Abstract Algebra (I) 242(Prequisite Math 151)

Level :6Creidt Hours: 3 hoursInstructor: Dr. Rola Assad Hijazi1421/ 2000

1. Groups and Subgroups:

1. a. Defdinitions of Binary Operation, Mathematical Systems, Commutative,Associative, Distributive Operations, Identity and Inverse element ,Examples.Definitions of Semi-group, Group, Examples. Finite and Infinitegroups, Group Tables.

b. Properties with proofs.c. Subgroups, Examples, Properties with proofs.d. Exersices.

2. Two Important Groups:

1. a. Group of integers modulo n :i. Definition of congruent modulo n, congruenss class modulo n. Define a binary

operations ⊕ ,⊙on Zn.ii. Theorem Zn,⊕ forms a commutative group( with proof).

iii. Theorem Zn,⊙ forms a commutative semi-group with identity 1 ∈ Zn.iv. Zp

∗,⊙ is a commutative group.v. Construct the operation tables for some examples of Zn,⊕, Zn,⊙.

1. b. Symmetric group of order n.

1. a. i. Definition of permutation. Theorem SG, ∘ forms a group which is called asymmetric group on the set G with proof). Definition of Symmetric group oforder n, Examples. Definition of order of Sn .

ii. Definitions of k −cycle, transposition, disjoint cycles, product of cycles, orderof the cycle, order of the product cycles, examples.

iii. Theorem every permutation can be written uniquely as a product of disjointcycles (without proof).

iv. Definitions of even and odd permutation, and an Alternating groups,Examples.

v. Theorem An, ∘ form a group of order n!2 (with proof).

1. c. Exercises.

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3. Cyclic Groups:

1. a. Definitions of Cyclic group, Generator, Order of an element g ∈ G, The cyclicsubgroup generated by g, Examples.

b. Theorems: Every cyclic group is commutative, and every subgroup of a cyclicgroup is cyclic (with proofs).

c. Finite and Infinite cyclic groups and applications .i. How to find all distinct subgroups of finite cyclic groups with examples.

ii. How to find all generators of the finite cyclic groups with examples.d. Exersices.

4. Normal subgroups, Lagrange’s theory, and Quotient groups:

1. a. Definitions of left and right cosets, the index of H in G , examples.b. Theorem the relation ≡l mod H is an equivalence relation on G, the

equivalence classes are left cosets of H in G (with proof).c. Lagrange Theorem (with proof), applications on lagrange theorem.d. Definition of normal subgroups (in different ways) with examples, properties with

proofs.e. Define the quotient set G/H with coset multiplication (coset addition), theorem

G/H, . forms a group which is called the quotient group of G by H with proof),examples. Definition of commutator, and properties (with proofs).

f. Exercises.

5. Homomorphisms and Cayley’s Theorem:

1. a. Definitions of Homomorphism, Isomorphism, Endomorphism, andAutomorphism, Examples.Theorems HomG,G, ∘ forms a semi-group withidentity ( with proof), AutG, ∘ forms a subgroup of SG (with proof), and thehomomorphism functions preserve identites and inverses (with proofs).

b. Definitions of kernel f, and image f and properties of kernel and image( withproofs). Theorem the natural map (with proof).

c. A finite cyclic group ≅ Zn and an infinite cyclic group ≅ Z (with proof).d. Cayley’s theorem (with proof), example.e. Exercises

6. Fundamental Homomorphisms Theorems.

1. a. The First Homomorphisms Theorem (with proof).b. The Second Homomorphisms Theorem (with proof).c. The Third Homomorphisms Theorem (with proof).d. Exercises.

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7. Rings.

1. a. Definition of a ring, examples. Properties with proofs. Definitions of division ring,field , an integral domain, zero divisors with examples.

b. Subring with examples.c. Theorems the relation between field and an integral domain (with proofs).

References:

1. Abstract and linear Algebra by David M. Bruton2. A First course in Abstract Algebra. by John B. Fraleigh3. Topics in Abstract Algebra by Herstein4. Elements of Modern Algebra by Gilbert5. Lecture Notes by Dr. Hamza Abu Jabal

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1. GROUPS AND SUBGOUPS

The idea of a group is one of the focal points of modern algebra. This idea isgeneral and abstract, and it is interesting and important because of the cumulative interestand importance of its many special cases. Roughly, a group is a set of elements that can becombined through some operation such that as addition or multiplication, subject to somedefinite rules like ordinary addition of numbers. The elements may be something other thannumbers, however, and the operation something other the usual operations of arithmetic.For instance, the elements of the groups used to study symmetry are things like rotations,translations, and reflections.

1.1 GROUPSThe definitions of sets, relations, functions (mappings), and binary operations are

essential for a study of an algebraic system. An algebraic system is a non empty set inwhich at least one or more binary operations are defined . The simplest cases occur whenthere is only one- binary operation, as in the case with the algebraic system known as agroup. Before beginig, we introduce the concept of a binary as follows:

Definition 1.1:A binary operation on a non-empty set S is a function (mapping) of the cartesianproduct S S into S.

: S S S given by a,b a b (1)

Thus, a binary operation assigns to each ordered pair S S a uniquely element of thesame set.

Remark 1.2:1. We generally use symbols as , −, , , ∘, , and , to represent binary

operations.

2. If is a binary operation on S then we say that S is closed under the operation or isa closed operation.

3. If : S S T is a function such thet T S, then we say that is not a closedoperation and if T ⊆ S then is a closed operation.

Definition 1.3:By a mathematical system or algebraic system, we shall mean a non-empty set of

elements S with one or more binary operations defined on this set. A mathematicalsystem with one operation is denoted by S, , with two operations is denoted by S,, ∘.

Example 1.4:1. Z, is an algebraic system.

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2. Z,−, Q, are not an algebraic systems.

3. Q∗, ,R, and Z,− are algebraic systems.

Definition 1.5:A binary operation : S S → S on the set S is

(a) Commutative: if x y y x ∀ x , y ∈ S.

(b) Associative : if x y z x y z ∀ x, y, z ∈ S.

(c) Left distributative: Let ∘ be another binary operation on S, then is a leftdistributive over ∘, if

x y ∘ z x y ∘ x z ∀ x,y, z ∈ S.

(d) Right distributative: is a right distributive over ∘ ,ify ∘ z x y x ∘ z x ∀ x,y, z ∈ S.

(e) Distributive: If is both left and right distributive over ∘ ,then is said to bedistributive over ∘.

Example 1.6:1. Addition and Multiplication of real numbers are commutative and associative

operations.

2. Union and Intersection of sets are commutative and associative operations.

3. The Algebraic systems Z,, , Q,, and R ,, are all algebraic systems withoperation is distributive over .

4. PA,∩,, and PA,,∩ are algebraic systems with operation is distributiveover ∩ and ∩ is distributive over .

5. Z,,− is a mathematical system , but the operation − is not distribute over theoperation .

Exercise :Study the distributive property of the following system PA,,−?

Definition 1.7:A Mathematical system S, is said to have an identity element if there exists e ∈ S

such that

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x e e x x ∀ x ∈ S. (2)

Example 1.8:1. In R, , zero is the identity element.

2. In PA, , ∅ is the identity element.

3. In PA,∩ , A (universal set) is the identity element.

4. Z, has no identity.

Definition 1.9:Let S, be a mathematical system with identity e ∈ S. Then an element x ∈ S is said

to have an inverse under the operation denoted by x−1 such that :

x x−1 x−1 x e (3)

Definition 1.10:A semi-group is a pair S, consisting of a non-empty set S together with an

associative binary operation defined on S.

Example 1.11:On the set Q of rational numbers, define the binary operation by

a b a b ab.

Then the system Q, forms a commutative semi-group with identity element 0. Eacha ∈ Q, other than −1 ,has an inverse , for :

a b 0 a b ab 0 a b1 a 0 b − a

1 a ,where a ≠ −1.

Similarly , b a 0 b a−1 − a1 a , a ≠ −1.

There is a mathematical system, known as a group, which displays most of theproperties we have so far discussed.

Definition 1.12:A group is a pair G, consisting of a non-emptey set G and a binary operation

defined on G ,satisfying the following four properties:

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1. G is closed under the operation .2. The operation is associative.3. G contains an identity element e for .4. Each a ∈ G has an inverse a−1 ∈ G.

Remark 1.13:If it happens that the group operation satisfies the commutativity property, then G,

is called a commutative group or an abelian group.

Example 1.14:Let a ≠ 0 be a real number. Define

G na : n ∈ Z.

Then G, is an abelian group, where 0 is the identity element, and −na ∈ G is theinverse element of na.

Example 1.15:Let G a,b : a,b ∈ R, a ≠ 0. Define on G by

a,b c,d ac, bc d.

Show that G, is a non-commutative group?

Solution1. Since a ≠ 0 ac ≠ 0 ac, bc d ∈ G. i.e., is a closed operation.2. is an associative operation, for

a,b c,d e, f ac, bc d e, f ace, bc de f

ace, bce de f

a,b ce, de f a,b c,d e, f

3. Consider

a,b e1,e2 a,b e1,e2 a,b ae1, be1 e2 a,b ae1 a and be1 e2 b e1 1 and e2 0.

Thus the identity element is 1,0 ∈ G.

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4. Consider

a,b a−1,b−1 a−1,b−1 a,b 1,0 aa−1, ba−1 b−1 a−1a, b−1a b 1,0 aa−1 a−1a 1 and ba−1 b−1 b−1a b 0

a−1 1a and b−1 − b

a .

i.e. The inverse of a,b is 1/a, − b/a ∈ G.

5. is not commutative for 1,2 3,4 3,10 ≠ 3,4 1,2 3,6.

Example 1.16:Let G f1, f2, f3, f4, f5, f6 consists of six functions from R − 0,1 into itself,

where

f1x x , f2x 1/x , f3x 1 − x ,

f4x 11 − x , f5x x − 1

x , f6x xx − 1 .

Then G, ∘, ( where ∘ is a functional composition ) is a non-commutative group.

Solution :1. Clearly ∘ is not commutative, for consider

f2 ∘ f4x f2f4x f2 11 − x 1 − x.

f4 ∘ f2x f4f2x f4 1x x.

2. To prove that G, ∘ is a group, consider the following operation table:

∘ f1 f2 f3 f4 f5 f6

f1 f1 f2 f3 f4 f5 f6

f2 f2 f1 f4 f3 f6 f5

f3 f3 f5 f1 f6 f2 f4

f4 f4 f6 f2 f5 f1 f3

f5 f5 f3 f6 f1 f4 f2

f6 f6 f4 f5 f2 f3 f1

Clearly from the above table G, ∘ is a group.

Theorem 1.17:Let G , be a group. Then

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1. The identity element is unique.2. The inverse of each element is unique.3. If a b a c, then b c (left cancellation).

Proof :1. Let G, has 2-identites elements e1, e2. Then

e2 e1 e1 e2 e2

e1 e2 e2 e1 e1

e1 e2.

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2. Let x be an element in G with two inverses x1 ,x2. Then

x x1 x1 x ex x2 x2 x e.

Considerx1 x1 e x1 x x2 x1 x x2 e x2 x2

3. Let a b c b, since G, is a group , then every element has inverse.i.e., b has inverse say b−1.

i.e., a b b−1 c b b−1. a e c e a c.

Example 1.18:Let V e, a , b , c . Define on V by

e a a e a , e b b e be c c e c , c b b c aa b b a c , a2 b2 c2 e

Construct the group table?

Solution :

e a b ce e a b ca a e c bb b c e ac c b a e

Thus V is an abelian group which is called Klien -4-group, denoted by K4.

Theorem 1.19:If G is a group, then

(a) a−1−1 a ∀ a ∈ G.(b) ab−1 b−1a−1 ∀ a , b ∈ G.Proof :(a) By definition, a−1−1 is the inverse of a−1.

a−1−1.a−1 a−1.a−1−1 e

But a.a−1 a.a−1 e.

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a is the inverse of a−1. But inverses of group elements are unigue. Thus a−1−1 a.

(b) To show that ab−1 b−1a−1, it is sufficient to show that

b−1a−1ab abb−1a−1 e ??

Consider

b−1a−1ab b−1a−1ab b−1a−1ab b−1eb b−1b e

A similar computation leads to abb−1a−1 e. Therefore ab−1 b−1a−1.

Remark 1.20:1. If S is a semi-group and a1,a2, ...,an are a set of n ≥ 3 elements of S (not necessarily

distinct), then all possible products of a1,a2, ...,an taken in this order and arrived at byreplacing parentheses in any meaningful positions, yield one and the same element.This called a General Associative Law.

2. If a1,a2, ...,an n ≥ 2 are elements of the commutative semi-group S, then the producta1,a2, ...,an is independent of the order of the factors.

Definition 1.21:Let G be a group and a ∈ G . We define the power of a as follows:

(i) a0 e is the identity in G.

(ii) an

n−times

a.a....a , for n 0, and n ∈ Z.

(iii) a−n a−1n

n−times

a−1.a−1....a−1 for n ∈ Z.

Definition 1.22:1. Let G be a group. Then the number of the elements in G is called the order of G, and

denoted by |G| .2. The group G is called finite if it has a finite number of elements and called infinite if

it has an infinite number of elements.

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Theorem 1.23:Let x, y be elements of the group G, and m ,n be integers. Then

(a) xn.x−n e(b) xm.xn xmn xm.xn.(c) xn−1 x−n x−1n

(d) xmn xmn.(e) If G is abelian, then xyn xnyn.

Proof :(a) Exercise.

(b) xmxn xmn (The following cases will examine all combination of m, n).

Case 1 : Either m or n is zero., i.e., suppose m 0; n ≠ 0. Then

xmn x0n xn e.xn x0.xn.Thus xmxn xmn.

Case 2 : Suppose m , n 0.

xmn

mn-times

x.x.x...xm−times

x.x...xn−times

x.x...x xm.xn

Case 3 : Suppose both m ,n 0. Let m −p , n −q, for some p ,q ∈ Z. Thenp ,q 0, therefore

xmn x−pq

x−1pq by definition 1.21 x−1px−1q by case 2 x−px−q by definition 1.21 xmxn

Case 4 : Suppose that m ,n have opposite signs. We will consider m 0, n 0. Wehave 2 cases :(i) m 0, n 0 and m n ≥ 0. (ii) m 0, n 0 and m n 0.

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(i) When m n ≥ 0 .Then

xmn xmne xmnx−nxn

xmnx−nxn

xmn−nxn from case 2, since − n 0 xmxn

(ii) m n 0 − m n 0 by case 2 , we get

x−mnxm x−m−nxm x−m−nm x−n

Consider

xmn xmnx−nxn since x−nxn e xmnx−m−nxmxn

xmnx−m−nxmxn

xmxn since x−n−mis the inverse of xmn

(c) x−n xn−1 x−1n . We will consider 3-cases for n ∈ Z.

Case 1 : For n 0, the statement is true.

Case 2 : Suppose n 0, by definition 1.21, x−n x−1n (use induction).

(i) For n 1, x−1 x1−1 x−11. i.e., the statement is true.

(ii) Suppose the statement is true for n k. Then

x−k xk−1 x−1k.

(iii) Consider n k 1

x−k1 x−1k1 by definition 1.21 x−1kx−1

xk−1x−1 by (ii)

This implies that

xk1x−k1 xxkxk−1x−1

xex−1 e.

That is, x−k1 xk1−1. Therefore the statement is true for n k 1. Hence it istrue for all n 0.

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Case 3 : Suppose n 0. Then m −n 0

x−n xm xm−1−1 since x−1−1 x x−m−1 since m 0 by case 2 xn−1 since − m −−n n

(1)

and

x−n x−1−1−n

x−1−−n by definition 1.21 x−1n since − −n n

(2)

Hence from equation (1) and (2) we get

x−n xn−1 x−1n

(d) xmn xmn. Again we will consider 3-cases.

Case 1 : If n 0 then the statement is true.

Case 2 : Suppose n 0, [use induction] :(i) For n 1 we have

xm1 xm1

So the statement is true for n 1.

(ii) Asume the the statment is true for n k 1. i.e.,

xmk xmk

(iii) Consider

xmk1 xmkm xmk.xm

xmk.xm

xmk1.

Case 3 : Suppose n 0 − n 0 and so

xmn x−mn−1 xm−n−1 xm−n−1 xmn.

(e) Exercise.

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1.2 SUBGROUPS:

Definition 1.24:Let H be a non-empty subset of the group G. Then H is said to be a subgroup of G if H

itself a group under the same operation as that of G, we denote it by H ≤ G.

Notation Any group has two trivial subgroups G and e.

Example 1.25:Let G Z, the group of integers under . Then Ze, Zo (The set of even and odd

integers, resp.) are subsets of Z. But Ze, is a subgroup of Z while Zo, is not asubgroup of Z (why?).

Theorem 1.26:Let G be a group and ∅ ≠ H ⊆ G. Then

H ≤ G a,b ∈ H ab−1 ∈ G.

Proof :Suppose H ≤ G, a,b ∈ H b−1 ∈ H a,b−1 ∈ H ab−1 ∈ H (closure condition.)

Suppose that ∅ ≠ H ⊆ G which contains ab−1 ∈ H.i.e., ,ab−1 ∈ H whenever a,b ∈ H.

i ∵ b,b ∈ H bb−1 ∈ H e ∈ H

ii ∵ e,b ∈ H eb−1 ∈ H b−1 ∈ H.

iii ∵ a, b−1 ∈ H ab−1−1 ∈ H ab ∈ H.

Since H inherits the associative law as a subset of G. Hence H ≤ G.

Theorem 1.27:Let Hi be a collection of subgroups of a group G . Then ∩Hi ≤ G.

Proof :(i) ∩Hi ≠ ∅, since each Hi contains the identity element of G.

(ii) Let a,b ∈ ∩Hi a,b ∈ Hi to each i ab−1 ∈ Hi for each i ab−1 ∈ ∩Hi ∩ Hi ≤ G.

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Theorem 1.28:Let H1, H2 be subgroups of a group G. Then

H1 H2 ≤ G H1 ⊆ H2 or H2 ⊆ H1.

Proof :Suppose a,b ∈ H1 H2

a,b ∈ H1 or a,b ∈ H2. ab−1 ∈ H1 or ab−1 ∈ H2.

ab−1 ∈ H1 H2 H1 H2 ≤ G.

Suppose to the contrary of this assertion were false, i.e.,

H1 H2 and H2 H1 , H1 H2 ≤ G.

∃ a ∈ H1 − H2 and b ∈ H2 − H1.

if ab ∈ H1 b a−1ab ∈ H1

if ab ∈ H2 a abb−1 ∈ H2

ab ∉ H1 and ab ∉ H2 ab ∉ H1 H2 .

Corollary 1.29:A group G cannot be the union of two of its proper subgroups.

Proof :For let G H1 H2 , then by theorem 1.28, either H1 ⊆ H2 or

H2 ⊆ H1 G H1 H2 H1 or G H1 H2 H2 .

Definition 1.30:The center of a group G , denoted by centG, is the set

centG g ∈ G : xg gx ∀ x ∈ G.

Theorem 1.31:Let G be a group. Then centG ≤ G.

Proof :(i) centG ≠ ∅, since ex xe ∀ x ∈ G e ∈ centG.

(ii) Let a,b ∈ centG. Then ax xa and bx xb ∀ x ∈ G. Considerab−1x ab−1

xb b−1 ab−1bxb−1

ax b−1 xab−1.

ab−1 ∈ centG.

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Definition 1.32:A subgroup H of a group G is a proper subgroup if H ≠ G. We denote it by H G.

Example 1.33:In Example 1.18, the subgroups of the Kelin 4 are V, e, e,a, e,b, e,c.

Theorem 1.34:let H be a non-empty subset of a group G. Then

H ≤ G i H is closed ab ∈ H ∀ a,b ∈ Hii a−1 ∈ H for each a ∈ H.

Proof : If H ≤ G it is trivial that i, ii satisfied.

Suppose that i, ii holds, we want to prove that H ≤ G i.e., H is a group itself.(1) ∵ i holds, then the closure condition is satisfied.(2) Since . is associative then

a.b.c a.b.c ∀ a,b,c ∈ H.

(3) ∵ H ≠ ∅, then at least one element say a ∈ H. But by ii a−1 ∈ H a.a−1 ∈ H by i i.e., e ∈ H.

(4) Condition ii means that every element has inverse. Thus H ≤ G.

Example 1.36:Let G be a group, a ∈ G. Let H x ∈ G : x an for n ∈ Z. Then H ≤ G.

Solution :(i) H ≠ ∅ since e a0 ∈ H.

(ii) If x,y ∈ H x an, y am for some m,n ∈ Z.Consider

xy an.am anm ∈ H where n m ∈ H.

(iii) Let x an ∈ H n ∈ Z. Then

x−1 a−n ∈ H where − n ∈ Z.

Thus H ≤ G. This subgroup is called subgroup generated by a and denoted by ⟨a.

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PROBLEMS (1)

1. Suppose that the system S, has an identity e ∈ S. If the equation

a b c d a c b d ∀ a,b,c,d ∈ S.Show that is associative and commutative.

2. Let S be a set containing more than one element. If the operation is defined by

a b b ∀ a,b ∈ S.

Verify that S, is a non-commutative semi-group which admits no identity.

3. In the following cases , determine whether the systems G, are commutative groups;for those systems failing to be so, indicate which axioms are not satified.

(a) G Z, a b 0.

(b) G Q − 1, a b a b ab.

(c) G Z Z, a,b c,d a c,b d.

(d) G R R, a,b c,d ac − bd, ad bc.

4. Let G be the set if mappings f1, f2, f3, f4 from R − 0 into itself, where

f1x x, f2x 1/x, f3x −x, f4x −1/x

for each x ∈ R − 0. Verify that the system G, ∘ forms a commutative group.

5. For any group G, prove that the following conditions are equivalent.

(a) G is commutative.(b) ab−1 a−1b−1 ∀ a,b ∈ G.(c) ab2 a2b2 ∀ a,b ∈ G.

6. If a2 e for each element of the group G, verfiy that G is a commutative group.

7. For each of the following sets H, establish that H, . is a subgroup of G, . :

(a) G 1,−1, i,−i, H 1,−1 where i2 −1.

(b) G Q − 0, H 2n : n ∈ Z.

(c) G R − 0, H a b 2 : a,b ∈ Q ,a2 b2 ≠ 0

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8. Let G be a group and a ∈ G . Define

Ca x ∈ G : xax−1 a.

Show that Ca ≤ G.

9. Let H be a subgroup of G, a be a fixed element of G, and K be the set of all elementsof the form

K x ∈ G : x aha−1 for some h ∈ H.

Show that K ≤ G.

10. Assume that H, K are subgroups of the abelian group G. Let

HK g ∈ G : g hk h ∈ H, k ∈ K.

Prove that HK ≤ G.

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2. TWO IMPORTANT GROUPS

This section is devoted to an examination of two important and frequently usedgroups: The group of integers modulo n and the group of permutations of the elements of aset, the so-called symmetric group. We begin with an investigation of the notion ofcongruence, in terms of which the group of integers modulo n will be formulated.

2.1 Group of integers modulo n :

Definition 2.1:Let n be a fixed positive integer. Two integers a and b are said to be

congruent modulo n, written

a ≡ b modn a − b is divisible by n.

i.e., a − b qn for some integer q. If a − b is not divisible by n, then we say that a isincongruent to b.

For example, 3 ≡ 24 mod7 and − 5 ≡ 2 mod7.

Clearly, ≡ is an equivalence relation. To find the equivalence classes we give thefollowing definition.

Definition 2.2:For an arbitrary integer a, let

a a x ∈ Z : x ≡ a modn x ∈ Z : x a qn q ∈ Z

We call a a the congruenss class modulo n, determined by a and refer to a as arepresentative of this class.

By the way of illustration, suppose we are dealing with congruence modulo 3. Thus

0 x ∈ Z : x ≡ 0 mod3 x ∈ Z : x 3q q ∈ Z ...,−6,−3,0,3,6, ....

1 x ∈ Z : x ≡ 1 mod3 x ∈ Z : x 3q 1 q ∈ Z ...,−5,−2,1,4,7, ....

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2 x ∈ Z : x ≡ 2 mod3 x ∈ Z : x 3q 2 q ∈ Z ...,−4,−1,2,5,8, ....

Remark :1. Every integer lies in one of these 3 classes.2. Integers in the same class are congruent modulo 3, while integers in different classes

are incongruent modulo 3.

We return to the general case of congruence modulo n, let

Zn Z/⟨n 0, 1, 2, ...,n − 1.

This is the set of all equivalence classes.

Theorem 2.3:Let n be a ve integer and Zn be defined as above. Then

1. a ≠ ∅ for each a ∈ Zn.2. If a ∈ Zn such that b ∈ a b a.3. For any a, b ∈ Zn where a ≠ b a ∩ b ∅.4. a : a ∈ Z Z.

Definition 2.4:A binary operation ⊕ may be defined on Zn as follows:

a ⊕ b a b for each a, b ∈ Zn.

Thus

a ⊕ b a b if a b na b − n if a b ≥ n

.

For example : in Z7, 3 ⊕ 2 5 and 3 ⊕ 6 2.

We are now in a position to prove one of the principal theorems of this section.

Theorem 2.5:For each ve integer n, the mathematical system Zn,⊕ forms a commutative group,

known as the group of integers modulo n.

Proof :1. ⊕ is well defined. (i.e., ⊕ is a function), i.e., if

a1,b1 a,b

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a1 ⊕ b1 a ⊕ b ? ?

Let a1,b1 a,b a1 a and b1 b. a1 − a qn and b1 − b q′n q,q′ ∈ Z. a1 − a b1 − b qn q′n a1 b1 − a b q q′n q′′n q′′ ∈ Z. a1 b1 ≡ a b a1 b1 a b a1 ⊕ b1 a ⊕ b.

2. Zn,⊕ is closed.Since Z, is closed, then ∀ a,b ∈ Z Z : a b ∈ Z. a b ∈ r where r ∈ Zn. a b r ∈ Zn.

3. ⊕ is commutative.

∀ a, b ∈ Zn : a ⊕ b a b b a b ⊕ a.

4. ⊕ is associative.∀ a, b, c ∈ Zn :

a ⊕ b ⊕ c a b ⊕ c a b c a b c a ⊕ b c a ⊕ b ⊕ c.

5. e ∈ Zn.Consider a ⊕ e a, a e a a e − a qn where q ∈ Z e qn e qn 0 ∈ Zn.

6. Inverse.Let b a−1 be the inverse of a. Then

a ⊕ b 0 a b 0 a b ≡ 0 modn a b − 0 qn where q ∈ Z. b qn − a where q ∈ Z. b n − a when q 1.

Thus the inverse of a is n − a.

Therefore Zn,⊕ is a commutative group.

Definition 2.6:Let n be a fixed ve integer. Define in Zn, ⊙ by

24

a ⊙ b a.b for each a, b ∈ Zn.

Theorem 2.7:Zn,⊙ is a commutative semi-group, with identity 1 ∈ Zn.

Proof :Homework.

Example 2.8:Construct the operation table for

(1) Z5,⊕ (2) Z3,⊙ (3) Z5∗,⊙.

1. Z5,⊕

⊕ 0 1 2 3 40 0 1 2 3 41 1 2 3 4 02 2 3 4 0 13 3 4 0 1 24 4 0 1 2 3

2. Z3,⊙

⊙ 0 1 20 0 0 01 0 1 02 0 2 1

3. Z5∗,⊙

⊙ 1 2 3 41 1 2 3 42 2 4 1 33 3 1 4 24 4 3 2 1

25

26

2. Symmetric Group Of Order n :

Definition 2.9:1. Let G be any non-empty set. A permutation on a set G is a bijection (1-1 and onto)

from G to itself. Denote SG the set of all permutations on G.

2. Take G N 1,2,3, ...,n. Then the set of all permutations on N will be denoted bySn .

Theorem 2.10:The set SG is a group under function composition ∘. This group is called

symmetric group on the set G.

Proof :(i) Let f,g ∈ SG. Then f,g are bijective f ∘ g is also bijective (see 151).

Thus f ∘ g ∈ SG i.e., ∘ is a closed operation.

(ii) Since function composition is associative , in general ∘ is associative.

(iii) The identity of SG is the permutation id : G G defined bya a ∀ a ∈ G.

(iv) Since every element f ∈ SG is bijctive , therefore f−1 exists and it is bijective (see151) i.e., f−1 ∈ SG which satisfies f ∘ f−1 f−1 ∘ f id.

Hence SG, ∘ is a group.

Notation : Sn will be called the symmetric group of order n.

Example 2.11:Consider the set N 1,2,3. Find all permutations of N .

Solution :

(1) 11 1 , 12 2 , 13 3(2) 21 2 , 22 3 , 23 1(3) 31 3 , 32 1 , 33 2(4) 41 1 , 42 3 , 43 2(5) 51 2 , 52 1 , 53 3(6) 61 3 , 62 2 , 63 1

Thus there are 6 permutations of N.

27

Remark :The order of Sn n!

Since the injective fumction can send the number 1 to any n −elements of1,2, ...,n; 2 can then be any one

of the elements of this set except 1, so there are n − 1 choices for 2; 3 canbe any element except

1; 2 , so there are n − 2 choices for 3, and so on. Thus there are precisely

n.n − 1.n − 2.n − 3...3.2.1 n!

possible injective functions from N to itself.

Any permutation ∈ Sn may be described as a set of ordered pairs

a 1,1, 2,2, ..., n,n.

Or we represent in a two-line form:

1 2 3 ... n

1 2 3 ... n.

The columns may be rearranged without affecting the nature of the function . Forinstance, the following two symbols both represent the same element of S4.

1 2 3 42 4 3 1

or2 1 4 34 2 1 3

Permutations, being functions , may be multiplied under the operation offunctional composition. The resulting product is not just a function; it is also a permutationof N (see 151). Thus for , ∈ Sn.

∘ 1 2 3 ... n

1 2 3 ... n∘

1 2 3 ... n1 2 3 ... n

1 2 3 ... n

1 2 3 ... n∘

1 2 3 ... n1 2 3 ... n

1 2 3 ... n

1 2 3 ... n.

What we have done is to rearrange the columns of the first (left) permutationuntil its top row is the same as the bottom row of the second (right) permutation; the prouct

28

∘ is then the permutation whose top row is the top row of the second factor and whosebottom row is the bottom row of the first factor.

29

Example 2.12:In example 2.11,

1 1 2 31 2 3

, 2 1 2 32 3 1

, 3 1 2 33 1 2

,

4 1 2 31 3 2

, 5 1 2 32 1 3

, 6 1 2 33 2 1

Consider

4 ∘ 6 1 2 31 3 2

∘1 2 33 2 1

1 2 32 3 1

∘1 2 33 2 1

1 2 32 3 1

2

6 ∘ 4 1 2 33 2 1

∘1 2 31 3 2

1 2 33 1 2

3

6 ∘ 4 ≠ 4 ∘ 6. ∘ is not commutative.

Theorem 2.13:The system Sn, ∘ forms a group, known as the symmetric group on n symbols, which

is non-commutative for n ≥ 3.

Note that : the identity element on Sn, ∘ is the permutation

id 1 2 3 ... n1 2 3 ... n

while the multiplicative inverse of any permutation ∈ Sn is described by

30

−1 1 2 3 ... n

1 2 3 ... n.

In the following definition, we introduce a special type of permutation called cycle.

Definition 2.14:An element ∈ Sn is a cycle or k −cycle of length k if there exists a set i1, i2, ...., ik

of distinct integers such that

i1 i2, i2 i3, ..., ik−1 ik, ik i1

while i i for i ∉ i1, i2, ..., ik. i.e., leaves all other elements fixed.

Lemma 2.15:A cycle of length k has order k.

Example 2.16:

Let 1 2 3 4 51 5 2 4 3

253. This is a cycle permutation of length 3.

253 532 325.

Definition 2.17:A cycle of length 2 is called a transposition.

Notation : All cycles of length 1 ( 1-cycles) are equal to the identity permutation.

Definition 2.18:(i) Two cycles a1a2....ak and b1b2...bs of a set N are called disjoint if they

have no integer in common. i.e.,

a1,a2, ...,ak ∩ b1,b2, ...,bs ∅.

(ii) A set Γ of cycles is disjoint if any two distinct cycles in Γ are disjoint.

Example 2.19:(i) Let A 1,2,3,4,5 and

1 2 3 4 52 1 4 3 5

and 1 2 3 4 53 4 2 1 5

Then 1234 , 1324.

31

(ii) In S7 , let 1324, and 1762. Then ∩ ≠ ∅. Their product is definedusing mapping composition.

∘ 13241762 176424 ∘ 17621324 132476

∘ ≠ ∘ . (Note: the product of two cycles need not be a cycle).

(iii) If 15362, then

∘ ∘ 1324176215362 1527634.

Lemma 2.20:Let ∈ Sn have its cycle decomposition into disjoint cycles of length m1, ...,mk. Then

the order of is the least common multiple of m1, ...,mk.

Note : Disjointness is very important in the above Lemma for instance :

If 1213 132 the order of 3 while the order of 12 2 and theorder of 13 2 L.c.m. 2.

Example 2.21:The order of ∘ 176423 is the least common multiple of 4 , 2 4.

Notation : Disjoint cycles commute, while not disjoint cycles are not commute.

Take 45, 123 ∈ S5, 45123 12345 12345.

The following theorem shows why cycles are so important in the theory ofpermutations.

Theorem 2.22:Every permutation ≠ identity permutation) in Sn can be written uniquely as a

product of disjoint cycles.(Without proof).

For example : 1 2 3 4 5 65 6 1 4 3 2

15326.

Remark 2.23:1. Every transposition is its own inverse, since ijij ij id.

2. Every cycle can be expressed as a product of transpositions. i.e.,

32

a1a2a3...ak a1aka1ak−1...a1a2.

For example, let ∈ S4 s.t. 1234 141312.

3. Thus every permutation in Sn can be expressed as a product of transposition.

4. The factorization into a product of transpositions is not unique, for

1234 12234 122334

Therfore :

123....k 122...k 12233...k.... 122334...k − 1k

1k1k − 11k − 2...12

Example 2.24:Let 123 ∈ S3 and 16253 ∈ S6. Then , can be factorized as

follows: 1223 or 1312 and 162553 or 162325

Definition 2.25:(i) A permutation which can be expressed as the product of an even number of

transpositions is called an even permutation.

(ii) A permutation which can be expressed as a product of an odd number oftranspositions is called an odd permutation.

Example 2.26:(i) Let 12346 ∈ S6. Then 16141312 is an even permutation.

(ii) Let 1245 ∈ S5 122445 is an odd permutation.

(iii) Let 132524345 ∈ S6 151213243534 is aneven permutation.

Remark 2.27:(a) The product of any two even or any two odd permutations is even.(b) The product of an odd permutation and an even permutation is odd.(c) The inverse of an even permutation is an even permutation, and the inverse of an odd

permutation is an odd permutation.(d) The identity permutation is even.

33

Definition 2.28:Define An to be the collection of all even permutations. i.e.,

An ∈ Sn : is an even permutation.

and Bn be the collection of all odd permutations. i.e.,

Bn ∈ Sn : is an odd permutation.

Theorem 2.29:Let n ≥ 2 be an integer. Then An, ∘ form a group of order n!

2 which is calledthe alternating group of n-letters.

Proof :Since the identity permutation is even id ∈ An. The other conditions are easy

straight forward.

Now, let Sn An Bn such that An ∩ Bn ∅. Therefore

|Sn | |An Bn | n!

For a fixed transposition , define the function

f : An Bn by f ∘ where ∈ An

Thus ∘ ∈ Bn. Moreover,

(i) ft is one-one, for let

f f ∘ ∘ Since ∈ Sn −1 ∈ Sn

−1 ∘ ∘ −1 ∘ ∘

(ii) f is onto

since ∀ ∈ Bn ∃ ∈ An s.t. f . ∘

−1 ∘ ∈ An

i.e., f−1 ∘

Hence we have a correspondence between An and Bn . Thus |An | |Bn | n!/2.

34

PROBLEMS(2)

1. If a ≡ b modn and m ∣ n. Prove that a ≡ b modm.

2. Construct the operation tables for the groups Z8 ⊕ and Z7∗,⊙.

3. Show that the system 0,4,8,12, ⊕ mod16 forms a group.

4. Express the following permutations as(a) Products of disjoint cycles(b) Products of transpositions.(c) Decide whether these permutations are even or odd.(d) Find the order of each permutation.

1 2 3 4 5 63 6 4 1 2 5

, 1 2 3 4 5 6 75 7 6 2 1 3 4

1 2 3 4 5 6 7 8 9 1010 5 6 4 9 7 8 3 2 1

.

5. Form the set G f, f2, f3, f4, f5, f6 where f is the permutation

f 1 2 3 4 5 62 3 4 5 6 1

Prove that the pair G, ∘ is a commutative group.

6. Compute gfg−1 for each pair f, g of the following:(a) f 1243 g 132(b) f 1356 g 2546(c f 13524 g 2534

7. Which of the following functions is a permutation:(a) f1 : R R defined by f1x x 2.(b) f2 : R R defined by f2x x2.

8. List all elements of S3 and then find the alternating group A3 written in cycle notation.

9. List all elements of S4 and then find the alternating group A4 written in cycle notation.

10. If a ≡ b modn and x is any integer, then a x ≡ b x modn and ax ≡ bx modn.

35

11. If a ≡ b modn and c ≡ d modn , then a c ≡ b d modn and ac ≡ bd modn.

12. If ax ≡ ay modn and a,n 1 ,then x ≡ y modn

13. An element a ∈ Zn has a multiplicative inverse iff a,n are relatively prime.

14. Calculate , , −1, −1, −1, −1 where

1 2 3 4 5 62 3 6 5 4 1

, 1 2 3 4 5 61 3 5 6 2 4

15. Construct the operation table of S3 ∘.

36

3. CYCLIC GROUPS

Let G be any group and a be any element of G. One way of forming a subgroupH of G is by letting H be the set of all integer (ve, -ve, and zero) powers of a (thisguarantees closure under inverses and products at least as far as a is concerned). In thissection we study groups which are generated by one element.

Definition 3.1:A group G is called a cyclic group if there exist at least one element g ∈ G such that

G gn : n ∈ Z , if the operation is .G ng : n ∈ Z , if the operation is .

The element g is called the generator of G. The cyclic group generated by g is denotedby ⟨g.

Example 3.2:Let G 1,−1, i,−i where i2 −1. Then G, . is a group .Moreover, it is a cyclic

group gnerated by i, − i, for

i1 i , i2 −1 , i3 −i , i4 1 ,−i1 −i , −i2 −1 , −i3 i , −i4 1.

Therefore G ⟨ i ⟨ − i .

Example 3.3: (Infinite cyclic)Consider the group Z,.Then Z, is a cyclic group generated by 1, − 1, for

Z 1 n.1 : n ∈ Z ...,−2.1,−1.1,0,1.1,2.1,3.1,4.1, ... ...,−2,−1,0,1,2,3,4, ....

and

Z −1 n−1 : n ∈ Z ..., 3, 2,1,0,−1,−2, .....

Example 3.4: (Finite cyclic)The additive group Zn, ⊕ is a cyclic group with generator 1 , for

Zn 1 n.1 : n ∈ Z 0,1,1 ⊕ 1, 1 ⊕ 1 ⊕ 1, ... 0,1,2,3, ...,n − 1.

37

Elements other than 1 may also be generators. To illustrate this , consider the particularcase

Z6 0, 1, 2, 3, 4, 5.

Then 5 is a generator of Z6, for

Z6 5 n.5 : n ∈ Z 0.5, 1.5, 2.5, 3.5, 4.5, ........ 0, 1, 2, 3, 4, 5

The cyclic subgroups generated by the other elements of Z6 under addition are asfollows:

0 n.0 : n ∈ Z 0 2 n.2 : n ∈ Z 0, 2, 4 3 n.3 : n ∈ Z 0, 3 4 n.4 : n ∈ Z 0, 4, 2

Thus 1, and 5 are the only elements that are generators of the entire group.

Example 3.5:Consider 3Z, . This is a group of the multiple of 3 . Then this group is cyclic

generated by 3 for

3Z 3 n.3 : n ∈ Z ...,−3, 0, 3, 6, ....

In general, nZ n , for n ∈ Z, which is called the group of the multiple of n.

Notation 6Z 3Z mZ nZ if m n.

Definition 3.6:(i) Let G be a group . The order of g ∈ G is the smallest ve integer n such that

gn e ng e if the operation is () .We denote it by og n 0.(ii) If there is no such integer we say that g has infinite order denoted by og .

Example 3.7:(i) In example 3.4, o5 6, o2 3, o3 2, o4 3.(ii) In example 3.3, every non-zero element in Z is of infinite order.

Theorem 3.8:Given a group G , and g ∈ G, then the following are true :

(i) H gn : n ∈ Z ≤ G.(ii) H is a smallest subgroup of G containing g. i.e., if K ≤ G containing g H ⊆ K.

38

Proof :

(i) H is closed under the group opration, for if

gr, gs ∈ H gr.gs grs ∈ H where r s ∈ Z.

Moreover, g0 e ∈ H.Lastly, if y gr ∈ H gr−1 g−r ∈ H where −r ∈ Z. Hence H contains inverses

of all its elements. H ≤ G.

(ii) Suppose H g gn : n ∈ Z and K ≤ G s.t. g ∈ K. g.g ∈ K.Inductively, it is true that g.g.g...g gn ∈ K for any n ∈ Z .That is ∀ x ∈ K s.t. x gn x gn ∈ K.Therefore H ⊆ K. i.e., H is the smallest subgroup of G containing g.

Definition 3.9:(i) Let G be a group, and let g ∈ G. Then

H gn : n ∈ Z

is a subgroup of G . This group is called the cyclic subgroup of G generated by g .We write H g .

(ii) If G is a group such that for a ∈ G,

G am : m ∈ Z

then G is a cyclic group. Denote it by G a .

Theorem 3.10: (Division Algorthim for ZIf a,b ∈ Z such that b 0, then ∃ unique integers q, r s.t.

a qb r 0 ≤ r b.

Proof (is not required.)

Theorem 3.11:

Let G be a cyclic group. Then(i) G is abelian (every cyclic group is abelian)(ii) If H ≤ G H is cyclic. (every subgroup of a cyclic group is cyclic).

Proof :(i) Suppose G a an : n ∈ Z. Let x, y ∈ G .

39

x an, y am where n,m ∈ Z. xy an.am anm amn am.an y.x

G is abelian.

(ii) Let G a an : n ∈ Z, H ≤ G.If H e, the theorem is trivially true, for e is a cyclic subgroup generated by the

identity element.Suppose H ≠ e. If am ∈ H a−m ∈ H since H ≤ G. H must contain a ve powers of a. Suppose n is the least ve integer such that

an ∈ H.We have to show that any element in H is a power of an, i.e.,

H an an t : t ∈ Z ??

Let ak be arbitrary element in H. Since n is the least ve integer such that an ∈ H,then we can divide k by n and by Division Algorithim we get,

k qn r where 0 ≤ r n.

Consider

ar ak−qn ak.a−qn ak.an−q.

∵ ak, an ∈ H ar ∈ H since n is the least ve integer s.t.an ∈ H, r n .

r 0, k qn, i.e., k is a multiple of n. ak anq ∈ an , H an .

Theorem 3.12:Let a be an element in a group G . If n is the least ve integer such that an e, i.e.,

oa n. Then

(i) a has order n, and a e, a, a2, ..., an−1.

(ii) If am e n ∣ m.

(iii) If ak as k ≡ s modn.

Proof :(i) Firstly, we have to show that e, a, a2, ......,an−1 are all distinct. o a n.

Secondly , we show that any power of a is equal to these elements.Firstly, suppose

ai aj , 0 ≤ i n, 0 ≤ j n, i ≥ j. ai.a−j aj.a−j e ai−j e 0 ≤ i − j n.

But n is the least ve integer s.t. an e i − j 0 i j.Thus a contains the n −distinct elements e, a, a2, ..., an−1.

40

Secondly, let ak be any arbitrary element. By Division Algorithim there exists q, r s.t.

k nq r where 0 ≤ r n, r 0, 1, 2, ..., n − 1.

Consider

ak anqr anq,ar anq.ar eq.ar e.ar ar.

It follows that a e, a, a2, ..., an−1 and a has order n.

(ii) Suppose that n ∣ m m nq, q ∈ Z.Consider

am anq anq eq e.

Suppose that am e, then by Division Algorithm ,∃ q, r s.t. m nq r , 0 ≤ r n. Then

am anqr anq.ar anq.ar eq.ar ar.

But am e ar e s.t., r n , r 0, m nq n ∣ m.

(iii) Let ak as

ak.a−s as.a−s ak−s e.By (ii) n ∣ k − s k − s nq where q ∈ Z.

k ≡ s modn

Conversely, Let k ≡ s modn

k − s is divisible by n k − s qn , q ∈ Z k nq s.

Consider

ak anqs anq.as e.as as.

Definition 3.13:The order of an element a oa of the group G is the order of the subgroup generated

by a. That is oa o a .

Theorem 3.14:Let G be a finite cyclic group of order n i.e., G a , |G| n with generator

a ∈ G. Then

41

(1) If d m,n is the greatest common divisor for the integer m,n, then the subgroupgenerated by am is the same as the subgroup generated by ad.

i.e., ad am .

(2) The distinct subgroup of G are those subgroups ad da if the operation is(), where d is a ve divisor of n.

(3) If 1 m,n i.e., m,n are relatively prime, then am is a generator of G.

i.e., G am if the operation is . ma if the operation is ()

.

42

Notation : Theorem 3.14, provides a systematic way to obtain all subgroups of a cyclicgroup of order n.

Example 3.15:Let G a be a cyclic group of order 12.

(a) Find all distinct subgroups of G.(b) Find all generators of G.

Solution :(a) The divisors of 12 are 1, 2, 3, 4, 6, and 12.

So apply theorem 3,14, the distinct subgroups of G are a G. a2 a2, a4, a6 , a8, a10 , a12 e. a3 a3, a6, a9, e a4 a4, a8 , e. a6 a6, e. a12 e.

(b) By theorem 3.14 (3) , all generators of G are:

∵ 5,12 1 G a a5

∵ 7,12 1 G a a7

∵ 11,12 1 G a a11 .

Note that theorem 3.14 (1), makes it easy to detremine which subgroup is generated byeach element of the group.

∵ 8,12 4 a4 a8

∵ 9,12 3 a9 a3

∵ 10,12 2 a10 a2

Example 3.16:Let G a be a cyclic group of order 10. Then by theorem 3.14 (1),

G a a3 a7 a9 .

Example 3.17:1. Consider the group Z7, ⊕, then Z7 is a cyclic group generated by 1. Since |Z7 | 7,

then by theorem 3.14(3), every element of Z7 except 0, generates Z7 under addition.

Z7 ma m1

where m is relatively prime with 7. i.e.,

43

Z7 1 2.1 3.1 4.1 5.1 6.1

Z7 1 2 3 4 5 6 .

2. Consider Z7∗, ⊙ . Then this is a cyclic group generated by 3, for

3 3n : n ∈ Z

31, 32, 33, 34, 35, 36, .. 3, 2, 6, 4, 5, 1.

Hence Z7∗ 3 of order 6. The by theorem 3.14(3), the other generators are:

Z7∗ 3 35 since 5,6 1.

But 35 5. Therefore Z7∗ 3 5 .

Example 3.18:Consider the cyclic group Z8,⊕.

(i) Find all generators of Z8.(ii) Find all subgroups of Z8.

Solution :(i) Z8, ⊕ is a cyclic group generated by 1. ∵ |Z8 | 8, then all numbers which are

relatively prime with 8 are 3, 5, 7. Therefore by theorem 3.14(3), the generators ofZ8 other than 1 are

Z8 1 3.1 5.1 7.1

Z8 1 3 5 7 .

(ii) The divisor of 8 are 1,2,4,8 so by theorem 3.14(2), we get :

1.1 Z8. 2.1 2 n.2 : n ∈ Z 0, 2, 4, 6. 4.1 4 n.4 : n ∈ Z 0, 4. 8.1 8 n.8 : n ∈ Z 0.

Exercise :As in example 3.18, find all generators and all distinct groups for the following cyclic

groups:(a) Z12, ⊕ (b) Z16, ⊕.

Example 3.19:Consider the group Z5

∗,⊙ .

44

(a) Show that Z5∗ is cyclic

(b) Find all generators .(c) Find all distinct subgroups.

Solution :(a) ∵ 2 2n : n ∈ Z 2, 4, 3, 1.

Therefore Z5∗ is a cyclic group generated by 2. i.e., Z5

∗ 2 and |Z5∗ | 4

(b) By theorem 3.14, 3 is the only number which is relatively prime with 4 therefore

Z5∗ 2 23 3 .

(c) The divisors of 4 are 1,2,4. Therefore the distinct subgroups are:

21 2 Z5∗

22 4 4n : n ∈ Z 4, 1

24 1 1n : n ∈ Z 1.

Do the same as in example 3.19, for a Z12∗ ,⊙ , b Z16

∗ ,⊙.

Example 3.20:Find the cyclic group generated by 1342 ∈ S4.

Solution :

1342 1342n n ∈ Z 1, 2, 3, 4.

2 13421342 1423,

3 134213421342 14231342 1243,4 12431342 1234 id.

, 1342, 1423, 1243.

Note that : o 4 o .

Example 3.21:Find the order of 12345 ∈ S5.

Solution :We want to find the smallest ve integer n s.t., n id.2 1234512345 13524,3 1352412345 14253,4 1425312345 15432,5 1543212345 12345 id. o 5 the order of subgroup generated by 5.

45

Example 3.22:Let 123 ∈ S6, 1245 ∈ S6. Find the cyclic subgroup generated by

−12, i.e., −12 .

Solution :

−1 132 , 2 1425, and −12 14325.

2 1432514325 13542,3 1354214325 12453,4 1245314325 15234,5 1523414325 12345 id.

, , 2, 3, 4.

Note that: o o , and the o length of the cycle .

Theorem 3.23:

Let G a be an infinte cyclic group. i.e., oa . Then1. If n ≠ m ∈ Z an ≠ am.2. a, a−1 are the only generators of G.

Proof :1. Let oa ak ≠ e for any ve integer k.

Suppose n ≠ m ∈ Z s.t. an am. Then

an.a−m am.a−m e

an−m e if n − m 0, or am−n e if m − n 0.

In either cases we get that the element a has finite order .Therefore n ≠ m ∈ Z an ≠ am.

2. We want to prove that if G a has infinite order then G a−1 ??By the way of contradiction , suppose that

G am where m ≠ 1.

a amk amk k ∈ Z amk−1 e mk − 1 ∈ Z.

Therefore the element a has finite order .Thus the only generators are a, a−1.

Example 3.24:In Z, , the only generators are 1, − 1. i.e., Z 1 −1 .

46

Ze, , has two generators 2,−2. i.e., Ze 2 −2 .

47

PROBLEMS (3)

1. Consider the cyclic groups Z15,⊕ and Z18,⊕.(i) Find all generators of Z15, Z18.(ii) Find all subgroups of Z15, Z18.

2. Consider the cyclic groups Z17∗ ,⊙.

(i) Show that Z17∗ ,⊙ is cyclic.

(ii) Find all generators .(iii) Find all distinct subgroups .

3. Suppose G a is a cyclic group of order 24. List all generators.

4. Find the order of each of the elements 4, 5, 6 in the group Z7∗,⊙.

5. Find the order of each of the elements 2, 3, 4 in the group Z8,⊕.

6. Let G be a group, and a,x ∈ G. Suppose b x−1ax. Prove by induction that(i) bn x−1anx ∀ n ∈ Z.(ii) If oa n obab−1 n.

7. Find the elements of Z12,⊕ of order(a) 2, (b) 3, (c) 4, (d) 6, (e) 12.

8. Show that the group S,⊙ ,where S is the subset 1, 3, 5, 7 of Z8 is not cyclic.

9. Prove that the group H,⊙ , where H is the subset H 2, 4, 6, 8 of Z10 is cyclic.Find the generators.

10. Let G be a group , and x,y ∈ G. Use induction to show that

xyn y−1yxny.

48

4. NORMAL SUBGROUPS, QUOTIENT GROUPSAND LAGRANGE’S THEORY

In this section, we narrow the field and focus attention on a restriced classof subgroups which we shall refer to as a normal subgroups. This groups is important forconstruction a new algebraic structure known as quotient groups. As a starting point, weprove several theorems leading to the conclusion that each subgroup induces adecomposition of the element of the parent group into disjoint subsets known cosets.

Definition 4.1:Let H ≤ G and a ∈ G. The left coset of H in G is defined as follows:

aH ah : h ∈ H if the operation is . a H a h : h ∈ H if the operation is .

The right coset of H in G is defined as :

Ha ha : h ∈ H if the operation is . H a h a : h ∈ H if the operation is ,

where a is called a representative of aH.

Remark 4.2:1. The left cosets are different from the right cosets.

2. If G is commutative aH Ha.

3. eH eh : h ∈ H H H is a left coset of H.

4. Moreover, a ∈ H since a ae ∈ aH since e ∈ H.

Example 4.3:Let G S3 , 123, 132, 12, 13, 23 and K , 12, K ≤ S3.

Let 123, 12, 23, then compute K, K, K, K, and K, K ?

Solution :

K k : k ∈ K K k : k ∈ K 123, 12312 123, 12123 123, 13 123, 23.

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aK ≠ Ka.

K k : k ∈ K K k : k ∈ K 12, 1212 12, 1212 12, 12, .

12K K12 K since 12 ∈ K.

K k : k ∈ K K k : k ∈ K 23, 2312 23, 1223 23, 132 23, 123.

Note that 123K 13K, 23K 132K, and 12K K.Therefore, we have 3 −distinct left cosets of K in S3.i.e.,

The left cosets of K in S3 K, 23K, 13K.

Remark :The number of distinct left cosets of K in S3 |S3 ||K| 6

2 .

Example 4.4:Consider the commutative group Z4,⊕ . Let H 0, 2 ≤ Z4 0, 1, 2, 3.Find the distinct left cosets of H in Z4.

Solution :

The number of distinct left cosets |Z4 ||H| 4

2 2.

To find these cosets consider 0 ∈ Z4. Then

0 ⊕ H 0 ⊕ h : h ∈ H 0 ⊕ 0, 0 ⊕ 2 0 , 2 H.

0 ⊕ H 2 ⊕ H.

Let 1 ∈ Z4, then

1 ⊕ H 1 ⊕ h : h ∈ H 1 ⊕ 0, 1 ⊕ 2 1 , 3 H.

1 ⊕ H 3 ⊕ H. Therefore

the left cosets of H in Z4 H, 1 ⊕ H.

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Theorem 4.5:Let H ≤ G. Then there is a one-one correspondence between H and aH, for a ∈ G.i.e., |H| |aH| .

Proof :Let a ∈ G. Define

f : H aH by fh ah ,h ∈ H.

(i) f is 1-1:

fh1 fh2

ah1 ah2

h1 h2 (by cancellation law)

(ii) f is onto:Let y ∈ aH ∃ h ∈ H s.t. fh y ah y h a−1y.Thus fa−1y aa−1y y.

Example 4.6:Consider the group Z6,⊕. Let H 0, 3 ≤ Z6, find the left cosets of H in Z6 ?

Solution :

0 ⊕ H 0, 3 3 ⊕ H1 ⊕ H 1, 4 4 ⊕ H2 ⊕ H 2, 5 5 ⊕ H

Hence there are 3-distinct left cosets of H in Z6.Note that

0 ⊕ H 1 ⊕ H 2 ⊕ H Z6

0 ⊕ H ∩ 1 ⊕ H ∩ 2 ⊕ H ∅.

Definition 4.7:Let H ≤ G, a,b ∈ G. Then a and b are left (or right )congruent modulo H denote it by

a ≡l b ≡r

a ≡l b modH ab−1 ∈ Ha ≡r b modH ba−1 ∈ H

Theorem 4.8:Let H ≤ G.

(i) Then the relation ≡l modH is an equivalence relation on G.

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(ii) The equivalence classes are left cosets of H in G.

Proof :(1) Reflexive: Since a−1a e ∈ H a ≡l a.

(2) Symmetric: Let a ≡l b modH a−1b ∈ H a−1b−1 ∈ H b−1a ∈ H b ≡l a modH.

(3) Transitive: Let a ≡l b , b ≡ a modH a−1b, b−1c ∈ H.

a−1b.b−1c ∈ H since H ≤ G a−1c ∈ H a ≡l c modH.

(ii) Next, since ≡ is an equivalence relation. Therefore, it has an equivalence classes

a b ∈ G : a ≡l b modH b ∈ G : a−1b ∈ H b ∈ G : a−1b h b ∈ G : b ah b ∈ G : b ∈ aH aH.

Therefore the equivalence classes are left cosets of H in G.

Remark 4.9:(1) The set of left cosets of H in G partition G. i.e., G is the union of all left cosets of H

in G and distinct left coets are disjoint.(2) If b ∈ a a b. i.e., if b ∈ aH aH bH

b ah aH bH a−1b h aH bH a−1b ∈ H aH bH

Corollary 4.10:Let H ≤ G . Then

aH bH a−1b ∈ H.

Inparticular, aH H a ∈ H.

As in example 4.6, since 0 ∈ H, 3 ∈ H 0 H H 3 H.

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Definition 4.11:Let H ≤ G. Then the index of H in G is defined to be the number of distinct left

cosets of H, which is denote it by G : H.

As in axample 4.6, G : H 3, in example 4.4, G : H 2, in example 4.3,S3 : K 3.

Theorem 4.12 (Lagrange Theorem):Let G be a group of finite order n, i.e., |G| n, H be a subgroup of G of order k.Then every subgroup of G is a divisor of |G|. i.e., |H| ∣ |G|.

Proof :Let |G| n and |H| k. Then by theorem 4.5, if a ∈ G |H| |aH| k.By remark 4.9(1), G is the union of all left cosets of H in G. i.e.,

G a1H a2H a3H ... arH ai ∈ G.

i.e., G : H r |G| |a1H| |a2H| ... |arH|

n kr k ∣ n |H| ∣ |G|.

Corollary 4.13:Let |G| n. Then the order of any element a ∈ G divides n, and an e.

Proof :Let a ∈ G is of order k. i.e., oa k. Then by definition 3.13, the cyclic subgroup

a must be of order k.

i.e., o a k oa.

By Lagrange Theorem, o a ∣ oG k ∣ n n kr, i.e., order of adivides n.

Moreover, an akr akr er e.

Corollary 4.14:Any finite group of prime order has no-non-trivial subgroup.

Proof :Let |G| p. Suppose H ≤ G. Then by Lagrange Theorem

|H| ∣ |G| |H| ∣ p

|H| 1 or |H| p.

If |H| 1 H e.If |H| p H G.Therefore there is no non-trivial subgroup.

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Corollary 4.15:Every group G of prime order is cyclic.

Proof :Let |G| p. Consider the cyclic subgroup a with a ≠ e. By Lagrange Theorem,

o a ∣ p.

o a 1 or o a p.

But a contains more than one element o a p G a . Gis cyclic.

Theorem 4.16:Any non-commutative group has at least six elements.

Proof :Any group of order 2, 3, and 5 must be cyclic and hence commutative.Suppose |G| 4. Then by Lagrange each element a ∈ G must divide 4.

oa ∣ 4 oa 2 or oa 4.

If oa 4 oa o a 4 |G| G is cyclic and hencecommutative.

If oa 2 i.e., a2 e a a−1 ab2 e abab e ab ba which is commutative .Therefore ay non-commutative group has at least six-element.

Definition 4.17:A subgroup H of a group G is said to be normal in G if every left coset of H in G is

also a right coset of H in G.i.e.,

H is normal aH Ha ∀ a ∈ G.

We denote it be by H G.

Remark 4.18:1. G , e are normal. (Trivial normal subgroups}

2. Every subgroup of a commutative group is normal.

3. Every subgroup of a cyclic group is normal.

4. aH Ha it does not require that ah ha ∀ h ∈ H.

Example 4.19:Let G S3 , 123, 132, 12, 13, 23 and A3 , 123, 132.Show that A3 S3.

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Solution :Let 12 ∈ S3, then

H 12H 12, 12123, 12132 12, 23, 13

H H12 12, 12312, 13212 12, 23, 13

12H H12.Note that : ah ≠ ha for 12123 ≠ 12312.A similar computations we get

H 123H 132H , 123, 132H H123 H132 , 123, 13212H 13H 23H , 13, 23H12 H13 H23 , 13, 23

Thus H H ∀ ∈ S3 A3 S3.Take K , 12 in S3. Show that K is not a normal subgroup of S3.

The definition of a normal subgroup can be formulated in several different ways,one of it state in the following theorem.

Theorem 4.20:The subgroup H is normal in G aHa−1 ⊆ H for each element a ∈ G. i.e.,

aH Ha aHa−1 ⊆ H ∀ a ∈ G.

Proof : Suppose H G, aH Ha ∀ a ∈ G.Let x ∈ aHa−1

x

ah a−1 , a ∈ G, h ∈ H.

x h1aa−1 , a ∈ G, h1 ∈ H x h1 ∈ H. aHa−1 ⊆ H.

Suppose aHa−1 ⊆ H, we want to prove that aH Ha ??Let x ∈ aH x ah

∈H

aha−1 a ∈ Ha aH ⊆ Ha.

Let x ∈ Ha x ha. Since a−1 ∈ G a−1Ha−1−1 ⊆ H a−1Ha ⊆ H.

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∵ x ha a∈Ha1ha∈ aH

Ha ⊆ aH. i.e.,aH Ha.The assertion aHa−1 ⊆ H ∀ a ∈ G is equivalent to aHa−1 H. Since

H aa−1Haa−1 ⊆ aHa−1.

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Theorem 4.21:Let G be a group. Then

(1) If H ≤ G s.t. H ⊆ centG H G.(2) centG G.(3) Every subgroup of a commutative group is normal.

Proof :(1) We want to prove that aHa−1 ⊆ H ??

Since H ⊆ centG ifh ∈ H ⊆ centG h ∈ centG ha ah ∀ a ∈ G.

Let x ∈ aHa−1 x aha−1 haa−1 h ∈ H.Therefore H G.

(2) Let K centG g : ga ag ∀ a ∈ G.We want to show that

acentGa−1 ⊆ centG i.e., aKa−1 ⊆ K ??

Let x ∈ aKa−1

x aka−1 where k ∈ centG x kaa−1 k ∈ centG K.

i.e., centG G.

(3) Easy, straightforward.

Theorem 4.22:If H ≤ G, then

H G aHbH abH ∀ a,b ∈ G.

Proof : Let H G, i.e., Ha aH ∀ a ∈ G. Consider

aHbH aHbH abHH abH.

Let x ∈ aHa−1 x aha−1 aha−1e ∈ aHa−1H aa−1H H. H G.

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QUOTIENT GROUPS:

Let H G. Denote the set of all left (right) distinct cosets of H in G, by :

G/H

a aH : a ∈ G a Ha a ∈ G if the operation is .ora a H : a ∈ G a H a : a ∈ G if the operation is .

In example 4.19, since A3 S3, then S3/A3 A3, 12A3.Note that : |G/H| G : H.Define on G/H a coset multiplication by aHbH abH. Then we have the following

theorem:

Theorem 4.23:Let H G. Then G/H forms a group with respect to the coset multiplication

aHbH abH or a Hb H a b H

Proof :1. The coset multiplication is well defined :, i.e.,

: G/H G/H G/H

if a, b, a1, b1 ∈ G/H G/H s.t. a, b a1, b1 a b a1 b1 ??i.e., if a a1 and b b1 ab a1b1 ??i.e., if aH a1H and bH b1H aHbH a1Hb1H ??i.e., if aH a1H and bH b1H abH a1b1H ??

Suppose aH a1H and bH b1H , then by corollary 4.10, a−1a1 ∈ H andb−1b1 ∈ H.

b−1a−1a1b ∈ H since H G, b ∈ G.

Consider ab−1a1b1 b−1a−1a1b1

∈H

b−1a−1a1b

∈H

b−1b1∈ H

abH a1b1H (by corollary 4.10).

2. Closed:Let a, b ∈ G/H we want to show that a.b ∈ G/H ??Since a, b ∈ G/H a aH, b bH a.b aH.bH abH ab ∈ G/H.

3. Associative:

a.b.c a.b.c ??

Consider

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a.b.c aH.bH.cH abH.cH ab.cH a.b.cH aH.b.cH aH.bH.cH a.b.c.

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4. Identity:The identity element of G/H is H since aH.H H.aH aH.

5. Inverse:The inverse of aH is a−1H since aH.a−1H aa−1H eH H a−1H.aH.

Therefore G/H, is a group.

Definition 4.24:The set G/H is called the quotient group of G by H or the factor group of G by H.

Example 4.25:Consider a commutative group Z, . Then any subgroup of Z must be a normal

subgroup.Therefore 3Z ≤ Z 3Z Z.We can construct a quotient group of Z by 3Z :

Z/3Z a 3Z a ∈ Z 0 3Z, 1 3Z, 2 3Z, 4 3Z, ....

Since a H H if a ∈ H 3 3Z 3Z, 3 ∈ 3Z, and 4 3Z 1 3Z, 4 ∈ 3Z,and so on ...

Z/3Z 3Z, 1 3Z, 2 3Z.

Z : 3Z |Z/3Z| 3.

Example 4.26:Let G S3 , 12, 13, 23, 123, 132. Then |S3 | 3! 6.Then any subgroup of S3 must be of order 1,2, 3, and 6.The subgroup of order 1 is A .The subgroup of order 6 is B S3.The subgroup of order 2, 3, must be cyclic since 2,3 are prime numbers.Consider 12, then 1212 o12 2 o 12 2i.e., C , 12.Consider 13 , then 1313 o13 2 o 13 2i.e., D , 13.Consider 23 , then 2323 o23 2 o 23 2i.e., E , 23.Consider 123 , then 123123123 o123 3 o 123 3i.e., F , 132, 123.Therefore there are 6-subgroups

A D , 13B S3 E , 23C , 12 F , 132, 123

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It is easy to check that the only normal subgroups are : A, B and F (evenpermutations).

Therefore we can construct quotient groups which are :

S3/A A : ∈ S3

A, 12A, 13A, 23A, 123A, 132A , 12, 13, 23, 123, 132.

S3/S3 S3 : ∈ S3 S3.

S3/F F : ∈ S3

F, 12F F, 12F.

Remark 4.27:Since every group has two trivial normal subgroups e , G, then G/e and G/G

are quotient groups, where

G/e a.e : a ∈ G ≅ G

and |G/e| |G|.

G/G a.G : a ∈ G G

and |G/G| 1.

Theorem 4.28:Let G be a commutative group and H G .Then G/H is a commutative group. (The converse is not true).

Proof :Let aH, bH ∈ G/H aH.bH abH baH bH.aH.

The converse is not true : For S3/A3 is of order 2, therefore it is a cyclic subgroup andhence it is commutative while S3 is not a commutative group.

Now our question is : under what conditions will a non-commutative group possesscommutative quotient groups ? Our analysis begins with a basic definition.

Definition 4.29:Given a group G and a pair a,b ∈ G the commutator of a and b is defined to be the

product aba−1b−1 denote it by a,b.

i.e., a,b aba−1b−1.

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Note : The product of two commutator fails to be a commutator. To solve thisdifficiulty we work instead with the subgoup generated by the set of all commutators. Thissubgroup is known either as the derived subgroup or commutator subgroup of G, anddenoted by G,G.

G,G ai, bi : ai, bi ∈ G (finite product with one or more terms).

Theorem 4.30:Let G be a group, and G′ be the set of all commutators aba−1b−1 of a group G. Then

1. G′ G.2. G/G′ is abelian group.3. If H G, then G/H is abelian G′ ⊆ H.

Proof :The commutators certinely generate a subgroup G′. We must show that G′ G.Note that :

a,b−1 aba−1b−1−1 bab−1a−1 b,a.

So the inverse of a commutator is again a commutator.Also e eee−1e−1 e,e is a commutator.(G′ consists precisely of all finite products of commutator).

1. To prove that G′ G, we have to show that(i) if x ∈ G′ gxg−1 ∈ G′ ∀ g ∈ G.(ii) if x is a product of commutators, so is gxg−1 ∀ g ∈ G.

(i) Let x a,b ∈ G′ , g ∈ G, then

gxg−1 gaba−1b−1g−1 gag−1gbg−1ga−1g−1gb−1g−1

gag−1gbg−1gag−1−1gbg−1−1

gag−1, gbg−1 ∈ G′.

G′ G.

(ii) Let x ∈ G′ where x x1x2x3...xn, where xi ∈ G′ i 1, ...,n. Then

gxg−1 gx1x2...xng−1 gx1g−1gx2g−1...gxng−1 ∈ G′

G′ G.

2. G/G′ is an abelian group.

G/G′ aG′ : a ∈ G.

Let aG′ , bG′ ∈ G/G′ aG′bG′ abG′.∵ G′ contains all commutators b−1, a−1 ∈ G′ b−1,a−1G′ G′. Hence

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aG′bG′ abb−1,a−1G′ abb−1a−1baG′ baG′ bG′aG′.

Therefore G/G′ is an abelian group.

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3. G/H is abelian G′ ⊆ H. Suppose that G/H is abelian, and a,b ∈ G′. Then

aba−1b−1H aHbHa−1Hb−1H aHa−1HbHb−1H aa−1Hbb−1H eHeH H2 H.

aba−1b−1 ∈ H G′ ⊆ H.

Conversely,

aHbH abH abb−1a−1baH baH bHaH.

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PROBLEMS (4)

1. Determine the left coset decomposition of the following groups with respect to theindicated

subgroup H.(a) The group Z24,⊕, H 0, 6, 12, 18.(b) The symmetric group S3, ∘, H A3, ∘.(c) Z8,⊕, H 0, 4.

2. If H, K are normal subgroups of a group G with H ∩ K e. Show thathk kh ∀ h ∈ H, k ∈ K.

3. Let H, K be subgroups of the group G , and H ≤ K ≤ G. Show that

G : H G : KK : H.

4. If Hi is an indexed collection of normal subgroups of the group G, prove that ∩Hiis also normal in G.

5. If G : H 2 H G.

6. Show that An Sn.

7. Given H, K are subgroups of G. Prove the following :(i) If K G HK ≤ G.(ii) If H, K are normal of G HK G.(iii) If H ≤ K ≤ G,where H G H K.

8. Describe the quotient group of the following:(a) 0, 4, 8, 12, ⊕ in Z16, ⊕ .(b) Ze, in Z, .

9. Show that every quotient group of a cyclic group is cyclic. i.e.,

If G a and H G G/H is cyclic.

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5. HOMOMORPHISMS AND CAYLEY’S THEOREM

This section begins with an analysis of a class of functions which preserve thegroup operation and hence the algebraic structure of the group.

Definition 5.1:Let G,∗ and G′, ∘ be two groups. A homomorphism from G,∗ into G′, ∘ is a

mapping

f : G G′ such that fa ∗ b fa ∘ fb.

If f is one - one and onto , then f is said to be isomorphism,or G is isomorphic to G′ , denote it by G G′.

Example 5.2:Let f : R, R, , defined by fx ex. Then

1. f is a homomorphism :

fx y exy ex ey fx fy.

2. f is 1-1 :

fx fy ex ey x y.

3. f is onto :If r ∈ R ∃ ln r ∈ R such that

fln r e ln r r.

Hence R is isomorphic to R, i.e., R R.

Notation :(i) A homomorphism from G into itself is called endomorphism.(iii) An isomorphism from G onto itself is called automorphism.

Example 5.3:Consider two additive groups Z, and Zn,⊕. Define

f : Z Zn by fa a.

1. f is a homomorphism:

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fa b a b a ⊕ b fa ⊕ fb.

2. f is onto:

∀ a ∈ Zn ∃ a ∈ Z such that fa a.

3. f is not 1-1:Let 1, n 1 ∈ Z then f1 1 and fn 1 n 1 1.Therefore Z ≇ Zn.

Notation : Denote HomG,G′ The set of all homomorphisms from the group Ginto G′.

Theorem 5.4:HomG,G, ∘ forms a semi-group with identity. (∘ is a composition)

Proof :1. Let f,g ∈ HomG,G. Then

f ∘ gab fgab fgagb fgafgb f ∘ gaf ∘ gb

f ∘ g ∈ HomG,G.2. The ∘ is associative (easy).3. I : G G is the identity map.

Notation : AutG{The set of all isomorphisms from G onto itself}.

Theorem 5.5:AutG, ∘ is a subgroup of the symmetric group . i.e., AutG ≤ SG (see Th. 2.10).

Proof :1. Let f,g ∈ AutG f,g ∈ SG f ∘ g ∈ SG (see th. 2.10) i.e., f ∘ g is 1-1 and

onto.Moreover, f ∘ g is a homomorphism (th. 5.4) f ∘ g ∈ AutG.

2. ∘ is associative.3. I : G G is the identity map which beloongs to AutG.4. It only remains to verify that whenever f ∈ AutG , its inverse f−1 ∈ AutG ??

Let f ∈ AutG f ∈ SG f−1 ∈ SG, i.e., f−1 is 1-1 and onto.We have to show that f−1 is a homomorphism . Let

fa x and fb y a,b,x,y ∈ G.

Then

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f−1xy f−1fafb f−1fab ab f−1xf−1y

f−1 is a homomorphism f−1 ∈ AutG.Therefore AutG is a group , since AutG ⊆ SG AutG ≤ SG.

The following theorem shows that a homomorphism functions preserve identities andinverses.

Theorem 5.6:If f is a homomorphism from (G,∗ into (G′, ∘ , then :

1. If e ∈ G is the identity fe e ′ is the identity of G′.2. If a,a−1 ∈ G where a−1 is the inverse of a fa−1 fa−1 is the inverse of fa

in G′.

Proof :1. We have to show that

∀ fa ∈ G′ : fe ∘ fa fa ∘ fe fa ??

Consider

fe ∘ fa hom. fe ∗ a fa.

Similarly the other side. Therefore e ′ fe is the identity element of G′.

2. We have to show that

fa ∘ fa−1 fa−1 ∘ fa e ′ ??

Consider

fa ∘ fa−1 fa ∗ a−1

fe e ′.

Similarly the other side, therefore fa−1 is the inverse of fa.

Definition 5.7:Let f : G G′ be a homomorphism. Then the set

fG fa : a ∈ G Im f

is called the homomorphic image of f under G, and the set

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ker f a ∈ G : fa e ′ f−1e ′

is called the kernel of f.

Theorem 5.8:Let f : G G′ be a homomorphism. Then

1. If H ≤ G fH ≤ G′.2. If H′ ≤ G′ f−1H′ ≤ G.3. If H′ G′ f−1H′ G.4. If H G and fG G′ (onto) fH G′.

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Proof :1. If H ≤ G fH ≤ G′.

fH fh : h ∈ H.

We want to prove that

if fa, fb ∈ fH fafb−1 ∈ fH ??

Let fa, fb ∈ fH where a,b ∈ H ≤ G.Consider fafb−1 fab−1 , since ab−1 ∈ H ≤ G , therefore

fafb−1 fab−1 ∈ fH.

2. If H′ ≤ G′ f−1H′ ≤ G.

f−1H′ a ∈ G : fa ∈ H′.

We want to prove that

if a,b ∈ f−1H′ ab−1 ∈ f−1H′ ??

Let a,b ∈ f−1H′ fa, fb ∈ H′ fafb−1 ∈ H′ (since H′ ≤ G′hom fab−1 ∈ H′ ab−1 ∈ f−1H′ f−1H′ ≤ G′.

3. If H′ G′ f−1H′ G.We want to prove that

if h′ ∈ f−1H′ then ah′a−1 ∈ f−1H′ ??

Let h′ ∈ f−1H′ fh′ ∈ H′.∵ H′ G′ fafh′fa−1 ∈ H′ where fa, fa−1 ∈ G′.Since f is a homomorphism

fafh′fa−1 fah′a−1 ∈ H′ ah′a−1 ∈ f−1H′.

4. If H G and fG G′ (onto) fH G′.We want to prove that

if fh ∈ fH, then a′fha′−1 ∈ fH ??

Since H G aha−1 ∈ H ∀ a ∈ G.Consider faha−1 fafhfa−1, since fG G′ fa a′ faha−1 a′fha′−1 ∈ fH.

Theorem 5.9:Let f : G G′ be a homomorphism . Then

1. ker f ≤ G.2. ker f G.3. f is 1-1 ker f e.

Proof :1. ker f ≤ G.

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ker f a ∈ G : fa e ′.

(i)ker f ≠ ∅ since fe e ′ e ∈ ker f.(ii)Let a,b ∈ ker f fa e ′ and fb e ′.Consider

fab−1 fafb−1 fafb−1 e ′.e ′ e ′

ab−1 ∈ ker f.

2. ker f G.We want to prove that

if b ∈ ker f fb e ′ then aba−1 ∈ ker f ??

Consider

faba−1hom. fafbfa−1

fae ′fa−1

fa.a−1

fe e ′

aba−1 ∈ ker f ker f G.

3. f is 1-1 ker f e. Suppose f is 1-1 we want to prove that ker f e

Let a ∈ ker f fa e ′ fef is 1-1 a e.

Suppose ker f e we want to prove that f is 1-1.Let fa fb , then

fab−1hom. fa.fb−1 fb.fb−1 fb.b−1 fe e ′

ab−1 ∈ ker f e ab−1 e a b.

Theorem 5.10:Let H G. Then the mapping

: G G/H , a aH where a ∈ G

is an onto homomorphism with ker H. is called a natural homomorphism)

Proof :1. is a homomorphism since

ab abH aHbH ab.

2. is onto for every element aH ∈ G/H ∃ a ∈ G : a aH.

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3. Consider

ker a ∈ G : a e ′ a ∈ G : aH H

a ∈ H ker ⊆ H.Let a ∈ H

a aH a∈H H e ′

a ∈ ker H ⊆ ker. Hence ker H.

Example 5.11Consider the group Z4,⊕ and the multiplicative group G 1,−1, i,−i where

i2 −1.The operation tables for these groups are:

⊕ 0 1 2 30 0 1 2 31 1 2 3 02 2 3 0 13 3 0 1 2

1 −1 i −i1 1 −1 i −i−1 −1 1 −i ii i −i −1 1−i −i i 1 −1

We shall show that the groups Zn and G are isomorphic Zn G.

To do this , we must produce a 1-1 homomorphism f : Znonto G.

1. In any homomorphism fe e ′, f0 1 .2. In any homomorphism the image of the inverse of an element a the inverse of its

image.So if we define f1 −1 , then f1−1 f1−1.But f1−1 −1 f1−1 −1 f3 −1 (since 1−1 3.Therefore f1 f3 −1 f is not 1-1.Hence we are going to define f1 i , then f1−1 f1−1, where

f1−1 i−1 −i f3 −i .

3. f must preserve modular addition.

f2 f1 ⊕ 1 f1 f1 i.i −1

f2 −1.

Therefore f is defined as

f0 1 , f1 i , f2 −1 , f3 −i

Clearly, this function is 1-1 and onto. To establish that f is a homomorphism,

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we must consider all possible sums of two elements of Z4 and show that each sum ismapped onto the product of images of f, i.e.,

f1 ⊕ 2 f3 −if1 f2 i.−1 −i

f1 ⊕ 2 f1 f2

Continuning like this , we get that ∀ a, b ∈ Z4 : fa ⊕ b fa fb.Therefore Z4 G.

Notation :To show that two groups are not isomorphic, it is enough to show that some property of

one not possessed by the other. For example, Z4 ≇Kelien 4-group, since Z4 g is acyclic group of order 4, where K4 is not a cyclic group.

Theorem 5.11:(i) Let G be a finite cyclic group of order n 0. Then

G Zn (Zn is the additive group).

(ii) Let G be an infinite cyclic group . Then

G Z uder addition).

Proof :(i) Let G a be a finite cyclic group of order n 0. Then bt theorem 3.12,

G e, a, a2, ...,an−1.

Define the

f : G Zn by fak k 0 ≤ k n.

1. f is a homomorphism:fak.aS faks k s k ⊕ s fak ⊕ fas.

2. f is 1-1:Let fak fas k s k ≡ s modn ak as (by th.3.12).

3. f is onto:∀ r ∈ Zn ∃ r ∈ Z s.t. ar ∈ G and far r.

(ii) Let G be an infinite cyclic group . Thus

G an : n ∈ Z.

Define

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f : G Z by fak k ∀ ak ∈ G.

1. f is a homomorphism:fak.aS faks k s fak fas.

2. f is 1-1:Let fak fas k s ak as .

3. f is onto:∀ n ∈ Z ∃ an ∈ G s.t. fan n.

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Corollary 5.12:Any two cyclic groups of the same order are isomorphic.

Theorem 5.13 (Cayley):(i) Any group G is isomorphic to a subgroup of the symmetric group SG.(ii) When G is finite .Then any finite group of order n is isomorphic

to a subgroup of the symmetric group Sn.

Proof :(A) Construct a subgroup H of SG :

Let G be an arbitrary group and a be a fixed element of G. Define

fa : G G by fax ax ∀ x ∈ G.

We want to show that fa ∈ SG. i.e., fa is a 1-1 and onto function.

1. fa is a well-defined :Let x y ax ay fax fay.

2. fa is 1-1 :Let fax fay ax ay a−1ax a−1ay x y.

3. fa is onto :Let y ∈ G ∃ x ∈ G s.t. fax y ax y x a−1y ∈ G and

faa−1y aa−1y y.

Hence fa is a permutation of the set G. i.e., fa ∈ SG.

Define

H fa : a ∈ G.

We want to show that H ≤ SG. (That is H is a permutation group).(i) H is closed, for let fa, fb ∈ H ,a,b ∈ G, then

fa ∘ fbx fafbx fabx abx fabx ∈ H ,ab ∈ G.

(ii) fe is the identity element of H, for

fe ∘ fax fefax feax eax ax fax.

Similarly, fa ∘ fex fax.(iii) To find the inverse, considerfa ∘ fbx fex fabx fex abx ex abx x b a−1.Thus the inverse element is fa−1x.(iv) The operation ∘ is associtive . Hence H ≤ SG.

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(B) Define a map from G H and show that this map is an isomorphism:

Define

: G H by a fa for a ∈ G.

1. is well-defined :Let a b ax bx fa fb a b.

2. is a homomorphism:

ab fab abx fabx fa ∘ fbx a ∘ b.

3. is 1-1 :Let a b fa fb fax fbx ax bx a b.

4. is onto:∀ fa ∈ H ∃ a ∈ G s.t.a fa.

Thus G H.

Example 5.14:Determine the finite group Klien 4-group K4. Then by Cayley’s theorem K4 is

isomorphicto a subgroup of S4.Recall that K4 e,a,b,c where c ab. Now as in theorem 5.13,

H fa : a ∈ K4

H fe, fa, fb, fc.

where fee ee e, fea ea a, ..., fec cfae ae a, faa a2 e, fab c, fac a2b b.A similar computation we get

fe e a b ce a b c

, fa e a b ca e c b

, fb e a b cb c e a

,

fc e a b cc b a e

fe I , fa eabc, fb ebac, fcecab.K4 H A4, where A4 I, 1234, 1324, 1423.

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PROBLEMS (5)

1. Determine whethere the indicated function f is a homomorphism from the first groupinto the second group.a. f : R, R, fa −a.b. f : R, R − 0, fa 2a.c. f : R 0, R, fa a2.d. f : Q − 0, Q, fa |a|.

2. Given the permutations1 I, 2 123, 3 132, 4 23, 5 12, 6 13.

Define f : S3 Z2 by f i 0 for i 1,2,3, f i 1 for i 4,5,6.Show that f is a homomorphism.

3. Let f : G G′ be a homomorphism. Prove the following:a. fcentG ⊆ centfG.b. If G′ is commutative , then G,G ⊆ ker f.

4. Let f be a homomorphism from the group G into itself and H a ∈ G : fa a.Show that H ≤ G.

5. Let f : G G defined by fa a−1. Show that

G is commutative f is a homomorphism from G onto G.

6. Let f : Z, G, be a homomorphism from Z into G defined byfn an. Describe, fZ, ker f.

7. Let f : G G′ be a homomorphism. Prove the following:1. a. If G is commutative fG is commutatuve.b. If G is cyclic fG is cyclic.

8. Prove that the permutation 1 I, 2 1234, 3 1324, 41432 form asubgroup of S4 which is isomorphic to the multiplicative group G 1,−1, i,−iwhere i2 −1.

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6. FUNDAMENTAL HOMOMORPHISMS THEOREMS

In this section , we discuss a number of results concerning the relationshipbetween homomorphisms and quotient groups. These theorems , commonly known as theFundamental Homomorphisms Theorems For Groups.

Theorem 6.1 (1st homomorphism theorem):If f : G G′ be a homomorphism from a group G into a group G′, then

G/ ker f fG.

If f is onto, then

G/ ker f G′.

Proof :Define g : G/ ker f fG by gaK fa , a ∈ G (where K ker f).

1. g is well - defined:

Let aK bK ab−1 ∈ K ker f fab−1 e ′ fafb−1 e ′

fafb−1 e ′ fa fb gaK gbK.

2. g is 1-1:

Let gaK gbK fa fb fafb−1 e ′fis homo. fab−1 e ′

ab−1 ∈ ker f K aK bK

3. g is a homomorphism:

gaK.bK gabK fab fafb gaKgbK.

Therefore G/ ker f fG.Moreover, when f is onto fG G′ G/ ker f G′.

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Definition 6.2:Let G be a group, and H, K be a non-empty subsets of G. The product of H and K is

defined to be

HK hk : h ∈ H, k ∈ K.

Theorem 6.3(2ed homomorphism theorem):Let H, K be subgroups of the group G, and K G. Then

H/H ∩ K HK/K .

Proof :Before starting proving the theorem we have first to verfiy that

(a) H ∩ K H (b) HK ≤ G (c) K HK.(a) H ∩ K H:

To prove (a), we have to show that if x ∈ H ∩ K, then hxh−1 ∈ H ∩ K ??.∵ K G gkg−1 ∈ K ∀ g ∈ G.Consider x ∈ H ∩ K x ∈ H and x ∈ K gxg−1 ∈ K ∀ g ∈ G.In particular, h ∈ H ⊆ G hxh−1 ∈ K .But h, x ∈ H hxh−1 ∈ H .Therefore hxh−1 ∈ H ∩ K . i.e., H ∩ K H.

(b) HK ≤ G :To prove (b) we have to show that if x, y ∈ HK xy−1 ∈ HK.Note that HK ≠ ∅, since e e.e ∈ HK.Let x,y ∈ HK x h1k1 and y h2k2 where hi ∈ H, ki ∈ K i 1,2.Consider xy−1 h1k1k2

−1h2−1 h1k3h2

−1 ∈ K K G . xy−1

∈H

e

∈K

h1k3h2−1 ∈ HK . Therefore H ≤ G.

(c) K HK :To prove (c) we have to show that xk1x−1 ∈ K ∀ x ∈ HK.Let x ∈ HK x hk, h ∈ H, k ∈ K.Consider xk1x−1 hkk1k−1h−1 hk2h−1 ∈ K K G. K HK.

Define a map : H HK/K by h hK ∀ h ∈ H.

1. is well-defined:Let h1 h2 h1K h2K since K G, gK Kg ∀ g ∈ G. h1 h2.

2. is a homomorphism:

h1h2 h1h2K h1Kh2K h1h2.

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3. is onto:Since ∀ hK ∈ HK/K ∃ h ∈ H s.t. h hK.

Therefore by theorem 6.1, H/ ker HK/K.But

ker h ∈ H : h K h ∈ H : hK K h ∈ H : h ∈ K H ∩ K

Thus H/H ∩ K HK/K.

Theorem 6.4(3ed homomorphism theorem):Let H,K be two normal subgroups of the group G with K ⊆ H. Then

(i) H/K G/K.(ii) G/K╱H/K G/H.

Proof :(i) H/K G/K.

Since K ⊆ H and K G, then gkg−1 ∈ K ∀ g ∈ G, in particular,hkh−1 ∈ K, K H.

To prove that H/K G/K, recall that from theorem 5.8(4), we have if

H G fH G′.

Consider the natural map

: G G/K , g gK ∀ g ∈ G.

Then this is a homomorphism map which is into with ker H (see theorem 5.10).Since H G H G/H (by th. 5.8), H/K G/H.

(ii) G/K╱H/K G/H.Define a map

: G/K G/H, gK gH.

1. is well-defined.Let g1K g2K g1g2

−1 ∈ K ⊆ H g1H g2H g1K g2K.

2. is a homomorphism:

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g1Kg2K g1g2K g1g2H g1Hg2H g1Kg2K.

3. is onto:For if g ∈ G ∃ gK ∈ G/K s.t. gK gH.

Thus by 1st homomorphism theorem we get

G/H╱ker G/H.

But

ker gK ∈ G/K s.t. gK H gK ∈ G/K s.t. gH H gK ∈ G/K s.t. g ∈ H H/K.

Hence G/K╱H/K G/H.

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PROBLEMS (6)

1. Assume H,K are subgroups of the group G, with K G and G HK. Show that

G/K H H ∩ K e.

2. Let H, K1, K2 ≤ G and K1,K2 G. If H ∩ K1 H ∩ K2, prove that

HK1/K1 HK2/K2.

3. Let H1, H2, K ≤ G and K G. If H1K H2K, prove that

H1/H1 ∩ K H2/H2 ∩ K.

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7. Rings

This section is devoted to a study of algebriac structures having two binaryoperations, denoted by S,∗, ∘ . Obvious examples are Z,, ., Q,, . and PA,∩,.We shall be concerned with the abstract system which is known as a ring. Since a ring is acombination of commutative group and a semi-group so many of the important concepts ofa group theory have natural generalizations to systems with two operations.

Definition 7.1:A ring is an ordered triple R,, . consisting of a non-empty set R and two operations

(), (.) defined on R such that :1. R, is a commutative group.2. R, . is a semi-group.3. ∀ a,b,c ∈ R : a.b c a.b a.c (left dist.) and b c.a b.a c.a (right

dist.).

Definition 7.2:1. A ring R is said to be a commutative ring if

a.b b.a ∀ a,b ∈ R.

2. A ring R is said to be a ring with identity (unity) , if R has an element 1 ∈ R s.t.

a.1 1.a a ∀ a ∈ R.

3. Let R. , . be a ring with unity 1. An element 0 ≠ a ∈ R is called a unit of R ,if ithas a multiplicative inverse in R. i.e.,

∃ b ∈ R s.t. ab ba 1.

Example 7.3:1. Z,, ., Q,, ., R,, ., and C,, . are rings with usual addition and and

multiplication.2. Zn,⊕,⊙ is a commutative ring with identity 1 ∈ Zn. (see ths. 2.5,2.7).3. N,, ., Z,, . and D,, . are not a ring, where D denote the set of odd integers.4. 0,, . is a commutative ring which is called trivial ring.5. In Q,, ., and R,, . every non-zero number is a unit.6. In Z,, . the units are 1, − 1.

Theorem 7.4:Let R,, . be a ring . Then ∀ a,b,c ∈ R, we have

1. 0.a a.0 0.

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2. a−b −ab −ab.3. ab − c ab − ac.4. a − bc ac − bc.5. −a−b ab.

Proof :1. 0.a a.0 0.

Consider a.0 a.0 0 a.0 a.0by cancellation law a.0 0.Similarly, for the other side.

2. a−b −ab −ab.a−b ab a−b b a.0 0 a−b −ab.−ab ab −a ab 0.b 0 −ab −ab.Thus a−b −ab −ab.

3. ab − c ab − ac.ab − c a.b −c a.b a.−c a.b − a.c.

4. a − bc ac − bc.a − b.c a −b.c a.c −b.c ac − bc.

5. −a−b ab−a−b −−ab −−ab ab.

Definition 7.5:1. A ring is R,, . called a division ring if every non-zero element has inverse.2. A field is a commutative division ring.

Example 7.6:Q,, ., R,, ., C,, ., and Zp,, . are expamples of field.

Definition 7.7:A ring R is called an integeral domain if

1. R is a commutative ring.2. R has unity 1.3. ∀ x,y ∈ R : if xy 0 x 0 or y 0.

Example 7.8:Z,, . is an integral domain ,while Z4 is not an integeral domain.

Definition 7.9:An element 0 ≠ a ∈ R (ring) is called a left (right) zero divisor

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if there exists 0 ≠ b ∈ R s.t. ab 0 ba 0.

An element is called a zero divisor if it is both right and left zero divisor.

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Example 7.10:1. (Z6,⊕,⊙ is not an integeral domain.2. Z,, . has no-zero divisors.

Definition 7.11:If R is a ring and ∅ ≠ S ⊆ R, we say that S is a subring of R denoted by S ≤ R if S

itself is a ring over the same operation defined on R.

Theorem 7.12:A non-empty subset S of a ring R is a subring iff ∀ a,b ∈ S we have

(i) a − b ∈ S(ii) ab ∈ S.

Proof : Suppose S,, . is a subring of R,, .. Then S,, . is a ring itself. Therefore

S, is a commutative group and S, . is a semi-group. Thus a − b ∈ S and ab ∈ S.

Conversely, suppose that a − b ∈ S and ab ∈ S. Then1. S, is a commutative group, for

a − a ∈ S 0 ∈ S 0 − a ∈ S −a ∈ S .Moreover, ∀ a,b ∈ S : a − −b ∈ S a b ∈ S .Lastly, is commutative S, is a commutative group.

2. S, . is a semi-group, for ab ∈ S and . is associative.3. Clearly . is distributive over , therefore S,, . is a ring.

Example : The set of even integers Ze,, . forms a subring of a ring Z,, ..

Definition 7.13:Let R,, . be a ring. Then the center of R denoted by ZR is defined by

ZR a ∈ R : ax xa ∀ x ∈ R.

Theorem 7.14:The center of R is a subring of R.

Proof :1. ZR ≠ ∅ for 0.a a.0 0 ∈ ZR.2. Let a,b ∈ ZR ax xa and bx xb ∀ x ∈ R.

Consider

a − bx ax − bx xa − xb xa − b

a − b ∈ ZR.

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3. Consider

abx abx axb axb xab xab

ab ∈ ZR.Thus ZR ≤ R.

Remark : R is commutative ZR R.

Definition 7.15:Let R,, . be a ring. The smallest ve integer n s.t. na 0 ∀ a ∈ R is called the

characteristic of the ring. If no such ve integer exists, then R is said to becharateristic zero, we shall denote it by char R.

Example : char Z 0, and char Z6 6.

Theorem 7.16:Let R be a ring with unity 1. Then

char R n 0 n is the smallest ve integer s.t. n.1 0.

Proof : Let charR n 0, then by definition , n is the smallest ve integer s.t.

n.a 0 ∀ a ∈ R.Inparticular , a 1 ∈ R n.1 0.

Suppose n is the smallest ve integer s.t. n.1 0. Consider

n.a

n−times

a a ... a where a ∈ R

a1 1 ... 1 an.1 a.0 0.

Hence na 0 charR n.

Theorem 7.17:Every field is an integral domain . The converse is not true.

Proof :Since F is a field F,, . is a commutative with unity.It remains to show that F has no zero divisors. i.e.,

if x,y ∈ F s.t. xy 0 x 0 or y 0 ??

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Suppose x,y ∈ F s.t. xy 0. If x ≠ 0, then every non-zero element has inverse.

x1 ∈ F x−1xy x−1.0 1.y 0 y 0.

Similarly if y ≠ 0 x 0. Therefore F is an integeral domain.

Notation The converse is not true, for Z,, . is an integral domain but it is not a field.

Theorem 7.18:Every finite integral domain is a field.

Proof :Let R be a finite integral domain. |R| n, i.e., R r1, r2, ..., rn.Let a ∈ R s.t. a ≠ 0. Then

aR ar1,ar2, ...,arn.

The elements of aR are distinct for if ars ard rs rd (cancellation law). aR R. But 1 ∈ R aR 1 ars for some rs ∈ R.Hence a is a unit. Thus every non-zero element has inverse R is a field.

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