1. ALGEBRAIC STRUCTURES
SMA 3033
SEMESTER 1 2023/2024
CHAPTER 4 : PERMUTATION GROUP
BY:
ASSOC. PROF. DR ROHAIDAH HJ MASRI
1
Chapter 4 SMA3033 Sem 1 2023/2024
2. 4.1 PERMUTATION GROUPS
Definition 1 (Permutation)
A permutation of a set A is a bijective function f : A → A.
1-1 function Onto function
This definition preserves the informal
idea of rearrangement & has advantage
of being applicable to infinite set.
We will
concentrate
on finite set
only
2
Chapter 4 SMA3033 Sem 1 2023/2024
3. 4.1 PERMUTATION GROUPS
Example 1
Let T = { 1, 2, 3 }. The permutation f whose rule is:
f(1) = 2, f(2) = 3, f(3) = 1
may represented by the array
.1
.2
.3
.1
.2
.3
T T
3
Chapter 4 SMA3033 Sem 1 2023/2024
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4.1 PERMUTATION GROUPS
Example 2
Let T = { 1, 2, 3, 4 }. Given
f(1) = 4, f(3) = 2, f(2) = 1 and f(4) = 2.
Determine whether f is a permutation. Give your reason.
Solution:
Then, this mapping is not a permutation since 2 appears twice while 3 doesn’t
appear at all at the right column.
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4.1 PERMUTATION GROUPS
Notes:
i. The function composition o is a binary operation on the collection of
all permutation of a set A. This operation is called permutation
multiplication.
ii. If f & g are permutations, then the composite function (g o f ) is
defined by
A → A → A
gives a mapping of A into A.
iii. ( g o f ) on A must be read in right-to-left order.
f g
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4.1 PERMUTATION GROUPS
Example 3
Suppose that A = {1, 2, 3}. Let f & g are the permutations given by:
Determine ( f o g ).
Solution:
( f o g ) is the function given by:
( f o g )(1) = f(g(1)) = f(2) = 2
( f o g )(2) =
( f o g )(3) =
f(g(2)) = f(1) = 3
f(g(3)) = f(3) = 1
Multiplying in right-to-left
order
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4.1 PERMUTATION GROUPS
Also can be represented by,
( f o g ) =
Example 4
By using functions f & g in example 3, find ( g o f ) and show that
( g o f ) ( f o g ).
Answer:
Then, (g o f ) ( f o g ).
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4.1 PERMUTATION GROUPS
Example 5
Let A be a set and let f and g be permutations of A. Prove that the composite
function ( g o f ) is a permutation.
Proof:
( To show ( g o f ) is bijective )
Let a, b A.
( g o f )(a) = ( g o f )(b)
g(f (a)) = g(f (b))
f (a) = f (b)
a = b.
Then, ( g o f ) is 1-1.
f 1-1
g 1-1
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4.1 PERMUTATION GROUPS
( To show (g o f) is onto )
Let yA.
(To show exist xA such that ( g o f )(x) = y.)
hypothesis : f onto For all zA , xA such that f(x) = z
g onto For all yA , zA such that g(z) = y
(g o f)(x) = g(f(x))
= g(z)
= y Since g is onto
Since f is onto
Then, (g o f) is onto.
Therefore, (g o f) is a permutation.
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4.1 PERMUTATION GROUPS
Example 6
Let G be a group. Prove that the function a : G → G where a(x) = xa for
aG and xG, is a permutation.
(Try proof this. )
Definition 2 (Symmetric Group)
Let A be the finite set { 1, 2, 3, …, n }. The group of all permutations of A is the
symmetric group on n letters, and denoted by Sn .
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4.1 PERMUTATION GROUPS
Note:
The order (the number if element) of Sn is n! .
n! = n ( n – 1)(n – 2) … (3)(2)(1)
Example 7
Let A = {1, 2, 3} and S3 is a group of 3! = 6 elements. The six permutations
of A are:
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4.1 PERMUTATION GROUPS
Notes:
1. Let B = {1, 2, 3, 4, 5}. Then S5 is a symmetric group of 5! = 125.
2. If n 3, then Sn is non abelian.
(see example 3 & example 4)
3. For a permutation of f, the inverse function f-1 is the permutation that
reverses the direction of mapping f .
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4.1 PERMUTATION GROUPS
Example 8
Let A = { 1, 2, 3, 4, 5, 6 }. Give three permutations in S6 :
Find:
i. f -1
ii. f -2 g
iii. f -1 g2 h
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4.1 PERMUTATION GROUPS
Answer:
i.
ii.
iii.
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4.1 PERMUTATION GROUPS
Definition 3 (Cycle)
Let a1, a2, …, ak (with k 1) be distinct elements of the set
{ 1, 2, 3, …, n }. Then,
( a1, a2, …, ak ) denotes the permutation in Sn that maps:
a1 to a2 , a2 to a3 , … , ak – 1 to ak , and ak to a1
and maps every other element of {1, 2, 3, …, n } to itself.
Here, ( a1, a2, …, ak ) is called a cycle of length k or k-cycle.
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4.1 PERMUTATION GROUPS
Example 9 (a)
In S4, (1 4 3) is a cycle of length 3 that maps:
1 to 4
4 to 3
3 to 1
2 to 2
1
4
3
It was written in the old notation.
Note that: (1 4 3) = (4 3 1) = (3 1 4)
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4.1 PERMUTATION GROUPS
Example 9 (b)
Write permutation in S5 as a cycle.
Solution:
= (1 3 5 4) = (3 5 4 1)
= (5 4 1 3)
= (4 1 3 5)
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4.1 PERMUTATION GROUPS
Example 9 (c) (Try!)
Express the following permutations of {1, 2, 3, 4, 5, 6, 7} as a cycle.
i.
ii.
Disjoint cycles
Two cycles are said to be disjoint if they have no elements in common.
Example:
(1 3)o (2 5 4 6) represents disjoint cycles but,
(1 3) & (3 4 5) are not disjoint cycles since 3 appears in both cycles.
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4.1 PERMUTATION GROUPS
Example 10
Express as a product of disjoint cycles.
i.
ii.
iii. (1 4)o (2 3 5 7)o (1 3 4 7 2)
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4.1 PERMUTATION GROUPS
Theorem 1
If = ( a1 a2 … ak ) and = ( b1 b2 … br ) are disjoint cycles in Sn ,
then
o = o .
Theorem 2
Every permutation of a finite set is a product of disjoint cycles.
21. Corollary 1
Any permutation of a finite set of at least two elements is a product of
transposition.
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4.1 PERMUTATION GROUPS
Even & Odd Permutations
Definition 4 (Transposition)
A cycle of length 2 is a transposition.
Note:
( a1 a2 …. an ) = ( a1 an ) o ( a1 an-1 ) o … o (a1 a3 ) o ( a1 a2 )
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4.1 PERMUTATION GROUPS
Example 11
Write each permutation as a product of transposition.
i.
ii.
iii.
= (1 2) o (4 5) o (6 7 9)
= (1 3 6 5 4 2)
= (1 2) o (1 4) o (1 5) o (1 6) o (1 3)
= (1 4 7 6 2 8 3)
= (1 3) o (1 8) o (1 2) o (1 6) o (1 7) o (1 4)
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4.1 PERMUTATION GROUPS
Definition 5 (Even/ Odd Permutation)
A permutation in Sn is said to be even if it can be written as the product of
an even number of transpositions, and
odd if it can be written as the product of an odd number of transpositions.
Example 12
Determine whether each permutation in example 11 is even or odd.
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Theorem 2
No permutation in Sn can be written as the product of an even number of
transpositions and also as the product of an odd transpositions.
Definition 6 (Alternating Group)
The set of all even permutations in Sn is called the alternating group and
denoted by An .
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4.1 PERMUTATION GROUPS
Example 13
Determine whether the following permutations are even or odd.
i. (1 2 3)
i. (1 3 2 4 )
= (1 3) o (1 2) even
= (1 4) o (1 2) o (1 3) odd
Theorem 3 (Cayley’s Theorem)
Every group is isomorphic to a group of permutation.