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 Detailed Study of groups is a fundamental concept in the study of abstract algebra. To define the notion of groups,we require the concept of binary operation or composition which is a type of function that associates two elements of the set to a unique element of that set. 2

A binary operation on a set is a rule for combining two elements of the set. More precisely, if S iz a nonempty set, a binary operation on S iz a mapping f : S  S  S. Thus f associates with each ordered pair (x,y) of element of S an element f(x,y) of S. IN OTHER WORDS, An operation which combine two elements of a set to give another elements of a set to give another elements of the same set is called a binary operation 3

 1. Ordinary addition ‘+’ is a binary operation on Z Consider +: (Z*Z)  Z +: (3,7)=3+7=10 ∈ Z 4

 A non-empty set G equipped with one or more binary operation defined on it is called an algebraic structure.  Suppose ‘*’ is a binary operation on G, then (G, *) is an algebraic structure. 5

A g roup (G, ・ ) is a set G together with a binary operation ・ satisfying the following axioms. (i) Closure property: a.b ∈ G for all a,b ∈ G. (ii) The operation ・ is associative; that is, (a ・ b) ・ c = a ・ (b ・ c) for all a, b, c ∈ G. (iii) There is an i dentity element e ∈ G such that e ・ a = a ・ e = a for all a ∈ G. (iv) Each element a ∈ G has an i nverse element a −1 ∈ G such that a -1 ・ a = a ・ a −1 = e. 6

A group (G,.) is said to be abelian or commutative if in addition to the above four postulates, the following postulate is also satisfied: COMMUTATIVE PROPERTY: If the operation is commutative, that is, if a ・ b = b ・ a for all a, b ∈ G, the group is called commutative or abelian group 7

Example: Let G be the set of complex numbers {1,−1, i,−i} and let ・ be the standard multiplication of complex numbers. Then (G, ・ ) is an abelian group. The product of any two of these elements is an element of G; thus G is closed under the operation. Multiplication is associative and commutative in G because multiplication of complex numbers is always associative and commutative. 8

The identity element is 1, and The inverse of each element a is the element 1/a. Hence 1 −1 = 1, (−1) −1 = −1, i −1 = −I, and (−i) −1 = i. 9

Solution. (i) CLOSURE PROPERTY: since the sum of two integers is also an integer i.e., a+b ∈ Z for all a,b ∈ Z Therefore the set Z is closed w.r.t. addition. Hence closure property is satisfied. (ii) ASSOCIATIVE LAW: since addition of integers obey associative law, therefore a+(b+c)=(a+b)+c for all a, b, c ∈ Z 10

Thus addition is an associative composition. (iii) EXISTENCE OF IDENTITY: the number 0 ∈ Z and a+0 = 0+a =a for all a ∈ Z The integer 0 is the identity for (Z, +) (iv) EXISTENCE OF INVERSE: for each a ∈ Z, There exists a unique element -a ∈ Z such that a+(-a)=0=(-a)+a Thus each integer possesses an additive inverse. 11

(v) COMMUTATIVE LAW: the commutative law holds good for addition of integers i.e. a+b=b+a for all a,b ∈ Z Thus (Z, +) is an abelian group. Also Z contains an infinite number of elements.therefore (Z, +) is an abelian group of infinite order. 12

 An algebraic structure (G, *) is called a semi- group, if only the first two postulates, i.e., closure and associative law are satisfied. 13

 The algebraic structure (N,+), (W,+), (Z, +), (R, +) and (C, +) are semi-groups, where N, W, Z, R and C have usual meanings. 14

 Definition: If the number of elements in the group G are finite, then the group is called a finite group otherwise it is an infinite group 15

 Example of Finite Group: {1,−1, i,−i} is an example of finite group. Example of Infinite Group: (Z, +) is an example of infinite group. 16

 Definition: The number of elements in a finite group is called the order of the group. An infinite group is said to be of infinite order. The order of a group G is denoted by the symbol o(G). 17

 If (G,.) is a group, then (i) the identity element of G is unique. (ii) every element has a unique inverse. 18

Proof:: If possible let e 1 and e 2 be two identities in the group (G,.) Since e 1 is identity and e 2 ∈ G therefore e 1. e 2 = e 2 = e 2. e 1 …..(1) Also since e 2 is the identity and e 1 ∈ G therefore e 1. e 2 = e 1 = e 2. e 1 …..(2) therefore from (1) and (2), e 1 = e 2 Hence the identity is unique 19

PROOF: Let a be any element of the group ( G,.) If possible, let b 1 and b 2 be two inverses of a under the binary operation ‘. ‘ and let e be the identity element in G. Then a. b 1 = e = b 1. a And a. b 2 = e = b 2. a Now b 1 = b 1. e =b 1. ( a. b 2 ) 20

b 1 =(b 1. a). b 2 [by associativity] =e. b 2 = b 2 Therefore b 1 = b 2. hence the inverse is unique. 21

PProposition. If a, b are elements of a group G, then (i) ( a −1 ) −1 = a. (ii) ( ab) −1 = b −1 a −1. i.e., the inverse of the product of two elements of a group is the product of their inverses in the reverse order. 22

Proof: for each a ∈ G, we have a. a −1 = e = a −1. a  Inverse of a −1 is a. Therefore (a −1 ) −1 = a 23

24 Proof:

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Subgroups:: It often happens that some subset of a group will also form a group under the same operation.Such a group is called a subgroup. If (G, ・ ) is a group and H is a nonempty subset of G, then (H, ・ ) is called a subgrou p of (G, ・ ) if the following conditions hold: (i) a ・ b ∈ H for all a, b ∈ H. ( closure ) (ii) a −1 ∈ H for all a ∈ H. ( existence of inverses ) Conditions (i) and (ii) are equivalent to the single condition: (iii) a ・ b −1 ∈ H for all a, b ∈ H. 26

Prove that: (i) The identity of the sub-group is same as that of the group. (ii) The inverse of any element of a sub-group is the same as the inverse of that element in the group. 27

(i) let H be a sub-group of the group G. if e is the identity element of G, then ea = ae = a for all a ∈ G As H is a sub-set of G therefore ea = ae= a for all a ∈ H [ since a ∈ H => a ∈ G] => e is an identity element of H. Hence identity of the sub-group is same as that of the group. 28

(ii) Let e be the identity of G as well as of H. Let a be any element of H  a is an element of group G suppose b is the inverse of a in H and c is the inverse of a in G.  ab = ba = e ….. (1) ac = ca = e …. (2) From (1) & (2), ab = ac Therefore b=c Hence, the result. 29

Solution. Let G = {…., -4, -3, -2, -1, 0, 1, 2, 3, 4, …..} i.e. G is additive group of integers. Let H = {…, -3k, -2k, -k, 0, k, 2k, 3k,….} Therefore H ≠ ф Here H is a subset of G. We have to show that H is a subgroup of G. 30

Let ak, bk be any two elements of H such that a, b are integers. Inverse of bk in G is –bk. Now ak - bk = (a - b)k, which is an element of H as (a – b) is some integer. Thus for ak, bk ∈ H, we have ak – bk ∈ H Hence H is a subgroup of G. 31

Solution :- H is a subset of R but H is not a subgroup of R, The reason being that the composition in H is different from the composition in R. 32

 Definition. Let G be a group and let a  G ane e be the identity element in G. If a k = e for some k  1, then the smallest such exponent k  1 is called the order of a; if no such power exists, then one says that a has infinite order. 33

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Definition. If G is a group and a  G, write = {a n : n  Z } = {all powers of a }. It is easy to see that is a subgroup of G. is called the cyclic subgroup of G generated by a. A group G is called cyclic if there is some a  G with G = ; in this case a is called a generator of G. 35

Proof:- Let G be a finite group and o(G)= s Let a  G such that o(a)= s If H = {a n : n  Z } Then o(H) = s = o(a) Therefore H is a cyclic subgroup of G. Also o(H) = o(G) implies G itself is a cyclic group and a is a generator of G. 36

Example:- If G = {0, 1, 2, 3, 4, 5 } and binary operation + 6 is Then prove that G is a cyclic group. Solution. we see that 1=1 1 = = 2 1 = = 3 1 = =

1 = = 5 1 = = 0 Therefore G = { 0, 1, 2, 3, 4, 5 } is a cyclic group. 1 is generator of given group. Therefore G = Hence proved

Definition: let G be a group and H be any subgroup of G. For any a  G, the set Ha = { ha : h  H} is called right coset of H in G generated by a. Similarly, the set aH = { ah : h  H} is called left coset of H in G generated by a. Obviously, Ha and aH are both subsets of G. H is itself a right and left coset as eH = H =He, where e is an identity of G. 39

Find the right cosets of the subgroup { 1, -1 } of the group { 1, -1, i, -i} w.r.t. usual multiplication. Solution. let G = { 1, -1, i, -i} be a group w.r.t. usual multiplication And H = { 1, -1 } be a subgroup of G. The right cosets of H in G are H(1) = {1(1), -1(1)} = {1, -1}= H 40

H(-1)={1(-1), -1(-1)}={-1, 1}=H H(i)={1(i), -1 (i)} ={i, -i} H(-i) = {1(-i), -1(-i)}= {-i, i} thus we have only two distinct right cosets of H in G. 41

1. Define group. Show that Z (the set of all integers) is an abelian group w.r.t. addition. 2. Show that the set of integers Z is an abelian grop w.r.t. binary operation ‘ * ‘ defined as a * b = a + b + 1 for a, b ∈ Z. 3. If (G,.) is a group, then (i) the identity element of G is unique. (ii) every element has a unique inverse. 42

4. If a, b are elements of a group G, then (i) (a −1 ) −1 = a. (ii) (ab) −1 = b −1 a −1. i.e., the inverse of the product of two elements of a group is the product of their inverses in the reverse order. 5. If every element of a group is its own inverse, then show that the group is abelian. 6.If a group has four elements, show that it must be abelian 43

7.The order of every element of a finite group is finite and is less than or equal to the order of the group. 8.Let G = { 0, 1, 2, 3, 4, 5 }. Find the order of elements of the group G under the binary operation addition modulo Define subgroup.Let G be the additive group of integers. Prove that the set of all multiples of integers by a fixed integer k ig a subgroup of G. 44

10. Prove that: (i) the identity of the subgroup is same as that of the group. (ii) the inverse of any element of a subgroup is the same as the inverse of that element in the group. 11.Let H be the multiplicative group of all positive real numbers and R the additive group of all real numbers. Is H a subgroup of R? 45

12.Let G be a group with binary operation denoted as multiplication. The set {h ∈ G : for all x ∈ G } is called the centre of the group G. Show that the centre of G is a subgroup of G. 13. Define cyclic group. If a finite group ‘ s ‘ contains an element of order ‘ s ‘, then the group must be cyclic. 46

14. If G = { 0, 1, 2, 3,4, 5 } and binary operation is addition modulo 6, then prove that G is a cyclic group. 15.Define cosets. Find the right cosets of the subgroup { 1, -1 } of the group { 1, -1, i, -i} w.r.t. usual multiplication. 47

 DO ANY THREE QUESTIONS. 1. If a, b are elements of a group G, then (i) (a −1 ) −1 = a. (ii) (ab) −1 = b −1 a −1. i.e., the inverse of the product of two elements of a group is the product of their inverses in the reverse order. 2. Define cyclic group. If a finite group ‘ s ‘ contains an element of order ‘ s ‘, then the group must be cyclic. 48

3. If G = { 0, 1, 2, 3,4, 5 } and binary operation is addition modulo 6, then prove that G is a cyclic group. 4.Define cosets. Find the right cosets of the subgroup { 1, -1 } of the group { 1, -1, i, -i} w.r.t. usual multiplication. 5. Define subgroup. Let G be the additive group of integers. Prove that the set of all multiples of integers by a fixed integer k ig a subgroup of G. 49