112_Math 3121Abstract Algebra I

Math 3121 Abstract Algebra I Lecture Sections and 10 Section • Section 9: Orbits, Cycles, and the Alternating Group – Definition: Orbits of a permutation – Definition: Cycle permutations – Theorem: Every permutation of a finite set is a product of disjoint cycles – Definition: Transposition – Definition/Theorem: Parity of a permutation – Definition: Alternating Group on n letters Orbits • Look at what happens to elements as a permutation is applied: • Example: 1 5  α =   4 Orbits Theorem: Let p be a permutation of a set S The following relation is an equivalence relation: a~b ⇔ b = pn(a), for some n in ℤ Proof: 1) reflexive: a = p0(a) ⇒ a~a 2) symmetric: a~b ⇒ b = pn(a), for some n in ℤ ⇒ a = p-n(b), with -n in ℤ ⇒ b~a 3) transitive: a~b and b~c ⇒ b = pn1(a) and c = pn2(b) , for some n1 and n2 in ℤ ⇒ c = pn2(pn1(a)) , for some n1 and n2 in ℤ ⇒ c = pn2+n1(a) , with n2 + n1 in ℤ ⇒ a~c Definition: An orbit of a permutation p is an equivalence class under the relation: a~b ⇔ b = pn(a), for some n in ℤ Example • Find all orbits of 1 5  α =   4 • Method: Let S be the set that the permutation works on 0) Start with an empty list 1) If possible, pick an element of the S not already visited and apply permutation repeatedly to get an orbit 2) Repeat step until all elements of S have been visited Cycles Definition: A permutation is a cycle if a most one of its orbits is nontrivial (has more than one element) Notation: Cycle notation: list each orbit within parentheses Example: Do this for (1, 2, 3)(4, 5) 1 5  α =   4 Cycle Multiplication • Examples: (without commas) ( 3) (1 3) = (1 3) (1 3) = (1 4)(1 3)(1 2) = Cycle Decomposition Theorem: Every permutation of a finite set is a product of disjoint cycles Proof: Let σ be a permutation Let B1, B2, … Br be the orbits Let μi be the cycle defined by μi (x) = σ(x) if x in Bi and x otherwise Then σ = μ1 μ2 … μr Note: Disjoint cycles commute Examples • Decompose S3 and make a multiplication table Transpositions Definition: A cycle of length is called a transposition: Lemma: Every cycle is a product of transpositions Proof: Let (a1, a2, …, an) be a cycle, then (a1, an) (a1, an-1) … (a1 a2) = (a1, a2, …, an) Theorem: Every permutation can be written as a product of transpositions Proof: Use the lemma plus the previous theorem Parity of a Permutation Definition: The parity of a permutation is said to be even if it can be expressed as the product of an even number of transpositions, and odd if it can be expressed as a product of an odd number of transpositions Theorem: The parity of a permutation is even or odd, but not both Parity of a Permutation Definition: The parity of a permutation is said to be even if it can be expressed as the product of an even number of transpositions, and odd if it can be expressed as a product of an odd number of permutations Theorem: The parity of a permutation is even or odd, but not both Proof: We show thatFor any positive integer n, parity is a homomorphism from Sn to the group ℤ2, where represents even, and represents odd (These are alternate names for the equivalence classes 2ℤ and 2ℤ+1 that make up the group ℤ2 Defining the Parity Map There are several ways to define the parity map They tend to use the group {1, -1} with multiplicative notation instead of {0, 1} with additive notation One way uses linear algebra: For the permutation π define a map from Rn to Rn by switching coordinates as follows Lπ(x1, x2, …, xn) = (x π(1), xπ(2), …, xπ(n)) Then Lπ is represented by a nxn matrix Mπ whose rows are the corresponding permutation of the rows of the nxn identity matrix The map that takes the permutation π to Det(Mπ) is a homomorphism from Sn to the multiplicative group {-1, 1} Another way uses the action of the permutation on the polynomial P(x1, x2, …, xn ) = Product{(xi - xj )| i < j } Each permutation changes the sign of P or leaves it alone This determines the parity: change sign = odd parity, leave sign = even parity Another way is to work directly with the cycles as in Proof2 in the book Alternating Group • Definition: The alternating group on n letters consists of the even permutations in the symmetric group of n letters HW Section • Don’t hand in: Pages 94-95: 19, 39 • Hand in Tues, Oct 28: Pages 94-95: 10, 24, 36 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 x ~R y ⇔ y = x h, for some h in H ⇔ x y-1 = h, for some h in H ⇔ x = h y, for some h in H Equivalence Modulo a Subgroup 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 nℤ are: nℤ, nℤ+1, nℤ+2, …, nℤ + (n-1 Note: Cosets in nonabelian case: left and right don’t always agree • In the book: H = { ρ0, μ1} in S3 has different left and right cosets Counting Cosets Theorem: For a given subgroup of a group, every coset has exactly the same number of elements, namely the order of the subset 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 La from H to aH by the formula La(g) = a g This is 11 and onto Define a map Ra from H to Ha by the formula Ra(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 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 – Pages 101-102: 8, 10 ... have been visited Cycles Definition: A permutation is a cycle if a most one of its orbits is nontrivial (has more than one element) Notation: Cycle notation: list each orbit within parentheses... disjoint cycles – Definition: Transposition – Definition/Theorem: Parity of a permutation – Definition: Alternating Group on n letters Orbits • Look at what happens to elements as a permutation...Section • Section 9: Orbits, Cycles, and the Alternating Group – Definition: Orbits of a permutation – Definition: Cycle permutations – Theorem: Every permutation of a finite set is a product