Symmetric Group S4

Let S = {1, 2, 3, 4}
The elements of S4 are:-
{ 𝑝₀ = 𝐼,𝑝₁ = (12),𝑝₂ = (13),𝑝₃ = (14),𝑝₄ = (23),𝑝₅ = (24),𝑝₆ = (34),
𝑟₀ = (12)(34),𝑟₁ = (13)(24),𝑟₂ = (14)(23),
 𝑎₀ = (123),𝑎₁ = (124),𝑎₂ = (132),𝑎₃ = (134),𝑎₄ = (142),a₅ = (143), 𝑎₆ = (234),𝑎₇ = (243),
 𝑏₀ = (1234),𝑏₁ = (1243),𝑏₂ = (1324),𝑏₃ = (1342),𝑏₄ = (1423),𝑏₅ = (1432)}
ORDERS:  
1. The order of each of p₁, . . ., p₆ and r₀, . . ., r₃ is 2.
2. The order of each of a₀, . . ., a₇ is 3.
3. The order of each of b₀, . . ., b₅ is 4.

INVERSES: 
1. The inverse of the elements 𝑝₀,p₁, . . . ,p₆ and r₀, . . ., r₃ is the element itself.
2. 𝑎₀-1=𝑎₂ , 𝑎₁-1= 𝑎₄ , 𝑎₃-1= 𝑎₅ , 𝑎₆-1= 𝑎₇
3. 𝑏₀-1= 𝑏₅, 𝑏₁-1= 𝑏₃, 𝑏₂-1=𝑏4.

NON-ABELIAN:  
As 𝑝₁𝑝₂= (132) and 𝑝₂𝑝₁ =(123)
Implies 𝑝₁𝑝₂ ≠ 𝑝₂𝑝₁
Thus, S4 is non-abelian group.

NON-CYCLIC:  
S4 is a non-cyclic group.  

CALAY TREE: 











SUBGROUPS 
Subgroups of order 24: 
S4 
Subgroups of order 12: 
A4 
Subgroups of order 8: 
There are just 3 subgroups of order 8 
{𝐼,(1234),(1432),(13)(24),(12)(34),(14)(23),(13),(24)}
{𝐼,(1243),(1342),(14)(23),(13)(24),(12)(34),(14),(23)}
 {𝐼,(1423),(1324),(12)(34),(13)(24),(14)(23),(12),(34)} 
Subgroups of order 6: 
There are just 4 subgroups of order 6
{𝐼,(12),(13),(23),(123),(132)}
{𝐼,(12),(14),(24),(124),(142)}
{𝐼,(13),(14),(34),(134),(143)}
{𝐼,(23),(24),(34),(234),(243)}
Subgroups of order 4: 
There are 7 subgroups of order 4 and all of these are cyclic
{𝐼,(1234),(1432),(13)(24)}
{𝐼,(1243),(1342),(14)(23)}
{𝐼,(1423),(1324),(12)(34)}
{𝐼,(12)(34),(13)(24),(14)(23)}
{𝐼,(12),(34),(12)(34)}
{𝐼,(13),(24),(13)(24)}
{𝐼,(14),(23),(14)(23)}
Subgroups of order 3: 
There are 4 subgroups of order 3 and all of these are cyclic
 {𝐼,(123),(132)}
 {𝐼,(134),(143)}
  {𝐼,(124),(142)}
  {𝐼,(234),(243)}
Subgroups of order 2: 
There are 9 subgroups of order 2 and all of these are cyclic
 {𝐼,(12)}
 {𝐼,(13)}
 {𝐼,(14)}
 {𝐼,(23)}
 {𝐼,(24)}
 {𝐼,(34)}
 {𝐼,(12)(34)}
 {𝐼,(13)(24)}
{𝐼,(14)(23)}
Subgroups of order 1:
 {𝐼}
Commutator Subgroup of S4 
Let G=S4and N=A4,
Then |S4/A4|= 2
⟹  S4/A4 is cyclic, so it is abelian.
By using theorem “let G´ be the commutator subgroup of a group G.
Then G/N is abelian iff G´⊆N.”
⟹  G´⊆A4
⟹ G´ = {e} or |G´|=2 or |G´|=3 or |G´|=4 or G´= A4, as |A4|=12
Since G= S4is non-abelian. So, G´≠ {e}.
If |G´|=2 or |G´|=3 or |G´|=4, then G´ is not normal in G for all these three cases.
 So
|G´|≠ 2, |G´|≠ 3, |G´|≠ 4.
Hence G´= A4.
Conjugacy classes of S4
n=4

Partitions of 4	Cycle Type	Representatives of conjugacy class	Number of elements in conjugacy class	Conjugacy class	Even/ Odd
1, 1, 1, 1	(1)	(1)	1	{(1)}	Even
1,1,2	(1,2)	(1)(2)(34)	6	{(12),(13),(14),(23),(24),(34)}	Odd
2,2	(2,2)	(12)(34)	3	{(12)(34),(13)(24),(14)(23)}	Even
1,3	(1,3)	(1)(234)	8	{(123),(124),(132),(134), (142),(143),234),(243)}	Even
4	(4)	(1234)	6	{(1234),(1432),(1243), (1342)(1423),(1324)}	Odd

Partitions of 4	Cycle Type	Representatives of conjugacy class	Number of elements in conjugacy class	Conjugacy class	Even/ Odd
1, 1, 1, 1	(1)	(1)	1	{(1)}	Even
1,1,2	(1,2)	(1)(2)(34)	6	{(12),(13),(14),(23),(24),(34)}	Odd
2,2	(2,2)	(12)(34)	3	{(12)(34),(13)(24),(14)(23)}	Even
1,3	(1,3)	(1)(234)	8	{(123),(124),(132),(134), (142),(143),234),(243)}	Even
4	(4)	(1234)	6	{(1234),(1432),(1243), (1342)(1423),(1324)}	Odd

N0rmal Subgroups of S4 
There are four normal subgroups of S4:
1. {(1)} 
2. V4 
3. A4 
4. S4 
Alternating Group A4 
The elements of A4 are:- 
{I,(12)(34),(13)(24),(14)(23),(123),(124),(132),(142),(234),(134),(143),(243)} 

Conjugacy Classes of A4 
No.                    Conjugacy Class 
1                              {(1)} 
2                 {(123),(214),(341),(432)} 
3                 {(132),(241),(314),(423)} 
4               {(12)(34),(13)(24),(14)(23)} 

Normal Subgroups of A4 
 There are three normal subgroups of A4:- 
1. {(1)} 
2. V4 
3. A4