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Math 3121 Abstract Algebra IPowerPoint Presentation

Math 3121 Abstract Algebra I

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

Math 3121 Abstract Algebra I

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Math 3121Abstract Algebra I

Lecture 8

Sections 9 and 10

- 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.

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

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

- 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.

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)

- Examples: (without commas)
( 1 2 5 6 3) (1 3) =

(1 2 3) (1 2 3) =

(1 4)(1 3)(1 2) =

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.

- Decompose S3 and make a multiplication table.

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.

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.

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.

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.

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

- Don’t hand in:
Pages 94-95: 19, 39

- Hand in Tues, Oct 28:
Pages 94-95: 10, 24, 36

- 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

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

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 .

- 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.

- 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.

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.

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.

- 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.

- Don’t hand in:
- Pages 101: 3, 6, 9, 15

- Hand in Tues, Nov 4
- Pages 101-102: 8, 10