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Symmetric Group

Chapter 8

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Arrangement of OrdersIf we have one object, then we only have one possible order to arrange it, which is {1}If we have two objects, then we have two possible orders to arrange them. They are

{1, 2} and {2, 1}If we have three objects, then we have six possible orders to arrange them. That are

{1,2,3} {1,3,2} {2,1,3} {2,3,1} {3,1,2} {3,2,1}From algebra, total n objects will have n! different orders ofarrangement. For example, if we have four objects, then we have 4!=24 possible orders to arrange them.

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PermutationsEach arrangement also defines a permutation among objects.Suppose we now have 3 objects on the plane, what ever the position or location they are. How many possible ways we could re-arrange them among their positions? They are(1) Any object stays to its previous position, that is

(2) Object 1 replaced by object 2

Object 2 replaced by object 3

Object 3 replaced by object 1

( )1 2 31 2 3

1 1, 2 2, 3 3, we use notation e→ → → =

( )1 2 32 3 1

1 2, 2 3, 3 1, we use notation α→ → → =

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(4) Object 1 replaced by object 3

Object 2 replaced by object 1

Object 3 replaced by object 2

(3) Object 1 stays in previous position

Object 2 replaced by object 3

Object 3 replaced by object 2

( )1 2 33 1 2

1 3, 2 1, 3 2, we use notation �→ → → =

( )1 2 31 3 2

1 1, 2 3, 3 2, we use notation �→ → → =

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(5) Object 1 replaced by of object 3

Object 2 stays in previous position

Object 3 replaced by object 1

(6) Object 1 replaced by object 2

Object 2 replaced by object 1

Object 3 stays in previous position

( )1 2 33 2 1

1 3, 2 2, 3 1, we use notation �→ → → =

( )( ) ( )2 1 3 3 1 21 2

1 2 3

3 3 2 1

2 1 3

We also could write � or �

But if top numbers

1 2, 2 1, 3 3,

keep increasing, t

we use n

he notati

otation �

ons are unique

→ →

= =

→ =

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Symmetric GroupDefinition: The set of all permutations among n objects is a group with respect to the operation composition. This group is denoted as Sn and called the symmetric group.Because every permutation is an 1−1 mapping from a set with nobjects to its self. The associative property is always true formapping compositions. The identity mapping exists. And the inverse mapping exists for every permutation.Every subgroup of symmetric group is called a permutation group

Lets see S3 = { e, α, β, γ, σ, δ }Identity is e.What is the inverse of α? Actually

( ) ( )1 2 3 1 1 2 31 2 3 3 1 2

�α − = = =

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( ) ( ) ( )( ) ( )( ) ( ) ( )( )( ) ( )( )

1 1 1 1

3

2 1 2 3 1 2 3 1 2 3 2 3 1 1 2 32 3 1 2 3 1 2 3 1 3 1 2 3 1 2

1 2 3 1 2 3 1 2 31 3 2 2 3 1 3 2 1

1 2 3 1 2 3 1 2 31 3 2 3 1 2 2 1 3

1 2 3 1 2 32 3 1

Similarly � , � =�, � =�, � =�Actually, every element of can be generated by and �

� �

�� �

�� �

Also ��

S

αα

− − − −=

= = = =

= = =

= = =

= ( ) ( ) 2

3 2

2 23

3 2 2

1 2 31 3 2 2 1 3

� = �� =��

We see that � � . has order 3, � has order 2

Therefore { , �, � , �, ��, �� }

with � and � = ��

e

S e

e

α

α α

= =

= ==

= =

8

β

β

β

eα

γ

γ

So γ = α2 , δ = βα, σ = βγ, also δ = γβ = α2β

Operation Picture

αα

β

δ σγ

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Group of size 6Theorem: Every group of size 6 is either generated by one element of order 6 or generated by two elements α and β such that order(α) =3, order(β) = 2 and αβ= βα2 which means it has the same structure of symmetric group S3

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2 2

� { , �

Proo Because that the size of is 6, from a theorem before

has element of order 3. Then is a subgroupof . Let element and T

, � }� � .

� { , �, �

hen we have

From co} { �, ��, �� }

f

se h

:

t t e

G

G H e

G G H

G H H e

=∈ ∉

= ∪ = ∪

2

2 2

2 2 2

orem, those six elements are all elements of .

Now consider element what is it in the above list of .

If wrong! If wrong!

If This is wrong t

� ,

� � � � �� � �

� �� � o� o.

G

G

e= � = = � == � =

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2 2

2 3 2 6 3

2 2 3 2 2 6 6

Therefore

Case 1: If , then and

Case 2: If , then and In those both cases the order of is 6, so is generated by

Case 3: If

� { , �, � }

� � � �� �� � �

� � � �� �� � �

� .

�

e

e e

e e

G

∈= = = ≠ = == = = ≠ = =

2

2

2

2

then the order of is 2.Now consider element where is it?

If then we will get this is wrong!

So

C

� = , �

��,

�� { , �, � } �

�� � {�, ��, �� }.

�ertainly, So it is � �. �� ��either �� �� .

�� ��

or

If

e

H e H

H

∈ = ∈∈ =

≠ = =

= ( ) ( ) ( )2 3 62

23

then , and

The order of is 6, so is generated by

�� � �� � ��

�� ��.

��If this is the case o� .� f

e e e

G

S

= ≠ = ≠ =

=

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Example: Let A={1, 2, 3, 4, 5} define 2 permutation α and βα(1) =2, α(2) =1, α(3) =3, α(4) =5, α(5) =4

� β(1) =1, β(2) = 4, β(3) =2, β(4) =3, β(5) =5Find α2, β2, αβ and βαSolution:

2

2

1 2 3 4 5 1 2 3 4 52 1 3 5 4 1 4 2 3 5

1 2 3 4 5 1 2 3 4 5 1 2 3 4 52 1 3 5 4 2 1 3 5 4 1 2 3 4 5

1 2 3 4 5 1 2 3 4 5 1 2 3 4 51 4 2 3 5 1 4 2 3 5 1 3 4 2 5

1 2 3 4 5 1 2 3 4 5 1 2 3 42 1 3 5 4 1 4 2 3 5 2 5 1

we can get �= �=

Then �

�

��=

� � � �� � � �� � � �

� �� � � �=� �� � � �

� �� � � �

� �� � � �=� �� � � �

� �� � � �

� �� �=� �� �

� �� �

=

=

53 4

1 2 3 4 5 1 2 3 4 5 1 2 3 4 51 4 2 3 5 2 1 3 5 4 4 1 2 5 3

��=

� �� �� �

� �� � � �=� �� � � �

� �� � � �

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Home works1. Find all possible subgroups of S3

2. Let A={1, 2, 3, 4, 5} define 2 permutation α and βα(1) =4, α(2) =3, α(3) =5, α(4) =1, α(5) =2

� β(1) =2, β(2) = 3, β(3) =1, β(4) =4, β(5) =5Find α2, β2, αβ and βα

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Subgroup of S3Because that the size of S3 is | S3 |=6, the size of subgroup is possibly 1, 2, 3 or 6

If size = 1, the subgroup is { ε}

If size = 2, the subgroup is { ε, β}, { ε, σ} , { ε, δ}

If size = 3, the subgroup is { ε, α, γ}

If size = 6, the subgroup is { ε, α, β, γ, σ, δ }