Download presentation

Presentation is loading. Please wait.

Published byLuke Alexander Booker Modified over 9 years ago

1
1

2
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

3
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

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

5
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

6
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

7
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

8
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

9
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

10
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

11
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

12
(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

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

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

15
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

16
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

17
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

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

19
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

20
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

21
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

22
PProposition. 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

23
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
24 Proof:

25
25

26
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

27
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

28
(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

29
(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

30
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

31
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

32
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

33
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

34
34

35
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

36
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

37
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 2 3 2 4 3 1

38
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 = Hence proved 38 54 6 5

39
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

40
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

41
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

42
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

43
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

44
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. 44

45
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

46
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

47
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

48
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

49
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

Similar presentations

© 2024 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google