Symmetric Group S3

                     Symmetric Group S3 

Let $A = \{1, 2, 3\}$
The elements of $S_{3}$ are:-
               $\{ p_{0} = 𝐼, p_{1} = (12), p_{2} = (13), p_{3} = (23), p_{4} = (123), p_{5} = (132)\}$
ORDERS: 
 1. The order of each of p₁, p₂, p₃, is 2.
 2. The order of each of p₄, p₅, is 3.

INVERSES:
1. The inverse of the elements $p_{0}, p_{1}, p_{2}, p_{3}$ is the element itself.
2.  $p_{4}^{-1} = p_{5},  p_{5}^{-1} = p_{4}$.

NON-ABELIAN:
                          As $p_{1}p_{2}= (132)$ and $p_{2}p_{1} =(123)$
                          Implies $p_{1}p_{2}\neq p_{2}p_{1}$
                          Thus, $S_{3}$  is non-abelian group.

NON-CYCLIC:
                       $S_{3}$ is a non-cyclic group.

CALAY TREE:
                       The subgroups of $S_{3}$ are of order 1, 2, 3, 6.

SUBGROUPS

Subgroup of order 6:

                                      The subgroup of order 6 is $S_{3}$ itself.

Subgroup of order 3:

                                     There is only one subgroup of order 3 in $S_{3}$_.

                                     $< (123)> = < (132)> = \{(1), (123), (132)\}$

Subgroups of order 2:  

   There are three subgroups of order 2 in $S_{3}$. 

1. $<(12)> = \{(1), (12)\}$ 

2. $<(13)> = \{(1), (13)\}$

3. $<(23)> = \{(1), (23)\}$ 

Subgroup of order 1:  

                                     The subgroup of order one is $\{𝐼\}$. 

Commutator Subgroup Of $S_{3}$

$S_{3}=\{(1),(12),(13),(23),(123),(132)\}$ 

$[(1), (1)] = [(1), (12)] = [(1), (13)] = [(1), (23)] = [(1), (123)] = [(1), (132)] = (1) $

$[(12), (1)] = [(12), (12)] = (1)$ 

$[(12), (13)] = [(12), (132)] = (123)$ 

$[(12), (23)] = [(12), (123)] = (132)$ 

$[(13), (1)] = [(13), (13)] = (1)$ 

$[(13), (12)] = [(13), (123)] = (132)$ 

$[(13), (23)] = [(13), (132)] = (123)$ 

$[(23), (1)] = [(23), (23)] = (1)$ 

$[(23), (12)] = [(23), (132)] = (123)$ 

$[(23), (13)] = [(23), (123)] = (132)$ 

$[(123), (1)] = [(123), (123)] = [(123), (132)] = (1)$ 

$[(123), (12)] = [(123), (13)] = [(123), (23)] = (123)$ 

$[(132), (1)] = [(132), (123)] = [(132), (132)] = (1)$ 

$[(132), (12)] = [(132), (13)] = [(132), (23)] = (132)$ 

The set of commutators is:- 

                                               $X = \{(1), (123), (132)\}$ 

Since, 
          $<X>=\{(1),(123),(132)\}=X$ 

Thus, X is commutator subgroup of $S_{3}$. 

Conjugacy Classes Of $S_{3}$ 

Here n=3

Partitions of 3	cycle type	Representative of conjugacy class	No. of elements in conjugacy class	conjugacy class	Even/Odd
1,1,1	$(1)$	$(1)(2)(3)$	1	$\{(1)\}$	Even
1,2	$(1,2)$	$(1)(23)$	3	$\{(12),(13)(23)\}$	Odd
3	$(3)$	$(123)$	2	$\{(123),(132)\}$	Even

Normal Subgroups of $S_{3}$ 

There are three normal subgroups of $S_{3}$:

 1. $\{(1)\}$

 2. $\{(1),(123),(132)\}$ 

3. $S_{3}$ 

Alternating group $A_{3}$ 

The elements of $A_{3}$ are:- 

                                  $\{(1),(123),(132)\}$ 

Conjugacy classes of $A_{3}$

There are three conjugacy classes of $A_{3}$:-

 1. $\{(1)\}$

 2. $\{(123)\}$ 

 3. $\{(132)\}$ 

Normal Subgroups of $A_{3}$ 

There are two normal subgroups of A3:- 

1. $\{(1)\}$ 

2. $A_{3}$