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The International Journal Of Engineering And Science (IJES)
||Volume||3 ||Issue|| 1||Pages|| 18-22||2014||
ISSN(e): 2319 – 1813 ISSN(p): 2319 – 1805

On Cycle Index and Orbit – Stabilizer of Symmetric Group
1
1
2

Mogbonju M. M., 2Bello G . G., 3Ogunleke I. A.,

Department of Mathematics, University of Abuja,P.M. B. 117 Abuja.Nigeria
Department of Mathematics, University of Abuja, P.M. B. 117 Abuja.Nigeria
3
Alvan Ikoku Federal College of Education. Department of Mathematics.
P.M. B. 1033. Owerri, Imo State . Nigeria.

-----------------------------------------------------ABSTRACT ----------------------------------------------------In this paper explicit formulas are used to find the cycle index of permutation group especially the symmetric
groups, the alternating group and the number of orbits of . Finally we presented the orbits – stabilizer of
symmetric group.

KEYWORDS: Cycle index, Orbit, Stabilizer, Permutation group, Symmetric group.
--------------------------------------------------------------------------------------------------------------------------------------Date of Submission: 20 August 2013
Date of Acceptance: 25 January 2014
---------------------------------------------------------------------------------------------------------------------------------------

I.

INTRODUCTION

Let be a permutation group with object set. It is well known that each permutation on can be written
uniquely as a product of disjoint cycles. Every permutation group has associated with polynomials called the
“cycle index”. The concept can be traced back to Frobenius [1] as a special case of a formulation in terms of a
group characters. Rudvalis and Snapper [1] point out the connection between these generalized characters and
the theorems of deBruijn [ 2] and Foullers [ 3]. Redfield [ 4]who discovered cycle indexes independently
devised a clever schemes which enable him to determine the number of classes of certain matrices by forming a
special product of cycle indexes.In this paper, we establised the connection between cycle index and orbitstabilizer of a symmetric group.

II.

PRELIMINARIES AND DEFINITIONS

2.1. The following notations are used:
 denotes the symmetric group.
 denotes the cycle index of symmetric group.
 is the number of orbits of group .
 denotes the alternating group of
2.2. Definition
[1] A cycle of length k is a permutation for which there exists an element in such that are the only elements
moved by .it is required that since with , the element itself would not be moved either.
[2] The cycle index of a permutation group, is the average of the cycle index monomials of all the permutation
[3] Consider a group on a set The orbit of a point is a set of to which can be moved by the elements of
[4] The orbits of is defined by: .
[5] The associated equivalent relation is defined by saying if and only if there exist with .
[6] The elements are equivalent if and only if their orbits are the same i.e.
[7] The set is called a subset of called the orbits of under
[8] For every , we define the stabilizer subgroup of as the set of all elements in that fix .
[9] This is a subgroup of . The action of is freeif and only if all stabilizer are all travial

III. CYCLE INDEX OF SYMMETRIC GROUP
Let be a permutation group with object set = . It is well known that each permutation in can be written
uniquely as a product of disjoint cycles and so for each integer from to , we let be the number of cycles of length
in the disjoint cycle decomposition of .Then the cycle index of denoted by (), ( from the word Zyklenzeiger) is
the polynomial in the variables defined by

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Page 18
On Cycle Index And Orbit – Stabilizer…
() = -1 (2.1)
or it is written as Z ( : ).
Consider the symmetric group on object. For = 3
() = ( !)
={,,,,,}
For =1 = = 3 i.e. the number of cycles of length
For 1 and 2 = = 1
For = 3 = = 1
This implies that
Z( ) = ! { + + }
Where
implies 3 cycle of length 1 occurring together i.e. in the same permutation or in one pair ( identity).
implies cycle of length 1 and 2 occurring together in 3 different pairs
implies cycle of length 3 occurring in 2 different pairs.
For we have
Z( ) = { + } i.e (1)(2) (12)
For = 1 = = 2
For = 2 = = 1
implies 2 cycle of length of one occurring together in one pair
implies 1 cycle of length 2 occurring once.
Therefore if the length are in one pair, it will be raised to the power of and if they are more than one we add
them together i.e. different pairs.
And means length 1, length 2, and length 3 respectively and so on.
Note that each permutation of objects can be associated with the partition of which has for each from 1 to ,
exactly parts equal to
Let be the number of permutation in , whose cycle decomposition will determine the partition (j), so that for
each , =
Then it is easy to see that
=!
(3.2)
= 3!
,,,,,}
For = 1
===
For = 2
===
For = 3 = =
Which gives
++
Also for = 2! i. e. (1)(2) (12)
For = 1
===
For = 2
===
Hence we have
+
For = ! = for =
===
Which gives
Thus the cycle index takes the form shown in the next theorem.
Theorem
The cycle index of the symmetric group is given by
= (3.3)
Where the sum is over all partition of , and is given by (3.2)
Corollar
The cycle index of the alternating group is given by
Z() = + (3.4)
To illustrate, we know that

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Page 19
On Cycle Index And Orbit – Stabilizer…
() = ( + + )
() = ( + + )
It implies
() = ( + )
Also for
() = ( + )
() = ( + )
Which gives
() = ( )
It is often convenient to express () in terms of ( with < . For this purpose we define () = and this give a recursive
formula, whose inductive proof is straight forward and can be stated as follows.
Theorem
The cycle index of the symmetric group satisfies the recurrence relation
() = 1/
Which is the general formula for the cycle index for .
Example
() = 1
() = 1! = 1,
() = 2! = 2,
() = { () + () }
= { () + ()}

= { () + (1)} = { { + }
Z(S3) = { () + ( + (}
= { () + ( +(}
= {s1( + S2) + + } = { + + + }
={++}
Z(S4) = { () + ( + ( + (}
= { () + ( + ( + (}
= { [ ( + + )] + [ ( + )] + + }
= { / 6 + / 6 + /2 + / 2 + / 6 + + }
={/6+/6+/2+/6+}
= { + + + } ...

IV. ORBIT - STABILIZER OF SYMMETRIC GROUP.
For each in , let = {| = }. Thus is called the stabilizer of .
Let be a group of permutation of a set , for each in , let stabilizer = . Then = { | }
Theorem
For any element of an orbit of , that is the elements in the orbit of is the index of the stabilizer of in .
=
(4.1)
Proof
We first express as a union of right cosets modulo

=
. . . (4.2)
It only remain to observe the natural - correspondence between these cosets and the elements of . For each = to
m, we associate the coset with the element in . For ≠ we have ≠because otherwise is an element of and
hence αi is an element of . Thus contradicting the fact = .
Therefore this correspondence is 1-1, for any object in , we have = for some permutation in . From the coset
decomposition of , it follows that
= with in
Hence = and thus every element of corresponds to some coset. Therefore is the number of elements in and (4.1)
is proved.
Now we prepared for the first lemma which proves a formula for the number of orbits of in terms of the average
number of fixed points of the permutations in

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Page 20
On Cycle Index And Orbit – Stabilizer…
Lemma . (Burnside's Lemma)
The number of orbits of is given by
= |A|-1
. . . (4.3)
Proof:
Let be the orbits of A and for each = 1 to , let be an element of the orbit . Then from (4.1) we have
=|
(4.4)
we have seen that if and are in the same orbits, then = . Hence the right side of (4.4) can be altered to obtain
=
Or in other notation, we have
N(A)|A| = 1 (4.5)
Now on interchanging the order of summation on the right side of (4.5) and modifying the summation indices
accordingly, we have
N(A)|A| = . . .
But is just . Thus the proof is completed on division by
= |A|-1 .
Example
Consider a graph below using the product notation, the group of may be expressed as = . Now has order 4 and
each permutation fixes the three points and
14
3
5

7

26
Figure 4.1 :A graph with three fixed points
Let the permutation be denoted by
= =,
= =
Then = , that is all the seven points are fixed, = , = , =
Thus = ( + + + ) =
And it is clear that the orbits are {3}, {5}, {7}, {1,2}and {4,6}
The number of orbits is precisely the number of ways in which can be rooted. To obtain all rooting of one
simply chooses one point from each of the orbits as a roots.
Definition 4.1. Let be a group of permutation of a set . for each in i.e.
= { | }.
The set is called a subset of s called the orbits of under
Examples on orbit
From the example of Figure 3.1
= { 1, 2} = { 1,2} = { 3} = { 4, 6} = { 5} = { 4,6} = { 7}
Therefore = { 1,2 }, { 3 }, { 4,6 }, { 5 }, { 7 }.
Let = { , , , , }
= { 1,2,3} = { 1,2,3 } = { 1, 2,3,4 }
= { 2, 4, 5 } = { 5,6 } = { 4,5,6,7 }
= { 6,7,8 } = { 7,8 }
These implies that = = = = =
=(++++)==
Therefore the orbits are {{1,2,3,4},{2,4.5},{4,5,6,7},{6,7,8}}
Example on stabilzer
Given a group = {, , , , }
Therefore stabilizers of are
=,=,=,=,=,
=,==,,
That is were all the points are fixed.
= { (1), (78)} { (1), (123)(67), (78)} { (1), (234)(56), (78)} { (1), (123)(67), (234)(56)} { (1), (234)(56)} = 5
Theorem
Let be a finite group of permutation of a set . then for any from , then
= and this is called the orbit-stabilizer theorem.

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The IJES

Page 21
On Cycle Index And Orbit – Stabilizer…
Example
For ={ , }
=,=,
=,=,
= and = ,
Hence = = =
For = , , , , }
=,=,=,
Also
= = =
Hence = i.e. =
4.5 Weighted form of burnside lemma
We provide a slight generalization of (1) called the weighted form of burnside lemma
= |A|-1 (4.6) . . .
Let be any commutative ring containing the rational and let be a function the weighted function, from the object
set of into the ring . In practice the weight function is constant on the orbits of . Hence in this case we can define
the weight of any orbit to be the weight of any element in the orbit.
For each orbit , we denote the weight of by and by definition, = for any in .
Theorem The sum of the weights of the orbits of is given by
(4.7)
The proof is similar to the proof of (4.1)
We consider again the Graph in figure 4.1
To illustrate this lemma and to display a sum of cycle indices in a way which will be used effectively,for each
point of . We define the weight (k) to be the cycle index of the stabilizer of in . Thus
= ( + ) . . . (4.8)
i.e. has two stable points , which implies that the identity of length is , and the cycle of length and 2 are
= ( + + ) . . . (4.9)
= ,, ,
Note that = = = i.e.
=(+)=(+)=(+)
Also = = i.e.
=(++)=(++)=(++)
Thus in particular is constant on the orbits.
We sketch the verification of () for this example by observing that the sum of the orbit weight is + + + + = 2 + 3
Since = and = =
Now the right side of () is the sum
=
=,
Where the last two terms in the above equation are found at once by inspecting α3 and α4 in the graph above and
the verification is finished.
Similarly, the cycle index sum for all the different rooted graphs obtained from any graph can be obtained in
terms of the weight of the weights of the fixed points of .

V. CONCLUSION
In this research, we discovered that the cycle index of symmetric group can be understand better by the
use of formula..The use of formula and examples is also used to established that their is strong relatioship
between orbits of a symmetric group and orbits stablizer of a symmetric group,

REFERENCES
[1]
[2]
[3]
[4]
[5]
[6]

N.G.dEbruin. Enumeration of triangulations of the disk. Proc.London Math. Soc.. 14,1964, 746 - 768.
N.G. dEbruin..Poly'stheory of countings in applied Combinatorial Mathematics (E. F. Beckeabach, ed).1964. 144 - 184.
H.O. Foulkes.On Redfields group reduction functions. Canad. J. Math, 15,1963, 272 - 234.
Frank Harary, Edgar, M. Palmer (1973) . Graphical eEnumeration. Academic press, New York London. 1973, 32 - 43.
J.H.Redfield .The theory of group - reduced distribution, Amer. J. Math.49,1927, 433 -455
A.Rudvalis .Numerical polynomials for arbitrary characterS. J. Combinatoral t heory 10 A, 1971, 145 -159.

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The IJES

Page 22

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  • 1. The International Journal Of Engineering And Science (IJES) ||Volume||3 ||Issue|| 1||Pages|| 18-22||2014|| ISSN(e): 2319 – 1813 ISSN(p): 2319 – 1805 On Cycle Index and Orbit – Stabilizer of Symmetric Group 1 1 2 Mogbonju M. M., 2Bello G . G., 3Ogunleke I. A., Department of Mathematics, University of Abuja,P.M. B. 117 Abuja.Nigeria Department of Mathematics, University of Abuja, P.M. B. 117 Abuja.Nigeria 3 Alvan Ikoku Federal College of Education. Department of Mathematics. P.M. B. 1033. Owerri, Imo State . Nigeria. -----------------------------------------------------ABSTRACT ----------------------------------------------------In this paper explicit formulas are used to find the cycle index of permutation group especially the symmetric groups, the alternating group and the number of orbits of . Finally we presented the orbits – stabilizer of symmetric group. KEYWORDS: Cycle index, Orbit, Stabilizer, Permutation group, Symmetric group. --------------------------------------------------------------------------------------------------------------------------------------Date of Submission: 20 August 2013 Date of Acceptance: 25 January 2014 --------------------------------------------------------------------------------------------------------------------------------------- I. INTRODUCTION Let be a permutation group with object set. It is well known that each permutation on can be written uniquely as a product of disjoint cycles. Every permutation group has associated with polynomials called the “cycle index”. The concept can be traced back to Frobenius [1] as a special case of a formulation in terms of a group characters. Rudvalis and Snapper [1] point out the connection between these generalized characters and the theorems of deBruijn [ 2] and Foullers [ 3]. Redfield [ 4]who discovered cycle indexes independently devised a clever schemes which enable him to determine the number of classes of certain matrices by forming a special product of cycle indexes.In this paper, we establised the connection between cycle index and orbitstabilizer of a symmetric group. II. PRELIMINARIES AND DEFINITIONS 2.1. The following notations are used:  denotes the symmetric group.  denotes the cycle index of symmetric group.  is the number of orbits of group .  denotes the alternating group of 2.2. Definition [1] A cycle of length k is a permutation for which there exists an element in such that are the only elements moved by .it is required that since with , the element itself would not be moved either. [2] The cycle index of a permutation group, is the average of the cycle index monomials of all the permutation [3] Consider a group on a set The orbit of a point is a set of to which can be moved by the elements of [4] The orbits of is defined by: . [5] The associated equivalent relation is defined by saying if and only if there exist with . [6] The elements are equivalent if and only if their orbits are the same i.e. [7] The set is called a subset of called the orbits of under [8] For every , we define the stabilizer subgroup of as the set of all elements in that fix . [9] This is a subgroup of . The action of is freeif and only if all stabilizer are all travial III. CYCLE INDEX OF SYMMETRIC GROUP Let be a permutation group with object set = . It is well known that each permutation in can be written uniquely as a product of disjoint cycles and so for each integer from to , we let be the number of cycles of length in the disjoint cycle decomposition of .Then the cycle index of denoted by (), ( from the word Zyklenzeiger) is the polynomial in the variables defined by www.theijes.com The IJES Page 18
  • 2. On Cycle Index And Orbit – Stabilizer… () = -1 (2.1) or it is written as Z ( : ). Consider the symmetric group on object. For = 3 () = ( !) ={,,,,,} For =1 = = 3 i.e. the number of cycles of length For 1 and 2 = = 1 For = 3 = = 1 This implies that Z( ) = ! { + + } Where implies 3 cycle of length 1 occurring together i.e. in the same permutation or in one pair ( identity). implies cycle of length 1 and 2 occurring together in 3 different pairs implies cycle of length 3 occurring in 2 different pairs. For we have Z( ) = { + } i.e (1)(2) (12) For = 1 = = 2 For = 2 = = 1 implies 2 cycle of length of one occurring together in one pair implies 1 cycle of length 2 occurring once. Therefore if the length are in one pair, it will be raised to the power of and if they are more than one we add them together i.e. different pairs. And means length 1, length 2, and length 3 respectively and so on. Note that each permutation of objects can be associated with the partition of which has for each from 1 to , exactly parts equal to Let be the number of permutation in , whose cycle decomposition will determine the partition (j), so that for each , = Then it is easy to see that =! (3.2) = 3! ,,,,,} For = 1 === For = 2 === For = 3 = = Which gives ++ Also for = 2! i. e. (1)(2) (12) For = 1 === For = 2 === Hence we have + For = ! = for = === Which gives Thus the cycle index takes the form shown in the next theorem. Theorem The cycle index of the symmetric group is given by = (3.3) Where the sum is over all partition of , and is given by (3.2) Corollar The cycle index of the alternating group is given by Z() = + (3.4) To illustrate, we know that www.theijes.com The IJES Page 19
  • 3. On Cycle Index And Orbit – Stabilizer… () = ( + + ) () = ( + + ) It implies () = ( + ) Also for () = ( + ) () = ( + ) Which gives () = ( ) It is often convenient to express () in terms of ( with < . For this purpose we define () = and this give a recursive formula, whose inductive proof is straight forward and can be stated as follows. Theorem The cycle index of the symmetric group satisfies the recurrence relation () = 1/ Which is the general formula for the cycle index for . Example () = 1 () = 1! = 1, () = 2! = 2, () = { () + () } = { () + ()}  = { () + (1)} = { { + } Z(S3) = { () + ( + (} = { () + ( +(} = {s1( + S2) + + } = { + + + } ={++} Z(S4) = { () + ( + ( + (} = { () + ( + ( + (} = { [ ( + + )] + [ ( + )] + + } = { / 6 + / 6 + /2 + / 2 + / 6 + + } ={/6+/6+/2+/6+} = { + + + } ... IV. ORBIT - STABILIZER OF SYMMETRIC GROUP. For each in , let = {| = }. Thus is called the stabilizer of . Let be a group of permutation of a set , for each in , let stabilizer = . Then = { | } Theorem For any element of an orbit of , that is the elements in the orbit of is the index of the stabilizer of in . = (4.1) Proof We first express as a union of right cosets modulo  = . . . (4.2) It only remain to observe the natural - correspondence between these cosets and the elements of . For each = to m, we associate the coset with the element in . For ≠ we have ≠because otherwise is an element of and hence αi is an element of . Thus contradicting the fact = . Therefore this correspondence is 1-1, for any object in , we have = for some permutation in . From the coset decomposition of , it follows that = with in Hence = and thus every element of corresponds to some coset. Therefore is the number of elements in and (4.1) is proved. Now we prepared for the first lemma which proves a formula for the number of orbits of in terms of the average number of fixed points of the permutations in www.theijes.com The IJES Page 20
  • 4. On Cycle Index And Orbit – Stabilizer… Lemma . (Burnside's Lemma) The number of orbits of is given by = |A|-1 . . . (4.3) Proof: Let be the orbits of A and for each = 1 to , let be an element of the orbit . Then from (4.1) we have =| (4.4) we have seen that if and are in the same orbits, then = . Hence the right side of (4.4) can be altered to obtain = Or in other notation, we have N(A)|A| = 1 (4.5) Now on interchanging the order of summation on the right side of (4.5) and modifying the summation indices accordingly, we have N(A)|A| = . . . But is just . Thus the proof is completed on division by = |A|-1 . Example Consider a graph below using the product notation, the group of may be expressed as = . Now has order 4 and each permutation fixes the three points and 14 3 5 7 26 Figure 4.1 :A graph with three fixed points Let the permutation be denoted by = =, = = Then = , that is all the seven points are fixed, = , = , = Thus = ( + + + ) = And it is clear that the orbits are {3}, {5}, {7}, {1,2}and {4,6} The number of orbits is precisely the number of ways in which can be rooted. To obtain all rooting of one simply chooses one point from each of the orbits as a roots. Definition 4.1. Let be a group of permutation of a set . for each in i.e. = { | }. The set is called a subset of s called the orbits of under Examples on orbit From the example of Figure 3.1 = { 1, 2} = { 1,2} = { 3} = { 4, 6} = { 5} = { 4,6} = { 7} Therefore = { 1,2 }, { 3 }, { 4,6 }, { 5 }, { 7 }. Let = { , , , , } = { 1,2,3} = { 1,2,3 } = { 1, 2,3,4 } = { 2, 4, 5 } = { 5,6 } = { 4,5,6,7 } = { 6,7,8 } = { 7,8 } These implies that = = = = = =(++++)== Therefore the orbits are {{1,2,3,4},{2,4.5},{4,5,6,7},{6,7,8}} Example on stabilzer Given a group = {, , , , } Therefore stabilizers of are =,=,=,=,=, =,==,, That is were all the points are fixed. = { (1), (78)} { (1), (123)(67), (78)} { (1), (234)(56), (78)} { (1), (123)(67), (234)(56)} { (1), (234)(56)} = 5 Theorem Let be a finite group of permutation of a set . then for any from , then = and this is called the orbit-stabilizer theorem. www.theijes.com The IJES Page 21
  • 5. On Cycle Index And Orbit – Stabilizer… Example For ={ , } =,=, =,=, = and = , Hence = = = For = , , , , } =,=,=, Also = = = Hence = i.e. = 4.5 Weighted form of burnside lemma We provide a slight generalization of (1) called the weighted form of burnside lemma = |A|-1 (4.6) . . . Let be any commutative ring containing the rational and let be a function the weighted function, from the object set of into the ring . In practice the weight function is constant on the orbits of . Hence in this case we can define the weight of any orbit to be the weight of any element in the orbit. For each orbit , we denote the weight of by and by definition, = for any in . Theorem The sum of the weights of the orbits of is given by (4.7) The proof is similar to the proof of (4.1) We consider again the Graph in figure 4.1 To illustrate this lemma and to display a sum of cycle indices in a way which will be used effectively,for each point of . We define the weight (k) to be the cycle index of the stabilizer of in . Thus = ( + ) . . . (4.8) i.e. has two stable points , which implies that the identity of length is , and the cycle of length and 2 are = ( + + ) . . . (4.9) = ,, , Note that = = = i.e. =(+)=(+)=(+) Also = = i.e. =(++)=(++)=(++) Thus in particular is constant on the orbits. We sketch the verification of () for this example by observing that the sum of the orbit weight is + + + + = 2 + 3 Since = and = = Now the right side of () is the sum = =, Where the last two terms in the above equation are found at once by inspecting α3 and α4 in the graph above and the verification is finished. Similarly, the cycle index sum for all the different rooted graphs obtained from any graph can be obtained in terms of the weight of the weights of the fixed points of . V. CONCLUSION In this research, we discovered that the cycle index of symmetric group can be understand better by the use of formula..The use of formula and examples is also used to established that their is strong relatioship between orbits of a symmetric group and orbits stablizer of a symmetric group, REFERENCES [1] [2] [3] [4] [5] [6] N.G.dEbruin. Enumeration of triangulations of the disk. Proc.London Math. Soc.. 14,1964, 746 - 768. N.G. dEbruin..Poly'stheory of countings in applied Combinatorial Mathematics (E. F. Beckeabach, ed).1964. 144 - 184. H.O. Foulkes.On Redfields group reduction functions. Canad. J. Math, 15,1963, 272 - 234. Frank Harary, Edgar, M. Palmer (1973) . Graphical eEnumeration. Academic press, New York London. 1973, 32 - 43. J.H.Redfield .The theory of group - reduced distribution, Amer. J. Math.49,1927, 433 -455 A.Rudvalis .Numerical polynomials for arbitrary characterS. J. Combinatoral t heory 10 A, 1971, 145 -159. www.theijes.com The IJES Page 22