2. Permutation
Let X be a non-empty set.
A bijective function f: X → X will be called a
permutation of X.
Consider the case when X is a finite set with n
elements
X = {1, 2, 3, …, n}
The collection of all permutations of this set X will be
called the symmetric group on n symbols and is
denoted by Sn.
3. S3
It is easier to study a smaller set first so we shatt
start with the case of n = 3.
A bijective function f: X → X will be called a
permutation of X.
Consider the case when X is a finite set with n
elements
X = {1, 2, 3}
The collection of all permutations of this set X will
be called the symmetric group on n symbols and is
denoted by S3
4. S3
A symmetry is an action that can be
performed on an object to leave it
looking the same as before.
S3 corresponds to the set of
symmetries of an equilateral triangle.
What symmetries can you see for an
equilateral triangle?
5. Symmetries of an equilateral Triangle
There are six symmetries in all.
6. The Six Symmetries
E- Identity
R - Rotation anticlockwise of 120
R2 - Rotation anticlockwise of 240
V- Reflection in the vertical
RV – reflection in left vertical
R2V - Reflection in right vertical
We can most easily see a bijective mapping
using the conventional two line notation.
For instance, R can be represented as:
7. Practice with a Cardboard model
Sometimes, we may prefer to use a physical model to see things. If
you want, cut out an equilateral triangle and place it on a flat surface
(board). Number the corners of the triangle as shown and also the
positions on the board.
8. R – Rotation of 120 degrees
We can most easily see a bijective mapping using the
conventional two line notation. For instance, R can be
represented as:
where the upper number represents the number on the triangle and the
lower the number on the board. Thus we have 1 → 2; 2 → 3 and 3 → 1.
9. E – The identity map
Using the notation described, what
would be the identity map?
10. R2 – Rotation of 240 degrees
What would be the
case for R2?
11. Naming the Permutations
Lets use the symbol ρ to denote these three
permutations:
E = ρ0 =
R = ρ1 =
R2 = ρ2 =
12. V – Reflection in the vertical
In this permutation, we just
switch 1 and 2. 3 remains
unchanged. This is an example
of a transposition.
13. RV – Reflection in the left vertical
In this permutation, we just switch
2 and 3, with 1 remaining
unchanged. This is another
example of a transposition.
14. Notation
We shall use the notation RV = R * V to mean R
followed by V.
Thus, R = (1, 2, 3) and V = (1, 2) and R * V = (2, 3)
R =
V = RV =
16. Notation
You may have been wondering why we have RV!
It is easy to see that not even knowing the name, a
reflection in the left vertical just results in an
exchange in the elements 2 and 3.
RV signifies that the permutation is equivalent to a
rotation of 120 followed by a reflection in the
vertical. Try this to see if you end up with the
result you expect.
20. R2V – Reflection in the right vertical
Complete this one as
an exercise.
21. Cycle Notation
So far, we have been using the notation:
This form is somewhat inefficient.
If we have permutations which leave the majority of
the numbers unchanged then we just need to say what
happens to the changed numbers.
We shall use instead what is known as cycle notation
for our permutations.
In this case, we see that only 1 and 3 change places so
we say (1 3) meaning 1 → 3 and 3 → 1.
22. Definition
Let x1, x2, . . . , xr be r distinct elements of Xn = {1, 2, . . .
, n} with 1 ≤ r ≤ n.
The r-cycle (x1 , x2 ,. . . , xr) is the permutation in Sn
which maps x1 → x2, x2 → x3, . . . , xr−1 → xr, xr → x1 and
fixes all other points in {1, 2, . . . , n}.
For instance, the permutation
written as a cycle is (1, 2, 3) which means 1 → 2, 2
→ 3 and 3 → 1.
This is the notation we shall use in SAGE notebook…
24. Group Properties
To show that a set G under a given operation is a
group, we need to show that it satisfies the group
axioms:
CLOSURE:Closure: x, y ∈ G ⇒ x * y ∈ G
ASSOSIATIVE: x, y, z ∈ G , x * (y * z) = (x * y) * z
IDENTITY: e ∈ G st g ∈ G3 , g * e = e * g = g
INVERSES: g ∈ G, g’ ∈ G st g * g’= g’* g = e
25. Symmetric Group
Let G = {E, R, R2, V, RV, R2V}. This set under the
permutations described form a group under
composition.
We need to show that:
any multiplications of elements results in
another element of the set.
The operation is associative.
There is an identity and each element has an
inverse.
The normal way of showing this is via the
multiplication table.
27. Group Table
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* E R R2 V RV R2V
E E R R2 V RV R2V
R R R2 E RV R2V V
R2 R2 E R R2V V RV
V V R2V RV E R2 R
RV RV V R2V R E R2
R2V R2V RV V R2 R E
28. Group Properties
Clearly the operation is closed as the
multiplication of any two elements results in
another element of the group.
E is the identity
Each element has an inverse.
We can also show that for all elements:
R.(V.RV) = R.R2 = E = (R.V).RV
That is the operation is associative.
We have a group!
29. The Six Symmetries
E - Identity
R - Rotation anticlockwise of 120
R2 - Rotation anticlockwise of 240
V- Reflection in the vertical
RV – reflection in left vertical
R2V - Reflection in right vertical
We need to be quite familiar with the permutations in
this symmetric set so take a minute to review them.
30. Transpositions
Note that:
V * R2V = (1, 2) * (1, 3) = (1, 2, 3) = R
Conversely,
R = (1, 2, 3) = (1, 2) * (1, 3)
So that a 3-cycle can be written as a
product of 2-cycles (transpositions).
31. Order of the Permutations
The order of a permutation is the
number of transpositions when the
permutation of expressed in this form.
Thus (1, 2, 3) = (1, 2) * (1, 3) is of order 2
R = (1, 2, 3) is an EVEN permutation.
The identity E is an even permutation.
32. Even and Odd permutations
Expressing a permutation as a product
of 2-cycles enables us to determine the
parity of the permutation.
If the permutation can be expressed as
an even number of 2-clycles it is an
even permutation else it is an odd
permutation.
33. Even and Odd
The identity permutation is an even
permutation.
An even permutation can be obtained
from the identity permutation by an
even number of exchanges (called
transpositions) of two elements, while
an odd permutation can be obtained by
an odd number of transpositions.
34. Rules
The following rules follow directly from the
corresponding rules about addition of integers:
the composition of two even permutations is even
the composition of two odd permutations is even
the composition of an odd and an even permutation is
odd
From these it follows that
the inverse of every even permutation is even
the inverse of every odd permutation is odd
37. A3 = AlternatingGroup(3)
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* E R R2
E E R R2
R R R2 E
R2 R2 E R
Note that if we explicitly want
to do a L-R evaluation, we
shall use Roman letters for the
permutations and insert the *
multiplication sign.
40. Naming the Permutations
Lets use the symbol μ to denote these last
three permutations:
V = μ1 =
RV = μ2 =
R2V = μ3 =
41. Functional Notation
Permutations are bijective functions.
Thus, we can use function notion when
dealing with permutations.
But we have to be careful!
43. Permutations as Bijections
A permutation on a set X= {1, 2, 3} is a bijection on
this set.
As such, we shall adopt the formal approach for
composition of functions.
Let ρ and τ be permutations on X. then, the
composition of ρ and τ, ρ o τ means (ρ o τ)(x) = (x)
That is: (ρ o τ)(x) = ρ(τ(x))
So, we apply τ and then ρ.
44. Composing permutations
We can now refer to S3 = {1, 2, 3} as the symmetric group.
Consider two permutations ρ1 and μ2 as follows:
ρ1 =
μ3 =
Then ρ1μ3 means ρ1(μ3(x)) using the conventional
multiplication of functions notation. This is a composition.
So we must apply μ3 and then ρ1. We go right to left.
45. Composition of functions
Recall that when dealing with functions, we wrote
f(g(x)) meaning f o g but just represented it as fg..
This was evaluated by first of all evaluating g and then
f.
If we adopt the same notation when dealing with
permutations, we should evaluate then the same way.
Thus ρ1 μ3 would mean we evaluate μ3 then ρ1 .
We shall also adopt the convention of not putting the
commas between the elements of the cycles so that μ =
(1 3).
Don’t let this confuse you with the way we worked
before!!
46. Composing permutations
So,
ρ1 μ3 =
Evaluating as functions from right to left we can see that this
results in the mapping:
1 → 3 → 1
2 → 2 → 3
3 → 1 → 2
The resulting map is
Thus ρ1μ3 = μ2.
47. Another Example
So,
ρ1 μ3 =
We can see that this results in the mapping:
μ3 ρ1
1 → 3 → 1
2 → 2 → 3
3 → 1 → 2
The resulting map is
Thus ρ1μ3 = μ2.
48. Multiplication of Permutations
To multiply two permutations, we shall
follow the convention mentioned by
Herstein (p 110) whereby the
multiplication follows the normal
composition of functions.
We apply the permutation on the right
and this followed by that on the left.
49. Working with cycles
Let us take two symmetries with which we are familiar:
μ2 = (2 3)
ρ1 = (1 2 3)
Then μ2ρ1 = (2 3)(1 2 3)
ρ1 μ2
1 → 2 → 3
2 → 3 → 2
3 → 1 → 1
So μ2 ρ1 = (1 3) = μ3
50. Another example
Let us take two symmetries with which we are familiar:
μ1 = (1 2)
μ3 = (1 3)
Then μ1 μ3 = (1 2)(1 3)
μ3 μ1
1 → 3 → 3
2 → 2 → 1
3 → 1 → 2
So μ1 μ3 = (1 3 2) = ρ2
52. Water Break for me…
Get out a clean sheet of paper and a pencil.
We are going to complete the group table using the
cycles
I recommend converting the cycle into the two line
format to help you see things clearly as you go…
Thus, for instance: (1 2 3)(1 3)=(2 3)
As you may have noted, I have worked out some
examples for you. Try the others now…
64. Transpositions
Note that with the following permutations, we
could effect the change with one transposition
or one 2-cycle:
μ1 = = (1 2)
μ2 = = (2 3)
μ3 = = (1 3)
65. Transpositions
In the case of these, we need two 2-cycles:
ρ0 = = (1) = (1 3) (1 3)
ρ1 = = (1 2 3) = (1 2)(2 3)
ρ2 = = (1 3 2) = (1 3)(3 2)
Exercise: Confirm the last two cases above.
66. Parity Group
Let P = {e, o} where e are
the even permutations
and o are the odd
permutations.
(P, *) is a group under
compositions.
* e o
e e o
o o e
67. Transpositions
Question: Show that the permutation μ = 34521 can be
expressed as a product of 2-cycles.
Proof: Consider e = 12345
τ1 = (1 5) τ1(e) = 52341
τ2 = (2 4) τ2(τ1(e))= 54321
τ3 = (1 3) τ3(τ2(τ1(e)))= 34521
τ3(τ2(τ1(e)))= μ
So μ = τ3τ2τ1
And our permutation μ can be expressed as a product of
three 2-cycles.
68. Check
Show that the permutation μ = 34521 can be expressed as a
product of 2-cycles.
Proof: μ = τ3τ2τ1 with τ1 = (1 5), τ2 = (2 4) and τ3 = (1 3)
μ = τ3τ2τ1 = (1 5)(2 4)(1 3)
(1 3) (2 4) (1 5)
1 → 3 → 3 → 3
2 → 2 → 4 → 4
3 → 1 → 1 → 5
4 → 4 → 2 → 2
5 → 5 → 5 → 1
So μ = τ3τ2τ1
And our permutation μ can be expressed as a product of
three 2-cycles. Note that this is an odd permutation.
69. Note
The permutation μ = 34521 can be expressed as a
product of 2-cycles in a number of ways.
For instance, μ = 34521 =(2 3)(1 2)(2 4)(3 5)(4 5)
We have 5 2-cyles and it is an odd permutation.
Recall that we had expressed this as 3 2-cycles.
So even though there may be many ways of
expressing this permutation as a product of 2-cycles,
we will always get an odd number. It will always be
an odd permutation.
73. Exercise
Consider the permutation on the set S =
{1, 2, 3, 4, 5} given by 𝛔(x) = 6 – x for x
𝜖 S, write out the permutation table and
express this permutation as a product of
distinct cycles.
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74. Solution
Consider the permutation on the set S = {1, 2, 3, 4, 5} given by 𝛔(x) = 6 –
x for x 𝜖 S, write out the permutation table and express this permutation
as a product of distinct cycles.
5/11/2020 74
S = {1, 2, 3, 4, 5}
𝛔(x) = 6 – x for x 𝜖 S
1 → 5
2 → 4
3 → 3
4 → 2
5 → 1
And the permutation is (1 5)(2 4)
75. Note that the product contains
(i)an even number of even permutations – even.
(ii)an odd number of even permutations – odd
(iii) an odd number of even permutations - odd
