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CONTENTS

CHAPETER I INTRODUCTION AND PRELIMINARIES GROUPS

1.1 Introduction1.2 Preliminaries

CHAPTER II MBIUS INVERSION THEOREM OVER A LATTICEOF SUB2.1 Definitions and some properties of group actions2.2 Mbius inversion theorem over a lattice of subgroups

CHAPTER III SOME CONGRUENCES ON BINOMIALCOEFFICIENTS

AND STIRLING NUMBERS OF BOTH KINDS IN

THE CASEG Cp

3.1 Mbius function defined on ( )PL C

3.2 Congruence relation on Binomial coefficients in the case pG C=3.3 Congruence relations on Stirling numbers of second kind in thecase

pG C=3.4 Congruence relations on Stirling numbers of first kind in the case

pG C=

CHAPTER IV SOME CONGRUENCES ON BINOMIALCOEFFICIENTS

AND STIRLING NUMBERS OF BOTH KINDS INTHE CASE

G Cp Cp

4.1 Mbius function defined on ( )p pL C C 4.2 Congruence relation on Binomial coefficients in the case

p pG C C= 4.3 Congruence relation on Stirling numbers of second kind in the

casep pG C C=

4.4 Congruence relation on Stirling numbers of first kind in the case

p pG C C=

CHAPTER V SOME CONGRUENCES ON BINOMIALCOEFFICIENTS

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AND STIRLING NUMBERS OF BOTH KINDS INTHE

GENERAL CASE rpG C=

5.1 Mbius function defined on ( )rpL C

5.2 Congruence relation on Binomial coefficients in the caser

pG C=5.3 Congruence relation on Stirling numbers of second kind in the

caser

pG C=

5.4 Congruence relation on Stirling numbers of first kind in the caser

pG C=

CHAPTER I

INTRODUCTION AND PRELIMINARIES

1.1 INTRODUCTION

In 1872, J. Peterson[19] used the action of the cyclic group oforder p to prove the congruences of Fermat and Wilson. In 1980, Rota

and Sagan [24] investigated congruences derived from group acting on

functions, Gessel[11] has studied congruences for groups acting on

graphs and Bruce E. Sagan[26] has derived congruences for group

acting on a set of subsets, partitions or permutations. Here we have

made a little attempt to survey some of congruences for binomial

coefficients, Stirling numbers of second kind, Stirling numbers of first

kind and some other related congruences by using action of finite

abelian groups in different cases.

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In this dissertation for notational convenience we let

[ ] { }1,2,...,n n= and [ ] { }1, 2,...,m n m m m n+ = + + + . We take G to be the

abelian subgroup of the symmetric group which is a product of cyclic

groups 1 2 ... rm m mC C C , where rmC is generated by the cycle

( )1, 2,..., j J J J m+ + + , ii j

J m = .

(4) 0n

k

=

, if 0, 0n k= > .

Property1.2.23 [21]: The general formula for computing the Stirling

numbers of the second kind is

( )0

1( 1)

!

kni

i

n kk i

k ik =

=

, for all positive integers n and k.

Proof: Finding an ordered partition of n elements into k blocks is

equivalent to finding onto functions from [ ]n to [ ]k . There are nk ways

to assign each of the distinct n elements a block. Then we subtract

and add ( )nk

k ii

respectively until we have counted all possible onto

functions, which is a standard application of the principle of inclusion-

exclusion and is exactly what the property states. This gives an

explicit formula forn

k .

Property1.2.24 (Recurrence relation) [21]:

Stirling numbers of the second kind obey the recurrence relation

given by

1 1

1

n n nk

k k k

= +

, if n > 0 , k > 0 .

Proof: Observe that a partition of the n objects into k nonempty

subsets either contains the subset { }n or it does not. The number of

ways that { }n is one of the subsets is given by

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1

1

n

k

.

Since we must partition the remaining 1n objects into the

available 1k subsets. The number of ways that { }n is not one of thesubsets (that is, n belongs to a subset containing other elements) is

given by

1nk

k

.

Since we partition all elements other than n into ksubsets, andthen are left with kchoices for inserting the element n.

Summing these two values gives the desired result.

Property1.2.25 [21]: The generating function for the Stirlingnumbers of the second kind is

0

( )n

n

k

k

nx x

k=

=

,

where ( )kx ( 1)( 2)...( 1) x x x k = + is falling factorial polynomial.

Traditionally, this is the way in which Stirling numbers of second kindare introduced.

Property1.2.26 []: 11

n =

, since there is only one way to partition a

set ofn elements into one block.

Property1.2.27 []:1 2

n n

n

=

, this is because dividing n elements

into 1n sets necessarily means dividing it into one set of size 2 and

2n sets of size 1. Therefore we need only to pick those two

elements from n elements.

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Property1.2.27 []: 1n

n

=

, since there is only one way to partition a

set of n elements into n blocks.

Definition 1.2.38 (Bell numbers) [26]: The Bell numbers are

denoted by ( )B n or nB and defined as ( )k

nB n

k

=

, which counts all

partitions of[ ]n , heren

k

is a Stirling number of second .

Examples1.2.35 []:

(1). (3) 5B = , since there are five ways the numbers { }1,2,3 can be

partitioned: { } { } { }{ }1 , 2 , 3 , { } { }{ }1, 2 , 3 , { } { }{ }1,3 , 2 , { } { }{ }1 , 2,3 ,and { }{ }1,2,3 .

(2). (4) 15B = , since there are 15 ways the numbers { }1,2,3,4 can be

partitioned:{ } { } { } { }{ } { } { }{ } { } { }{ }1 , 2 , 3 , 4 , 1, 2,3 , 4 , 1,2, 4 , 3 ,

{ } { }{ } { } { }{ }1,3,4 2 , 2,3, 4 1 { } { }{ } { } { }{ } { } { }{ }1,3 , 2,4 , 1, 4 , 2,3 , 1, 2 , 3, 4 , { } { } { }{ }1 , 2 , 3,4 ,

{ } { } { }{ }1 , 3 , 2,4 , { } { } { }{ }1 , 4 , 2,3 , { } { } { }{ } { } { } { }{ }2 , 3 , 1, 4 , 2 , 4 , 1,3 , { } { } { }{ }3 , 4 , 1,2 ,

and { }{ }1,2,3,4 .

The diagram below shows the constructions giving (3) 5B = and(4) 15B = , with line segments representing elements in the same subset

and dots representing subsets containing a single element.

(3) 5B =

(4) 15B =

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Definition 1.2.36[]: The sign less Stirling numbers of the first kind

count the numbers of permutations of n elements with k disjoint

cycles. Singles Stirling numbers of first kind are denoted by ( , )c n k [] or

n

k

[] or 1( , )S n k [].

That is [ ]{ }:n

isthe permutationof n with k disjoint cyclesk

=

, for ,k n and

1 k n .

Example 1.2.37[]:

(1) The list {1, 2, 3, 4} can be permuted into two cycles in the

following ways:

{{1,3,2},{4}}

{{1,2,3},{4}}

{{1,4,2},{3}}

{{1,2,4},{3}}

{{1,2},{3,4}}

{{1,4,3},{2}}

{{1,3,4},{2}}

{{1,3},{2,4}}

{{1,4},{2,3}}

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{{1},{2,4,3}}

{{1},{2,3,4}}

There are 11 such permutations, thus4

112

=

.

(2) Here are some illegible diagrams showing the cycles for

permutations of a list with five elements.

524

1 = :

5

353

=

:

5

104

=

:

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5

1

5

=

:

Remark 1.2.38: The following is a table for the first few values of sign

less Stirling numbers of the first kind.

n \ k 0 1 2 3 4 5 6 7 8 9

0 1

1 0 1

2 0 1 1

3 0 2 3 1

4 0 6 11 6 1

5 0 24 50 35 10 1

6 0 120 274 225 85 15 17 0 720 1764 1624 735 175 21 1

8 0 5040 13068 13132 6769 1960 322 28 1

9 0 4032010958

411812

46728

42244

9 4536 546 36 1

Definition 1.2.39[]: The normal (or signed) Stirling numbers of the

first kind is denoted by ( , )s n k [] or ( )knS [] ork

nS [], and is denoted as,

( , ) ( 1)n k nk

s n k =

.

Example 1.2.40: 4 2(4,2) ( 11)4

12

s

=

= .

51

5

=

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Remark 1.2.41: The following is a table for the first few values of

normal Stirling numbers,

n \ k 0 1 2 3 4 5 6 7 8 9

0 1

1 0 1

2 0 1 1

3 0 2 3 1

4 0 6 11 6 1

5 0 24 50 35 10 1

6 0 120 274 225 85 15 1

7 0 720 1764 1624 735 175 21 1

8 0

504

0 13068

1313

2 6769

196

0 322 28 1

9 0 40320

109584

118124

67284

22449

4536 546

36 1

Properties1.2.42 []: The following are some basic results on signed

and sign less Stirling numbers of first kind.

(1). 00

n = for 1n .

Since, there are no ways to arrange n objects in zero cycles.

(2). 1n

n

=

for 0n .

Since, there is only one way to assign n objects into n cycles, each

object in its own cycle.

(3). ( 1)1

nn

= !

, for 1n .

Since, if S = { : is a permutation on [ ]n with 1 cycle}

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= { : is a cycle on [ ]n and its length n},

Then S = the number of n length cycles on[ ]n

= ( 1)n ! .

Therefore by the definition 1.2.34, we have

1

nS

=

Hence,1

n

( 1)n= !.

(4).1 2

n n

n

=

.

Since, if S = { : is a permutation on [ ]n with ( 1)n disjoint

cycles}, then it follows that every element is given its own cycle

except for one pair of elements, which is placed in a 2-cycle. Order is

irrelevant for entire case, so we only need to select the two elements

of the same cycle. Therefore we select2

n

ways to select permutation

of S.

That is S =2

n

.

By the definition 1.2.34, we have

1

nS

n

=

Hence, 1 2

n n

n

= .

(5). 0n

k

=

if k n> .

Since there are no ways to arrange n objects in k cycles ifk n> .

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(6). The generating function for stirling numbers of second kind is

given by

0( )

nk

n

k

n

x xk=

= ,

where, ( ) ( 1)( 2)...( 1)n x x x x x n= + is falling factorial polynomial.

That is, the signed stirling numbers of the first kind are the

coefficient of kx in the expansion of the falling factorial polynomial ( )nx

.

(7). The recurrence relation for stirling numbers of the first kind is

given by1 1

( 1)1

n n nn

k k k

= +

for all 1n k ,

(8). 00

kk

=

, where 00 0

1 0k

if k

if k

= =

is kronecker delta.

.

Definition 1.2.42[]: A relation on a set L is said to be a partial

orderon L if it is reflexive, antisymmetric and transitive.

That is, (i) a a , for all a L (reflexive),

(ii) if a b and b a , then a b= , for a,b L

(antisymmetric), and

(iii) if a b and b c then a c , for a, b, c L (transitive).

Definition 1.2.43[]: If the relation on a set L is a partial order

then ( , )L is called a partially ordered setor simply a poset.

Definition 1.2.44[]: A partial order on a set L is said to be total

orderif for every pair of elements ,x y L , we have either x y or y x

.

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Definition 1.2.45[]: If the relation is a total on a set L , then we

say that ( , )L is a totally ordered setor chain.

Definition 1.2.46[]: A partially ordered set ( , )L is called a latticeif

every pair of elements in L has a least upper bound x y (x join y) and

greatest lower bound x y (x meat y).

That is , (i) For every ,x y L , there is a unique element denoted by

x y called as least upper bound, with the property that z x and z y

u implies z x y , and

(ii) For every,x y L

, there is a unique element denoted byx y called as greatest lower bound, with the property that z x and

z y implies z x y .

Example1.2.47: ( )L G , the lattice of subgroupsof a group G is the

lattice whose elements are the subgroups ofG, with the partial order

relation being set inclusion. In this lattice, the join of two subgroups is

the subgroup generated by their union, and the meet of two subgroupsis their intersection.

Definition 1.2.48[]: The Hasse diagram (or Lattice diagram) of

partially ordered set (or lattice) L is the directed graph with vertex set

L and an edge form xup to yify covers x(i.e.,x< yand there is no z

such that x z y< < ).

Example1.2.49: The following diagrams are lattice diagrams of two

lattices namely, the devisors of 35 and the set of all subsets of three

element set { }1,2,3 .

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CHAPTER II

MOBIUS INVERSION THEOREM OVER A

LATTICE OF SUBGROUPS

2.1. Definitions and some properties of action ofgroups:

In this section, the definitions of group actions, stabilizer, orbit

and apereodic elements are given. Some basic properties are studies

from [].

Definition 2.1.1: Let G be a finite group with identity element e, and

S is a finite set. Then G is said to be acts on S if there is a mapping

: G S S , with ( , )g s g s = such that,

(i) ( ) ( ) g h s gh s = , for all , ,g h G s S and

(ii) e s s = , for all s S .

Example 2.1.2: Let ( ),.G be a finite group. Define .g s g s = ,

,g G s G . Then G acts on G it self.

Because, (i) ( ) *( . ) .( . ) ( . ). ( . )g h s g h s g h s g h s g h s = = = = , for all

, ,g h s G ,

(ii) * .e s e s s= = , for all s G .

This action of the group G on itself is called translation.

Example 2.1.3: Let ( ),.G be a finite group. Define 1. .g s g s g = ,

,g G s G . Then G acts on G it self.

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Because, (i)

1 1 1 1( ) *( . . ) .( . . ). ( . ). .( . ) ( . )g h s g h s h g h s h g g h s g h g h s = = = = , for all , ,g h s G .

(ii) 1* . .e s e s e s= = , for all s G .

This action of the group G on itself is called conjugation.

Definition 2.1.4: Let G be a finite group with identity element e and

S is a finite set on which G acts. Givens S , the set { }: *sG g G g s s= =

is called the stabilizerofs.

Lemma 2.1.5: sG , the stabilizer ofs, is a subgroup ofG.

Proof: Clearly, sG . Since se G , and sG is a subset of finite group G.

Let , sg h G . Then by using condition (i) of the definition 2.1.1,

we have,

( )* *( * ) gh s g h s=

* ,g s= since sh G

,g= since sg G .

Hence s gh G .

This proves that sG is a subgroup ofG.

Definition 2.1.6: Let G be a finite group with identity element e andS is a finite set on which G acts. Given s S , the set { : * ,sO t S t g s= =

for some }g G is called the orbitofs.

Remark 2.1.7: sO , the orbit ofs S , is a subset ofS.

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Theorem 2.1.8: Let G be a finite group with identity element e, and

let S be a finite set on wish G acts. Then

s

s C

S O

= U (disjoint),

where C is any sub set ofS containing exactly one element from each

orbit.

Proof: For any ,r s S , we define a relation : on S by r s: iff r g s=

, for some g G .

Now by using (i) and (ii) conditions of definition 2.1.1, we have,

(1) * s e s= , for all s S .

Which imply that s s: for alls S .

Therefore the relation : is reflexive.

(2) If r g s= , for some g G , then

1 1* *( * )g r g g s =

1( )*g g s=

*e s=

s= .

That is, if r g s= , then 1 * s g r = , for some g G .

Which imply that, if r s: then s r: .

Therefore the relation is symmetric.

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(3) If r g s= and s h t = , then

*( )r g h t =

( )* gh t = , for any ,g h G .

That is, if r s: and s t: , then r t: .

Therefore the relation is transitive.

Hence the relation : is an equivalence relation on S, and hence

this equivalence relation partition S into mutually disjoint equivalence

classes. The equivalence class of an element s S is the orbit sO , the

orbit ofs S . Because, the equivalence class of s S ,

[ ] { }:s t S t s= : { : * ,t S t g s= = for some }g G = sO .

Hence, the set of all orbits is a partition ofS.

Therefore, ss C

S O

= U (disjoint), where C is any sub set ofS

containing exactly one element from each orbit.

Theorem 2.1.9: Let G be a finite group with identity element e, and S

be a finite set on which G acts.

Then ss

GO

G= , where | . |, denoted cardinality.

Proof: For a given s S , define a mapping s s

G

O G : by

( * ) s g s gG = , for all g G .

First we prove that is well defined.

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Suppose, for any ,g h G , * *g s h s= . Then by using (i) and (ii)

conditions of definition 2.1.1, we have 1 1( )* *( * )g h s g h s =

1 *( * ) g g s=

*e s=

s= .

Hence, 1 sg h G .

This shows that, s s gG hG= .

Therefore is well defined.

Now we prove is injective.

Suppose for any ,g h G , we have ( * ) ( * ) g s h s = . Then we get

s s gG hG= or1

sg h G . That is 1( )*g h s s = . Then in view of definition

2.1.1 we have,

( )1* * *g s g g h s =

( )1( ) *g g h s =

( ) *eh s=

*h s= .

This shows that is injective.

Since, for any left coset sgG , there exists an element such that

* sg s O such that ( * ) s g s gG = .

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This shows that is surjective.

Therefore is bijective.

Hence, ss

GOG

=

s

G

G=

.

Definition 2.1.10: Let G be a finite group with identity element e,

and S is a finite set on which G acts. An element s S is said to be an

aperiodic elementif the stabilizer ofs S contains only e. That

is { }sG e= .

Theorem 2.1.11: An element s S is aperiodic if and only if s sO G= .

Proof: Suppose s S is aperiodic element. Then by definition 2.1.10

we have, { }sG e= , that is sG = 1 . Then by theorem 2.1.9 we obtain

s sO G= .

Conversely suppose that s sO G= , then by theorem2.1.9, we

obtain sG = 1 , that is { }sG e= , therefore s S is aperiodic element.

Theorem 2.1.12: The number of aperiodic elements in S is divisible

by G .

Proof: Let G be a finite group with identity element e and S is a finite

set on which G acts.

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We first prove If an element s S is aperiodic then so is *g s , for

everyg G .

Suppose s S is an aperiodic element. Then by definition 2.1.10

we have { }sG e= . That is *g s s= only ifg e= .

Now *( * ) ( . )*h g s h g s e= = only if h e= , so *g s is an aperiodic

element.

Hence if s S is aperiodic element then we get that every

element in sO is also aperiodic.

Hence if { }1 2, ,..., nB s s s= is the set of all aperiodic elements ofS

which are not related to each other then we get,

ss B

P O

=U (disjoint)

Hence in view of theorem 2.1.11 we get

s

s BP O

=

s BG

=

n G=

.

Which prove that number of aperiodic elements in S is divisible by G .

Definition 2.1.13: Let ( )L G be the lattice of subgroups ofG under

partial ordering by inclusion. For ( )H L G , ( )H is the number of

elements in S, which are stabilized by H.

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That is, { }( ) : sH s S G H = = .

Theorem 2.1.14: { }( ) ( )0 mode G .

Proof: By using definition 2.1.13 we have,

{ } { }{ }( ) : se s S G e = =

i.e., { }( )e denotes the number of aperiodic elements in S.

Therefore, by using theorem 2.1.12 we have, { }( )G e| |

i.e., { }( ) ( )0 mode G .

Definition 2.1.15: Let ( )L G be the lattice of subgroups ofG under

partial ordering by inclusion. For ( )H L G , ( )H be the number of

elements in S, whose stabilizer contains H.

That is, ( ) { }: sH s S G H = .

Remark 2.1.16: It is obvious that ( ) ( )H H .

Theorem 2.1.17 (The relation between ( )H and ( )H ):

( )( )

( )K L GK H

H K

=

Proof: Let{ }

( )

: sK L G

K H

s S G K

= =Uand { }: sB s S G H =

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We now show that A B= .

Suppose, t A , then t S and tG K= , for some ( )K L G , K H .

i.e., t S and tG H

i.e., t B .

Conversely suppose that t B .

Then t S and tG H

i.e., t S and tG K= , for some ( )K L G , K H .

i.e., t A .

Hence A B=

(1)

We now show that the union defined in A is disjoint.

Suppose { }:i s i

A s S G K = = and{ }

: j s j

A s S G K = = , for some

, ( )i jK K L G and iK H , jK H , with i j .

If possible assume that i jt A A I

i.e., it A and jt A

i.e., t S , t iG K= and t jG K= , for some , ( )i jK K L G and ,iK H jK H .

i.e., i jK K= , for some , ( )i jK K L G and ,iK H jK H

Which is a contradiction, since i j .

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Hence i jA A = I , for i j .

(2)

Therefore the union defined inA is disjoint.

By using the results (1) and (2) we obtain,

( ) { }: sH s S G H =

B=

A=

{ }( )

: sK L GK H

s S G K

= =U

{ }( )

: sK L GK H

s S G K

= =

( )( )K L G

K H

H

=

2.2. Mbius inversion theorem over a lattice of

subgroups:

In this section we define Mbius function on lattices and we

prove the Mbius inversion theorem over a lattice of subgroups which

is given in []. Throughout this section 0

is the minimal element and 1

is maximal element and the relation H G means H is a subgroup ofG.

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Definition 2.2.1(Mbius function): Let L be a finite lattice with

unique minimal 0

element, unique maximal element 1

and Mbius

function be the set of all integers. The Mbius function :L is

defined recursively by,

(i) ( ) 1a = , if 0a

= and

(ii) ( ) ( )b a

a b

, where p is a prime number.

Proof: We prove the theorem by induction on k.

By (3) of property 1.2.42, we get1

1

nn

+ = ! (1)

Also since, 0 (mod ), n pn p for ! (2)

Therefore from (1) and (2) we get ,

1

0 (mod ), n p1

n p for

+

(or) 0 (mod ), -11

n p for n p

i.e., 0 (mod ),1

n p for n p

>

Therefore the theorem is true for 1k= .

Assume t he theorem is true for k m= ,

That is, 0 (mod )n

p if n mpm

>

(3)

Taking 1k m= + in theorem 3.4.1 we obtain,

( 1) (mod )1 1

n p n np p

m m p m

+ + + +

(4)

By using (3) in (4) we get,

(mod )1 1

n p np if n mp

m m p

+ > + +

(5)

Since,2 ( 1 )n mp mp p p m p p> > + = + , so using (3) in (5) we get,

0 (mod )1

n pp if n mp

m

+ > +

(or) 0 (mod ) 1)1

n pp if n p m p

m

+ + > ( + +

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i.e., 0 (mod ) 1)1

n p if n m p

m

> ( + +

.

Which shows that the theorem is true for 1k m= + .

Hence by induction, we obtain

0 (mod ),n

p n kpk

>

.

CHAPTER IV

SOME CONGRUENCES ON BINOMIAL

COEFFICIENTS AND STIRLING NUMBERS OF

BOTH KINDS IN THE CASE G Cp Cp

4.1 Mbius function defined on L (Cp Cp):

In this section, we find Mbius function defined on ( )P PL C C , and hence we

prove a congruence relation from corollary 2.2.7.

Lemma 4.1.1: The Mbius function defined on ( )P PL C C is given by,

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2

1 1

( ) 1

if H

H if H p

p if H p

=

= = =

Proof: Let P PG C C= acts on a set S. For notational convenience let , denote the

permutations given by (1,2,..., )p = , ( 1, 2,..., ) p p p p = + + + and forA G let A

denotes the sub group generated by A. Let ( )P PL C C =

{ }{ }2 2 1, , , . , . ,..., . , . , ,p pe G = . Then ( )( ),P PL C C is a

lattice and its lattice diagram is,

If { }H e= , then 1H = . Since { }e is the unique minimal element in ( )P PL C C ,

then by definition 2.2.1 we get,

{ }( ) 1e = , i.e., ( ) 1H = if 1H = .

IfH = , then H p= = , and hence,

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1

1

1

( )

( ) ( ) ( )

P P

HH L C C

H H