example 1.2.18. Let f : Z → N be the map defined by f (x) = |x|. The map
h = iN : N → Z is a right inverse of f . Also g : N → Z defined by g(x) = −x
is a right inverse of f .
Corollary 1.2.19. Let f : X → Y be a map. The following are equivalent
(1) f is bijective.
(2) f has a left inverse and a right inverse.

(3) f has a two-sided inverse.
Moreover, if f satisfies one of the above conditions, then
(i) every left inverse of f is a two-sided inverse of f .
(ii) every right inverse of f is a two-sided inverse of f .
(iii) f has a unique two-sided inverse.
Such an inverse of f is called the inverse of f and it is denoted by f −1 .

Also f −1 is bijective and (f −1 )−1 = f .
Proof. We have (1) =⇒ (2) by Theorem 1.2.16.
(2)=⇒(3) f ◦ h = 1y and g ◦ f = 1x . So that g = g ◦ 1y = g ◦ (f ◦ h) =

(g ◦ f ) ◦ h = 1x ◦ h = h. Hence, g coincides with h and then it is a two-sided
inverse of f . By means of this equality (i),(ii),(iii) can be easily proved.
(3)=⇒(1) follows from Theorem 1.2.16.
Finally, since f ◦ f −1 = 1Y and f −1 ◦ f = 1X , we have that f is the

two-sided inverse of f −1 . Thus f −1 is bijective and (f −1 )−1 = f.
Corollary 1.2.20. Let f : X → Y and g : Y → Z be bijections. Then

g ◦ f : X → Z is a bijection and (g ◦ f )−1 = f −1 ◦ g −1 .
Proof. Firstly, we have f −1 ◦ g −1 : Z → X. Moreover,
(g ◦ f ) ◦ (f −1 ◦ g −1 ) = (g ◦ (f ◦ f −1 )) ◦ g −1 = g ◦ g −1 = 1Z .
Also
(f −1 ◦ g −1 ) ◦ (g ◦ f ) = (f −1 ◦ (g −1 ◦ g)) ◦ f = f −1 ◦ f = 1X .
Hence f −1 ◦ g −1 is an inverse of g ◦ f , that is
(g ◦ f )−1 = f −1 ◦ g −1 .
The formula (g ◦ f )−1 = f −1 ◦ g −1 has been called the ”dressing-undressing

principle” or ”socks and shoes principle”. what goes on in dressing comes off
in the reverse order in undressing.
Theorem 1.2.21. A map is invertible if and only if it is both injective (one-
to-one) and surjective (onto).
Proof. Suppose that f : X → Y is invertible with inverse g : Y → X. Then

g ◦ f = 1X . If x1 , x2 ∈ X with f (x1 ) = f (x2 ), then
x1 = g(f (x1 )) = g(f (x2 )) = x2 ,
f is injective. Suppose that y ∈ Y , then f (g(y)) = y by f invertible. Thus

g(y) ∈ X. Let x = g(y), then f (x) = y, f is surjective.
Conversely, let f be bijective. Let y ∈ Y . There exist x ∈ X such that

f (x) = y. This x is unique by f is injective. Define g by letting g(y) = x and
g ◦ f = 1X , f ◦ g = 1Y . g is the inverse of f .
Proposition 1.2.22. Let f : X −→ Y be a map and (Ai )i∈I be a family of

subsets of X. Then
[ [
f ( Ai ) = f (Ai ), (1.1)
i∈I i∈I
\ \
f( Ai ) ⊆ f (Ai ). (1.2)
i∈I i∈I
S S
Proof. Let y ∈ f ( i∈I Ai ). there exists an a ∈ i∈I Ai such that y = f (a).
S
Since a ∈ Ai0 . It follows that y = f (a) ∈ f (Ai0 ) and hence y ∈ i∈I f (Ai ).
S S
Therefore f ( i∈I Ai ) ⊆ i∈I f (Ai ).
S
Conversely, let y ∈ i∈I f (Ai ). There exists an i0 ∈ I such that y ∈ f (Ai0 ).
S
Therefore y = f (ai0 ) for some ai0 ∈ Ai0 . since Ai0 ⊆ i∈I Ai , we have that
S S S S
ai0 ∈ i∈I Ai so that y ∈ f ( i∈I )Ai . Hence f ( i∈I Ai ) ⊆ i∈I f (Ai ). Thus
(1.1) holds.
T T
Let y ∈ f ( i∈I Ai ), there exists an a ∈ i∈I Ai such that y = f (a). Since
a ∈ Ai for every i ∈ I, we have that y = f (a) ∈ f (Ai ) for every i ∈ I, thus
T
y ∈ i∈I f (Ai ). Thus (1.2) holds.
Example 1.2.23. Let f : Z −→ Z be the constant map equal to 0. Let
A1 = {x ∈ Z|x ≥ 0},
A2 = {x ∈ Z|x < 0}.

T T
Then A1 A2 = Ø, while f (A1 ) = {0} = f (A2 ), so that f (A1 ) f (A2 ) = {0}.
Proposition 1.2.24. Let f : X −→ Y be an injective map and (Ai )i∈I be a

family of subsets of X. Then
\ \
f( Ai ) = f (Ai ).
i∈I i∈I
T T T
Proof. We only need to show that f ( i∈I Ai ) ⊇ i∈I f (Ai ). Let y ∈ i∈I f (Ai ),
then y ∈ f (Ai ) for every i. Since f is injective, then there exists unique
T
x0 ∈ Ai , i ∈ I such that f (x0 ) = y, so y ∈ f ( i∈I Ai ).
1.3 Equivalence relations and equivalence classes
Definition 1.3.1. Given sets A and B, we can define a new set
X × Y = {(x, y)|x ∈ X, y ∈ Y }.
X × Y is called the Cartesian product of X and Y .
Example 1.3.2. Let X = {x, y}, Y = {1, 2, 3}, and Z = Ø, then
X × Y = {(x, 1), (x, 2), (x, 3), (y, 1), (y, 2), (y, 3)},
and X × Z = Ø.
Cartesian product of n sets to be
X1 × X2 × · · · × Xn = {(x1 , x2 , · · · , xn )|xi ∈ Xi , i = 1, · · · n}.
and X n = X × X × · · · × X. Especially, Rn = R × R × · · · × R.
Definition 1.3.3. Let X be a set. We say X admits a binary operation if

there exists a map f : X × X → X.
Let X and Y be sets. A relation R between X and Y is a subset of the

set product X × Y . The definition will be clarified if we use a more suggestive
notation: if (x, y) ∈ R, then x is said to be related to y by R and we write
xRy.
Example 1.3.4. let X be a set, define
4(x) = {(x, y)|x, y ∈ X, x = y} ⊆ X × X.
4(x) is a binary relation of X × X. We call 4(x) as diagonal relation of X.
Example 1.3.5. Let P be the set of prime numbers. Define
R = {(x, y) ∈ P × N|x divides y},
R is a binary relation of P × N. R is called divisibility relation.
The binary relation Ro opposite to R is the binary relation between Y and

X given by
Ro = {(y, x) ∈ Y × X|(x, y) ∈ R}.
Definition 1.3.6. Let X be a set. A relation R is called reflexive if (x, x) ∈ R

for all x ∈ X; R is called symmetric if (x, y) ∈ R implies (y, x) ∈ R; R is
called transitive if (x, y)(y, z) ∈ R imply (x, z) ∈ R. If a relation is reflexive,
symmetric and transitive, it is called an equivalence relation.
Given an equivalence relation R on a set X, we write x ∼ y instead of

(x, y) ∈ R ⊂ X × X.
Example 1.3.7. Let A and B be n × n matrix. We define A ∼ B if there

exists an invertible matrix P , such that P AP −1 = B. Let E be the identity
matrix. It is easy to check that EAE −1 A = A. Suppose that P −1 AP = B by a
invertible matrix P , then A = P −1 B(P −1 )−1 . So the relation is reflexive and
symmetric. Suppose that P −1 AP = B, QBQ−1 = C, where P, Q invertible

matrixes. We have
C = QBQ−1 = QP AP −1 Q−1 = (QP )A(QP )−1 .
Thus the relation is transitive. So it is an equivalence relation. This relation

is said to be similar in matrix theory.
Example 1.3.8. Suppose that f (x) and g(x) are differentiable functions on
R. We can define f (x) ∼ g(x) if f 0 (x) = g 0 (x). It is clear that the relation is
reflexive, symmetric and transitive. This is an equivalence relation.
Example 1.3.9. For (x1 , y1 ) and (x2 , y2 ) ∈ R2 , define (x1 , y1 ) ∼ (x2 , y2 ) if

x21 + y12 = x22 + y22 . It is obvious that this is an equivalence relation.
Example 1.3.10. The binary relation R on Q given by xRy ⇐⇒ x − y ∈ Q

is an equivalence relation.
Definition 1.3.11. A partition P of a set X is a collection of sets X1 , X2 , · · · , Xn ,

such that Xi ∩ Xj = Ø, i 6= j and ∪i Xi = X.
Let ∼ be an equivalence relation on a set X, and x ∈ X,
[x] = {y ∈ X|y ∼ x}
is called the equivalence class of x.
Theorem 1.3.12. Let ∼ be an equivalence relation on a set X. Then the

equivalence classes of X from a partition of a set X. Conversely, if P = {Xi }
is a partition of a set X, then there is an equivalence relation on X with
equivalence classes Xi .
Proof. Suppose that there exists an equivalence relation ∼ on a set X. For

x ∈ X, we have x ∼ x, i.e. x ∈ [x], so [x] is nonempty. It is clear that
S
[x] = X. Next, we will show that Xi ∩ Xj = ∅ or Xi =Xj . Assume that
x∈X
z ∈ Xi ∩ Xj , write Xi = [x], Xj = [y], let z ∈ [x] ∩ [y] i.e z ∼ x and z ∼ y. So
x ∼ y, [x] ⊆ [y], [y] ⊆ [x]. Hence [y] = [x],
S
Conversely, if P is a partition of X, X = Xi , define x ∼ y if x ∈ Xi , y ∈
i
Xi . We have x ∼ x, then ∼ is reflexive. If x ∼ y, that means x ∈ Xi and
y ∈ Xi , then y ∼ x, ∼ is symmetric. If x ∼ y, y ∼ z, then x ∼ z. In a word,
∼ is an equivalence relation.
Corollary 1.3.13. Two equivalence classes of an equivalence relation are ei-

ther disjoint or equal.
Let p, q, r and s be integers, where q and s are nonzero. Define p/q ∼ r/s
if ps = qr. Clearly ∼ is reflexive and symmetric. It is also transitive, suppose
that p/q ∼ r/s and r/s ∼ t/u, with q, s and u all nonzero. Then ps = qr and
ru = st. Therefore,
psu = qru = qst.
Since s 6= 0, pu = qt. Consequently, p/q ∼ t/u. Thus, two pairs of integers,

(p, q) and (r, s) are in the same equivalence class when they reduce to the same
fraction in its lowest terms.
Let r and s be two integers and suppose that n ∈ N. We say that r is

congruent to s module n if r − s is divisible by n, i.e. r − s = nk for some
k ∈ Z. We write r ≡ s( mod n). For example,
5 ≡ 11( mod 3),
because 5 − 2 = 3, and −1 − 2 = 3 × (−1). Thus 5 and −1 have the same re-

mainder, that is 5( mod 3) = 11( mod 3). In this case, we call 5 is congruent
to −1 module 3. Write [2] = {· · · , −1, 0, 2, 5, · · · }. Sometimes, we also write 2̄
instead [2].
Proposition 1.3.14. Congruence modulo n forms an equivalence relation of

Z.
Proof. Define r ∼ s is congruent to s module n. We have r − r = 0 × n, ∼ is

reflexive. If r ∼ s, i.e. r − s = nk for some k ∈ Z. Then s − r = n(−k), that
means s ∼ r. If r ∼ s, s ∼ t, then r − s = nk, s − t = nl for some k, l ∈ Z,
then r − t = n(k + l). Transitive property is true.
Considering the equivalence relation by the integers modulo 3, we have
[0] = {· · · , −3, 0, 3, 6, · · · },
[1] = {· · · , −2, 1, 4, 7, · · · },
[2] = {· · · , −1, 2, 5, 8, · · · }.
Then Z = [0] ∪ [1] ∪ [2]. And [0], [1] and [2] are disjoint. The sets [0], [1]and
[2] form a partition of the integers. In fact, the integers modulo n are a very
important example in the study of abstract algebra. It will become quite useful
in groups and rings.
Definition 1.3.15. A factor set of X about an equivalence relation R is a set

which elements are equivalence classes.
Let X = {1, 2, 3, 4}. Define an equivalence relation R by (x, y) ∼ (u, v) if

x + y = u + v. There are seven equivalence class on X × X. They are
[(1, 1)] = {(1, 1)}; [(1, 2)] = {(1, 2), (2, 1)};
[(1, 3)] = {(1, 3), (3, 1), (2, 2)}; [(1, 4)] = {(1, 4), (4, 1), (2, 3), (3, 2)};
[(3, 3)] = {(3, 3), (2, 4), (4, 2)}; [(3, 4)] = {(3, 4), (4, 3)};
[(4, 4)] = {(4, 4)}.
The factor set is
X × X/R = {[(1, 1)], [(1, 2)], [(1, 3)], [(1, 4)], [(3, 3)], [(3, 4)], [(4, 4)]}.
Let Z be the set of integers. We have a factor set Zn given by integers

module n. Factor set
Zn = Z/nZ = {0̄, 1̄, · · · , n − 1},
where
0 = {· · · , −n, 0, n, 2n, · · · },
1 = {· · · , −n + 1, 1, n + 1, · · · },
······
n − 1 = {· · · , −1, n − 1, 2n + 1, · · · }.
For example, let n = 2, then
Z2 = Z/2Z = {0̄, 1̄},
where
0̄ = {· · · , −2, 0, 2, 4, · · · },
the set of even integers;
1̄ = {· · · , −1, 1, 3, · · · },
the set of odd integers.
It is interesting that one can define addition ” + ” and multiplication ” · ”

on a factor set Zn .
Define
a( mod n) + b( mod n) = a + b( mod n), (1.3)

(a( mod n))(b( mod n)) = ab( mod n). (1.4)
Example 1.3.16. Consider Z3 , 2, 7 ∈ Z3 . Then
2+7=2 (mod 3) + 7 (mod 3) = (2 + 7) (mod 3) = 9( mod 3) = 0.
We also write this as 2 + 7 = 0 in Z3 .
Example 1.3.17. Take 7̄, 11 ∈ Z5 , we have
7 + 11 = 7( mod 5 + 11( mod 5 = (7 + 11)( mod 5) = 3( mod 5) = 3.

7 · 11 = (7( mod 5))(11( mod 5)) = (7 · 11)( mod 5) = 77( mod 5) = 2.
6 + 9 = 6( mod 5) + 9( mod 5) = (6 + 9)( mod 5) = 0( mod 5) = 0.
6 · 9 = (6( mod 5)) · (9( mod 5)) = (6 · 9)( mod 5) = 54( mod 5) = 4.
One can write them as
7 + 11 = 3, 7 · 11 = 2, 6 + 9 = 0, 6 · 9 = 4.
Next proposition give properties of addition and multiplication on Zn .

Proposition 1.3.18. Let Zn be the set of equivalence classes of the integer

module n, and a, b, c ∈ Zn . Then
a + b = b + a,
ab = ba,
(a + b) + c = a + (b + c),
(ab)c = a(bc),
a + 0 = a,
a1 = a,
a(b + c) = ab + ac,
a + (−a) = 0.
1.4 Arithmetic
Definition 1.4.1. We say a relation R ⊆ X × X is proper or antisymmetric

if xRy and yRx implies x = y. If R is reflexive, antisymmetric and transitive,
then R is called a partial ordering relation. Most of time, a partial ordering
set is denoted by (X, ≤).
Definition 1.4.2. A ordered pair (X, R) is called a partial ordered set if X

is a set and R is a partial relation on X. If for all x, y ∈ X, either xRy or
yRx, that is whenever any two elements are comparable. In this case, (X, R)
is called a totally ordering set.
Example 1.4.3. Let X be a nonempty set. Let R be a binary relation on

P (X) given by ARB if and only if A ⊆ B. Then relation R is reflexive,
transitive, antisymmetric. R is partial ordering relation on P (X).
It is a total ordering set if and only if X = {a}.
If X = {a, b}, then P (X) = {Ø, {a}, {b}, X}. Then there is no relation
{a} and {b}. So X is not a totally ordering set.
Example 1.4.4. The binary relation R on Z given by xRy if and only if |x| ≤
|y|, x, y ∈ Z is reflexive, transitive, it is neither symmetric nor antisymmetric.

Since if |x| ≤ |y| and |y| ≤ |x|, then |x| = |y|. We can not have x = y.
Example 1.4.5. The binary relation ≤ on Z, Q, R are totally ordering rela-

tion. It is not an equivalent relation since if x ≤ y, but y 6≤ x.
In general, if X be a partial ordered set, x, y ∈ X, then x < y or x = y, or

y < x, or x and y are not comparable. If X is a totally ordered set, x, y ∈ X,
then x < y or x = y, or y < x.
Definition 1.4.6. A nonempty subset S of Z is well-ordered if S contains a

least element.
Example 1.4.7. N, Z+ are well-ordered sets. But Z is not a well-ordered set.
Theorem 1.4.8. Principle of well-ordering: Every nonempty subset of

the natural numbers is well-ordered.
A chain in a partially ordered set (S, ≤) is a subset C which is linearly

ordered by the partial order ≤. An upper bound for C is an element s of S
such that c ≤ s is valid for all c ∈ C. Finally, a maximal element of S is an
element m such that m ≤ s and s ∈ S imply that s = m. Note that in general
a partially ordered set may contain several maximal elements or none at all.
Next conclusion is the famous axiom, Zorn’s Lemma.
Lemma 1.4.9. Zorn’s Lemma: If S is a nonempty partially ordered set

such that every chain of S has an upper bound in S, then S contains at least
one maximal element.
The element x is an upper bound of the set A if a ≤ x for every a ∈ A.

Note that x need not belong to A, but in the statement of Zorn’s lemma, we
require that if A is a chain of S then A has an upper bound that belongs to
S.
Theorem 1.4.10. Division Algorithm: Let a and b be integers, with b > 0.

Then there exists an unique integer q and r such that a = bq + r, where
0 6 r < b.
Proof. Existence of q and r. Let
S = {a − bk | k ∈ Z and a − bk > 0} ⊆ N ⊆ Z.
a
If 0 ∈ S, ∃k 0 ∈ Z, such that 0 = a − bk 0 , i.e. a = bk 0 . Let q = and r = 0,
b
a
then a = b · + 0. If 0 ∈ / S, we will show that S is nonempty. If a > 0, then
b
a − b · 0 ∈ S. If a < 0, then a − b(2a) = a(1 − 2b) ∈ S.
In either case, S is nonempty. S is a nonempty subset of N, there exists

a smallest number in S. Let r = a − bq ∈ S be the smallest number in S.
Therefore, a = bq + r, r > 0. We now show that r < b. Suppose that r > b.
Then a − b(q + 1) = a − bq − b = r − b > 0. We have a − b(q + 1) ∈ S. But
a − b(q + 1) < a − bq, which contradict the fact that r = a − bq is the smallest
number in S. So r ≤ b. Since 0 ∈ / S, r 6= b and so r < b.
Uniqueness of q and r. Suppose that there exist integers r, r0 , q, q 0 , such

that
a = bq + r, 0 6 r < b.
a = bq 0 + r0 , 0 6 r0 < b.
Assume that r0 > r. From the last equation we have
bq − bq 0 = b(q − q 0 ) = r0 − r.
Therefore b | r0 − r and 0 6 r0 − r < r0 < b. This is possible only if r0 − r = 0.

Hence r = r0 and q = q 0 .
Definition 1.4.11. The greatest common divisor of integers a and b is a

positive integer d such that d is a common divisor of a and b and if d0 is any
other common divisor of a and d, then d0 |d. We write d = gcd(a, b).
Definition 1.4.12. p is a prime number if the positive numbers that divide p

are 1 and p itself. An integer n > 1 that is not prime is said to be composite.
Theorem 1.4.13. Let a and b be nonzero integers. Then there exist integers r
and s such that gcd(a, b) = ar + bs. Furthermore, the greatest common divisor
of a and b is unique.
Proof. Let
S = {am + bn|m, n ∈ Z, am + bn > 0}.
Clearly, S is nonempty, hence, S must have a smallest member d = ar + bs by
well-ordering principle 1.4.8. We claim that d = gcd(a, b). Write
a = dq + r0 , 0 ≤ r0 < d.
If r0 > 0, then
r0 = a − dq = a − (ar + bs)q = a − arq − bsq = a(1 − rq) + b(−sq) ∈ S.
It is contradict the fact that d is the smallest member of S. Hence, r = 0

and d divides a. A similar argument shows that d divides b. Therefore, d is a
common divisor of a and b. Suppose that d0 is another common divisor of a
and b, and we want to show that d0 |d. If we let a = d0 h and b = d0 k, then
d = ar + bs = d0 hr + d0 ks = d0 (hr + ks).
So d0 must divide d. Hence, d must be the unique greatest common divisor of

a and b.
Corollary 1.4.14. Let a and b be two integers that are relatively prime. Then
there exist integers r and s such that ar + bs = 1.
Lemma 1.4.15. Let a and b be integers and p be a prime number. If p | ab,

then either p | a or p | b.
Proof. Suppose that p - a, then we will show that p must divide b. Since
gcd(a, p) = 1, there exists integers r and s such that ar + ps = 1. Then
b = b(ar + ps) = (ab)r + p(bs). Since p | ab and p|ps, p must divide b.
Theorem 1.4.16. There exist infinite numbers of primes.
Proof. Suppose that there are only finite number of primes, say p1 , · · · , pn .
Let p = p1 · · · pn + 1. Next we will show that p is a different prime number,
which contradicts the assumption that we have only n primes. If p is not a
prime, then p can be divided by some pi , 1 ≤ i ≤ n. In this case, pi must
divide p1 · · · pn + 1. pi | p1 · · · pn . Thus pi | 1. This is a contradiction.
Theorem 1.4.17. Fundamental theorem of Arithmetic: Let n be an

integer and n > 1. Then
n = p1 p2 · · · pk ,
where p1 , p2 , · · · , pk are primes(not necessary distinct). Furthermore, this fac-

torization is unique, that is if n = q1 q2 · · · ql , then k = l, and the qi ’s are just
the pi ’s rearranged.
Proof. We will use induction on n to show uniqueness. It is true for n = 2.

Assume that the result holds for all integers m, 1 ≤ m < n and
n = p1 p2 · · · pk = q1 q2 · · · ql ,
where p1 ≤ p2 ≤ · · · ≤ pk , and q1 ≤ q2 ≤ · · · ≤ qk . By lemma 1.4.15, p1 |qi

for some i and q1 |pj for some j. Because qi0 s and Pj0 s are primes. So p1 = qi ,
q1 = pj . Hence, p1 = q1 , since
p1 ≤ pj = q1 ≤ qi = p1 .
By the induction hypothesis, n0 = p2 · · · pk = q2 · · · ql has a unique factoriza-

tion. Hence k = l and qi = pi for i = 1, · · · , k.
For existence, we suppose that there is some integers that cannot be written
as the product of primes. Let S be the set of all such numbers. S ⊆ N, S has
a smallest number, say a. If the only positive factors of a are a and 1, then
a is prime, which is a contradiction. Hence a = a1 a2 , where 1 < a1 < a and
1 < a2 < a. Neither a1 ∈ S nor a1 ∈ S, since a is the smallest number of S.
So a1 = p1 · · · ps and a2 = q1 · · · ql . Therefore a = p1 · · · ps q1 · · · ql . So a ∈
/ S.
It’s contradiction to the definition of S.
Chapter 2
Group
2.1 Definitions and examples
Definition 2.1.1. A binary operation of a set G is a function
G × G → G,
for any (a, b) ∈ G × G a unique element a ◦ b or ab in G.
Definition 2.1.2. A group (G, ◦) is a set with a binary operation, G×G → G,

that satisfies the following axioms:
(1) The binary operation is associative. That is (a ◦ b) ◦ c = a ◦ (b ◦ c).
(2) There exists an element e, called the identity element, such that for
any element a ∈ G, a ◦ e = a = e ◦ a.
(3) For each element a ∈ G, there exists an inverse element in G, denoted

by a−1 , such that a ◦ a−1 = e = a−1 ◦ a.
Then (G, ◦) is called a group.
A group G with the property that a ◦ b = b ◦ a for all a, b ∈ G is called

abelian or commutative. Groups are not satisfying this property are said to
be nonabelian or noncommutative.
23
CHAPTER 2. GROUP 24
Example 2.1.3. (Z, +) is a group with identity 0, the inverse of n ∈ Z is −n,

”+” is associative and commutative. So (Z, +) is an abelian group.
Example 2.1.4. Let R∗ be the set of nonzero elements of R. Then R∗ is a

group with multiplication. Multiplication is associative. The identity of this
group is 1 and the inverse of any element a ∈ R∗ is just a1 .
Example 2.1.5. Let M2 (R) be the set of 2 × 2 matrices over R. GL2 (R) be
the subset of M2 (R) consisting of invertible matrices. Then (GL2 (R), ·) is a
group with matrices product, called the general linear group.
Example 2.1.6. (Z5 , +) is a group.
Cayley table is a table which describe a group in terms of an addition or

multiplication table. The Cayley table of (Z5 , +) is the following.
Table 2.1: Caley Table (Z5 , +)

+ 0 1 2 3 4
0 0 1 2 3 4
1 1 2 3 4 0
2 2 3 4 0 1
3 3 4 0 1 2
4 4 0 1 2 3
Example 2.1.7. Let n ∈ N, define a set
U (n) = {a| gcd(a, n) = 1} .
Then (U (n), ·) is associative with identity. Moreover, an element in U (n) have

an inverse from gcd(a, n) = 1. Then (U (n), ·) is a group called the group of
units of Z.
For example, U8 = {1, 3, 5, 7} is a group under multiplication.
Cayley table for (U8 , ·) is

CHAPTER 2. GROUP 25
Table 2.2: Caley Table (U (8), ·)

+ 1 3 5 7
1 1 3 5 7
3 3 1 7 5
5 5 7 1 3
7 7 5 3 1
Example 2.1.8. Let

! ! ! !
1 0 0 1 0 i i 0
1= ,I = ,J = ,K = ,
0 1 −1 0 i 0 0 −i
where i2 = −1 the imaginary number, and
I 2 = J 2 = K 2 = −1,
IJ = K, JK = I, KI = J,
JI = −K, KJ = −I, IK = −J.
The set Q8 = {±1, ±I, ±J, ±K} is a group under multiplication, which called
the quaternion group. Notice that Q8 is unabelian.
Denote Q = {a + bI + cJ + dK|a, b, c, d ∈ R}. Note that
(i) If c = d = 0, then Q = C.
(ii) If b = c = d = 0, then Q = R
Especially, the conjugate of an element q ∈ Q is q = a − bI − cJ − dK, and

p p
|q| = qq = a2 + b2 + c2 + d2 .
Let
q1 = a1 + b1 I + c1 J + d1 K,
q2 = a2 + b2 I + c2 J + d2 K.
Define ” + ” and ” · ” of Q as following:
q1 + q2 = (a1 + a2 ) + (b1 + b2 )I + (c1 + c2 )J + (d1 + d2 )K, (2.1)

CHAPTER 2. GROUP 26
q1 q2 = (a1 a2 − b1 b2 − c1 c2 − d1 d2 ) + (a1 b2 + b1 c2 + c1 d2 − c2 d1 )I
(2.2)
+(a1 c2 + a2 c1 + b2 d1 − d2 b1 )J + (a1 d2 + d1 a2 + b1 c2 − d1 c2 )K.
Since q1 · q2 6= q2 q1 , Q is noncommutative under ” · ”.
Definition 2.1.9. A group is finite or has finite order if it contains a finite

number of elements. Otherwise, the group is called infinite.
The order of a finite group is the number of elements that it contains. If

G is a group containing n elements, we write |G| = n. The group Z3 is a finite
group of order 3; the integers Z form an infinite group under addition, and we
sometimes write |Z| = ∞.
Proposition 2.1.10. The identity element in a group G is unique. i.e. there

exists a unique element e such that eg = ge = e, ∀g ∈ G.
Proof. Assume that e and e0 are identities of G. If e is the identity of G, for

e0 ∈ G, then ee0 = e0 . If e0 is the identity of G, then ee0 = e, so e = e0 .
Proposition 2.1.11. The inverse of g ∈ G is unique.
Proof. For g ∈ G, assume that g 0 and g 00 are inverses of g, we have gg 0 = g 0 g = e

and gg 00 = g 00 g = e. Then g 0 = g 0 e = g 0 (gg 00 ) = (g 0 g))g 00 = eg 00 = g 00 .
Proposition 2.1.12. Let G be a group. For any a ∈ G, then (a−1 )−1 =a.
Proof. Observe that a−1 (a−1 )−1 = e. Consequently, multiplying both sides of
this equation by a, we have (a−1 )−1 = e(a−1 )−1 = aa−1 (a−1 )−1 = ae = a.
Proposition 2.1.13. Let G be a group and a and b be any two elements in

G, then the equations ax = b and ya = b have unique solution in G.
Proof. Suppose that ax = b, then x = ex = a−1 ax = a−1 b. Suppose that x1

and x2 are both solutions of ax = b, then ax1 = b = ax2 . So
x1 = a−1 ax1 = a−1 ax2 = x2 .

CHAPTER 2. GROUP 27
The proof for the existence and uniqueness of the solution of xa = b is

similar.
The next proposition tell us the right and left cancellation laws are true in
groups. The proof leave as an exercise.
Proposition 2.1.14. If G is a group, and a, b, c ∈ G, then ba = ca implies

b = c and ab = ac implies b = c. Here, ba = ca implies b = c is called right
cancellation. ab = ac implies b = c is called left cancellation.
Theorem 2.1.15. In a group, denote
gn = g ◦ g ◦ · · · ◦ g
and
g −n = g −1 ◦ g −1 ◦ · · · ◦ g −1 ,
then we have usual laws of exponents hold.
g m g n = g m+n ,
(g m )n = g mn ,
(gh)n = ((gh)−1 )−n = ((h−1 )(g −1 ))−n .
But, in general gh 6= hg, and (gh)n 6= g n hn .
If the operation of a group is ”+”, m, n ∈ Z, g, h ∈ G, then
ng = g + · · · + g, mg + ng = (m + n)g,
m(ng) = (mn)g, m(g + h) = mg + mh.
2.2 Subgroups
Definition 2.2.1. A subgroup H of a group (G, ◦) to be a subset H of G such

that H is a group under the group operation of G.
CHAPTER 2. GROUP 28
Observe that every group G with at least two elements will always have
at least two subgroups, the subgroup consisting of the identity element alone
and the entire group itself. The subgroup H = {e} of a group G is called the
trivial subgroup. A subgroup (H, ◦) that is a proper subset of G is called a
proper subgroupof G.
Example 2.2.2. 1. Let Q∗ = { pq |p, qare nonzero integers}. The identity of

Q∗ is 1 = 11 . Given two elements pq , rs ∈ Q∗ , their product pr
qs is also in Q .
∗
The inverse of any element pq is again in Q∗ since ( pq )−1 = pq . Multiplication

in Q∗ is associative. Then Q∗ is a proper subgroup of R∗ .
Example 2.2.3. Recall that C∗ is the multiplicative group of nonzero complex

numbers. Let H = {1, −1, i, −i} ⊆ C∗ . H is a subgroup of C∗ .
Example 2.2.4. Let SLn (R) = {A|A ∈ Gln (R), |A| = 1} be the subset
of GLn (R). Note that the product of two matrices of determinant one also
has determinant one. The group SL2 (R) is called the special linear group.
SLn (R) ⊆ GLn (R) is a subgroup of GLn (R).
Note that a subset H of a group G can be a group without being a subgroup

of G. For H to be a subgroup of G it must inherit the binary operation of G.
The set of all Mn (R), forms a group under the operation of addition. GLn (R)
is a subset of Mn (R), and is a group under matrix multiplication, but it is not
a subgroup of Mn (R).
Next, we will examine some criteria for determining exactly when a subset
of a group is a subgroup.
Proposition 2.2.5. A subset H of G is a subgroup if and only if it satisfies

the following conditions.
(1). The identity e of G is in H.
(2). If h1 , h2 ∈ H, then h1 h2 ∈ H.
(3). If h ∈ H, then h−1 ∈ H.
Proof. Suppose that H is a subgroup of G. Let e is the identity of G.

CHAPTER 2. GROUP 29
(1). Since H is a group, it must have an identity eH . We must show

that eH = e. We know that eH eH = eH and that eeH = eH e = eH ; hence,
eeH = eH eH . By right-hand cancellation, e = eH .
(2) Since a subgroup H is a group. For any h1 , h2 ∈ H, then h1 h2 ∈ H.
(3) Let h ∈ H. Since H is a group, there is an element g ∈ H such that

hg = gh = e. By the uniqueness of the inverse in G, g = h−1 .
Conversely, if (1)-(3) hold, we must show that H is a group under the same
operation as G; however, these conditions plus the associativity of the binary
operation are exactly the axioms stated in the definition of a group.
Proposition 2.2.6. Let H be a subset of a group G. Then H is a subgroup

of G if and only if H is nonempty, and whenever g, h ∈ H, then gh−1 ∈ H.
Proof. Assume that H is a subgroup of G. Since h is in H, its inverse h−1 ∈ H.

Because of the closure of the group operation, gh−1 ∈ H.
Conversely, suppose that H ⊂ G such that H is nonempty and gh−1 ∈ H

whenever g, h ∈ H. If g ∈ H, then gg −1 = e ∈ H. If g ∈ H, then eg −1 =
g −1 ∈ H. Now let h1 , h2 ∈ H. We must show that their product is also in H.
However, h1 (h−1
2 )
−1 = h h ∈ H. Hence, H is a subgroup of G.
1 2
2.3 Cyclic groups
In this section, we will study the properties of cyclic groups and cyclic
subgroups, which play a fundamental part in the classification of all abelian
group.
Theorem 2.3.1. Let G be a group, g ∈ G. Then the set hgi = {g k |k ∈ Z}

is the smallest subgroup of G. Furthermore, hgi is the smallest subgroup of G
contains g.
Proof. Since a0 = e, thus the identity in G. If h and f are any two elements
in hgi, then by the definition of hgi, we can write h = g m and f = g n for some
CHAPTER 2. GROUP 30
integers m and n. So hf = g m g n = g m+n is again in hgi. Finally, if h = g n

in hgi, then the inverse h−1 = g −n is also in hgi. Clearly, any subgroup H of
G containing g must contain all the powers of g by closure property of group,
hence H contains hgi. Therefore, hgi is the smallest subgroup of G containing
g.
Definition 2.3.2. hgi is called the cyclic subgroup generated by g. If G = hgi,

then G is called a cyclic group, g is a generator of G.
If G =< g > is a cyclic group, then
G = {· · · , g −2 , g −1 , g 0 , g 1 , g 2 , · · · }.
If g ∈ G, we define the order of g to be the smallest positive number n, such

that g n = e, and we write |g| = n. If there is no such integer n, we say that
the order of g is infinite and write |a| = ∞ to denote the order of g.
Example 2.3.3. The groups Z and Zn are cyclic groups. And Z = h1i,
Zn = h1i. Moreover, Z = h−1i, Zn = h−1i. Notice that a cyclic group can
have more than a single generator.
Example 2.3.4. Z6 is a cyclic group generated by 1 or 5. Not every element

in a cyclic group is necessarily a generator of the group. 2 ∈ Z6 , the cyclic
subgroup generated by 2 is {0, 2, 4}.
Example 2.3.5. U (9) = {1, 2, 4, 5, 7, 8} = h2i.

1 2 3
2 = 2, 2 = 4, 2 = 8,
4 5 6
2 = 7, 2 = 5, 2 = 1.
Example 2.3.6. U (16) = {1, 3, 5, 7, 9, 11, 13, 15} is not a cyclic group, we can
not find a generator in it. Every element of U (16) generate Z(16), that is Z16 .
For example, Z16 = h3i.
1 · 3 = 3, 2 · 3 = 6, 3 · 3 = 9, 4 · 3 = 12,
5 · 3 = 15, 6 · 3 = 2, 7 · 3 = 5, 8 · 3 = 8,
9 · 3 = 11, 10 · 3 = 14, 11 · 3 = 1, 12 · 3 = 4,
13 · 3 = 7, 14 · 3 = 10, 15 · 3 = 13, 16 · 3 = 0.
CHAPTER 2. GROUP 31
Theorem 2.3.7. Every cycle group is abelian.
Proof. Suppose G = hai, g1 , g2 ∈ G, then g1 = am , g2 = ak . Since
g1 g2 = am ak = am+k = ak+m = ak am = g2 g1 ,
G is abelian.
Theorem 2.3.8. Every subgroup of a cycle group is cycle.
Proof. Let G = hai be a cycle group, H be a subgroup of G. If H = {e}, then

H is cycle. If H 6= {e}, then H contains some others element am distinct from
the identity e. Since H must contains (am )−1 , we may assume that m > 0.
Let m be the smallest positive integer such that h = am ∈ H.
Assume that h0 = ak ∈ H, k ∈ Z. By the division Algorithm, there exist

integers q, r such that k = mq + r, 0 ≤ r < m, then
h0 = ak = amq+r = amq ar = hq ar .
So ar = ak a−mq ∈ G. Since ak , amq ∈ H, ar must also be in H. However, m

was the smallest positive number such that am was in H. Consequently, r = 0
and so k = mq. Therefore, h0 = ak = amq = hq , and H is generated by h.
Example 2.3.9. The group (Z, +) is among the most familiar and easily
understood group, which generated by 1 or −1.
Example 2.3.10. It is obvious that 2Z = {. . . , −4, −2, 0, 2, 4, . . . } is a sub-

group of Z. All subgroups of Z are nZ, n ∈ N.
Proposition 2.3.11. Let G be a cyclic group of order n and G = {a|an = e}.

Then ak = e if and only if n | k.
Proof. Suppose that ak = e. By the division Algorithm, k = nq + r, where

0 ≤ r < n, hence
e = ak = anq+r = anq ar = ear = ar .

CHAPTER 2. GROUP 32
Since the smallest positive integer m such that am = e is n, r = 0.
Conversely, if n divides k, then k = ns for some integer s. Consequently,
ak = ans = (an )s = es = e.
The next result tells us how to set about constructing the subgroups of a
given cyclic group.
Theorem 2.3.12. Let G be a cyclic group of order n and suppose that a ∈ G

is a generator of the group. If b = ak , then the order of b is nd , where d =
gdc(k, n).
Proof. Let m be the order of b = ak , then bm =(ak )m = akm . Since n is the

order of G, n | km. Thus nd prime to kd . Hence, if nd divides km
d , we must have
n n
d | m. Thus, this kind smallest m is d .
Corollary 2.3.13. The generator of Zn are the integers r so that gcd(r, n) =

1.
For example, we have Z3 = h1̄i = h2̄i, and Z8 = h1i = h3i = h5i = h7i.
Proposition 2.3.14. Let G = hai be a cyclic group.
(1) If G is infinite, then each subgroup of G has the form Gm = ham i

where m ≥ 0. Furthermore, the Gm are all distinct and Gm has infinite order
if m > 0.
(2) If G has finite order n, then it has exactly one subgroup of order d for
n
each positive divisor d of n, namely < a d >.
Proof. Assume first that G is infinite and let H be a subgroup of G. By

Proposition2.4.5 H is cyclic, say H = ham i where m ≤ 0. Thus H = Gm . If
am had finite order s, then ams = 1, since a has infinite order. This can only
mean that m = 0 and H = {e}. Thus H is certainly infinite cyclic if m > 0.
CHAPTER 2. GROUP 33
Next Gm = Gs implies that am ∈ has i and as ∈ ham i, that is, m|s and s|m,
so that m = s. Thus all the Gm ’s are different.
Next let G have finite order n and suppose d is a positive divisor of n. Now
n n nl
(a )d = an = e, so the order l of a d must divide d. But also a d = e, and
d
n
hence n divides nl
d , i.e., d divides l. It follows that l = d and thus hx i has
d
order exactly d.
To complete the proof, suppose that H = har i is another subgroup with

order d. Then xrd = 1, so n divides rd and nd divides r. This shows that
n n n
H = har i ⊆ ha d i. But |H| = |ha d i| = d, from which it follows that H = ha d i.
Consequently there is exactly one subgroup of order d.
2.4 Permutation groups
A permutation of a set X is a bijection from X to X. Assume that

X = {x1 , x2 , · · · , xn }, σ be a permutation of X. Since σ is injective, then
σ(x1 ), σ(x2 ), · · · , σ(xn ) are all different and therefore constitute all n elements
of the set X, but in some different order from x1 , · · · , xn . So a permutation
can be take as a rearrangement of the order x1 , x2 , · · · , xn . A convenient way
to denote the permutation σ is
!
x1 x2 ··· x4
σ=
σ(x1 ) σ(x2 ) · · · σ(x4 )
where the first line consists elements of X, the second row consists of the
images under σ of the elements of the first row.
It should be clear that nothing essential is lost if we take X to be the set

{1, 2, · · · , n}. We can write σ as
!
1 2 3 4
σ=
σ(1) σ(2) σ(3) σ(4)
! !
1 2 3 4 1 2 3 4
For any two permutations σ = ,δ = of X,
2 3 4 1 3 1 2 4
CHAPTER 2. GROUP 34
We can compute the composition of them:

! ! !
1 2 3 4 1 2 3 4 1 2 3 4
σδ = =
2 3 4 1 2 4 1 3 3 1 2 4
and ! ! !
1 2 3 4 1 2 3 4 1 2 3 4
δσ = =
2 4 1 3 2 3 4 1 4 1 3 2
Let |X| = n, denote the set of all permutations of X as Sn . Next, we will

show that Sn is a group.
Theorem 2.4.1. Sn is a group with n! elements. The operation of Sn is the

composition of bijections.
Proof. Let X be a set. It is obvious that Sn × Sn → Sn is a binary operation

of Sn since the composition of bijections is still a bijection of X. And the
composition of bijections is associative. The identity of Sn is identity map, we
write it as e. If f ∈ Sn , then f −1 ∈ Sn since f is a bijection.
Consider the number of ways of constructing the second row of a permu-

tation !
x1 x2 ··· xn
σ=
σ(x1 ) σ(x2 ) · · · σ(xn )
There are n choices for σx1 , but only n − 1 choices for σ(x2 ). Next we cannot
choose σx1 or σ(x2 ), so there are n − 2 choices for σ(x3 ), and so on. Finally,
there is just one choice for σ(xn ). Each choice of σ(xi ) can occur with each
choice of σ(xj ). Therefore the number of different permutations of X is
n(n − 1)(n − 2) · · · 1 = n!.
Note that although it is natural to multiply elements in a group from left

to right, functions are composed from right to left. For example, σδ means
that we do δ first, then σ.
CHAPTER 2. GROUP 35
Definition 2.4.2. Sn is called a symmetric group. A subgroup of Sn is called

a permutation group.
Let us take a look at symmetric group S3 . There are 6 elements:

! ! !
1 2 3 1 2 3 1 2 3
, , ,
1 2 3 1 3 2 2 1 3
! ! !
1 2 3 1 2 3 1 2 3
, , .
3 2 1 2 3 1 3 1 2
Definition 2.4.3. Let x1 , x2 , · · · , xt be distinct elements of the set {x1 , x2 , · · · , xn }.

A cycle of length t is a permutation σ if
!
x1 x2 x3 · · · xt
σ= ,
x2 x3 x4 · · · x1
while leaving all the remaining elements of {x1 , x2 , · · · , xn } fixed. We write

σ = (x1 x2 · · · xt ). Two cycles σ = (x1 · · · xs ), τ = (y1 · · · yt ) in Sn are disjoint
if xi 6= yj for all i and j.
In S6 ,
! !
1 2 3 4 5 6 1 2 3 4 5 6
(135) = , (26) = .
3 2 5 4 1 6 1 6 3 4 5 2
(135) and (26) are disjoint. And length(135) = 3, length(26) = 2.
Proposition 2.4.4. Let σ and τ be two disjoint cycles in SX . Then στ = τ σ.
Proof. Let σ = (x1 x2 · · · xs ), τ = (y1 y2 · · · yt ), we want to show στ (x) =

τ σ(x) for any x ∈ X. If x is neither x1 · · · xs nor y1 · · · yt , then both σ and τ
fix x. That is σ(x) = x and tau(x) = x. Hence,
στ (x) = σ(τ (x)) = σ(x) = x = τ (x) = x = τ (σ(x)) = τ σ(x).

CHAPTER 2. GROUP 36
Suppose that x ∈ {x1 , · · · , xs }, then σ(xi ) = xi mod s+1 . However, τ (xi ) =

xi since σ and τ are disjoint. Therefore
στ (xi ) = σ(τ (xi )) = σ(xi )

= xi mod s+1
= τ (xi mod s+1 )
= τ σ(xi ).
Similarly, if x ∈ {y1 , y2 , · · · , yt }, then σ and τ also commute.
Theorem 2.4.5. Every permutation in Sn is expressible as a product of dis-

joint cycles and the cycles appearing in the product are unique.
Proof. We deal with the existence of the expression first. Assume that X =
{1, 2, · · · , n}. If σ is the identity (1), then obviously σ = (1)(2) · · · (n). Assume
that σ ∈ Sn and σ 6= (1), define X1 = {σ(1), σ 2 (1), · · · } and |X1 | < ∞ since
|X| = n < ∞. Now let i be the first integer in X that is not in X1 and
define X2 by {σ(i), σ 2 (i), · · · } and |X2 | < ∞. Continuing in this manner,
we can define finite disjoint sets X3 , X4 , · · · . Since X is a finite set, we are
guaranteed that this process will end, and there will be only a finite number
of these sets r. If σi is the cycle defined by

σ(x), x ∈ X
i
σi (x) =
x, x∈/X, i
then σ = σ1 σ2 · · · σr . Since the sets X1 , X2 , · · · , Xr are disjoint, the cycles

σ1 , σ2 , · · · , σr must also be disjoint.
Now to establish uniqueness. Assume that there are two expressions for σ
as a product of disjoint cycles, say
(x1 x2 · · · )(y1 y2 · · · ) · · ·
and
(a1 a2 · · · )(b1 b2 · · · ) · · · .
CHAPTER 2. GROUP 37
By Proosition2.4.4, disjoint cycles commute. Thus without loss of generality

we can assume that x1 occurs in the cycle (a1 a2 · · · ). Since any element of
a cycle can be moved up to the initial position, it can also be assumed that
x1 = a1 . Then x2 = σ(x1 ) = σ(a1 ) = a2 ; similarly x3 = a3 , and so on. The
other cycles are dealt with in the same way. Therefore the two expressions for
σ are identical.
! !
1 2 3 4 5 6 1 2 3 4 5 6
Example 2.4.6. Let σ = ,τ = ,
6 4 3 1 5 2 3 2 1 5 6 4
then σ = (1624) is a cycle of length 4, τ = (1 3)(4 5 6) is product of cycles.
Moreover, στ = (1 3 6)(2 4 5), τ σ = (1 4 3)(2 5 6).
Definition 2.4.7. A permutation with length 2 is called a transposition.
Corollary 2.4.8. Every element of Sn is expressible as a product of transpo-

sitions.
Proof. Because of Theorem2.4.5, it is sufficient to show that each cyclic per-

mutation is a product of transpositions. This is true since
(x1 x2 · · · xr ) = (x1 xr )(x1 xr−1 ) · · · (x1 x3 )(x1 x2 ).
Let σ is a cycle in Example2.4.6, then
σ = (1 6 2 4) = (1 4)(1 2)(1 6),
τ = (1 3)(4 5 6) = (1 3)(4 6)(4 5).

Note that not every permutation in Sn is expressible as a product of disjoint
transpositions. There is no unique way to represent permutation as the prod-
uct of transpositions. For instance, we can write the identity permutation as
(1) = (1 2)(1 2), or as (1 3)(2 4)(1 3)(2 4), and in many other ways.
Lemma 2.4.9. If the identity is written as the product of transposition,
id = σ1 σ2 · · · σr ,
then r is an even number.

CHAPTER 2. GROUP 38
Proof. We will show the conclusion by induction on r. A transposition cannot

be the identity (1). Hence, r > 1. If r = 2, then we have done.
Suppose that r > 2. Let σr−1 σr be the last two transpositions, then σr−1 σr
must be one of the following cases:
(x y)(x y) = id,
(y z)(x y) = (x z)(y z),
(z w)(x y) = (x y)(z w),
(x z)(x y) = (x y)(y z),
where x, y, z, w are distinct. The first equation simply says that a transposition
is its own inverse. If this case occurs, delete σr−1 σr from the product to obtain
id = σ1 σ2 · · · σr−2 . By induction r − 2 is even; hence, r must be even.
In each of the other three cases, we can replace σr−1 σr with the righthand
side of the corresponding equation to obtain a new product of r transpositions
for the identity. In this new product the last occurrence of x will be in the
next-to-the-last transposition. We can continue this process with σr−2 σr−1 to
obtain either a product of r − 2 transpositions or a new product of r transpo-
sitions where the last occurrence of x is in σr−2 . If the identity is the product
of r − 2 transpositions, then again we are done. Otherwise, we will repeat the
procedure with σr−3 σr−2 .
Either we will have two adjacent, identical transpositions canceling each

other out or x will be shuffled so that it will appear only in the first trans-
position. However, the latter case cannot occur, because the identity would
not fix x in this instance. Therefore, the identity permutation must be the
product of r − 2 transpositions and, again by our induction hypothesis, we are
done.
Definition 2.4.10. A permutation is called even permutation if the permuta-

tion can be written as even number of transpositions. Similarly, odd permuta-
tion is defined the same way.
Example 2.4.11. Let us consider S3 , the even permutations are (1), (1 2 3), (1 3 2),
CHAPTER 2. GROUP 39
while the odd permutations are (1 2), (2 3), (1 3). What is more, in S7 ,
(1 6)(2 5 3) = (1 6)(2 3)(2 5)
is odd permutation,
(1 7)(2 5 3 4) = (1 7)(2 4)(2 3)(2 5)
is even permutation.
One of the most important subgroups of Sn is the set of all even permuta-
tions.
Definition 2.4.12. The alternating group An is the set of all even permuta-
tions of Sn .
Theorem 2.4.13. The alternating group An is a subgroup of Sn .
Proof. Firstly, operation of An ⊆ Sn is closed because the product of two even

permutations must also be an even permutation. An is associative since Sn is
associative. Secondly, the identity (1) is an even permutation by Lemma2.4.9.
Let σ = σ1 σ2 · · · σs with s even. Then
−1
σ −1 = (σ1 σ2 · · · σs )−1 = σs−1 σs−1 · · · σ1−1
is an even transpositions.
It is easy to see that the number of even permutation in Sn (n ≥ 2) is equal

to the number of odd permutations, and |An | = n!/2.
2.5 Dihedral groups
The regular n-gon (n ≥ 3) is a polygon in the plane with n equal edges,

label the vertices 1, 2, · · · , n.
Definition 2.5.1. The n-th dihedral group is a group of rigid motions of a

regular n-gon. We will denote this group by Dn .
CHAPTER 2. GROUP 40
Notice that there are exactly n choices to replace the first vertex. If we
replace the first vertex by k, then the second vertex must be replaced either
by vertex k + 1 or by vertex k − 1; hence, there are 2n possible rigid motions
of the n-gon.
There are two kinds of rigid motions: rotation and reflection. There are
n−1 clockwise rotations about the line perpendicular to the plane and through
the centroid, through angles i 2π
n , for i = 1, 2, · · · , n − 1.
1 n
n 2 n−1 1
n−1 3 n−2 2
··· ···
Figure 2.1
π
For example, the rotation through 2n is represented by the n-cycle (1 2 3, · · · n);
other rotations correspond to powers of this n-cycle. Write these rotation by
r, r2 , · · · , rn = (1).
There are n reflections in axes of symmetry in the plane. There are two
cases of reflections, depending on whether n is even or odd. If n is an even
number, there are two types of reflections, in an axis joining a pair of opposite
vertices, for example, (2 n)(3 n − 1) · · · ( n2 n2 + 2);
CHAPTER 2. GROUP 41
1 1
n 2 2 n
n−1 3 3 n−1
··· ···
Figure2.2
and in an axis joining midpoints of opposite edges, for example (1 n)(2 n −

1) · · · ( n2 n2 + 1). Thus in all there are n reflections. If there are odd number
of vertices, then only a single vertex is left fixed by a reflection. For example,
(1)(2 n)(3 n − 1) · · · corresponds to one such reflection.
n 1 1 n
n−1 2 2 n−1
··· ···
Figure2.3
CHAPTER 2. GROUP 42
In either case, the order of each reflection is two. Denote these rotation by
s, s2 = (1).
Since all axes of symmetry of the n-gon have now been exhausted, we
conclude that the order of Dn is 2n. We summarize these results in the
following theorem.
Theorem 2.5.2. The dihedral group Dn , is a subgroup of Sn of order 2n.

And
Dn =< s, r|s2 = 1.rn = 1, srs = r−1 > .
Proof. The possible motions of a regular n-gon are either reflections or rota-
tions. There are exactly n possible rotations:
360◦ 360◦ 360◦
Id, ,2 · , · · · , (n − 1) .
n n n
360◦
We will denote the rotation n by r. The rotation r generates all of the other
rotations. That is,
360◦
rk = k .
n
Label the n reflections s1 , s2 , · · · , sn , where sk is the reflection that leaves
vertex k fixed. There are two cases of reflections, depending on whether n is
even or odd. If there are an even number of vertices, then two vertices are left
fixed by a reflection, and s1 = s n2 +1 , s2 = s n2 +2 , · · · , s n2 = sn . If there are an
odd number of vertices, then only a single vertex is left fixed by a reflection
and s1 , s2 , · · · , sn are distinct. In either case, the order of each sk is two. Let
s = s1 . Then s2 = Id and rn = Id. Since any rigid motion t of the n-gon
replaces the first vertex by the vertex k, the second vertex must be replaced
by either k + 1 or by k − 1. If the second vertex is replaced by k + 1, then
t = rk . If the second vertex is replaced by k − 1, then t = srk . Hence, r
and s generate Dn . That is, Dn consists of all finite products of r and s,
Dn = {1, r, r2 , · · · , rn?1 , s, sr, sr2 , · · · , srn−1 }.
Example 2.5.3. Let us consider D3 and D4 .
D3 =< r, s | s2 = id = r3 , srs = r−1 >= {1, s, r, r2 , rs, r2 s}.

CHAPTER 2. GROUP 43
Let r = (1 2 3), s = (2 3),
1 3
(1 2 3)
3 2 2 1
1 1
(2 3)
3 2 2 3
Figure2.4
then we can write
D3 = {(1), (2 3), (1 2 3), (1 3 2), (1 2), (1 3)} ⊆ S3 .
In this case, |S3 | = |D3 | = 6, that means D3 = S3 .
We have
D4 =< r, s | s2 = id = r4 , srs = r−1 >= {1, s, r, r2 , r3 , rs, r2 s, r3 s}.
Assume that r = (1 2 3 4), s = (1 2)(3 4),

CHAPTER 2. GROUP 44
1 2 4 1
(1 2 3 4)
4 3 3 2
Figure2.5
1 2 2 1
(1 2)(3 4)
4 3 3 4
Figure2.6
therefore
r2 = (1 3)(2 4),
r3 = (1 4 3 2),
rs = (1 2 3 4)(1 2)(3 4) = (1 3),
r2 s = (1 3)(2 4)(1 2)(3 4) = (1 4)(2 3),
r3 s = (1 4 3 2)(1 2)(3 4) = (2 4).
Thus
D4 = {(1), (1 2 3 4), (1 3)(2 4), (1 4 3 2), (1 2)(3 4), (1 3), (1 4)(2 3), (2 4)}.
CHAPTER 2. GROUP 45
2.6 Cosets and Lagrange’s Theorem
In this section, we will introduce Lagrange’s Theorem, one of the most

important results in finite group theory. Lagrange’s Theorem provides a pow-
erful tool for analyzing finite groups. Using this theorem, we can know what
type of subgroups a finite group to possess.
Definition 2.6.1. Let G be a group, and H be a subgroup of G. Define a left

coset of H with representative g ∈ G to be the set of
gH = {gh|h ∈ H},
define right coset to be

Hg = {hg|h ∈ H}.
gH and Hg are subsets of G. If the left coset is equal to the right coset, it is
called coset.
Example 2.6.2. Recall that G = (Z6 , +) is a commutative group, H = {0̄, 3̄}

is a subgroup of G. We have
0̄ + H = H = H + 0̄ = {0̄, 3̄}
1̄ + H = H = H + 1̄ = {1̄, 4̄}
2̄ + H = H = H + 2̄ = {2̄, 5̄}
3̄ + H = H = H + 3̄ = {3̄, 0̄}
4̄ + H = H = H + 4̄ = {4̄, 1̄}
5̄ + H = H = H + 5̄ = {5̄, 2̄}
CHAPTER 2. GROUP 46
Example 2.6.3. Consider S3 , H = {(1), (123), (132)}. Then
(1)H = H = H(1) = H,
(123)H = H = H(123),
(132)H = H = H(132),
(12)H = {(12), (23), (13)},
H(12) = {(12), (23), (13)},
(13)H = {(12), (23), (13)},
H(13) = {(12), (23), (13)},
(23)H = {(12), (23), (13)},
H(23) = {(12), (23), (13)},
If K = {(1), (12)}, then
(1)K = K(1) = (12)K,

(13)K = {(13), (123)} = (123)K,
(23)K = {(23), (132)} = (132)K,
K(13) = {(13), (132)},
K(23) = {(23), (123)} = K(123),
In this case, left cosets of K are not right cosets.
Lemma 2.6.4. Let H be a subgroup of G, and suppose that g1 , g2 ∈ G. The

following conditions are equivalent:
(1) g1 H = g2 H.
(2) Hg1−1 = Hg2−1 .
(3) g2 ∈ g1 H.
(4) g1−1 g2 ∈ H.
Proof. (1)⇐⇒(2): Let g1 H = g2 H for g1 , g2 ∈ G. Then g1−1 g1 H = g1−1 g2 H.

We have H = g1−1 g2 H. And we know that e ∈ H, g1−1 g2 e ∈ H, then g1−1 g2 ∈
CHAPTER 2. GROUP 47
H. And (g1−1 g2 )−1 ∈ H, so g2−1 g1 ∈ H. Then
H = He = Hg1−1 g1 = Hg2−1 g1 ,
thus Hg1−1 = Hg2−1 . Conversely, If Hg1−1 = Hg2−1 , we have g2−1 g1 ∈ H, so

g2−1 g1 ∈ H and g1−1 g2 ∈ H. Then
H = eH = g1−1 g1 H = g1−1 g2 H.
Multiply g1 from the left side, we have g1 H = g2 H.
(1)⇐⇒(3): It is obvious that g2 e ∈ g1 H, that is g2 ∈ g1 H. Conversely, if

g2 ∈ g1 H, there exist h ∈ H, such that g2 = g1 h, then g1 = g2 h−1 . We have
g1 H = g2 h−1 H, that is g1 H = g2 H.
(3)⇐⇒(4): If g2 ∈ g1 H, there exist h ∈ H, such that g2 = g1 h, then

g1−1 g2= h. On the other hand, since g1−1 g2 ∈ H, there exist h ∈ H such that
g1−1 g2= h. Then g2 = g1 h ∈ g1 H.
Theorem 2.6.5. Let H be a subgroup of a group G, then the group G is the

disjoint union of the left coset of H in G.
Proof. Let G = {g1 , g2 . . . gn · · · }. We have

[ [
G = g1 H g2 H . . . gn H ··· .
T T
Assume that g1 H g2 H 6= ∅. Let a ∈ g1 H g2 H. Then we have a = g1 h1 =
g2 h2 for h1 , h2 ∈ H. Then g2 = g1 h1 h−1
2 ∈ g1 H, thus g1 H = g2 H by 2.6.4.
Definition 2.6.6. Let G be a group and H be a subgroup of G. Define the

index of H in G to be the number of left cosets of H in G. Denote [G : H].
Example 2.6.7. Let H = {0, 3} be a subset of Z6 . Then [Z6 : H] = 3. That

means
0 + H = 3 + H = H,
1 + H = 4 + H = {1, 4},
2 + H = 5 + H = {2, 5}.
CHAPTER 2. GROUP 48
Theorem 2.6.8. Let H be a subgroup of G, the number of left cosets of H in

G is the same as the number of right cosets.
Proof. Let LH denote the set of left cosets of H in G, RH the set of right cosets
of H in G. Define φ : LH −→ RH , by φ(gH) = Hg −1 . Then φ is well-defined
since ∀gH ∈ LH , there exists an image in RH . And if g1 H ∈ LH , there exists
only one element in RH , such that φ(g1 H) = Hg1−1 , i.e. the image is unique.
Therefore, φ is well-defined (φ is a map). Moreover, φ is a bijection. Assume
that
Hg1−1 = φ(g1 H) = φ(g2 H) = Hg2−1 ,
we have g1 H = g2 H by 2.6.4, thus φ is injective. φ is surjective since ∀Hg −1 ∈

RH , there exists gH ∈ LH such that φ(gH) = Hg −1 .
Proposition 2.6.9. Let H be a subgroup of G, g ∈ G, and define a map

φ : H −→ gH by φ(h) = gh. The map φ is bijective, the number of elements
in H is the same as the number of elements in gH.
Proof. Suppose that φ(h1 ) = φ(h2 ) for h1 , h2 ∈ H. We must show that h1 =

h2 , but φ(h1 ) = gh1 and φ(2) = gh2 . So gh1 = gh2 and by left cancellation
h1 = h2 . Every alement of gH is of the form gh for some h ∈ H and φ(h) =
gh.
Theorem 2.6.10. Lagrange’s Theorem: Let G be a finite group and let H be

|G|
a subgroup of G, Then |H| = [G : H] is the number of distinct left cosets of
Hin G. In particular, the number of elements in H must divide the number
of elements in G.
Proof. The group G is partitioned into [G : H] distinct left cosets. Each left
cosets has |H| elements, therefore |G| = [G : H]|H|.
Corollary 2.6.11. Suppose that G is a finite group and g ∈ G. Then the

order of g must divide the number of elements in G.
Corollary 2.6.12. Let |G| = p with p a prime number. Then G is cyclic and
any g ∈ G such that g 6= e is a generator.
CHAPTER 2. GROUP 49
Proof. By Corollary2.6.11, the order of g must divide the order of the group.
Since |hgi| > 1, it must be p. Hence g generate G.
Corollary 2.6.13. Let H and K be subgroups of a finite group G, such that

G ⊃ H ⊃ K. Then
[G : K] = [G : H][H : K].
Proof. Observe that
G |G| |H|
[G : K] = = = [G : H][H : K].
K |H| |K|
The inverse of Lagrange’s Theorem is false. We will give a counterexample.
Theorem 2.6.14. Two cycles τ and µ in Sn have the same length if and only
if there exists a δ ∈ Sn , such thah µ = δτ δ −1 .
Proof. Suppose that τ = (a1 a2 · · · ak ) and µ = (b1 b2 · · · bk ). Define δ ∈ Sn ,

such that
δ(a1 ) = b1 , δ(a2 ) = b2 · · · , δ(an ) = bn .
Then µ = δτ δ −1 since δτ δ −1 (b1 ) = δτ (a1 ) = δ(a2 ) = b2 = µ(b1 ).
Conversely, suppose τ =(a1 a2 · · · ak ) is a cycle with length k and δ ∈ Sn .

If δ(ai ) = b and δ(ai+1 ) = b0 , then µ(b) = b0 , hence µ=(δ(a1 )δ(a2 )· · · δ(an )).
Since that δ is bijection, µ is a cycle of the same length as τ .
Proposition 2.6.15. The group A4 has no subgroup of order 6.
Proof. Assume that H be a subgroup of A4 and |H| = 6. Since [A4 : H] = 2,

there are only two cosets of H in A4 . In as much as one of the cosets is H
itself right and left cosets must coincide, therefore gH = Hg or gHg −1 = H
for every g ∈ A4 . Since there are eight 3 − cycles in A4 , at least one 3 − cycle
CHAPTER 2. GROUP 50
must in H. Assume that (123) ∈ H. Then (123)−1 = (132) ∈ H. Since

gHg −1 ∈ H for h ∈ H and
(124)(123)(124)−1 = (124)(123)(142) = (243),

(243)(123)(243)−1 = (243)(123)(234) = (142).
We can conclude that H must have at least seven elements:
(1), (123), (123)−1 = (132), (243), (243)−1 = (234), (142), (142)−1 = (124).
Therefore A4 has no subgroup of order 6.
Definition 2.6.16. The Euler function is the map Φ : N → N defined by

Φ(1) = 1 and Φ(n) is the number of positive integers m with 1 ≤ m < n and
gcd(m, n) = 1.
Theorem 2.6.17. Let U (n) be the group of unite in Zn . Then |U (n)| = Φ(n).
Theorem 2.6.18. Euler’s Theorem: Let a and n be integers such that n > 0
and gcd(a, n) = 1. Then aΦ(n) = 1( mod n)
Proof. By Theorem2.6.17, |U (n)| = Φ(n), so |a| | Φ(n). Since aΦ(n) = 1, then

aΦ(n) − 1|n. Therefore aΦ(n) = 1( mod n).
Theorem 2.6.19. Fermat’s little Theorem: Let p be any prime numbers and
suppose that P - a, then ap−1 ≡ 1( mod p). Furthermore, for any integer b,
bp ≡ b( mod p).
4
For example, U (5) = {1, 2, 3, 4}, |U (n)| = Φ(5) = 4. Then 4 = 1( mod 5),
since |4| = 2.
2.7 Normal subgroups and factor groups
Definition 2.7.1. A subgroup H of G is called a normal group in G if
gH = Hg
for all g ∈ G.
CHAPTER 2. GROUP 51
Example 2.7.2. If G is an abelian group, then every subgroup H of G is

normal.
Example 2.7.3. Let S3 = {(1), (12), (13), (23), (123), (132)}, then H1 = {(1)}
is normal, and H = {(1), (123), (132)} is normal. But H2 = {(1), (12)} is not
normal.
Theorem 2.7.4. Let G be a group and let N be a subgroup of G. Then the

following statements are equivalent.
(1) The subgroup N is normal in G.
(2) For all g ∈ G, gN g −1 ⊂ N.
(3) For all g ∈ G, gN g −1 = N.
Proof. (1) =⇒ (2): N is normal, we have gN = N g, for g ∈ G. Then there

exist n1 , n2 ∈ N such that gn1 = n2 g, n2 = gn1 g −1 ∈ N . Therefore, gN g −1 ⊂
N.
(2) =⇒ (3) Because gN g −1 ⊂ N , there exist n ∈ N , we have g −1 ng =

g −1 n(g −1 )−1 ∈ N . Thus g −1 ng = n1 ∈ N , and gn1 = ng. Then n = gn1 g −1 ,
i.e. n ∈ gN g −1 . Therefore, N ⊂ gN g −1 . Then N = gN g −1 .
(3) =⇒ (1) Since gN g −1 = N , there exists n1 , n2 ∈ N such that gn1 g −1 =

n2 , then gn1 = n2 g. Therefore, gN ⊂ N g. Similarly, we have N g ⊂ gN . In
a word, gN = N g, N is normal.
Definition 2.7.5. If N is a subgroup of a group G, then the cosets of N in

G form a group G/N under the operation
(aN )(bN ) = abN.
This group is called the factor group or quotient group of G on N.
Note that N is the identity of G/N . For any bN ∈ G/N , we have

(eN )(bN ) = ebN = bN .
CHAPTER 2. GROUP 52
Example 2.7.6. Let S3 = {(1), (12), (13), (23), (123), (132)}, then N = {(1), (123), (132)}
be normal in S3 . In fact, we have
N = (1)N = (123)N = (132)N
and
(12)N = N (12) = (13)N = N (13) = (23)N = N (23) = {(12), (13), (23)}.
Then
S3 /N = {N, (12)N }.
Theorem 2.7.7. Let N be a subgroup of G. The coset of N in G form a group

G/N of order [G : N ].
Proof. Note that G/N is a set with operation G/N × G/N → G/N . The
operation is well-defined since if aN = bN, cN = dN , then aN cN = bN dN .
Let aN = bN and cN = dN , there exist a = bn1 , n1 ∈ N , and c = dn2 , n2 ∈ N .
Thus
aN cN = acN = bn1 dn2 N = bn1 dN = bn1 N d = bN d = bdN.
It is easy to check that the operation is closed and associative since G is

associative . N is the identity. Let g ∈ G, the inverse of gN is g −1 N . In a
word, G/N is a group.
2.8 Isomorphisms
Definition 2.8.1. Let (G, ◦) and (H, ∗) are two groups. If there exists a
bijection φ : G → H such that φ(g1 ◦ g2 ) = φ(g1 ) ∗ φ(g2 ), then φ is called an
isomorphism from G to H. We call G is isomorphic to H, denoted by G ∼ = H.
Example 2.8.2. (Z4 , +) be a group and (hii , ·), where hii = {i, 1, −1, −i} .
CHAPTER 2. GROUP 53
There is an map given by
Z4 −→ hii ,
0 7−→ 1,
1 7−→ i,
2 7−→ −1,
3 7−→ −i.
It is easy to check that
φ(1 + 2) = φ(3) = −i = i(−1) = φ(1)φ(2),

φ(1 + 3) = φ(4) = 1 = i(−i) = φ(1)φ(3),
φ(2 + 3) = φ(5) = i = (−1)(−i) = φ(2)φ(3).
Thus φ(m + n) = φ(m)φ(n), m, n ∈ Z4 .

Example 2.8.3. Let φ : (R, +) → (R+ , ·) be a map given by x 7→ ex . Then
φ is an injective by ex = ey implies x = y. For any ex , there exists a x ∈ R
such that φ(x) = ex Then φ is surjective. Moreover, we have
φ(x + y) = ex+y = ex · ey = φ(x) · φ(y).
Note that Z8 is not isomorphic to Z12 since |Z8 | 6= |Z12 |. But there is an
isomorphism between U (8) and U (12), the isomorphism is given by
Z8 −→ Z12 ,
1 7−→ 1,
3 7−→ 5,
5 7−→ 7,
7 7−→ 11.
Next, we will show that φ preserve the operation. It is easy to check that
φ(1) = 1. We have
φ(3) · φ(5) = 5 · 7 = 11, φ(3 · 5) = φ(15) = φ(7) = 11,

φ(3) · φ(7) = 5 · 11 = 7, φ(3 · 7) == φ(21) = φ(5) = 7,
φ(5) · φ(7) = 7 · 11 = 5, φ(5 · 7) == φ(35) = φ(3) = 5.
CHAPTER 2. GROUP 54
Although |S3 | = 6 = |Z6 |, there is no isomorphism between them. Recall

that (S3 , ·) is a nonabelian group, but (Z6 , +) is an abelian group. If φ is
an isomorphism of Z6 to S3 , then we should have φ(m + n) = φ(n + m) =
φ(n) · φ(m), but φ(m) · φ(n) 6=. Then φ is not an isomorphism.
Theorem 2.8.4. Let φ : G → H be an isomorphism of two groups (G, ·) and

(H, ∗). Then the following statements are true
(1) φ−1 : H → G is an isomorphism.
(2) |G| = |H|.
(3) If G is abelian, then H is abelian.
(4) If G is cyclic, then H is cyclic.
(5) If G has a subgroup of order n, then H has a subgroup of order n.
Proof. (1) It is obvious that φ−1 is a bijection by φ bijective. If φ(g) = h,

then φ−1 (h) = g. So φ−1 preserve the operation.
(2) Since φ is a bijection, then |G| = |H|.
(3) Let G be abelian, g1 , g2 ∈ G, and φ(g1 ) = h1 , φ(g2 ) = h2 . Then
φ(g1 · g2 ) = φ(g1 ) ∗ φ(g2 ) = h1 ∗ h2 .
On the other hand, we have
φ(g1 · g2 ) = φ(g2 · g1 ) = φ(g2 ) ∗ φ(g1 ) = h2 ∗ h1 .
That is h1 ∗ h2 = h2 ∗ h1 , H is abelian.
(4) If g is cyclic with generator g, then for any g m ∈ G, m ∈ N , we have
φ(g m ) = φ(g) ∗ φ(g) ∗ ... ∗ φ(g) = [φ(g)]m ∈ H.
Then H = hφ(g)i.
(5) Let G1 be a subgroup of G, with |G1 | = n. Then φ(G1 ) is subset of

H since φ is an isomorphism. We well show that (φ(G1 ), ·) is a group. It is
CHAPTER 2. GROUP 55
obvious that φ(G1 ) is closed because G is closed, and
φ(g1 · g2 ) = φ(g1 ) ∗ φg2 ∈ φ(G1 ).
The identity of φ(G1 ) is φ(eG ) = eH . Let φ(g) ∈ φ(G1 ), then φ(g)−1 ∗ φ(g) =
eH . And
φ(eg ) = φ(g −1 · g) = φ(g −1 ) ∗ φ(g) = [φ(g)]−1 ∗ φ(g).
Then φ(g ( − 1)) = φ−1 (g). At last, we have
φ [(g1 · g2 ) · g3 ] = φ(g1 · g2 ) ∗ φ(g3 ) = φ(g1 ) ∗ φ(g2 ) ∗ φ(g3 ),
and
φ [g1 · (g2 · g3 )] = φ(g1 ) ∗ φ(g2 ) ∗ φ(g3 ).
Then (φ(G1 ), ·) is associative. In a word, φ(G1 ) is a group of H.
Theorem 2.8.5. All cyclic groups of infinite order are isomorphic to Z+ .
Proof. Let G = hgi. Assume that
φ : (Z+ , +) → (G, ◦) : m 7→ g m
is a map. We will show φ is an isomorphism. φ is an injection since g m 6= g n

if m 6= n. For any g m ∈ G, there exists m ∈ Z + such that φ(m) = g m .
So φ is a surjection. Thus φ is a bijection. Moreover, let m, n ∈ Z+ , then
φ(m + n) = g m+n = g m ◦ g n = φ(m) · φ(n).
Theorem 2.8.6. If G is a cyclic group of order n, then G is isomorphic to

Z+
n.
Proof. Let φ : (Z+ m

n , +) → (G, ◦) : m 7→ g . Assume that g
m = g l , then
g m g −l = g m−l = e. Thus m − l = 0 ∈ Z+ +
n . So m = l in Zn . Then φ is a
injection. φ is a surjection since ∀g m ∈ G, there exists m ∈ Z+n , such that
φ(m) = g . m
Corollary 2.8.7. If G is a group of order p, where p is a prime number, then

G is isomorphic to Zp+ .
CHAPTER 2. GROUP 56
Theorem 2.8.8. Cayley Theorem: Every group is isomorphic to a group

of permutations.
Proof. Let G be a group. Define a map λg : G → G by a 7→ ga. Next, we will

show λg is a permutation of G, i.e. λg is a bijection of G. If λg (a) = λg (b),
then
λg (a) = ga = λg (b) = gb.
We have a = b. For ∀a ∈ G, there exists g −1 a ∈ G, such that λg (g −1 a) =

gg −1 a = a. So λg is a permutation.
Denote G = {λg |g ∈ G} a set. In fact, G is a group with operation of

composition of maps. Firstly, G is closed since
(λg ◦ λh )(a) = λg (ha) = gha = (gh)a = λgh (a).
Let λg , λh , λw ∈ G, then
((λg ◦ λh ) ◦ λw )(a) = (λg ◦ λh )(wa)

= λg (λh (wa))
= λg (hwa)
= g(hw)a
= ghwa
and
(λg ◦ (λh ◦ λw ))(a) = λg (λh ◦ λw )(a)

= λg (λh (wa))
= λg (h(wa))
= g(hwa)
= ghwa.
So the operation of G is associative. Note that λe ◦ λg = λeg = λg , so λe

is the identity of G. Moreover, ∀λg ∈ G, the inverse of λg is λg−1 since
λg ◦ λg−1 = λg◦g−1 = λe and λg−1 ◦ λg = λg−1 g = λe .
CHAPTER 2. GROUP 57
Finally, we will show that G is isomorphic to G as group. Define φ : G →

G : g 7→ λg . We have
φ(gh) = λgh = λg ◦ λh = φ(g)φ(h).
If φ(g)(a) = φ(h)(a), then ga = λg (a) = λh (a) = ha, so g = h. φ is surjective

since λg = φ(g) for any λg ∈ G, g ∈ G. φ is an isomorphism.
Example 2.8.9. Let Dn =< r, s|rn = 1 = s2 , srs = r−1 > be a dihedral

group, Rn =< r|rn = 1 >be a normal group of Dn . Then
Dn /Rn = {Rn , sRn } ∼

= Z2 .
Example 2.8.10. We have Z3 ∼

= {(1), (123), (132)}. In fact, define
φ(0̄) = (1), φ(1̄) = (123), φ(2̄) −→ (132).
Then φ is an isomorphism since
φ(1̄ + 2̄) = φ(0̄) = (1)
and
φ(1̄ + 2̄) = (123) ◦ (132) = (1).
2.9 Homomorphism and Isomorphism Theorems.
Definition 2.9.1. A homomorphism between group (G, ·) and (H, ◦) is a map

φ : G → H, such that φ(g1 · g2 ) = φ(g1 ) ◦ φ(g2 ). Images of φ in H is called
the homomorphic image of φ.
Every isomorphism is a homomorphism.
Example 2.9.2. Let G be a group and g ∈ G. Define a map φ : Z → G by

φ(n) = g n . Then φ is a homomorphism since
φ(m + n) = g m+n = g m g n φ(m)φ(n).
The homomorphism maps Z to the cyclic subgroup of G generated by g.

CHAPTER 2. GROUP 58
!
a b
Example 2.9.3. Let G = GL2 (R). If A = ∈ G, then det(A) =
c d
ad − bc 6= 0. For A, B ∈ G, we have det(AB) = det(A)det(B). Define φ :
GL2 (R) → R∗ by φ(A) = det(A), then
φ(AB) = det(AB) = det(A)det(B) = φ(A)φ(B).
Thus φ is a homomorphism.
Proposition 2.9.4. Let φ : G1 → G2 be a homomorphism, then
(1) If e is the identity of G1 , then φ(e) is the identity of G2 .
(2) For any elements g ∈ G1 , φ−1 (g) = φ(g −1 ).
(3) If H1 is a subgroup of G1 , then φ(H1 ) is a subgroup of G2 .
(4) If H2 is a subgroup of G2 , then φ−1 (H2 ) = {g ∈ G|Φ(g) ∈ H2 } is a

subgroup of G1 . Furthermore, if H2 is normal in G2 , then φ−1 (H2 ) is normal
in G1 .
Proof. (1) Suppose that e and e0 are the identities of G1 and G2 , respectively;
then e0 φ(e) = φ(e) = φ(ee) = φ(e)φ(e). By cancellation law, φ(e) = e0 .
(2) This statement follows from the fact that
φ(g −1 φ(g) = φ(g −1 g) = φ(e) = e0 .
(3) The set φ(H1 ) is nonempty since the identity of G2 is in φ(H1 ). Suppose
that H1 is a subgroup of G1 and let x and y be in φ(H1 ). There exist elements
a, b ∈ H1 such that φ(a) = x, φ(b) = y. Since
xy −1 = φ(a)(φ(b))−1 = φ(a) · φ(b−1 ) = φ(ab−1 ).
Because ab−1 ∈ H1 , therefore φ(ab−1 ) ∈ φ(H1 ). φ(H1 ) is a subgroup of G2 .
(4) Let H2 be a subgroup of G2 and define H1 to be φ−1 (H2 ); that is,

H1 is the set of all g ∈ G1 such that φ(g) ∈ H2 . The identity is in H1 since
φ(e) = e0 . If a and b are in H1 , then φ(ab−1 ) = φ(a)φ−1 (b) ∈ H2 since H2
CHAPTER 2. GROUP 59
is a subgroup of G2 . Therefore, ab−1 ∈ H1 and H1 is a subgroup of G1 . If

H2 is normal in G2 , we must show that g −1 hg ∈ H1 for h ∈ H1 and g ∈ G1 .
But φ(g −1 hg) = φ−1 (g)φ(h)φ(g) ∈ H2 since H2 is a normal subgroup of G2 .
Therefore, g −1 hg ∈ H1 . H1 is normal in G1 .
Definition 2.9.5. Let φ : G → H be a group homomorphism and suppose

that e is the identity of H. The kernel of φ is the subgroup {φ−1 (e)}. Denote
Kerφ = {x|φ(x) = eH , x ∈ G}.
Example 2.9.6. Let φ : Z → G : n 7→ g n be a map. If |G| =∝, Then

Kerφ = {0}. If |G| = n, then Kerφ = nz since (g n̄ = e).
Example 2.9.7. Let us consider the homomorphism
φ : GL2 (R) → R+ : A 7→ detA.
Because Φ(A) = e, A ∈ SL2 (R), Kerφ = SL2 (R).
Theorem 2.9.8. Let φ : G → H be a homomorphism. Then the kernel of φ

is a normal subgroup of G.
Proof. Since eH is the identity of H, eH is a normal subgroup of H. Kerφ =

φ−1 eH is also normal subgroup of G.
Definition 2.9.9. Let H be a normal subgroup of G. Define the natural or

canonical homomorphism. φ : G → G/H : g 7→ gH.
Remark 2.9.10. The natural homomorphism ψ is a homomorphism. Since
ψ(g1 g2 ) = ψ(g1 )ψ(g2 ) = g1 Hg2 H = g1 g2 H.
And Kerψ = H. Because the identity of G/H is H = eH = He. And

ψ −1 (eH) = ψ −1 (H) = H.
Example 2.9.11. Let N = {(1), (123), (132)} be the normal subgroup of S3 ,

ψ : S3 → S3 /H be the natural homomorphism. Then Kerψ = N .
CHAPTER 2. GROUP 60
Theorem 2.9.12. First Isomorphism Theorem for groups: If φ : G →

H is a group homomorphism with K = Kerφ, then K is normal in G. Let
ψ : G → G/H be the natural homomorphism. There exist an isomorphism
η : G/K → φ(G) such that φ = ηψ.
φ
G φ(G)
ψ η
G/K
Proof. K is normal by the Theorem2.9.8. Define
η : G/K → φ(G)
gK 7→ φ(g).
Let g1 K = g2 K. For any g1 k ∈ g1 K, there exists a g2 such that g2 = g1 k, and
η(g1 K) = φ(g1 ) = φ(g1 )φ(k) = φ(g1 k) = φ(g2 ) = η(g2 K).
Thus η is well defined.
For any φ(g) ∈ φ(G), there exists some gK ∈ G/K such that η(gK) =
φ(g), hence η is a surjection. Let η(g1 K) = η(g2 K). Then φ(g1 ) = φ(g2 ), and
e = φ−1 (g1 )φ(g2 ) = φ(g1−1 )φ(g2 ) = φ(g1−1 g2 ),
which means g1−1 g2 ∈ Kerφ. So we have g1−1 g2 K = K, and g1 K = g2 K.

Therefore, η is injective.
For any g1 K and g2 K in G/K,
η(g1 Kg2 K) = η(g1 g2 K) = φ(g1 g2 ).
Since φ is a homomorphism, φ(g1 g2 ) = φ(g1 )φ(g2 ) = η(g1 K)η(g2 K), which

shows that η preserves the operation. Now we have that η a well-defined
bijection that preserves the operation, thus satisfies the definition of isomor-
phism.
CHAPTER 2. GROUP 61
Let G be a group generated by g. Define a surjective homomorphism
ψ:Z→G
n 7→ g n
If the order of g is infinite. Kerψ = {0}. And Z/Kerψ = Z/0 = Z. If g has a

finite order m. Then Kerψ = {ml|l ∈ Z}. And Z/Kerψ = Z/mZ = Zm . We
have (
Z , |g| = ∞
ψ(Z) = G ∼ = Z/Kerψ =
Zm , |g| = m
Theorem 2.9.13. Second Isomorphism Theorem for groups: Let H be
a subgroup of a group G (not necessarily normal in G) and N be a normal
subgroup of G. Then HN is a subgroup of G, H ∩ N is a normal subgroup of
H, and
H/H ∩ N ∼ = HN/N.
Proof. Take h1 n1 and h2 n2 in N . Then
(h1 n1 )(h2 n2 ) = h1 h2 h−1 −1

2 n1 h2 h2 = h1 h2 (h1 n1 h2 )n2 .
Since N is a normal subgroup of G, h1−1 n1 h2 belongs to N . Thus, (h1 n1 )(h2 n2 )

is in HN , and HN is closed.
HN satisfies the associative law by the definition of group. The identity

of G is also the identity of HN . For any hn ∈ HN , the inverse
(hn)−1 = n−1 h−1 = h−1 hn−1 h−1 = h−1 (hn−1 h−1 )
is also contained in HN . Therefore, HN is a subgroup of G.
Let
ψ : H → HN/N
h 7→ hN
be a map. Since
ψ(h1 h2 ) = h1 h2 N = h1 h2 N N = (h1 N )(h2 N ) = ψ(h1 )ψ(h2 ),

CHAPTER 2. GROUP 62
ψ is surjective, which is to say HN/N = ψ(H). Kerψ ⊆ N and ψ is defined

on H, so Kerψ = H ∩ N . Consequently, we have H/H ∩ N ∼= HN/N by the
first isomorphism theorem.
Theorem 2.9.14. Correspondence Theorem for groups: Let N be a nor-

mal subgroup of a group G. Then H → H/N is a one-to-one correspondence
between {the set of subgroup of H containing N } and {the set of subgroup of
G/N } Furthermore, the normal subgroups of H correspond to normal subgroup
of G/N .
Let n be a positive integer. Then subgroups of Zn = Z/nZ are of the

form K/nZ, where nZ ≤ K ≤ Z. There is an integer m > 0 such that K =
hmi = mZ, and further m divides n since nZ ≤ K. Thus the Correspondence
Theorem tells us that the subgroups of Zn correspond to the positive divisors
of n.
Theorem 2.9.15. Third Isomorphism Theorem for groups: Let G be a

group and N and H be normal group of G with N ⊂ H. Then
G/N
G/H ∼
= .
H/N
Proof. Define φ : G/N −→ G/H by the rule φ(xN ) = xH. It is easy to verify
that φ is a well-defined homomorphism. Also Im(φ) = G/H and Ker(φ) =
H/N .
Example 2.9.16. Let m, n be positive integers. Then
(mZ + nZ)/nZ ∼
= mZ/(mZ ∩ nZ).
What does this tell us about the integers m, n? Fairly obviously mZ∩nZ = kZ,
where k is the least common multiple of m and n. Now mZ+nZ consists of all
ma + nb, where a, b ∈ Z. We see that this is just dZ where d = gcd(m, n). So
the assertion is that dZ/nZ ∼ = mZ/kZ. Now dZ/nZ ∼ = Z/ nd Z via the mapping
dx + nZ −→ x + nd Z. Similarly mZ/kZ ∼ = Z/ mk
Z. Thus Z/ nd Z ∼ k
= Z/ m Z. It
n k
follows that d = m or mn = dk since isomorphic groups have the same order.
CHAPTER 2. GROUP 63
2.10 Endmorphism and automorphism of groups
Definition 2.10.1. An automorphism of a group G is an isomorphism from

G to itself. An endmorphism of a group G is an homomorphism from G to
itself.
It is obvious that the set of all automorphisms of G is a subset of symmetric

group. Denote this subset as Aut(G).
Proposition 2.10.2. If G is a group, then Aut(G) is a subgroup of the sym-

metric group Sym(G).
Proof. Identity map Id is an automorphism, so Id ∈ Aut(G). If f, g ∈ Aut(G),

then (f g)(xy) = f (g(x)g(y)) = (f g)(x)(f g)(y), which shoes that f g ∈ Aut(G)
and Aut(G) is closed. If f ∈ Aut(G), then f −1 ∈ Aut(G) by f is an isomor-
phism.
Example 2.10.3. Let (G, +) be a group, define f : G −→ G : x 7→ −x. Then

f is an isomorphism since
f (x + y) = −(x + y) = −x − y = f (x) + f (y).
And f −1 = f .
Let f be an automorphism of group (Z, +). Now, we have f (1) = n for

some integer n. Note that f is completely determined by n since f (m) =
f (m1) = mf (1) = mn. Since f is surjective, f (x) = 1 for some integer x
. Then 1 = f (x) = f (x1) = xf (1) = xn. It follows that n = ±1. Hence
there are just two automorphism, namely, identity and f (x) = −x. Therefore
|Aut(Z)| = 2.
Let a be an element of a group G, the conjugate of x ∈ G by a is defined

by axa−1 . Define a map δa on G by the rule
δ(a)(x) = ax−1 , x ∈ G.
CHAPTER 2. GROUP 64
Since
δa (xy) = a(xy)a−1 = (axa−1 )(aya−1 ) = δa (x)δa (y).
We see that δa is a homomorphism. It is clearly that the inverse of δa is δa−1 .

Therefore δa is an automorphism of G. We call the automorphism as inner
automorphism induced by a. Thus we have a function
δ: G −→ Aut(A)
a 7−→ δa
We will show that δ is a homomorphism, called the conjugation homomorphis-

m, indeed
δab (x) = abx(ab)−1 = abxb−1 a−1 = (δa δb )(x).
Thus δ(ab) = δ(a)δ(b).
Proposition 2.10.4. Let G is a group. Then the subgroup Inn(G) is normal

in Aut(G).
Proof. Let a ∈ G, we will claim that f δa f −1 ∈ Inn(G) for any f ∈ Aut(G).

Assume x ∈ G,
(f δa f −1 )(x) = (f δa )(f −1 (x)) = f (af −1 (x)a−1 ) = f (a)xf (a−1 = δf (a) (x).
Let G be a group, c is called a center element if for any element x ∈ G,

cx = xc. Denote the set of center elements as Center(G). Applying the
First Isomorphism Theorem to the homomorphism δ. We will discover some
interesting conclusion about inner automorphism and center of group in the
following proposition.
Proposition 2.10.5. Let G be a group and δ : G −→ Aut(A) be the con-

jugation homomorphism. Then Ker(δ) = Center(G) and Im(δ) = Inn(G).
Hence Inn(G) ∼
= G/Center(G).
CHAPTER 2. GROUP 65
Proof. The image set of δ is the set of all inner automorphisms of G. Note
that a ∈ Ker(δ) if and only if δa (x) = x for all x ∈ G, i.e., axa−1 or ax = xa.
Thus the kernel of δ is the center of G.
Example 2.10.6. Let S3 be the symmetric group of 3 elements. Then Aut(S3 ) ∼ =

∼
S3 . Note that Center(S3 ) = {(1)}, then S3 = S3 /Center(S3 ) = Inn(S3 ). We
have |Inn(S3 )| = 6, hence |Aut(S3 )6| ≥ 6. Let a = (12), b = (13), c = (23),
X = {a, b, c}. Define a map
φ : Aut(S3 ) −→ SX ,
σ 7−→ fσ
!
a b c
where fσ = . Note that a, b are generators, so φ is injec-
σ(a) σ(b) σ(c)
tive. Thus |Aut(S3 )| ≤ |SA | = 6. Hence Aut(S3 ) = Inn(S3 ) =∼ = S3 .
Example 2.10.7. Let (Q)∗ be the set Q\{0}. We have (Q, +), (Q∗ , ·) are
groups. Then Aut(Q, +) ∼
= (Q∗ , ·). Let a ∈ Q∗ , define a map
φa : (Q, +) −→ (Q, +)
x 7−→ ax.
Then φa ∈ Aut(Q, +). Assume that Φ ∈ Aut(Q, +) such that Φ(1) = a ∈ Q∗ .

Then
Φ(m) = Φ(1) + · · · + Φ(1) = am, m ∈ N.
Then Φ(−m) = −am, m ∈ N. So Φ(n) = an, n ∈ Z. Let x = pq , p, q ∈ Z, we

have Φ(p) = ap and Φ(qx) = Φ(x) + · · · + Φ(x) = qΦ(x). So Φ(x) = ap
q = rx.
∗
Hence Φ = φa . That is Aut(Q, +) = {φa |a ∈ Q }. Define
η : Aut(Q, +) −→ (Q∗ , ·)
φa 7−→ a.
It is obvious that η is a bijection by above definition, and
η(φa φb ) = η(φab ) = ab = η(φa )η(φb ).
Therefore Aut(Q, +) ∼
= (Q∗ , ·).
Chapter 3
Rings
3.1 Rings
Definition 3.1.1. If a non-empty set R has two closed binary operations,

addition and multiplication, satisfying the following conditions, for a, b, c ∈ R
(1) a + b = b + a
(2) (a + b) + c = a + (b + c)
(3) There is an element 0 ∈ R such that 0 + a = a.
(4) There exists an element −a ∈ R such that a + (−a) = 0.
(5) (ab)c = a(bc).
(6) a(b + c) = ab + ac; (a + b)c = ac + bc.
(R, +, ·) is called a ring. If ab = ba, ∀a, b ∈ R, we call R an abelian ring.
In fact, a ring is an abelian group (R, +) together with a second binary

operation satisfying
(ab)c = a(bc), a(b + c) = ab + ac, (a + b)c = ac + bc.
It is obvious that Z, Q, R are rings.
66
CHAPTER 3. RINGS 67
Definition 3.1.2. If there is an element 1R ∈ R such that 1R 6= 0 and 1R a =

a1R , ∀a ∈ R, we say R is a ring with unit or identity.
Example 3.1.3. The continuous real-valued functions on an interval [a, b]

form a commutative ring by defining (f + g)(x) = f (x) + g(x) and (f g)(x) =
f (x)g(x).
Definition 3.1.4. A nonzero element a ∈ R is called a zero divisor if there

is a non-zero element b, such that ab = 0.
For example, 2, 3 ∈ Z6 , we have 2 · 3 = 0. So 2 and 3 is a zero divisor.

There is no zero divisor in Z, Q, R.
Definition 3.1.5. A commutative ring with identity is called an integral do-

main if for any a, b ∈ R such that ab = 0, either a = 0 or b = 0.
We have Z, Q, R and C are integral domains. Z6 is not a integral domain.
Definition 3.1.6. A divisor ring R is a ring with identity in which every

nonzero element in R is a unit, that is for each nonzero a ∈ R, there exists
an unique element a−1 ∈ R such that aa−1 = a−1 a = 1.
A commutative divisor ring is called a field.
It is well know that Q, R and C are fields. (Zn , +, ·) is a ring, but it is

neither an integral domain nor a divisor ring.
Example 3.1.7. (M2×2 (R), +, ·) is a ring with identity E2 , which is not

commutative. Moreover it is neither an integral domain nor a divisor ring,
thus it is not a field.
Example 3.1.8. Let Q8 be the quaternion group in Example2.1.8. We have
Q = {a + bI + cJ + dK|a, b, c, d ∈ R}
is a noncommutative divisor ring. Q is a ring under addition in (2.1) and

multiplication in (2.2), called the ring of quaternions.
CHAPTER 3. RINGS 68
Q is a divisor ring. In fact, for every a + bI + cJ + dK, we have a − bI −

cJ − dK such that
(a + bI + cJ + dK)(a − bI − cJ − dK) = a2 + b2 + c2 + d2 .
This element can be zero only if a, b, c and d are all zero. So if a+bI+cJ+dK 6=
0,
a − bI − cJ − dK
(a + bI + cJ + dK)( )=1
a2 + b2 + c2 + d2
Proposition 3.1.9. Let R be a ring with a and b in R. Then
(1) a0 = 0a = 0;
(2) a(−b) = (−a)b = −ab;
(3) (−a)(−b) = ab;
(4) ∀a ∈ R, −a = −1R a = a(−1R );
(5) ∀a, b ∈ R and n ∈ Z, n(ab) = (an)b = a(nb);
(6) (n1R )a = a(n1R ) = na;
(7) Given a1 , a2 , · · · , am and b1 , b2 , · · · , bn ∈ R,

m
X n
X
(a1 + a2 + · · · + am )(b1 + b2 + · · · + bn ) = ai bj .
i=1 j=1
Proof. (1) a0 = a(0 + 0) = a0 + a0. Solving the equation, we have a0 = 0.

Similarly, 0a = 0.
(2) Since ab + a(−b) = a(b + (−b)) = a0 = 0, a(−b) = −ab.
(3) Following from (2), (−a)(−b) = −(a(−b)) = −(−ab) = ab.
(4) a(−1R ) = −(a1R ) = −a = −(1R a).
(5) Prove by the mathematical induction. If n = 0, 0(ab) = 0 = (0a)b =

a(0b). Assume that it is true for m, which is to say m(ab) = (ma)b = a(mb).
Then (m + 1)ab = mab + ab = (ma + a)b = a((m + 1)b).
CHAPTER 3. RINGS 69
(6) (n1R )a = a(1R n) = na = n(a1R ) = na.
(7) Prove by the mathematical induction. When n = 1,
(a1 + a2 + · · · + am )b = (a1 + (a2 + · · · + am ))b

=a1 b + (a2 + · · · + am )b
=a1 b + a2 b + (a3 + · · · + am )b
=···
=a1 b + a2 b + · · · + am b
m
X
= ai b
i=1
Assume that it is true for s − 1. When n = s,
(a1 + a2 + · · · + am )(b1 + b2 + · · · + bs )
=(a1 + a2 + · · · + am )(b1 + b2 + · · · + bs−1 ) + (a1 + a2 + · · · + am )bs
m
X s−1
X m
X
= ai bj + ai bs
i=1 j=1 i=1
Xm Xs
= ai bj .
i=1 j=1
We have cancellation law in a group. Next proposition says there is can-

cellation law in an integral domain.
Proposition 3.1.10. Let D be a commutative ring with identity. The D is

an integral domain if and only if for all nonzero elements a ∈ D with ab=ac,
we have b=c.
Proof. let D be an integral domain, then D has no zero divisors. Suppose that
ab = ac and a 6= 0, then 0 = a · b − a · c = a · (b − c). Then b − c = 0, i.e. b = c.
Conversely, suppose that the cancellation law is true in D. That is, suppose
that ab = ac implies b = c. Let ab = 0. If a 6= 0, then ab = 0 = a · 0 implies
CHAPTER 3. RINGS 70
b = 0. Therefore, a can not be a zero divisor. So there is nonzero division, D

is an integral domain.
Proposition 3.1.11. Every finite integral domain is a field.
Proof. Let D be a finite integral domain, D∗ = D/{0}. For each a ∈ D∗ ,

define a map λa : D∗ → D∗ by b 7−→ ab for a ∈ D. Firstly, λa is well-defined
since if a 6= 0, b 6= 0, then ab 6= 0 by D is integral domain. Next, λa is injective.
Assume that λa (b) = λa (c), i.e. ab = ac, then b = c by the cancellation law.
It is obvious that λa is surjective since D∗ is a finite set, the map λa must
also be surjective. Hence, for some d ∈ D∗ , λa (d) = ad = 1, Therefore, a
has a left inverse. Since D is commutative, d must be a right inverse. for a.
Consequently, D is a field.
Definition 3.1.12. The characteristic of a ring R is the least positive inte-

ger n such that nr = 0, for any r ∈ R. If no such integer exists, then the
characteristic of R is defined to be 0. Denote as char(R)
Example 3.1.13. Zp is a field if p is prime. Then pa = 0, for a ∈ Zp , then

char(Zp ) = p. Moreover, we have char(Q) = char(R) = char(C) = 0.
Theorem 3.1.14. The characteristic of an integral domain is either prime or

zero.
Proof. Let D be an integral domain, suppose that char(D) = n, n 6= 0. If n

is not prime, then n = a · b, where 1 < a < n, 1 < b < n. Since 0 = n · 1 =
(a · b)1 = (a · 1)(b · 1), note that D is integral domain, a · 1 = 0 or b · 1 = 0 If
a · 1 = 0, then char(D) = a < n. Its contradiction. The same for b. Hence, n
must be prime.
3.2 Subrings and Ideals
Definition 3.2.1. Let (R, +, ·)be a ring. A subring S of R is a subset S of R

such that (S, +, ·) is a ring.
CHAPTER 3. RINGS 71
Example 3.2.2. The ring nZ is a subring of Z. Notice that even though the
original ring may have an identity, we do not require that its subring have an
identity. We have the following chain of subrings: Z ⊂ Q ⊂ R ⊂ C.
The following proposition gives us some easy criteria for determining whether
or not a subset of a ring is indeed a subring.
Proposition 3.2.3. Let R be a ring and S a subset of R. Then S is a subring

of R if and only if the following conditions are satisfied.
(1) S 6= Ø.
(2) r − s ∈ S for all r, s ∈ S.
(3) rs ∈ S for all r, s ∈ S.
Proof. If S is a subring, then (S, +) is a subgroup of (R, +), thus (1) and (2)
hold. (3) is clear because S is a subring.
If S satisfy (1), (2) and (3), we have S is subgroup under ”+” by (1) and
(2). (3) means that S is closed under multiplication. S is a subset of R, so S
is associative and distributive. Thus (S, +, ·) is a subring of R.
Example 3.2.4. Let

!
n a b o
R = M2 (R) = | a, b, c, d ∈ R
c d
and !
n p q o
T ri = | p, q, r ∈ R .
0 r
! !
u v p q
T ri is a nonempty set. Let , ∈ T , then
0 w 0 r
! ! !
u v p q up uq + vr
= ∈ T,
0 w 0 r 0 wr
! ! !
u v p q u−p v−q
− = ∈T
0 w 0 r 0 w−r
CHAPTER 3. RINGS 72
Then T is a subring of R.
Definition 3.2.5. A subring I of R is an ideal if ar, ra ∈ I, for a ∈ I, r ∈ R,

or equivalent rI ⊂ I, Ir ⊂ I.
Example 3.2.6. {0} and R are ideals of R. These two ideals are called trivial
ideals.
Proposition 3.2.7. A nonempty set I of R is an ideal of R if
(1) a, b ∈ I, then a − b ∈ I.
(2) a ∈ I, r ∈ I, then ar, ra ∈ I.
Proof. We have I is a commutative subgroup under ”+” by (1). I is a subring

by (1) and (2), and ar, ra ∈ I. Thus I of R is an ideal of R.
Example 3.2.8. (Z, + ,0). Take n 6= 0, n ∈ Z.
I = {rn | r ∈ Z} = {· · · , −2n, −n, 0, n, 2n · · · }
is an ideal of Z since ZI, IZ⊆I.
We have all ideals of (Z, +, 0) are nZ for every n 6= 0.
Example 3.2.9. Let

!
n a b o
R= |a, b, c ∈ Z ,
0 c
then
!
n 0 d o
I= |d ∈ Z
0 0
is an ideal of R.
Note that
!
n a b o
J= |a, b ∈ Z
0 0
CHAPTER 3. RINGS 73
is not an ideal of R. Let a1 , b1 , c, d, a2 , b2 ∈ Z, we have

! ! !
a1 b1 a2 b2 aa1 a2 b1 b2
= 6 J.
=
c d 0 0 ca2 db2
Example 3.2.10. R is a commutative, identity ring, a ∈ R,
I = hai = {ar | r ∈ R}
is an ideal of R. Since 0 = a0 ∈ hai and a = a1 ∈ hai, I is nonempty. If

ar1 ∈ I, ar2 ∈ I, then ar1 − ar2 = a(r1 − r2 ) ∈ I. If ar ∈ hai, s ∈ R, we have
s(ar) = as(r) ∈ hai. Thus < a > is an ideal.
Let R be a ring, ∀a ∈ R. Define
U = {x1 ay1 + x2 ay2 + · · · + xm aym + sa + at + na|xi , yi , s, t ∈ R, n ∈ Z},(3.1)
where i = 1, · · · m. U is an ideal. In fact, if u1 , u2 ∈ U, u1 ± u2 ∈ U. For

r ∈ R, u = x1 ay1 + x2 ay2 + · · · + xm aym + sa + at + na, then
r(x1 ay1 + x2 ay2 + · · · + xm aym + sa + at + na)

= rx1 ay1 + rx2 ay2 + · · · + rxm aym + rsa + rat + rna.
Note that rx1 , rx2 , · · · , rxm , rs, r, rn ∈ R, ru ∈ U. Similarly, ur ∈ U.
Definition 3.2.11. Let R be a ring, a ∈ R, an ideal of the form
hah= {x1 ay1 + x2 ay2 + · · · + xm aym + sa + at + na|xi , yi , s, t ∈ R, n ∈ Z}
is called a principal ideal.
Remarks 3.2.12. Let R be a ring, a ∈ R, then
(1) If R is a commutative ring, then hai = {ra + na|r ∈ R, n ∈ Z}.

P
(2) If R is a ring with identity, then hai = { i xi ayi |xi , yi ∈ R}.
(3) If R is a commutative ring with identity , then hai = {ra|r ∈ R}.

CHAPTER 3. RINGS 74
Let R be a ring, ∀a1 , a2 , · · · , am ∈ R. Define
I = {s1 + s2 + · · · + sm |si ∈ hai i}, (3.2)
Let u = s1 + s2 + · · · + sm , u0 = s01 + s02 + · · · + s0m ∈ I, then
u − u0 = (s1 − s01 ) + (s2 − s02 ) + · · · + (sm − s0m ) ∈ I.
Let r ∈ R, then rsi , rs0i ∈ hai i and
ru = rs1 + rs2 + · · · + rsm ∈ I, ur = s1 r + s2 r + · · · + sm r ∈ I
Thus I is an ideal of R. Denote I = ha1 , a2 , · · · , am i.
Theorem 3.2.13. There are only two ideals in a divisor ring i.e. trivial
ideals.
Proof. Assume that I is an ideal of R. For a 6= 0, a ∈ I, there exists a−1 ∈ R

such that a−1 a = aa−1 = 1. Thus for any b ∈ R, b = b1 ∈ I, that means
I = R.
Theorem 3.2.14. Every ideal in the ring of integers Z is a principle ideal.
Proof. If I = h0i = {0}, then I is a principle ideal. If I 6= {0}, then I ⊆ Z. I

is consisted by some integers in Z. Let n be the smallest positive integer of I
by the principle of wee-ordering. For all a ∈ I, there exist q, r ∈ Z, 0 ≤ r < n,
such that a = nq + r. That is r = a − nq ∈ I, but r = 0 since n is the least
positive element in I. Therefore a = nq. So I = hni.
Remark 3.2.15. Let n ∈ Z. Note that nZ is an ideal of Z. If na ∈ nZ, then

nab ∈ nZ, b ∈ Z. By [?], all ideals of Z are nZ. For example
h4i = {· · · − 8, −4, 0, 4, 8, · · · },
h2i = {· · · − 4, −2, 0, 2, 4, · · · }.
And h4i ∈ h2i.

CHAPTER 3. RINGS 75
We are fairly familiar with polynomials by the time we begin to study

algebra. Next, let us consider polynomial ideals. Let R be a commutative ring
with identity. Any expression of the form
f (x) = a0 + a1 x + a2 x2 · · · + an xn ,
where ai ∈ R and an 6= 0, is called a polynomial over R with indeterminate

x. The elements a0 , a1 , · · · , an are called the coefficients of f . The coefficient
an is called the leading coefficient. A polynomial is called monic if the leading
coefficient is 1. If n is the largest nonnegative number for which an 6= 0, we
say that the degree of f is n and write degf (x) = n. If no such n exists, that
is, if f = 0 is the zero polynomialthen the degree of f is defined to be −∞.
We will denote the set of all polynomials with coefficients in a ring R by R[x].
Example 3.2.16. Let p(x) = 3 + 3x3 and q(x) = 4 + 4x2 + 4x4 be polynomials
in Z12 [x]. Then
p(x) + q(x) = 7 + 4x2 + 3x3 + 4x4

p(x)q(x) = 0.
This example tells us that we can not expect R[x] to be an integral domain if
R is not an integral domain.
Let F be a field. Recall that a principal ideal in F [x] is an ideal hp(x)i

generated by some polynomial p(x), that is,
hp(x)i = {p(x)q(x)|q(x) ∈ F [x]}.
For example, the polynomial x2 ∈ F [x] generates the ideal hx2 i consisting of
all polynomials with no constant term or term of degree 1.
Theorem 3.2.17. Let F be a field. Then every ideal in F [x] is a principal

ideal.
Proof. Let I be an ideal of F [x]. If I is the zero ideal, the theorem is easily
true. Suppose that I is a nontrivial ideal in F [x], and let p(x) ∈ I be a nonzero
CHAPTER 3. RINGS 76
element of minimal degree. If degp(x) = 0, then p(x) is a nonzero constant

and 1 must be in I. Since 1 generates all of F [x], h1i = I = F [x] and I is
again a principal ideal. Now assume that degp(x) ≥ 1 and let f (x) be any
element in I. By the division algorithm there exist q(x) and r(x) in F [x] such
that f (x) = p(x)q(x) + r(x) and degr(x) < degp(x). Since f (x), p(x) ∈ I and
I is an ideal, r(x) = f (x) − p(x)q(x) is also in I. However, since we chose
p(x) to be of minimal degree, r(x) must be the zero polynomial. Since we can
write any element f (x) ∈ I as p(x)q(x) for some q(x) ∈ F [x], it must be the
case that I = hp(x)i.
Theorem 3.2.18. Let F be a field and suppose that p(x) ∈ F [x]. Then the
ideal generated by p(x) is maximal if and only if p(x) is irreducible.
Proof. Suppose that p(x) generates a maximal ideal of F [x]. Then hp(x)i is
also a prime ideal of F [x]. Since a maximal ideal must be properly contained
inside F [x], p(x) cannot be a constant polynomial. Let us assume that p(x)
factors into two polynomials of lesser degree, say p(x) = f (x)g(x). Since
hp(x)i is a prime ideal one of these factors, say f (x), is in hp(x)i and therefore
be a multiple of p(x). But this would imply that hp(x)i ⊂ hf (x)i, which is
impossible since hp(x)i is maximal.
Conversely, suppose that p(x) is irreducible over F [x]. Let I be an ideal

in F [x] containing hp(x)i. By Theorem 3.2.18, I is a principal idea, hence,
I = hf (x)i for some f (x) ∈ F [x]. Since p(x) ∈ I, it must be the case that
p(x) = f (x)g(x) for some g(x) ∈ F [x]. However, p(x) is irreducible; hence,
either f (x) or g(x) is a constant polynomial. If f (x) is constant, then I = F [x]
and we are done. If g(x) is constant, then f (x) is a constant multiple of I and
I = hp(x)i. Thus, there are no proper ideals of F [x] that properly contain
hp(x)i.
CHAPTER 3. RINGS 77
3.3 Ring homomorphisms
Definition 3.3.1. Let R and S be rings, a map φ : R −→ S is a ring homo-

morphism, if for all a, b ∈ R,
φ(a + b) = φ(a) + φ(b),

φ(ab) = φ(a)φ(b).
The kernel of a ring homomorphism to be the set
Kerφ = {a | φ(a) = 0, a ∈ R}. (3.3)
If φ : R −→ S is a bijection, then φ is called an isomorphism of rings. If there

is an isomorphism φ : R −→ S, we say R is isomorphic to S, denote R ∼ = S.
Example 3.3.2. Let φ : Z −→ Zn : a 7−→ a( mod n) be a map. Then φ is a

ring homomorphism, since
φ(a + b) = φ(a) + φ(b)

= (a + b)( mod n)
= a( mod n) + b( mod n)
= φ(a) + φ(b)
and
φ(ab) = ab( mod n) = a( mod n)b( mod n) = φ(a)φ(b).
Example 3.3.3. Let C[a, b] be the ring of continuous real-valued functions on

an interval [a, b]. For a fixed α ∈ [a, b], we define a map
φα : C[a, b] −→ R,
f 7−→ f (α).
This is a ring homomorphism since
φα (f + g) = (f + g)(α) = f (α) + g(α) = φα (f ) + φα (g),

φα (f g) = (f g)(α) = f (α)g(α) = φα (f )φα (g).
Ring homomorphisms of the type φα are called evaluation homomorphism.

CHAPTER 3. RINGS 78
Example 3.3.4. The zero homomorphism 0 : R −→ S : r 7−→ 0.

!
n a b o
Example 3.3.5. Let R = |a, b ∈ R be a subring of the matrix
−b a
ring M2 (R). Now define a map
φ : R −→ C,
!
a b
−→ a + bi.
−b a
Then φ is a ring homomorphism. In fact

! ! !
a b1 a2 b2 a a −b b a1 b2 + a2 b1
1 1 2 1 2
φ = φ
−b1 a1 −b2 a2 −a1 b2 − a2 b1 a1 a2 − b1 b2
= a1 a2 − b1 b2 + (a1 b2 + a2 b1 )i
= (a1 + b1 i)(a2 + b2 i)
! !
a b1 a b2
1 2
= φ )φ ,
−b1 a1 −b2 a2
! ! !
a1 b1 a2 b2 a1 + a2 b1 + b2
φ + = φ
−b1 a1 −b2 a2 −b1 − b2 a1 + a2
= a1 + a2 + (b1 + b2 )i
= (a1 + b1 i) + (a2 + b2 i)
! !
a b a b
1 1 2 2
= φ )+φ
−b1 a1 −b2 a2
Certainly φ is surjective; it is also injective since a + ib = 0 implies that

a = 0 = b. Therefore φ is an isomorphism and we obtain the interesting and
amazing fact that
R∼= C.
Thus complex numbers can be represented by real 2 × 2 matrices. In fact this
is one way to define complex numbers
Proposition 3.3.6. Let char(S) 6= 0, φ : R −→ S be a ring homomorphism.
(1) If R is a commutative ring, then φ(R) is commutative.

CHAPTER 3. RINGS 79
(2) φ(0R ) = 0S .
(3) Let 1R , 1S be the identities for R and S. If φ is surjective, then φ(1R ) =

1S .
(4) If R is a field and φ(R) 6= 0, then φ(R) is a field.
Proof. (1) Let a, b ∈ R and ab = ba. Then
φ(a)φ(b) = φ(ab) = φ(ba) = φ(b)φ(a).
S is commutative.
(2) We have
φ(0R ) = φ(0R + 0R ) = φ(0R ) + φ(0R ) = 2φ(0R ),
then φ(0R ) = 0S .
(3) For any a ∈ R, then
φ(a) = φ(a1R ) = φ(a)φ(1R ),
hence φ(1R ) = 1S by φ being surjective.
(4) Let a ∈ R, φ(a) ∈ S, then
φ(a)φ(a)−1 = 1S = φ(1R ) = φ(aa−1 ) = φ(a)φ(a−1 ).
Thus, for every φ(a) ∈ S, φ(a)−1 = φ(a−1 ).
Proposition 3.3.7. The kernel of any ring homomorphism φ : R → S is an

ideal in R.
Proof. Note that kerφ is a subgroup of R. For all a ∈ kerφ, r ∈ R, we have
φ(ra) = φ(r)φ(a) = φ(r)0 = 0,

φ(ar) = φ(a)φ(r) = 0φ(r) = 0.
So ra, ar ∈ kerφ.
CHAPTER 3. RINGS 80
Theorem 3.3.8. Let I be an ideal of R. The factor group R/I is a ring with
multiplication defined by
(r + I)(s + I) = rs + I. (3.4)
Proof. Let r + I, s + I ∈ R/I, then R/I is an abelian group since
(r + I)(s + I) = r + s + I.
We will show that the multiplication is well-defined. That is the product

(r + I)(s + I) = r + s + I is independent of the choice of coset. If r0 ∈
r + I, s0 ∈ s + I, then r0 s0 ∈ rs + I. Since r0 , s0 ∈ r + I, there exist a, b ∈ I such
that r0 = r + a and s0 = s + b. The multiplication
r0 s0 = rs + as + rb + ab.
is in r + I since as + rb + ab ∈ I.
The multiplication satisfy associative law and distributive laws because
((r + I)(s + I))(t + I) = (rs + I)(t + I) = (rs)t + I = rst + I,

(r + I)((s + I)(t + I)) = (r + I)(st + I) = r(st) + I = rst + I.
and
((r + I) + (s + I))(t + I) = (r + s + I)(t + I) = (r + s)t + I,

(r + I)((s + I) + (t + I)) = (r + I)(s + t + I) = r(s + t) + I.
Definition 3.3.9. Let R be a ring, and I be an ideal of R, ring R/I is called

factor ring or quotient ring.
Theorem 3.3.10. Let I be an ideal of R. The map ψ : R → R/I defined by

ψ(r) = r + I is a ring homomorphism of R onto R/I with kernel I.
CHAPTER 3. RINGS 81
Proof. It is obvious that ψ : R → R/I is a group homomorphism by 2.9.10.

Let r, s ∈ R, then
ψ(rs) = ψ(r)ψ(s) = (r + I)(s + I) = rs + I.
Thus ψ is a ring homomorphism. If r ∈ I, then
ψ(r) = r + I = I = 0 + I = 0̄ ∈ R/I.
Definition 3.3.11. The map ψ : R → R/I is called natural homomorphism

or canonical homomorphism.
Theorem 3.3.12. First Isomorphism Theorem for Rings: Let φ : R →

S be a ring homomorphism. Then Kerφ is an ideal of R. If ψ : R → R/kerφ is
the canonical homomorphism, then there exists an isomorphism η : R/Kerφ →
φ(R) such that φ = ηψ.
Proof. By the First Isomorphism Theorem for groups, there exist a well-
defined additive group homomorphism η : R/Kerφ → φ(R) defined by η(r +
Kerφ) = φ(r). And η preserve multiplication since
η((r + I)(s + I)) = η(rs + I) = φ(rs) = φ(r)φ(r),

η(r + I)η(s + I) = φ(r)φ(s).
We can use the proofs of the isomorphism theorems for groups. The iso-
morphisms constructed in these proofs still stand if we allow for the additive
notation. One has only to check that they are isomorphisms of rings.
Theorem 3.3.13. Second Isomorphism Theorem for rings: Let I be a

subring of a ring R and J an ideal of R. Then I ∩ J is an ideal of I, and
I/I ∩ J ∼
= (I + J)/J. (3.5)
CHAPTER 3. RINGS 82
Theorem 3.3.14. Third Isomorphism Theorem: Let R be a ring and I

and J be ideals of R where J ⊂ I. Then
R/J
R/I ∼
= . (3.6)
I/J
Theorem 3.3.15. Correspondence Theorem of rings: Let I be an ideal

of a ring R. Then S −→ S/I is a one-to-one correspondence between the set
of subrings S containing I and the set of subrings of R/. Furthermore, the
ideals of R containing I correspond to ideals of R/.
Proof. The correspondence between subgroups applies here. All one has to do
is verify that S is a subring if and only if S/I is. Assume that S is a subring
containing I as an ideal, then there is a factor ring S/I. Conversely, if S/I is
a subring of R/I, then S is a subring of R.
3.4 Maximal ideals and prime ideals
Definition 3.4.1. A proper ideal M of a ring R is a maximal ideal of R if

the ideal M is not a proper subset of any ideal of R except R itself.
That is, M is a maximal ideal if for any ideal I properly containing M ,

I = R.
The following theorem completely characterizes maximal ideals for com-

mutative rings with identity in terms of their corresponding factor rings.
Theorem 3.4.2. Let R be a commutative ring with identity and M an ideal

in R. Then M is a maximal ideal of R if and only if R/M is a field.
Proof. Let M be a maximal ideal in R. If R is a commutative ring, then R/M

must also be a commutative ring. Clearly, 1 + M acts as an identity for R/M .
We must also show that every nonzero element in R/M has an inverse. If
a + M is a nonzero element in R/M , then a ∈ / M . Define I to be the set
CHAPTER 3. RINGS 83
{ra + m|r ∈ R, m ∈ M }. We will show that I is an ideal in R. The set I is

nonempty since 0a + 0 = 0 ∈ I. If r1 a + m1 , r2 a + m2 ∈ I, then
(r1 a + m1 ) − (r2 a + m2 ) = (r1 − r2 ) + (m1 − m2 ) ∈ I.
Also, for any r ∈ R, rI ⊂ I. Hence, I is closed under multiplication and

satisfies the necessary conditions to be an ideal. Therefore, I is an ideal
properly containing M . Since M is a maximal ideal, I = R. consequently, by
the definition of I there must be an m ∈ M and an element b ∈ R such that
1 = ab + m. Therefore,
1 + M = ab + M = ba + M = (a + M )(b + M ).
Conversely, suppose that M is an ideal and R/M is a field. Since R/M is a

field, it must contain at least two elements: 0 + M = M and 1 + M . Hence,
M is a proper ideal of R. Let I be any ideal properly containing M . We
need to show that I = R. Choose a ∈ I but a ∈ / M . Since a + M is a
nonzero element in a field, there exists an element b + M ∈ R/M such that
(a + M )(b + M ) = ab + M = 1 + M . Consequently, there exists an element
m ∈ M such that ab + m = 1 and 1 ∈ I. Therefore, r1 = r ∈ I for all r ∈ R.
Consequently, I = R
Example 3.4.3. Let p be prim, pZ be an ideal in Z. Then pZ is a maximal

ideal since Z/pZ ∼
= Zp is a field.
Definition 3.4.4. A proper ideal P in a commutative ring R is called a prime

ideal if whenever ab ∈ P , then either a ∈ P or b ∈ P .
Example 3.4.5. It is easy to check that the set P = {0, 2, 4, 6, 8, 10} is an

ideal in Z12 . This ideal is prime and maximal ideal.
Proposition 3.4.6. Let R be a commutative ring with identity 1R . Then P

is a prime ideal in R if and only if R/P is an integral domain.
Proof. Assume that P is an ideal in R and R/P is an integral domain. Suppose

that ab ∈ P . If a + P, b + P ∈ R/P such that
(a + P )(b + P ) = 0 + P = P,
CHAPTER 3. RINGS 84
then either a + P = P or b + P = P . This means that either a ∈ P or b ∈ P ,

which shows that P must be prime.
Conversely, suppose that P is prime and
(a + P )(b + P ) = ab + P = 0 + P = P.
Then ab ∈ P . If a ∈
/ P , then b must be in P by the definition of a prime ideal.
Hence, b + P = 0 + P and R/P is an integral domain.
Example 3.4.7. Every ideal in Z is of the form nZ. The factor ring Z/nZ ∼ =
Zn is an integral domain only when n is prime. It is actually a field. Hence,
the nonzero prime ideals in Z are the ideals pZ for p a prime.
Since every field is an integral domain, we have the following corollary.
Corollary 3.4.8. Every maximal ideal in a commutative ring with identity is

also a prime ideal.
3.5 Extension fields
It is natural to have the rational numbers inside the real numbers, while in
turn, the real numbers live inside the complex numbers. We can also study the
fields between Q and R. For example, if we are given a polynomial p(x) ∈ Q[x],
we can ask whether or not we can find a field E containing Q such that p(x)
factors into linear factors over E[x]. For example, if we consider the polynomial
p(x) = x2 − 2. It is obvious that p(x) = x2 − 2 is irreducible in Q[x]. If we
wish to find a zero of p(x), we must go to a larger field. Certainly the field of
√ √
real numbers will work, since p(x) = (x − 2)(x + 2). It is possible to find
a smaller field in which p(x) has a zero, namely
√ √
Q( 2) = {a + b 2|a, b ∈ Q}.
Definition 3.5.1. Let E is a field. A subfield F is a subset of E and F is a

field. A field E is an extension field of a field F if F is a subfield of E. The
field F is called the base field. We write F ⊂ E or E/F .
CHAPTER 3. RINGS 85
Given a field extension E/F , the larger field E can be considered as a

vector space over F . The elements of E are the vectors and the elements of F
are the scalars. For example, C = {a + bi|a, b ∈ Q, i2 = −1} is a vector space
of Q, and C is a extension field of Q.
Definition 3.5.2. If an extension field E of a field F is a finite dimensional

vector space over F of dimension n, then we say that E is a finite extension
of degree n over F . We write [E : F ] = n to indicate the dimension of E over
F.
Example 3.5.3. Let

√ √
Q( 2) = {a + b 2|a, b ∈ Q}.
√ √ √
Then Q( 2) is a extension field of Q, the basis of Q( 2) over Q is {1, 2},
and
√
[Q( 2) : Q] = 2.
Example 3.5.4. If we consider the polynomial p(x) = x4 − 5x2 + 6 = (x2 −

2)(x2 − 3) ∈ Q. (x2 − 2) and (x2 − 3) are irreducible in Q. Let
√ √ √ √
E = Q( 2, 3) = {a + b 3|a, b ∈ Q( 2)}.
Then p(x) is reducible in E. In this case, E is a extension field of Q, the basis

√ √ √ √
of Q( 2, 3) over Q( 2) is {1, 3}, and
√ √ √
[Q( 2, 3) : Q( 2)] = 2.
Furthermore,
√ √ √ √ √ √ √
Q( 2, 3) = Q( 2)( 3) = {a + b 2 + c 3 + d 6|a, b, c, d ∈ Q},
then
√ √
[Q( 2, 3) : Q] = 4.
Example 3.5.5. Let p(x) = x2 + x + 1 ∈ Z2 [x]. Since neither 0 nor 1 is a root

of this polynomial, we know that p(x) is irreducible over Z2 . We will construct
a field extension of Z2 containing an element α such that p(α) = 0. By
CHAPTER 3. RINGS 86
Theorem3.2.18, the ideal hp(x)i generated by p(x) is maximal; hence, Z2 /hp(x)i

is a field. Let f (x) + hp(x)i be an arbitrary element of Z2 /hp(x)i. By the
division algorithm,
f (x) = (x2 + x + 1)q(x) + r(x),
where the degree of r(x) is less than the degree of x2 + x + 1. Therefore,
f (x) + hx2 + x + 1i = r(x) + hx2 + x + 1i.
The only possibilities for r(x) are then 0, 1, x and 1 + x. Consequently, E =

Z2 /hp(x)i is a field with four elements and must be a field extension of Z,
containing a zero α of p(x). The field Z2 (α) consists of elements
0 + 0α = 0,
1 + 0α = 1,
0 + 1α = α,
1 + 1α = 1 + α.
Notice that α2 + α + 1 = 0. Hence, if we compute
(1 + α)2 = (1 + α)(1α) = 1 + α + α + (α)2 = α.
Other calculations are accomplished in a similar manner. We summarize these

computations in the following tables, which tell us how to add and multiply
elements in E.
Table 3.1: Addition Table of Z2 (α)

+ 0 1 α 1+α
0 0 1 α 1+α
1 1 0 1+α α
α α 1+α 0 1
1+α 1+α α 1 0
CHAPTER 3. RINGS 87
Table 3.2: Multiplication Table of Z2 (α)

· 0 1 α 1+α
0 0 0 0 0
1 0 1 α 1+α
α 0 α 1+α 1
1+α 0 1+α 1 α
Theorem 3.5.6. Telescope Formula: If E is a finite extension of F and K

is a finite extension of E, then K is a finite extension of F and
[K : F ] = [K : E][E : F ]. (3.7)
Proof. Let {α1 , · · · , αn } be a basis for E as a vector space over F and {β1 , · · · , βm }
be a basis for K as a vector space over E. We claim that {αi βj } is a basis for
K over F . We will first show that these vectors span K. Let u ∈ K. Then
u= m
P Pn
j=1 bj βj and bj = i=1 aij αi , where bj ∈ E and aij ∈ F . Then
m X
X n X
u= ( aij αi )βj = aij (αi βj ).
j=1 i=1 i,j
So the mn vectors αi βj must span K over F .
We must show that {αi βj } are linearly independent. Suppose that there
exist cij ∈ F such that
X X
u= cij (αi βj ) = (cij αi )βj ) = 0.
i,j i,j
Since the βj0 s are linearly independent over E, it must be the case that
X
cij αi = 0,
i,j
for all j. However, the αj are also linearly independent over F . Therefore,
cij = 0 for all i and j, which completes the proof.
CHAPTER 3. RINGS 88
Corollary 3.5.7. If Fi is a field for i = 1, · · · , k and Fi+1 is a finite extension

of Fi , then Fk is a finite extension of F1 and
[Fk : F1 ] = [Fk : Fk−1 ] · · · [F2 : F1 ] (3.8)
Definition 3.5.8. An element α in an extension field E over F is algebraic

over F if f (α) = 0 for some nonzero polynomial f (x) ∈ F [x]. An element in
E that is not algebraic over F is transcendental over F . An extension field E
of a field F is an algebraic extension of F if every element in E is algebraic
over F .
If E is a field extension of F and α1 , · · · , αn are contained in E, we denote

the smallest field containing F and α1 , · · · , αn by F (α1 , · · · , αn ). If E = F (α)
for some α ∈ E, then E is a simple extension of F .
√
Both 2 and i are algebraic over Q since they are zeros of the polynomials
x2 − 2 and x2 + 1, respectively. Clearly π and e are algebraic over the real
numbers. However, it is a nontrivial fact that they are transcendental over Q.
Numbers in R that are algebraic over Q are in fact quite rare. Almost all real
numbers are transcendental over Q. In many cases we do not know whether
or not a particular number is transcendental; for example, it is still not known
whether π + e is transcendental or algebraic.
A complex number that is algebraic over Q is an algebraic number. A

transcendental number is an element of C that is transcendental over Q.
p √
Example 3.5.9. We will show that 2 + 3 is algebraic over Q. If α =
p √ √ √
2 + 3, then α = 2 + 3. Hence, α − 2 = 3 and (α2 − 2)2 = 3.
2 2
Since α4 − 4α2 + 1 = 0, it must be true that α is a zero of the polynomial

x4 − 4x2 + 1 ∈ Q[x].
3.6 Field automorphisms
Proposition 3.6.1. The set of all automorphisms of a field F is a group

under composition of functions.
CHAPTER 3. RINGS 89
Proof. If φ and ψ are automorphisms of F , then so are φψ and φ−1 . The

identity is certainly an automorphism, hence, the set of all automorphisms of
a field F is indeed a group.
Proposition 3.6.2. Let E be a field extension of F . Then the set of all

automorphisms of E that fix F is a group; that is, the set of all automorphisms
φ : E → E such that φ(α) = α for all α ∈ F is a group.
Proof. We need only show that the set of automorphisms of E that fix F is
a subgroup of the group of all automorphisms of E. Let φ and ψ be two
automorphisms of E such that φ(α) = α and ψ(α) = α for all α ∈ F . Then
φψ(α) = φ(α) = α and φ−1 (α) = α. Since the identity fixes every element of
E, the set of automorphisms of E that leave elements of F fixed is a subgroup
of the entire group of automorphisms of E.
Proposition 3.6.3. Q be the field of rational field. Then Aut(Q) = {Id}.
Proof. Let σ be an automorphism of Q. Under isomorphism, the identity el-

ement corresponds to identity element and so for the inverse element and the
negative element. So we have σ(1)=1, σ(2)=σ(1+1)=σ(1)+σ(1)=2. Gener-
ally, σ(m)=m, σ(−m)=−m (m ∈ N+). Obviously, σ(m−1 )=m−1 , σ( m n n
)= m
(m 6= 0), which means σ is the identity element of Q. So Q only has identical
automorphism.
Proposition 3.6.4. Let Q be the field of rational number. Show that the field
Q(i) = {a + bi|a, b ∈ Q} has and only has two automorphisms.
Proof. Obviously, identity transformation and transformation σ : a + bi −→

a − bi are two automorphisms for Q(i). Now let τ be another automorphism
of Q(i). According to 2, for ∀ a ∈ Q, τ (a) = a. Moreover, let τ (i)=c + di
(d 6= 0), then −1=τ (i2 )=(c + di)2 =c2 − d2 + 2cdi. There exist c = 0, d = ±1.
If d = 1, since τ (a)=a, τ (i)=i, hence τ (a + bi)=a + bi, which means that τ
is an identity automorphism. Conversely, if d = −1, τ (a + bi)=a − bi, that is
τ =σ. So Q(i) has and only has two automorphisms.
CHAPTER 3. RINGS 90
Proposition 3.6.5. Show that automorphisms of the field of real numbers

only exist identity automorphisms.
Proof. Suppose that τ is an arbitrary automorphism of the field of real num-

n n
bers R. According to the second question, τ ( m )= m . Then suppose that a
is a real number and a>0. Hence τ (a)>0. Since we have supposed a = b2 ,
τ (a)=τ (b2 )=τ (b)2 >0. If a>c, a−c>0, so τ (a−c)=τ (a)−τ (c)>0, so τ (a)>τ (c).
Now suppose that a is an arbitrary real number. Then there exist rational
numbers r and s, such that r<a<s. From the previous proof, τ (r)<τ (a)<τ (s),
that is r<τ (a)<τ (s). In other words, for all rational number r and s satisfying
r<a<s, there must exist r<τ (a)<s. So τ (a)=a, which means τ is an arbitrary
identity automorphism.
Let E be a field extension of F . We will denote the full group of automor-

phisms of E by Aut(E). We define the Galois group of E over F to be the
group of automorphisms of E that fix F elementwise; that is,
G(E/F ) = {φ ∈ Aut(E)|φ(α) = α, α ∈ F }.

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Chapter 1

A set is a well-defined collection of objects. The objects that belong to a

We will denote sets by capital letters such as A, B, C. We write a ∈ A to

A set X is usually given by listing all elements in it, such as

If there are infinite elements in X, we write |X| = ∞

N is the set of all natural numbers {0, 1, 2, 3, · · · }.

Z is the set of all integers {· · · , −2, −1, 0, 1, 2, · · · }.

Z+ is the set of all integer {1, 2, 3, · · · }.

R is the set of all real numbers.

C is the set of all complex numbers.

A is contained in B or that B contains A if every element of A is also an

We have A = B if and only if A ⊆ B and B ⊆ A. If B is a subset of A

Remark 1.1.1. (1) A and Ø are subsets of a set A.

(2) Ø is a proper subset of a set A if and only if A 6= Ø.

Proposition 1.1.2. Let A and B be sets. Then

(1) A ∪ B = A if and only if B ⊆ A.

(2) A ∪ B = Ø if and only if A = Ø and B = Ø.

Proposition 1.1.3. Let A and B be sets. Then

(1) Commutative law of union:

(2) Commutative law of intersection:

(3) Associative law of union:

(4) Associative law of intersection:

(5) Distributive law of intersection with respect to union:

(6) Distributive law of union with respect to intersection:

Proposition 1.1.4. Let A be a subset of set X. Then

Proposition 1.1.5. Let A, B and C be sets. Then

(1) A\(B ∪ C) = (A\B) ∩ (A\C).

(2) A\(B ∩ C) = (A\B) ∪ (A\C).

(3) (B ∪ C)\A = (B\A) ∪ (C\A).

(4) (B ∩ C)\A = (B\A) ∩ (C\A).

Proof. We will prove (1) and leave (2)-(4) to readers.

(1) Let x ∈ A\(B ∪ C). That means x ∈ A and x ∈ / (B ∪ C). That is x ∈ A

On the other hand, let x ∈ (A\B)∩(A\C), i.e., x ∈ (A\B) and x ∈ (A\C).

(2) and (3) in next proposition are called De Morgan’s laws.

Proposition 1.1.6. Let A and B be subsets of a set X. Then

(1) A ⊆ B if and only if B 0 ⊆ A0 .

Example 1.1.10. If X = Ø, then P (X) = {Ø}. If X = {Ø}, then P (X) =

Example 1.1.11. If X = {1, 2, 3}, then

Definition 1.2.1. A map f consists of a nonempty set X, of a nonempty set

X is called the domain of map f , Y is called the codomain of map f , f (x) is

It is obvious that the set of images is contained in Y . The set of preimages

Example 1.2.2. We have f : N → N : x 7→ x2 is a map. f : N → Z : x 7→ −x

Let us consider f : Q → Z, where Q = {x|x = pq |p, q ∈ Z}, define f ( pq ) = p,

There are some particular maps here.

Let X be a nonempty set. The identity map on X, denoted by 1x (or idx , or

If A is a nonempty subset of X, the conclusion map of A in X is the function

Note that if A is a nonempty proper subset of X. 1a and ia have different

Let f : X → Y be a map and A be a nonempty subset of X. The map

Let X be a nonempty set and let y be a fixed element of Y . The map

Example 1.2.3. Let f1 : N → N : x 7→ 2x. Im(f1 ) = {y|∃ x ∈ N , such that y =

f2 : N → N : x 7→ 2x + 1. Im(f2 ) = {y|∃ x ∈ N , such that y = 2x + 1} is

Example 1.2.4. Let f3 : Z → Z : x 7→ 3x, f4 : Z → Z : x 7→ −3x. Then

If f5 : Q → Q : x 7→ 3x, then Im(f5 ) = Q .

Definition 1.2.5. Let f : X → Y be a map, f is said to be a surjective map

It is equivalent that f is surjective if and only if for any y ∈ Y , ∃x ∈ X,

Definition 1.2.6. Let f : X → Y be a map, f is said to be injective or

Definition 1.2.7. Let f : X → Y is called bijective if f is surjective and

We have that f is bijective if and only if for any y ∈ Y , there exists an

If there exists a bijective map f : X → Y , we call X is isomorphic to Y ,

Example 1.2.8. Let