International Journal Of Scientific & Engineering Research, Volume 7, Issue 7, July-2016 ISSN

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1 1510 A Cayley graphs for Symmtric group on Degree four S.V IJAY AKUMAR C.V.R.HARINARAY ANAN Research Scholar Department of Mathematics PRIST University ThanjavurTamilnadu India. Research Supervisor Assistant ProfessorDepartment of Mathematics Government Arts College ParamakudiTamilnaduIndia. July Abstract In this paper we determine all of subgroups of symmetric group S 4. First we observe the multiplication table of S 4 then we determine all possibilities of every subgroup of order n with n is the factor of order S 4. We found 30 subgroups of S 4. The diagram of Cayley graphs of S 4 is then presented. Keywords Perumutation - symmetric Group - Cayley graph. 1 Introduction For an arbitrary nonempty set S define A(S) to be the set of all one-to-one mapping of the set S onto itself. The set A(S) with composition function operation is a group.if the set S contains n elements then group A(S) are denoted by S n. Group S n has n! elements and will be called the symmetric group.there are many references on subgroups of S 2 and S 3. In this paper we determine all subgroups of S 4 and then draw diagram of Cayley graphs of S 4. The number of subgroup of cyclic groups of order p n where p is a prime number and this subgroups are finite cyclic groups. The subgroups of non abelian symmetric groups are S 2 S 3 S 4 and etc

2 1511 Thereforethe result of this paper that is a diagram of cayley graphs of S 4 is very important to determine the number of subgroup of S 4. 2 Preliminary Definition 2.1 A nonempty subset H of a group G is said to be a subgroupof G if under the product in G H itself forms a group. Theorem 2.2 If G is a finite group and H is a sub-group of G then order of H is a divisor of order G. Theorem 2.3 If G is a finite group and a G then order of a is a divisor of order G. Theorem 2.4 Let G be a finite group and let G = p n m where n 1 p is a prime number and (p m) = 1. Then G contains a subgroup of order p i for each i where 1 i n. Definition 2.5 Let G be a finite group and let G = p n m where n 1 p is a prime number and (p m) = 1. The subgroup of G of order p n is called the sylow p subgroup of G. Theorem 2.6 Let G be a finite group and let G = p n m where n 1 p is a prime number and (p m) = 1. Then the number of Sylow p subgroup is of the form (1 + kp) where k is a non-negative integer and (1 + kp) divides the order of G

3 1512 Definition 2.7 A subgroup N of G is said to be a normal subgroup of G if for every g G and n N gng 1 N. Theorem 2.8 There is a unique Sylow p -subgroup of the finite group G if only if it is normal. Theorem 2.9 Let G be a group of order pq where p and q are distinct primes and p < q. Then G has only one subgroup of order q. This subgroup of order q is normal in G. 3 Elements of Symmtric group Let A ( = { } Then S 4 consists of ) ( ) ( ) ( ) e = P 01 = P 02 = P 03 = P 04 = P 05 = P 06 = P 07 = P 08 = P 09 = P 10 = P 11 = P 12 = P 13 = P 14 = P 15 = P 16 = P 17 = P 18 = P 19 = P 20 = P 21 = P 22 = P 23 = In this group e is the identity element. Thus S 4 is a group containing 4! = 24 elements. 4 Cayley Table

4 1513 e P01 P02 P03 P04 P05 P06 P07 P08 P09 P10 P11 P12 P13 P14 P15 P16 P17 P18 P19 P20 P21 P22 P23 e e P01 P02 P03 P04 P05 P06 P07 P08 P09 P10 P11 P12 P13 P14 P15 P16 P17 P18 P19 P20 P21 P22 P23 P01 P01 e P03 P02 P05 P04 P07 P06 P09 P08 P11 P10 P19 P18 P21 P20 P23 P22 P13 P12 P15 P14 P17 P16 P02 P02 P05 P04 P01 e P03 P19 P18 P21 P20 P23 P22 P07 P06 P09 P08 P11 P10 P12 P13 P14 P15 P16 P17 P03 P03 P04 P05 e P01 P02 P12 P13 P14 P15 P16 P17 P06 P07 P08 P09 P10 P11 P19 P18 P21 P20 P23 P22 P04 P04 P03 e P05 P02 P01 P13 P12 P15 P14 P17 P16 P18 P19 P20 P21 P22 P23 P07 P06 P09 P08 P11 P10 P05 P05 P02 P01 P04 P03 e P18 P19 P20 P21 P22 P23 P13 P12 P15 P14 P17 P16 P06 P07 P08 P09 P10 P11 P06 P06 P07 P08 P09 P10 P11 e P01 P02 P03 P04 P05 P15 P16 P17 P12 P13 P14 P23 P20 P19 P22 P21 P18 P07 P07 P06 P09 P08 P11 P10 P01 e P03 P02 P05 P04 P20 P23 P22 P19 P18 P21 P16 P15 P12 P17 P14 P13 P08 P08 P11 P10 P07 P06 P09 P20 P23 P22 P19 P18 P21 P01 e P03 P02 P05 P04 P15 P16 P17 P12 P13 P14 P09 P09 P10 P11 P06 P07 P08 P15 P16 P17 P12 P13 P14 e P01 P02 P03 P04 P05 P20 P23 P22 P19 P18 P21 P10 P10 P09 P06 P11 P08 P07 P16 P15 P12 P17 P14 P13 P23 P20 P19 P22 P21 P18 P01 e P03 P02 P05 P04 P11 P11 P08 P07 P10 P09 P06 P23 P20 P19 P22 P21 P18 P16 P15 P12 P17 P14 P13 e P01 P02 P03 P04 P05 P12 P12 P13 P14 P15 P16 P17 P03 P04 P05 e P01 P02 P09 P10 P11 P06 P07 P08 P22 P21 P18 P23 P20 P19 P13 P13 P12 P15 P14 P17 P16 P04 P03 e P05 P02 P01 P21 P22 P23 P18 P19 P20 P10 P09 P06 P11 P08 P07 P14 P14 P17 P16 P13 P12 P15 P21 P22 P23 P18 P19 P20 P04 P03 e P05 P02 P01 P09 P10 P11 P06 P07 P08 P15 P15 P16 P17 P12 P13 P14 P09 P10 P11 P06 P07 P08 P03 P04 P05 e P01 P02 P21 P22 P23 P18 P19 P20 P16 P16 P15 P12 P17 P14 P13 P10 P09 P06 P11 P08 P07 P22 P21 P18 P23 P20 P19 P04 P03 e P05 P02 P01 P17 P17 P14 P13 P16 P15 P12 P22 P21 P18 P23 P20 P19 P10 P09 P06 P11 P08 P07 P03 P04 P05 e P01 P02 P18 P18 P19 P20 P21 P22 P23 P05 P02 P01 P04 P03 e P14 P17 P16 P13 P12 P15 P11 P08 P07 P10 P09 P06 P19 P19 P18 P21 P20 P23 P22 P02 P05 P04 P01 e P03 P08 P11 P10 P07 P06 P09 P17 P14 P13 P16 P15 P12 P20 P20 P23 P22 P19 P18 P21 P08 P11 P10 P07 P06 P09 P02 P05 P04 P01 e P03 P14 P17 P16 P13 P12 P15 P21 P21 P22 P23 P18 P19 P20 P14 P17 P16 P13 P12 P15 P05 P02 P01 P04 P03 e P08 P11 P10 P07 P06 P09 P22 P22 P21 P18 P23 P20 P19 P17 P14 P13 P16 P15 P12 P11 P08 P07 P10 P09 P06 P02 P05 P04 P01 e P03 P23 P23 P20 P19 P22 P21 P18 P11 P08 P07 P10 P09 P06 P17 P14 P13 P16 P15 P12 P05 P02 P01 P04 P01 e

5 Subgroups According to the nontrivial subgroups of S 4 must have order or 12. We will determine all of the subgroups of S 4. Clearly the subgroup of S 4 of order 1 is the trivial subgroup H 1 = {e}. Subgroups of order 2: Let H be an arbitrary subgroup of S 4 of order 2. Since 2 is a prime number then H is cyclic. Therefore H is generated by an element of S 4 of order 2. Thus all subgroups of S 4 of order 2 are H 2 = {e P 01 } H 3 = {e P 03 } H 4 = {e P 05 } H 5 = {e P 06 } H 6 = {e P 07 } H 7 = {e P 14 } H 8 = {e P 15 } H 9 = {e P 22 } H 10 = {e P 23 }. Subgroups of order 3: The subgroups of S 4 of order 3 is generated by an element of S 4 of order 3. Thus all subgroups of S 4 of order 3 are H 11 = {e P 02 P 04 } H 12 = {e P 08 P 13 } H 13 = {e P 09 P 12 } H 14 = {e P 10 P 19 } H 15 = {e P 11 P 18 } H 16 = {e P 16 P 20 } H 17 = {e P 17 P 21 }. Subgroups of order 4: Let H be an arbitrary subgroup of S 4 of order 4. then H is cyclic. Therefore H is generated by an element of S 4 of order 4. Thusall subgroups of S 4 of order 4 are H 18 = {e P 01 P 06 P 07 } H 19 = {e P 03 P 22 P 23 } H 20 = {e P 05 P 14 P 15 } H 21 = {e P 07 P 14 P 22 } H 22 = {e P 07 P 17 P 21 } H 23 = {e P 08 P 13 P 22 } H 24 = {e P 10 P 14 P 19 }. Subgroups of order 6: Let H be an arbitrary subgroup of S 4 of order 6. then H is cyclic. Therefore H is generated by an element of S 4 of order 6. Thus all subgroups of S 4 of order 6 are H 25 = {e P 01 P 02 P 03 P 04 P 05 } H 26 = {e P 01 P 15 P 16 P 20 P 23 }. Subgroups of order 8: Let H be an arbitrary subgroup of S 4 of order 8. then H is cyclic. Therefore H is generated by an element of S 4 of order 8. Thus all subgroups of S 4 of order 8 are

International Journal Of Scientific & Engineering Research Volume 7 Issue 7 July-2016 1515 H27 = {e P01 P06 P07 P14 P17 P21 P22 } H28 = {e P03 P07 P08 P13 P14 P22 P23 } H29 = {e P05 P07 P10 P14 P15

6 International Journal Of Scientific & Engineering Research Volume 7 Issue 7 July H27 = {e P01 P06 P07 P14 P17 P21 P22 } H28 = {e P03 P07 P08 P13 P14 P22 P23 } H29 = {e P05 P07 P10 P14 P15 P19 P22 }. Subgroups of order 12: Obviously the alternating group A4 = H30 = {e P02 P04 P07 P09 P11 P12 P14 P16 P18 P20 P22 }. is a subgroup of S4 of order 12. We will prove that A4 is the unique subgroup of S4 of order 12. According to this result we have the diagram of cayley graphs diagram is figure 1 below. Figure 1: Cayley graphs of S4 References [1] I.N.Herstein. Topic in Algebra ( John Wiley and SonsNew York1975). [2] J.B.Fraleigh. A First Course in WesleyLondon1992) Abstract Algebra (Addison-

Lagrange s Theorem. Philippe B. Laval. Current Semester KSU. Philippe B. Laval (KSU) Lagrange s Theorem Current Semester 1 / 10

Lagrange s Theorem. Philippe B. Laval. Current Semester KSU. Philippe B. Laval (KSU) Lagrange s Theorem Current Semester 1 / 10 Lagrange s Theorem Philippe B. Laval KSU Current Semester Philippe B. Laval (KSU) Lagrange s Theorem Current Semester 1 / 10 Introduction In this chapter, we develop new tools which will allow us to extend