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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.

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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

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1. Ordinary addition ‘+’ is a binary operation on Z

Consider +: (Z*Z) Z +: (3,7)=3+7=10 ∈Z

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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.

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A group (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 identity element e ∈ G such that e ・ a = a ・ e = a for all a ∈ G.(iv) Each element a ∈ G has an inverse element

a−1 ∈ G such that a-1 ・ a = a ・ a−1 = e.

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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 group7

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.

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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.

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Solution.(i) CLOSURE PROPERTY: since the sum of

two integers is also an integer i.e. , a+b ∈ Z for all a,b ∈ ZTherefore 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 ∈ ZThe 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)+aThus each integer possesses an additive inverse.

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(v) COMMUTATIVE LAW: the commutative law holds good for addition of integers

i.e. a+b=b+a for all a,b ∈ ZThus (Z , +) is an abelian group. Also Z

contains an infinite number of elements.therefore (Z , +) is an abelian group of infinite order.

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

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

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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

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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.

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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).

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

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Proof:: If possible let e1 and e2 be two identities in

the group (G, .) Since e1 is identity and e2 ∈ G therefore e1 . e2 = e2 = e2. e1 …..(1) Also since e2 is the identity and e1 ∈ G therefore e1 . e2 = e1 = e2 . e1 …..(2) therefore from (1) and (2), e1 = e2 Hence the identity is unique

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PROOF: Let a be any element of the group ( G , .) If possible , let b1 and b2 be two inverses

of a under the binary operation ‘ . ‘ and let e be the identity element in G. Then

a . b1 = e = b1 . a

And a . b2 = e = b2 . a

Now b1 = b1 . e

=b1 . ( a . b2)

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b1=(b1 . a) . b2 [by

associativity] =e . b2 = b2

Therefore b1 = b2.

hence the inverse is unique.

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Proposition . If a, b are elements of a group G , then

(i) (a−1)−1 = a.(ii) (ab)−1 = b−1a−1.i.e. , the inverse of the product of two

elements of a group is the product of their inverses in the reverse order.

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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

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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 agroup and H is a nonempty subset of G, then (H, ・ ) is called a subgroup of (G, ・ ) if thefollowing 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.

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(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 = acTherefore b=cHence, the result.

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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 ≠ фHere H is a subset of G.We have to show that H is a subgroup of G.

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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.

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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.

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Definition. Let G be a group and let a G ane e be the identity element in G. If ak = 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.

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Definition. If G is a group and a G, write <a > = {an : n } = {all powers of a } . It is easy to see that <a > is a subgroup of G . < a > is called the cyclic subgroup of G

generated by a. A group G is called cyclic if there is some a G

with G = < a >; in this case a is called a generator

of G.

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Proof:-Let G be a finite group and o(G)= sLet a G such that o(a)= sIf H = {an : n } 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.

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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 = 1+ 6 1 = 2

1 = 1 + 6 1 = 3

1 = 1 +6 1 = 4 37

23 2

4 3

1

1 = 1 + 6 1 = 5

1 = 1 + 6 1 = 0

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

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5 4

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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.

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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

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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.

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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.

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4. If a, b are elements of a group G , then(i) (a−1)−1 = a.(ii) (ab)−1 = b−1a−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

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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 6.

9. 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.

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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?

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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.

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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.

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DO ANY THREE QUESTIONS.1. If a, b are elements of a group G , then (i) (a−1)−1 = a. (ii) (ab)−1 = b−1a−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.

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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.

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