0 Sets and Induction. Sets

Transcription

1 0 Sets and Induction Sets A set is an unordered collection of objects, called elements or members of the set. A set is said to contain its elements. We write a A to denote that a is an element of the set A. The notation a / A denotes that a is not an element of the set A. 1

2 Defining a Set The roster method is a way of defining a set by listing all of its members. Examples S = {a, b, c} S = {1, 2, 3, 4} S = {1, 4, 7, 10, 13, 16,...} 2

3 Defining a Set Another way to describe a set is using set builder notation. Examples S = {1, 2, 3, 4}, or S = {x x is a positive integer 4} S = {1, 4, 7, 10, 13, 16,...}, or S = {x x = 1 + 3k for some nonnegative integer k} 3

4 Defining a Set Example: Express the following set using set builder notation. S = {..., 12, 8, 4, 0, 4, 8, 12,...} 4

5 Important Sets Notation Z = {..., 2, 1, 0, 1, 2,...}, the set of integers Z + = {1, 2, 3,...}, the set of positive integers Q = { p q p Z, q Z, and q 0}, the set of rational numbers R, the set of real numbers R +, the set of positive real numbers C, the set of complex numbers 5

6 Special Sets A set with no elements is called the empty set, or null set, and is denoted by Ø. The empty set can also be denoted by { }. A set with one element is called a singleton set. For example, S = {1} is a singleton set containing only the number 1. 6

7 Subsets The set A is a subset of B if and only if every element of A is also an element of B. That is, A is a subset of B iff x (x A x B). We use the notation A B to indicate that A is a subset of B. 7

8 Proper Subsets We say that A is a proper subset of B if and only if A B and A B. That is, A is a proper subset of B iff x (x A x B) x (x B x / A). We use the notation A B to indicate that A is a proper subset of B. 8

9 Subsets Examples If S = {1, 2, 3} and T = {1, 2, 3, 4, 5}, then S T. If S = {π, 2} and T = {5, 2}, then S T. Z + Z 9

10 Special Subsets For every set S, (i) Ø S (ii) S S 10

11 Equality of Sets Two sets are equal if and only if they have the same elements. Therefore, if A and B are sets, then A = B if and only if x (x A x B). That is, A = B iff A B and B A. 11

12 Equality of Sets Example The sets {1, 3, 5} and {3, 5, 1} are equal, because they have the same elements. Note that the order in which the elements of a set are listed does not matter. 12

13 Intersection and Union Let S and T be sets. The intersection of S and T, denoted S T, is the set of elements which belong to both S and T. That is, S T = {x x S and x T } The union of S and T, denoted S T, is the set of elements which belong to S or T. That is, S T = {x x S or x T } 13

14 Intersection and Union Example Let S = {1, 2, 3, 4, 5} and T = {2, 4, 6}. Then, S T = {2, 4}. S T = {1, 2, 3, 4, 5, 6}. 14

15 Sets and Logic Statements involving sets and their logical equivalents. x / S (x S) x S T (x S) (x T ) x S T (x S) (x T ) S T x (x S x T ) S = T x (x S x T ) 15

16 Sets and Proofs To show A B Assume x A, then show x B. To show A B Show there exists an x A such that x B. 16

17 Sets and Proofs To show A = B Step 1. Assume x A, then show x B. Step 2. Assume x B, then show x A. 17

18 Sets and Proofs Example: Let S and T be sets. Prove that S T S T. 18

19 Sets and Proofs Example: Let S and T be sets. Prove that S T if and only if S T = S. 19

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21 Mathematical Induction Let P (n) is a propositional function with domain Z +. To prove that P (n) is true for all positive integers n, we complete two steps: 1. (Base Case) Verify that P (1) is true. 2. (Inductive Step) Prove the conditional statement P (k) P (k + 1) is true for all positive integers k. To complete the inductive step, we assume P (k) is true (this assumption is called the induction hypothesis), then show P (k+1) must also be true. 20

22 Mathematical Induction Example: Show that if n is a positive integer, then n = n(n + 1) 2 21

23 Mathematical Induction Proof: Let P (n) be the proposition n(n + 1) n =. 2 We want to show P (n) is true for all n 1. Base Step: 22

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25 Mathematical Induction Example: Conjecture a formula for the sum of the first n positive odd integers. Then prove your conjecture using mathematical induction. 1 = = = = = 23

26 Mathematical Induction Conjecture: For all positive integers n, the following proposition P (n) holds n 1 = Proof: 24

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28 Mathematical Induction Example: Use mathematical induction to prove that 2 divides n 2 + 5n for all n 1. 27

29 Mathematical Induction Proof: Let P (n) be the proposition 2 (n 2 + 5n). We want to show P (n) is true for all n 1. Base Step: 28

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31 1 Binary Operations Cartesian Product Let A and B be sets. The Cartesian product of A and B, denoted by A B, is the set of all ordered pairs (a, b), where a A and b B. Hence, A B = {(a, b) a A and b B}. 1

32 Cartesian Product Example: If A = {1, 2} and B = {a, b, c}, then A B = {(1, a), (1, b), (1, c), (2, a), (2, b), (2, c)}. 2

33 Binary Operations If S is a set, then a binary operation on S is a function that associates to each ordered pair (s 1, s 2 ) S S an element of S which we denote by s 1 s 2. That is, is a function from S S to S. 3

34 Binary Operations Examples Addition defines a binary operation on Z + since n + m Z + for all (n, m) Z + Z +. Multiplication defines a binary operation on Z + since n m Z + for all (n, m) Z + Z +. Subtraction does not define a binary operation on Z + since there exists (n, m) Z + Z + such that n m / Z +. Division does not define a binary operation on Z + since there exists (n, m) Z + Z + such that n/m / Z +. 4

35 Binary Operations Examples Subtraction is a binary operation on each of the sets Z, R, and Q. Division is a binary operation on each of the sets Q + and R +. 5

36 Binary Operations Example Let X be a set, and let P(X) be the set of all subsets of X. Then the operations of union and intersection are binary operations on P(X). For example, if X = {1, 2}, then P(X) = {Ø, {1}, {2}, {1, 2}} and the operations of union and intersections are binary operations on S. The set P(X) is called the power set of X. 6

37 Binary Operations Example Let X be a set, and let P(X) be the set of all subsets of X. Then the operation defined by A B = (A B) (B A) is a binary operation on P(X). The set A B is called the symmetric difference of A and B. 7

38 Binary Operations Let S be the set of all 2 2 matrices with real entries. Then, the operation, defined by ( ) ( a 11 a 12 a 21 a 22 b 11 b 12 b 21 b 22 ) = a 11b 11 + a 12 b 21 a 11 b 12 + a 12 b 22 a 21 b 11 + a 22 b 21 a 21 b 12 + a 22 b 22 is a binary operation on S. The corresponds to multiplication of matrices. 8

39 Commutativity and Associativity A binary operation on S is called commutative iff a b = b a for all a, b S. A binary operation on S is called associative iff (a b) c = a (b c) for all a, b, c S. 9

40 Commutativity and Associativity Subtraction on Z is neither commutative nor associative. For example, (1 2) 3 1 (2 3) Division on R + is neither commutative nor associative. For example, 1/2 2/1 (1/2)/3 1/(2/3) Addition and multiplication are both associative and commutative on the sets Z, Q, R. 10

41 Commutativity and Associativity Example: Let be a binary operation on Z defined by a b = 2(a + b) is commutative? is associative? 11

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43 Commutativity and Associativity Example: Multiplication of 2 2 matrices is associative. (( a c b d ) ( e g f h )) ( i k j l ) = ( a c b d ) (( e g f h ) ( i k j l )) is not commutative. ( ( ) ) ( ( ) ) = = ( ( ) ) 12

44 Commutativity and Associativity Example: Let be the symmetric difference operator given by A B = (A B) (B A) is commutative? is associative? 13

45 2 Groups Definition of a group A set G together with a binary operation is called a group if the following conditions hold (i) (closure) x y G for all x, y G. (ii) (associativity) (x y) z = x (y z) for all x, y, z G. (iii) (identity element) There is an element e G such that x e = e x = x for all x G. (iv) (inverse elements) For each element x G there is an element y G such that x y = y x = e. 14

46 Definition of a group Remarks We use the notation (G, ) to represent the group with elements in G under the operation. Condition (i) states that is a binary operation on G. We say G is closed with respect to. The element e in condition (iii) is called an identity element of G. In particular, this means that all groups are nonempty. Condition (iv) states that every element x in G has an inverse element y. 15

47 Abelian Group A group (G, ) for which is commutative is called an abelian group. 16

48 Groups Example: (Z, +) The set Z under addition is a group. (i) x + y Z for all x, y Z. (ii) (x+y)+z = x+(y+z) for all x, y, z Z. (iii) x + 0 = 0 + x = x for all x Z. (iv) for each x Z there is an element x Z such that x+( x) = ( x)+x = 0. 17

49 Groups Example: (Q +, ) The set Q + under multiplication is a group. (i) x y Q + for all x, y Q +. (ii) (x y) z = x (y z) for all x, y, z Q +. (iii) x 1 = 1 x = x for all x Q +. (iv) for each x Q + there is an element 1 x Q + such that x 1x = 1 x x = 1. 18

50 Groups Example: (R n, +) The set of ordered n-tuples (a 1, a 2,..., a n ) of real numbers forms a group under the operation given by (a 1, a 2,..., a n ) (b 1, b 2,..., b n ) = (a 1 +b 1, a 2 +b 2,..., a n +b n ) identity element: inverse element: 19

51 Groups Example: (GL(2, R), ) The set of invertible 2 2 matrices with real entries forms a group under matrix multiplication. This group is called the general linear group of degree 2 over R, and is denoted by (GL(2, R). identity element: inverse element: 20

52 Groups Example: ({f f : R R}, +) The set of real-valued functions with domain R forms a group under the operation of pointwise addition of functions. (f + g)(x) = f(x) + g(x) identity element: inverse element: 21

53 Groups Example: (P(X), ) The set of subsets of a set X forms a group under the operation of symmetric difference of sets: A B = (A B) (B A) identity element: inverse element: 22

54 Groups Example: Consider the set Z together with the binary operation given by a b = 2(a + b). Does (Z, ) form a group? 23

55 Groups Example: Consider the set G = R {0} together with the binary operation given by a b = 2ab. Does (G, ) form a group? 24

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57 Additive group of integers modulo n Theorem (The Division Algorithm) Let a be an integer and n a positive integer. Then there are unique integers q and r, with 0 r < n, such that a = qn + r. Notation: The integer r in the division algorithm is called the remainder of a mod n, and will be denoted by a. 25

58 Additive group of integers modulo n Lemma: Assume n Z +. Then, a = b if and only if a b = nk for some k Z. Proof: 26

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61 Additive group of integers modulo n Example: (Z n, ) Let n be a positive integer, and consider the set Z n = {0, 1, 2,..., n 1}. Z n forms a group under the binary operation defined by x y = x + y where x + y is the unique number in Z n such that x + y x + y (mod n) This group is called the additive group of integers modulo n. identity element: inverse element: 27

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65 3 Fundamental Theorems About Groups Uniqueness of the Identity Element Theorem 3.1 If (G, ) is a group, then there is only one identity element in G. Proof: Assume e 1 and e 2 are identity elements of G. Since e 1 is an identity element, we know e 1 e 2 = e 2. Since e 2 is an identity element, we know e 1 e 2 = e 1. This proves e 1 = e 2. 1

66 Uniqueness of Inverses Theorem 3.2 If (G, ) is a group and x is any element of G, then x has only one inverse in G. Proof: Assume x G and assume y 1 and y 2 are inverses of x. Then, y 1 = y 1 e = y 1 (x y 2 ) = (y 1 x) y 2 = e y 2 = y 2 This proves y 1 = y 2, and we conclude that the inverse of an element is unique. 2

67 Inverse Element Notation If (G, ) is a group, then the inverse of an element x G is denoted by x 1. 3

68 Inverse Element Theorem 3.3 If (G, ) is a group and x G, then (x 1 ) 1 = x. Proof: Let y = (x 1 ) 1. Then, y is the unique element of G such that x 1 y = y x 1 = e. Note that the previous equation is satisfied when y = x, since x 1 is the inverse of x. Therefore, (x 1 ) 1 = y = x. 4

69 Alternate Proof: Assume x G. We have, (x 1 ) 1 = (x 1 ) 1 e = (x 1 ) 1 (x 1 x) = ((x 1 ) 1 x 1 ) x = e x = x.

70 Inverse Element Examples Consider the group (Z, +). all n Z, we have ( n) = n. Then, for Consider the group GL(2, R). Then, for all ( ) a b c d GL(2, R), we have ( ( ) ) a b 1 1 ( ) c d = a b c d. 5

71 Inverse of a Product Theorem 3.4 If (G, ) is a group and x, y G, then (x y) 1 = y 1 x 1. Proof: Assume x, y G. Let z = y 1 x 1. Then, (x y) z = (x y) (y 1 x 1 ) = x (y (y 1 x 1 )) = x ((y y 1 ) x 1 ) = x (e x 1 ) = x x 1 = e. A similar argument shows z (x y) = e. Therefore, (x y) 1 = z = y 1 x 1. 6

72 Alternate proof: (x y) 1 = (x y) 1 e = (x y) 1 (x x 1 ) = (x y) 1 ((x e) x 1 ) = (x y) 1 ((x (y y 1 )) x 1 ) = (x y) 1 (((x y) y 1 ) x 1 ) = (x y) 1 ((x y) (y 1 x 1 )) = ((x y) 1 (x y)) (y 1 x 1 ) = e (y 1 x 1 ) = y 1 x 1

73 Inverse Element Theorem 3.5 Assume (G, ) is a group and let x, y G. If either x y = e or y x = e, then y = x 1. Proof: First assume x y = e. We will solve for y by multiplying both sides of the equation on the left by x 1. We have, x 1 (x y) = x 1 e (x 1 x) y = x 1 e y = x 1 y = x 1 Similarly, if y x = e, then right multiplication by x 1 shows y = x 1. 7

74 Cancellation Laws Theorem 3.6 Assume (G, ) is a group and let x, y G. Then: (i) if x y = x z, then y = z; and (ii) if y x = z x, then y = z. Proof: See homework. 8

75 Identities and Inverses Notation Let G be a set, and assume is an associative binary operation on G. Then: if e G and x e = x for all x G, then we say e is a right identity with respect to. if x G and e x = x for all x G, then we say e is a left identity with respect to. if x, y G and x y = x, then we say y is a right inverse of x. if x, y G and y x = x, then we say y is a left inverse of x. 9

76 Sufficient Conditions for a Group Theorem 3.8 Let G be a set, and assume is an associative binary operation on G. If there exists a right identity e G with respect to, and if every element x G has a right inverse, then (G, ) is a group. 10

77 Proof: First we will show that right identity e G is also a left identity. Given x G, let y denote the right inverse of x, and let z denote the right inverse of y. Then, e x = (e x) e = (e x) (y z) = e (x (y z)) = e ((x y) z) = e (e z) = (e e) z = e z = (x y) z = x (y z) = x e = x This shows that e x = x for any x G. Hence, e is an (two-sided) identity element.

78 Finally we will show that for any x G, if y is the right inverse of x, then y is also a left inverse of x. If z denotes the right inverse of y, we have y x = (y x) e = (y x) (y z) = y (x (y z)) = y ((x y) z) = y (e z) = (y e) z = y z = e. This shows that for all x G there exists a y G such that x y = y x = e. Therefore, (G, ) is a group.

79 Insufficient Conditions for a Group Example: Consider the set Z + with the binary operation given by x y = x. is associative since (x y) z = x y = x = x (y z) the element 1 Z + is a right identity element since x 1 = x for all x Z +. (In fact, any element y Z + is an right identity element.) every element y Z + has the element 1 Z + as a left inverse element since 1 y = 1 for all y Z +. 11

80 However these conditions are not sufficient to make Z + a group with respect to, since there is no left identity element: y x x unless y = x Note that multiplication table for Z + with respect to has repeated elements in its rows (See Homework 3, Problem 3):

81 4 Powers of an Element; Cyclic Groups Notation When considering an abstract group (G, ), we will often simplify notation as follows x y will be expressed as xy (x y) z will be expressed as xyz x (y z) will be expressed as xyz In other words, we will omit the symbol, and use product notation, unless a more appropriate notation is called for (e.g., the symbol + will be used in the case of the group (Z, +)). Also, when associativity allows, we can omit parentheses. 1

82 Powers of an Element Notation Let G be a group, and let x be an element of G. We define powers of x as follows: x 0 = e x n = xxx x }{{} n factors x n = x 1 x 1 x 1 x 1 }{{} n factors 2

83 Powers of an Element Theorem 4.1 Let G be a group and let x G. Let m, n be integers. Then: (i) (ii) (iii) x m x n = x m+n (x n ) 1 = x n (x m ) n = x nm = (x n ) m 3

84 Order of an Element Definitions If G is a group and x G, then x is said to be of finite order if there exists a positive integer n such that x n = e. If such an integer exists, then the smallest positive n such that x n = e is called the order of x and denoted by o(x). If x is not of finite order, then we say that x is of infinite order and write o(x) =. 4

85 Order of an Element Examples Let G = (Z 3, ). Then o(1) = 3, since = 0 Let G = (Z, +). Then o(1) =, since

86 Order of an Element Examples Let G = (Q +, ). Then o(2) =, since Let G = GL(2, R), then o (( )) = 2, since ( ) ( ) ( ) ( ) 0 1 =

87 Greatest Common Divisor Let a and b be integers, not both zero. The largest integer d such that d a and d b is called the greatest common divisor of a and b. The greatest common divisor of a and b is denoted by gcd(a, b). 7

88 Greatest Common Divisor Examples gcd(24, 40) = gcd(32, 100) = gcd(12, 91) = 8

89 Relatively Prime The integers a and b are relatively prime if their greatest common divisor is 1. Examples The integers 12 and 25 are relatively prime, since gcd(12, 25) = 1 The integers 27 and 75 are not relatively prime, since gcd(27, 75) =

90 The Euclidean Algorithm Suppose that a and b are positive integers with a b. Successively applying the division algorithm, we obtain a = b q 0 + r 1, 0 r 1 < b, b = r 1 q 1 + r 2, 0 r 2 < r 1, r 1 = r 2 q 2 + r 3, 0 r 3 < r 2, r 2 = r 3 q 3 + r 4, 0 r 4 < r 3, r n 2 = r n 1 q n 1 + r n, 0 r n < r n 1, r n 1 = r n q n. Theorem: gcd(a, b) = r n, where r n is the last nonzero remainder in the sequence above. 10

91 The Euclidean Algorithm Lemma Let a = bq + r, where a, b, q, and r are integers. Then gcd(a, b) = gcd(b, r). Proof: Assume d a and d b. Therefore d r where r = a bq. This shows that all common divisors of a and b are common divisors of b and r. Next, assume d b and d r. Therefore d a where a = bq + r. This shows that all common divisors of b and r are common divisors of a and b. Therefore the common divisors of a and b are the same as the common divisors of b and r. Therefore, their greatest common divisors are the same. 11

92 The Euclidean Algorithm Examples Find gcd(198, 252) using the Euclidean algorithm Find gcd(414, 662) using the Euclidean algorithm 12

93 Greatest Common Divisor Theorem 4.2 If a and b are integers, not both zero, then there exist integers x and y such that gcd(a, b) = ax + by. Proof: Solving for the last nonzero remainder in the Euclidean Algorithm, we have r n = r n 2 r n 1 q n 1 Using the preceding step of the Euclidean Algorithm, we may express the term r n 1 in terms of r n 2 and r n 3, thereby obtaining r n = r n 2 (r n 3 r n 2 q n 2 ) q n 1 After a sequence of n 1 substitutions, we obtain an expression for gcd(a, b) = r n in terms of a and b. 13

94 Greatest Common Divisor Example: Express gcd(198, 252) = 18 as a linear combination of 252 and

95 Greatest Common Divisor Theorem 4.3 If a bc and gcd(a, b) = 1 and, then a c. Proof: Since gcd(a, b) = 1, there exist integers x and y such that ax + by = 1. Multiplying on both sides by c yields axc + byc = c a(xc) + bc(y) = c Since a divides each term on the left-hand side, a also divides the sum of the two terms. Therefore, a divides c. 15

96 Powers of an Element Theorem 4.4 Let G be a group and let x G. Let m, n, d be integers. Then: (i) o(x) = o(x 1 ) (ii) (iii) If o(x) = n and x m = e, then n m If o(x) = n and gcd(m, n) = d, then o(x m ) = n/d. 16

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100 Cyclic Groups Definition A group G is called cyclic if there is an element x G such that G = x = {x n n Z}. The element x is called a generator for G, and the cyclic group generated by x is denoted by x. 17

101 Cyclic Groups Example The group G = (Z 1, ) is the trivial group {0} consisting of one (identity) element. It is the cyclic group generated by x = 0: (Z 1, ) = 0 For all n 2, the group G = (Z n, ) is a finite cyclic group generated by x = 1: (Z n, ) = 1 = {0, 1, 1 1,..., } 1 1 1{{ 1} } n 1 terms 18

102 Cyclic Groups Example The group G = (Z, +) is an infinite cyclic group generated by x = 1: (Z, +) = 1 = {..., ( 1) + ( 1), ( 1), 0, 1, 1 + 1, ,...} The group G = (Q, +) is not cyclic, since there does not exist q Q such that ever rational number r Q has the form r = nq for some n Z. 19

103 Cyclic Groups Theorem 4.5 Let G = x. If o(x) =, then x j x k for j k, and consequently G is infinite. If o(x) = n, then x j = x k iff j k (mod n), and consequently the distinct elements of G are e, x, x 2,..., x n 1. Proof: 20

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105 Cyclic Groups Definition The order of a group G, denoted by G, is the number of elements in G. Corollary 4.6 If G = x, then G = o(x). 21

106 Cyclic Groups Theorem 4.7 If G is a cyclic group, then G is abelian. Proof: 22

107 5 Subgroups Definition of a Subgroup A subset H of a group (G, ) is called a subgroup of G if the elements of H form a group under. 1

108 Subgroups Examples (2Z, +) is a subgroup of (Z, +) (Z, +) is a subgroup of (Q, +) (Q, +) is a subgroup of (R, +) (Q +, ) is a subgroup of (R +, ) {( a b b a ) a, b R and a 2 + b 2 0 } is a subgroup of G = GL(2, R). {( a b 0 d ) a, b, d R and ad 0 } is a subgroup of G = GL(2, R). 2

109 Subgroups Theorem If H is a subgroup of G and e is the identity element for G, then e H and e is the identity element for H. 3

110 Proof: Assume H is a subgroup of G. Then H is group and must have an identity element e H. We claim e H = e where e is the identity element for G. Since e H identity element for H, we know, e H e H = e H. is an Since H G, we know e H G. Therefore, e H has an inverse element e 1 H G. Multiplying on both sides of the previous equation by e 1 H yields: e 1 H (e 1 H (e H e H ) = e 1 H e H) e H = e e e H = e e H e H = e

111 Trivial Subgroups For any group G, the subgroups H = {e} and H = G are called trivial subgroups. 4

112 Groups of Order 1 There is only one group of order 1, namely G = {e}. The group table for G is given by e e e The group G has only one subgroup H = G = {e}. The group G is cyclic. The element e is a generator for G. 5

113 Groups of Order 2 There is only one group of order 2. If we write G = {e, a}, then the group table for G is given by e a e e a a a The group G has two subgroups, namely H = {e} and H = G = {e, a}. The group G is cyclic. The element a is a generator for G. 6

114 Groups of Order 3 There is only one group of order 3. If we write G = {e, a, b}, then the group table for G is given by e a b e e a b a b a b The group G has two subgroups, namely H = {e} and H = G = {e, a, b}. The group G is cyclic. Both elements a and b are generators for G. 7

115 Groups of Order 4 There are two groups of order 4. If we write G = {e, a, b, c}, then the possible group tables for G are e a b c e e a b c e a b c e e a b c a a a a b b b b c c c c e a b c e e a b c e a b c e e a b c a a a a b b b b c c c c 8

116 Cyclic Group of Order 4 If we write G = {e, a, b, c}, then the following group table for G represents a cyclic group of order 4. e a b c e e a b c a b c a b c The group G has three subgroups, namely H = {e}, H = {e, b}, and H = {e, a, b, c}. The group G is cyclic. Both elements a and c are generators for G. 9

117 Klein s 4-group If we write G = {e, a, b, c}, then the following group table for G represents a noncyclic group of order 4 called Klein s 4- group. e a b c e e a b c a b c a b c Klein s 4-group has five subgroups: H = {e}, H = {e, a}, H = {e, b}, H = {e, c}, and H = {e, a, b, c}. 10

118 Subgroup Lattice A subgroup lattice is a diagram which depicts the subgroups of a group G in a way that indicates all subset relations among subgroups. Example (Klein s 4-group) 11

119 Groups of Order n The previous examples beg the question: How many groups of order n exist for a given integer n? The answer for n 99 is given below

120 Subgroups Theorem 5.1 Let H be a nonempty subset of a group G. Then H is a subgroup of G if and only if the following two conditions are satisfied: (i) for all a, b H, ab H, and (ii) for all a H, a 1 H. Remark: Condition (i) is expressed by saying that H is closed under the binary operation on G, and condition (ii) is expressed by saying that H is closed under inverses. 13

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123 Subgroups Theorem Let G be a group, and let a G. Then a, the set of elements generated by integer powers of a, is a subgroup of G. Proof: Let H = a. Clearly H is nonempty since a H. Also for all a j, a k H, we have a j a k = a j+k H. Also, for any a j H, we have (a j ) 1 = a j H. Therefore, by Theorem 5.1, H = a is a subgroup of G. 14

124 Subgroups Examples Let G = (Z, +) and let n Z. H = n = nz is a subgroup of G. Then, Let G = (Z 6, ). Then the following are subgroups of G: 0 = {0} 1 = {0, 1, 2, 3, 4, 5} 2 = {0, 2, 4} 3 = {0, 3} 4 = {0, 2, 4} 5 = {0, 1, 2, 3, 4, 5} 15

125 Subgroups Theorem 5.2 If G is a cyclic group, then every subgroup of G is cyclic. Proof: 16

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127 Subgroups Theorem 5.4 Let H and K be subgroups of a group G. Then: (i) H K is a subgroup of G; and (ii) H K is a subgroup of G if and only if H K = H or H K = K. Proof: 17

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129 Subgroups Theorem 5.5 Let G = x be a cyclic group of order n. Then: (i) For any positive integer m, G has a subgroup of order m if and only if m divides n. (ii) If m divides n, then G has a unique subgroup of order m. (iii) The elements x r and x s generate the same subgroup of G if and only if (r, n) = (s, n). 18

130 Group of Unit Quaternions Consider the general linear group of degree two over the complex numbers GL(2, C) = {( z1 z 2 z 3 z 4 ) } z k C and z 1 z 4 z 2 z 3 0 The group of unit quaternions, denoted Q 8, is the subgroup of GL(2, C) consisting of the following 8 elements: { ( 1 0 ), 0 1 ( ( i 0 0 i ), ), ( ( i 0 0 i ), ), ( ( 0 i i 0 ) ), ( 0 i i 0 )} 19

131 Group of Unit Quaternions Let J = ( i 0 ) ( 0 1 ) ( 0 i, K =, L = 0 i 1 0 i 0 The group of unit quaternions has the form Q 8 = {I, I, J, J, K, K, L, L}, ). and the following relations hold J 2 = K 2 = L 2 = I JK = L KL = J LJ = K KJ = L LK = J JL = K 20

132 6 Direct Products Direct Product of Groups Let G and H be groups. Then the set of ordered pairs G H = {(g, h) g G, h H} forms a group under componentwise multiplication: (g 1, h 1 ) (g 2, h 2 ) = (g 1 g 2, h 1 h 2 ) The group G H is called the direct product of G and H. 1

133 Direct Product of Groups Remarks The identity element of G H is (e G, e H ) where e G and e H are the respective identity elements of G and H. The inverse of (g, h) G H is the ordered pair (g 1, h 1 ). 2

134 Generalized Direct Product Let G 1, G 2, G 3,..., G n be groups. Then the set of ordered n-tuples G 1 G n = {(g 1, g 2,..., g n ) g i G i } forms a group under componentwise multiplication. The identity element of G 1 G n is (e G1, e G2,..., e Gn ) where e Gi is the identity element for G i. The inverse of (g 1, g 2,..., g n ) is the element (g 1 1, g 1 2,..., g 1 n ). 3

135 Direct Products Example Let G = H = (Z 2, ). Then, G H = Z 2 Z 2 is the group of ordered pairs {(0, 0), (0, 1), (1, 0), (1, 1)} under componentwise addition mod 2. Z 2 Z 2 is a finite, non-cyclic group of order 4 with the following non-trivial subgroups (0, 1) = {(0, 0), (0, 1)} (1, 0) = {(0, 0), (1, 0)} (1, 1) = {(0, 0), (1, 1)} 4

136 Direct Products Example Let G = (Z 2, ) and H = (Z 3, ). Then, G H = Z 2 Z 3 is the group of ordered pairs {(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2)} under componentwise addition. Z 2 Z 3 is a finite, cyclic group of order 6 with generator (1, 1), and with the non-trivial subgroups (0, 1) = {(0, 0), (0, 1), (0, 2)} (1, 0) = {(0, 0), (1, 0)} 5

137 Direct Products Example Let G 1 = G 2 = = G n = (R, +). Then, G 1 G n = R R }{{} n-factors = R n is the group of ordered n-tuples of real numbers under componentwise addition. The identity element of R n is (0, 0,..., 0). The inverse of (x 1, x 2,..., x n ) is the element ( x 1, x 2,..., x n ). 6

138 Direct Products Theorem 6.1 Let G = G 1 G n. (i) If g i G i for 1 i n, and each g i has finite order, then o((g 1,..., g n )) is the least common multiple of the numbers o(g 1 ), o(g 2 ),..., o(g n ). (ii) If each G i is a cyclic group of finite order, then G is cyclic iff G i and G j are relatively prime for i j. 7

139 Proof:

140

141

142 Direct Products Examples Consider the element (1, 3) Z 6 Z 8. Since o(1) = 6 in the group Z 6, and o(3) = 8 in the group Z 8, we have o((1, 3)) = lcm(6, 8) = 24 The groups Z 14 Z 15 and Z 8 Z 9 Z 5 are cyclic. The groups Z 14 Z 16 and Z 8 Z 9 Z 6 are not cyclic. 8

143 7 Functions Function, Domain, and Codomain Let A and B be nonempty sets. A function f from A to B, denoted f : A B, is a subset of the Cartesian product A B such that for each a A, there is a unique element b B such that (a, b) belongs to f. We say that A is the domain of f and B is the codomain of f. 1

144 Range of a Function If f : A B and the ordered pair (a, b) belongs to f, then we write f(a) = b and we say b the image of a under f. The range of f, denoted f(a), is the set of all images of elements of A. 2

145 Examples f : R R, defined by f(x) = x 2 f : GL(2, R) R, defined by f (( )) a b c d = ad bc f : G G, defined by f(x) = a x where G is a group and a G. F : C(R, R) R, defined by F (g) = 1 0 g(x) dx where C(R, R) is the set of continuous functions from R to R. 3

146 Image and Preimage If f : A B and S A, then the image of S under f is the set f(s) = {b B b = f(s) for some s S} If T B, then the preimage of T is the set f 1 (T ) = {a A f(a) T } 4

147 Example Let f : R R be defined by f(x) = x 2. domain of f: codomain of f: range of f: If S = [ 2, 3), then f(s) = If T = ( 1, 4], then f 1 (T ) = 5

148 Example Let f : R R be defined by f(x) = x 2. The domain of f is R. The codomain of f is R. The range of f is R + {0}. If S = [ 2, 3), then f(s) = [0, 9). If T = ( 1, 4], then f 1 (T ) = [ 2, 2]. 6

149 One-To-One Function A function f : A B is said to be one-toone, or injective, if and only if the following condition holds: x y (x y f(x) f(y)) or equivalently x y (f(x) = f(y) x = y) where the domain for x and y is the set A. If f is one-to-one, we say f is an injection. 7

150 Onto Function A function f : A B is said to be onto, or surjective, if and only if the following condition holds: b a (f(a) = b) where the domain for b is the set B, and the domain for a is the set A. If f is onto, we say that f is a surjection. Note that f is onto if and only if f(a) = B. That is, f is onto if and only if the range of f is equal to the codomain of f. 8

151 Examples Let A = {1, 2, 3, 4} and B = {a, b, c}. Determine if the given functions are oneto-one, onto, both or neither. 1. f : A B defined by f(1) = a f(2) = b f(3) = c f(4) = b 2. f : B A defined by f(a) = 3 f(b) = 1 f(c) = 4 9

152 Examples f : R R, defined by f(x) = x 2 f : GL(2, R) R, defined by f (( )) a b c d = ad bc f : G G, defined by f(x) = a x where G is a group and a G. F : C(R, R) R, defined by F (g) = 1 0 g(x) dx where C(R, R) is the set of continuous functions from R to R. 10

153 Proof Strategy To show f is injective Show that if f(x) = f(y), then x = y. To show f is not injective Show that there exist x and y such that f(x) = f(y) and x y. To show f is surjective Show that for each element y in the codomain there exists an element x in the domain such that f(x) = y. To show f is not surjective Show there exists an element y in the codomain such that y f(x) for any x in the domain. 11

154 Example Prove or disprove: The function f : Z Z defined by is injective. f(n) = 2n 3 12

155 Example Prove or disprove: The function f : Z Z defined by is surjective. f(n) = 2n 3 13

156 Bijection A function f : A B is said to be a bijection, or one-to-one correspondence if f is both one-to-one and onto. Examples f : Z Z defined by f(n) = n + 1 f : R R defined by f(x) = 2x f : Z + Z + defined by f(n) = { n + 1 if n is odd n 1 if n is even 14

157 Inverse Function Let f : A B be a bijection. The inverse function of f, denoted by f 1, is the function that assigns to an element b B the unique element a A such that f(a) = b. That is, f 1 (b) = a if and only if f(a) = b If f has an inverse function, we say that f is invertible. Note that we use the notation f 1 (T ) to denote the preimage of a set T B, even when f is not invertible. 15

158 Example Determine if each function invertible. If so, find f f : Z Z defined by f(n) = n f : R R defined by f(x) = 2x f : R R defined by f(x) = x 2 16

159 Composition of Functions Let g : A B and f : B C be functions. The composition of the functions f and g, denoted by f g, is a function from A to C, defined by (f g) (a) = f(g(a)) 17

160 Examples Let A = {1, 2, 3, 4}, B = {a, b, c}, C = {w, x, y, z}, and let f and g be the functions below g : A B f : B C defined by defined by g(1) = a g(2) = b g(3) = c g(4) = b f(a) = x f(b) = w f(c) = z Then, f g : A C is given by (f g)(1) = (f g)(2) = (f g)(3) = (f g)(4) = 18

161 Composition of Functions Example Let g : R R and f : R R be defined by g(x) = 2 1 (x 3) and f(x) = 2x + 3. Find (a) (f g)(x) = (b) (g f)(x) = 19

162 Inverse Functions and Composition Let f : A B and g : B A. Then, g is the inverse function for f if and only if the following two conditions hold. (i) (f g)(x) = x for all x B. (ii) (g f)(x) = x for all x A. In particular, if f is invertible, then (i) f(f 1 (x)) = x for all x B. (ii) f 1 (f(x)) = x for all x A. 20

163 Identity Function Let A be a set. The identity function on A is the function i A : A A defined by i A (x) = x That is, i A is the function that assigns each element of A to itself. The identity function i A is one-to-one and onto, so it is a bijection. 21

164 Set of Bijections from X to X Let X be a nonempty set, and define S X = {f : X X f is a bijection} Theorem 7.1: (S X, ) is a group. Proof: (i) (Closure) See Homework 5, Problem 7 (ii) (Associativity) for all x X, For all f, g, h S X, and ((f g) h)(x) = (f g)(h(x)) = f(g(h(x))) = f((g h)(x)) = (f (g h))(x) That is, (f g) h = f (g h), which proves is associative. 22

165 (iii) (identity element) The identity function i X : X X is a bijection, hence belongs to S X, and has the property (f i X )(x) = f(i X (x)) = f(x) = i X (f(x)) = (i X f)(x) for all x X. This proves i X is an identity element for S X. (iv) (inverse elements) For all f S X, there is an element g S X, namely g = f 1, such that (f f 1 )(x) = f(f 1 (x)) = i X (x) = f 1 (f(x)) = (f 1 f)(x)

166 for all x X. This proves that every element of S X has an inverse. Thus, (S X, ) is a group.

167 8 Symmetric Groups Permutations If X is a nonempty set, then a permutation of X is a bijection f : X X. Recall that the set S X of all permutations of X forms a group under composition of functions. This group, denoted (S X, ), is called the symmetric group on X. 1

168 Symmetric Group of Degree n Let X = {1, 2, 3,..., n}. The symmetric group on X is called the symmetric group of degree n, and is denoted by S n. Given a permutation f S n, we represent f in array form as follows: n f(1) f(2) f(3) f(n) 2

169 Composition of Permutations Example Consider f, g S 4 defined by f(1) = 2 g(1) = 3 f(2) = 4 g(2) = 2 f(3) = 1 g(3) = 4 f(4) = 3 g(4) = 1 In array form we have: f = ( ) f(1) f(2) f(3) f(4) = ( ) g = ( ) g(1) g(2) g(3) g(4) = ( )

170 To compute the composition f g, the following diagram is helpful Therefore, f g = ( )

171 Observe that (f g)(1) = (( ) ( )) (1) = ( ) (( ) ) (1) = ( ) (3) = 1 (f g)(2) = (( ) ( )) (2) = ( ) (( ) ) (2) = ( ) (2) = 4

172 Also, we have (f g)(3) = (( ) ( )) (3) = ( ) (( ) ) (3) = ( ) (4) = 3 (f g)(4) = (( ) ( )) (4) = ( ) (( ) ) (4) = ( ) (1) = 2

173 Composition of Permutations Example: Determine f g where f, g S 7 are given by f = ( ) g = ( ) Solution:

174 Cycles An element f S n is called a cycle if there exist distinct integers i 1, i 2,..., i r {1, 2,..., n} such that f(i 1 ) = i 2 f(i 2 ) = i 3 f(i r 1 ) = f(i r ) f(i r ) = i 1 and f(j) = j otherwise. In this case, we express f in cycle notation as follows: f = (i 1, i 2, i 3,..., i r ) 5

175 Cycle Notation Example: Consider f, g S 7 given by f = ( ) g = ( ) Then, f and g are both cycles of length 4. In cycle notation, we have f = (1, 3, 2, 4), g = (1, 7, 6, 2). 6

176 Cycle Notation A cycle of length r is called an r-cycle. A 2-cycle is called a transposition. The identity permutation is usually denoted by the 1-cycle (1). 7

177 Disjoint Cycles Two cycles (x 1, x 2,..., x r ) and (y 1, y 2,..., y s ) are said to be disjoint if r 2, s 2, and {x 1, x 2,..., x r } {y 1, y 2,..., y s } = Ø Examples (1, 3), (2, 4) S 4 (1, 5, 2), (3, 7, 8, 4) S 8 (2, 6, 9, 4), (5, 7) S 9 8

178 Product of Cycles While not all permutations f S n are cycles, all permutations can be expressed as a product (composition) of disjoint cycles. Theorem 8.1: Let f S n. Then, there exist disjoint cycles f 1, f 2,..., f m S n such that f = f 1 f 2 f m. 9

179 Product of Cycles Example: Consider f S 9 given by f = ( ) Note that f is composed of three disjoint cycles as indicated below f(1) = 4 f(3) = 7 f(5) = 6 f(4) = 9 f(7) = 8 f(6) = 5 f(9) = 2 f(8) = 3 f(2) = 1 Therefore, we write f = (1, 4, 9, 2)(3, 7, 8)(5, 6) 10

180 Order of a Permutation Theorem: Assume f = f 1 f 2 f m where f 1, f 2,..., f m are disjoint cycles. Then o(f) is the least common multiple of o(f 1 ), o(f 2 ),..., o(f m ). That is, o(f) = lcm (o(f 1 ), o(f 2 ),..., o(f m )). Note: If f is an r-cycle, then o(f) = r. 11

181 Order of a Permutation Example: Consider f S 9 given by f = ( ) As shown in the previous example, f can be expressed as a composition of disjoint cycles as follows: f = (1, 4, 9, 2)(3, 7, 8)(5, 6). Therefore, o(f) = lcm (4, 3, 2) = 12 12

182 Cycles and Transpositions Every cycle f = (i 1, i 2,..., i r ) can be expressed as a product of transpositions as follows: f = (i 1, i r ) (i 1, i r 1 ) (i 1, i 3 ) (i 1, i 2 ) Examples (1, 3, 2, 4) = (1, 4) (1, 2) (1, 3) (3, 7, 9, 5, 8) = (3, 8) (3, 5) (3, 9) (3, 7) 11

183 Permutations and Transpositions Theorem 8.3: Every permutation f S n, where n 2, can be expressed as a product of transpositions. Proof: By Theorem 8.1, f = f 1 f 2 f m where f 1, f 2,..., f m S n are disjoint cycles. Since each cycle f i can be expressed as a product of transpositions (as described above), this proves f is a product of transpositions. 12

184 Even/Odd Permutations We say that a permutation f S n is even if it can be written as the product of an even number of transpositions, and we say f is odd if it can be written as the product of an odd number of transpositions. Theorem 8.4: even and odd. No permutation is both 13

185 Product of Cycles Example: Consider the permutation f = (1, 3, 2, 4)(1, 7, 6, 2) S 7 Express f as a product of disjoint cycles. 14

186 Even/Odd Permutations Example: Consider the permutation f = (1, 3, 2, 4)(1, 7, 6, 2) S 7 Express f as a product of transpositions, then determine if f is even or odd. 15

187 Alternating Groups The alternating group of degree n, denoted A n, is defined to be the subset of S n consisting of all even permutations. 16

188 Alternating Groups Example: List all elements in S 3 using cycle notation, then express each element as a product of transpositions. Finally, determine the elements of A n. 17

189 Alternating Groups Theorem 8.5: If n 2, then A n is a subgroup of S n, and A n = n! 2 = S n 2 18

190 The symmetric group S 3 (1) (1, 2) (1, 3) (2, 3) (1, 2, 3) (1, 3, 2) (1) (1) (1, 2) (1, 3) (2, 3) (1, 2, 3) (1, 3, 2) (1, 2) (1, 2) (1) (1, 3, 2) (1, 2, 3) (2, 3) (1, 3) (1, 3) (1, 3) (1, 2, 3) (1) (1, 3, 2) (1, 2) (2, 3) (2, 3) (2, 3) (1, 3, 2) (1, 2, 3) (1) (1, 3) (1, 2) (1, 2, 3) (1, 2, 3) (1, 3) (2, 3) (1, 2) (1, 3, 2) (1) (1, 3, 2) (1, 3, 2) (2, 3) (1, 2) (1, 3) (1) (1, 2, 3) 19

191 S 3 as a group of symmetries If we associate the elements of the set X = {1, 2, 3} with the vertices of an equilateral triangle, then each element of the permutation group S 3 can be associated with a symmetry of the triangle as follows. 22

192 S 3 as a group of symmetries Rotations: Reflections: 23

193 Dihedral Groups In general, the symmetries a regular n-gon can be associated with a subgroup of S n called the dihedral group of order 2n. This group of symmetries consists of n rotations and n reflections, and is denoted by D n. If n = 3, the dihedral group D 3 is identical to S 3. If n 4, the dihedral group D n is a nontrivial subgroup of S n. We consider the case n = 4 below. 24

194 The dihedral group D 4 Rotations: Reflections: 25

195 The dihedral group D 4 Therefore, the elements of D 4 are {(1), (1, 2, 3, 4), (1, 3)(2, 4), (1, 4, 3, 2), (1, 2)(3, 4), (2, 4), (1, 4)(2, 3), (1, 3)} 26

196 9 Equivalence Relations; Cosets Relations A relation on a nonempty set S is a nonempty subset R of S S. Notation If (x, y) R, we write x R y. 1

197 Relations Example Let S be a nonempty set, then any function f S S is a relation. For example, if S = R, then f = {(x, x 2 ) x R} R R is the relation corresponding to the function f(x) = x 2. 2

198 Relations Example Let S = {1, 2, 3}. Then, R = {(1, 2), (1, 3), (2, 3)} S S is a relation on S corresponding to the less than relation. That is, for x, y S x R y iff x < y. 3

199 Equivalence Relation A relation R on S is called an equivalence relation if the following three properties hold: 1. (Reflexivity) x R x, for all x S. 2. (Symmetry) if x R y, then y R x. 3. (Transitivity) if x R y and y R z, then x R z. 4

200 Equivalence Relation Example Let R be the relation on S = Z defined by a R b iff a b (mod n) where n is a fixed positive integer. Prove that R is an equivalence relation. 5

201 Proof: Recall that a b (mod n) if and only if b a = nk for some integer k. (Reflexivity) (i) x R x, since x x = 0 = n 0. (Symmetry) (ii) Assume x R y. Then, x y = nk for some integer k. Therefore, y x = n( k), where k is an integer. Therefore, y R x. (Transitivity) (iii) Assume x R y and y R z. Then, x y = nk for some integer k, and y z = nj for some integer j. Then, x z = (x y) + (y z) = nk + nj That is, x z = n(k + j) where k + j is an integer. Therefore, x R z.

202 Equivalence Classes Let R be an equivalence relation on a nonempty set S. Given s S, we define [s] = {x S x R s}. That is, [s] is the subset of elements in S which are related to s. The set [s] is called the equivalence class of s under R. An element x S is called a representative of [s] if x [s]. 6

203 Equivalence Classes Let R be the relation on S = Z defined by a R b iff a b (mod 4) Then, [0] = {..., 8, 4, 0, 4, 8,...} [1] = {..., 7, 3, 1, 5, 9,...} [2] = {..., 6, 2, 2, 6, 10,...} [3] = {..., 5, 1, 3, 7, 11,...} where 0, 1, 2, and 3 are representatives of their respective equivalence classes. 7

204 Equivalence Classes Theorem 9.1 Let R be an equivalence relation on S. Then, every element of S is in exactly one equivalence class under R. That is, the equivalence classes partition S into a family of mutually disjoint nonempty subsets. Conversely, given any partition of S into mutually disjoint nonempty subsets, there is an equivalence relation on S whose equivalence classes are precisely the subsets in the given partition of S. 8

205 Proof: For every s S, we have s [s]. This shows every element of S is in at least one equivalence class. Next, suppose s [s 1 ] and s [s 2 ]. We want to show [s 1 ] = [s 2 ]. Since s 1 R s (by symmetry) and s R s 2, it follows by transitivity that s 1 R s 2. Therefore, by a homework exercise, [s 1 ] = [s 2 ]. This shows that each element in S is contained in exactly one equivalence class.

206 Equivalence Relation Theorem 9.2: Assume G is a group, and let H be a subgroup G. Then the relation H defined by x H y iff xy 1 H is an equivalence relation. 9

207 Proof: For any x G, we have xx 1 = e H. Therefore, x H x, which means H is reflexive. Next, assume x H y. Then, xy 1 H, and since H is closed under inverses, we also have yx 1 = (y 1 ) 1 x 1 = (xy 1 ) 1 H That is, yx 1 H, which means y H x. This proves H is symmetric. Finally, assume x H y and y H z. This means, xy 1 H and yz 1 H. Since H is closed under multiplication, we also have xz 1 = (xy 1 )(yz 1 ) H Therefore, x H z, which proves H is transitive.

208 Cosets Assume G is a group, and let H be a subgroup of G. Given any fixed element a G, the set ah = {x G x = ah for some h H} is called a left coset of H in G. Similarly, the set Ha = {x G x = ha for some h H} is called a right coset of H in G. 10

209 Cosets Example: Let G = S 3, and consider the subgroup H = {(1), (1, 2)}. Then, for the fixed element a = (1, 2, 3) G, we have the left coset ah = {(1, 2, 3)(1), (1, 2, 3)(1, 2)} and the right coset Ha = {(1)(1, 2, 3), (1, 2)(1, 2, 3)} 11

210 Cosets Theorem 9.3: Assume G is a group, and let H be a subgroup of G. If a G and [a] denotes the equivalence class of a under the equivalence relation H. Then, [a] = Ha That is, the equivalence classes of H precisely the right cosets of H. are 12

211 Proof: Assume G is a group, and let H be a subgroup of G. If a G, we have x [a] iff x H a iff xa 1 H iff xa 1 = h for some h H iff x = ha for some h H iff x Ha. Therefore, [a] = Ha.

212 Cosets Corollary 9.4: Assume G is a group, and let H be a subgroup of G. Then for all a, b G, Ha = Hb if and only if ab 1 H. 13

213 Proof: Assume G is a group, and let H be a subgroup of G. For all a, b G, we have Ha = Hb iff [a] = [b] iff iff a H b ab 1 H That is, Ha = Hb if and only if ab 1 H.

214 Cosets Theorem 9.5: Assume G is a group, and let H be a subgroup of G. If a G and [a] denotes the equivalence class of a under the equivalence relation H, defined by x H y iff x 1 y H, then [a] = ah. That is, the equivalence classes of H are precisely the left cosets of H. Furthermore, we have [a] = [b] iff ah = bh iff a 1 b H. 14

215 Cosets Example: Let G = S 3, and consider the subgroup H = {(1), (1, 2)}. Then, the left cosets of H are (1)H = {(1)(1), (1)(1, 2)} (1, 2)H = {(1, 2)(1), (1, 2)(1, 2)} (1, 3)H = {(1, 3)(1), (1, 3)(1, 2)} (2, 3)H = {(2, 3)(1), (2, 3)(1, 2)} (1, 2, 3)H = {(1, 2, 3)(1), (1, 2, 3)(1, 2)} (1, 3, 2)H = {(1, 3, 2)(1), (1, 3, 2)(1, 2)} 15

216 Cosets Example: Let G = S 3, and consider the subgroup H = {(1), (1, 2)}. Then, the left cosets of H are (1)H = {(1)(1), (1)(1, 2)} = {(1), (1, 2)} (1, 2)H = {(1, 2)(1), (1, 2)(1, 2)} = {(1), (1, 2)} (1, 3)H = {(1, 3)(1), (1, 3)(1, 2)} = {(1, 3), (1, 2, 3)} (2, 3)H = {(2, 3)(1), (2, 3)(1, 2)} = {(2, 3), (1, 3, 2)} (1, 2, 3)H = {(1, 2, 3)(1), (1, 2, 3)(1, 2)} = {(1, 3), (1, 2, 3)} (1, 3, 2)H = {(1, 3, 2)(1), (1, 3, 2)(1, 2)} = {(2, 3), (1, 3, 2)} 16

217 Cosets Example: Let G = S 3, and consider the subgroup H = {(1), (1, 2)}. Then, the right cosets of H are H(1) = {(1)(1), (1, 2)(1)} H(1, 2) = {(1)(1, 2), (1, 2)(1, 2)} H(1, 3) = {(1)(1, 3), (1, 2)(1, 3)} H(2, 3) = {(1)(2, 3), (1, 2)(2, 3)} H(1, 2, 3) = {(1)(1, 2, 3), (1, 2)(1, 2, 3)} H(1, 3, 2) = {(1)(1, 3, 2), (1, 2)(1, 3, 2)} 17

218 Cosets Example: Let G = S 3, and consider the subgroup H = {(1), (1, 2)}. Then, the right cosets of H are H(1) = {(1)(1), (1, 2)(1)} = {(1), (1, 2)} H(1, 2) = {(1)(1, 2), (1, 2)(1, 2)} = {(1), (1, 2)} H(1, 3) = {(1)(1, 3), (1, 2)(1, 3)} = {(1, 3), (1, 3, 2)} H(2, 3) = {(1)(2, 3), (1, 2)(2, 3)} = {(2, 3), (1, 2, 3)} H(1, 2, 3) = {(1)(1, 2, 3), (1, 2)(1, 2, 3)} = {(2, 3), (1, 2, 3)} H(1, 3, 2) = {(1)(1, 3, 2), (1, 2)(1, 3, 2)} = {(1, 3), (1, 3, 2)} 18

219 Cosets Remark: If G is an abelian group, and H is a subgroup of G. Then the left cosets and right cosets of H are identical. That is ah = Ha for all a G. Example: Let G = (Z, +), and consider the subgroup H = 4Z. The left and right cosets of H are given by 0 + 4Z = {..., 8, 4, 0, 4, 8,...} = 4Z Z = {..., 7, 3, 1, 5, 9,...} = 4Z Z = {..., 6, 2, 2, 6, 10,...} = 4Z Z = {..., 5, 1, 3, 7, 11,...} = 4Z + 3 Also note that a+4z = b+4z iff a b 4Z iff a b (mod 4). 19

220 10 Counting the Elements of a Finite Group Lagrange s Theorem Theorem 10.1 (Lagrange s Theorem) Let G be a finite group, and assume H is a subgroup of G. Then, H divides G. 1

221 Lagrange s Theorem Lemma 10.2: Let G be a group, and assume H is a subgroup of G. Then, for all a, b G, there is a one-to-one correspondence between the elements of the right coset Ha and the right coset Hb. In particular, if G is a finite group then we write Ha = Hb. 2

222 Lagrange s Theorem Proof: Let H be a subgroup of G, and assume a, b G. We consider the function f : Ha Hb defined by f(ha) = hb We want to show f is one-to-one and onto. First, observe that f(h 1 a) = f(h 2 a) iff iff h 1 b = h 2 b (h 1 b)b 1 = (h 2 b)b 1 iff h 1 = h 2 iff h 1 a = h 2 a This proves f is one-to-one. Next given any element hb Hb, we have f(x) = hb when x = ha. Hence, f is onto. 3

223 Lagrange s Theorem Proof of Lagrange s Theorem: Assume H is a subgroup of a finite group G. Let H denote the equivalence relation given by a H b iff ab 1 H. By Theorem 9.1, the equivalence classes under H partition G into mutually disjoint nonempty subsets, and by Theorem 9.3, these equivalence classes are just the right cosets of H. Since G is finite, there are finitely many distinct right cosets which we denote by Ha 1, Ha 2,..., Ha k. Thus, G = Ha 1 Ha 2 Ha k = Ha 1 + Ha Ha k = H + H + + H }{{} k terms = k H. 4

224 Note that the third equality uses the fact (Lemma 10.2) that all right cosets have the same size; that is, Ha i = He = H for all i = 1, 2,..., k. Since we ve shown G = k H, we conclude that H divides G.

225 Index of a Subgroup If G is any group (not necessarily finite) and H is any subgroup, then the number of distinct right cosets of H in G is called the index of H in G. We denote this number by [G : H]. In particular, if G is a finite group, the proof of Lagrange s Theorem shows that G = [G : H] H 5