451082075 algebra2 pdf

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... Prove the following: a finite group with an ev

CAO HUY LINH

ABSTRACT ALGEBRA

COLLEGE OF EDUCATION - HUE UNIVERSITY

Huế, Febrary 05, 2020.

nior Lecturer of Department of Mathematics , lege of Education-Hue University This lecture note

Col-is used for teaching and learning the course Algebrafor the Class Val de Loire 2019-2020

N0 = N ∪ {0}

NOTATIONS ii

1 Binary Operations 1

2 Groups 5

3 Subgroups 11

4 cyclic groups 15

5 Normal subgroups and quotient groups 19

6 Group homomorphisms 25

7 Direct product of groups 33

2 Rings 37 1 Rings and Fields 37

2 Subrings, Ideals and quotient rings 42

3 Ring homomorphisms 47

4 Characteristic of rings 52

5 Field of fractions of a integral domain 52

3 Application 53 References 55

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 multipli- cation, x.y or xy is called the product of x and y.

(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

(4) Addition and multiplication of matrices also provide binary operations

on the set M n (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

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 x 0 ∈ X such that x 0 ∗ x = e (resp.

x ∗ x 0 = 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 x 0 ∈ X such that

x 0 ∗ x = x ∗ x 0 = e Then the such element x 0 is called the inverse of x and denote by x −1

(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 tion (+) and multiplication (·).

addi-(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.

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 x l , , x n be elements of G (where n is an integer

> 1) We define their product inductively:

n

Y

i=1

x i = x1· · · x n = (x1· · · x n−1 ) · x n

x n = x1 · · · x n which x i = x for i = 1 n and x0 = e.

If (X, +) be a monoid, then we define

nx = x1+ · · · + x n where x i = · · · x n 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.

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), ∩)

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 1 G 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.

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, Z m , a group under multiplication (¯ x.¯ y = ¯ 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 GL n (K) the set of invertible matrices of the size n over a field

K Then GL n (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, , i n of X determines a function σ : X → X, namely, σ(1) = i1, σ(2) =

i2, , σ(n) = i n 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

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 ample, we obtain 1 → 3 → 4 → 1, 2 → 5 → 2 We express this result by

ex-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 nodifference which mapping is applied first

The above analysis illustrates the fact that any permutation can be

ex-pressed 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; theproduct 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

Proposition 2.4 The set of all the permutations of a set X, denoted by S X , along with composition is a group and it is called the symmetric group on X.

When X = {1, 2, , n}, S X is usually denoted by S n, 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 10

G are two identity elements of G Then

1G= 1G 1 0

G = 10

G

(ii) Assume that y and z are two inverses of x Then

y = y.1 G = y.(x.z) = (y.x).z = 1 G 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 1 G y = 1 G z.

Hence y = z; that is, x satifies the left cancellation law Similarly, x also

satifies the right cancellation law

§ 2 Groups 9

Proposition 2.8 Let (G, ·) be a group Then

(i) (1 G)−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

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 1 G 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 = 1 G and a2 is a solution of the equation a.x = 1 G Then

a1 = a1 1 G = a1 (a.a2) = (a1.a).a2 = 1G a2 = a2

Hence a −1 = a1 = a2 is the inverse of a So, G is a group.

2.1 Draw a multiplication table of S3

2.2 Let M m×n (K) be the set of m by n matrices over a field K Does M m×n (K)

form a group under matrix addition?

2.3 Let M n (K) ∗ be the set of nonzero square matrices of the size n over a

field K Does M n (K) ∗ form a group under matrix multiplication?

(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 x 0 of x; that is, x 0 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

con-tains 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

is not a group

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 1 G Then H1 = {1 G }

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

(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, 1 G = 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 H i are subgroups of G for i ∈ I Denote

H =Ti∈I H i Suppose that x, y ∈ H Then x, y ∈ H i for all i ∈ I Because H i are subgroup of G, xy −1 ∈ H i for all i ∈ I Hence xy −1 ∈ Ti∈I H i = H This implies that H is a subgroup of G.

§ 3 Subgroups 13

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

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 {1 G }, and so h∅i = {1 G }.

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, , m k } ⊂

Z, then hZi = dZ, where d = (a1, , a k) the greatest common divisor of

a1, , a k

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 = x e1

1 x e n

n , where x i ∈ X and

e i = ±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.

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 1 G = 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 , (a n)−1 = a −n , and a m a n = a m+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

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, a 0 )(b, b 0 ) = (ab 0 + b, a 0 b 0))

§ 4 cyclic groups 15

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

In another word, a cyclic subgroup (or cyclic group) of G is a subgroup

generated by only one element So,

hai = {a n | n ∈ Z}.

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

non-negative 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 ¯4 is 3, the order of ¯5 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 = {1 G }, then it is evident that H is cyclic generated

by 1G If G 6= {1 G } and H 6= {1 G }, there exists an integer m 6= 0 such that

a m ∈ H Since H is a subgroup of G, a −m = (a m)−1 ∈ H Therefore, there is

an positive number h > 0 such that a h ∈ H Suppose that n is the smallest

§ 4 cyclic groups 17

positive integer such that a n ∈ H We need to prove that H is a cyclic subgroup

generated by a n Indeed, for all x = a m ∈ H Then, there exist q, r ∈ Z such

that

m = nq + r for 0 ≤ r < n.

As a r = a m (a n)−q ∈ H, r = 0 Hence m = nq and then

x = a nq = (a n)q = b q ,

where b = a n 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 a m1 6= a m2.

(2) a n= 1G

(3) If a m = 1G then m = nq, for q ∈ Z.

Proof (1) Without loss of generality, we assune that m1 < m2 and a m1 = a m2.

Hence a m2−m1 = 1G Put k = m2−m1 Then 0 < k < n Since a k+l = a k a l = a l,

for l ∈ Z, |G| ≤ k < n This is a contradiction.

(2) From (1), we have G = {1 G , a, a2, , a n−1 } Suppose that a n 6= 1 G

Then, there exists k ∈ Z such that 0 < k < n and a n = a k Hence a n−k = 1G

But 0 < n − k < n This is a contradiction.

(3) From (1), we have G = {1 G , a, a2, , a n−1 } Assume a m = 1G Hence

a −m = (a m)−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 a m = a nq a r = (a n)q a r = a r = 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

0 ≤ r < n, the order of a r is n

(n, r) , where (n, r) is the greatest common

divisor.

Proof Home work

Corollary 4.10 Let G = hai be a finite cyclic group of order n For 0 ≤ r < n,

the element a r 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 a r such that (r, 12) = 1, for 1 ≤ r < 12 Hence r = 1, 5, 7, 11.

4.1 In additional group Z18, determine h¯3i, h¯4i, h¯5i

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 ha r i is n

(n, r).b) a r is a generating element of G if and only if (n, r) = 1.

4.4 Is the symmeteric group S n cyclic? Why?

§ 5 Normal subgroups and quotient groups 19

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 finitegroup which has at most two elements

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 shallsoon 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 cosetsare Ỏparallel; that is, distinct (right) cosets are disjoint

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 whofirst 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 {a1H, a2H, , a t H} be the family of all the distinct cosets of H in

G Then

G = a1H ∪ a2H ∪ ∪ a t H,

because each g ∈ G lies in the left coset gH, and gH = a i H for some i

Moreover, Lemma 5.3 (ii) shows that the left cosets partition G into pairwise

disjoint subsets It follows that

|G| = |a1H| + |a2H| + · · · + |a t H|.

But |a i 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 arethe 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.

§ 5 Normal subgroups and quotient groups 21

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 a n = 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 a n= 1G

Proof If a has order d, then |G| = dm for some integer m, by the previous

corollary, and so a |G| = a dm = (a d )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 = Nx 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 {1 G } and the non-proper

are normal subgroups of G.

(3) Define the center of a group G, denoted by Z(G), to be

(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 (xNx −1 ⊆ N) for all x ∈ G, h ∈ N.

Proof Assume that N / G Then for all x ∈ G, and for any a ∈ xNx −1, there

exists h ∈ N such that a = xhx −1 Since xN = Nx, there is an element

h 0 ∈ N such that xh = h 0 x Therefore xhx −1 = h 0 ∈ N Hence xNx −1 ⊆ N.

Conversely, suppose that xNx −1 ⊆ N for all x ∈ G It follows xN ⊆ Nx.

Moreover, take y = x −1 , we have x −1 Nx ⊆ N This implies Nx ⊆ xN So,

xN = Nx; 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.

§ 5 Normal subgroups and quotient groups 23

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

a) Show that ∼ is an equivalent relation on G.

b) Show that ¯a = aH for all a ∈ G.

5.2 a) In the symmetric group S3, let σ2 = (12), σ5 = (123) and H =

hσ2i, K = hσ5i Find all left cosets and right cosets for H and K.

b) In Z12, let H1 = h¯2i, H2 = h¯3i Find all left cosets and right cosets for

H1 and H2

5.3 Recall GL n (R), set of square matrices A in M n (R) such that det(A) 6= 0,

along with matrix multiplication is a group and it is called a general lineargroup

a) Define the special linear group by

SL(n, R) = {A ∈ GL n (R)| det(A) = 1}.

Prove that SL(n, R) is a normal subgroup of GL n(R)

b) Prove that GL n (Q) is a subgroup of GL n(R) Is the general linear group

GL(n, Q) a normal subgroup of GL n(R)?

5.4 Let G be a finite group with subgroups H and K Prove that if H 5 K,

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

a p = 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

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.