Math 3121 Abstract Algebra I

1. Math 3121Abstract 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 = 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

5. Definition: An orbit of a permutation p is an equivalence class under the relation: a ~ b ⇔ b = pn(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 2 5 6 3) (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 σbe a permutation. Let B1, B2, … Br be the orbits. Let μi be the cycle defined by μi (x) = σ(x) if x in Biand x otherwise Then σ = μ1μ2 … μr Note: Disjoint cycles commute.

10. Examples • Decompose S3 and make a multiplication table.

11. S3 = Symmetric Group on 3 Letters

12. Transpositions Definition: A cycle of length 2 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.

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 Sn to the group ℤ2, where 0 represents even, and 1 represents odd. (These are alternate names for the equivalence classes 2ℤ and 2ℤ+1 that make up the group ℤ2.

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 Rn to Rn by switching coordinates as follows Lπ(x1, x2, …, xn) = (xπ(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.

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 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.

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 La from H to aH by the formula La(g) = a g. This is 1-1 and onto. Define a map Ra from H to Ha by the formula Ra(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 101-102: 8, 10