SECTION 9 Orbits, Cycles, and the Alternating Groups Given a set A, a relation in A is defined by :...
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Transcript of SECTION 9 Orbits, Cycles, and the Alternating Groups Given a set A, a relation in A is defined by :...
SECTION 9 Orbits, Cycles, and the Alternating Groups
Given a set A, a relation in A is defined by :
For a, bA, let ab if and only if b = n(a) for some nZ. (1)We can check that the relation is indeed an equivalent relation.
(Reflexive, Symmetric, Transitive)
Since an equivalence relation on a set yields a natural partition of the set, then we have the following
DefinitionLet be a permutation of a set A. The equivalence classes in A
determined by the equivalence relation (1) are the orbits of .
Example
Example:• Let A={a, b, c, d}, and be the identity permutation of A. Find the
orbits of in A.Solution: The orbits of are: {a}, {b}, {c}, {d}
• Find the orbits of the permutation in S8.
Solution: The orbits containing 1 is {1, 3, 6} The orbits containing 2 is {2, 8} The orbits containing 4 is {4, 7, 5}Since these three orbits include all integers from 1 to 8, the complete list
of orbits of is {1, 3 ,6}, {2, 8}, {4, 7, 5}
1 2 3 4 5 6 7 83 8 6 7 4 1 5 2
Cycles
For the remainder of this section, we suppose that A= {1, 2, 3, , n} and that we are dealing with the elements of the symmetric group Sn.
Recall has orbits : {1, 3 ,6}, {2, 8}, {4, 7, 5}, which
can be indicated graphically by using circles. Such a permutation, described graphically be a single circle, is called a cycle.
Note: we also consider the identity permutation to be a cycle.
Here is the term cycle in a mathematically precise way:
1 2 3 4 5 6 7 83 8 6 7 4 1 5 2
Cycle
DefinitionA permutation Sn is a cycle if it has at most one orbit containing
more than one element. The length of a cycle is the number of elements in its largest orbit.
We also introduce a cyclic notation for a cycle. For example
Given . It has orbits {1, 3, 6}, {2}, {4}, {5}, {7}, {8}.
We can denote =(1, 3, 6).
Note: an integer not appearing in this notation for is left fixed by .
1 2 3 4 5 6 7 83 2 6 4 5 1 7 8
Example
ExampleIn S5, we see that
Example
1 2 3 4 5(1,3,5,4)
3 2 5 1 4(3,5,4,1)(5,4,1,3)(4,1,3,5)
1 2 3 4 5 6 7 83 8 6 7 4 1 5 2
(1,3,6)(2,8)(4,7,5)
Note: these cycles are disjoint, meaning that any integer is moved by atmost one of these cycles; thus no one number appears in the notations ofTwo cycles.
Theorem
TheoremEvery permutation of a finite set is a product of disjoint cycles.
Note: • Permutation multiplication is in general not commutative, but the
multiplication of disjoint cycles is commutative.
Example:Given the permutation , write it as a product of disjoint
cycles.
Solution:
1 2 3 4 5 66 5 2 4 3 1
1 2 3 4 5 6(1,6)(2,5,3)
6 5 2 4 3 1
Example
Consider the cycles (1, 4, 5, 6) and (2, 1, 5) in S6.
Find (1, 4, 5, 6)(2, 1, 5) and (2, 1, 5)(1, 4, 5, 6).
Solution: (1, 4, 5, 6)(2, 1, 5)=
(2, 1, 5)(1, 4, 5, 6)=
1 2 3 4 5 66 4 3 5 2 1
1 2 3 4 5 64 1 3 2 6 5
Even and Odd Permutations
DefinitionA cycle of length 2 is a transposition.
A computation shows that (a1,a2,…, an)=(a1,an)(a1,an-1)…(a1,a3)(a1,a2), therefore any cycle is a product of transpositions.
CorollaryAny permutation of a finite set of at least two elements is a product of
transpositions.
Examples
Example:• (1, 6)(2, 5, 3)=(1, 6)(2, 3)(2, 5)• In Sn for n2, the identity permutation is the product (1, 2)(2, 1).
Note: a representation of the permutation in this way is not unique, but the number of transposition must either always be even or always be odd.
Theorem: No permutation is Sn can be expressed both as a product of an even
number of transpositions and as a product of an odd number of transpositions.
Odd/Even permutation
DefinitionA permutation of a finite set is even if it can be expressed as a product
of an even number of transpositions. A permutation of a finite set is odd if it can be expressed as a product of an odd number of transpositions.
Example:• The identity permutation in Sn is an even permutation since
=(1, 2)(2, 1).
• The permutation (1, 4, 5, 6)(2,1, 5) in S6 is odd since
(1, 4, 5, 6)(2,1, 5)=(1, 6)(1, 5)(1, 4)(2, 5)(2, 1)
The Alternating Groups
TheoremIf n2, then the collection of all even permutations of {1, 2, 3,…, n}
forms a subgroup of order n! / 2 of the symmetric group Sn.
DefinitionThe subgroup of Sn consisting of all even permutations of n letters is
the alternating groups An on n letters.