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Math 3121 Abstract Algebra I Lecture 8 Sections 9 and 10.

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Presentation on theme: "Math 3121 Abstract Algebra I Lecture 8 Sections 9 and 10."— Presentation transcript:

1 Math 3121 Abstract Algebra I Lecture 8 Sections 9 and 10

2 Section 9 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.

3 Orbits Look at what happens to elements as a permutation is applied: Example:

4 Orbits Theorem: Let p be a permutation of a set S. The following relation is an equivalence relation: a ~ b ⇔ b = p n (a), for some n in ℤ Proof: 1) reflexive: a = p 0 (a) ⇒ a~a 2) symmetric: a~b ⇒ b = p n (a), for some n in ℤ ⇒ a = p -n (b), with -n in ℤ ⇒ b~a 3) transitive: a~b and b~c ⇒ b = p n1 (a) and c = p n2 (b), for some n1 and n2 in ℤ ⇒ c = p n2 (p n1 (a)), for some n1 and n2 in ℤ ⇒ c = p n2+n1 (a), with n2 + n1 in ℤ ⇒ a~c

5 Definition: An orbit of a permutation p is an equivalence class under the relation: a ~ b ⇔ b = p n (a), for some n in ℤ

6 Example Find all orbits of 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 1 until all elements of S have been visited.

7 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)

8 Cycle Multiplication Examples: (without commas) ( ) (1 3) = (1 2 3) (1 2 3) = (1 4)(1 3)(1 2) =

9 Cycle Decomposition Theorem: Every permutation of a finite set is a product of disjoint cycles. Proof: Let σ be a permutation. Let B 1, B 2, … B r be the orbits. Let μ i be the cycle defined by μ i (x) = σ(x) if x in B i and x otherwise Then σ = μ 1 μ 2 … μ r Note: Disjoint cycles commute.

10 Examples Decompose S 3 and make a multiplication table.

11 S 3 = Symmetric Group on 3 Letters ρ0ρ0 ρ1ρ1 ρ2ρ2 μ1μ1 μ2μ2 μ3μ3 ρ0ρ0 ρ0ρ0 ρ1ρ1 ρ2ρ2 μ1μ1 μ2μ2 μ3μ3 ρ1ρ1 ρ1ρ1 ρ2ρ2 ρ0ρ0 μ3μ3 μ1μ1 μ2μ2 ρ2ρ2 ρ2ρ2 ρ0ρ0 ρ1ρ1 μ2μ2 μ3μ3 μ1μ1 μ1μ1 μ1μ1 μ2μ2 μ3μ3 ρ0ρ0 ρ1ρ1 ρ2ρ2 μ2μ2 μ2μ2 μ3μ3 μ1μ1 ρ2ρ2 ρ0ρ0 ρ1ρ1 μ3μ3 μ3μ3 μ1μ1 μ2μ2 ρ1ρ1 ρ2ρ2 ρ0ρ0

12 Transpositions Definition: A cycle of length 2 is called a transposition: Lemma: Every cycle is a product of transpositions Proof: Let (a 1, a 2, …, a n ) be a cycle, then (a 1, a n ) (a 1, a n-1 ) … (a 1 a 2 ) = (a 1, a 2, …, a n ) Theorem: Every permutation can be written as a product of transpositions. Proof: Use the lemma plus the previous theorem.

13 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.

14 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 S n to the group ℤ 2, where 0 represents even, and 1 represents odd. (These are alternate names for the equivalence classes 2 ℤ and 2ℤ+1 that make up the group ℤ 2.

15 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 R n to R n by switching coordinates as follows L π (x 1, x 2, …, x n ) = (x π(1), x π(2), …, x π(n) ). Then L π is represented by a nxn matrix M π whose rows are the corresponding permutation of the rows of the nxn identity matrix. The map that takes the permutation π to Det(M π ) is a homomorphism from S n to the multiplicative group {-1, 1}. Another way uses the action of the permutation on the polynomial P( x 1, x 2, …, x n ) = Product{(x i - x j )| 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.

16 Alternating Group Definition: The alternating group on n letters consists of the even permutations in the symmetric group of n letters.

17 HW Section 9 Don’t hand in: Pages 94-95: 19, 39 Hand in Tues, Oct 28: Pages 94-95: 10, 24, 36

18 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

19 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

20 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.

21 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.

22 Examples Cosets of n ℤ are: n ℤ, n ℤ+1, n ℤ+2, …, n ℤ + (n-1 Note: Cosets in nonabelian case: left and right don’t always agree. In the book: H = { ρ 0, μ 1 } in S 3 has different left and right cosets.

23 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 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.

24 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.

25 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.

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


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