Vol. 9, No. 1, 2016, 84-113

ISSN 1307-5543 – www.ejpam.com

**Counting Symmetries with Burnside’s Lemma and**

**Pólya’s Theorem**

### Md. Taufiq Nasseef

*Department of Mathematics, School of Mathematics,Statistics and Actuarial Science , University of*
*Kent, Kent, United Kingdom*

**Abstract.** Counting concerns a large part of combinational analysis. Burnside’s lemma, sometimes also
called Burnside’s counting theorem, the Cauchy-Frobenius lemma or the orbit-counting theorem[5], is
often useful in taking account of symmetry when counting mathematical objects. The Pólya’s theorem
is also known as the Redfield-Pólya Theorem which both follows and ultimately generalizes Burnside’s
lemma on the number of orbits of a group action on a set. Pólya’s Theory is a spectacular tool that
allows us to count the number of distinct items given a certain number of colors or other
characteris-tics. Sometimes it is interesting to know more information about the characteristics of these distinct
objects. Pólya’s Theory is a unique and useful theory which acts as a picture function by producing a
polynomial that demonstrates what the different configurations are, and how many of each exist.The
dynamics of counting symmetries are the most interesting part. This subject has been a fascination for
mathematicians and scientist for ages. Here 16 Bead Necklace, patterns of*n*tetrahedron with 2 colors,
patterns of*n*cubes with 3 and 4 colorings and so on have been solved.

**2010 Mathematics Subject Classifications**: 14L,14P,14Q

**Key Words and Phrases**: Burnside’s Lemma, Pólya’s Theorem, Counting Symmetries,Group

**1. Introduction and Preliminaries**

**1.1. Group Action**

Let*G*be a group and*A*is a set which is nonempty. Suppose that*G* is a permutation group
acting on a set*A, and consider the setBA*of all functions

*f* :*A*→*B.*

Then G acts naturally on*BA*as follows: if*g*∈*G*and *f* ∈*BA*then define the function *f g*by[8]

*f g*(*a*) = *f*(*a g*−1).

*Email address:*taufiq278@gmail.com

**1.2. The Cycle Index of a Permutation Group**

Let us consider *G* is a permutation group acting on a set*A*which contains *p* number of
elements. Suppose *X* ∈*G* which can be written as the product of *t*_{1} disjoint cycles of length
1,*t*_{2} disjoint cycles of length 2, . . . ,*t _{p}*disjoint cycles of length

*p*then

*X*is called of cycle type

(*t*1,*t*2, . . . ,*tp*).

**Example 1.** *Let G*={*X*1,*X*2,*X*3}*be a cyclic group of order 3 and the set A*={1, 2, 3}*. Then we*
*have*

*X*1=

1 2 3 1 2 3

= (1)(2)(3)

*X*_{2}=

1 2 3 2 3 1

= (123)

*X*3=

1 2 3 3 1 2

= (132)

*Here X*_{1}*has 3 cycle type of length 1. It is of type*(3, 0, 0)*, known as identity mapping. X*_{2}*and X*_{3}
*each has 1 cycle type of length 3 and are of type (0,0,1)[6].*

Note that if we split*A*into*k*_{1}cycles of length 1,*k*_{2}cycles of length 2, . . . ,*k _{p}*cycles of length

*p*then we have

*k*1+2k2+3k3+. . .+

*pkp*=

*p, the sum of lengths of the cycles are the total*number of elements in

*A*[2].

**Definition 1.** *Let G be a group and its elements are the permutations of the set A. Consider the*
*group operation be multiplication and let us define a polynomial in p variables t*_{1},*t*_{2},*t*_{3}, . . . ,*t _{p}*

*where the coefficients are non negative. Then for every g*∈

*G we can form the product*

*tk*1

1 ,*t*
*k*2

2 ,*t*
*k*3

3 , . . . ,*t*
*kp*

*p* *. So the cycle index G is the polynomial defined by*

*Z*(*G,t*_{1},*t*_{2},*t*_{3}, . . . ,*t _{p}*) = 1

|*G*|{

X

*g*∈*G*
*tk*1

1 *t*
*k*2

2 *t*
*k*3

3 . . .*t*
*kp*

*p* }

**Example 2.** *Consider G*=*S*_{3}={*X*_{1},*X*_{2},*X*_{3},*X*_{4},*X*_{5},*X*_{6}}*where*

*X*_{1}=

1 2 3 1 2 3

= (1)(2)(3)

*X*_{2}=

1 2 3 2 3 1

= (123)

*X*3=

1 2 3 3 1 2

= (132)

*X*_{4}=

1 2 3 3 2 1

= (2)(13)

*X*_{5}=

1 2 3 1 3 2

*X*6=

1 2 3 2 1 3

= (12)(3)

*We observe that the permutations of each X*_{4}*, X*_{5} *and X*_{6} *has one cycle of length 2 and one cycle*
*of length 1 and are of type (1,1,0). X*1,*X*2*and X*3*are known to us by the Example 1. The order*
*of this group is*3!=6. Hence by the definition the cycle index of S_{3}*is[6]*

*Z*(*S*3,*t*1,*t*2,*t*3) =
1
6{*t*

3

1+3t1*t*2+2t3}.

**Theorem 1.** *The cycle index of the symmetric group Z*(*S _{p}*)

*can be defined by*

*Z*(*S _{p}*) = 1

*p!*{

X

*t*(*tk*1

1 *t*
*k*2

2 *t*
*k*3

3 . . .*t*
*kp*

*p* )}

*where*

*t* = *p!*

(1*k*1!)(2*k*2!)(3*k*3!). . .(*pkp*_{!}_{)}

*and the summation is over p-tuple*(*k*1,*k*2,*k*3, . . . ,*kp*)*which are nonnegative integers kisatisfying*
1k_{1}+2k_{2}+3k_{3}+. . .+*pk _{p}*=

*p[6].*

There are some formulas for cycle indices. In this study we will use indices for symmetric groups, cyclic groups, and dihedral groups.

**Definition 2.** *The totient functionφ(n*)*, also called Euler’s totient function or the Euler-phi *
*func-tion, is defined as the number of positive integers*≤ *n that are relatively prime to n, where 1 is*
*counted as being relatively prime to all numbers*∗*.*

*For example, there are four numbers relatively prime to 12 that are less than 12 (1,5,7, and 11),*
*soφ(*12) =4.

For cyclic groups and dihedral groups we have the following formulas:
*Z*(*Cn*) =

1
*n*

X

*k/n*

*φ(k*)*tn _{k}/k*, (1)

and

*Z*(*Dn*) =
1

2*Z*(*Cn*) +

¨_{1}

2*t*1*t*

(*n*−1)*/*2

2 if *n*odd,
1

4(*t*
*n/*2
2 +*t*

2
1*t*

(*n*−2)*/*2

2 ) if *n*even.

*Z*(*D _{n}*)is divided into two cases by imagining a necklace of four beads versus a necklace of five
beads. With four beads, there are two types of flips: those over the line through the centers
of opposite edges, and those over the line connecting opposite vertices. However, with five
beads, there is only one type of flip: those over lines connecting a vertex with the center of an
opposite edge. This can be extended to any even or odd number of beads, respectively[4].

∗_{K. R. Rumery, Twelve-Tone Composition, Part One,} _{http://jan.ucc.nau.edu/~krr2/12tone/12tone1.}

**Example 3.** *Let us use this formula to compute the cycle index of C*_{4}*,*

*ZC*4=

1 4

X

*l/*4

*φ(l*)*t*4_{l}/l

=1

4(φ(1)*t*
4*/*1

1 +*φ(*2)*t*
4*/*2

2 +*φ(*4)*t*
4*/*4
4 )

=1

4(*t*
4
1+*t*

2 2+2t4)

**Example 4.** *Here we will illustrate the formula by computing the cycle index of D*4*,*

*Z _{D}*

4=

1
2*ZC*4+

1
4(*t*

4 2

2 +*t*
2
1*t*

4−2 2

2 )

=1

2(
1
4(*t*

4
1+*t*

2

2+2t4)) +
1
4(*t*

4 2

2 +*t*
2
1*t*

4−2 2

2 )

=1

8(*t*
4
1+*t*

2

2+2t4) +
1
4(*t*

2
2+*t*

2
1*t*2)

=1

8(*t*
4
1+2t

2

1*t*2+3t22+2t4)

Now we have already got the cycle index of the symmetric groups from Theorem 1. The
following example will find the coefficient on*n-th term.*

**Example 5.** *Let A be a set of n elements and let G be the group of all permutations of A. That is*
*G is a symmetric group of degree n. Its cycle index turns out to be equal to the coefficient of xn* *in*
*the development of*

*e x p*(*x t*_{1}+ *x*

2* _{t}*
2

2 +

*x*3*t*_{3}
3 +. . .)
*as a power series of in x. This expression can be written as*

∞ X

*g*1=0

*xg*1* _{t}g*1

1
*g*_{1}!

∞ X

*g*1=0

*x*2g2* _{t}g*2

2
*g*_{2}!2*g*2 . . .

*The coefficient of xnis obtained by summing the expression*

*tg*1

1 *t*
*g*2

2 . . .(*g*1!*g*2!2*g*2. . .)−1
*over all possible g*_{1},*g*_{2}, . . .*satisfying g*_{1}+2g_{2}+. . .=*n[2].*

**1.3. Equivalence Relations**

Let*A*be a set and∼be a binary relation of*A*. Then∼is an**equivalence relation**if and
only if for all*p,q,r* ∈*A*the following conditions are satisfied:

**Reflexivity** *p*∼*p*

**Symmetry** If*p*∼*q*then*q*∼*p*

**Transitivity** If*p*∼*q*and*q*∼*r* then*p*∼*r*

The**equivalence class** of*p* under∼is the set{*p* ∈*A*|*p*∼*q*}[1]. These equivalence classes
are called**patterns.**

**1.4. Weights and Inventory**

**Definition 4.** *Given a set of colors C we want to associate a weight w _{c}for all c*∈

*C. Then we will*

*define the*∈

**weight**of a coloring a*A to be the product of the weights of the colored elements[1].*

*w*(*a*) =Y

*i*∈*N*
*wa*(*i*)

*This is also called the inventoryof A.*

**Example 6.** *Consider 4 bead necklace to illustrate with C={Red(r), Yellow(y)}. Let a* = *r r y y*
*where*

*a*(1) =*Red*, *a*(2) =*Red,* *a*(3) =*Yel l ow and a*(4) =*Yel l ow*

*Let the colors Red and Yellow have weights R and Y respectively. Then the coloring of a as follows*

*w*(*a*) =

Y

*i*∈*N*
*w _{a}*(

*i*)

=*w _{a}*

_{(}

_{1}

_{)}

*w*

_{a}_{(}

_{2}

_{)}

*w*

_{a}_{(}

_{3}

_{)}

*w*

_{a}_{(}

_{4}

_{)}

=*RRY Y*

=*R*2*Y*2

**Definition 5.** *Let G be a permutation group on a set A. Suppose that*∼*defines an equivalence*
*relation on A. Then the equivalence class containing the element a, denoted by Or b*(*a*)*is called*
*the G- orbitof a. It is defined as[6]*

*Or b*(*a*) ={*π(a*):*π*∈*G*}

**Example 7.** *Consider 4 bead necklace where A*= {1, 2, 3, 4}*, B* = {*Red*(*R*),*Yel l ow*(*Y*)}*, C be*
*the group of coloring with f* :*A*→*B and G*={*e,*(12),(34),(12)(34)}*be the permutation group*
*[1]. Then*

*Or b*(*RRY Y*) ={*RRY Y*}*Or b*(*RY RR*) =*Or b*(*Y RRR*) ={*RY RR,Y RRR*}

*Or b*(*Y Y Y Y*) ={*Y Y Y Y*}*Or b*(*Y Y Y R*) =*Or b*(*Y Y RY*) ={*Y Y Y R,Y Y RY*}

*Or b*(*Y Y RR*) ={*Y Y RR*}*Or b*(*Y RY Y*) =*Or b*(*RY Y Y*) ={*Y RY Y,RY Y Y*}

*Or b*(*RY RY*) =*Or b*(*Y RY R*) =*Or b*(*RY Y R*) =*Or b*(*Y RRY*) ={*RY RY,Y RY R,RY Y R,Y RRY*}

The problem of determining the equivalent objects in *A*is to count all *G-orbits. *
How-ever this method looks impractical and even more boring in complex situations. Fortunately,
Burnside’s Lemma gives an analytical formula for such counting of *G-orbits. It is a powerful*
technique and particularly efficient when the order of the group is small. It is also considered
one of the essential parts in development of the Pólya Theory.

**Definition 6.** *Let G be a permutation group on a set A. An element a in A is said to be invariant*
*under a permutationπ*∈*G if and only ifπ(a*) =*a. To each a*∈*A, the stabilizerof a, denoted*

*by S t a b*(

*a*)

*is defined to be the set*

*S t a b*(*a*) ={*π*∈*G*:*π(a*) =*a*}

*and*∀*a*∈*G. Letη(π)denotes the number of elements of A that are invariant underπthat is[6]*

*η(π) =*{*a*∈*A*:*π(a*) =*a*}

**Example 8.** *Following the Example 7 the stabilizer sets for each coloring in C are given below*
*S t a b*(*RRRR*) =*S t a b*(*Y Y Y Y*) =*S t a b*(*RRY Y*) =*S t a b*(*Y Y RR*) ={*e,*(12),(34),(12)(34)}
*S t a b*(*RRRY*) =*S t a b*(*Y Y Y R*) =*S t a b*(*RRRY*) =*S t a b*(*Y Y Y R*) ={*e,*(12)}

*S t a b*(*RY RR*) =*S t a b*(*Y RY Y*) =*S t a b*(*RY Y Y*) =*S t a b*(*Y RRR*) ={*e,*(34)}

*S t a b*(*Y RRY*) =*S t a b*(*RY Y R*) =*S t a b*(*RY RY*) =*S t a b*(*Y RY R*) ={*e*}

**2. Burnside’s Lemma and Pólya’s Theorem**

**2.1. Orbit-Stabilizer Formula**

Let*G* be a group, acting on a set*A*, then For all*a*∈*A*,|*G*|=|*S t a b*(*a*)||*Or b*(*a*)|[6].
**Example 9.** *By following the Example 7 and 8 we have*

*S t a b*(*Y Y RR*) ={*e,*(12),(34),(12)(34)}*S t a b*(*Y RRR*) ={*e,*(34)}

*Or b*(*Y Y RR*) ={*Y Y RR*}*Or b*(*Y RRR*) ={*RY RR,Y RRR*}

*Therefore*

|*S t a b*(*Y Y RR*)|=4 |*S t a b*(*Y RRR*)|=2
|*Or b*(*Y Y RR*)|=1 |*Or b*(*Y RRR*)|=2

**2.2. Burnside’s Lemma**

**Lemma 1.** *Let G be a finite group that acts on a set A. For each g* ∈*G, let Ag* *denote the set of*
*elements in A that are fixed by g. Burnside’s lemma asserts the following formula for the number*
*of orbits, denoted*|*A/G*|*:*

|*A/G*|= 1

|*G*|

X

*g*∈*G*
|*Ag*|

*Thus the number of orbits (a natural number or*+∞*) is equal to the average number of points*
*fixed by an element of G[3].*

*Proof.* The proof uses orbit-stabilizer theorem and the fact that*A*is the disjoint union of
the orbits:

X

*g*∈*G*

|*Ag*|=|{(*g,a*)∈*G*×*A*|*g*·*a*=*a*}|=X

*a*∈*A*

|*S t a b*(*a*)|=X

*a*∈*A*
|*G*|
|*Or b*(*a*)|

=|*G*|X
*a*∈*A*

1

|*Or b*(*a*)| =|*G*|

X

*P*∈*A/G*

X

*a*∈*P*
1
|*P*| =|*G*|

X

*P*∈*A/G*

1=|*G*| · |*A/G*|

Therefore[6]

1
|*G*|

X

*g*∈*G*

|*Ag*|=|*A/G*|

This proves the lemma.

**Example 10.** *Continuing With the Example 7 and 8, We have A*(*e*) =*C*

*A*(12) ={*RRRR,RRY R,RRRY,RRY Y,Y Y RR,Y Y Y R,Y Y RY,Y Y Y Y*}
*A*(34) ={*RRRR,Y RRR,RY RR,RRY Y,Y Y RR,RY Y Y,Y RY Y,Y Y Y Y*}
*A*((12)(34)) ={*RRRR,RRY Y,Y Y RR,Y Y Y Y*}

*By applying Burnside’s Lemma, we have*

1
|*G*|

X

*g*∈*G*

|*Ag*|= 1

4(16+8+8+4) =9

**Example 11.** *Let G be the group of rotations of a cube induced on the set of 6 faces (Fig. 1). The*
*rotations of the cube which leaves it invariant are*

*(i) 6 rotations of 90 degree(clockwise or anti-clockwise) about the axes joining the centers of*
*the opposite faces;*

*(ii) 3 rotations of 180 degree each of the above axes;*

*(iv) 6 rotations of 180 degree about the axes joining the midpoints of the opposite edges and*

*(v) the identity.*

*To determine the cyclic index of its action on the set of faces it is observed that*

*(i) has rotations of order 4 and cycle type t*_{1}2*t*_{4}*;*

*(ii) has rotations of order 2 and cycle type t*_{1}2*t*2_{2}*;*

*(iii) has rotations of order 3 and cycle type t*_{3}2

*(iv) has rotations of order 2 and cycle type t*_{2}3

*(v) has rotations of order 1 and cycle type t*_{1}6

*Therefore the cycle index of G is*

*Z*(*G*,*t*_{1},*t*_{2}, . . . ,*t*_{6}) = 1

24(6t 2

1*t*4+3t12*t*
2
2+8t

2 3+6t

3
2+*t*

6 1).

Figure 1: Faces of the cube

(a) Axis (b) Corner (c) Midpoint

Figure 2: Types of Rotations of the Cube

*the number of integers*1≤*p*≤*d with g cd*(*p,d*) =1, which is already given by Euler’s function

*φfrom the elementary theory (Definition 4). Thus from*(*Eq1*)*we obtain*

*Z*(*G,t*1,*t*2, . . . ,*tn*) =
1
*n*

X

*d/n*

*φ(d*)*tn _{d}/d*

*whereφ(pn*) = (*p*−1)*pn*−1 *for prime powers andφ(m*) = Π*k _{i}*

_{=}

_{1}

*φ(pni*

*i* )*for m*= Π
*k*
*i*=1*p*

*ni*

*i* *[3].*

**Example 12.** *Again let G be the group of rotations of a cube induced on the set of 8 vertices. The*
*rotations of the cube which leaves it invariant are the same as Example 11. It is now necessary to*
*see what the rotations do with the vertices. To determine the cycle index of its action on the set of*
*faces it is observed that*

*(i) has rotations of order 4 and cycle type t*_{4}2*;*

*(ii) has rotations of order 2 and cycle type t*_{2}4*;*

*(iii) has rotations of order 4 and cycle type t*_{1}2*t*2_{3}

*(iv) has rotations of order 2 and cycle type t*_{2}4

*(v) has rotations of order 1 and cycle type t*_{1}8

*Therefore the cycle index of G is[2]*

*Z*(*G*,*t*_{1},*t*_{2}, . . . ,*t*_{8}) = 1

24(6t 2 4+9t

4 2+8t

2
1*t*

2 3+6t

3
2+*t*

8 1).

**2.3. Pólya’s Substitution**

Let*σ(c*_{1},*c*_{2}, . . . ,*c _{k}*)be the polynomial obtained from the cycle index

*Z*(

*G,t*

_{1},

*t*

_{2}, . . . ,

*t*)by substituting

_{n}*t*_{1}7−→*c*_{1}+*c*_{2}+. . .+*c _{k}*,

*t*27−→

*c*12+

*c*

2

2+. . .+*c*
2
*k*,
. . .

*t _{n}*7−→

*c*

_{1}

*n*+

*c*

_{2}

*n*+. . .+

*c*,

_{k}nThen the number of non-equivalent*k*colorings of A pattern[*p*_{1},*p*_{2}, . . . ,*p _{k}*]is equal to the
co-efficient of the term

*cp*1

1 *c*
*p*2

2 . . .*c*
*pk*

*k* in*σ(c*1,*c*2, . . . ,*ck*)[3]. Then we find the generating function

*Z _{G}*

*k*

X

*i*=1
*c _{i}*,

*k*

X

*i*=1
*c _{i}*2, . . . ,

*k*

X

*i*=1
*cn _{i}*

!

,

**Problem 1** (Edge Coloring of a Tetrahedron). A tetrahedron has 6 edges. How many
in-equivalent ways are there to color it with

(a) 2 colors?

(b) 3 black and 3 yellow edges?

*Solution: First we need to find out the symmetries of the tetrahedron with permutations of the*
edges. Let us consider the set*A*of the edges of a tetrahedron and*G* be the set of permutations
of*A*which will be produced by rotating the tetrahedron. Their cyclic structures are given as
follows

(i) The identity leaves all 6 edges fixed with structure *t*6_{1}.

(ii) There are four 120◦rotations about a corner and the middle of the opposite face (Fig.
3). They are two cycles of length three. They cyclically permute the edges incident
to that corner and also the edges bounding the opposite face, so the cycle structure
representation is *t*2_{3}.

(iii) There are four 240◦rotations as like the 120◦rotations (Fig. 3). So the cycle structure is also same.

(iv) There are three 180◦rotations about opposite edges leave the two edges fixed. The other
four edges are left in cycles of length 2. Thus we have the cycle structure *t*2_{1}*t*_{2}2[7].

Then by Burnside’s Lemma we have

*Z*(*G,t*_{1},*t*_{2}, . . . ,*t*_{6}) = 1

12(*t*
6
1+8t

2 3+3t

2
1*t*

2 2)

(a) The number of distinct 2 colorings on tetrahedron edge will be
*Z _{G}*= 1

12(2

6_{+}_{8}_{(}_{2}2_{) +}_{3}_{(}_{2}2_{)(}_{2}2_{)) =}_{12}

(b) Now we will apply Pólya’s Substitution method. Let the colors black and yellow be denoted
by *x* and *y* respectively. Then substituting *t _{i}* →(

*xi*+

*yi*)for each

*i*={1, 2, 3, 4, 5, 6}in the previous equation, we get

*Z _{G}*= 1

12((*x*+*y*)

6_{+}_{8}_{(}* _{x}*3

_{+}

*3*

_{y}_{)}2

_{+}

_{3}

_{(}

_{x}_{+}

_{y}_{)}2

_{(}

*2*

_{x}_{+}

*2*

_{y}_{)}2

_{)}By Maple we get,

*ZG*=*x*6+*x*5*y*+2x4*y*2+4x3*y*3+2x2*y*4+*x y*5+*y*6

The term 4*x*3*y*3contains tetrahedron with 3 black and 3 yellow edges. The coefficient of
this term is 4. Hence there are 4 ways to color the edge of tetrahedron.

**2.4. Pólya’s Fundamental Theorem**

**Theorem 2.**

*The Pattern inventory is*

*P I*=*Z _{G}*(X

*c*∈*C*
*w _{c}*,X

*c*∈*C*

*w*2* _{c}*,X

*c*∈

*C*

*w*3* _{c}*, . . . ,X

*c*∈

*C*

*wp _{c}*)

*where Z _{G}is the cycle index and p*=|

*D*|

*[2].*

*Proof.* Let*X* =*X*_{1}∪*X*_{2}∪. . .∪*X _{m}*be the equivalence classes of

*X*under the relation

*x*∼

*y*if and only if

*W*(

*x*) =

*W*(

*y*)

By Orbit-Stabilizer Formula,*g*∗*x* ∼*x* for all*x* ∈*X*,*g*∈*G* and so we can think of*G*acting on
each*X _{i}* individually. We use the fact that

*x*∈

*X*implies

_{i}*g*∗

*x*∈

*X*for all

_{i}*i*∈[

*m*],

*g*∈

*G. We*use the notation

*g*(

*i*)∈

*G*(

*i*)when we restrict attention to

*X*.

_{i}Let*mi* denote the number of orbits. Then

*P I*=

*m*

X

*i*=1

*m _{i}W_{i}* =

*m*

X

*i*=1
*W _{i}*( 1

|*G*|

X

*g*∈*G*

|*F i x*(*g*(*i*))|),

By Burnside’s Lemma, we have

*P I*= 1

|*G*|

X

*g*∈*G*
*m*

X

*i*=1

|*F i x*(*g*(*i*))|*Wi*=
1
|*G*|

X

*g*∈*G*

Note that (1) follows *F i x*(*g*) =S*m _{i}*

_{=}

_{1}

*F i x*(

*g*(

*i*))since

*x*∈

*F i x*(

*g*(

*i*))if and only if

*x*∈

*W*and

_{i}*g*∗

*x*=

*x*.

Suppose that *Z _{G}*=

*xk*1

1 *x*
*k*2

2 . . .*x*
*kp*

*p* . Then we can claim that
*W*(*F i x*(*g*)) = (X

*c*∈*C*
*w _{c}*)

*k*1

_{(}

X

*c*∈*C*

*w _{c}*)

*k*2

_{. . .}

_{(}X

*c*∈*C*

*w _{c}*)

*kp*

_{(3)}

Substituting (2) and (3) yields the theorem.

To verify (2) we use the fact that if *x*∈*F i x*(*g*)), then the elements of a cycle of *g*must be
given the same color. A cycle of length*i*will then contribute a factorP_{c}_{∈}* _{C}wi_{c}*where the term

*wi*comes from the choice of

_{c}*c*for every element of the cycle[8].

**Example 13.** *Let A*={1, 2, 3, 4, 5, 6}*and B*={*whi t e,bl ack,g r a y*}*. Consider a function f* ∈*BA*
*which is a 6-tuple of colors and can be viewed as necklace representation.*

w 1

w 2

b 3

b 4

g 5

b 6

Figure 4: Colored Necklace

*Now we are interested to assign different weights to the beads. Suppose that bl ack*(*b*) =1,
*whi t e*(*w*) =10*and g r a y*(*g*) =100. Then the weight of the necklace is

*W*(*f*) =10×10×1×1×100=104
*Note that rotation does not change the weight.*

Table 1: Fixed Points
Element *g* |*f i x*(*g*)|

*e* 36

(1, 2, 3, 4, 5, 6) 3

(1, 3, 5)(2, 4, 6) 32

(1, 4)(2, 5)(3, 6) 33

(1, 5, 3)(2, 6, 4) 32

(1, 6, 5, 4, 3, 2) 3

*Now we have to consider counting the sums of the weights of the fixed points of each element*
*of C*_{6}*. Let us calculate Wfix(g) for the element g*= (1, 4)(2, 5)(3, 6)*. For a function f to be fixed*
*by g we must have*

*f*(1) = *f*(4) *f*(2) = *f*(5) *f*(3) =*f*(6)

*and hence*

*W*(*f*) =*W*(*f*(1))2*W*(*f*(2))2*W*(*f*(3))2

*Now f*(1)*, f*(2)*and f*(3)*can be any white, black or gray and that is why*
*W f i x*(*g*) = (*W*(*whi t e*)2+*W*(*bl ack*)2+*W*(*g r a y*)2)3

*. By repeating this process we can find out the other elements of C*6*. Then we get*

Table 2: Weighted Fixed Points
Element *g* Wfix(*g)*

*e* (*W*(*whi t e*) +*W*(*bl ack*) +*W*(*g r a y*))6
(1, 2, 3, 4, 5, 6) *W*(*whi t e*)6+*W*(*bl ack*)6+*W*(*g r a y*)6
(1, 3, 5)(2, 4, 6) (*W*(*whi t e*)3_{+}_{W}_{(}_{bl ack}_{)}3_{+}_{W}_{(}_{g r a y}_{)}3_{)}2

(1, 4)(2, 5)(3, 6) (*W*(*whi t e*)2_{+}_{W}_{(}_{bl ack}_{)}2_{+}_{W}_{(}_{g r a y}_{)}2_{)}3

(1, 5, 3)(2, 6, 4) (*W*(*whi t e*)3+*W*(*bl ack*)3+*W*(*g r a y*)3)2
(1, 6, 5, 4, 3, 2) *W*(*whi t e*)6+*W*(*bl ack*)6+*W*(*g r a y*)6

*For any given values of W*(*whi t e*),*W*(*bl ack*) *and W*(*g r a y*)*we can calculate this sum and*
*find the sum of the weights of the weighted*(6, 3)−*necklaces.*

*At this moment we would like to apply Pólya’s Fundamental Theorem. First we need to assign*
*the weights of the three colors. Let W*(*whi t e*) = *x, W*(*bl ack*) = *y and W*(*g r a y*) =*z. Then the*
*weight of a necklace representative is not a number, but a multivariate polynomial in the variables*
*x, y and z* *[8].*

x 1

x 2

y _{3}
y _{4}
z

5

y 6

Figure 5: Weighted Necklace

*By using Burnside’s Lemma we get that*

*Z*(*C*_{6},*t*_{1},*t*_{2}, . . . ,*t*_{6}) = 1

6(*t*
6
1+*t*

3 2+2t

2
3+2t6)
*According to Pólya’s Substitution we have*

*t*17−→*x*+*y*+*z,*

*t*_{2}7−→*x*2+*y*2+*z*2,
*t*37−→*x*3+*y*3+*z*3,
*t*_{4}7−→*x*4+*y*4+*z*4,
*t*57−→*x*5+*y*5+*z*5,
*t*67−→*x*6+*y*6+*z*6,
*Now*

*P I* =1

6((*x*+ *y*+*z*)

6_{+ (}* _{x}*2

_{+}

*2*

_{y}_{+}

*2*

_{z}_{)}3

_{+}

_{2}

_{(}

*3*

_{x}_{+}

*3*

_{y}_{+}

*3*

_{z}_{)}2

_{+}

_{2}

_{(}

*6*

_{x}_{+}

*6*

_{y}_{+}

*6*

_{z}_{))}

*Using Maple the result becomes*

*P I* =16*x*2*y*2*z*2+10*x*3*y*2*z*+5x4*yz*+10*x*3*yz*2+10*x*2*y*3*z*+10x y3*z*2+10*x y*2*z*3

+5x y4*z*+*x*5*y*+*x*5*z*+3x4*y*2+3x4*z*2+4x3*y*3+4x3*z*3+**3x2y4**+3x2*z*4+*x y*5

x 1

y 2

y _{3}
y 4
x

5

y 6

(a)

y 1

x 2

y _{3}
y _{4}
x

5

y 6

(b)

x 1

x 2

y _{3}
y _{4}
y

5

y 6

(c)

Figure 6: 3 Necklaces with 2 White Beads and 4 Black Beads

**Problem 2**(Face Coloring of a Cube). A cube has 6 faces and the faces are to be colored. How
many inequivalent ways are there to color it with namely

(a) 4 red and 2 green faces?

(b) 3 red, 2 green and 1 blue faces?

(c) 2 red, 1 green, 1 blue and 2 yellow faces?

*Solution: We have to color the faces of the cube. So recall Example 11. Then the cycle index*
is

*Z*(*G,t*_{1},*t*_{2}, . . . ,*t*_{6}) = 1

24(6t 2

1*t*4+3t12*t*
2
2+8t

2 3+6t

3
2+*t*

6 1)

Now we will apply Pólya’s Substitution method. Let the colors red, green, blue and yellow
be denoted by *x*,*y,z* and *p* respectively. Then substituting *t _{i}* →(

*xi*+

*yi*+

*zi*+

*pi*)for each

*i*={1, 2, 3, 4, 5, 6}in the previous equation, we get

*Z _{G}*= 1

24(6(*x*+*y*+*z*+*p*)

2_{(}* _{x}*4

_{+}

*4*

_{y}_{+}

*4*

_{z}_{+}

*4*

_{p}_{) +}

_{3}

_{(}

_{x}_{+}

_{y}_{+}

_{z}_{+}

_{p}_{)}2

_{(}

*2*

_{x}_{+}

*2*

_{y}_{+}

*2*

_{z}_{+}

*2*

_{p}_{)}2

+8(*x*3+*y*3+*z*3+*p*3)2+6(*x*2+*y*2+*z*2+*p*2)3+ (*x*+*y*+*z*+*p*)6)

Using Maple we find that,

*Z _{G}*=3

*y*3

*p x*2+3

*y*3

*z p*2+3

*yz*3

*p*2+3

*y*3

*pz*2+3

*y p*3

*z*2+3x p3

*y*2+3x3

*p y*2+3x3

*pz*2

+3x p3*z*2+6x2*z*2*p*2+2*y p x*4+2z p x4+2yz p4+2x pz4+3z p3*y*2+2*yz x*4

+8x yz2*p*2+8xz y2*p*2+8x p y2*z*2+**8yzx2p2**+8*y p x*2*z*2+8z p x2*y*2+5x3*yz p*

+5x y3*z p*+5x yz p3+5x yz3*p*+3*yz*3*x*2+3*y p*3*x*2+3x y3*z*2+2x yz4+6x2*y*2*z*2

+3x3*yz*2+3x y3*p*2+2z p y4+2xz y4+3*y*3*z x*2+**3x3zy2**+3z3*p x*2+6*y*2*z*2*p*2

+3z3*p y*2+3z p3*x*2+3xz3*y*2+2*y*2*z*4+*y*5*z*+*yz*5+2z2*x*4+2z2*y*4+*x*5*z*

+*xz*5+**2y2x4**+2x2*y*4+2x2*z*4+*x*5*y*+*x y*5+*x*6+ *y*6+*z*6+*p*6+2x3*y*3+2x3*z*3

+2*y*3*z*3+2x2*p*4+*x*5*p*+*x p*5+2*y*2*p*4+ *y*5*p*+*y p*5+2z2*p*4+*z*5*p*+*z p*5+2p2*x*4

+2p2*y*4+2p2*z*4+2x3*p*3+2y3*p*3+2z3*p*3

Each term of this expression on the right hand side corresponds to a fixed number of red, green, blue and yellow colors.

(a) The term 2x4*y*2 says that there are 2 possible ways to color the cube with 4 red and 2
green faces.

(b) The term 3x3*y*2*z*says that there are 3 possible ways to color the cube with 3 red, 2 green
and 1 blue faces.

(c) The term 8x2*yz p*2 says that there are 8 possible ways to color the cube with 2 red, 1
green, 1 blue and 2 yellow faces.

**Problem 3**(Vertex Coloring of a Cube). A cube has 8 vertices and the vertices are to be colored.
How many inequivalent ways are there to color it with

(a) 6 black, 1 yellow and 1 green vertices? (b) 1 black, 1 red, 4 blue and 2 yellow vertices?

(c) 3 black, 1 red, 1 green 1 blue and 2 yellow vertices?

*Solution: We have to color the vertices of the cube. So recall Example 12. Then the cycle index*
is

*Z*(*G*,*t*_{1},*t*_{2}, . . . ,*t*_{8}) = 1

24(6t 2 4+9t

4 2+8t

2
1*t*

2 3+6t

3
2+*t*

8 1).

Now we will apply Pólya’s Substitution method. Let the colors black, red, green, blue and
yellow be denoted by*x*,*y,z,p*and*v*respectively.Then substituting*t _{i}*→(

*xi*+

*yi*+

*zi*+

*pi*+

*vi*)

for each*i*={1, 2, 3, 4, 5, 6, 7, 8}in the previous equation, we get
*Z _{G}*= 1

24(6(*x*

4_{+}* _{y}*4

_{+}

*4*

_{z}_{+}

*4*

_{p}_{+}

*4*

_{v}_{)}2

_{+}

_{9}

_{(}

*2*

_{x}_{+}

*2*

_{y}_{+}

*2*

_{z}_{+}

*2*

_{p}_{+}

*2*

_{v}_{)}4

+8(*x*+*y*+*z*+*p*+*v*)2(*x*3+*y*3+*z*3+*p*3+*v*3)2
+6(*x*2+*y*2+*z*2+*p*2+*v*2)3+ (*x*+*y*+*z*+*p*+*v*)8).

We use Maple to expand and find that the expression on the right hand side is very long. Each term of this expression corresponds to a fixed number of black, red, green, blue and yellow colors.

(b) In this case the term is 35p4*v*2*x y* which contains cube 1 black, 1 red, 4 blue and 2 yellow
vertices. So there are 35 ways to color the cube.

(c) For this we need to find out the coefficient of the term 140pv2*x*3*yz*which contains cube
with 3 black, 1 red, 1 green 1 blue and 2 yellow vertices. There are 140 ways to color the
cube.

**Problem 4** (16 Bead Necklace). Count the number of ways to arrange beads on a necklace,
where 16 total beads arranged on the necklace with

(a) 2 different colors;

(b) 14 red and 2 black colors; (c) 4 different colors;

*Solution: With a necklace, we will obviously rotate it around and flip over it. We will also use*
the Definition 2 with the formula for the cyclic and dihedral groups to solve this problem. The
cycle index of this group*C*16from (1) is

*Z*(*C*16,*t*1, . . . ,*t*16) =
1
16

X

*l/*16

*φ(l*)*t*16_{l}*/l*

= 1

16(φ(1)*t*
16*/*1

1 +*φ(*2)*t*
16*/*2

2 +*φ(*4)*t*
16*/*4

4 +*φ(*8)*t*
16*/*8

8 +*φ(*16)*t*
16*/*16
16 )

= 1

16(*t*
16
1 +*t*

8 2+2t

4 4+4t

2 8+8t

1 16)

Now we will illustrate the formula by computing the cycle index of*D*_{16}
*Z*(*D*16,*t*1, . . . ,*t*16) =

1

2*Z*(*C*16,*t*1, . . . ,*t*16) +
1
4(*t*

8
2+*t*

2
1*t*
7
2)
=1
2(
1
16(*t*

16
1 +*t*

8 2+2t

4 4+4t

2 8+8t

1 16)) +

1
4(*t*

8
2+*t*

2
1*t*

7 2)

= 1

32(*t*
16
1 +9t

8 2+8t

2
1*t*

7 2+2t

4 4+4t

2 8+8t

1 16)

(4)

Let us consider 2 colors red and black denoted by *x* and *y* respectively. Now we will
ap-ply Pólya’s Substitution Method. Substituting *ti* →(*xi*+ *yi*)for each *i* ={1, 2, 3, . . . , 16}in
equation (4), we get

*Z*(*D*_{16},*t*_{1}, . . . ,*t*_{16}) = 1

32((*x*+ *y*+)

16_{+}_{9}_{(}* _{x}*2

_{+}

*2*

_{y}_{)}8

+8(*x*+*y*)2(*x*2+*y*2)7+2(*x*4+*y*4)4
+4(*x*8+*y*8)2+8(*x*16+ *y*16)1)

(ii) Expanding the same equation we have by Maple

*Z _{D}*

_{16}=

*x*15+

**8x14y2**+21

*x*13

*y*3+72

*x*12

*y*4+147x11

*y*5

+280x10*y*6+375x9*y*7+440x8*y*8+375*x*7*y*9

+280x6*y*10+147x5*y*11+72*x*4*y*12+21*x*3*y*13

+8x2*y*14+*x y*15+*x*16+*y*16

The term 8x14*y*2 contains 14 red and 2 black colors. The coefficient of this term is 8.
So there are 8 ways to decorate a 16 beaded Necklace with 14 red and 2 black colors.
(iii) For this case consider 4 colors *x*,*y,z, and* *p. We will again apply Pólya’s Substitution*

Method. Substituting *t _{i}* →(

*xi*+

*yi*+

*zi*+

*pi*)for each

*i*= {1, 2, 3, . . . , 16}in equation (4), we get

*Z*(*D*16,*t*1, . . . ,*t*16) =
1

32((*x*+*y*+*z*+*p*)

16_{+}_{9}_{(}* _{x}*2

_{+}

*2*

_{y}_{+}

*2*

_{z}_{+}

*2*

_{p}_{)}8

+8(*x*+*y*+*z*+*p*)2(*x*2+ *y*2+*z*2+*p*2)7+2(*x*4+*y*4+*z*4+*p*4)4
+4(*x*8+*y*8+*z*8+*p*4)2+8(*x*16+*y*16+*z*16+*p*16)1)

Let *x*= *y*=*z*=*p*=1. Then using Maple by the above equation we get the total count
134301715.

**3. Generalization of Pólya’s Theorem**

In the preceding sections we considered mappings of*X*into*Y*, introduced by a permutation
group *G* of*X*. We are now going to move on more general situation. Let us consider two
mappings *f*1∈*YX* and *f*2∈*YX*. They are equivalent if there exist elements*g*∈*G* and*h*∈*H*
such that *f*1*g*=*h f*2, that is *f*1(*g x*) =*h f*2(*x*)for all*x* ∈*X*.

Next we assume that each *f* ∈*YX* has a certain weight*W*(*f*). Then we can also assume
that equivalence functions have the same weight:

*f*_{1}∼ *f*_{2} implies*W*(*f*_{1}) =*W*(*f*_{2}).

If*F*denotes a pattern, we define its weight*W*(*F*)as the common value of all*W*(*f*)with*f* ∈*F*

[2].

**Lemma 2.** *The Pattern inventory is*

X

*W*(*F*) = 1

|*G*|
1
|*H*|

X

*g*∈*G*

X

*h*∈*H*

(*g,h*)
X

*f*

*W*(*f*)

**3.1. Patterns of One-to-one Mappings**

We already have the finite sets*X* and*Y*, subject to the permutation groups*G*and*H*. Now
we define the weight*W*(*f*)of any *f* ∈*YX* by

*W*(*f*) =

¨

1 if *f* is a one-to-one mapping
0 otherwise

If *g* ∈ *G* and *h* ∈ *H*, then the mapping*h f g*−1 is one-to-one if and only if *f* is one-to-one
and thus*W*(*f*1) =*W*(*f*2). The inventory

P

*W*(*F*)is just the number of patterns of one-to-one
function.

Now let*g*be of type{*g*_{1},*g*_{2}, . . .}where*g _{i}*shows the number of cycles of length

*i. Similarly*let

*h*be of type {

*h*1,

*h*2, . . .}. We will find the number of one-to-one mappings

*f*of

*Y*into

*X*that satisfies

*f g*=

*h f*. Let

*x*be any element in

*X*, belonging to a cycle of length

*j. This cycle*consists of elements

*x*,

*g x,g*2

*x*, . . . ,

*gj*−1

*x*and we have

*gjx*=

*x*. Now

*f g*=

*h f*implies

*f g*2= *f g g*=*h f g*=*hh f* =*h*2*f*,*f g*3=*h*3*f*, . . .

Hence we have*hjf x* =*f gjx*= *f x*. It follows that the length of cycle of*Y* to which *f x*belongs
is a divisor of *j. The number of elements of one-to-one mappings of a set* *g _{j}* elements into a
set of

*hj*elements equals

*hj*(

*hj*−1)(

*hj*−2). . .(

*hj*−

*gj*+1)which is zero if

*hj<gj*. Therefore we have

(*g,h*)
X

*f*

*W*(*f*) =

Y

*j*
*jgj _{h}*

*j*(*hj*−1)(*hj*−2). . .(*hj*−*gj*+1).

This product runs over all*j*for*g _{j}>*0; but if

*g*=0 we can take the over all values

_{j}*j*=1, 2, 3, . . . as well.

Next we can write *jgh*(*h*−1)(*h*−2). . .(*h*−*g*+1)as*gth derivative of*(1+*jz*)*h*at the point
*z*=0. As the result of a number of partial differentiations with respect to variables*z*1,*z*2, . . .
at the points*z*_{1}=*z*_{2}=. . .=0, we have

*∂*
*∂z*_{1}

*g*_{1}

*∂*
*∂z*_{2}

*g*_{2}

*∂*
*∂z*_{3}

*g*_{3}

. . .(1+*z*_{1})*h*1_{(}_{1}_{+}_{2z}

2)*h*2(1+3z3)*h*3. . .

The differential operator is obtained from a term of cycle index*Z _{G}*(

*t*1,

*t*2, . . .)of

*G*upon sub-stitution of

*t*

_{1}=

_{∂}∂_{z}1, *t*2=

*∂*

*∂z*2, . . . and the operand is obtained upon substitution of*t*1=1+*z*1,

*t*2=1+2z2, . . . into a term of the index*ZH*(*t*1,*t*2, ..)of*H* [2].

**Theorem 3.** *The number of patterns of one-to-one mappings of X into Y equals*

*ZG*

*∂*
*∂z*_{1},

*∂*
*∂z*_{2},

*∂*
*∂z*_{3}, . . .

*ZH*(1+*z*1, 1+2z2, 1+3z3, . . .)

*If*|*X*|=|*Y*|*, then we always have*P* _{j}gj*=

P

*jhj. So that we have either g*1=*h*1*, g*2=*h*2, . . .
*or at least once g _{j}*

*>h*

_{j}. Thus we have[2]

*∂*
*∂z*_{1}

*g*_{1}

*∂*
*∂z*_{2}

*g*_{2}

*∂*
*∂z*_{3}

*g*_{3}

. . .(*z*_{1})*h*1_{(}_{2z}

2)*h*2(3z3)*h*3. . . . .
**Theorem 4.** *If*|*X*|=|*Y*|*, then the number of patterns is equal to*

*Z _{G}*

*∂*
*∂z*1

, *∂*

*∂z*2
, *∂*

*∂z*3
, . . .

*Z _{H}*(

*z*

_{1}, 2z

_{2}, 3z

_{3}, . . .)

*evaluated at z*_{1}=*z*_{2}=*z*_{3}=. . .=0.

*The inverse mappings are one-to-one mappings of Y onto X . In this case we can easily find*
*that the number of patterns is equal to*

*Z _{H}*

*∂*
*∂z*1

, *∂*

*∂z*2
, *∂*

*∂z*3
, . . .

*Z _{G}*(

*z*

_{1}, 2z

_{2}, 3z

_{3}, . . .)

*evaluated at z*_{1}=*z*_{2}=*z*_{3}=. . .=0*[2].*

**Problem 5.** How many geometrically different ways can the faces of a cube be arranged in
cyclic order[2]?

*Solution: We need to find out the number of patterns of one-to-one mappings. LetX* be the set
of faces of a cube and*G* be the cube induced by rotations(Example 11);*Y* be the 6th roots of
unity and again*H* is the group of permutation induced by the rotations in the complex plane
(Definition 2 with cycle index formula*Z*(*C _{n}*)). Then|

*X*|=|

*Y*|.

The cycle indexes are

*Z _{G}*= 1

24(**t**

**6**
**1**+**6t**

**3**
**2**+**8t**

**2**
**3**+3t

2
1*t*

2 2+6t

2
1*t*4)

*Z _{H}* =1

6(**t**

**6**
**1**+**t**

**3**
**2**+**2t**

**2**
**3**+2t6)
The number of patterns is equal to

*Z _{G}*

*∂*
*∂z*_{1},

*∂*
*∂z*_{2},

*∂*
*∂z*_{3}

*Z _{H}*(

*z*

_{1}, 2z

_{2}, 3z

_{3}) = 1

24 · 1 6

*∂*
*∂z*_{1}

6

(*z*_{1})6+

*∂*
*∂z*_{2}

3

(2z_{2})3+

*∂*
*∂z*_{3}

2

(3z_{3})2

= 1

24 · 1

6(6!+6·2

3_{·}_{3!}_{+}_{16}_{·}_{3}2_{·}_{2!}_{)}

=9

*Solution: Here we also need to find out the number of patterns of one-to-one mappings. Again*
let*X* be the set of edges of a tetrahedron and*G* be the tetrahedron induced by rotations;Y be
the 6th roots of unity and again*H* is the group of permutation induced by the rotations in the
complex plane (Definition 2 with cycle index formula *Z*(*C _{n}*)). Then|

*X*|=|

*Y*|.

The cycle indexes are

*ZG*=
1
12(**t**

**6**
**1**+**8t**

**2**
**3**+3t

2
1*t*

2 2)

*Z _{H}* =1

6(**t**

**6**
**1**+*t*

3
2+**2t**

**2**
**3**+2t6)
The number of patterns is equal to

*Z _{G}*

*∂*
*∂z*_{1},

*∂*
*∂z*_{3}

*Z _{H}*(

*z*

_{1}, 2z

_{2}) = 1

12· 1 6

*∂*
*∂z*_{1}

6

(*z*_{1})6+

*∂*
*∂z*_{3}

2

(3z_{3})2

= 1

12· 1

6(6!+16·3
2_{·}_{2!}_{)}

=14

**Example 14.** *We have 6 colored labels l*1,*l*2,*l*3,*l*4,*l*5,*l*6 *to be pasted on the faces of a cube, one*
*on each side. Label l*_{1} *and l*_{2}*are yellow and black respectively. The labels l*_{3}*and l*_{4}*are both violet*
*and can not be distinguished from each other. The same condition holds for l*_{5} *and l*_{6} *which are*
*both purple. We are interested to find out the number of patterns.*

*Now X be set of 6 labels and Y is the set of 6 faces. The group G consists of 8 permutations,*
*characterized by the conditions that l*_{1}*and l*_{2} *are fixed and the subset*{*l*_{3},*l*_{4}}*mapped onto either*
*itself or the set*{*l*5,*l*6}*[2]. The cycle index of this group is*

*Z _{G}*= 1

8(**t**

**6**
**1**+2t

4

1*t*2+**3t21t**
**2**
**2**+**2t**

**2**
**1t4**)

*and also given by*

*ZH*=
1
24(**t**

**6**
**1**+6t

3 2+8t

2
3+**3t**

**2**
**1t**

**2**
**2**+**6t**

**2**
**1t4**)

*For the number of patterns we obtain*

*Z _{H}*

*∂*
*∂z*_{1},

*∂*2

*∂z*_{1}*∂z*_{2},

*∂*2

*∂z*_{1}*∂z*_{4}

*Z _{G}*(

*z*1, 2z2, 4z4)

= 1
24·
1
8
*∂*
*∂z*_{1}

6

(*z*1)6+

*∂*2
*∂z*_{1}*∂z*_{2}

2

(*z*12z2)2+

*∂*
*∂z*_{1}

2

*∂*
*∂z*_{1}

(*z*_{1}24z_{4})

= 1

24· 1

8(6!+3·3·2!2

2_{·}_{2!}_{+}_{6}_{·}_{2}_{·}_{2!4}_{·}_{1!}_{)}

**Example 15.** *Consider 6 colors l*_{1},*l*_{2},*l*_{3},*l*_{4},*l*_{5},*l*_{6}*to be colored on the edges of a tetrahedron, one*
*on each edge by following the same conditions with Example 14. Then the cycle index of this group*
*is*

*Z _{G}*= 1

8(**t**

**6**
**1**+2t

4

1*t*2+**3t21t**
**2**
**2**+2t

2
1*t*4)

*and also given by*

*Z _{H}* = 1

12(**t**

**6**
**1**+8t

2
3+**3t**

**2**
**1t**

**2**
**2**)

*For the number of patterns we obtain*

*ZH*

*∂*
*∂z*_{1},

*∂*2
*∂z*_{1}*∂z*_{2}

*ZG*(*z*1, 2z2)

= 1
12·
1
8
*∂*
*∂z*_{1}

6

(*z*_{1})6+

*∂*2
*∂z*_{1}*∂z*_{2}

2

(*z*_{1}2z_{2})2

= 1

12· 1

8(6!+3·3·2!2
2_{·}_{2!}_{)}

=9

**3.2. The Total Number of Patterns**

Again consider finite sets *X* and *Y* with permutation groups *G* and *H* of *X* and *Y*
re-spectively. We will find the number of patterns. Again we will apply Lemma 2 to evaluate

P(*g,h*)

*f* *W*(*f*). We will use the subsection 3.1. So we can easily compute the number of
possi-bilities for *f*. For each cycle of*X* we will select an element, called the selected element. The
number of possibilities for the element of *Y* onto which is selected element can be mapped
by an *f* isP_{j}_{/}_{i}*jc _{j}*, where

*i*is the length of the cycle of

*X*to which the selected item belongs,

*c*′s are of type{

*c*

_{1},

*c*

_{2}, . . .}of

*h*and the sum is over all divisors

*j*of

*i. Because the choices of*function values for the various selected elements are independent and determine

*f*completely. The number of

*f*equals the product ofP

_{j}_{/}_{i}*jc*taken over all selected elements. Since there are

_{j}*g*cycles of length

_{i}*i, then we get*

(*g,h*)
X

*f*

*W*(*f*) =Y(X

*j/i*
*jc _{j}*)

*gi*

=(*c*1)*g*1(*c*1+2c2)*g*2(*c*1+3c3)*g*3(*c*1+2c2+4c4)*g*4. . .

(5)

Note that a power with exponent 0 has to be interpreted as 1 in this context even if the base is zero. Therefore is no difficulty about the interpretation of the infinite product.

As in subsection 3.1, we can interpret the previous equation as a derivative. A power*ab*
can be written as the *bth derivative ofeaz* at*z* =0; if*a* = *b* =0, this still gives the desired
value 0◦=1. So the previous expression can be written as

*∂*
*∂z*_{1}

*g*_{1}

*∂*
*∂z*_{2}

*g*_{2}

*∂*
*∂z*_{3}

*g*_{3}

. . .*e x p*(X

*i*
*zi*

X

evaluated at*z*_{1}=*z*_{2}=. . .=0. The exponent can be expressed as

(X

*i*
*z _{i}*

X

*j/i*

*jc _{j}*) =X

*j*

*jc _{j}*(

*z*+

_{j}*z*

_{2}

*+*

_{j}*z*

_{3}

*+. . .).*

_{j}Now the differential operator is obtained from*tg*1

1 *t*
*g*2

2 *t*
*g*3

3 . . . on substitution of *t*1=
*∂*
*∂z*1,

*t*_{2}= _{∂}∂_{z}

2, *t*3=

*∂*

*∂z*34, . . . and the operand is obtained from*t*

*c*1

1 *t*
*c*2

2 *t*
*c*3

3 . . . on substitution of[2]
*t*1=*ez*1+*z*2+*z*3+...,*t*2=*e*2(*z*2+*z*4+*z*6+...),*t*3=*e*3(*z*3+*z*6+*z*9+...), . . . .

**Theorem 5.** *The total number of patterns of mapping of X into Y equals*

*ZG*

*∂*
*∂z*_{1},

*∂*
*∂z*_{2},

*∂*
*∂z*_{3}, . . .

*ZH*[*ez*1+*z*2+*z*3+...,*e*2(*z*2+*z*4+*z*6+...),*e*3(*z*3+*z*6+*z*9+...), . . .]

*evaluated at z*_{1}=*z*_{2}=*z*_{3}=. . .=0.

*There is a second expression for the number of patterns, which sometimes may turn out to*
*be simpler to handle. Note that*(5)*is obtained by substituting t*1 =*c*1*, t*2 = *c*1+2c2*,* . . .*into*
*tg*1

1 *t*
*g*2

2 . . .. Summing for g and dividing by|*G*|*, we obtain*

*Z _{G}*(

*c*1,

*c*1+2c2,

*c*1+3c3,

*c*1+2c2+4c4, . . .),

*in which the ith argument is*P_{j}_{/}_{i}jcj. Thus again using Lemma 2 for the number of patterns we*have*

|*H*|−1X
*h*∈*H*

*Z _{G}*(

*c*

_{1},

*c*

_{1}+2c

_{2},

*c*

_{1}+3c

_{3},

*c*

_{1}+2c

_{2}+4c

_{4}, . . .)

*Here it has to be remembered that*{*c*1,*c*2,*c*3, . . .}*is of type h[2].*

*Now we are going to discuss some of the examples. In all cases, m and n stand for the number*
*of elements of X and Y respectively.*

**Example 16.** *Let G be a group consisting of the identity only, where Z _{G}*=

*tm*

_{1}

*. Then the patterns*

*are called patterns of variation with repetition. Now let*{1, 2, . . . ,

*m*}

*be a sequence of X then for*

*each f*∈

*XY, the sequence*{

*f*(1),

*f*(2), . . . ,

*f*(

*m*)}

*shows repetitions. For the number of patterns,*

*theorem 4 gives the expression*

*d*
*dz*

*m*

*Z _{H}*(

*ez*, 1, 1, . . .)

*z*=0

(6)
*A second specialization of* (6)*can be obtained by keeping Y and H general but X has only one*
*element d*_{1}*. Now two functions f*_{1}*and f*_{2}*are equivalent if and only if f*_{1}(*d*_{1})*are mapped to f*_{2}(*d*_{1})*.*
*So the number of patterns is the number of transitive sets. If we take m*=1*in*(6)*then we get*
|*H*|−1*c*_{1}*and c*_{1} *is the number of elements of Y that are invariant under the permutation h[2].*

*whereas patterns are formed with respect to H acting on Y . Let the weights of P be*1,*w*1,*w*2, . . ..
*If we apply theorem 4, we have as differential operator the coefficient of wm* *in*

*e x p*(*w* *∂*

*∂z*_{1}+
1
2*w*

2 *∂*

*∂z*_{2}+. . .)

*by Example 5. The effect of this operator on a function* *ψ(z*_{1},*z*_{2}, . . .) *at z*_{1} = *z*_{2} = . . . = 0*is*

*ψ(w,*1_{2}*w*2,_{3}1*w*3, . . .)*. Now by substituting z*_{1}=*w,z*_{2}= 1_{2}*w*2,*z*_{3}= 1_{3}*w*3, . . .*in theorem 4 we get*

*e x p*(*z*_{1}+*z*_{2}+. . .) =*e x p*(*w*+1

2*w*
2_{+}1

3*w*

3_{+}_{. . .}_{) = (}_{1}_{−}_{w}_{)}−1
*e x p*(2(*z*2+*z*4+. . .)) =*e x p*(*w*2+

1
2*w*

4_{+}1
3*w*

6_{+}_{. . .}_{) = (}_{1}_{−}* _{w}*2

_{)}−1

*etc. Thus we obtain ZH*[(1−*w*)−1,(1−*w*2)−1, . . .]*[2].*

**Example 18.** *Next let us take for H the symmetric group of all permutations of Y and for G take*
*the identity only. Now the patterns turn out to be class partition of X into at most m parts, where*
*a class partition is defined as a set of disjoint subsets(classes) of X whose union is X . The previous*
*result follows from the fact that if f* ∈*YX, then f defines a class partition by putting into one class*
*all d that are mapped by f onto one and the same element of Y . According to*(*a*)*and Example*
*5 we find that the number of patterns is the coefficient of wmin*

*d*
*dz*

*m*

*e x p*(*wez*+1

2*w*
2_{+}1

3*w*
3_{+}_{. . .}_{)}

*z*=0

*and we can easily find to be m!times the coefficient of zmwnin the expression of*(1−*w*)−1*ew*(*ez*−1)*.*
*Here the number of partitions can be at most n parts. So the total number of patterns equals m!*
*times the coefficients of zm* *in eez*−1 *[2].*

**Example 19.** *Now we are interested for H the symmetric group of X without any specialization*
*of G. We get partitions of class partitions of X . It seems to us impossible to simplify the result of*
*theorem 4. If we take G to the symmetric group of X , the partitions become class partitions of a*
*set of unidentifiable objects. So we are only concern about the size of the classes. Our patterns*
*can be brought into one-to-one correspondence with the partitions of m into at most n parts. A*
*partition of m is a solution*{*g*_{1},*g*_{2}, . . .}*of the equation*

*g*_{1}+2g_{2}+3g_{3}+. . .=*m*

*in nonnegative integers g*1,*g*2, . . .*So m has been partitioned into g*1*1’s, g*2*2’s,*. . .*and accordingly*
*g*_{1}+*g*_{2}+*g*_{3}+. . .*is called the number of parts of the partition.*

*The number of partitions can be obtained from Example 17 by specializing H to be the *
*sym-metric group. We obtain the coefficient of zm* *in*

*which becomes in this case, the coefficient of zmwnin*

*e x p*[*w*(1−*z*)−1+1

2*w*

2_{(}_{1}_{−}* _{z}*2

_{)}−1

_{+}1 3

*w*

3_{(}_{1}_{−}* _{z}*3

_{)}−1

_{+}

_{. . .}

_{]}

_{,}

*and this expression can be reduced to*

*e x p*[*l o g*(1−*w*)−1+*l o g*(1−*wz*)−1+*l o g*(1−*wz*2)−1+. . .] =

∞ Y

*k*=1
1
1−*wzk*.

*This is a well-know result for the generating function of the number of partitions of a given number*
*into a given number of components[2].*

**Example 20.** *Let H be a symmetric group and let G be specialized by taking n* = 2. So the
*patterns of class partitions into at most two parts. By*(5)*the number of these is*

1

2*ZG*(2, 2, 2, . . .) +
1

2*ZG*(0, 2, 0, 2, . . .),
*since ZH* =12*t*

2 1+

1
2*t*2*.*

*If we compare this to the number of patterns of partitions into two labeled classes, Z _{G}*(2, 2, 2, . . .)

*,*

*we observe that the term Z*(0, 2, 0, 2, . . .)

_{G}*represents the number of symmetric patterns of X into*

*two classes X*

_{1}

*and X*

_{2}

*. That is, Z*(0, 2, 0, 2, . . .)

_{G}*represents the number of symmetric patterns of*

*subsets X*1

*that are equivalent to there complement, where equivalence is defined by the*

*permuta-tions of G and patterns are defined by this equivalence.*

*Let X be the set of 10 roots of unity and G be the group of 10 rotations. The cycle index (see*
*Definition 2 with cyclic groups formula*(1)) is

*ZG*=
1
10(*t*

10
1 +*t*

5 2+4t

2

5+4t10)
*and therefore[2]*

*Z _{G}*(0, 2, 0, 2, . . .) = 1

10((2)

5_{+}_{4.2}_{) =}_{4.}
**Theorem 6**(Pólya). *We have*

*Z _{G}*[

*H*](

*t*1,

*t*2, . . .) =

*ZG*[

*ZH*(

*t*1,

*t*2,

*t*3, . . .),

*ZH*(

*t*2,

*t*4,

*t*6, . . .), . . .]

*where Z _{G}*

_{[}

_{H}_{]}

*is in Kranz group and the the right hand side is obtained on substitution of*

*xj*=

*ZH*(

*tj*,

*t*2

*j*,

*t*3

*j*, . . .)

*into ZG*(

*x*1,

*x*2,

*x*3, ...)

*[2].*

**Example 21.** *Consider n cubes and we want to color the faces with Red and Yellow. We are*
*interested for the number of ways to do this when equivalences are defined by permutations of the*
*sets of cubes and rotations of the separate cubes.*

*The group is considered as Sn*[*G*] *where Sn* *is the symmetric group of degree n and G is the*
*cube-face group of Example 11. Then we have to substitute t*_{1}=*t*_{2}=. . .=2*in to its cycle index*
*to find the required number. If we make this substitution in any of the polynomials*

*we always get Z _{G}*(2, 2, 2, . . .)

*and hence*

*Z*(2, 2, 2, . . .) = 1

_{G}24(6(2)

2_{(}_{2}_{) +}_{3}_{(}_{2}_{)}2_{(}_{2}_{)}2_{+}_{8}_{(}_{2}_{)}2_{+}_{6}_{(}_{2}_{)}3_{+}_{2}6_{) =}_{10}

*Thus the answer of the question is S _{n}*(10, 10, 10, . . .)

*. This number equals the coefficient of wn*

*in*

*the development of*

*e x p*(10w+1

210w
2_{+}1

310w

3_{+}_{. . .}_{) = (}_{1}_{−}_{w}_{)}−10
*therefore*

*S _{n}*(10, 10, 10, . . .) =(

*n*+9)!

*n!9!*

*How many of the preceding patterns have the property that they do not change if we interchange*
*the color?*

*According to Example 20 this number is found by substituting*

*t*_{1}=*t*_{3}=*t*_{5}=....=0,
*t*_{2}=*t*_{4}=*t*_{6}=...=2

*into the cycle index. Under the substitution the polynomials* (7) *become ZG*(0, 2, 0, 2, . . .) *and*
*Z _{G}*(2, 2, 2, . . .)

*, alternately. As Z*(0, 2, 0, 2, . . .) =2. We obtain

_{G}*Z _{S}*

*n*[*G*](0, 2, 0, 2, . . .) =*ZSn*(2, 10, 2, 10, . . .)

*This is the coefficient of wnin*

*e x p*(2w+1

210w
2_{+}1

32w
3_{+}1

410w

4_{+}_{. . . .}_{)}

=*e x p*[2l o g(1−*w*)−1+8l o g(1−*w*2)−1*/*2]
=(1−*w*)−2(1−*w*2)−4

=(1+2w+*w*2)(1−*w*2)−6

=1+2w+7w2+12w3+27w4+42w5

+77w6+112w7+182w8+252w9+. . .
*For example if n*=9*the required number of patterns is 252[2].*

**Example 22.** *Consider n tetrahedrons and we want to color the edges with Red and Yellow. By*
*following problem 1 we get from edge coloring of a Tetrahedron*

*ZG*(2, 2, 2, . . .) =12

*Thus the answer of this question is S _{n}*(12, 12, 12, . . .)

*. This number equals the coefficient of wn*

*in*

*the development of*

*e x p*(12w+1

212w
2_{+}1

312w

*therefore*

*S _{n}*(12, 12, 12, . . .) = (

*n*+11)!

*n!11!*

*Now we want to find the number of the preceding patterns having the property that they do not*
*change if we interchange the color. Again following problem 1 we get*

*Z _{S}_{n}*[

*G*](0, 2, 0, 2, . . .) =

*ZSn*(2, 12, 2, 12, . . .)

*This is the coefficient of wnin*

*e x p*(2w+1

212w
2_{+} 1

32w
3_{+} 1

412w

4_{+}_{. . . .}_{)}

=*e x p*[2l o g(1−*w*)−1+10l o g(1−*w*2)−1*/*2]
=(1−*w*)−2(1−*w*2)−5

=(1+2w+*w*2)(1−*w*2)−7

=(1+2w+*w*2)(1+7w2+28w4+. . .)
=1+2w+8w2+14w3+35w4+. . .
*For example if n*=4*the required number of patterns is 35.*

**Problem 7.** Let us take*n*cubes and we want to color the faces with Red, Yellow and green. We
are interested for the number of ways to do this when equivalences are defined by permutations
of the sets of cubes and rotations of the separate cubes.

*Solution: The group is considered asSn*[*G*]where*Sn*is the symmetric group of degree *n*and
*G* is the cube-face group of Example 11. Then we have to substitute*t*_{1}=*t*_{2}=. . .=3 in to its
cycle index to find the required number. If we make this substitution in any of the polynomials

*Z _{G}*(

*t*

_{1},

*t*

_{2},

*t*

_{3},]. . .),

*Z*(

_{G}*t*

_{2},

*t*

_{4},

*t*

_{6}, . . .),

*Z*(

_{G}*t*

_{3},

*t*

_{6},

*t*

_{9}, . . .), . . . , we always get

*Z*(3, 3, 3, . . .)and hence

_{G}*Z _{G}*(3, 3, 3, . . .) = 1

24(6(3)

2_{(}_{2}_{) +}_{3}_{(}_{3}_{)}2_{(}_{3}_{)}2_{+}_{8}_{(}_{3}_{)}2_{+}_{6}_{(}_{3}_{)}3_{+}_{3}6_{) =}_{57}

Thus the answer of the question is*Sn*(57, 57, 57, . . .). This number equals the coefficient of*wn*
in the development of

*e x p*(57w+1

257w
2_{+}1

357w

3_{+}_{. . .}_{) = (}_{1}_{−}_{w}_{)}−57

therefore

*Sn*(57, 57, 57, . . .) =

(*n*+56)!
*n!*(56)!

By (5), this number is found by substituting

*t*_{1}=*t*_{2}=*t*_{4}=*t*_{5}. . .=0,
*t*_{3}=*t*_{6}=*t*_{9}=. . .=3

into the cycle index. Under the substitution the polynomials become*Z _{G}*(0, 0, 3, 0, 0, 3, . . .)and

*ZG*(3, 3, 3, . . .), alternately. As

*ZG*(0, 0, 3, 0, 0, 3, . . .) =3. We obtain

*ZSn*[*G*](0, 0, 3, 0, 0, 3, . . .) =*ZSn*(3, 3, 57, 3, 3, 57, . . .)

This is the coefficient of*wn* in
*e x p*(3w+1

23w
2_{+}1

357w
3_{+} 1

43w
4_{+}_{....}_{)}

=*e x p*[3l o g(1−*w*)−1+54l o g(1−*w*3)−1*/*3]
=(1−*w*)−3(1−*w*3)−18

=(1+3w+6w2+10w3. . .)(1+18w3+171w6+. . .) =1+3w+6w2+28w3+. . .

For example if*n*=3 the required number of patterns is 28.

**Problem 8.** Let us take*n*cubes and we want to color the faces with Red, black, Yellow and
green. How many of the patterns have the property that they do not change if we interchange
the color?

*Solution: With the help of (5), this number is found by substituting*

*t*1=*t*3=*t*5=*t*7. . .=0,
*t*2=*t*6=*t*10=. . .=4,
*t*4=*t*8=. . .=4

into the cycle index. Under the substitution the polynomials become*ZG*(0, 0, 0, 4, 0, 0, 0, 4, . . .),
*Z _{G}*(0, 4, 0, 4, 0, 4, 0, 4, . . .)and

*Z*(4, 4, 4, . . .), alternately. Now

_{G}*Z _{G}*(4, 4, 4, . . .) = 1

24(6(4)

2_{(}_{4}_{) +}_{3}_{(}_{4}_{)}2_{(}_{4}_{)}2_{+}_{8}_{(}_{4}_{)}2_{+}_{6}_{(}_{4}_{)}3_{+}_{4}6_{) =}_{240}
and

*Z _{G}*(0, 4, 0, 4, . . .) = 1

24(6(0)

2_{(}_{4}_{) +}_{3}_{(}_{0}_{)}2_{(}_{4}_{)}2_{+}_{8}_{(}_{0}_{)}2_{+}_{6}_{(}_{4}_{)}3_{+}_{0}6_{) =}_{16}
As*Z _{G}*(0, 0, 0, 4, 0, 0, 0, 4, . . .) =0. We obtain

*Z _{S}*

*n*[*G*](0, 0, 0, 4, 0, 0, . . .) =*ZSn*(0, 16, 0, 240, 0, 16, 0, 240, . . .)

This is the coefficient of*wn* in
*e x p*(0w+ 1

216w
2_{+}1

30w
3_{+} 1