The Structure of the Lattice of Subgroups of the Symmetric Group S5

International Journal of Statistics and Applied Mathematics 2018; 3(2): 652-663

A Vethamanickam A Vethamanickam and C Krishna Kumar Associate Professor, Department of Mathematics Rani Anna Government College Abstract for Women, Tirunelveli, In this paper we determine all the subgroups of the symmetric group S5 explicitly by applying Lagrange’s Tamil Nadu, India theorem, Sylow’s theorem. We give the diagram of the lattice of subgroups of S5. We also study some lattice theoretic properties of the lattice of subgroups of Sn, when n≤5. C Krishna Kumar Assistant Professor, Keywords: Symmetric group, Lagrange’s theorem, Sylow’s theorem, p-Sylow subgroup, lattice Department of Mathematics, St. Jerome’s College of Arts and 1. Introduction Science, Ananthanadarkudy, Nagercoil, Tamil Nadu, India In mathematics, the notion of permutation relates to act of arranging all the members of a set into some sequence or order or the set is already ordered or rearranging its elements a process called permuting. For any non-empty set S, define A(S) be the set of all bijections (permutations) on S. A(S) together with the binary operation of composition of functions is a group. If the set S contain n

elements, then the set A(S) is denoted by Sn. The group Sn has n! Elements and is called the symmetric group of degree n. Lattice of subgroups of S2, S3 and S4 have been studied already [2] [3] [5] [12] eg. Refer to , , and . In this paper we determine all the subgroups of S5, and then draw the diagram of the lattice subgroups of S5, also study some lattice theoretic properties. For the preliminaries and the basic definitions we refer to [2], [3], [4], [5] and [8].

2. Preliminary We recall some definitions and results that will be used later.

Definition 2.1

The symmetric group S5 is defined to be the group of all permutations on a set of five elements, ie, the symmetric group of degree five. In particular, it is a symmetric group of prime degree and it is denoted by S5.

Definition 2.2

A group generated by a single element is called cyclic and we know that cyclic groups are abelian.

Definition 2.3 A subgroup of G is said to be a normal subgroup of G if for every g∈G and n∈N, gng-1∈N.

Definition 2.4 A partial order on a non-empty set P is a binary relation denoted by ≤ on P that is reflexive, anti- symmetric and transitive. If ≤ is a partial order on P, the pair (푃, ≤) is called a partially ordered set or poset. A Poset (푃, ≤) is called totally ordered or a chain. If any two elements x, Correspondence C Krishna Kumar y ∈ P are comparable, that is either x ≤ y or y ≤ x. A non-empty subset S of P is a chain in P, Assistant Professor, if S is totally ordered by ≤. Department of Mathematics, St. Jerome’s College of Arts and Science, Ananthanadarkudy, Nagercoil, Tamil Nadu, India ~652~ International Journal of Statistics and Applied Mathematics

Definition 2.5 Let (푃, ≤) be a poset and let S⊆P. An upper bound for S is an element x ∈ P for which s≤x for all s ∈ S. The least upper bound of S is called the supremum or join of S. A lower bound for S is an element x ∈ P for which x ≤ 푠 for all s∈ S. The greatest lower bound of S is called the infimum or meet of S. Poset (푃, ≤) is called the lattice if every pair x, y elements of P has a Supremum and an Infimum, and denoted by x∨y and x∧y respectively.

Definition 2.6 The set of all subgroups of a given group G is a partially ordered set under ‘⊆’ inclusion relation. It is also a lattice denoted by L (G).

Definition 2.7

In the poset (P,≤), a covers b or b is covered by a (in notation, a≻b or b≺a) if and only if b < a and, for no x ∈ P, b < x < a.

Definition 2.8 Let G be a finite group and let |퐺| = pnm where n≥1, and p is a prime number such that (p, m) = 1. The subgroup of G of order pn is called a p-sylow subgroup of G.

Definition 2.9 An element ‘a’ is an atom, if a≻0 and a dual atom, if a≺1.

Definition 2.10 A lattice L is said to be modular lattice, if a ∨ (b ∧ c) = (a ∨ b) ∧ c for every a, b, c in L and a ≤ c.

Definition 2.11 A Lattice L is said to be semi-modular if whenever a covers a ⋀ b, then a ⋁ b covers b, for all a, b ∈ L.

Definition 2.12 An element of a lattice L is called join irreducible if x ⋁ y = a ⇒ x = a or y = a.

Definition 2.13 A Lattice L is said to be consistent if whenever j is a join-irreducible element in L, then for every x in L, x ⋁ j is join irreducible in the upper interval [x, 1].

Theorem 3.1 If G is a finite group and H is a subgroup of G, then the order of H is a divisor of the order of G.

Theorem 3.2 If G is a finite group and a ∈ G, then order of a is a divisor of order of G.

Theorem 3.3 Let G be a finite group and let p be any prime number that divides the order of G. Then G contains an element of order p.

Theorem 3.4 If P is a prime number and 푝훼 ∖o (G), 푝훼+1 ∤o (G), then G has a subgroup of order푝훼.

Theorem 3.5 If G is a finite group, p is a prime number and 푝푛 ∖o (G), but 푝푛+1 ∤o (G), then any two subgroups of G of order pn are conjugate.

Theorem 3.6 The number of p-sylow subgroups in G, for a given prime p, is of the form 1+kp.

Theorem 3.7 If a group G has a unique sylow p-subgroup, then it is normal.

Theorem 3.8 Let G be a group of order pq, where p and q are distinct primes and p

Theorem 3.9 If N is a normal subgroup of G and H is any subgroup of G, then NH is a subgroup of G.

Theorem 3.10 If M and N are two normal subgroups of G, then NM is also normal subgroup of G.

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Theorem 3.11 Sn has a normal subgroup of index 2 the alternating group An, consisting of all even permutation.

Notation If G is a group, then we denote the lattice of all subgroups of G by the notation L (G).

Result A non-empty subset H of the group G is a subgroup of G iff a, b ∈ H⇒ab∈H and a ∈H ⇒ a-1∈H. We produce below the structure of the lattice of subgroups of the symmetric groups S2, S3 and S4.

Fig 1: Lattice Structure of L (S2)

Fig 2: Lattice Structure of L (S3)

Fig 3: Lattice Structure of L (S4)

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4. Main Result The symmetric group S5 is the group of all permutations of the set S = {1, 2, 3, 4, 5}, we know that the order of S5 is 120. We name the elements according to their orders.

Let S5 = {i, 훼1, 훼2,… …, 훼10, 휎1, 휎2,… … , 훼20, 휏1, 휏2,……, 휏30, 훾1, 훾2……,훾15, 훽1, 훽2,……, 훽24, 훿1, 훿2,……, 훿20}, where ‘i’ is

The identity permutation. The cycles of length 2 are,

훼1=(4 5), 훼2=(3 5), 훼3=(3 4), 훼4=(2 5), 훼5=(2 3), 훼6=(2 4), 훼7=(1 5), 훼8=(1 4), 훼9= (1 3), 훼10= (1 2).

The cycles of length 3 are, 휎1=(1 2 3), 휎2=(1 3 2), 휎3=(1 2 4), 휎4=(1 4 2), 휎5=(1 2 5), 휎6=(1 5 2), 휎7=(1 3 4), 휎8=(1 4 3), 휎9=(1 4 5), 휎10=(1 5 4),

휎11=(1 3 5), 휎12=(1 5 3), 휎13=(2 3 4), 휎14=(2 4 3), 휎15= (2 3 5), 휎16=(2 5 3), 휎17= (2 4 5), 휎18= (2 5 4), 휎19= (3 4 5), 휎20= (3 5 4).

The cycles of length 4 are,

휏1=(2 3 4 5), 휏2=(2 5 4 3), 휏3=(2 3 5 4), 휏4=(2 4 5 3), 휏5=(2 4 3 5), 휏6=(2 5 3 4), 휏7=(1 2 3 4), 휏8=(1 4 3 2), 휏9=(1 2 3 5),

휏10=(1 5 3 2), 휏11=(1 2 4 3), 휏12=(1 3 4 2), 휏13=(1 2 4 5), 휏14=(1 5 4 2), 휏15=(1 2 5 3), 휏16=(1 3 5 2), 휏17=(1 2 5 4),

휏18=(1 4 5 2), 휏19=(1 3 4 5), 휏20=(1 5 4 3), 휏21=(1 3 5 4), 휏22=(1 4 5 3), 휏23=(1 3 2 4), 휏24=(1 4 2 3), 휏25=(1 3 2 5),

휏26=(1 5 2 3), 휏27=(1 4 3 5), 휏28=(1 5 3 4), 휏29=(1 4 2 5), 휏30=(1 5 2 4).

The elements which are products of two transpositions are,

훾1=(2 4)(3 5), 훾2=(2 5)(3 4), 훾3=(2 3)(4 5), 훾4=(1 3)(2 4), 훾5=(1 3)(2 5), 훾6=(1 4)(2 3), 훾7=(1 4)(2 5), 훾8=(1 5)(2 3),

훾9=(1 5)(2 4), 훾10=(1 4)(3 5), 훾11= (1 5)(3 4), 훾12=(1 2)(3 4), 훾13=(1 2)(3 5), 훾14=(1 3)(4 5), 훾15=(1 2)(4 5).

The cycles of length 5 are,

훽1=(1 2 3 4 5), 훽2=(1 3 5 2 4), 훽3=(1 4 2 5 3), 훽4=(1 5 4 3 2), 훽5=(1 2 3 5 4), 훽6=(1 3 4 2 5), 훽7=(1 5 2 4 3), 훽8=(1 4 5 3 2),

훽9=(1 2 4 5 3), 훽10=(1 4 3 2 5), 훽11=(1 5 2 3 4), 훽12=(1 3 5 4 2), 훽13=(1 2 4 3 5), 훽14=(1 4 5 2 3), 훽15=(1 3 2 5 4),

훽16=(1 5 3 4 2), 훽17=(1 2 5 4 3), 훽18=(1 5 3 2 4), 훽19=(1 4 2 3 5), 훽20=(1 3 4 5 2), 훽21=(1 2 5 3 4), 훽22=(1 5 4 2 3),

훽23=(1 3 2 4 5), 훽24=(1 4 3 5 2).

The elements which are the products of a 3 cycle and a transposition are,

훿1=(1 2 3)(4 5), 훿2=(1 3 2)(4 5), 훿3=(1 2 4)(3 5), 훿4=(1 4 2)(3 5), 훿5=(1 2 5)(3 4), 훿6=(1 5 2)(3 4), 훿7=(1 3 4)(2 5),

훿8=(1 4 3)(2 5), 훿9=(1 4 5)(2 3), 훿10=(1 5 4)(2 3), 훿11=(1 3 5)(2 4), 훿12=(1 5 3)(2 4), 훿13=(2 3 4)(1 5), 훿14=(2 4 3)(1 5),

훿15=(2 3 5)(1 4), 훿16=(2 5 3)(1 4), 훿17=(2 4 5)(1 3), 훿18=(2 5 4)(1 3), 훿19=(3 4 5)(1 2), 훿20=(3 5 4)(1 2).

In the following table we list the elements according to their order,

Table 1

Order Elements 1 i 2 훼1, 훼2,… …, 훼10, 훾1, 훾2……,훾15 3 휎1, 휎2,… … , 훼20 4 휏1, 휏2,……, 휏30 5 훽1, 훽2,……, 훽24 6 훿1, 훿2,……, 훿20

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5. Subgroups of S5 According to Lagrange’s theorem, the order of any non-trivial subgroups of S5 divides the order of S5. So we must have the subgroups of S5 of orders 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60 and 120.

Subgroup of order 1 Obviously, the only subgroup of S5 of order 1 is the trivial subgroup G1 = {i}.

Subgroups of order 2 Let H be an arbitrary subgroup of S5 of order 2, since 2 is a prime number, H is cyclic, Hence H is generated by an element of order 2. Thus all the subgroups of S5 of order 2 are,

H2 = {i, 훼1} = < 훼1 >, H3 = {i, 훼2}=< 훼2 >, H4 = {i, 훼3}=< 훼3 >, H5 = {i, 훼4} = < 훼4 >, H6 = {i, 훼5} = < 훼5 >, H7 = {i, 훼6} =

< 훼6 >, H8 = {i, 훼7} = < 훼7 >, H9 = {i, 훼8} = < 훼8 >, H10 = {i, 훼9}=< 훼9 >, H11 = {i, 훼10}= < 훼10 >, H12 = {i, 훾1}=< 훾1 >,

H13 = {i, 훾2}=< 훾2 >, H14 = {i, 훾3}=< 훾3 >, H15 = {i, 훾4}=< 훾4 >, H16 = {i, 훾5}=< 훾5 >, H17 = {i, 훾6}=< 훾6 >, H18 = {i, 훾7}=<

훾7 >, H19 = {i, 훾8}=< 훾8 >, H20 = {i, 훾9}=< 훾9 >, H21 = {i, 훾10}=< 훾10 >, H22 = {i, 훾11}=< 훾11 >, H23 = {i, 훾12}=< 훾12 >,H24 =

{i, 훾13}=< 훾13 >, H25 = {i, 훾14}=< 훾14 >, H26 = {i, 훾15}=< 훾15 >.

Subgroups of order 3 Let K be an arbitrary subgroup of order 3, since 3 is a prime number, K is cyclic and generated by an element of order 3. Since 3∖o(G), 32 ∤o(G), G has a 3- sylow subgroup of order 3. The number of 3-sylow subgroups is of the form 1+3k and 1+3k∖o(G), that is, 1+3k∖23×3× 5. Therefore 1+3k∖23×5, the possible values for k = 0, 1, 3. Therefore the maximum number of 3-sylow subgroups of order 3 is 10 when k = 3. Thus all the subgroups of S5 of order 3 are,

K27 = {i, 휎1, 휎2}=< 휎1 >=< 휎2 >, K28 = {i, 휎3, 휎4}=< 휎3 >=< 휎4 >,

K29 = {i, 휎5, 휎6}=< 휎5 >=< 휎6 >, K30 = {i, 휎7, 휎8}=< 휎7 >=< 휎8 >,

K31 = {i, 휎9, 휎10}=< 휎9 >=< 휎10 >, K32 = {i, 휎11, 휎12}=< 휎11 >=< 휎12 >,

K33 = {i, 휎13, 휎14}=< 휎13 >=< 휎14 >, K34 = {i, 휎15, 휎16}=< 휎15 >=< 휎16 >,

K35 = {i, 휎17, 휎18}=< 휎17 >=< 휎18 >, K36 = {i, 휎19, 휎20}=< 휎19 >=< 휎20 >.

Subgroups of order 4 Let L be the arbitrary subgroup of S5 of order 4, then the elements of L must have order 1, 2 or 4. If L contains an element of order 4, then L is generated by an element of order 4. Therefore the subgroups generated by an element of order 4 are,

L37 = {i, 휏1, 휏2, 훾1 }=< 휏1 >=< 휏2 >, L38 = {i, 휏3, 휏4, 훾2 }=< 휏3 >=< 휏4 >,

L39 = {i, 휏5, 휏6, 훾3 }=< 휏5 >=< 휏6 >, L40 = {i, 휏7, 휏8, 훾4 }=< 휏7 >=< 휏8 >,

L41 = {i, 휏9, 휏10, 훾5 }=< 휏9 >=< 휏10 >, L42 = {i, 휏11, 휏12, 훾6 }=< 휏11 >=< 휏12 >,

L43 = {i, 휏13, 휏14, 훾7 }=< 휏13 >=< 휏14 >, L44 = {i, 휏15, 휏16, 훾8 }=< 휏15 >=< 휏16 >,

L45 = {i, 휏17, 휏18, 훾9 }=< 휏17 >=< 휏18 >, L46 = {i, 휏19, 휏20, 훾10 }=< 휏19 >=< 휏20 >,

L47 = {i, 휏21, 휏22, 훾11 }=< 휏21 >=< 휏22 >, L48 = {i, 휏23, 휏24, 훾12 }=< 휏23 >=< 휏24 >,

L49 = {i, 휏25, 휏26, 훾13 }=< 휏25 >=< 휏26 >, L50 = {i, 휏27, 휏28, 훾14 }=< 휏27 >=< 휏28 >,

L51 = {i,휏29, 휏30, 훾15 }=< 휏29 >=< 휏30 >.

If L contains an element of order 2, then L does not contain an element of order 4. Therefore, the order of all the elements of L is 2, except the identity element. The table below shows the multiplications of all the combinations of elements of order 2.

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Table 2

∙ 훼1 훼2 훼3 훼4 훼5 훼6 훼7 훼8 훼9 훼10 훾1 훾2 훾3 훾4 훾5 훾6 훾7 훾8 훾9 훾10 훾11 훾12 훾13 훾14 훾15 훼1 i 휎20 휎19 휎18 훾3 휎17 휎10 휎9 훾14 훾15 휏5 휏6 훼5 훿17 훿18 훿9 휏29 훿10 휏30 휏27 휏28 훿19 훿20 훼9 훼10 훼2 휎19 i 휎20 휎16 휎15 훾1 휎12 훾10 휎11 훾13 훼6 휏2 휏1 훿11 휏25 훿15 훿16 휏26 훿12 훼8 휏20 훿20 훼10 휏19 휏19 훼3 휎20 휎19 i 훾2 휎13 휎14 훾11 휎8 휎7 훾12 휏4 훼4 휏3 휏23 훿7 휏24 훿8 훿13 훿14 휏22 훼7 훼10 훿19 휏21 훿20 훼4 휎17 휎15 훾2 i 휎16 휎18 휎6 훾7 훾5 휎5 휏3 훼3 휏4 훿18 훼9 훿16 훼8 휏10 휏14 훿15 훿6 훿5 휏9 훿17 휏13 훼5 훾3 휎16 휎14 휎15 i 휎13 훾8 훾6 휎2 휎1 휏6 휏5 훼1 휏12 휏16 훼8 훿15 훼7 훿13 훿16 훿14 휏11 휏15 훿2 훿1 훼6 휎18 훾1 휎13 휎17 휎14 i 훾9 휎4 훾4 휎3 훼2 휏1 휏2 훼9 훿17 휏8 휏18 훿14 훼7 훿4 훿13 휏7 훿3 훿18 휏17 훼7 휎9 휎11 훾11 휎5 훾8 훾9 i 휎10 휎12 휎6 훿11 훿5 훿9 훿12 휏15 훿10 휏17 훼5 훼6 휏21 훼3 훿6 휏16 휏22 휏18 훼8 휎10 훾10 휎7 훾7 훾6 휎3 휎9 i 휎8 휎4 훿3 훿7 훿10 휏11 훿8 훼5 훼4 훿9 휏13 훼2 휏19 휏12 훿4 휏20 휏14 훼9 훾14 휎12 휎8 훾5 휎1 훾4 휎11 휎7 i 휎2 훿12 훿8 훿1 훼6 훼4 휏7 훿7 휏9 훿11 휏28 휏27 휏8 휏10 훼1 훿2 훼10 훾15 훾13 훾12 휎6 휎2 휎4 휎5 휎3 휎1 i 훿4 훿6 훿2 휏24 휏26 휏23 휏30 휏25 휏29 훿3 훿5 훼3 훼2 훾14 훼1 훾1 휏6 훼6 휏3 휏4 휏5 훼2 훿12 훿4 훿11 훿3 i 훾3 훾2 휎11 훽23 훽24 훽8 훽7 휎12 휎4 훽22 훽5 휎3 훽6 훽21 훾2 휏5 휏1 훼4 훼3 휏6 휏2 훿6 훿8 훿7 훿5 훾3 i 훾1 훽15 휎7 훽3 휎8 훽16 훽4 훽14 휎6 휎5 훽1 훽2 훽13 훾3 훼5 휏2 휏4 휏3 훼1 휏1 훿10 훿9 훿2 훿1 훾2 훾1 i 훽20 휎12 휎9 훽19 휎10 훽11 훽10 훽18 훽9 훽17 휎2 휎1 훾4 훿18 훿12 휏24 훿17 휏11 훼9 훿11 휏12 훼6 휏23 휎12 훽14 훽17 i 휎17 훾12 훽20 훽13 휎11 훽16 훽19 훾6 훽18 휎18 훽15 훾5 훿17 휏26 훿8 훼9 휏15 훿18 휏16 훿7 훼4 휏25 훽22 휎8 훽9 휎18 i 훽21 휎7 훾13 훽12 훽11 훽24 훽10 훾8 휎17 훽23 훾6 훿10 훿16 휏23 훿15 훼8 휏7 훿9 훼5 휏8 휏24 훽21 훽2 휎10 훾12 훽24 i 휎15 휎9 훽1 휎16 훽23 훾4 훽3 훽4 훽22 훾7 휏30 훿15 훿7 훼8 훿16 휏17 휏18 훼4 훿8 휏29 훽5 휎7 훽18 훽17 휎8 휎16 i 훽8 훾15 휎15 훽20 훽6 훽19 훽7 훾9 훾8 훿9 휏25 훿14 휏9 훼7 훿13 훼5 훿10 휏10 휏26 훽6 훽13 휎9 훽16 훾13 휎10 훽5 i 휎13 훽15 휎14 훽7 훾5 훽8 훽14 훾9 휏29 훿11 훿13 휏13 훿14 훼7 훼6 휏14 훿12 휏30 휎11 훽1 훽10 휎12 훽9 훽4 훾15 휎14 i 훽12 휎13 훽11 훽2 훽3 훾7 훾10 휏28 훼8 휏21 훿16 훿15 훿3 휏22 훼2 휏27 훿4 휎3 훽15 훽11 훽13 훽10 휎15 휎16 훽14 훽9 i 훾14 훽12 휎4 훾11 훽11 훾11 휏27 휏19 훼7 훿5 훿13 훿14 훼3 휏20 휏28 훿6 훽23 휎5 훽19 훽18 훽21 훽22 훽17 휎13 휎14 훾14 i 휎6 훽20 훾10 훽24 훾12 훿20 훿19 훼10 훿6 휏12 휏8 훿5 휏11 휏7 훼3 훽8 휎6 훽12 훾6 훽11 훾4 훽7 훽6 훽10 훽9 휎5 i 휎19 훽5 휎20 훾13 훿19 훼10 훿20 휏10 휏16 훿4 휏15 훿3 휏9 훼2 휎4 훽4 훽20 훽19 훾8 훽2 훽18 훾5 훽3 휎3 훽17 휎20 i 훽1 휎19 훾14 훼9 휏20 휏22 훿18 훿1 훿17 휏21 휏19 훼1 훿2 훽7 훽3 휎1 휎17 휎18 훽1 훽6 훽5 훽2 훾11 훾10 훽8 훽4 i 휎2 훾15 훼10 훿20 훿19 휏14 훿2 휏18 휏17 휏13 훿1 훼1 훽24 훽16 휎2 훽14 훽22 훽23 훾9 훽15 훾7 훽13 훽21 휎19 휎20 휎1 i

We observe row by row of table-2. From the second row we conclude that

{훼1, 훼2}, {훼1, 훼3}, {훼1, 훼4}, {훼1, 훼6}, {훼1, 훼7}, {훼1, 훼8}, {훼1, 훾1}, {훼1, 훾2}, {훼1, 훾4}, {훼1, 훾5}, {훼1, 훾6}, {훼1, 훾7}, {훼1, 훾8},

{훼1, 훾9}, {훼1, 훾10}, {훼1, 훾11}, {훼1, 훾12}, {훼1, 훾13}

Are not subsets of L. Using similar argument to the other rows, we obtain the subgroups of order 4 as,

L52 = {i, 훼2, 훼6, 훾1}, L53 = {i, 훼3, 훼4, 훾2}, L54 = {i, 훼1, 훼5, 훾3}, L55 = {i, 훼6, 훼9, 훾4}, L56 = {i, 훼4, 훼9, 훾5}, L57 = {i, 훼5, 훼8, 훾6},

L58 = {i, 훼4, 훼8, 훾7}, L59 = {i, 훼5, 훼7, 훾8}, L60 = {i, 훼6, 훼7, 훾9}, L61 = {i, 훼2, 훼8, 훾10}, L62 = {i, 훼3, 훼7, 훾11}, L63 = {i, 훼3, 훼10,

훾12}, L64 = {i, 훼2, 훼10, 훾13}, L65 = {i, 훼1, 훼9, 훾14}, L66 = {i, 훼1, 훼10, 훾15}, L67 = {i, 훾4, 훾6, 훾12}, L68 = {i, 훾1, 훾2, 훾3}, L69 = {i,

훾5, 훾8, 훾13}, L70 = {i, 훾7, 훾9, 훾15}, L71 = {i, 훾10, 훾11, 훾14}.

Subgroups of order 5 Let M be an arbitrary subgroup of S5, since 5 is a prime number, the subgroup M is cyclic and is generated by an element of order 5. Since |퐺| = 23×3×5 and 5∖o (G), 52 ∤o (G), therefore G has 5- sylow subgroup of order 5. The number of 5-sylow subgroups of G of order 5 is of the form 1+5k and 1+5k∖o(G), that is, 1+5k∖23×3× 5. Therefore 1+5k∖23×3, the possible values for k = 0 and 1. Therefore, the maximum number of 5-sylow subgroup of order 5 is 6 when k = 1. Thus all the subgroups of S5 of order 5 are,

M72 = {i, 훽1, 훽2, 훽3, 훽4}= < 훽1 > = < 훽2 > = < 훽3 > = < 훽4 >

M73 = {i, 훽5, 훽6, 훽7, 훽8}= < 훽5 > = < 훽6 > = < 훽7 > = < 훽8 >

M74 = {i, 훽9, 훽10, 훽11, 훽12}= < 훽9 > = < 훽10 > = < 훽11 > = < 훽12 >

M75 = {i, 훽13, 훽14, 훽15, 훽16}= < 훽13 > = < 훽14 > = < 훽15 > = < 훽16 >

M76 = {i, 훽17, 훽18, 훽19, 훽20}= < 훽17 > = < 훽18 > = < 훽19 > = < 훽20 >

M77 = {i, 훽21, 훽22, 훽23, 훽24}= < 훽21 > = < 훽22 > = < 훽23 > = < 훽24 >.

~657~ International Journal of Statistics and Applied Mathematics

The subgroups of order 6 Let N be any arbitrary subgroup of S5 of order 6, since |푁| = 2×3, according to the third sylow’s theorem, N has exactly one subgroup of order 3 and this group is normal in N. Now we generate the elements of S5 of order 6, If N contains an element of order 6 then it generates N, there are 10 subgroups of order 6 contain the elements of order 2, order 3 and order 6 and is isomorphic to the cyclic group Z6, they are,

N78 = {i, 휎1, 휎2, 훿1, 훿2, 훼1}, N79 = {i, 휎3, 휎4, 훿3, 훿4, 훼2}, N80 = {i, 휎5, 휎6, 훿5, 훿6, 훼3}, N81 = {i, 휎7, 휎8, 훿7, 훿8, 훼4}, N82 = {i,

휎9, 휎10, 훿9, 훿10, 훼5}, N83 = {i, 휎11, 휎12, 훿11, 훿12, 훼6}, N84 = {i, 휎13, 휎14, 훿13, 훿14, 훼7}, N85 = {i, 휎15, 휎16, 훿15, 훿16, 훼8}, N86 = {i,

휎17, 휎18, 훿17, 훿18, 훼9}, N87 = {i, 휎19, 휎20, 훿19, 훿20, 훼10}.

If N does not contain an element of order 6, we get the other 10 subgroups of S5 of the elements order 2 and the elements of order 3. Therefore the subgroups are as follows,

N88 = {i, 휎1, 휎2, 훼5, 훼9, 훼10}, N89 = {i, 휎3, 휎4, 훼6, 훼8, 훼10}, N90 = {i, 휎5, 휎6, 훼4, 훼7, 훼10}, N91 = {i, 휎7, 휎8, 훼3, 훼8, 훼9}, N92 = {i,

휎9, 휎10, 훼1, 훼7, 훼8}, N93 = {i, 휎11, 휎12, 훼2, 훼7, 훼9}, N94 = {i, 휎13, 휎14, 훼3, 훼5, 훼6}, N95 = {i, 휎15, 휎16, 훼2, 훼4, 훼5}, N96 = {i,

휎17, 휎18, 훼1, 훼4, 훼6}, N97 = {i, 휎19, 휎20, 훼1, 훼2, 훼3}.

Also the other subgroups of order 6 of S5 by taking the product of the elements of S5 of order 2 and the elements of order 3. Therefore such subgroups are,

N98 = {i, 휎1, 휎2, 훾3, 훾14, 훾15}, N99 = {i, 휎3, 휎4, 훾1, 훾10, 훾13}, N100 = {i, 휎5, 휎6, 훾2, 훾11, 훾12}, N101 = {i, 휎7, 휎8, 훾2, 훾5, 훾7}, N102 = {i,

휎9, 휎10, 훾3, 훾6, 훾8}, N103 = {i, 휎11, 휎12, 훾1, 훾4, 훾9}, N104 = {i, 휎13, 휎14, 훾8, 훾9, 훾11}, N105 = {i, 휎15, 휎16, 훾6, 훾7, 훾10}, N106 = {i,

휎17, 휎18, 훾4, 훾5, 훾14}, N107 = {i, 휎19, 휎20, 훾12, 훾13, 훾15}.

Subgroups of order 7 3 3 4 Let Q be an arbitrary subgroup of S5, Since |퐺| = 2 ×3×5 and 2 ∖o(G), 2 ∤o(G), therefore the number of 2-sylow subgroups of G of order 8 is of the form 1+2k and 1+2k∖o(G), that is, 1+2k∖23×3× 5. Therefore 1+2k∖5×3, the possible values for k = 0, 1, 2 and 7. Therefore the maximum number of 2-sylow subgroup of order 8 is 15 when k = 7. Since there is no element of order 8, so every element of Q must have order 1, 2 and 4, such subgroups are got by the product of the elements of S5 of order 2 and the elements of order 4 and they are,

Q108 = { i, 휏1, 휏2, 훾1, 훾2, 훾3, 훼2, 훼6}, Q109 = { i, 휏3, 휏4, 훾1, 훾2, 훾3, 훼3, 훼4},

Q110 = { i, 휏5, 휏6, 훾1, 훾2, 훾3, 훼1, 훼5}, Q111 = { i, 휏7, 휏8, 훾4, 훾6, 훾12, 훼6, 훼9},

Q112 = { i, 휏9, 휏10, 훾5, 훾8, 훾13, 훼4, 훼9}, Q113 = { i, 휏11, 휏12, 훾4, 훾6, 훾12, 훼5, 훼8},

Q114 = { i, 휏13, 휏14, 훾7, 훾9, 훾15, 훼4, 훼8}, Q115 = { i, 휏15, 휏16, 훾5, 훾8, 훾13, 훼5, 훼7},

Q116 = { i, 휏17, 휏18, 훾7, 훾9, 훾15, 훼6, 훼7}, Q117 = { i, 휏19, 휏20, 훾10, 훾11, 훾14, 훼2, 훼8},

Q118 = { i, 휏21, 휏22, 훾10, 훾11, 훾14, 훼3, 훼7}, Q119 = { i, 휏23, 휏24, 훾4, 훾6, 훾12, 훼3, 훼10},

Q120 = { i, 휏25, 휏26, 훾5, 훾8, 훾13, 훼2, 훼10}, Q121 = { i, 휏27, 휏28, 훾10, 훾11, 훾14, 훼1, 훼9},

Q122 = { i, 휏29, 휏30, 훾7, 훾9, 훾15, 훼1, 훼10}.

Subgroups of order 8 Let P be any arbitrary subgroup of S5 of order 10, since |푃| = 2×5, according sylow’s theorem, P has exactly one subgroup of order 5, there is no element of order 10 in S5. Therefore P is not cyclic and P contains only elements of order 2 and order 5, by taking the product of subgroups of order 2 and order 5, we get the following subgroups of order 10 and they are,

P123 = {i, 훽1, 훽2, 훽3, 훽4, 훾2, 훾6, 훾9, 훾13, 훾14},

P124 = {i, 훽5, 훽6, 훽7, 훽8, 훾1, 훾7, 훾8, 훾12, 훾14, },

P125 = {i, 훽9, 훽10, 훽11, 훽12, 훾3, 훾5, 훾9, 훾10, 훾12},

P126 = {i, 훽13, 훽14, 훽15, 훽16, 훾2, 훾4, 훾8, 훾10, 훾15}, ~658~ International Journal of Statistics and Applied Mathematics

P127 = {i, 훽17, 훽18, 훽19, 훽20, 훾3, 훾4, 훾7, 훾11, 훾13},

P128= {i, 훽21, 훽22, 훽23, 훽24, 훾1, 훾5, 훾6, 훾11, 훾15}.

Subgroups of order 9 2 Let R be any arbitrary subgroup of S5 of order 12, since |푅| = 2 ×3, the elements of R must be of order 2, 3 or 6 as there is no element of order 12 in S5. We obtain 10 subgroups of S5 of order 12 is the combination of the elements of order 2, order 3 and order 6 are,

R129 = {i, 휎1, 휎2, 훿1, 훿2, 훾3, 훾14, 훾15, 훼1, 훼5, 훼9, 훼10},

R130 = {i, 휎3, 휎4, 훿3, 훿4, 훾1, 훾10, 훾13, 훼2, 훼6, 훼8, 훼10},

R131= {i, 휎5, 휎6, 훿5, 훿6, 훾2, 훾11, 훾12, 훼3, 훼4, 훼7, 훼10},

R132 = {i, 휎7, 휎8, 훿7, 훿8, 훾2, 훾5, 훾7, 훼3, 훼4, 훼8, 훼9},

R133 = {i, 휎9, 휎10, 훿9, 훿10, 훾3, 훾6, 훾8, 훼1, 훼5, 훼7, 훼8},

R134 = {i, 휎11, 휎12, 훿11, 훿12, 훾1, 훾4, 훾9, 훼2, 훼6, 훼7, 훼9},

R135= {i, 휎13, 휎14, 훿13, 훿14, 훾8, 훾9, 훾11, 훼3, 훼5, 훼6, 훼7},

R136= {i, 휎15, 휎16, 훿15, 훿16, 훾6, 훾7, 훾10, 훼2, 훼4, 훼5, 훼8},

R137 = {i, 휎17, 휎18, 훿17, 훿18, 훾4, 훾5, 훾14, 훼1, 훼4, 훼6, 훼9},

R138 = {i, 휎19, 휎20, 훿19, 훿20, 훾12, 훾13, 훾15, 훼1, 훼2, 훼3, 훼10},

Also we obtain the other 5 subgroups of S5 of order 12 of the combination of the elements of order 2 and order 3. Therefore the 5 subgroups of order 12 are,

R139 = {i, 휎1, 휎2,휎3, 휎4, 휎7, 휎8, 휎13, 휎14, 훾4, 훾6, 훾12},

R140 = {i, 휎1, 휎2, 휎5, 휎6, 휎11, 휎12, 휎15, 휎16, 훾6, 훾7, 훾10},

R141 = {i, 휎3, 휎4, 휎5, 휎6, 휎9, 휎10, 휎17, 휎18, 훾4, 훾5, 훾14},

R142 = {i, 휎7, 휎8, 휎9, 휎10, 휎11, 휎12, 휎19, 휎20, 훾12, 훾13, 훾15},

R143 = {i, 휎13, 휎14, 휎15, 휎16, 휎17, 휎18, 휎19, 휎20, 훾8, 훾9, 훾11}.

Subgroups of order 10 Subgroups of order 15 do not exist, since 15 = 3×5 and as 5≢1(mod3), a group of order 15 is cyclic, therefore it must be generated by an element of order 15, hence a subgroup of G of order 15 does not exist.

Subgroups of order 11 2 Let T be any arbitrary subgroup of S5 of order 20, since |푇| = 2 × 5, the number of 2-sylow subgroups of order 4 in T is 1+2k and 1+2k∖ 5, the possible values of k are 0 and 2. Hence the number of subgroups of T of order 4 is either 1 or 5, thus the maximum number of subgroups of order 4 is 5, similarly the number of 5-sylow subgroups of T of order 5 is 1+5k and 1+5k∖4, the possible values of k is 0 only. Hence the number of subgroups of order 5 is 1. Therefore the number of subgroups of T of order 4 is 5 and the number of subgroups of order 5 is 1. Therefore each subgroup of order 20 contains a unique subgroup of order 5 and each contains a subgroup of order 10. Hence there are 6 subgroups of S5 of order 20 are,

T144 = {i, 훽1, 훽2, 훽3, 훽4, 훾2, 훾6, 훾9, 훾13, 훾14, 휏3, 휏4, 휏11, 휏12, 휏17, 휏18, 휏25, 휏26, 휏27, 휏28},

T145 = {i, 훽5, 훽6, 훽7, 훽8, 훾1, 훾7, 훾8, 훾12, 훾14, 휏1, 휏2, 휏13, 휏14, 휏15, 휏16, 휏23, 휏24, 휏27, 휏28},

T146 = {i, 훽9, 훽10, 훽11, 훽12, 훾3, 훾5, 훾9, 훾10, 훾12, 휏5, 휏6, 휏9, 휏10, 휏17, 휏18, 휏19, 휏20, 휏23, 휏24},

T147 = {i, 훽13, 훽14, 훽15, 훽16, 훾2, 훾4, 훾8, 훾10, 훾15, 휏3, 휏4, 휏7, 휏8, 휏15, 휏16, 휏19, 휏20, 휏29, 휏30},

T148 = {i, 훽17, 훽18, 훽19, 훽20, 훾3, 훾4, 훾7, 훾11, 훾13, 휏5, 휏6, 휏7, 휏8, 휏13, 휏14, 휏21, 휏22, 휏25, 휏26},

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T149= {i, 훽21, 훽22, 훽23, 훽24, 훾1, 훾5, 훾6, 훾11, 훾15, 휏1, 휏2, 휏9, 휏10, 휏11, 휏12, 휏21, 휏22, 휏29, 휏30}.

Subgroups of order 12 3 Let V be any arbitrary subgroup of S5 of order 24, since |푉| = 2 × 3, the number of 2-sylow subgroups of order 8 in V is 1+2k and 1+2k∖3, the possible values of k are 0 and 1. Hence the number of subgroups of V of order 8 is either 1 or 3, thus the maximum number of subgroups of order 8 is 3, similarly the number of 3-sylow subgroups of order 3 in V is 1+3k and 1+3k∖8, the possible values of k are 0 and 1. Hence the number of subgroups of order 3 in V is either 1 or 4, thus the maximum number of subgroups of order 3 is 4, Therefore by multiplying a subgroups of order 8 with subgroups of order 3, we determine 5 subgroups of S5 of order 24 are,

V150 = {i, 훼3, 훼5, 훼6, 훼8, 훼9, 훼10, 휎1, 휎2,휎3, 휎4, 휎7, 휎8, 휎13, 휎14, 훾4, 훾6, 훾12, 휏7, 휏8, 휏11, 휏12, 휏23, 휏24},

V151 = {i, 훼1, 훼2, 훼3, 훼4, 훼5, 훼6, 휎13, 휎14,휎15, 휎16, 휎17, 휎18, 휎19, 휎20, 훾1, 훾2, 훾3, 휏1, 휏2, 휏3, 휏4, 휏5, 휏6},

V152 = {i, 훼1, 훼2, 훼3, 훼7, 훼8, 훼9, 휎7, 휎8,휎9, 휎10, 휎11, 휎12, 휎19, 휎20, 훾10, 훾11, 훾14, 휏19, 휏20, 휏21, 휏22, 휏27, 휏28},

V153 = {i, 훼1, 훼4, 훼6, 훼7, 훼8, 훼10, 휎3, 휎4,휎5, 휎6, 휎9, 휎10, 휎17, 휎18, 훾7, 훾9, 훾15, 휏13, 휏14, 휏17, 휏18, 휏29, 휏30},

V154 = {i, 훼2, 훼4, 훼5, 훼7, 훼9, 훼10, 휎1, 휎2,휎5, 휎6, 휎11, 휎12, 휎15, 휎16, 훾5, 훾8, 훾13, 휏9, 휏10, 휏15, 휏16, 휏25, 휏26}.

Subgroup of order 13 The subgroup of order 30 cannot exist since 30 = 2×3×5, by multiplying a subgroup of order 3 and subgroup of order 10 or multiplying a subgroup of order 5 and subgroup of order 6, ie) by finding KiPj or MiNj for all i and j we get in each case an element of order 4 which cannot exist in a subgroup of order 30.

Subgroup of order 14 The subgroup of order 40 also cannot exist, since 40 = 23×5 = 4 × 10 by multiplying a subgroup of order 5 and subgroup of order 8 that is by finding MiQj for all i and j we get elements of order 3 or elements of order 6 which cannot exist in a subgroup of order 40. Also multiplying a subgroup of order 4 and a subgroup of order 10, that is by finding LiPj for all i and j, we get elements of order 3 or 6 which cannot exist in a subgroup of order 40.

Subgroup of order 15 The only subgroup of S5 of order 60 is the alternating group A5 consisting of all even permutations in S5. Therefore the only one subgroup of order 60 is,

A5 = {i, 휎1, 휎2, 휎3, … … , 훼20, 훾1, 훾2, 훾3, ……, 훾15, 훽1, 훽2, 훽3, ……, 훽24}.

Subgroup of order 16 Every group is a subgroup of itself; hence the whole group S5 is a subgroup of S5 of order 120. According these results, we have the diagram of lattice subgroups of S5 is shown below,

Fig 3 ~660~ International Journal of Statistics and Applied Mathematics

Row I: (Left to right): R129 to R138, V150 to V154, R139 to R143 and T144 to T149 Row II: (Left to right): N78 to N87, N98 to N107, N88 to N97, Q108 to Q122 and P123 to P128 Row III: (Left to right): K27 to K36, L37 to L51, L52 to L66, L67 to L71 and M72 to M77 Row IV: (Left to right): H2 to H11 and H12 to H26.

6. Lattice Identities in the Subgroup Lattices of the Symmetric Group Sn 6.1 Consistency in Lattices of Subgroups of the Symmetric Group Sn Lemma: 6.1 L (Sn) is consistent if n = 2 and n = 3 and L (Sn) is not consistent if n = 4 and n = 5. Proof: From the Fig.1 and Fig.2, it is easily seen that whenever j is a join-irreducible element in L(Sn), then x ∨ j is join-irreducible in the upper interval [x,1] for every x∈L(Sn) when n ≤ 3. So, L (Sn) is consistent, when n ≤ 3. When n = 4, We choose the join irreducible element K2 ∈ L (S4), and now K2⋁ K5 = S4= M21∨N24 in the upper interval [K5, S4]. So L (S4) is not consistent.

Fig 4

Fig 5 ~661~ International Journal of Statistics and Applied Mathematics

When n = 5, Now we choose join irreducible element L27 ∈ L (S5), and K27⋁P97 = S5 = T138∨V151 in the upper interval [P97, S5]. So L (S5) is not consistent.

Fig 6

Fig 7

6.2 Modularity in Lattices of Subgroups of Symmetric Group Sn Lemma: 6.2 L (Sn) is modular if n = 2 and n = 3 and L (Sn) is not modular if n = 4 and n = 5. Proof: Since L (S2) is isomorphic to B1, the 2-element chain, it is modular. So that L (S2) is modular. Now L (S3) is isomorphic to M4, it is modular, so that L (S3) is modular. But L (S4) is not modular, since we know that N5 is not modular and it contains the sublattice {{e}, K4, M18, L11, A4} which is isomorphic to the pentagon N5. Hence L (S4) is not modular. Also, L (S5) is not modular, since we know that N5 is not modular and it contains the sublattice {{e}, K27, H23, L67, R139} which is isomorphic to the pentagon N5. Hence L (S5) is not modular.

~662~ International Journal of Statistics and Applied Mathematics

6.3 Semi-Modularity in Lattices of Subgroups of Symmetric Group Sn Lemma: 6.3 L (Sn) is semi-modular if n = 2 and n = 3 and L (Sn) is not semi-modular if n = 4 and n = 5

Proof: We know that L (S2) and L (S3) are modular, so they are also semi- modular. When n = 4,

In L (S4), there are two elements K5 and L11 such that K5 ∧L11 = {e} which is covered by K5 while K5 ∨ L11 = S4 which does not cover L11. Therefore, L (S4) is not semi-modular. When n = 5,

In L (S5), there are two elements K27 and M77 such that K27 ∧ M77 = {e} which is covered by K27 while K27 ∨ M77 = S4 which does not cover M77. Therefore, L (S5) is not semi-modular.

7. Conclusion Consistency of L (Sn), Modularity of L (Sn), Semi-Modularity of L (Sn) for n ≤ 5 were investigated in this section. But to decide the properties for n > 5 seems to be difficult.

8. References 1. Baumslag B. Theory and Problems of Group Theory: Schaum’s outline Series, McGraw-Hill, New York, 1968. 2. Fraleigh JB. A First Course in Abstract Algebra, Addison-Wesley, London, 1992. 3. Gardiner CF. A First Course in Group Theory, Springer-Verlag, Berlin, 1997. 4. Gratzer G. Lattice Theory: Foundation. Birkhauser Veslag, Basel, 1998. 5. Herstien IN. Topics in Algebra, John Wiley and sons, New York, 1975. 6. Herstien IN. Abstract Algebra, Third Edition, Prentice Hall Inc. Upper Saddle River, 1996. 7. Jebaraj Thiraviam D. A Study on some special types of lattices, Ph.D thesis, Manonmaniam Sundaranar University, 2015. 8. Ralph Mckenzie N, George Mcnulty F, Walter Taylor F. Algebras Lattices, Varieties, Wadsworth and Brooks/Cole, Monterey California, 1987, I. 9. Roman S. Lattice and Ordered set, Springer, New York, 2008. 10. Samila D. Counting the Subgroups of the One Headed Group S5 upto Automorphism, IOSR Journal of Mathematics. 2013; 8(3):87-93. 11. Scott WR. Group Theory, New York, Dover, 1987. 12. Sulaiman R. Subgroups Lattice of Symmetric Group S4; International Journal of Algebra. 2012; 6(1):29-35. 13. Vethamonickam A, Jebaraj Thiraviam. On Lattices of Subgroups, Int. Journal of Mathematical Archiv. 2015; 6(9):1-11.

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