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Math 3121Abstract Algebra I
Lecture 8Sections 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.– 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:
45132
54321
OrbitsTheorem: 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~a2) symmetric: a~b ⇒ b = pn(a), for some n in ℤ⇒ a = p-n(b), with -n in ℤ⇒ b~a3) 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
• 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.
45132
54321
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)
45132
54321
Cycle Multiplication
• Examples: (without commas)( 1 2 5 6 3) (1 3) =(1 2 3) (1 2 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.
S3 = Symmetric Group on 3 Letters
ρ0 ρ1 ρ2 μ1 μ2 μ3
ρ0 ρ0 ρ1 ρ2 μ1 μ2 μ3
ρ1 ρ1 ρ2 ρ0 μ3 μ1 μ2
ρ2 ρ2 ρ0 ρ1 μ2 μ3 μ1
μ1 μ1 μ2 μ3 ρ0 ρ1 ρ2
μ2 μ2 μ3 μ1 ρ2 ρ0 ρ1
μ3 μ3 μ1 μ2 ρ1 ρ2 ρ0
312
321
123
321
231
321
213
321
132
321
321
321
3
2
1
2
1
0
Transpositions
Definition: A cycle of length 2 is called a transposition:Lemma: Every cycle is a product of transpositionsProof: 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 PermutationDefinition: 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 PermutationDefinition: 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.
Defining the Parity MapThere 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 9
• 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 SubgroupDefinition: 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
Equivalence Modulo a SubgroupTheorem: 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 followsDefinition: 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-1Note: 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 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.
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 4– Pages 101-102: 8, 10