CAO HUY LINH
ABSTRACT ALGEBRA
COLLEGE OF EDUCATION - HUE UNIVERSITY
Huế, May 05, 2013.
This lecture note is written by Cao Huy Linh, Senior Lecturer of Department of Mathematics , College of Education - Hue University. This lecture
note is used for teaching and learning the course
Algebra of Val de Loire program.)
NOTATIONS
Ký hiệu
N
N0 = N ∪ {0}
Z
Q
R
C
Zn
Z∗n
Z(G)
Card S
cấp G
(G : H)
¢
⊕
<P >
< x 1 , . . . , xn >
G/N
Nghĩa ký hiệu
Set of natural numbers
Set of integers
Set of rational numbers
Set of real numbers
Set of complex numbers
integers modulo m n
Group of invertible elements of Zn
Center of a group G
The number of elements of S
Order of a group G
Index of G over H
normal subgroup
Direct sum
Submodule generated by P
Submodule generated by x1 , . . . , xn
Quotient group of G over N
ii
PREFACE
iii
iv
CONTENTS
NOTATIONS
ii
PREFACE
iii
1 Binary relation
1
1
Binary Operations: Definitions and Examples . . . . . . . . . .
1
2
equivalence relations . . . . . . . . . . . . . . . . . . . . . . . .
4
3
order relations . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
2 Chapter 1
9
1
Binary Operations . . . . . . . . . . . . . . . . . . . . . . . . .
2
Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
3
Subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
4
cyclic groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
5
Normal subgroups and quotient groups . . . . . . . . . . . . . . 27
6
Group homomorphisms . . . . . . . . . . . . . . . . . . . . . . . 33
7
Direct product of groups . . . . . . . . . . . . . . . . . . . . . . 41
3 Rings
9
45
1
Rings and Fields . . . . . . . . . . . . . . . . . . . . . . . . . . 45
2
Subrings, Ideals and quotient rings . . . . . . . . . . . . . . . . 50
3
Ring homomorphisms . . . . . . . . . . . . . . . . . . . . . . . . 56
4
Characteristic of rings . . . . . . . . . . . . . . . . . . . . . . . 61
5
Field of fractions of a integral domain. . . . . . . . . . . . . . . 64
v
4 Application
67
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69
vi
Chương 1
Binary relation
§1
BINARY OPERATIONS: DEFINITIONS AND EXAMPLES
A relationship between two objects is something like
- "x likes y";
- "x is greater than y";
- "x and y have the same color";
- "x2 + y 2 = 4.
Exactly, we have the following definition.
Definition 1.1. A binary relation from a non-empty set X to a non-empty
set Y is a subset R of the product X × Y ; that is R ⊂ X × Y . The set X is
called the domain of the relation and Y is called the codomain.
A binary relation R on a non-empty set X is a subset of the product
X × X. For simplicity, we write (X, R) for the binary relation R on X.
Let R be a binary relation from X to Y and (x, y) ∈ X × Y . If (x, y) ∈ R,
we say x is related to y by R and write xRy. If (x, y) ∈
/ R, we write xR̄y.
Definition 1.2. (i) Let R be a binary from X to Y and S a binary from Y
to Z. The binary relation R ◦ S from X to Z is defined by
R ◦ S = {(x, z) ∈ X × Z | ∃y ∈ Y such that xRy and ySz}.
1
2
Chương 1. Binary relation
We denote by R2 the binary relation R ◦ R , i.e. the product of the relation R
with itself.
(ii) Let R be a binary from X to Y . The converse relation (or reverse
relation) of R is the binary relation R−1 ⊂ Y × X defined by
yR−1 x ⇐⇒ xRy.
Example 1.3. (1) Let X = {Duc, T ri, Dung} and Y = {M ai, Lan, Cuc, Hue}.
For x ∈ X, y ∈ Y , xRy if x likes y. Then, R is a binary relation from X to Y .
In this example, if
R = {(Duc, Lan), (Duc, Cuc), (T ri, M ai), (Dung, Lan), (Dung, Cuc)},
then Duc likes Lan and Cuc, Tri likes Mai, Dung likes Lan, Dung likes Cuc.
(2) Let X be a set of students, say X = {Ann, Bev, Carl, Dougg} and
Y be a set of courses, say Y = {M ath, History, English}. Then X × Y has
12 elements. An example of a binary relation R ⊂ X × Y is the set of pairs
(x, y) ∈ X × Y for which "x is enrolled in y." Another example is the relation
S defined by "xSy if x received an A grade in Y". In this example, we have
S ⊂ R , i.e., xSy ⇒ xRy.
Example 1.4. (1) Let X = {1, 2, 3, 4, 5, 6}. Define a binary relation R on X
as follows: xRy if x − y is divisible by 3. Then,
R = {(1, 4), (4, 1), (2, 5), (5, 2), (3, 6), (6, 3)}.
.
(2) Given m is a positive integer and define xRy if (x − y)..m; that is,
x ≡ y mod m, for all x, y ∈ Z. Then, (Z, R) is a binary relation and it is said
to be a congruence relation.
Example 1.5. (1) (N, ≤), (Z, ≤), (Q, ≤), (R, ≤) are binary ralations.
(2) (N, <), (Z, <), (Q, <), (R, <) are binary ralations.
(3) (N, =), (Z, =), (Q, =), (R, =) are binary ralations.
(4) (N, ≥), (Z, ≥), (Q, ≥), (R, ≥) are the converse ralations of ≤.
Example 1.6. Let denote P(X) by the set of subsets of X. Then, (P(X), ⊂)
is a binary relation on P(X). The converse relation of ⊂ is (P(X), ⊃).
. Cao
Huy Linh -College of Education-Hue University
§ 1. Binary Operations: Definitions and Examples
3
Example 1.7. Let a and b be integers. We say that a divides b is denoted
a | b , provided we have an integer m such that b = am . In this case we can
also say the following:
• b is divisible by a;
• a is a factor of b;
• a is a divisor of b;
• b is a multiple of a.
Then (Z, |) is a binary relation on Z.
Example 1.8. Let E be a K-vector space and F a subspace of E. For x, y ∈ E,
we define x ∼ y if x − y ∈ F . Then, (E, ∼) is a binary relation on E.
Example 1.9. Let f : X −→ Y be a map.
(1) Define G as follows: G := {(x, y) ∈ X × Y | y = f (x)}. The set G is
said to be a graph of f . Then G is a binary relation from X to Y , where xGy
means that y = f (x).
(2) Define x ∼ x0 to mean f (x) = f (x0 ). Then, ∼ is a binary relation on
X.
Definition 1.10. A binary relation R on a set X is said to be
(i) reflexive if ∀x ∈ X : xRx;
(ii) symmetric if ∀x, y ∈ X : xRy ⇒ yRx;
(iii) antisymmetric if ∀x, y ∈ X : [xRy ∧ yRx] ⇒ x = y;
(iv) transitive if ∀x, y, z ∈ X : [xRy ∧ yRz] ⇒ xRy;
(v) complete if ∀x, y ∈ X : xRy ∨ yRx.
Example 1.11. Let X be a set of people. Each of the following is a binary
relation on X:
(1) xN y if x lives next door to y. Then N would typically be symmetric,
and not transitive;
(2) xBy if x lives on the same block as y. Then B would typically be
reflexive, symmetric, and transitive.
. Cao
Huy Linh -College of Education-Hue University
4
Chương 1. Binary relation
.
Example 1.12. Let (Z, R) be a binary relation with xRy if (x − y)..m for all
x, y ∈ Z. Then R satisfies reflexive, symmetric and transitive properties.
Example 1.13. Let (R, ≤) be a binary relation as Exxamlpe 1.5. Then ≤
satisfies reflexive, antisymmetric and transitive properties
§2
EQUIVALENCE RELATIONS
Definition 2.1. A binary relation R on a non-empty set X is said to be an
equivalence relation if R is reflexive, symmetric and transitive.
.
Example 2.2. (1) Let (Z, R) be a binary relation with xRy if (x − y)..m for
all x, y ∈ Z. Then R is an equivalence relation on Z.
(2) Let f : X → Y be a map. For all x, y ∈ X, define x ∼ y if f (x) = f (y).
Then, (X, ∼) is an equivalence relation on X.
(3) Let E be a K-vector space and F a subspace of E. For all x, y ∈ E,
x ∼ y if x − y ∈ F . Clearly, (E, ∼) is an equivalence relation on E.
(4) For x = (a, b), y = (c, d) ∈ Z × Z∗ , define x ∼ y if ad = bc. It is not
dificult to check that ∼ is an equivalence relation on Z × Z∗ .
Definition 2.3. Let R be an equivalence relation on X and a ∈ X.Then
ā = {x ∈ X | aRx} ⊂ X
is said to be an equivalence class of a.
Proposition 2.4. Let R be an equivalence relation on X and a, b ∈ X. Then
(i) If b ∈ ā, then ā = b̄;
(ii) If ā ∩ b̄ 6= ∅, then ā = b̄.
S
(iii) X = x∈X x̄.
Proof. (i) For any x ∈ ā, one has xRa. Since b ∈ ā, aRb. From R is an
equivalence relation on X, it follows xRb; that is x ∈ b̄. Hence ā ⊂ b̄. Similarly,
we can prove b̄ ⊂ ā. So, ā = b̄.
(ii) Since ā ∩ b̄ 6= ∅, there exists x ∈ ā ∩ b̄. By (i), ā = x̄ and b̄ = x̄. This
implies ā = b̄.
. Cao
Huy Linh -College of Education-Hue University
5
§ 3. order relations
From Proposition 2.4, if R is an equivalence relation on X and a, b ∈ X,
then we have ā = b̄ or ā ∩ b̄ = ∅. This leads to the following definition.
Definition 2.5. Let R be an equivalence relation on X. Set of equivalence
classes on X is said to be a quotient set of X with respect to R, and it is
denoted by X/R.
Example 2.6. (1) Consider an equivalence relation R on Z with xRy if
.
(x − y)..2 for all x, y ∈ Z. Then
.
0̄ = {x ∈ Z | x..2} = 2Z = 2̄ = 4̄ = · · ·
.
1̄ = {x ∈ Z | (x − 1)..2} = 3̄ = 5̄ = · · ·
Thus the quotient set
X/R = {0̄, 1̄}.
.
(2) In general, if R is an equivalence relation on Z with xRy if (x − y)..m.
Then
ā = {a + λm | λ ∈ Z}.
The quotient set of X with respect to R is
X/R = {0̄, 1̄, 2̄, ..., m − 1}.
In this case, X/R is denoted by Zm . Zm is said to be the set of integers modulo
m. For example,
Z2 = {0̄, 1̄}; Z3 = {0̄, 1̄, 2̄}; Z6 = {0̄, 1̄, 2̄, 3̄, 4̄, 5̄};
Zm = {0̄, 1̄, 2̄, ..., m − 1}.
§3
ORDER RELATIONS
Definition 3.1. A binary relation R on a non-empty set X is said to be an
order relation if R is reflexive, antisymmetric and transitive.
. Cao
Huy Linh -College of Education-Hue University
6
Chương 1. Binary relation
Example 3.2. (1) (N, ≤), (Z, ≤), (Q, ≤), (R, ≤) is order relations;
(2) Let X be a set. Then (P(X), ⊂) is an order relation on P(X);
(3) Let X ⊂ N. Then (X, | ) is an order relation on X.
Definition 3.3. Let (X, ≤) be an order relation and A ⊂ X.
(i) An element a ∈ A is said to be the maximum of A if ∀x ∈ A : x ≤ a,
denote max(A) = a;
(ii) An element a ∈ A is said to be the minimum of A if ∀x ∈ A : a ≤ x,
denote min(A) = a;
(iii) An element a ∈ X is said to be an upper bound of A if ∀x ∈ A : x ≤ a;
(iv) An element a ∈ X is said to be a lower bound of A if ∀x ∈ A : a ≤ x;
(v) The maximum of upper bounds is said to be a supremum of A, denote
Sup(A);
(vi) The minimum of lower bounds is said to be a infimum of A, denote
Inf(A);
(vii) An element a ∈ A is said to be the maximal of A if
∀x ∈ A : a ≤ x ⇒ a = x;
(viii) An element a ∈ A is said to be the minimal of A if
∀x ∈ A : x ≤ a ⇒ a = x.
Example 3.4. Consider an oreder relation ≤ on R.
(1) If A = [1, 2) ⊂ R, then
- the minimal element of A = min A = Inf(A) = 1;
- there doesn’t exist any maximal element of A;
- there doesn’t exist any maximum element of A;
- Sup(A) = 2.
(2) If B = [1, 2) ∪ {3}, then
- the minimal element of A = min A = Inf(A) = 1;
- the maximal element of A = max A = Sup(A) = 3;
Example 3.5. (1) Consider an order relation (N, | ) and A = {1, 2, 3, 4, 5, 6}.
- The minimal element of A = minimum of A = Inf(A) = 1;
. Cao
Huy Linh -College of Education-Hue University
§ 3. order relations
- Maximal elements of A are 5 and 6;
- There doen’t exist a maximum of A;
- Sup(A) = 30.
(2) Consider an order relation (N, | ) and B = {1, 2, 3, 4, 6, 12}.
- The minimal element of A = the minimum of A = Inf(A) = 1;
- The maximal element of A = the maximum of A = Sup(A) = 12;
Example 3.6. Let X = {a, b, c} and consider an oreder relation (P(X), ⊂).
(1) If take A = P(X), then
- the minimal element of A = min(A) = Inf(A) = ∅;
- the maximal element of A = max(A) = Sup(A) = X.
(2) If take B = {{a}, {b}, {c}, {a, b}, {a, c}, {b, c}}, then
- the minimal elements B are {a}, {b}, {c};
- there doesn’t exist a minimum of B;
- Inf(B) = ∅;
- the maximal elements of B are {a, b}, {a, c}, {b, c};
- there doesn’t exist a maximum of B;
- Sup(B) = X.
. Cao
Huy Linh -College of Education-Hue University
7
8
Chương 1. Binary relation
. Cao
Huy Linh -College of Education-Hue University
Chương 2
Groups
§1
BINARY OPERATIONS
Definition 1.1. Let X be a non-empty set. A binary operation ∗ on X is a
map
∗: X ×X →X
(x, y) 7→ x ∗ y,
For binary operations ∗ : X × X → X, two notations are in widely used:
the additive notation and the multiplicative notation. In the additive notation,
x ∗ y is denoted by x + y; then ∗ is an addition. In the multiplicative notation,
x ∗ y is denoted by x.y or by xy; then ∗ is a multiplication. In this chapter we
mostly use the multiplicative notation.
Addition (+) and multiplication (.) are two familiar binary operations on
appropriate sets. For addition, x + y is called the sum of x and y; for multiplication, x.y or xy is called the product of x and y.
Example 1.2.
(1) The ordinary addition on N (or Z, Q, R, C...) is a binary operation on
N (or Z, Q, R, C...)
(2) The ordinary multiplication on N (or Z, Q, R, C...) is a binary operation
on N (or Z, Q, R, C...).
(3) Denote X X by the set of the maps from X to X. For f, g ∈ X X , the
composition map f ◦ g of g and f is a binary operation on X X .
9
10
Chương 2. Chapter 1
(4) Addition and multiplication of matrices also provide binary operations
on the set Mn (R) of all n × n matrices with coefficients in a field K, for any
given integer n > 0.
Definition 1.3. Let ∗ be a binary operation on X.
(1) The binary operation ∗ is said to be associative if for all elements
x, y, z ∈ X we have (x ∗ y) ∗ z = x ∗ (y ∗ z).
(2) The binary operation ∗ is said to be commutative if for all elements
x, y ∈ X we have x ∗ y = x ∗ y.
(3) An element x ∈ X is said to be cancelable on the left (right) if for all
y, z ∈ X,
x ∗ y = x ∗ z =⇒ y = z
(resp. y ∗ x = z ∗ x =⇒ y = z).
(4) An element x ∈ X is said to be cancelable if it is cancelable on the left
and right.
(5) The binary operation ∗ is said to be cancelable on X if for all x ∈ X,
x is cancelable on X.
The ordinary addition and multiplication on N (or Z, Q, R, C...) in Example
1.2 are associative and commutative. The operation ◦ on X X is associative but
it is not commutative. By definition, associativity states that products with
three terms, or more, do not depend on the placement of parentheses.
Definition 1.4. Let ∗ be a binary operation on X.
(1) An element e of X is called an identity of ∗ on X if e ∗ x = x = x ∗ e
for all x ∈ X.
(2) Let e be an identity of ∗ on X. An element x of X is said to be left
invertible (or right invertible) if there exists x0 ∈ X such that x0 ∗ x = e (resp.
x ∗ x0 = e).
(3) Let e be an identity of ∗ on X. An element x of X is said to be invertible
if x is left invertible and right invertible; that is, there exists x0 ∈ X such that
x0 ∗ x = x ∗ x0 = e. Then the such element x0 is called the inverse of x and
denote by x−1 .
. Cao
Huy Linh -College of Education-Hue University
§ 1. Binary Operations
11
Readers will easily show that an identity element, if it exists, is unique.
The identity of additional operation is usually called the neutral element.
In the multiplicative (additional) notation, we usually denote the identity
(resp.neutral) element, if it exists, by 1 (resp. 0).
Definition 1.5.
(1) A non-empty set X along with a binary operation ∗ on its is said to be
a semi-group if the operation ∗ is associative, written by (X, ∗).
(2) A semi-group (X, ∗) is said to be a monoid if the operation ∗ has an
identity e.
(3) A monoid (X, ∗) is said to be a group if every element of X is invertible.
A group X is said to be an Abel group if the operation ∗ is commutative.
Example 1.6.
(1) (N, +) is a semi-group.
(2) (N0 , +) and (Z, ·) are monoids but they are not groups.
(3) (Z, +) is a group; more exactly, (Z, +) is a Abel group.
Definition 1.7. A non-empty set X along with two binary operations addition (+) and multiplication (·).
(1) X is said to be a ring if it satisfies three properties
(1) (X, +) is a Albel group.
(2) (X, ·) is a semi-group.
(3) The multiplication · is distributive to the addition + on the left and
right; that is, x(y + z) = xy + xz and (x + y)z = xz + yz.
If the operation · on the ring X is commuatative (resp has identity), we say
X is a commutative ring (resp with identity).
(2) A ring X with identity (with the neutral element denotes by 0) is said
to be a skew-field if every non-zero element x in X is invertible.
(3) A commutative skew-field X is said to be a field. Another way to say
that a commutative ring X with identity is said to be a field if every non-zero
element x in X is invertible.
. Cao
Huy Linh -College of Education-Hue University
12
Chương 2. Chapter 1
Example 1.8. Let + and · be the ordinary addition and multiplication on
the number sets N, N0 , Z, Q, R, C. Then
(1) (Z, +, ·) is a commutative ring.
(2) (Q, +, ·), (R, +, ·) and (C, +, ·) are fields.
Let X be a monoid, and xl , ..., xn be elements of G (where n is an integer
> 1). We define their product inductively:
n
Y
xi = x1 · · · xn = (x1 · · · xn−1 ) · xn .
i=1
xn = x1 · · · xn
which xi = x for i = 1...n and x0 = e.
If (X, +) be a monoid, then we define
nx = x1 + · · · + xn
where xi = · · · xn
for i = 1...n
EXERCISES
1.1. Let X X denote the set of the maps from X to X and ¦ the composite
operation of the maps. Show that
a) f ∈ X X satisfies the cancellation law on the left if and only if f is
injective.
b) f ∈ X X satisfies the cancellation law on the right if and only if f is
surjective.
c) f ∈ X X satisfies the cancellation law if and only if f is bijective.
1.2. Let ∗ be a binary operation on X such that x ∗ y = y for all x, y ∈ X.
An element e of X is called the left (resp. right) identity if e ∗ x = x (resp.
x ∗ e = x for all x ∈ X. Show that
a) (X, ∗) is a simi-group which having the left identity.
b) If X has at least two elements, then it doesn’t have the right identity.
. Cao
Huy Linh -College of Education-Hue University
13
§ 2. Groups
1.3. Let ∗ be a binary operation on X and
S = {x ∈ X | x ∗ (y ∗ z) = (x ∗ y) ∗ z, ∀y, z ∈ X}.
Show that
a) For all x1 , x2 ∈ S, x1 ∗ x2 ∈ S.
b) (S, ∗) is a semi-group.
1.4. Find all invertible elements to the multiplication of the monoids N, Z, Q, R, C.
1.5. Find all invertible elements to the multiplication of the monoids Z7 and
Z12 .
1.6. Let denote by P(X) the set of subsets of X.
a) Show that (P(X), ∪) and (P(X), ∩) are monoids.
b) Find invertible elements of monoids (P(X), ∪) and (P(X), ∩)
§2
GROUPS
It is necessary to remind the notion of group.
Definition 2.1. A monoid (G, ·) is said to be a group if every element of
X is invertible. In other words, (G, ·) is a group if it satifies three following
properties
(1) The operation · is associative on G.
(2) There exists the identity element for the operation · on G.
(3) Every element of G is invertible.
The identity element of the group G is denoted by 1G . In fact, the identity
of a group G is unique. If G is a finite group (the number of elements of G is
finite), then the number of elements in G, denoted by |G|, is called the order
of G. If G is a infinite group, then G is called a group of infinite oder.
. Cao
Huy Linh -College of Education-Hue University
14
Chương 2. Chapter 1
Example 2.2.
(1) Z (Q, R, C) along with the ordinary addition is a group.
(2) If we put Q∗ = Q \ {0}, R∗ = R \ {0}, C∗ = C \ {0}, then Q∗ , R∗ and C∗
along with the ordinary multiplication are groups.
(3) Set of integers modulo m, Zm , forms a group under addition.
(4) Z along with the ordinary multiplication is not a group.
Question: Is set of integers modulo m, Zm , a group under multiplication
(x̄.ȳ = xy)?
¯
Example 2.3.
(1) Let E be a vector space over a field K and GL(E) be the set of linear
automorphisms of E. Then GL(E) is a group under composition of maps.
(2) Denote by GLn (K) the set of invertible matrices of the size n over a field
K. Then GLn (K) forms a group under the multiplication of matrices.
A permutation of a set X is a bijection from X to itself. In high school
mathematics, a permutation of a set X is defined as a rearrangement of
its elements. For example, there are six rearrangements of X = {1, 2, 3} :
123; 132; 213; 231; 312; 321.
Now let X = {1, 2, ..., n}. A rearrangement is a list, with no repetitions, of
all the elements of X. All we can do with such lists is count them, and there
are exactly n! permutations of the n-element set X. Now a rearrangement
i1 , i2 , ..., in of X determines a function σ : X → X, namely, σ(1) = i1 , σ(2) =
i2 , ..., σ(n) = in . For example, the rearrangement 213 determines the function
σ with σ(1) = 2, σ(2) = 1, and σ(3) = 3. We use a two-rowed notation to
denote the function corresponding to a rearrangement; if σ(j) is the j-th item
on the list, then
Ã
!
1
2
···
n
σ=
.
σ(1) σ(2) · · · σ(n)
. Cao
Huy Linh -College of Education-Hue University
15
§ 2. Groups
For example, if S = {1, 2, 3, 4, 5}, then
Ã
σ=
1 2 3 4 5
3 5 4 1 2
!
is the permutation such that σ(1) = 3, σ(2) = 5, σ(3) = 4, σ(4) = 1, σ(5) = 2.
If we start with any element x ∈ S and apply σ repeatedly to obtain
σ(x), σ(p(x)), σ(σ(σ(x))), and so on, eventually we must return to x, and there
are no repetitions along the way because σ is one-to-one. For the above example, we obtain 1 → 3 → 4 → 1, 2 → 5 → 2. We express this result by
writing
σ = (1, 3, 4)(2, 5).
where the cycle (1, 3, 4) is the permutation of S that maps 1 to 3, 3 to 4 and
4 to 1, leaving the remaining elements 2 and 5 fixed. Similarly, (2, 5) maps 2
to 5, 5 to 2, 1 to 1, 3 to 3 and 4 to 4. The product of (1, 3, 4) and (2, 5) is
interpreted as a composition, with the right factor (2, 5) applied first, as with
composition of functions. In this case, the cycles are disjoint, so it makes no
difference which mapping is applied first.
The above analysis illustrates the fact that any permutation can be expressed as a product of disjoint cycles, and the cycle decomposition is unique.
A permutation σ is said to be even if it can be decomposed further into a
product of an even number of transpositions; otherwise it is odd. For example,
σ = (1, 2, 3, 4, 5) = (1, 5)(1, 4)(1, 3)(1, 2);
so, σ is a even permutation.
It is easy to see that the product of two even permutations is even; the
product of two odd per- mutations is even; and the product of an even and an
odd permutation is odd. To summarize very compactly, define the sign of the
permutation σ as
(
1 if σ is even;
sgn(σ) =
−1 if σ is odd.
Recall that if σ is a cycle of the length k then sgn(σ) = (−1)k−1 .
. Cao
Huy Linh -College of Education-Hue University
16
Chương 2. Chapter 1
Proposition 2.4. The set of all the permutations of a set X, denoted by SX ,
along with composition is a group and it is called the symmetric group on X.
When X = {1, 2, ..., n}, SX is usually denoted by Sn , and it is called the
symmetric group on n letters.
Example 2.5. S3 is a symmetric group on 3 letters with identity 1S is the
identification map Id. We have
S3 = {σ1 = Id, σ2 = (12), σ3 = (13), σ4 = (23), σ5 = (123), σ6 = (132)}.
Example 2.6. (Klein four-group) Let V4 = {1, a, b, c} be a set of four elements.
We define a multiplication on V as follows
1.1 = 1, 1.a = a, 1.b = b, 1.c = c, a.1 = a, a.a = 1, a.b = c, a.c = b,
b.1 = b, b.a = c, b.b = 1, b.c = a, c.1 = 1, c.a = b, c.b = a, c.c = 1.
Readers will verify that V4 is indeed a group and this group is called the Klein
four-group
Proposition 2.7. Let (G, ·) be a group. Then
(i) The identity element of G is unique.
(ii) For all x ∈ G, the inverse of x is unique.
(iii) Every x ∈ G sstifies the cancellation law.
Proof.
(i) Assume 1G and 10G are two identity elements of G. Then
1G = 1G .10G = 10G .
(ii) Assume that y and z are two inverses of x. Then
y = y.1G = y.(x.z) = (y.x).z = 1G z = z.
(iii) For any x ∈ G and for all y, z ∈ G, assume x.y = x.z. Then x−1 .(x.y) =
x−1 .(x.z). It follows that (x−1 .x).y = (x−1 .x).z. This implies 1G .y = 1G .z.
Hence y = z; that is, x satifies the left cancellation law. Similarly, x also
satifies the right cancellation law.
. Cao
Huy Linh -College of Education-Hue University
17
§ 2. Groups
Proposition 2.8. Let (G, ·) be a group. Then
(i) (1G )−1 = 1G .
(ii) For all x ∈ G, (x−1 )−1 = x.
(iii) For all x, y ∈ G, (x.y)−1 = (y)−1 .(x)−1 .
Let (G, ·) be a group and a ∈ G. For n ∈ Z, we define


if n > 0;
 a.a...a
n
a =
1G
if n = 0;

 −1
a ....a−1 if n < 0.
Theorem 2.9. Let (G, ·) be a semi-group. Then, G is a group if and only if
for any a, b ∈ G, the equations a.x = b and y.a = b have a solution in G.
Proof. Assume that G is a group with the identity element 1G . It is easily to
see that x1 = a−1 .b is a solution of the equation a.x = b and x2 = b.a−1 is a
solution of the equation y.a = b.
Conversely, since G 6= ∅, there exists a ∈ G. Assume that x0 is a solution
of the equations a.x = a. We need to prove that b.x0 = b for all b ∈ G. Indeed,
suppose that y0 is a solution of the equation y.a = b. We have
b.x0 = (y0 .a)x0 = y0 .(ax0 ) = y0 .a = b.
Hence x0 is the right identity element of G. Similarly, we also prove that if x1
is a solution of the equation y.a = a., then x1 is a left identity element. As x1
is the left identity element of G, x0 = x1 .x0 . But x0 is a right identity element
of G, so x1 .x0 = x1 . It follows that x0 = x1 . This implies that x0 = 1G is the
identity of G.
On the other hand, for all a ∈ G assume a1 is a solution of the equation
y.a = 1G and a2 is a solution of the equation a.x = 1G . Then
a1 = a1 .1G = a1 .(a.a2 ) = (a1 .a).a2 = 1G .a2 = a2 .
Hence a−1 = a1 = a2 is the inverse of a. So, G is a group.
. Cao
Huy Linh -College of Education-Hue University
18
Chương 2. Chapter 1
EXERCISES
2.1. Draw a multiplication table of S3 .
2.2. Let Mm×n (K) be the set of m by n matrices over a field K. Does Mm×n (K)
form a group under matrix addition?
2.3. Let Mn (K)∗ be the set of nonzero square matrices of the size n over a
field K. Does Mn (K)∗ form a group under matrix multiplication?
2.4. Let Z∗m = Zm \ {0̄}. Does Z∗m form a group under multiplication (x̄.ȳ =
xy)?
¯
2.5. Show that a non-empty set G along with a multiplication · is a group if
and only if
(i) (G, ·) is a semi-group.
(ii) There exists an left identity element e; that is, e.x = x for all x ∈ G.
(iii) For all x ∈ G, there exists a left inverse x0 of x; that is, x0 .x = e.
2.6. Let G be a non-empty set such that |G| < ∞. Prove that (G, .) is a
group if and only if (G, .) is a semi-group and every element of G satifies
the cancellation law. Give an example of an infinite semigroup in which the
cancellation laws hold, but which is not a group.
2.7. Prove the following: a finite group with an even number of elements contains an even number of elements x such that x−1 = x. State and prove a
similar statement for a finite group with an odd number of elements
2.8. Let G be a finite group. Prove that the inverse of an element is a positive
power of that element.
2.9. Let (G, .) be a group and a ∈ G. Prove that for m, n ∈ N we have
a) am .an = am+n ;
b) (am )n = amn .
. Cao
Huy Linh -College of Education-Hue University
19
§ 3. Subgroups
§3
SUBGROUPS
A subgroup of a group G is a subset of G that inherits a group structure
from G. This section contains general properties.
Definition 3.1. A subgroup of a group G (written multiplicatively) is a nonempty subset H of G such that
(1) For all x, y ∈ H implies x.y ∈ H;
(2) 1G ∈ H;
(3) x ∈ H implies x−1 ∈ H.
A non-empty subset H of a group G satifies the condition (1) of Definition
3.1 is said to be a stable subset of G. So, we can say that a subgroup H of
a group G is a stable subset of G such that H along with the multiplication
from G becomes a group. Therefore, a subgroup H of G is a group and the
identity element of H is that of G.
Remark 3.2. If G is an addition group, then a subgroup of a group G is a
subset H of G such that for all x, y ∈ H implies x + y ∈ H 0 ∈ H , and x ∈ H
implies −x ∈ H.
Example 3.3. Let G be a group with the identity element 1G . Then H1 = {1G }
is a subgroup of G and this subgroup is called the trivial subgroup of G. It
is easy to see that G is also a subgroup of itself and this subgroup is called
the improper subgroup of G. A subgroup H is called a proper subgroup of G
if H 6= G.
Example 3.4. (Z, +) is a subgroup of (Q, +); (Q, +) is a subgroup of (R, +);
(R, +) is a subgroup of (C, +). On the other hand, (N, +) is not a subgroup
of (Z, +) (even though N is closed under addition).
Example 3.5.
(1) It is well known that (Z, +) is a group. Denote by
mZ = {mx | x ∈ Z} ⊆ Z.
Then mZ is a subgroup of the additional group Z.
. Cao
Huy Linh -College of Education-Hue University
20
Chương 2. Chapter 1
(2) Let Q+ be the set of positive rational numbers. Then Q+ is a subgroup
of the multiplicative group Q∗ . Similaly, set of positive real numbers R+
is a of the multiplicative group R∗ .
(3) Let
H = {z ∈ C | | z |= 1}.
Then H is a subgroup of the multiplicative group C∗ .
Example 3.6. The multiplicative table of V4 = {1, a, b, c} shows that {1, a}
is a subgroup of V4 ; so are {1, b} and {1, c}.
Question: Find all subgroups of the group S3 .
We denote the relation H is a subgroup G by H ≤ G and H < G to show
that H is a proper subgroup of G.
Proposition 3.7. A subset H of a group G is a subgroup if and only if H 6= ∅,
and x, y ∈ H implies xy ∈ H, and if x ∈ H implies x−1 ∈ H.
Proof. "only if:" It is evident from Definition 3.1.
"If:" We need to prove that 1G ∈ H. Since H 6= ∅, there exists x ∈ H. By
hypothesis, x−1 ∈ H. Then, 1G = xx−1 ∈ H. Hence H is a subgroup of G.
Proposition 3.8. A subset H of a group G is a subgroup if and only if H 6= ∅
and x, y ∈ H implies xy −1 ∈ H.
Proof. These conditions are necessary by (1), (2), and (3). Conversely, assume
that H 6= ∅ and x, y ∈ H implies xy −1 ∈ H. Then there exists h ∈ H and
1G = hh−1 ∈ H. Next, x ∈ H implies x−1 = 1G .x−1 ∈ H. Hence x, y ∈ H
implies y −1 ∈ H and xy = x(y −1 )−1 ∈ H. Therefore H is a subgroup.
Proposition 3.9. The intersection of any non-empty family of subgroups of
a group G is again a subgroup of G. In particular, if H and K are subgroups
of G, then H ∩ K is a subgroup of G.
Proof. Assume that I 6= ∅ and Hi are subgroups of G for i ∈ I. Denote
T
H = i∈I Hi . Suppose that x, y ∈ H. Then x, y ∈ Hi for all i ∈ I. Because Hi
T
are subgroup of G, xy −1 ∈ Hi for all i ∈ I. Hence xy −1 ∈ i∈I Hi = H. This
implies that H is a subgroup of G.
. Cao
Huy Linh -College of Education-Hue University
21
§ 3. Subgroups
Corollary 3.10. If X is a subset of a group G, then there is a subgroup hXi of
G containing X that is smallest in the sense that hXi 5 H for every subgroup
H of G that contains X.
Proof. There exist subgroups of G that contain X; for example, G itself contains X. Define
\
hXi =
H,
X⊆H5G
the intersection of all the subgroups H of G that contain X. The set of all
the subgroups of G is non-empty, because G is an element of this set. By
Proposition 3.9, hXi is a subgroup of G; of course, hXi contains X because
every H contains X. Finally, if H is any subgroup containing X, then H is
one of the subgroups whose intersection is hXi; that is, hXi 5 H.
Note that there is no restriction on the subset X in the last corollary; in
particular, X = ∅ is allowed. Since the empty set is a subset of every set, we
have ∅ ⊂ H for every subgroup H of G. Thus, h∅i is the intersection of all the
subgroups of G; in particular, h∅i 5 {1G }, and so h∅i = {1G }.
Definition 3.11. Let X be a subset of a group G. Then hXi is called the
subgroup generated by X.
Example 3.12. See Z as a additional group, m ∈ Z, and X = {m}. Then
hXi = mZ. If Y = {4, 6} ⊂ Z, then hY i = 12Z. Generally, if Z = {a1 , ..., mk } ⊂
Z, then hZi = dZ, where d = (a1 , ..., ak ) the greatest common divisor of
a1 , ..., ak
Question: Let (G, .) be a group and a ∈ G. Determine h{a}i.
Definition 3.13. Let X is a nonempty subset of a group G, define a word
on X to be an element g ∈ G of the form g = xe11 ...xenn , where xi ∈ X and
ei = ±1 for all i = 1...n.
Proposition 3.14. If X is a nonempty subset of a group G, then hXi is the
set of all the words on X.
. Cao
Huy Linh -College of Education-Hue University
22
Chương 2. Chapter 1
Proof. Denote by W (X) the set of all the words on X. First, we need to prove
that W (X) is a subgroup of G. If x ∈ X, then 1G = xx−1 ∈ W (X); the product
of two words on X is also a word on X; the inverse of a word on X is a word
on X. Therefore, W (X) is a subgroup of G containing X; that is hXi. On the
other hand, any subgroup of G containing X must also contain W (X). Hence
W (X) ⊆ hXi, and so hXi = W (X).
Thus, H = hXi when every element of H is a product of elements of X
and inverses of elements of X.
Corollary 3.15. Let G be a group and let a ∈ G. The set of all powers of a
is a subgroup of G; in fact, it is the subgroup generated by {a}.
Proof. That the powers of a constitute a subgroup of G follows from the parts
a0 = 1G , (an )−1 = a−n , and am an = am+n . Also, nonnegative powers of a are
products of a, and negative powers of a are products of a−1 , since a−n = (a−1 )n
.
EXERCISES
3.1. Prove that a subset H of a finite group G is a subgroup if and only if
H 6= ∅ and x, y ∈ H implies xy ∈ H.
3.2. Let (G, ·) be a group. Show that the subset
Z(G) = {z ∈ G | zx = xz
for all x ∈ G}
is a subgroup of G, Z(G) is called the center of group.
3.3. Let (G, ·) be a Abel group. Show that the subset
H = {x ∈ G | x2 = 1G }
is a subgroup of G.
3.4. Let G = R × R∗ . We define a multiplication on G as follow:
(a, a0 )(b, b0 ) = (ab0 + b, a0 b0 ))
. Cao
Huy Linh -College of Education-Hue University
23
§ 4. cyclic groups
a) Show that (G, ·) is a group.
b) Show that the subset
H = {(a, 1) | a ∈ R}
is a subgroup of the group G.
3.5. Let G be a finite group. Show that
a) The inverse of an element in G is a positive power of that element.
b) The subgroup hXi of G generated by a subset X of G is the set of all
products in G of elements of X
3.6. Find a group with two subgroups whose union is not a subgroup
3.7. Let A and B be subgroups of a group G. Prove that A ∪ B is a subgroup
of G if and only if A ⊆ B or B ⊆ A.
3.8. Find all subgroups of V4
3.9. A subgroup M of a finite group G is maximal when M 6= G and there is
no subgroup M $ H $ G. Show that every subgroup H 6= G of a finite group
is contained in a maximal subgroup.
§4
CYCLIC GROUPS
In this lesson, we will study a special group which generated by only one
element.
Definition 4.1. Let G be a group and a ∈ G. The subgroup h{a}i of G is
called the cyclic subgroup of G generated by a; denoted by hai. A group G is
called cyclic if there exists a ∈ G with G = hai, in which case a is called a
generator of G.
In another word, a cyclic subgroup (or cyclic group) of G is a subgroup
generated by only one element. So,
hai = {an | n ∈ Z}.
. Cao
Huy Linh -College of Education-Hue University
24
Chương 2. Chapter 1
Example 4.2. G = Z is an additive group and m ∈ Z. The subgroup of Z
generated by m is mZ. In particular, Z is a cyclic group generated by 1.
Corollary 4.3. A cyclic subgroup is a Abel group.
Proposition 4.4. Every subgroup of Z is cyclic, generated by a unique nonnegative integer.
Proof. The proof uses integer division. Let H be a subgroup of (the additive
group) Z. If H = {0}, then H is cyclic, generated by 0. Now assume that
H 6= {0}, so that H contains an integer m 6= 0. If m < 0, then −m ∈ H;
hence H contains a positive integer. Let n be the smallest positive integer
that belongs to H . Every integer multiple of n belongs to H . Conversely, let
m ∈ H. Then m = nq + r for some q, r ∈ Z, 0 ≤ r < n . Since H is a subgroup,
qn ∈ H and r = m − qn ∈ H. Now, 0 < r < n would contradict the choice of
n ; therefore r = 0, and m = qn is an integer multiple of n . Thus H is the set
of all integer multiples of n and is cyclic, generated by n > 0. (In particular,
Z itself is generated by 1. Moreover, n is the unique positive generator of H,
since larger multiples of n generate smaller subgroups.
Definition 4.5. Let G be a group and a ∈ G. The order of the group hai is
called order of a.
Example 4.6. (1) In the Klein-four group V4 = {1, a, b, c}, the order of a,
the order of b, the order of c are the same and equal 2.
(2) In the additive group Z12 , the order of 4̄ is 3, the order of 5̄ is 12.
(3) In the additive group Z, the order of an element m ∈ Z is ∞.
Proposition 4.7. Every subgroup of a cyclic group is cyclic.
Proof. Assume that G = hai is a cyclic group generated by a and H is a
subgroup of G. If H = {1G }, then it is evident that H is cyclic generated
by 1G . If G 6= {1G } and H 6= {1G }, there exists an integer m 6= 0 such that
am ∈ H. Since H is a subgroup of G, a−m = (am )−1 ∈ H. Therefore, there is
an positive number h > 0 such that ah ∈ H. Suppose that n is the smallest
. Cao
Huy Linh -College of Education-Hue University
25
§ 4. cyclic groups
positive integer such that an ∈ H. We need to prove that H is a cyclic subgroup
generated by an . Indeed, for all x = am ∈ H. Then, there exist q, r ∈ Z such
that
m = nq + r for 0 ≤ r < n.
As ar = am (an )−q ∈ H, r = 0. Hence m = nq and then
x = anq = (an )q = bq ,
where b = an . This implies H is cyclic subgroup generated by b.
Proposition 4.8. Let G = hai be a finite cyclic group with order n. Then
(1) If 0 ≤ m1 , m2 ≤ n − 1 and m1 6= m2 , then am1 6= am2 .
(2) an = 1G .
(3) If am = 1G then m = nq, for q ∈ Z.
Proof. (1) Without loss of generality, we assune that m1 < m2 and am1 = am2 .
Hence am2 −m1 = 1G . Put k = m2 −m1 . Then 0 < k < n. Since ak+l = ak al = al ,
for l ∈ Z, |G| ≤ k < n. This is a contradiction.
(2) From (1), we have G = {1G , a, a2 , ..., an−1 }. Suppose that an 6= 1G .
Then, there exists k ∈ Z such that 0 < k < n and an = ak . Hence an−k = 1G .
But 0 < n − k < n. This is a contradiction.
(3) From (1), we have G = {1G , a, a2 , ..., an−1 }. Assume am = 1G . Hence
a−m = (am )−1 = 1G . Without loss of generality, assume m ≥ 0. By Euclidean
algorithm, there exist q, r ∈ Z such that
m = nq + r for 0 ≤ r < n.
Then am = anq .ar = (an )q ar = ar = 1G . But r < n, so r = 0; that is m = nq.
Proposition 4.9. Let G = hai be a finite cyclic group with order n. For
n
, where (n, r) is the greatest common
0 ≤ r < n, the order of ar is
(n, r)
divisor.
. Cao
Huy Linh -College of Education-Hue University
26
Chương 2. Chapter 1
Proof. Home work
Corollary 4.10. Let G = hai be a finite cyclic group of order n. For 0 ≤ r < n,
the element ar of G is generator og G if and only if (n, r) = 1.
Example 4.11. Let G = hai be a finite cyclic group of order 12. A generator
of G has the form ar such that (r, 12) = 1, for 1 ≤ r < 12. Hence r = 1, 5, 7, 11.
So,
G = hai = ha5 i = ha7 i = ha11 i.
12
= 6. A generator
(12, 2)
of ha2 i is (a2 )r such that (r, 6) = 1. Thus r = 1, 5. This implies
By Proposition 4.7, the subgroup ha2 i has the order
ha2 i = ha10 i.
Similarly, we have
ha3 i = ha9 i, ha4 i = ha8 i.
So, there are 6 different subgroup of G:
G, {1G }, ha2 i, ha3 i, ha4 i, ha6 i.
EXERCISES
BÀI TẬP
4.1. In additional group Z18 , determine h3̄i, h4̄i, h5̄i .
4.2. In symmetric group S4 , denote α = (1234), β = (12)(34). Determine hαi,
hβi.
4.3. Let G be a cyclic group of order n generated by a. Prove that
a) For 1 ≤ r ≤ n − 1, order of the subgroup har i is
n
.
(n, r)
b) ar is a generating element of G if and only if (n, r) = 1.
4.4. Is the symmeteric group Sn cyclic? Why?.
. Cao
Huy Linh -College of Education-Hue University
27
§ 5. Normal subgroups and quotient groups
4.5. Find all generating elements of Z18 .
4.6. Find all subgroups of Z18 and draw a inclusive diagram between them.
4.7. Show that every group of prime order is cyclic.
4.8. Prove that an infinite cyclic group has exact two generated elements.
4.9. Show that a cyclic group which has only one generated element is a finite
group which has at most two elements.
§5
NORMAL SUBGROUPS AND QUOTIENT GROUPS
Definition 5.1. Let H be a subgroup of a group G and a ∈ G, then the left
coset aH (or the right coset Ha) is the subset of G, where
aH = {ah : h ∈ H} (or Ha = {ha : h ∈ H}).
In general, the left cosets and right cosets may be different, as we shall
soon see.
Example 5.2. If G = S3 and H = h(12)i, there are exactly three left cosets
of H, namely
H = {(1), (12)}
= (12)H
(13)H = {(13), (123)} = (123)H
(23)H = {(23), (132)} = (132)H.
Consider the right cosets of H = h(12)i in S3 :
H = {(1), (12)}
= H(12)
H(13) = {(13), (132)} = H(132)
H(23) = {(23), (123)} = H(123).
Again, we see that there are exactly 3 (right) cosets. Note that these cosets
are Ỏparallel; that is, distinct (right) cosets are disjoint.
. Cao
Huy Linh -College of Education-Hue University
28
Chương 2. Chapter 1
Lemma 5.3. Let H be a subgroup of a group G, and let a, b ∈ G.
(i) aH = bH if and only if b−1 a ∈ H. In particular, aH = H if and only if
a ∈ H.
(ii) If aH ∩ bH 6= ∅ , then aH = bH.
(iii) |aH| = |H| for all a ∈ G.
Proof. The first two statements follow from observing that the relation on G,
defined by a ∼ b if b−1 a ∈ H, is an equivalent relation whose equivalent classes
are the left cosets. Therefore, (i) and (ii) are clear. The third statement is true
because h 7−→ ah is a bijection form H to aH.
The next theorem is named after J. L. Lagrange, who saw, in 1770, that
the order of certain subgroups of Sn are divisors of n!. The notion of group
was invented by Galois 60 years afterward, and it was probably Galois who
first proved the theorem in full.
Theorem 5.4. (Lagrange’s Theorem) If H is a subgroup of a finite group
G, then |H| is a divisor of |G|.
Proof. Let {a1 H, a2 H, ..., at H} be the family of all the distinct cosets of H in
G. Then
G = a1 H ∪ a2 H ∪ ... ∪ at H,
because each g ∈ G lies in the left coset gH, and gH = ai H for some i .
Moreover, Lemma 5.3 (ii) shows that the left cosets partition G into pairwise
disjoint subsets. It follows that
|G| = |a1 H| + |a2 H| + · · · + |at H|.
But |ai H| = |H| for all i , by Lemma 5.3 (iii), so that |G| = t|H|, as desired.
As we see that the number of left cosets and the number of right cosets are
the same. This leads to the following notion.
Definition 5.5. The index of a subgroup H in G, denoted by [G : H], is the
number of left cosets of H in G.
. Cao
Huy Linh -College of Education-Hue University
§ 5. Normal subgroups and quotient groups
29
Corollary 5.6. If H is a subgroup of a finite group G, then
|G| = [G : H]|H|.
Corollary 5.7. If G is a finite group and a ∈ G, then the order of a is a
divisor of |G|.
By Proposition 4.8, if G = hai is a cyclic group of order n, then an = 1G .
What happen if G is an arbitrary group of order n.
Corollary 5.8. If G is a finite group of order n, then an = 1G .
Proof. If a has order d, then |G| = dm for some integer m, by the previous
corollary, and so a|G| = adm = (ad )m = 1.
Corollary 5.9. If p is a prime, then every group G of order p is cyclic
Proof. If a ∈ G and a 6= 1, then a has order d > 1, and d is a divisor of p.
Since p is prime, d = p, and so G = hai.
As we have seen that left cosets and right cosets of a subgroup H might be
different. This leads to the following notion.
Definition 5.10. A subgroup N of a group G is normal when xN = N x for
all x ∈ G. If N is a normal subgroup of G, we write N / G.
Example 5.11.
(1) In an Abel group G, every subgroup H of G is a normal subgroup of G
(2) In an arbitrary group G, the trivial subgroup {1G } and the non-proper
are normal subgroups of G.
(3) Define the center of a group G, denoted by Z(G), to be
Z(G) = {z ∈ G : zg = gz for all g ∈ G};
that is, Z(G) consists of all elements commuting with everything in G. It is
easy to see that Z(G) is a subgroup of G; it is a normal subgroup because if
z ∈ Z(G) and g ∈ G, then
gzg −1 = zgg −1 = z ∈ Z(G).
. Cao
Huy Linh -College of Education-Hue University
30
Chương 2. Chapter 1
(4) The four-group V is a normal subgroup of S4 . Recall that the elements
of V are
V = {(1), (12)(34), (13)(24), (14)(23)}.
It is well known that every conjugate of a product of two transpositions is
another such. But only 3 permutations in S4 have this cycle structure, and so
V is a normal subgroup of S4 .
Proposition 5.12. A subgroup N of a group G is normal if and only if
xhx−1 ∈ N (xN x−1 ⊆ N ) for all x ∈ G, h ∈ N .
Proof. Assume that N / G. Then for all x ∈ G, and for any a ∈ xN x−1 , there
exists h ∈ N such that a = xhx−1 . Since xN = N x, there is an element
h0 ∈ N such that xh = h0 x. Therefore xhx−1 = h0 ∈ N . Hence xN x−1 ⊆ N .
Conversely, suppose that xN x−1 ⊆ N for all x ∈ G. It follows xN ⊆ N x.
Moreover, take y = x−1 , we have x−1 N x ⊆ N . This implies N x ⊆ xN . So,
xN = N x; that is, N ¢ G.
Another special kind is constructed as follows from normal subgroups. Let
G be a group and N be a normal subgroup. Then the left cosets and the right
cosets are the same. So, we write
G/N = {xN |x ∈ G},
and define a multiplication on G/N as follows:
xN.yN = (xy)N,
for all x, y ∈ G.
Proposition 5.13. If N is a normal subgroup of a group G, then G/N along
with the above multiplication becomes a group.
Proof. This proof left for readers.
Definition 5.14. Let N be a subgroup of a group G. Then G/N, .) is called
quotient group G by N ; when G is finite, its order |G/N | is the index [G : N ] =
|G|/|N | (presumably, this is the reason why quotient groups are so called.
. Cao
Huy Linh -College of Education-Hue University
31
§ 5. Normal subgroups and quotient groups
Note that if (G, +) is a group and N is a normal subgroup of G. Then
elements of the quotient group G/N has the form x + N for all x ∈ G and the
addition on G/N as follows
(x + N ) + (y + N ) = (x + y) + N
for all x, y ∈ G.
Example 5.15. We take G is the additional group Z and N = mZ for m > 1.
Then Z/mZ is the group of integers modulo m; that is, Z/mZ = Zm .
EXERCISES
5.1. Let G be a group and H a subgroup of G. Define a binary relation ∼ on
G as follows:
∀a, b ∈ G, a ∼ b ⇔ b−1 a ∈ H.
a) Show that ∼ is an equivalent relation on G.
b) Show that ā = aH for all a ∈ G.
5.2.
a) In the symmetric group S3 , let σ2 = (12), σ5 = (123) and H =
hσ2 i, K = hσ5 i. Find all left cosets and right cosets for H and K.
b) In Z12 , let H1 = h2̄i, H2 = h3̄i. Find all left cosets and right cosets for
H1 and H2 .
5.3. Recall GLn (R), set of square matrices A in Mn (R) such that det(A) 6= 0,
along with matrix multiplication is a group and it is called a general linear
group.
a) Define the special linear group by
SL(n, R) = {A ∈ GLn (R)| det(A) = 1}.
Prove that SL(n, R) is a normal subgroup of GLn (R).
b) Prove that GLn (Q) is a subgroup of GLn (R). Is the general linear group
GL(n, Q) a normal subgroup of GLn (R)?
. Cao
Huy Linh -College of Education-Hue University
32
Chương 2. Chapter 1
5.4. Let G be a finite group with subgroups H and K. Prove that if H 5 K,
then
[G : H] = [G : K][K : H].
5.5. Let H and K be subgroups of a group G such that |H| and |K| relatively
primes (coprime). Prove that H ∩ K = {1G }.
5.6. Let G be a group of order 4. Prove that either G is cyclic or x2 = 1G for
every x ∈ G.
5.7. If H is a subgroup of a group G, prove that the number of left cosets for
H in G is equal to the number of right cosets for H in G.
5.8. (Small Fecmart’s Theorem:) Prove that if p is a prime and a ∈ Z,
then
ap = a mod p.
5.9. (Euler’s Theorem:) Let us denote φ the Euler function; that is, φ(n) is
the number of integers k such that 1 ≤ k ≤ n and coprime to n. Prove that if
(a, m) = 1, then
aφ(m) = 1 mod m.
5.10. (Wilson’s Theorem:) Prove that an integer p is a prime if and only if
(p − 1)! ≡ −1 mod p.
5.11. Prove that a cyclic group of order n has a unique subgroup of order d,
for each divisor d of n.
5.12. Prove that a group G of order n is cyclic if and only if, for each divisor
d of n, there is at most one cyclic subgroup of order d.
5.13. Prove that if G is an abelian group of order n having at most one cyclic
subgroup of order p for each prime divisor p of n, then G is cyclic.
5.14. Prove that the intersection of any non-empty family of normal subgroups
of a group G is itself a normal subgroup of G.
. Cao
Huy Linh -College of Education-Hue University
§ 6. Group homomorphisms
33
5.15. Denote by An the set of even permutations of order n. Show that An is
a normal subgroup of Sn .
5.16. Prove that
a) If H is a subgroup of index 2 in a group G, then g 2 ∈ H for every g ∈ G.
b) If H is a subgroup of index 2 in a group G, then H is a normal subgroup
of G.
5.17. Show that the alternating group A4 is a group of order 12 having no
subgroup of order 6.
5.18. Let H1 and H2 be normal subgroups of a group G such that H1 ∩ H2 =
{1G }. Prove that for h1 ∈ H1 and h2 ∈ H2 , h1 h2 = h2 h1 .
5.19. Find the order of the quotient group Sn /An .
5.20. Let G be a group. Let [G, G] be a subgroup of G generated by elements
of the form xyx−1 y −1 for all x, y ∈ G; that is,
[G, G] = h{xyx−1 y −1 | x, y ∈ G}i.
a) Show that [G, G] ¢ G and G/[G, G] is a Abel group.
b) Show that for any normal subgroup N of G, the quotient group G/N is
Abel if and only if [G, G] ⊆ N .
§6
GROUP HOMOMORPHISMS
Group homomorphisms are mappings that preserve products. They allow
different groups to relate to each other.
Definition 6.1. Let (G, .) and (H, .) be groups. A map f : G −→ H is called
a group homomorphism if
f (x.y) = f (x).f (y) for all x, y ∈ G.
. Cao
Huy Linh -College of Education-Hue University
34
Chương 2. Chapter 1
If G is written additively, then f (x.y) becomes f (x + y); if H is written
additively, then f (x).f (y) becomes f (x) + f (y).
Example 6.2. (1) Let (G, .) and (H, .) be groups. A map
f : G −→ H
x 7−→ 1H
is a group homomorphism, and this homomorphism is called a trivial group
homomorphism.
(2) Let G be a group and H be a subgroup of G. The map
ι : H −→ G
x 7−→ x
is a group homomorphism, and this homomorphism is called a inclusive homomorphism. In particular, if H = G then ι = IdG the identification homomorphism.
(3) Let G be a group and H be a normal subgroup of G. The map
p : G −→ G/H
x 7−→ xH
is a group homomorphism, and this homomorphism is called a canonical projection(or canonical epimorphism).
(4) Let G = Z and H be a multiplicative group (for example H = Sn ).
Given an element a in H. Then a map
f : Z −→ H
n 7−→ an
is a group homomorphism.
(5) Let G = R be a additional group and H = C∗ be multiplicative group
of non-zero complexes numbers. The map
p : R −→ C ∗
x 7−→ cos x + i sin x
. Cao
Huy Linh -College of Education-Hue University
35
§ 6. Group homomorphisms
is a group homomorphism.
(6) Let G = R be a additional group and H = R∗ be multiplicative group
of non-zero complexes numbers. The map
p : R −→ R∗
x 7−→ ex
is a group homomorphism.
Group homomorphisms preserve identity elements, inverses, and powers.
In particular, homomorphisms of groups preserve the constant and unary operation as well as the binary operation.
Proposition 6.3. Let f : G −→ H be a group homomorphism. Then
(1) f (1G ) = 1H ,
(2) f (x−1 ) = (f (x))−1 for all x ∈ G,
(3) f (xn ) = (f (x))n for all x ∈ G and n ∈ Z.
Proof. (1) We have 1G = 1G .1G . So, f (1G ) = f (1G .1G ) = f (1G ).f (1G ).. This
implies f (1G ) = 1H .
(2) From x.x−1 = 1G , we have 1H = f (1G ) = f (x.x−1 ) = f (x).f (x−1 ).
Hence f (x−1 ) = (f (x))−1 .
(3) Use induction to show that f (xn ) = f (x)n for all n > 0. Then observe
that x−n = (x−1 )n , and use part (2).
In algebraic topology, continuous mappings of one space into another induce
homomorphisms of their fundamental groups at corresponding points.
Proposition 6.4. If f : G −→ H and g : H −→ K are group homomorphisms,
then so is the composition map gf : G −→ K.
Proof. For all x, y ∈ G,
gf (xy) = g(f (xy)) = g(f (x)f (y)) (as f is a group homomorphism)
= g(f (x))g(f (y)) (as g is a group homomorphism)
= gf (x).gf (y).
. Cao
Huy Linh -College of Education-Hue University
36
Chương 2. Chapter 1
Definition 6.5. Let f : G −→ H be a group homomorphism. Then,
(1) f is said to be a group monomorphism (resp. epimorphism, or isomorphism) if f is injective (resp. surjective, or bijective)
(2) Two groups G and H are said to be isomorphic to each other if there
exists a group isomorphism from G to H. If G is isomorphic to H, then we
write G ∼
= H.
Example 6.6.
(1) Let H be a subgroup of a group G. Then the inclusive map ι : H −→ G
is a group monomorphism.
(2)Let N be a normal subgroup of a group G. Then the canonical projection
p : G −→ G/N is a group epimorphism.
A group homomorphism is isomorphic, so is the inverse.
Proposition 6.7. If f : G −→ H is a group isomorphism, so is the inverse
f −1 .
Proof. The proof is left for students as an exercise.
Proposition 6.8. Let f : G −→ H be a group homomorphism. Then,
(1) If G1 is a subgroup of G, then f (G1 ) is a subgroup of H.
(2) If H1 is a subgroup of H, then f −1 (H1 ) is a subgroup of G.
Proof. (1) Assume that G1 is a subgroup of G. Since 1G ∈ G1 , 1H = f (1G ) ∈
f (G1 ). Thus f (G1 ) 6= ∅. Moreover, for all y1 , y2 ∈ f (G1 ), there exist x1 , x2 ∈ G1
such that y1 = f (x1 ) and y2 = f (x2 ). Since G1 is a subgroup of G, x1 x−1
2 ∈ G1 .
Then,
−1
y1 y2−1 = f (x1 )[f (x2 )−1 ] = f (x1 )f (x−1
2 ) = f (x1 x2 ) ∈ f (G1 ).
Hence f (G1 ) is a subgroup of H.
(2) Assume that H1 is a subgroup of H. Since f (1G ) = 1H ∈ H1 , 1G ∈
−1
f (H1 ). Thus f −1 (H1 ) 6= ∅. On the other hand, for all x1 , x2 ∈ f −1 (H1 ), we
have
−1
f (x1 .(x2 )−1 ) = f (x1 )f (x−1
2 ) = f (x1 ).f (x2 ) .
. Cao
Huy Linh -College of Education-Hue University
§ 6. Group homomorphisms
37
But f (x1 ), f (x2 ) ∈ H1 and H1 5 H, f (x1 ).f (x2 )−1 ∈ H1 . Therefore f (x1 .(x2 )−1 ) ∈
−1
H1 ; that is x1 .x−1
(H1 ). So, we conclude that f −1 (H1 ) is a subgroup of
2 ∈ f
G.
Question: Let f : G −→ H be a group homomorphism, G1 and H1 normal
subgroups of G and H, respectively.
1. Is f (G1 ) a normal subgroup of H?
2. Is f −1 (H1 ) a normal subgroup of G?
Definition 6.9. Let f : G −→ H be a group homomorphism. The image of f
is
Im f = f (G) = {f (x)|x ∈ G}.
The kernel of f is
Ker f = f −1 ({1H }) = {x ∈ G | f (x) = 1H }.
Remark: It is clear that Im f is a subgroup of H and Ker f is a normal
subgroup of G.
Theorem 6.10. Let f : G −→ H be a group homomorphism.
(1) f is monomophic if and only if Ker f = {1G }.
(2) f is epimophic if and only if Im f = H.
Proof. (1) Assume that f is monomophic. Since f is injective, Ker f = {1G }.
Conversely, suppose that ker f = {1G }. We need to prove f is injective. Indeed,
for any x1 , x2 ∈ G, assume that f (x1 ) = f (x2 ). It follows f (x1 )[f (x2 )−1 ] =
−1
f (x1 .x−1
∈ Ker f . By hypothesis, x1 x−1
= 1G . This
2 ) = 1H . Hence x1 x2
2
implies that x1 = x2 .
(2) Obviously.
Corollary 6.11. A group homomorphism f : G −→ H is isomorphic if and
only if Ker(f ) = {1G } and Im(f ) = H.
. Cao
Huy Linh -College of Education-Hue University
38
Chương 2. Chapter 1
Theorem 6.12. Let f : G −→ H be a group homomorphism and g : G −→ K
be a group epimorphism such that Ker g ⊆ Ker f . There exists a unique group
homomorphism h : K −→ H such that f = hg. Moreover,
(1) If Ker g = Ker f , then h is monomorphic.
(2) If f is epimorphic, then h is is epimorphic.
Proof. For y ∈ K, there exist x ∈ G such that y = g(x). Then we set h(y) =
f (x). First, we need to show that h is a map from K to H. Indeed, for all
y1 , y2 ∈ K, there are x1 , x2 ∈ G such that y1 = g(x1 ) and y2 = g(x2 ). Assume
−1
that y1 = y2 . Then g(x1 ) = g(x2 ). Therefore g(x1 x−1
2 ) = 1H . Hence x1 x2 ∈
Ker g ⊆ Ker f . Thus f (x1 ) = f (x2 ); that is, h(y1 ) = h(y2 ).
Second, for all y1 , y2 ∈ K, there are x1 , x2 ∈ G such that y1 = g(x1 )
and y2 = g(x2 ). We have y1 y2 = g(x1 )g(x2 ) = g(x1 x2 ). It follows h(y1 y2 ) =
f (x1 x2 ) = f (x1 )f (x2 ) = h(y1 )h(x2 ). That means h is a group homomorphism.
It is easy to see that f = hg.
Third, suppose that there is a group homomorphism h0 from K to H such
that f = h0 g. Then, for all y ∈ K, there exists x ∈ G such that y = g(x). Then
h0 (y) = h0 (g(x)) = h0 g(x) = f (x) = h(y). This implies h = h0 .
Next, assume that Ker g = Ker f . For any y ∈ Ker h, there exist x ∈ G
such that y = g(x). We have h(y) = 0 = f (x). Hence x ∈ Kerf = Ker g.
Therefore y = g(x) = 0. This implies Ker h = 0. So. h is monomorphic.
Finally, suppose that f is epimorphic. Then, Im h = h(K) = h(g(G)) =
hg(G) = f (G) = H. Hence h is epimorphic.
Now, we consider a special case of g = p : G −→ G/ Ker f is a canonical
projection, that is, K = G/ Ker f . Then g is a epimorphism and Ker g = Ker f .
The following corollary is implied from Theorem 6.12.
Corollary 6.13. Let f : G −→ H be a group homomorphism.
(1) If f is epimorphic then
G/ Ker f ∼
= H,
(2)
G/ Ker f ∼
= Im(f ).
. Cao
Huy Linh -College of Education-Hue University
39
§ 6. Group homomorphisms
Corollary 6.14. Let G be a cyclic group.
(1) If G is infinite, then
G∼
= Z,
(2) If G is finite, then
G∼
= Zm ,
for some m ∈ N.
Proof. Assume that G = hai. Consider a map
f : Z −→ G
m 7−→ am .
It is clear that f is a group epimorphism. If G is infinite, then Ker f = {0}.
By Corollary 6.14, Z ∼
= G. If G is finite, we suppose that |G| = m. Then
Ker f = mZ. Applying Corollary 6.14, we have Z/ Ker f ∼
= G. It follows that
G∼
= Zm .
EXERCISES
6.1. Prove that a group G is abelian if and only if the map f : G −→ G, given
by f (a) = a−1 , is a homomorphism.
6.2. Show that every group G with |G| ≤ 5 is abelian.
6.3. Let G = {f : R −→ R| f (x) = ax + b, where a 6= 0}. Prove that G
is a group under composition that is Ãisomorphic
! to the subgroup of GL2 (R)
a b
consisting of all matrices of the form
.
0 1
6.4. For a positive integer n, prove that the additional group Z is isomorphic
to the subgroup nZ of itself.
6.5. Show that every group homomorphism from (Q, +) to (Z, +) is trivial.
. Cao
Huy Linh -College of Education-Hue University
40
Chương 2. Chapter 1
6.6. Let f : G −→ H be a group homomorphism.
a) Suppose that G1 ¢ G. Is f (G1 ) a normal subgroup of H?
b) Show that if f is epimorphic and G1 ¢ G, then f (G1 ) ¢ H.
6.7. Let f : G −→ H be a group homomorphism. Show that
a) If H1 ¢ H then f −1 (H1 ) ¢ G;
b) If f is epimorphic and H1 ¢ H, then
G/f −1 (H1 ) ∼
= H/H1 .
6.8. Let A be a group and let B, C be normal subgroups of A. Prove that if
C ⊆ B, then C is a normal subgroup of B, B/C is a normal subgroup of A/C,
and
A/B ∼
= (A/C)/(B/C).
6.9. Let A and B be subgroups of a multiplicative group G. Denote
AB = {a.b | a ∈ A, b ∈ B}.
Is AB a subgroup of G? Why? Give an example to illustrate that.
6.10. Let A be a subgroup of a group G, and let N be a normal subgroup of
G. Show that
a) AN = {an | a ∈ A, n ∈ N } is a subgroup of G;
b) N is a normal subgroup of AN ;
c) A ∩ N is a normal subgroup of A;
d) AN/N ∼
= A/(A ∩ N ).
6.11. Denote Aut(G) by the set of automorphisms of G; that is, Aut(G) is
the set of isomorphisms f : G −→ G. For all f, g ∈ Aut(G), define f.g is the
composition of two maps g and f . Prove that
. Cao
Huy Linh -College of Education-Hue University
§ 7. Direct product of groups
41
a) (Aut(G), .) is a group;
b) If G is a cyclic group, then Aut(G) is an Abel group;
c) If G is a cyclic group of order prime p, then Aut(G) is a cyclic group of
order prime p − 1.
6.12. Let f : G −→ H be a group homomorphism.
a) Show that if |G| = n then | Im(G)| is a divisor of n;
b) Let K ≤ G such that [G : K] = n. Show that if Ker(f ) ⊆ K and f is
epimorphism, then [H : f (K)] = n;
§7
DIRECT PRODUCT OF GROUPS
Let H, K be groups. Let us denote by H × K the set of all ordered pairs
(h, k) with h ∈ H and k ∈ K. We define a binary operation on H × K as
follows
(h1 , k1 )(h2 , k2 ) = (h1 h2 , k1 k2 ) for all (h1 , k1 ), (h2 , k2 ) ∈ H × K.
It is easy to check that H × K along with the above binary operation is a
group with the identity (1H , 1K ) and (h, k)−1 = (h−1 , k −1 ).
Definition 7.1. The group H × K is called a direct product of two group H
and K.
If H = K = G, we write G × G = G2 .
Example 7.2. (1) Z2 = Z × Z is a direct product of two copies Z. Note that
Z2 is a additional group with neutral element (0, 0) and the symetric element
of (x, y) is (−x, −y).
(2) Z2 × Z3 is a additional group of 6 elements.
(3) Consider a additional group Zm and a multiplicative symmetric group
S3 . Then the direct product of Zm and S3 is the group Zm × S3 of 6m elements
with the law of multiplication:
(ā, α)(b̄, β) = (ā + b̄, αβ) for all ā, b̄ ∈ Zm , α, β ∈ S3 .
. Cao
Huy Linh -College of Education-Hue University
42
Chương 2. Chapter 1
We now apply the first isomorphism theorem to direct products.
Proposition 7.3. Let G and G0 be groups, and let K ¢ G and K 0 ¢ G0 be
normal subgroups. Then K × K 0 ¢ G × G0 , and there is an isomorphism
(G × G0 )/(K × K 0 ) ∼
= G/K × G0 /K 0 .
Proof. Let π : G −→ G/K and π 0 : G0 −→ G0 /K 0 be the canonical projections.
It is routine to check that f : G × G0 −→ (G/K) × (G0 /K 0 ), given by
f (g, g 0 ) = (π(g), π 0 (g)) = (gK, g 0 K 0 )
is a surjective homomorphism with Ker f = K × K 0 . The first isomorphism
theorem now gives the desired isomorphism.
Let H, K be subgroups of a group G and denote
HK = {hk | h ∈ H, k ∈ K}.
By Exercise 6.10, if H (or K) is a normal subgroup of G, then HK is a subgroup
of G. The following Proposition give us a case H × K = HK.
Proposition 7.4. If G is a group containing normal subgroups H and K with
H ∩ K = {1G } and HK = G, then G ∼
= H × K.
Proof. We show first that if g ∈ G, then the factorization g = hk, where h ∈ H
and k ∈ K, is unique. If hk = h0 k 0 , then h0−1 h = k 0 k −1 ∈ H ∩ K = {1G }.
Therefore, h0 = h and k 0 = k. We may now define a function ϕ : G −→ H × K
by ϕ(g) = (h, k), where g = hk, h ∈ H, and k ∈ K. To see whether ϕ is a
homomorphism, let g 0 = h0 k 0 , so that gg 0 = hkh0 k 0 . Hence, ϕ(gg 0 ) = ϕ(hkh0 k 0 ),
which is not in the proper form for evaluation. If we knew that if h ∈ H and
k ∈ K, then hk = kh, then we could continue:
ϕ(hkh0 k 0 ) = ϕ(hh0 kk 0 )
= (hh0 , kk 0 )
= (h, k)(h0 , k 0 )
= ϕ(g)ϕ(g 0 ).
. Cao
Huy Linh -College of Education-Hue University
43
§ 7. Direct product of groups
Let h ∈ H and k ∈ K. Since K is a normal subgroup, (hkh−1 )k −1 ∈ K;
since H is a normal subgroup, h(kh−1 k −1 ) ∈ H. But H ∩ K = {1G }, so that
hkh−1 k−1 = 1G and hk = kh. Finally, we show that the homomorphism ϕ is
an isomorphism. If (h, k) ∈ H × K, then the element g ∈ G defined by g = hk
satisfies ϕ(g) = (h, k); hence ϕ is surjective. If ϕ(g) = (1, 1), then g = 1, so
that Ker(ϕ) = {1G } and ϕ is injective. Therefore, ϕ is an isomorphism.
Remark. We must assume that both subgroups H and K are normal. For
example, S3 has subgroups H = h(123)i and K = h(12)i. Now H ¢S3 , H ∩K =
{1G }, and HK = S3 , but S3 H × K (because the direct product is abelian).
Of course, K is not a normal subgroup of S3 .
Proposition 7.5. Let G be a group, and let a, b ∈ G be commuting elements
of orders m and n, respectively. If (m, n) = 1, then ab has order mn.
Proof. Since a and b commute, we have (ab)r = ar br for all r , so that (ab)mn =
amn bmn = 1. It suffices to prove that if (ab)k = 1, then mn|k. If 1 = (ab)k =
ak bk , then ak = b−k . Since a has order m, we have 1 = amk = b−mk . Since b
has order n, Proposition 4.8 gives n|mk. As (m, n) = 1, n|k; a similar argument
gives m|k. It follows that mn|k. Therefore, mn ≤ k, and mn is the order of
ab.
EXERCISES
7.1. Show that if m and n are relatively prime, then
Zmn ∼
= Zm × Zn .
7.2. Prove that Z4 Z2 × Z2 .
7.3. Let ϕ be Euler function (ϕ(n) is the number of relatively primes to n).
Prove that if (m, n) = 1 then ϕ(mn) = ϕ(m)ϕ(n).
7.4. Let U (Zm ) denote the set of invertible elements of Zm . It is well known
that U (Zm ) along with multiplication is a group. Show that U (Z9 ) ∼
= Z6 and
U (Z15 ) ∼
= Z4 × Z2 .
. Cao
Huy Linh -College of Education-Hue University
44
Chương 2. Chapter 1
7.5. Let G be a group of order 4. Prove that G ∼
= Z4 or G ∼
= Z2 × Z2 .
7.6. Let G be a group of order 6. Prove that G ∼
= Z6 or G ∼
= Z2 × S3 .
7.7.
a) Let H and K be groups. Prove that H ∗ = {(h, 1) : h ∈ H} and
K ∗ = {(1, k) : k ∈ K} are normal subgroups of H × K with H ∼
= H∗
and K ∼
= K ∗.
b) Prove that K ∗ ¢ H × K and that
(H × K)/K ∗ ∼
= H.
7.8. If G is a group for which Aut(G) = {1}, prove that |G| ≤ 2.
7.9. Prove that if G is a group for which G/Z(G) is cyclic, where Z(G) denotes
the center of G, then G is abelian
7.10. If H and K are normal subgroups of a group G with HK = G, prove
that
G/(H ∩ K) ∼
= (G/H) × (G/K).
. Cao
Huy Linh -College of Education-Hue University
Chương 3
Rings
Rings are our second major algebraic structure; they marry the complexity of semigroups and the good algebraic properties of abelian groups. Gauss
(1801) studied the arithmetic properties of complex numbers a + bi with
a, b ∈ Z, and of polynomials with integer coefficients. From this start ring
theory expanded in three directions. Sustained interest in more general numbers and their properties finally led Dedekind (1871) to state the first formal
definition of rings, fields, ideals, and prime ideals, though only for rings and
fields of algebraic integers. The quaternions, discovered by Hamilton (1843),
were generalized by Pierce (1864) and others into another type of rings: vector
spaces with bilinear multiplications. Growing interest in curves and surfaces
defined by polynomial equations led Hilbert (1890-1893) and others to study
rings of polynomials. Modern ring theory began in the 1920s with the work of
Noether, Artin, and Krull. This chapter contains general properties of rings,
with some emphasis on arithmetic properties.
§1
RINGS AND FIELDS
Definition 1.1. A ring R is a set with two binary operations, addition and
multiplication, such that
(i) R is an abelian group under addition;
(ii) R is a semi-group under multiplication;
45
46
Chương 3. Rings
(iii) The multiplication is distributive in both side to the addition; that is,
a(b + c) = ab + ac and (b + c)a = ba + ca for every a, b, c ∈ R.
The neutral element of a ring R under addition usually is denoted by 0R
or 0. If the mutiplication on R has the identity element, then we say that R is
a ring with identity and the identity element is usually written by 1R . If the
mutiplication on R is commutative, then we say R is a commutative ring.
Definition 1.2.
(i) A ring R with identity is said to be a skew-field if every non-zero
element of R is invertible.
(ii) A commutative ring R with identity is said to be a field if every
non-zero element of R is invertible; another word, a commutative skew-field is
called a field.
Example 1.3. (1) The set of integer numbers Z (resp. Q, R, C) along with
ordinary addition and multiplication is a commutative ring with identity. This
ring is called the ring of integers (resp. ring of rational numbers, ring of real
numbers, ring of complex numbers). Moreover, Q, R, C are fields and they are
said to be field of rational (real, complex, respectively) numbers.
¯ b and
(2) On the set of integers modulo m, Zm , we define ā + b̄ = a +
¯ for all ā, b̄, c̄, d¯ ∈ Zm . Then (Zm , +, ·) is a commutative ring with
āb̄ = ab
identity. Espectialy, if p is a prime then Zp is a field.
(3) Let Z[i] = {a + ib | a, b ∈ Z}. We define two binary operation as follow
(a + ib) + (c + id) = a + c + i(b + d) for all a, b, c, d ∈ Z,
and
(a + ib)(c + id) = (ac − bd) + i(ad + bc) for a, b, c, d ∈ Z.
It is easy to verify that Z[i] along with above addition and multiplication is a
commutative ring with identity. This is called ring of Gauss integers.
(4) Let EndK (E) be the set of linear endomorphisms of vector space E.
We define two binary operations on EndK (E) as follow:
(f + g)(~x) = f (~x) + g(~x) for all ~x ∈ E,
. Cao
Huy Linh -College of Education-Hue University
47
§ 1. Rings and Fields
f g(x) = f (g(x)) for all ~x ∈ E.
Then, EndK (E) along with two above operations is a ring with identity IdE
and it is called the ring of linear endomorphisms.
(5) Let us denote by Mn (K) the set of square matries of the size n. Then,
Mn (K) is a ring with identity In under two operations matrix addition and
matrix multiplication. This ring is called the ring of matrices.
(6) Let K[x1 , ..., xn ] be the set of polynomials with coefficients over a commutative ring K with identity. Then K[x1 , ..., xn ] along with polynomial addition and polynomial multiplication is a commutative ring with identity. This
ring is called the ring of polynomials.
(7) On the additional group Zm × Zn , we define a multiplication as follows:
(a1 , b1 ).(a2 , b2 ) = (a1 a2 , b1 b2 ),
for all (a1 , b1 ), (a2 , b2 ), ∈ Zm × Zn . Then, (Zm × Zn , +, ·) is a commutative ring
with identity.
Generally, let R, S be rings. Define two binary operations on R × S as
follows:
(a1 , b1 ) + (a2 , b2 ) = (a1 + a2 , b1 + b2 );
(a1 , b1 ).(a2 , b2 ) = (a1 a2 , b1 b2 ),
for all (a1 , b1 ), (a2 , b2 ) ∈ R × S. Then, R × S along with two above operations
is a ring. This ring is called a product of rings R and S.
(8) Let X ⊆ R and RX be the set of functions from X to R. For all
f, g ∈ RX , we define f + g and f g as follow
(f + g)(x) = f (x) + g(x) for all x ∈ X,
f g(x) = f (x).g(x) for all x ∈ X.
Then, RX is a ring under two above operations.
Proposition 1.4. For all x, y of a ring R, we have
(i) x.0R = 0R .x = 0R ;
(ii) (−x).y = x.(−y) = −xy;
(iii) (−x).(−y) = x.y.
. Cao
Huy Linh -College of Education-Hue University
48
Chương 3. Rings
Proof. The proof is left as an exercise for the reader.
Corollary 1.5. For any integer m and for all x, y in a ring R, we have
m(xy) = (mx)y = x(my).
Definition 1.6. Let R be a ring and a ∈ R \ {0R }.
(1) The element a is called a zero divisor on the left (res. on the right) if
there is an element b ∈ R \ {0R } such that ab = 0R (res. ba = 0).
(2) If a is not a zero divisor on the left and right, then a is called a non-zero
divisor or regular element of R.
Definition 1.7. A commutative ring R with identity is said to a domain (or
integral domain) if every element of R is a non-zero divisor. Another word, a
domain is a commutative ring R with identity satisfies
∀a, b ∈ R : ab = 0R ⇒ a = 0R ∨ b = 0R .
Example 1.8.
(1) Z, Q, R and C are domains.
(2) In Z6 , the elements 2̄, 3̄ are zero divisors. The ring Z6 is not a domain.
(3) If p is a prime, then Zp is a domain.
Proposition 1.9. If R is a domain, then every non-zero element satisfies the
cancelation law under multiplication.
Proof. Assume a ∈ R \ {0R } and ab = ac for all b, c ∈ R. Then a(b − c) = 0R .
Because R is a domain and a 6= 0R , b − c = 0R . This implies b = c.
Proposition 1.10. Every field is a domain.
Proof. Suppose that F is a field. Assume a, b ∈ F and a.b = 0F . If a 6= 0F
then a is invertible. Then, a−1 (a.b) = a−1 0F . Hence (a−1 .a).b = 1R .b = 0F .
This implies b = 0R . So, F is a domain.
Notice that Z is a domain, but it is not a field. The following theorem tells
us whenever a domain is a field.
. Cao
Huy Linh -College of Education-Hue University
49
§ 1. Rings and Fields
Theorem 1.11. Every finite domain is a field.
Proof. Suppose that D is a finite domain and |D| = n. For any a ∈ D \ {0D },
consider a map
f : D −→ aD
x 7−→ ax.
It is clear that f is bijective. Hence |D| = |aD|. Since aD ⊂ D and |D| < ∞∞,
D = aD. It follows that 1D ∈ aD. So, there exists b ∈ D such that ab = 1R ;
this means the element a is invertible. This implies that D is a field.
EXERCISES
1.1. In the definition of a ring with identity, show that one may omit the
requirement that the addition be commutative.
1.2. Let A be an additional Abel group. Denote by End(A) the set of group
homomorphisms from A to itself. We define two binary operation on End(A)
that, for all f, g ∈ End(A),
(f + g)(x) = f (x) + g(x) for all x ∈ A,
f g(x) = f (g(x)) for all x ∈ A.
Show that End(A) along with two above operations is a ring.
1.3. A unit of a ring R with identity is an element u of R such that uv = vu = 1
for some v ∈ R. Show that v is unique (given u). Show that the set of all units
of R is a group under multiplication.
1.4. Let R be a ring with identity. Show that u is a unit of R if and only if
xu = uy = 1R for some x, y ∈ R .
1.5. Show that an element x ∈ Zn is a unit of Zn if and only if x and n are
relatively prime.
1.6. Find all units of the ring of Gauss integer numbers.
. Cao
Huy Linh -College of Education-Hue University
50
Chương 3. Rings
√
1.7. Show that complex numbers of the form a + ib 2, where a and b are
integers, constitute a ring. Find the units.
1.8. Prove that Zm is a field if and only if m is prime.
1.9. Let R be a ring. Show that R1 = R × Z, with operations (x, m) + (y, n) =
(x + y, m + n), and (x, m)(y, n) = (xy + nx + my, mn), is a ring with identity.
1.10. A ring R is said to be von Neumann regular if for all a ∈ R, there exists
an element x ∈ R such that axa = a. Prove that the ring R1 in Exercise 1.9 is
not von Neumann regular.
1.11. Let R be a ring such that every element x in R satisfies x2 = x. Prove
that R is commutative.
1.12. Let R be a ring such that every element x in R satisfies x2 = −x. Prove
that R is commutative.
1.13. Let R be a ring such that every element x in R satisfies x3 = x. Prove
that R is commutative.
1.14. Let R be a non zero ring such that the equation ax = b has a solution
in R for all a ∈ R \ {0R } and b ∈ R. Prove that R is a skew-field.
1.15. Let R be a non-zero ring such that for all a ∈ R \ {0R } there exists a
unique b ∈ R such that aba = a. Prove that R is a skew-field.
1.16. Prove that a finite ring R with identity such that every non-zero element
of R is a non-zero divisor is a skew-field.
§2
SUBRINGS, IDEALS AND QUOTIENT RINGS
Subrings of a ring R are subsets of R that inherit a ring structure from R.
Definition 2.1. A subring of a ring R is a non-empty subset S of R such
that S is a subgroup of (R, +), is closed under multiplication (x, y ∈ S implies
xy ∈ S).
. Cao
Huy Linh -College of Education-Hue University
§ 2. Subrings, Ideals and quotient rings
51
Notice that if S is a subring of R, then S along with addition and multiplication induces from R is a ring.
Example 2.2. (1) If R is a ring, {0R } and R are subrings of R. We say {0R }
is called the trivial subring of R and R is called the improper subring of R.
(2) The ring of integers Z is a subring of the ring of rational numbers Q.
(3) If m ∈ Z, mZ is a subring of Z.
Proposition 2.3. Let R be a ring and S be a non-empty subset of R. The
following conditions are equivalent
(i) S is a subring of R.
(ii) For all x, y ∈ S: x + y ∈ S, −x ∈ S, and xy ∈ S.
(ii) For all x, y ∈ S: x − y ∈ S and xy ∈ S.
Proof. The proof is left as an exercise for the reader.
Example 2.4. Let R be a ring and the center of R is defined by
Z(R) = {x ∈ R | xa = ax for all a ∈ R}.
Then, Z(R) is a subring of R. Indeed, as 0R x = x0R = 0, 0R ∈ Z(R). It
follows Z(R) 6= ∅. Next, we take any x, y ∈ Z(R). Then, for all a ∈ R,
(x − y)a = xa − ya = ax − ay = a(x − y). Thus x − y ∈ Z(R). In addition,
(xy)a = x(ya) = x(ay) = (xa)y = (ax)y = a(xy). This implies xy ∈ Z(R). So,
Z(R) is a subring of R.
Definition 2.5. Let R be a ring. A subring I of (R) is said to be a left (right)
ideal of R if for all r ∈ R and for all x ∈ I, we have rx ∈ I (resp. xr ∈ I). A
subset I of R is said to be an ideal of R if I is both left and right ideal of R.
From now, for short we say an ideal instead of a left ideal.
Example 2.6. (1) If R is a ring, {0R } and R are ideals of R. We say {0R } is
called the trivial ideal of R and R is called the improper ideal of R.
(2) If m ∈ Z, mZ is a subring of Z.
(3) In the ring of rational numbers Q, Z is not an ideal of Q.
. Cao
Huy Linh -College of Education-Hue University
52
Chương 3. Rings
Proposition 2.7. Let R be a ring and ∅ 6= I ⊆ R. Then the following conditions are equivalent
(1) I is an ideal of R.
(2) For all x, y ∈ I and for all a ∈ R: x + y ∈ I, −x ∈ I, ax ∈ R.
(3) For all x, y ∈ I and for all a ∈ R: x − y ∈ I, ax ∈ R.
Proof. The proof is left as an exercise for the reader.
Example 2.8. Let R be a ring and z ∈ R. Then
Rz = {xz | x ∈ R}
is a left ideal of R. Indeed, since 0R = 0.z ∈ Rz, Rz 6= ∅. For all x, y ∈ Rz
and for all a ∈ R, there exists x1 , y1 ∈ R such that x = x1 z and y = y1 z. We
have x − y = x1 z − y1 z = (x1 − y1 )z ∈ Rz. On the other hand, ax = a(x1 z) =
(ax1 )z ∈ Rz. This implies that Rz is a left ideal of R.
Similarly, zR is a right ideal of R.
Proposition 2.9. The intersection of a non-empty family of left (right) ideals
is a left (right) ideal of R.
Proof. We need to show for one side ideal and another side is similar. Assume
T
that Λ 6= ∅ and Ij are left ideals of R for all j ∈ Λ. Set I = j∈Λ Ij . We need
to prove that I is a left ideal of R. Indeed in Chapter 1, I is a subgroup of
the additional group (R, +). Now, for all a ∈ R and x ∈ I, then x ∈ Ij for all
j ∈ Λ. It follows ax ∈ Ij for all j ∈ Λ. Therefore ax ∈ ∩j∈J Ij = I. This implies
that I is a left ideal of R.
Now, let X be a subset of a ring R. The family of left ideals of R that contain
X is non-empty, because R is an element of this family. So, the intersection of
the family of left ideals of R that contain X is a left ideal of R.
Definition 2.10. Let X be a subset of a ring R. The intersection of the family
of left (right) ideals of R that contain X is called the left (reght) ideal of R
generated by X; this ideal is denoted by (Xi (res. hX)).
Remark 2.11.
. Cao
Huy Linh -College of Education-Hue University
53
§ 2. Subrings, Ideals and quotient rings
Let X be a subset of a ring R.
(i) (Xi is the smallest left ideal of R that contains X.
(ii) (∅i = h∅) = {0R }.
(iii) If X = {x1 , ..., xr }, we can write (x1 , ..., xr i instead by ({x1 , ..., xr }i
and hx1 , ..., xr ) instead by h{x1 , ..., xr })
Example 2.12. Let R be a ring with identity, z ∈ R and X = {z}. Then
(zi = Rz and hz) = zR.
Question: Let R be a ring without identity and z ∈ R. Determine
(zi and hz).
Definition 2.13.
(i) A left (res. right) ideal I of a ring R is said to be cyclic if there exists
z ∈ R such that I = (zi (res. I = hz));
(ii) An integral domain D is said to be a principle ideal domain (PID) if
every ideal of R is cyclic.
Example 2.14.
(1) Z is a principle ideal domain;
(2) Let K be a field. Then, the polynomial ring K[x] is a principle ideal
domain.
Definition 2.15. Let R be a ring.
(1) If I1 , ..., Ir are left ideals of R, then we define
r
[
I1 + · · · + Ir = (
Ii i.
i=1
(2) If I1 , ..., Ir are right ideals of R, then we define
I1 + · · · + Ir = h
r
[
i=1
. Cao
Huy Linh -College of Education-Hue University
Ii ).
54
Chương 3. Rings
Example 2.16. Let R be a ring with identity, z1 , ..., zr ∈ R. Then
(z1 , ..., zr i = Rz1 + · · · + Rzr
and
hz1 , ..., zr ) = z1 R + · · · + zr R.
Now, let R be a ring and I both sides ideal of R. Then, I is a normal
subgroup of additional group R. We have a quotient group (R/I, +) with
elements x̄ = x + I for x ∈ R and x̄ + ȳ = x + y. Notice that x̄ = ȳ iff
x − y ∈ I.
Define a multiplication on R/I as follows
x̄.ȳ = x.y (that means (x + I)(y + I) = xy + I).
Proposition 2.17. The additional group (R/I, +) along with above multiplication becomes a ring.
Proof. The proof is left as an exercise for readers.
Definition 2.18. The ring R/I as in Proposition 2.17 is called a quotient ring
of R for I.
Example 2.19. Take R = Z and I = mZ is both sides ideal of Z. Then
Z/mZ = Zm .
Question: Let I be an ideal of a ring R. What happens if I contains a unit
or identity?
EXERCISES
2.1. Show that every intersection of subrings of a ring R is a subring of R.
2.2. Show that intersection of the family of subrings of R that contains a
subset X of R is the smallest subring of R. This subring is called the subring
of R generated by X, it is denoted by [X].
2.3. Show that the subring [X] of R generated by X is the set of all sums of
products of elements of X and opposites of such products.
. Cao
Huy Linh -College of Education-Hue University
§ 2. Subrings, Ideals and quotient rings
55
2.4. Let A be a ring and P ⊆ A. Put
Z(P ){x ∈ A | xp = px for all p ∈ P }.
Prove that
a) Z(P ) is a subring of R
b) If P1 ⊂ P2 , then Z(P2 ) ⊂ Z(P1 ).
c) Z(Z(Z(P ))) = Z(P ) for all P ⊂ A.
d) Z(P ) = Z([P ]) for all P ⊂ A.
2.5. Let I and J be ideals of a ring R . Show that union of I and J is an ideal
of R if and only if I ⊆ J or J ⊆ I.
2.6. Let R be a ring (without identity) and z ∈ R. Show that
(zi = Rz + Zz and hz) = zR + zZ.
2.7. Prove that if I is a left (res. right) ideal of a ring R contains a unit, then
I = R. In particular, if an ideal I contains identity, then I = R.
2.8. Find an example of a ring R and a left ideal I of R but I is not a right
ideal of R.
2.9. An element x of a ring is nilpotent when xn = 0 for some n > 0. Show
that the nilpotent elements of a commutative ring R constitute an ideal of R.
2.10. Let R be a ring and m ∈ Z. Show that the subset
I = {x ∈ A | mx = 0R }
is an ideal of R.
2.11. Let R be a ring and
X = {xy − yx | x, y ∈ R}.
Show that R/(X) is a commutative ring.
. Cao
Huy Linh -College of Education-Hue University
56
Chương 3. Rings
§3
RING HOMOMORPHISMS
In this section, we will study maps from a ring R into a ring S that preserves
two binary operations between R and S.
Definition 3.1. Let R and S be rings. A map f : R −→ S is said to be a ring
homomorphism if
(i) f (x + y) = f (x) + f (y) for all x, y ∈ R;
(ii) f (x.y) = f (x).f (y) for all x, y ∈ R.
From Definition 3.1, if f : R −→ S is a ring homomorphism then f is
a homomorphism of additional groups from R to S. Therefore, we get the
following proposition.
Proposition 3.2. Let f : R −→ S be a ring homomorphism.
(i) f (0R ) = 0S ;
(ii) f (−x) = −f (x) for all x ∈ R;
(iii) f (mx) = mf (x) for all x ∈ R and m ∈ Z;
(iv) f (x − y) = f (x) − f (y).
Proof. The proof is left as an exercise for the reader.
Example 3.3. (1) Let R, S be rings. The map
f : R −→ S
x 7−→ f (x) = 0S
is a ring homomorphism and it is called a trivial homomorphism.
(2) Let S be a subring of a ring R. The map
j : S −→ R
x 7−→ j(x) = x
is a ring homomorphism and it is called a inclusive homomorphism. In particular, if S = R then j = IdR is identification homomorphism.
. Cao
Huy Linh -College of Education-Hue University
57
§ 3. Ring homomorphisms
(3) Let I be a both sides ideal of a ring R. The map
p : R −→ R/I
x 7−→ p(x) = x̄ = x + I.
is a ring homomorphism and it is called a canonical projection (or canonical
epimorphism). .
Proposition 3.4. Let f : R −→ S be a ring homomorphism.
(i) If R1 is a subring of R, then f (R1 ) is a subring of S;
(ii) If S1 is a subring of S, then f −1 (S1 ) is a subring of R.
Proof. (i) Since 0R ∈ R1 , 0S = f (0R ) ∈ f (R1 ). It follows f (R1 ) 6= ∅. For all
y1 , y2 ∈ f (R1 ), there exist x1 , x2 ∈ R1 such that y1 = f (x1 ) and y2 = f (x2 ).
Then,
y1 − y2 = f (x1 ) − f (x2 ) = f (x1 − x2 )
and
y1 .y2 = f (x1 ).f (x2 ) = f (x1 .x2 ).
Since R1 is a subring of R, x1 − x2 ∈ R1 and x1 .x2 ∈ R1 . Hence y1 − y2 ∈ f (R1 )
and y1 .y2 ∈ f (R1 ). This implies that f (R1 ) is a subring of S
(ii) It is clear that f −1 (S1 ) 6= ∅. Suppose that x1 , x2 ∈ f −1 (S1 ). Then
f (x1 ) ∈ S1 and f (x2 ) ∈ S1 . Since S1 is a subring of S,
f (x1 − x2 ) = f (x1 ) − f (x2 ) ∈ S1
and
f (x1 .x2 ) = f (x1 ).f (x2 ) ∈ S1 .
Hence x1 −x2 ∈ f −1 (S1 ) and x1 .x2 ∈ f −1 (S1 ). This implies f −1 (S1 ) is a subring
of R.
The above proposition tell us that ring homomorphisms preserve structure
of subrings.
Remark 3.5.
. Cao
Huy Linh -College of Education-Hue University
58
Chương 3. Rings
Let f : R −→ S be a ring homomorphism.
(i) If I is a left ideal of R, then f (I) is not necessary a left ideal of S. For
example, taking the inclusive homomorphism j : Z −→ Q. Then Z is an ideal
of Z, but j(Z) = Z is not an ideal of Q.
(ii) If J is a left of S, then f −1 (J) is a left ideal of R.
Proof. By Proposition 3.4, f −1 (J) is a subring of R. For all a ∈ R and x ∈
f −1 (J), we have f (x) ∈ J. Since J is a left ideal of S, f (ax) = f (a).f (x) ∈ J.
Hence ax ∈ f −1 (J). So, f −1 (J) is a left ideal of R.
Definition 3.6. Let f : R −→ S be a ring homomorphism. Then,
(i) Im(f ) = f (R) is called the image of f ;
(ii) Ker(f ) = f −1 (0S ) = {x ∈ R | f (x) = 0S } is called the kernel of f .
Notice that Im(f ) is a subring of S and Ker(f ) is a two-sides ideal of R.
Definition 3.7.
(i) A ring homomorphism f : R −→ S is said to be monomorphic (res.
epimorphic, isomorphic) if f is injective (res. surjective, bijective).
(ii) Two rings R and S are called isomorphic to each other if there exists a
ring isomorphism f : R −→ S; then denote by R ∼
= S.
The following theorem give a criteria for a ring homomorphism to be
monomorphic (res. epimorphic, isomorphic).
Theorem 3.8. Let f : R −→ S be a ring homomorphism.
(i) f is monomorphic if and only if Ker(f ) = {0R }.
(ii) f is epimorphic if and only if Im(f ) = S.
Proof. (i) Suppose that the ring homomorphism f is monomorphic. Then For
all x ∈ Ker(f ), we have f (x) = 0S = f (0R ). Since f is injective, x = 0R . It
follows Ker(f ) = {0R }.
On the other hand, suppose that Ker(f ) = {0R }. Taking any x1 , x2 ∈ R
such that f (x1 ) = f (x2 ). Then f (x1 −x2 ) = f (x1 )−f (x2 ) = 0S . Thus x1 −x2 ∈
Ker(f ) By hypothesis Ker(f ) = {0R }, hence x1 − x2 = 0R . This implies that
x1 = x2 . So, f is injetive; that means f is monomorphism.
(ii) Easy and the proof is left for the reader.
. Cao
Huy Linh -College of Education-Hue University
§ 3. Ring homomorphisms
59
Corollary 3.9. Let f : R −→ S be a ring homomorphism. Then, f is isomorphic if and only if Ker(f ) = {0R } and Im(f ) = S.
Theorem 3.10. Let f : R → S be a ring homomorphism and g : R −→ T
be a ring epimorphism such that Ker(g) ⊆ Ker(f ). There exists a unique ring
homomorphism h : T −→ S such that f = hg. Moreover,
(i) If Ker(g) = Ker(f ), then h is monomorphic;
(ii) If f is epimorphic, then h is epimorphic.
Proof. For any y ∈ T , there exists x ∈ R such that y = g(x) (by hypothesis g
is epimorphic). Then, set h(y) = f (x). This defines a map h : T → S. Indeed,
for any y1 , y2 ∈ T , there exist x1 , x2 ∈ R such that y1 = g(x1 ), y2 = g(x2 ).
Suppose that y1 = y2 , then g(x1 ) = g(x2 ). It follows g(x1 − x2 ) = 0T . Hence
x1 − x2 ∈ Ker(g). By hypothesis Ker(g) ⊆ Ker(f ), x1 − x2 ∈ Ker(f ). That
means f (x1 − x2 ) = f (x1 ) − f (x2 ) = 0S . This implies that h(y1 ) = f (x1 ) =
f (x2 ) = h(y2 ). So, h is a map.
Next, we need to prove h is a ring homomorphism and f = hg. Indeed,
for all y1 , y2 ∈ T . Suppose y1 = g(x1 ) and y2 = g(x2 ). One has y1 + y2 =
g(x1 ) + g(x2 ) = g(x1 + x2 ) and y1 .y2 = g(x1 ).g(x2 ) = g(x1 .x2 ). Then,
h(y1 + y2 ) = f (x1 + x2 ) = f (x1 ) + f (x2 ) = h(y1 ) + h(y2 )
and
h(y1 .y2 ) = f (x1 .x2 ) = f (x1 ).f (x2 ) = h(y1 ).h(y2 ).
Hence h is a ring homomorphism. Moreover, for any x ∈ R, set y = g(x). We
have
hg(x) = h(g(x)) = h(y) = f (x).
It follows that hg = f .
Finally, soppose that there is a ring homomorphism h0 : T → S such that
f = h0 g. Then, for all y ∈ T , there exists x ∈ R such that y = g(x). we have
h0 (y) = h0 (g(x)) = h0 g(x) = f (x) = h(y).
This implies that h0 = h.
. Cao
Huy Linh -College of Education-Hue University
60
Chương 3. Rings
(i) Now, suppose that Ker(g) = Ker(f ). For all y ∈ Ker(h) and y = g(x),
h(y) = f (x) = 0S . Hence x ∈ Ker(f ) = Ker(g). This implies y = g(x) = 0T . It
follows that h is monomorphic.
(ii) Suppose that f is epimorphic. Then,
Im(h) = h(T ) = h(g(R)) = hg(R) = f (R) = Im(f ) = S.
So, h is epimorphic.
From the above theorem, we obtain the following corollary.
Corollary 3.11. d Let f : R −→ S be a ring homomorphism. Then,
(i) If f is epimorphic, then R/ Ker(f ) ∼
= S.
(ii) We always have
R/ Ker(f ) ∼
= S.
Proof.
EXERCISES
3.1. Let f : R −→ R be a ring homomorphism (it is called a ring endomorphism) and
B = {x ∈ R | f (x) = x}.
Show that B is a subring of R.
3.2. Let Mn (R) be the ring of n by n matrices with coefficients in R. If Ck is
the subset of Mn (R) = (aij ) consisting of matrices such that all entries aij are
0 except perhaps in the k-th column. Show that Ck is a left ideal of Mn (R).
Similarly, if Rk consists of matrices such that all entries aij are 0 except perhaps
in k-th row, then Rk is a right ideal of Mn (R).
3.3. Let R be a commutative ring with identity whose only ideals are {0} and
R. Show that R is a field.
3.4.
a) Let f : R −→ S be an epimorphism of rings and let I be a left ideal of
R. Show that f (I) is a left ideal of S.
. Cao
Huy Linh -College of Education-Hue University
61
§ 4. Characteristic of rings
b) Find a ring homomorphism f : R −→ S of commutative rings and a left
ideal I of R such that f (I) is not a left ideal of S.
3.5. Let f : R −→ S be a homomorphism of rings and let J be an ideal of S.
Show that f −1 (J) is an ideal of R.
3.6. Let R be a ring and let I be a both-sides ideal of R. Show that every
ideal of R/I is the quotient J/I of a unique ideal J of R that contains I.
3.7. Let I ⊆ J be both-sides ideals of a ring R . Show that
(R/I)/(J/I) ∼
= R/J
.
3.8. Let R be a ring and I, J be ideals of R. Denote by I + J is the smallest
ideal of R that contains I ∪ J. Show that
I + J = {x + y | x ∈ I, y ∈ J.}
3.9. Let R be a ring and I, J be both-sides ideals of R. Show that
I + J/J ∼
= I/(I ∩ J.
3.10. Let S be a subring of a ring R and let I be an ideal of R . Show that
S + I = {x + y | x ∈ S, y ∈ I}
is a subring of R , I is an ideal of S + I , S ∩ I is an ideal of S , and
(S + I)/I ∼
= S/(S ∩ I).
§4
CHARACTERISTIC OF RINGS
Definition 4.1. Let R be a ring. If there exists a positive integer n such that
nx = 0R for all x ∈ R, then the smallest such positive integer is called the
characteristic of R, and denoted by Char(R). If no such positive integer exists,
then R is said to be characteristic zero.
. Cao
Huy Linh -College of Education-Hue University
62
Chương 3. Rings
Example 4.2.
(1) Char(Zm ) = 0.
(2) Char(Z) = Char(Q) = Char(R) = Char(C) = 0.
If the ring has unity, the following theorem is useful to find the characteristic of a ring.
Proposition 4.3. Let R be a ring with unity (identity). .x (i) If n1R = 0R for
some n ∈ Z+ , then the smallest such integer n is the characteristic of R; that
is,
min{n ∈ Z+ |n1R = 0R }.
(ii) If n1R 6= 0R for all n ∈ Z+ , then R has characteristic zero.
Proof. (i) Suppose that n1R = 0R for some n ∈ Z+ . Then, for all x ∈ R we
have
n.x = n.(1R .x) = (n.1R ).x = 0R .x = 0R
(?).
Therefore k = Char(R) > 0. Let k 0 = min{n ∈ Z+ |n1R = 0R }. From (?), it
follows that k ≤ k 0 . Since k.1R = 0R end the smallest of k 0 , k 0 ≤ k. Hence
k = k0.
(ii) Evidently.
Example 4.4.
(i) Char(Zm × Zn ) = lcm(m, n).
(ii) Char(Z × Zm ) = 0.
Theorem 4.5. The characteristic of an integral domain D is either zero or a
prime.
Proof. Suppose that D is an integral domain and Char(D) = k > 0. Assume
that k = k1 .k2 . Then
0D = k.1D = (k1 .k2 ).1D = (k1 .1D ).(k2 .1D ).
Since D is an integral domain, k1 .1D = 0 or k2 .1D = 0. From the smallest of
k, it follows k1 = k or k2 = k. This implies that k is prime.
. Cao
Huy Linh -College of Education-Hue University
63
§ 4. Characteristic of rings
Corollary 4.6. The characteristic of a field F is either zero or a prime.
EXERCISES
4.1. Find the characteristic of the rings: Z5 × Z7 , Z4 × Z6 and Z2 × Z.
4.2. Prove that the characteristic of a finite ring R divides |R|.
4.3. Let F be a field of order 2n . Prove that Char(F ) = 2.
4.4. Find an infinite ring R such that Char(R) > 0.
4.5. Let R be a ring with identity. Show that
a) If Char(R) = n > 0, then R contains a subring isomorphic to Zn ;
b) If Char(R) = 0, then R contains a subring isomorphic to Z;
c) Every field F contains a subfield isomorphic to either Zp or Q.
4.6. Let K be a subfield of F . Show that Char(K) = Char(F ).
4.7. Show that if Fq is a finite field of characteristic p, then |Fq | = pn for some
positive integer n.
4.8. Let X be a set and P(X) the set of subsets of P . Define the two operations
+ and . on P(X) as follows:
A + B := (A ∪ B) \ (A ∩ B);
A.B := A ∩ B,
for A, B ∈ P(X).
a) Prove that (P(X), +, .) is a commutative ring with identity.
b) Find Char(P(X)).
. Cao
Huy Linh -College of Education-Hue University
64
Chương 3. Rings
§5
FIELD OF FRACTIONS OF A INTEGRAL DOMAIN.
A ring R is said to be embedded in a ring S if there exists a ring monomorphism of R into S. From this definition, any ring R can be embedded in a ring
S if there exist a subring of S which is isomorphic to R, i.e., R ∼
= f (R) ⊂ S.
Motivated by the construction of Q from Z, here we show that any integral
domain D can be embedded in a field F .
Let D be an integral domain, denoted D∗ = D \ {0D }. We define a binary
ralation ∼ on D × D∗ as follows:
(a, b) ∼ (c, d) if ad = bc for all (a, b), (c, d) ∈ D × D∗ .
Proposition 5.1. The binary ralation ∼ as above is an equivalence relation
on D × D∗ .
Proof. The proof is left as an exercise for readers.
a
The equivalence class := (a, b) = {(c, d) ∈ D × D∗ | ad = bc}. Let denote
b
by FD = D × D∗ the quotient set of D × D∗ with respect to ∼; that is,
a
FD = { | (a, b) ∈ D × D∗ }.
b
a c
For , ∈ FD , define
b d
a c
ad + bc
+ =
;
b d
bd
a c
ac
· = .
b d
bd
Theorem 5.2. The quotient set FD along with two binary operations as above
is a field.
Proof. The proof is left as an exercise for readers.
Definition 5.3. The field FD is said to be a field of fractions of an integral
domain D.
Remark 5.4.
0D
1D
(1)The field FD has the zero element 0FD =
and the identity 1FD =
.
1D
1D
a
b
(2) If x = ∈ FD \ {0FD }, then x−1 = .
b
a
. Cao
Huy Linh -College of Education-Hue University
§ 5. Field of fractions of a integral domain.
65
Example 5.5. (1) FZ = Q.
(2) Let K[x] be a polynomial ring over a field K. It is well known that K[x]
is a domain. Then
f
FK[x] = K(x) = { |f ∈ K[x], g ∈ K[x]∗ }.
g
Now, let D be an integral domain and FD be its field of fractions. Consider
a map
f : D −→ FD
a
a 7−→
.
1D
Proposition 5.6. The map f as above is a ring monomorphism.
Proof. Home work for students.
Theorem 5.6 gives us D ∼
= Im(f ). This allows us identifying D with Im(f ).
a
a
That means an element a ∈ D can be seen as
; that is, a =
.
1D
1D
Corollary 5.7. Any integral domain D can be embedded in a field F .
From Corollary 5.7, if FD is the field of fractions of D, then there exists an
integral subdomain D0 of FD such that D ∼
= D0 . Therefore, the domain D can
be seen as a subring of the field FD .
EXERCISES
5.1. Let D be an integral domain. Show that the field of fractions FD of D is
the smallest field containing D. That is, no field K such that D ⊂ K ⊂ F .
5.2. Any field of fractions of a field K is isomorphic to K; that is, FK ∼
= K.
5.3. Let
Z[i] = {a + bi | a, b ∈ Z}
be the ring of Gaussian integers. Show that
a) Z[i] is an integral domain;
b) FZ[i] = {x + yi | x, y ∈ Q}.
5.4. Find the fields of fractions of Z2 and Z2 [x].
5.5. Find an infinite field F such that Char(F ) > 0.
. Cao
Huy Linh -College of Education-Hue University
66
Chương 3. Rings
. Cao
Huy Linh -College of Education-Hue University
Chương 4
Application
67
68
Chương 4. Application
. Cao
Huy Linh -College of Education-Hue University
BIBLIOGRAPHY
[1] Bùi Huy Hiền - Phan Doãn Thoại - Nguyễn Hữu Hoan, Bài tập đại
số và số học, tập II. Nhà xuất bản Giáo dục, 1985.
[2] Lê Thanh Hà, Các cấu trúc đại số cơ bản . Nhà xuất bản Giáo dục,
1999.
[3] Lê Thanh Hà, Đa thức và Nhân tử hóa. Nhà xuất bản Giáo dục, 2000.
[4] Lê Thanh Hà, Môđun và Đại số. Nhà xuất bản Giáo dục, 2002.
[5] Nguyễn Hữu việt Hưng, Đại số đại cương. Nhà xuất bản Giáo dục,
1998.
[6] Nguyễn Xuân Tuyến - Lê văn Thuyết, Đại số trừu tượng. Nhà xuất
bản Giáo dục, 2005.
[7] M.F.Atiyah-I.G.Macdonald. Introduction to Commutative Algebra.
Addison-wesley Publishing Company, 1969.
[8] M.Hall, The Theory of Groups. New York: Macmillan, 1959.
[9] S.Mac Lane, G.Birkhoff, A survey of Modern Algebra. The Macmillan
Company, New York, 1967.
[10] S.Lang, Algebra. Addison-Wesley, 1971.
[11] A.I.Kostrikin. Introduction à d’algèbre. Mir-Moscou, 1976.
69