Math 3121 Abstract Algebra I Lecture 9 Finish Section 10 Section 11

HW Due Today Hand in: Due Tues, Oct 28: Page 73: 12, 14, 16 Page 84: 18 Pages 94-95: 10, 24, 36 Do not hand in: Pages 94-95: 19, 39

Section 10 Section 10: Cosets and the Theorem of Lagrange – Modular relations for a subgroup – Definition: Coset – Theorem of Lagrange: For finite groups, the order of subgroup divides the order of the group. – Theorem: For finite groups, the order of any element divides the order of the group

Modulo a Subgroup Definition: Let H be a subgroup of a group G. Define relations: ~ L and ~ R by: x ~ L y ⇔ x -1 y in H x ~ R y ⇔ x y -1 in H We will show that ~ L and ~ R are equivalence relations on G. We call ~ L left modulo H. We call ~ R right modulo H. Note: x ~ L y ⇔ x -1 y = h, for some h in H ⇔ y = x h, for some h in H x ~ R y ⇔ x y -1 = h, for some h in H ⇔ x = h y, for some h in H

Modulo a Subgroup is an Equivalence Relation Theorem: Let H be a subgroup of a group G. The relations: ~ L and ~ R defined by: x ~ L y ⇔ x -1 y in H x ~ R y ⇔ x y -1 in H are equivalence relations on G. Proof: We show the three properties for equivalence relations: 1) Reflexive: x -1 x = e is in H. Thus x ~ L x. 2) Symmetric: x ~ L y ⇒ x -1 y in H ⇒ ( x -1 y) -1 in H ⇒ y -1 x in H ⇒ y ~ L x 3) Transitive: x ~ L y and y ~ L z ⇒ x -1 y in H and y -1 z in H ⇒ ( x -1 y )( y -1 z) in H ⇒ ( x -1 z) in H ⇒ x ~ L z Similarly, for x ~ R y.

Cosets The equivalence classes for these equivalence relations are called left and right cosets modulo the subgroup. Recall: x ~ L y ⇔ x -1 y = h, for some h in H ⇔ y = x h, for some h in H Cosets are defined as follows Definition: Let H be a subgroup of a group G. The subset a H = { a h | h in H } is called the left coset of H containing a, and the subset H a= { a h | h in H } is called the right coset of H containing a.

Examples Cosets of n ℤ in ℤ are : n ℤ, n ℤ +1, n ℤ +2, …, n ℤ + (n-1) For example 2 ℤ in ℤ has two cosets. Cosets of {0, 3} in ℤ 6 Cosets of {0, 2, 4} in ℤ 6

Abelian versus Nonabelian Note: In abelian groups the left cosets are the right cosets. In nonabelian case: left and right don’t always agree. H = { e, μ 1 } in G = S 3 has different left and right cosets. For left cosets, make a multiplication table G×H. For right cosets, make a multiplication table H×G. (See slides after this one). H = { ρ 0, ρ 1, ρ 2 } in S 3 has same left and right cosets.

Multiplication Table for S 3 in Cycle Notation e(1 2 3)(1 3 2)(2 3)(1 3)(1 2) ee(1 2 3)( 13 2)(2 3)(1 3)(1 2) (1 2 3) (1 3 2)e(1 2)(2 3)(1 3) (1 3 2) e(1 2 3)(1 3)(1 2)(2 3) (1 3)(1 2)e(1 2 3)(1 3 2) (1 3) (1 2)(2 3)(1 3 2)e(1 2 3) (1 2) (2 3)(1 3)(1 2 3)(1 3 2)e

Left Cosets for S 3 in Cycle Notation e(2 3) ee (1 2 3) (1 2) (1 3 2) (1 3) (2 3) e (1 3) (1 3 2) (1 2) (1 2 3) Distinct left cosets of H = {e, (2 3)} {e, (2 3)} = e H = (2 3) H {(1 2 3), (1 2)} = (1 2 3) H = (1 2) H {(1 3 2), (1 3)} = (1 3 2) H = (1 3) H

Right Cosets of for S 3 in Cycle Notation e(1 2 3)(1 3 2)(2 3)(1 3)(1 2) ee(1 2 3)( 13 2)(2 3)(1 3)(1 2) (2 3) (1 3)(1 2)e(1 2 3)(1 3 2) Distinct right cosets of H = {e, (2 3)} {e, (2 3)} = H = H (2 3) {(1 2 3), (1 3)} = H (1 2 3) = H (1 3) {(1 3 2), (1 2)} = H (1 3 2) = H (1 2)

Counting Cosets Theorem: For a given subgroup of a group, every coset has exactly the same number of elements, namely the order of the subgroup. Proof: Let H be a subgroup of a group G. Recall the definitions of the cosets: aH and Ha. a H = { a h | h in H } H a= { a h | h in H } Define a map L a from H to aH by the formula L a (g) = a g. This is 1-1 and onto. Define a map R a from H to Ha by the formula R a (g) = g a. This is 1-1 and onto.

Lagrange Theorem (Lagrange): Let H be a subgroup of a finite group G. Then the order of H divides the order of G. Proof: Let n = number of left cosets of H, and let m = the number of elements in H. Then m is the number of elements in any left coset. Thus n m = the number of elements of G. Here m is the order of H, and n m is the order of G.

Orders of Cycles The order of an element in a finite group is the order of the cyclic group it generates. Thus the order of any element divides the order of the group.

HW Section 10 Don’t hand in: – Pages 101: 3, 6, 9, 15 Hand in Tues, Nov 4 – Pages : 6, 8, 10, 36, 40

Section 11 (as time permits) Direct Products and Finitely Generated Abelian Groups Cartesian Product of sets Direct product of groups Structure of Z n  Z m Structure of products of cyclic groups Next time: Structure of Finitely Generated Abelian Groups

Cartesian Products Definition: The Cartesian product of a finite collection of sets S k, for k = 1 to n is the set of all n-tuples (s 1, s 2, …, s n ), with s k in S k. The Cartesian product is denoted by S 1 ×S 2 × … ×S n or by product notations such as

Projection Maps The map p i : S 1 ×S 2 × … ×S n  S i that takes the n-tuple s = (s 1, s 2, …, s n ) to its ith component s i is called the ith projection map. In other words: For any s in S 1 ×S 2 × … ×S n the result of applying the ith projection map to s is called the ith component of s and denoted by s i. Note: Two elements a and b of the product are equal if and only if a i = b i, for all i = 1,…,n.

Direct Product of Groups Theorem: Let G 1, G 2, …, G n be groups with multiplicative notation. Define a binary operation on the Cartesian product G 1 ×G 2 ×… ×G n 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 ), then G 1 ×G 2 ×… ×G n is a group with this binary operation. The set G 1 ×G 2 ×… ×G n with this binary operation is called the direct product of the groups G 1, G 2, …, G n. Proof: Note that the binary operation is defined component- wise. That is, if a and b are in the product, then (a b) i = a i b i. 1) Associativity follows because each component binary operation is associative. 2) The identity is e the n-tuple e = (e 1, e 2, …, e n ), where each e i is the identity of its own component group G i. 3) For each a = (a 1, a 2, …, a n ) in the product, the n- tuple a -1 = ((a 1 ) -1, (a 2 ) -1, …, (a n ) -1 ) is the inverse of a.

Direct Sums of Groups Sometimes we call the direct product a direct sum, especially if we use additive notation. A direct product is characterized by its projection maps. These turn out to be homomorphisms. On the other hand, the direct sum is characterized by injection maps j i : G i  G 1 ×G 2 ×… ×G n that each take a i in G i to the n-tuple (e 1, e 2, …, a i, …, e n ) that has identities in all components except for the ith, which has a i. These also turn out to be homomorphisms.

Examples ℤ 2 × ℤ 3 ℤ 3 × ℤ 5 ℤ 2 × ℤ 2 ℤ 3 × ℤ 3 ℤ 2 × ℤ 6 ℤ 9 × ℤ 6

Internal Products Each component group of a direct product can also be considered a subgroup by injection. All of these subgroups commute with each other. Thus any element g of the product G 1 ×G 2 ×… ×G n can be uniquely written in the form g = g 1 g 2 … g n with g i in G i. When this happens, G is said to be an internal product of the subgroups G i.

LCM and GCD of two numbers Let x and y be integers, then LCM(x, y) is the least multiple of x and y: LCM(x, y) = min{ m in ℤ + | m is a multiple of x and m is a multiple of y} GCD(x, y) is the greatest common divisor of x and y: GCD(x, y) = max{ m in ℤ + | m divides x and m divides y} LCM(x, y) = x y /GCD(x, y).

Methods of finding LCM and GCD Euclidean Algorithm to find GCD in the form a x + b y: Start with positive integers x and y. Set r 0 = x, r 1 = y. Given r k-1 and r k, find q k and r k+1 such that: r k-1 = q k r k + r k+1, with 0 ≤ r k+1 < r k. Continue until r k+1 = 0. Then GCD(x, y) = r k. Then work backward to write GCD in the form a x + b y. Example: Find GCD of 64 and 58. For small numbers use prime factorization.