Permutations

1
Permutations

• Introduction Lecture 2

2
Permutations

• As we know a permutation p is a bijective
  bijektivna mapping of a set A onto itself p A
  ? A. Permutations may be multiplied and form the
  symmetric group Sym(A) SA, that has n!
  elements, where n A.

3
Permutation as a product of disjoint cycles

• A permutation can be written as a product of
  disjoint cycles in a uniqe way.

4
Example
1

• Example
• p(1) 2, p(2) 6,
• p(3) 5, p(4) 4,
• p(5) 3, p(6) 1.
• Permutation p can be written as
• p 2,6,5,4,3,1
• p (1 2 6)(3 5)(4) 2 S6

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2
3
5
4
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Positional Notation

• p 2,6,5,4,3,1
• Each row of the configuration table of Pappus
  configuration is a permutation.

1 2 3 4 5 6 7 8 9
8 7 2 1 6 9 3 4 5
6 1 8 9 7 3 4 5 2
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Example

• Each row of the configuration table of Pappus
  configuration can be viwed as permutation
  (written in positional notation.)
• a1 (1)(2)(3)(4)(5)(6)(7)(8)(9)
• a2 (184)(273)(569)
• a3(16385749)

7
Cyclic Permutation

• A permutation composed of a single cycle is
  called cyclic permutation.
• For example, a3 is cyclic.

8
Polycyclic Permutation

• A permutation whose cycles are of equal length is
  called semi-regular or polycyclic.
• Hence a1, a2 and a3 are polycyclic.

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Identity Permutation

• A polycyclic pemrutation with cycles of length 1
  is called identity permutation.

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Fix(p)

• Let Fix(p) x 2 A p(x) x denote the set of
  fdixed points of permutation p.
• fix(p) Fix(p).
• Fixed-point free premutation is called a
  derangment.

11
Order of p.
12
Involutions

• Permutation of order 2 is called an involution.
• Later we will be interseted in fixed-point free
  involutions.

13
Homework

• H1. Determine the number of polycylic
  configurations of degree n and order k.
• H2. Show that an ivolution is fixed-point free if
  and only if it is a polycyclic permutation
  composed of cycles of length 2.
• H3. Determine the number of involutions in Sn.