The document defines symmetric groups and discusses their properties. Some key points: - A symmetric group is the group of all permutations of a finite set under function composition. - Symmetric groups of finite sets behave differently than those of infinite sets. - The symmetric group Sn of degree n is the set of all permutations of the set {1,2,...,n}. - Sn is a finite group under permutation composition. Subgroups include the alternating group An of even permutations. - Examples discussed include S2, the Klein four-group, and S3, which is non-abelian with cyclic subgroups.

1. Symmetric groups
Introduction-
In mathematics the symmetric group on a set is the group consisting of all permutations
of the set i.e., all bijections from the set to itself with function composition as the group
operation.
The notion of set of permutation of finite set and notion of finite set and notion of the
productof two permutation was defined by KUFFINI and CAYLEY and gave the
conceptof a finite group of permutation .
The symmetric group of degree n is the symmetric group on the set X={1,2,3…,n}.
Symmetric groups on infinite sets behave quite differently than symmetric groups on
finite sets.
The symmetric group is important to diverse areas of mathematics such as Galois theory,
invariant theory , the representation theory of Lie-group and combinatories CAYLEY’S
theorem states that the every group G is isomorphic to subgroup of the symmetric groups
on G.
In this section , we shall focus on the finite symmetric group ,their application , their
elements ,their conjugacy classes ,a finite representation ,their subgroups ,their
automorphism groups and their representation theory.
Some Definitions-
Let S be a non empty finite set ,then one-one onto mapping f: S →S is called prmutation.

2. Let S={a1,a2,a3,…an} be finite set containing n elements if f: S →S is one-one onto
mapping,
then f is a permutation of degree n.
(a1)=b1
f(a2)=b2
f(a3)=b3
Let f(a1)=b1 , f(a2)=b2 … f(an)=bn
Symbolicaly a permutation f can be written as:
f = (
a1 a2
b1 b2
… an
… bn
) since f is one –one & onto map, therefore secend row is
only rearrangement (permutation ) of a1 , a2 , a3 . . .an.
Eg:- let S = {1,2} be the finite set containing 2 elements then
f1= (
1 2
1 2
) ; f2 = (
1 2
2 1
) are permutation of degree 2 .
f1
a1
a2
a3
…
an
n
f(a1)=b1
f(a2)=b2
f(a3)=b3
…
f(an)=bn
(an)=bn
1
2
1
2

3. f2
Remark:
If S be a set containing n elements, then the number of arrangement of elements of S are
nPn=n! .
Equality of Permutations :
Two permutation f and g are said to be equal, i.e. f=g iff f(a)=g(a) for every a ∈ S.
Identity Permutation :
If a permutation I of degree n is such that the I image of every element is the same
element, i.e. I(x)=x for every x ∈ S.
Ex. I = (
𝑎 𝑏 𝑐
𝑎 𝑏 𝑐
) is the permutation of degree 3.
Product or Composition of two Permutation :
Let S = { 1,2,3 . . .n} and f & g are two permutation on S , i.e.
f : S→ S & g : S→S are one- one & onto maps then there composition maps
gof : S→S & fog : S→S are also one- one & onto maps. Thus the productof two
permutations
1
2
2
1

4. f=(
1 2
𝑓(1) 𝑓(2)
3… 𝑛
𝑓(3) 𝑓(𝑛)
) and g = (
1 2
𝑔(1) 𝑔(2)
3… 𝑛
𝑔(3) 𝑔(𝑛)
) on S
defined as :
gf ≝ (
1 2
𝑔(𝑓(1)) 𝑔(𝑓(2))
3… 𝑛
𝑔(𝑓(3)) 𝑔(𝑓(𝑛))
)
Ex. f= (
1 2 3
2 1 3
) , g = (
1 2 3
1 3 2
)
gf = (
1 2 3
1 3 2
)(
1 2 3
2 1 3
) = (
1 2 3
3 1 2
)
Inverse of permutation:
Let S = { a1, a2, a3, . . . ,an} and f be permutation on S ,i.e.
f : S →S is one –one onto map then f-1 exist . Thus ,every permutation f on set S has
unique inverse permutation denoted by f-1.
Therefore if f = (
1 2
𝑓(1) 𝑓(2)
3… n
f(3) f(n)
) .
f-1 =(
𝑓(1) 𝑓(2)
1 2
f(3)… f(n)
3 n
).
cyclic notation:
A permutation f which moves cyclically a set of elements a1, a2, a3, . . .,ar in the sense
f(a1) = a2 , f(a2) = , f(a3) = a4, . . .f(ar-1) = f(ar) , f(ar) = a1 and keeps fix the other elements
,is called a cycle of length r or r cycle.
The cycle f is written as –
f = (a1 a2 a3 . . . ar) ,this is called cycle of length r.

5. Ex: 𝑓 = (
1 2
2 3
3 4
4 1
),
Cyclic form : (1 2 3 4)
Disjoint Cycles Commute:
If the pair of cycles α=(a1,a2,…am) and β= (b1, b2 ,…bn) have no entries in common, then
αβ=βα.
Transposition :
A cycle of length two is called transposition .
Ex: ( 1 3) , ( 1 2) are transpositions.
Order of a Permutation(Ruffini-1799) :
The order of a permutation of a finite set written in disjoint cycle form is the least
common multiple of the lengths of the cycles.
Ex: (1 2 3 4) has order 4, and ( 1 2)(3 4 5) has order l.c.m(2,3) = 6.
Group of permutations:
Theorem:
The set Sn of all permutations on n symbols is a finite group w.r.to permutation
composition.
Proof:

6. Let S = {1,2,3,. . .n} be a finite set having n distinct elements and Sn be set of all
permutation on n symbols, i.e.
Sn = {f : f is permutation on S}
1. Closure Property :
Let f , g ∈ Sn then f & g are permutation on S ,i.e.
f:S→S and g:S→S are one-one onto map.
 gof :S→S is also one-one onto map.
 gof is permutation on S.
 gof ∈ Sn.
 gf ∈ Sn. (for our convenience say , gof = gf )
also fg ∈ Sn.
2. Associative Property :
Since the productof two permutation on a set S is a productof two one-one onto
mapping on S and the productofmapping being associative , the productof
permutation also obeys associative law.
Hence.
(fg)h = f(gh) ∀ f,g,h ∈ Sn.
3. Existence of Identity :
Identity permutation I ∈ Sn is identity of Sn because
I.f = f.I = f ∀ f ∈ Sn.
4. Existence of inverse :

7. Let f ∈ Sn then f is permutation on S i.e. f is one –one onto map and hence it is
invertible.
Hence f-1 is also one-one onto.
∴ f-1∈ Sn s.t.
ff-1 = f-1f = I .
f-1 is inverse of f.
hence Sn forms the group under permutation product.
Even and Odd permutations:
A permutation that can be expressed as a productof an even number of 2-cycles is called
an even permutation.
A permutation that can be expressed as the productof an odd number of 2-cycles is
called an odd permutation.
Even permutations forms a group:
The set of even permutations in Sn forms a subgroup of Sn.
Alternating group of degree n (An) :
The group of n symbols is called the Alternating group of degree n.
i.e., denoted by An (a group of even permutations).
 An is symmetric group.
 |An| = n!/2 ;n>1.
 [Sn:An] = index of An in Sn
=|Sn|/|An| = n!/n!/2 = 2.
Therefore An is normal subgroup of Sn where n≥2 .

8.  For n ≥ 3 the subgroup generated by 3-cycles is A3.
Particular Cases:
1. Symmetric group S2 :
Let S = {1,2} and S2 = (
1 2
1 2
) be the set of all permutations on S .
We write as S2 = { I , f } , where I = (
1 2
1 2
) & f = (
1 2
2 1
)
Some observations:
We have S2 = { I, f }
1. Order of S2:
order of S2 = 2. Denoted by |S2| = 2.
2. Cayley diagram of S2:
3. Commutative property :
Since If = fI = f hence S2 is Abelian group.
4. Subgroups of S2 :
Since |S2| = 2 which is prime ,and order of subgroup divides order of group
 It has only improper subgroups.

9. Which are given below.
H1 = { I } & H2 = S2.
5. S2 is cyclic group :
S2 forms cyclic group with generator <f>
Hence
S2 = { f , f2 = I }.
6. Normal subgroups of S2 :
{ I } & S2 are only normal subgroups of S2 since S2 is Abelian & every subgroup
of Abelian group is NORMAL.
7. Simple group:
Since S2 has no proper normal subgroup therefore,S2 is SIMPLE GROUP.
8. Centre of S2 :
Z(S2) = S2 ,Since it is Abelian group .
9. Normal series :
Normal series is given below .
S2 ⊇ { I } .
10.Solvable group :
The normal series of S2 with Abelian factor ;
S2/I is Abelian.
Hence S2 is SOLVABLE GROUP.
11.Quotient group :
Since A2 = { I } is normal subgroup of S2 , then Quotient group ( S2/A2 ) exist.

10. Now
| S2/A2 | = |S2|/ | A2| = 2 , i.e. prime.
 Quotient group S2/A2 is cyclic .
 Quotient group S2/A2 is Abelian.
2. Symmetric group S3 :
Consider ( cyclic form )-
S3 = { I = (1 2) , f1 = (1 2)(3), f2 = (1 3)(2) , f3 = (2 3 )(1) , f4 = (1 2 3), f5 =
(1 3 2) }
Composition table for S3-
Observations :
1. Order of S3 :
order of S3 is 6 & denoted by |S3| = 6.
O I f1 f2 f3 f4 f5
I I f1 f2 f3 f4 f5
f1 f1 f4 f5 f2 f3
f2 f2 f5 f4 f3 f1
f3 f3 f4 f5 f1 f2
f4 f4 f3 f1 f2 f5
f5 f5 f2 f3 f1 f4
I
I
I
I
I
I

11. 2. Cayley diagram of S3:
3. S3 is non Abelian group :
Since
f1. f2 = (1 2)(1 3)
= (1 3 2 )
= f5
f2. f1 = (1 3)(1 2)
= (1 2 3)
= f4
Hence f4 ≠ f5
This shows f1. f2 ≠ f2. f1
Hence S3 is non Abelian group .
4. Subgroup of S3 :
Improper Subgroups : I & S3 are improper subgroups of S3.
ProperSubgroups :

12. let H1 = { I, (1 2) }
I ∈ H1 , (1 2) ∈ H1.
I.f-1 = (1)(2)(3)(1 2)-1
= (
1 2 3
1 2 3
)(
1 2 3
2 1 3
)-1
= (
1 2 3
2 1 3
)
= ( 1 2) ∈ H1 .
Similarly, H2 = { I , f2 } & H3 = { I , f3 } are sub group of S3
& H4 = { I , f4 , f5 } is also a subgroup of S3.
Remark:
H4 is the set of even permutations & hence it forms alternating subgroup A3 of S3 .
5. Specialtypes of permutations in S3:
The symmetric group on 4 elements S3 contains the following permutation.
Permutations Type Number
I 1-cycle 1
(1 2),(1 3),(2 3) 2-cycle 3
(1 2 3),(1 3 2) 3-cycle 2
6. There are four elements in S3 which satisfies the equation x2 = e (identity).
7. There are four elements in S3 which satisfies the equation y3 = e (identity).
8. The permutations contained in S3:
PERMUTATION TYPE NUMBER
I 1-cycle 1
(1 2),(1 3),( 2 3) 2-cycle 3

13. (1 2 3),(1 3 2 ) 3- cycle 2
9. Order of eachelement:
|I| = 1,
f1.f1 = (1 2).(1 2)
= (1)(2)
 |f1| = 2.
Similarly,
|f1| = |f2| = |f3| = 2.
& |f4| = |f5| = 3.
10.S3 is not cyclic group :
S3 has no element with its order equal to the order of group.
Hence it is not cyclic.
11.S3 has cyclic proper subgroups :
The proper subgroups ofS3 are-
H1 = { I , f1 } is cyclic & generated by f1.
H1 = <f1> = { f1 , f1
2=I }
H2 = { I , f2 } is cyclic & generated by f2.
H2 = <f2> = { f2 , f2
2 =I}
H3 = { I , f3 } is cyclic & generated by f3.
H3 = <f3> = { f3 , f3
2 =I}
H4 = { I , f4, f5 } is cyclic & generated by f4.
H4 = <f4> = { f4 , f4
2 , f4
3 =I }
Hence S3 has four cyclic subgroup in which three are of order 2 & one is of order 3

14. Remark:
S3 is non cyclic group whose all propersubgroups are cyclic.
12.Normal subgroups of S3 :
There are two improper normal subgroups –
N1 = {I } ,N2 = { S3 }
Consider [S3:A3] = index of A3 in S3
= |S3|/|A3| = 6/3 = 2.
Alternating group A3 is of index 2 in S3
 A3 is normal subgroup of S3
Hence A3 is only non-trivial (proper ) normal subgroup of S3.
13.Simple group :
Since S3 has propernormal subgroups A3 therefore it is not Simple group.
14.Quotient group :
Since A3 is normal subgroup of S3 therefore quotient group exist
Now |S3/A3| = |S3|/|A3| = 2 which is prime number.
 Quotient group S3/A3 is cyclic.
 Quotient group S3/A3 is Abelian (every cyclic group is Abelian).
And the map S3 S3/A3 is natural homomorphism where S3/A3 is Abelian but
S3 is not Abelian.
Hence we observed that-
(a)- Quotient group S3/A3 is cyclic while S3 is not cyclic.
(b)- Quotient group S3/A3 is Abelian while S3 is not Abelian.
(c)- Homomorphic image of S3 is Abelian while S3 is not Abelian.
The above facts shows that the converse of following theorems is not true-

15. 1- Quotient group of cyclic group is cyclic.
2- Quotient group of Abelian group is Abelian.
3- Homomorphic image of Abelian group is Abelian.
15.Syllows-subgroup of S3 :
we know that |S3| = 6 , since 2│|S3|
 The number of syllow -2 subgroup of the form (1+k.2) and (1+k.2) │|S3| this is
possible for k=1.
 Number of syllow-2 subgroups = 1+1.2 = 3 ,i.e.
H1 = { I, (1 2) }
H2 = { I, (1 3) }
H3 = { I, (2 3) } are three syllow-2 subgroup of S3.
Also since 3│|S3|,
 By syllow’s theorem-
 The number of syllow -3 subgroup of the form (1+k.3) and (1+k.3) │|S3| this is
possible for k=0.
 Number of syllow-3 subgroups = 1+0.3 = 1 ,i.e.
A3 = {I , (1 2 3), (1 3 2 )} is unique syllow -3 subgroup of S3.
16.Normal & compositionseries :
The series –
S3 ⊇ A3 ⊇ I is normal as well as composition series.
17.Solvable group :
Since normal series –
S3 ⊇ A3 ⊇ I is with Abelian factor because S3/A3 & A3/I are Abelian.

16.  S3 is solvable.
 Hence S3 is non-Abelian group which is SOLVABLE.
18.Centre of S3 :
Z(S3) = I .
19.Nilpotent group :
S3 is not nilpotent group .
20.Frattini subgroups :
The intersection of all maximal subgroup is called Frantini subgroup–
the maximal subgroups are given below-
H1 = { I, (1 2) }
H2 = { I, (1 3) }
H3 = { I, (2 3) }
H4= {I , (1 2 3), (1 3 2 )}
I(S3) = intersection of all maximal subgroups
= { I }.
21.Homomorphism :
Consider group (Z2,+) where Z2 = {0 ,1}
Define a map –
Ф:S3→Z2. Such that
Ф(f) = 0 : f is even permutation.
1 : f is odd permutation.
Claim : Ф is onto – homomorphism.
Ф is homomorphism :
Case (i) = let f1 , f2 ∈ S3 be even permutation.
22.Ф(f1) = 0 ,& Ф(f2) = 0.

17. If Ф(f1 o f2) = 0 =0+0 = Ф(f1) +Ф(f2).
Where “ o” is function composition.
Case (ii) = let f1 , f2 ∈ S3 be odd permutation.
23. Ф(f1) = 1 ,& Ф(f2) = 1.
Since productof two odd permutation is even then
Ф(f1 o f2) = 0 =2 mod(2) =1+1 = Ф(f1) +Ф(f2).
Case (iii) = let f1 , f2 ∈ S3 be even & odd permutation respectively.
24.Ф(f1) = 0 ,& Ф(f2) = 1.
Since productof two even & odd permutation is odd then
Ф(f1 o f2) = 1 = 0+1 = Ф(f1) +Ф(f2).
Ker Ф = { f ∈ S3 : Ф(f) = 0 (identity of (Z2, +)) }
= A3.
Ф is onto : for both 0, 1 of (Z2 ,+) ∃ f ∈ S3.
Hence we get –
Ф : S3
𝑜𝑛𝑡𝑜
→ Z2 homomorphism.
By fundamental theorem –
S3/ ker Ф ≅ Z2 .
25.S3/A3 ≅ Z2 .
3. Symmetric group S4:
Let S = {1, 2, 3, 4}

18. Consider S4 = set of all permutations on S
=
{ (
1 2
1 2
3 4
3 4
), (
1 2
2 1
3 4
3 4
), (
1 2
3 2
3 4
1 4
), (
1 2
4 2
3 4
3 1
), (
1 2
1 4
3 4
3 2
),
(
1 2
1 2
3 4
4 3
), (
1 2
2 1
3 4
4 3
), (
1 2
3 2
3 4
4 1
), (
1 2
4 2
3 4
1 3
), (
1 2
1 4
3 4
2 3
),
(
1 2
1 3
3 4
2 4
), (
1 2
2 3
3 4
1 4
), (
1 2
3 1
3 4
2 4
), (
1 2
4 3
3 4
2 1
), (
1 2
1 3
3 4
4 2
),
(
1 2
4 1
3 4
3 2
), (
1 2
4 1
3 4
2 3
), (
1 2
4 3
3 4
1 2
), (
1 2
3 1
3 4
4 2
), (
1 2
2 4
3 4
3 1
),
(
1 2
2 4
3 4
1 3
), (
1 2
3 4
3 4
1 2
), (
1 2
3 4
3 4
2 1
), (
1 2
2 3
3 4
4 1
)}
= { I,(1 2),(1 3), (1 4),(2 3), (2 4), (3 4), (1 2)(3 4),
(1 3)(2 4),(1 4)(2 3),(1 2 3),(1 2 4),(1 3 2),(1 3 4),(1 4 2),(1 4 3),(2 3 4),
(2 4 3),(1 2 3 4),(1 2 4 3 ),(1 3 2 4),(1 3 4 2),(1 4 2 3),(1 4 3 2) }
S4 forms a non-Abelian group of order 24.
Observations:
1. Order of S4:
The order of S4 = 4! = 24.
2. Cayley diagram of S4:

19. 3. S4 is non-Abelian group:
(1 2).(1 3) = (1 3 2)
(1 3).(1 2) = (1 2 3)
26.(1 2).(1 3) ≠ (1 3).(1 2)
Because commutative property fails
Hence S4 is not an Abelian group.
4. Specialtypes of permutations in S4:
The symmetric group on 4 elements S4 contains the following permutation.
Permutations Types Number
I 1-cycle 1
(1 2),(1 3),(1 4),(2 3),(2 4),(3 4) 2-cycle 6
(1 2)(3 4),(1 3)(2 4),(1 4)(2 3) Product of
2-cycle
3
(1 2 3),(1 2 4),(1 3 2),(1 3 4),(1 4 2),(1 4 3),
(2 3 4),(2 4 3)
3-cycle 8

20. (1 2 3 4),(1 2 4 3),(1 3 2 4),(1 3 4 2),(1 4 2 3),
(1 4 3 2)
4-cycle 6
5. Subgroups of S4:
There are 30 subgroups of S4 .Except {I} and S4 the 28 subgroups are given below
in the following table.
Subgroups- Order- Isomorphic
to-
{ I , (1 2)} 2 Z2
{I , (1 3)} 2 Z2
{ I , (1 4)} 2 Z2
{ I , (2 3)} 2 Z2
{ I , (2 4)} 2 Z2
{ I , (3 4)} 2 Z2
{ I , (1 2)(3 4)} 2 Z2
{ I , (1 3)(2 4)} 2 Z2
{ I , (1 4)(2 3)} 2 Z2
{ I , (1 2 3),(1 3 2)} 3 Z3
{ I , (1 2 4),(1 4 2)} 3 Z3
{ I , (1 3 4),(1 4 3)} 3 Z3
{ I , (2 3 4),(2 4 3)} 3 Z3
{ I , (1 2)(3 4),(1 3 2 4),(1 4 2 3)} 4 Z4
{ I , (1 3)(2 4),(1 2 3 4),(1 4 3 2)} 4 Z4
{ I , (1 4)(2 3),(1 2 4 3),(1 3 4 2)} 4 Z4
{ I , (1 2)(3 4),(3 4),(1 2)} 4 V4
{ I , (1 3)(2 4),(2 4),(1 3)} 4 V4

21. { I , (1 4)(2 3),(1 4),(2 3)} 4 V4
{ I , (1 2)(3 4),(1 3)(2 4),(1 4)(2 3)} 4 V4
{ I , (1 2),(1 3),(2 3),(1 3 2),(1 2 3)} 6 S3
{ I , (1 4),(1 3),(3 4),(1 3 4),(1 4 3)} 6 S3
{ I , (2 3),(2 4),(3 4),(2 3 4),(2 4 3)} 6 S3
{ I , (1 2),(1 4),(2 4),(1 2 4),(1 4 2)} 6 S3
{ I , (1 2),(3 4), (1 2)(3 4),(1 3)(2 4),(1 4)(2 3),
(1 3 2 4),(1 4 2 3), }
8 D4
{ I , (1 3),(2 4),( (1 2)(3 4),(1 3)(2 4),
(1 4)(2 3), (1 2 3 4),(1 4 3 2), }
8 D4
{ I , (1 4),(2 3), (1 2)(3 4),(1 3)(2 4),
(1 4)(2 3), (1 2 4 3),(1 3 4 2)}
8 D4
{ I,(1 2 4),(1 4 2),(1 2 3), (1 3 2),(1 2)(3 4),
(1 3)(2 4),(1 4)(2 3),(1 3 4),(1 4 3),(2 3 4 ),(2 4 3)}
12 A4
From above discussion we observed that-
(1)- Number of subgroup of order 2 in S4 = 9.
(2)- Number of subgroup of order 3 in S4 = 4.
(3)- Number of subgroup of order 4 in S4 = 7.
(4)- Number of subgroup of order 6 in S4 = 4.
(5)- Number of subgroup of order 8 in S4 = 3.
(6)- Number of subgroup of order 12 in S4 = 1.
(7)- Converse of Lagrange’s theorem hold in S4 because there is a
subgroup of order 12 (A4) which is divisor of order of S4.
6. S4 is non –Abelian group:
Since no element in S4 is of order equal to

22. order of S4 , therefore S4 cannot be cyclic group.
7. Cyclic subgroup of S4:
(1) All subgroup of order 2 in S4 are cyclic .
(2) All subgroup of order 3 in S4 are cyclic.
(3) 3 subgroup of order 4 are cyclic which are given below –
(i) H1 = { I , (1 2)(3 4),(1 3 2 4),(1 4 2 3)}
(ii) H2 = { I , (1 3)(2 4),(1 2 3 4),(1 4 3 2)}
(iii) H3 = { I , (1 4)(2 3),(1 2 4 3),(1 3 4 2)}
(4) Total number of cyclic subgroups = 16.
8. Normal subgroup of S4:
{I} and S4 are trivial normal subgroup of S4.
To find non – trivial normal subgroup of S4 , we divide it into class of
Permutations of the form.
C1 = {I}
C2 = {(1 2 3 4),(1 4 3 2),(1 3 2 4),(1 4 2 3),(1 2 4 3),(1 3 4 2)} the set of
cycle of length 4.
C3 = {(1 2 3),(1 3 2),(2 3 4),(2 4 3),(1 3 4),(1 4 3),(1 2 4),(1 4 2)} the set of
cycle of length 3.
C4 = {(1 2)(3 4),(1 4)(2 3),(1 3)(2 4)} the set of product of disjoint
transposition.
C5 = {(1 2),(1 3),(1 4),(2 3),(2 4),(3 4)} the set of transpositions.
Let H be a non – trivial normal subgroup of S4. Then clearly
C1 ⊆ H.

23. Further , H must be complete union of some classes Ci .
By the Lagrange’s theorem |H| divides |S4| = 24.
Therefore, the only possibilities are –
H = C1⋃ C3 ⋃ C4 and H = C1⋃ C4.
Thus, the non – trivial normal subgroups of S4 are –
(a) { I , (1 2 4),(1 4 2),(1 2 3), (1 3 2),(1 2)(3 4),
(1 3)(2 4),(1 4)(2 3),(1 3 4),(1 4 3),(2 3 4 ),(2 4 3)}
(b) { I , (1 2)(3 4),(1 3)(2 4),(1 4)(2 3)}.
Remark:
(I)- The above first normal subgroup is isomorphic to S4.
(II)- The second normal subgroup is isomorphic to Kelien’s four group
V4 = {e , a , b, c} ,the map –
e → I , a → (1 2)(3 4) , b → (1 3)(2 4) , c → (1 4)(2 3) is isomorphism .
Hence, V4 = {I , (1 2)(3 4) , (1 3)(2 4) , (1 4)(2 3)}
Thus V4 , A4 are two normal non – trivial
subgroups of S4.
(III)- V4 is normal subgroup of S4 i. e., V4 ⊲ S4 and {I , (1 2)(3 4)}⊲ V4
but {I , (1 2)(3 4)} is not normal in S4. Thus, H⊲ K and K⊲ G need
not imply that H⊲ G.
(IV)- S4/V4 ≅ S3 we have S3 is a subgroup of V4 and V4 ⊲ S4.

24. Since, |S3.V4| =
|S3|⋃|V4|
|S3⋂V4|
=
6×4
1
= 24.
Thus, S3V4 = S4
Now, by third isomorphism theorem
S3/S3⋂V4 ≅S3V4/V4 = S4/V4
Or S3/{I} ≅ S4/V4
Thus, S3 ≅ S4/V4
i.e., S4/V4 ≅ S3
Converse ofLagrange’s theorem:
The converse of lagrange’s theorem is not true. Consider the alternating group A4
such that |A4|=12.
Claim:
A4 contains no subgroup of order 6.
SupposeH be a subgroup of A4 of order 6 then,
[A4:H]= index of H in A4
= |A4|/|H|
= 12/6
= 2.
27.H is normal in A4
Since, H is of even order. It contains an element of order 2. The element of
order 2 are: (1 2)(3 4),(1 3)(2 4),(1 4)(2 3)
Supposethat, (1 2)(3 4)∈H.

25. Since H is normal.
(1 2)(3 4) = (1 2 3)(1 2)(3 4)(1 3 2)
= (1 2 3)(1 2)(3 4)(1 2 3)-1 belongs to H.
Similarly,
(1 3)(2 4) ∈ H
In fact, all the elements of order 2 in A4 are conjugate in A4. This shows that –
V4 is a subgroup of H.
This is a contradiction to Lagrange’s theorem (4 does not divides 6).
Thus, there is no subgroup of order 6 in A4.
Hence, converse of Lagrange’s theorem is not true in group A4.
9. S4 is not simple group:
Since, S4 contains propernormal subgroup V4 and A4.
28.S4 is not simple group.
10.Syllow -P subgroup of S4:
|S4| = 24 = 23 × 3
Since, 3│24
29.The number of syllow -3 subgroup is of the form (1+k.3) this possible for k =1
30.There are (1+1.3) syllow -3 subgroups.
There are 4 syllow -3 subgroups.
They are –
H1 = {I , (1 2 3), (1 3 2)}
H2 = { I , (1 2 4), (1 4 2)}
H3 = { I , (1 4 3), (1 3 4)}
H4 = { I , (2 3 4), (2 4 3)}

26. For the syllow-2 subgroup consider the Kelien’s four group V4 which is normal
in S4.
Let H = {I,(1 2)} , then HV4 is subgroup of order 8, thus it is syllow-2
subgroup.
Similarly we take K = {I,(1 3)} and L = {I,(2 3)} , then KV4 and LV4 are syllow
-2 subgroup of S4.
Hence number of syllow -2 subgroup = 3, and , number of syllow -3
subgroup = 4
11.Normaland compositionseries of S4:
We have {I} ,S4 , A4 , V4 are normal subgroup of S4 and {I ,(1 2)(3 4)} ,
{I ,(1 3)(2 4)} , {I ,(1 4)(2 3)} are normal subgroup of V4
Then normal series of S4 are –
 S4 ⊇ {I}
 S4 ⊇ A4 ⊇ {I}
 S4 ⊇ V4 ⊇ {I}
 S4 ⊇ V4 ⊇ {I , (1 2)(3 4)} ⊇ {I}
 S4 ⊇ V4 ⊇ {I , (1 3)(2 4)} ⊇ {I}
 S4 ⊇ V4 ⊇ {I , (1 4)(2 3)} ⊇ {I}
 S4 ⊇ A4 ⊇ V4 ⊇ {I}
 S4 ⊇ A4 ⊇ V4 ⊇ {I , (1 2)(3 4)} ⊇ {I}
 S4 ⊇ A4 ⊇ V4 ⊇ {I , (1 3)(2 4)} ⊇ {I}
 S4 ⊇ A4 ⊇ V4 ⊇ {I , (1 4)(2 3)} ⊇ {I}

27. Composition series-
The composition series of S4-
 S4 ⊇ A4 ⊇ V4 ⊇ {I , (1 2)(3 4)} ⊇ {I}
 S4 ⊇ A4 ⊇ V4 ⊇ {I , (1 3)(2 4)} ⊇ {I}
 S4 ⊇ A4 ⊇ V4 ⊇ {I , (1 4)(2 3)} ⊇ {I}
12.Commutator Subgroup of S4:
S4’ = A4
Proof:
Since S4/A4 ≈ {1 , -1} is Abelian
Therefore S4’ ⊆ A4 , the subgroup of A4 which are normal in S4 are A4 , V4 ,{I}
Since S4 is non –Abelian therefore S4’ ≠ { I }
Further since S4/V4 ≈ S3 is non- Abelian
S4’ ≠ V4.
S4’ = A4.
i.e., commutator subgroup of S4 is A4.
13.Solvable group:
Since S4 ⊇ A4 ⊇ V4 ⊇ {I} is normal series of S4 with Abelian factor.
Therefore S4 is Solvable .
14.Frattini subgroup: The maximal subgroup of S4 are –
A4 , V4 , { I , (1 2)} , {I , (1 3)}, {I , (1 4)}, {I , (2 3)}, {I ,(2 4)} , {I , (3 4)}
I(S4) = Frattini subgroup of S4.
= Intersection of all maximal subgroups.
= {I}

28. Vector space:
 V4 can be made vector space over field Z2 = {0 , 1}
Consider Kelien’s four group (V4 ,o) where -
V4 = {I , (1 2)(3 4) , (1 3)(2 4) , (1 4)(2 3)}
Say, e = I , a = (1 2)(3 4) , b = (1 3)(2 4) , c = (1 4)(2 3)
We define internal binary operation –
o: V4×V4 → V4 such that –
o e a b c
e (e) a b c
a a (e) c b
b b c (e) a
c c b a (e)
Now we define external binary operation –
Such that –
*: Z2 × V4 → V4.
0*a = e, ∀ a ∈ V4.
1*a = a, ∀ a ∈ V4.
By using these operations the structure (V4 , o, *) forms a vector space.

29. Groups
Binary operation:
A binary operation on a non empty set G is a function o: G×G → G such that
To each pair (x ,y) ∈ G×G gives the unique element x o y or xy in G ,called the
composition of x & y.
Group:
A group G consistof a non-empty set G together with a binary operation ‘o’ for
which the following properties are satisfied.
1. Associative law:
(x o y) o z =x o (y o z) ∀ x,y & z of G.
2. Existence of identity:
There exist an element ‘e’ of G such that
e o a = a o e = a for each ‘a’ ∈ G. then e is called identity of the group G.
3. Existence of inverse:
For every x ∈ G, ∃ an element x-1 of G. such that x o x-1 = x-1 o x = e
(i.e., identity of G). A group G is said to be Abelian (or commutative)
if x o y = y o x ∀ x , y ∈ G.
One can usually adopts multiplicative notation for groups , where the product x o y of
two elements x and y of a group G is denoted by xy. The associative property then
requires that (xy)z = x(yz) for all elements x, y & z of G. the identity element is often
denoted by e (by eG when it is necessary to specify explicitly the group to which it
belongs) and the inverse of an element x of G is then denoted by x-1.

30. It is sometimes convenient to use additive notation for certain groups. Here the group
operation is denoted by + ,the identity element of the group is denoted by 0, the inverse
of an element ‘x’ is denoted by ‘-x’. By convention, additive notation is rarely used for
non-Abelian groups. When express in additive notation the axioms for a Abelian group
require that (x+ y) + z = x+(y + z), (x + y) = (y + z), x+0 = 0+x = x, and
x + (-x) = (-x) + x = 0 for all elements x , y and z of the group.
Elementary property of groups:
 In a group G, there is only one identity element.
 For each element ‘a’ in a group G, there is a unique element ‘b’ in G such that
ab = ba = e.
 Let x and y be elements of a group G. then (xy)-1 = y-1 x-1.
Order of a group:
The number of elements of a group (finite or infinite) is called its order. We will use |G|
to denote the order of G.
Order of an element:
The order of an element g in a group G is smallest positive integer ‘n’ such that gn = e
(In addition notation, gn = ng = 0) if no such integer exist, we say that g has infinite
order. The order of an element is denoted by |g|.
Subgroup:
Let G be a group, let H be a subset of G. We say that H is a subgroup of G if the
following condition are satisfied-
 The identity element of G is an element of H.
 The productof any two elements of H is itself an element of H.
 The inverse of any element of H is itself an element of H.
Two subgroup namely the set containing {e} and G are called Trivial or Improper

31. Subgroup of G.
If there exist a subgroup of a group other than trivial subgroup will be called as
non-trivial or propersubgroup of the group.
Centre of a group:
The centre Z(G) of a group G is the subset of elements in G that commutes with
every element of G.
Mathematically-
Z(G) = { x ∈ G | ax = xa ∀ a ∈ G }.
Remark: Centre of group G is a Subgroup of G.
Centralizer of an element‘a’ in G:
Let ‘a’ be a fixed element of a group G. The centralizer of ‘a’ in G is the set of all
elements in G that commutes with ‘a’.
Mathematically-
C(a) = { x ∈ G| ax = xa}.
Remark: Centralizer of group G is a Subgroup of G.
Cyclic group:
Let G be a group of order ‘n’,is said to be cyclic group if ∃ an element ‘a’ ∈ G ,
Such that an = e ,(a ≠ e) then ‘a’ is called generator of group G & denoted by
G = <a>.
Cosets:
Let H be a subgroup of a group G. A left cosetof H in G is a subsetof G that is of
the form aH, where a ∈ G and
aH = { x ∈ G | x = ah for some h ∈ H }
similarly a right cosetof H in G is a subsetof G that is of the form ‘Ha’, where
a ∈ G and
Ha = { x ∈ G | x = ha for some h ∈ H }

32. Note that a subgroup G is itself a left cosetof H in G have the following
properties:-
1. a ∈ aH , ∀ a ∈ G.
2. If a and b are elements of G , and if b = ah for some h ∈ H , then aH = bH.
3. If a and b are elements of G , and if aH ⋂ bH is non-empty then aH = bH.
Lagrange’s Theorem-
Let G be a finite group and let H be a subgroup of G. then |G|/|H| = [G:H] is the
number of distinct left cosets of H in G. In particular, the number of elements in H
must divide the number of elements in G. where [G:H] is index of H in G.
n.|H| = |G|, where ,n = (The number of distinct left cosetH in G).
or [G:H].|H| = |G|.
Corollary:
1. Supposethat G is a finite group and g ∈ G. Then the order of g must divide
the number of elements in G.
2. Let |G| = p with (p is prime number). Then G is cyclic and any g ∈ G such
that g ≠ e is a generator.
3. The converse of Lagrange’s theorem is false we are not guaranteed that
subgroups of every possible order exist.
Normal subgroups:
A subgroup H of a group G is normal in G if gH = Hg for all g ∈G.
i.e, A normal subgroup of a group G is one in which the left and right cosets
are precisely same.
Or
A subgroup H of group G is said to be normal iff
aha-1 ∈ H , ∀ a ∈ G and h ∈H
a-1ha ∈ H , ∀ a ∈ G and h ∈H

33.  Every subgroup of Abelian group is normal.
 Every cyclic group is Abelian so that every subgroup of cyclic group
is normal.
 Intersection of any two normal subgroups of a group G is normal
subgroup.
 The centre of group G is normal subgroup of G.
Factorgroup:
If N is a normal subgroup of a group G, then the cosetof N in G form a group G/N
under the operation (aN)(bN) = abN. This group is called the factor or quotient
group of G and N.
Simple group:
In mathematics , a Simple group is a nontrivial group whose only normal
subgroups are the trivial group itself .A group that is not simple can be broken into
two smaller groups ,a normal subgroup and quotient group ,and the process canbe
repeated.
Solvable group:
We say that a group G is Solvable if G has a series of subgroups-
Such that- {e} = H0 ⊂ H1 ⊂ H2 ⊂ … ⊂ Hk = G.
Where for each 0≤ i < k, Hi is normal in Hi+1 and Hi+1/Hi is Abelian.
Nilpotent group:
In mathematics ,more specifically in the field of a group theory, a nilpotent group
is a group that is “almost Abelian”. This idea is motivated by the fact that nilpotent

34. groups are solvable, and for finite nilpotent groups, two elements having relatively
prime order must commute. It is also true that finite nilpotent groups are super
solvable.
For a nilpotent group, the smallest n such that G has a central series of length nis
called the nilpotency class of G and G is said to be nilpotent group of class n(By
definition ,the length is n if n+1 different subgroups in the series, including the
trivial subgroup and the whole group).Every Abelian group is Nilpotent.
Maximal subgroup:
A subgroup M of a group to be a Maximal subgroup of a group (G, o) if-
 If M ≠ G.
 There is no propersubgroup of G containing M properly.
P-Groups:
A group G is said to be a P-group iff |G| = Pn forsome n.
Syllow’s P-subgroup:
A maximal P-subgroup of G is called a syllow’s P-subgroup of G.
Syllow’s theorem:
Let G be a finite group and Pr divides |G| where P is prime then G has a subgroup
of order Pr.

35. Introduction
The notion of the group originated from the theory of substitution, which was developed
by LAGRANGE VANDEMODE and GAUSS in the process ofsolving algebraic
equations, LAGRANGE introduced the notion of commutative binary operation in the set
of quadratic forms of the same discriminant. On the other hand in the set of permutations
of a finite set the notion of productof two permutations was defined by KUFFINI and
later by CAUCHY and gave conceptof a finite group of permutations. GALOIS are some
fundamental notions of group theory like Normal Groups and Isomorphism of group[1].
C.JORDAN developed the work of GALOIS in his famous book[3].
He made a deep study of some special groups such as linear groups. He has also the
credit of giving the notion of quotient groups. The study of infinite groups was also
initiated by JORDAN which was further developed few years later by SOPHUSLIE and
H.POINCARE. But it was CAYLEY in 1854 who gave for the first time the notion of
Abstract groups[2]. The theory of Abstract infinite groups was developed towards 1880, a
good account of which is found in BURNSIDE’s book [4].
References:
1. E Galois, Mathematical works (in French) ,Gauthier, Villars Paris 1897.
2. Cayley, collected Mathematical papers t ,II, Cambridge University , 1889.
3. C-Jordan Treatise of substitutions and algebraic equations (In French), Gauthier,
Villars Paris 1879
4. W. Burnside, Theory of groups of finite order, Cambridge 1911.

36. Acknowledgement
The present project report entitled “Studyof Symmetric Group” is the outcome of the
work done by me during May 2015 to July 2015 as a M.Sc. student at the “Uttar Pradesh
Textile Technology Institute (Kanpur)” under the guidance of Dr. Mukesh Kumar Singh,
(Head of Department of Basic Science).
I have no words to express my heartfelt gratitude to my guide Dr.
Mukesh Kumar Singh for his most valuable guidance and utmost care which he took
throughout my work. It is all due to his kind guidance , sympathetic attitude , and timely
advice , encouragement and inspiration that the project work undertaken by me has been
completed .
I am highly indebted to my project guide Dr. Mukesh Kumar Singh for
inspiring me to enhance my knowledge by perusing the innovation. His advice and
supporthelped me to complete the project.
I am grateful to family and friends for their constant supportand
encouragement in perusing my goals.
PRASHANT PATEL
M.Sc. –Istyear
Mathematics
Enrolment No. - 16768
VNIT, NAGPUR
(maharashtra , 440010).

37. STUDENT’S DECLARATION
We hereby declare that the work which is presented in the M.Sc. report , entitled “Study
of symmetric Groups” in partial fulfillment of the requirements for the award of Degree
of Master of Science in Industrial Mathematics and Informatics submitted to Department
of Mathematics , VNIT, Nagpur is the authentic record of our own work carried out
under the Guidance of Dr. MukeshKumar Singh , Head (Department of Basic
Science), Uttar PradeshTextile TechnologyInstitute , Kanpur.
Dated: …….
Place: …….. PrashantPatel.