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Noah Nall
Dr. Hoover
Special Topics Essay
December 4, 2014
Polya Enumeration Theorem
George Polya was a Hungarian mathematician who made fundamental contributions to
the fields of combinatorics, number theory, numerical analysis, and probability theory. Many of
these contributions stemmed from the solutions to practical everyday problems, such as the
problem of enumerating the isomers of a chemical compound. It is his results to this problem that
have been applied to a number of counting problems and will be the driving point of what
follows. The basis of Polya’s theory is the ability for one to count under the idea of equivalence,
that is to count to equivalence classes of an equivalence relation rather than the total number of
objects one is counting. Say we have a straight line of ten colored beads, which can be either red
or blue, and we want to find the total number of way in which they could be arranged differently.
That is fairly easy and straightforward, but if we now connect the two ends of the line of beads to
form a necklace that is now allowed to freely move, the computation of different arrangements
becomes much more complicated rather quickly, especially if we add five more beads and an
additional color. Polya’s theory can be applied to this situation, greatly simplifying the
calculation, because it would count certain arrangements as equivalent.
Throughout the following explanation, we are going to be working with groups. It is key
to understand what a group is and a few properties of groups. A group is a nonempty set G
equipped with a binary operation * that satisfies the following axioms1:
1. Closure: If a, b are elements of G, then a*b is also an element of G.
2. Associativity: If a, b, c are elements of G then a*(b*c) = (a*b)*c.
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3. Identity: The group contains an identity, denoted I, such that a*I = a = I*a for all a that
are elements of G.
4. Inverse: For each a that is an element of G, there is an inverse of a, denoted a-1, such that
a*a-1 = I = a-1*a.
Next it is useful to discuss a lemma involving permutation contained within a group and the
concept of a permutation group. The latter is especially important because the group can be used
to describe how an object moves in space. For example, labeling the vertices of a square with the
numbers one to four, we can map the movements of a rotation of ninety degrees using a
permutation. This example will be explained in detail following the lemma, as it will be the main
example referred to as we progress through the building blocks of the larger picture of Polya’s
theorem.
Lemma One:
For any two permutations σi and σj in a group G, there exists a unique permutation σk = σi-1 * σj
in G such that σi * σk = σj.2 Since we are working with permutations contained within a group, let
* represent composition.
Proof:
~We first want to show that σi * σk = σj is true.
-Let σk = σi-1 * σj and we compose both sides by σi
-Reaching σi * σk = σi * (σi-1 * σj)
-By associativity, the right side equals (σi * σi-1) * σj = σI * σj = σj, because
σI is the identity.
-Thus, σi * σk = σj.
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~Now we must prove uniqueness.
-Suppose there exists σk’ such that σi * σk’ = σj
-By first part of proof we reach σi * σk’ = σi * σk
-Compose both sides by σi-1 and using associativity to simplify yields the
result that σk’ = σk, thus σk is unique. QED.
The following is a discussion of a very simple example which offers a glimpse of Poyla’s
original motivation in developing his theory as it could be seen as a chemical compound. Think
about how many ways one could arrange four balls where each ball is either black or white.
However, let us say that these balls are actually connected in such a way that they form a square.
We are going to refer to each different construction as a coloring of the corners of the square.
Locking the square in a fixed position makes each individual coloring a function that maps {1, 2,
3, 4} to {black, white} where each of the vertices of the square are labeled one to four so our
function maps each corner to black or white. Thus, there are 24 possible colorings. These
colorings along with the idea of a function can be seen below in figure one that displays the
sixteen possible colorings and labels them fi where i is a number from one to sixteen.
However, we now let the square freely rotate and notice that the number of different
colorings shrinks as we consider colorings to be equivalent if we can reach a different coloring
by simply rotating or flipping the square. These actions are known as the symmetries of the
square and will be expanded on further in the discussion of a dihedral group. It is important to
note that these actions do not rely on the colorings but on the square itself.
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Figure 1.3
Allowing the square to rotate we conclude that there are actually only six different
coloring and our list of sixteen colorings seen in figure one above reduces to the six colorings
seen below in figure two. Each coloring in figure one is equivalent to exactly one in figure two
below, further they can all reach one the one they are equivalent to by a rotation or flip of the
square. This idea seems relatively simple for this example but it is of great use when the number
of initial different colorings becomes much larger or a third color is added.
Figure 2.4
In its simplest form, a permutation of a set, S, of objects, is a rearrangement of its
elements. In our case, the elements of the set are the numbers 1 to n depending on the size of the
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set, where the number of possible permutations is n!. For example, the set S = {1, 2, 3} has six
possible permutations since n=3 and 3! = 6. The six possible permutations follow:
123
132
213
231
321
312
However, notice that each of these possible orderings takes each element of S and maps it
to another element of S, thus we define a permutation of the set S to be a bijective function from
S to S. Sn is the permutation group that is comprised of the set of all possible permutations of a
set S paired with the binary operation of composition. Any subgroup of Sn is also a permutation
group. The following theorem and proof show that the set of all permutations paired with the
function of composition is indeed a group.
Theorem (Symmetric Group): For any integer n > 0, (Sn, o) is a group. That is, the set of all
permutations of the set S is a group under the operation of function composition5.
Proof: Let n be a positive integer. Recall the four axioms necessary for a group.
Closure: We know that a permutation is a bijection. Also the composition of two
bijective functions results in a bijective function so Sn is closed under
composition.
Associativity: We know the function of composition is associative, thus Sn has the
associative property.
Identity: We define e as a bijective function such that e(j)=j for all j є the set {n}. Since e
is a bijection, it is clear e is an element of the set Sn. Further e ᵒ f = f = f ᵒ e
because f(e(j)) = f(j) and e(f(j)) = f(j) for all j є the set {n}. Therefore Sn has an
identity permutation that maps each element of {n} to itself.
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Inverse: We know that a permutation is a one to one function since every object
contained within the set {n} gets mapped to an object of {n} with no repeats.
Thus the inverse relation f-1 is also a bijection, and is contained within the set Sn
as it is all the possible permutations of the set {n}. Further f ᵒ f-1 = e = f-1 ᵒ f.
Therefore each element of Sn has an inverse in Sn.
Therefore, the set S paired with the binary operation of composition is a group, specifically
called the symmetric group, as it satisfies the four axioms necessary to be a group. QED.
Figure 3.6
It was previously mentioned that a permutation allows us to map the movement of an
object. These movements can be thoughts of as symmetries of the object and are what we model
using groups. The group Sn is useful because it allows us to work within the confines of groups
and group theory, but when it comes to finding a way to simplify the counting it is relatively not
useful. Instead, we will consider another specialized group known as the dihedral group. This
also allows us to generalize the end result to more than just a square, even though we will
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continue working with the corners of a square. For simplicity we are going to be working with a
regular n-gon. The dihedral group is a subgroup of the symmetric group, Sn, and is denoted Dn
where n is greater than or equal to three. The dihedral group consists of the symmetries of the
regular n-gon where the symmetries of the object are thought of as the rotations and flips of the
object, as can be seen for a square in the Figure 3 above. The order, or number of total
symmetries is found by taking n and multiplying by two. For example, the order of a hexagon
would be twelve as there are six possible rotations and six possible reflections across the lines of
symmetry. When finding the elements of Dn first label the vertices of the regular n-gon you are
working with, with the numbers one to n. Then write the permutations for each of the rotations or
flips across axis of symmetry. An example of this can be seen below in table one. The motions
match those that are shown in figure three at the beginning of this discussion. It is important to
remember that each permutation is a function which maps {1, 2, 3, 4} to {black, white}.
Table 1.
Motion
Permutation
Motion
Permutation
(
1
3
2 3 4
)
4 1 2
F1,2
(
1
4
2 3 4
)
1 2 3
F2,3
(
(
F2
R3
2 3 4
)
3 4 1
(
1
2
R2
(
(
F1
R1
2 3 4
)
2 3 4
(
1
1
I (R0)
1
1
2 3 4
)
2 3 2
1
3
2 3 4
)
2 1 4
1
2
2 3 4
)
1 4 3
1
4
2 3 4
)
3 2 1
We now have all the necessary components needed to begin a discussion leading towards
the Cauchy-Frobenius-Burnside theorem (CFB). That is the basis for Poyla’s enumeration
theorem so it is logical to build toward the CFB and later expand to Poyla’s theory. To begin we
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will define group action in order to understand how our group of permutations acts on the set of
functions (colorings).
Definition (Group acting on functions)7: Let S and C be finite sets, and let G be a group of
permutations of S. For the set CS of functions f: SC, the actions of G on CS is defined
by:
(θ(f))(s) := f(θ-1(s)) for each θ є G and s є S.
The purpose of this definition is to give us a mathematical way to find the result of an element of
the group G acting on a given function f. For basic scenarios, we could use our intuition or
investigate pictorially to find what a given group element does as it acts on a function but for
larger situations, the result may not be so intuitive.
Example: Recall the example discussed previously. Take G to be the dihedral group D4 since n
equals 4 in the set S which equals {1, 2, 3, 4} and C equals the set of colorings, {black,
white}. Below it is worked out showing that R3(f6) equals f9 which verifies our intuition
that says rotating f6 270 degrees results in f9. Recall, R3 describes how the group acts on
the corners and not the colorings.
R3(f6)(1) = f6(R3-1(1)) = f6(R1(1)) = f6(2) = white = f9(1)
R3(f6)(2) = f6(R3-1(2)) = f6(R1(2)) = f6(3) = black = f9(2)
R3(f6)(3) = f6(R3-1(3)) = f6(R1(3)) = f6(4) = black = f9(3)
R3(f6)(4) = f6(R3-1(4)) = f6(R1(4)) = f6(1) = white = f9(4)
So it is true that R3(f6) equals f9 because f9 is the coloring that labels the corners 1-4 of the square
white-black-black-white.
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This same procedure could be applied for any two finite sets, C and S, and group of
permutations of S. We now begin the discussion which will lead us to the Cauchy-FrobeniusBurnside Theorem. The discussion begins with an introduction of a couple of concepts which we
will put together in order to reach our goal of the CFB Theorem.
Definition: Orbit
Let S and C be finite sets and G be a permutation group of the set S. For any f є Cs, the orbit of f
under G is the set
orbG(f) = {σ(f) : σ є G}.8
That is the orbit of a given function is the set of functions which are reachable by applying each
of the elements of the group to the function. This is done using the previous definition of a group
acting on a function.
Example: Find the orbit of f12 using the same two color square example.
Given that G = D4 we need to simply apply each of the eight group elements to f12.
I(f12) = f12
R1(f12) = f13
R2(f12) = f14
R3(f12) = f15
F1(f12) = f12
F2(f12) = f14
F1,2(f12) = f13 F2,3(f12) = f15
Thus, orbD4(f12) = {f12, f13, f14, f15}.
There are a few interesting observations that can be drawn from looking a little deeper at
the concept of an orbit. The first observation is that every orbit is nonempty. This comes from
the fact that by definition every function f is contained within its own orbit. By the definition of
orbit, we are working with a group, and in order for something to be considered to be a group,
recall, it must have an identity element. To find the orbit of a function we have to apply all group
elements to the given function. Since one of those elements will always be the identity, we will
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always have at least the original function reachable and thus in the orbit, making every orbit
nonempty. A second observation stems from the first. That is, that every element of Cs is in some
orbit which we know because every function is in its own orbit. Further, the union of the orbits
equals the entire set of functions, CS. The last observation to be discussed is addressed formally
below in the theorem and accompanying proof.
Theorem: Orbits Partition the set CS
Let S and C be finite sets, and G be a permutation group of S. Then the orbits of CS partition CS.
That is, the set O = {orbG(f) : f є CS} is a partition of CS.9
Proof: It is important to first understand what this theorem is saying. For any set S, a
partition of S is a set of nonempty, disjoint, subsets of S whose union is S.
Therefore, we need to show that the set of orbits satisfy those three things.
~Recall, the previous discussion about every orbit being nonempty because each
function is contained within its own orbit and that every function is in an orbit.
~Thus, we can easily conclude that the set of orbits is a set of nonempty sets
whose union equals CS, leaving us to show that the orbits are disjoint, which
follows.
~First assume that orbG(f1) and orbG(f2) are two orbits that are not disjoint.
~Let t be any function belonging to both orbits.
~By definition of orbits this means that t = σ1(f1) and t = σ2(f2) for some σ1, σ2 є G
~We prove that orbG(f1) = orbG(f2) by showing that each is a subset of the other.
-Let h є orbG(f1)
-Hence, h = σ(f1) for some σ є G.
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-But since σ1(f1) = σ2(f2) we can rewrite f1 = σ1-1(σ2(f2)). This is able to be
done because of the Lemma 1 above.
-Thus, h = σ(f1) = σ(σ1-1(σ2(f2))) = (σ ᵒ σ1 ᵒ σ2)(f2) = ϕ(f2) by closure
component of a group, where ϕ is an element of G.
-Therefore, h = σ(f2) so h є orbG(f2) and hence orbG(f1) is a subset of
orbG(f2).
~Similarly the reverse can be shown the same way, resulting in the two orbits
being a subset of one another, showing they are equal to each other. This results
in a contradiction. Therefore we can conclude that the orbits are disjoint, and thus
the set of orbits partitions the set CS. QED.
Definition: Fixed Point Set
Let S and C be finite sets and G be a group of permutations of S. For any σ є G, the fixed point
set of σ is the set
fixG(σ) = {f є CS : σ(f) = f}.10
That is the fixed point set of a given group element σ is the set of functions which is unchanged
by the given group element. Further, the fixed point set of the group element, identity, will
always yield a set which equals the entire set of functions, CS.
Example: Find the fixed point set of F1,2.
Working with the same square example, we have G = D4.
To find fixD4(F1,2) we need to apply F1,2 to the collection of functions Cs and determine
which functions remain the same after F1,2 has been applied.
Thus, fixD4(F1,2) = {f1, f6, f8, f16}.
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Definition: Stabilizer
Let S and C be finite sets and G be a permutation group of S. For any f є CS, the stabilizer of f
under G is the set
stabG(f) = {σ є G : σ(f) = f}.11
That is the stabilizer of a given function is the set of group elements which when acting on the
said function, returns the original function.
Example: Find the stabilizer of f2.
Working with the same two color square example, we need to apply the elements of D4 to
f2 and keep the elements which return f2.
Thus, stabD4(f2) = {I, F1}.
We now have a working definition of orbits, fixed point sets, and stabilizers which allows us to
state the Cauchy-Frobenius-Burnside (CFB) theorem, use it to finish our two color square
example that we have been working with, and prove the theorem using three lemmas. The CFB
will allow us to determine the number of distinct colorings once we allow our square to rotate
freely around its center axis. It could also be applied to larger groups or more than two colors.
Theorem: Cauchy-Frobenius-Burnside
Let S and C be finite sets, G be a permutation group of the set S, and O be the set of orbits of CS.
Then
|O| =
1
∑|𝑓𝑖𝑥𝐺 (𝜎)| 12
|𝐺|
𝜎𝜖𝐺
Thus, determining the cardinality of the set of orbits amounts to determining the size of each
fixed point set. Intuitively this seems justifiable as the set of orbits partition the set of functions,
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and an individual orbit tells us the functions which are equivalent in terms of which we can reach
by allowing the square to rotate or flip. Let’s now apply this theorem to our two colored square
example.
Example: Find the cardinality of the set of orbits.
We need to find the cardinality of each of the fixed point sets. At this point we do not care about
the individual functions of each fixed point set, but just the number contained within each set.
Recall, fixed point set of identity equals the set CS thus |fixD4(I)| =24.
|fixD4(R1)| =21
|fixD4(R2)| =22
|fixD4(F2)| =23
|fixD4(R3)| =21
|fixD4(F1,2)| =22
|fixD4(F1)| =23
|fixD4(F2,3)| =22
|G| = 8
We can now use the equation given to use within CFB:
|O| = (1/8)(16+2+4+2+8+8+4+4) = (1/8)(48) = 6
Thus we know that there are six different colorings once the square is given the ability to rotate
around its center axis and flip across its axis of symmetries, confirming what we were able to
find pictorially. The proof of CFB is going to be broken into three different stages that allow us
to link all of the topics we have just discussed together.
Proof of CFB:
Assume S and C are finite sets, G is a permutation group of the set S, and O is the set of orbits of
CS. We want to show that
|O| =
1
∑|𝑓𝑖𝑥𝐺 (𝜎)|
|𝐺|
𝜎𝜖𝐺
Our first step is to link orbits and stabilizers together.
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-We do so using a lemma 2. The lemma states that the cardinality of the orbit of a
function multiplied by the cardinality of the stabilizer equals the cardinality of the group.
That is,
|𝑜𝑟𝑏𝐺 (𝑓)||𝑠𝑡𝑎𝑏𝐺 (𝑓)| = |𝐺|
Our next step is to derive a formula for the number of orbits. We do so using Lemma 3.
-Lemma 3: Let S and C be finite sets, G be a permutation group of S, and O be the set of
orbits of CS. Then
|O| =
1
∑ |𝑠𝑡𝑎𝑏𝐺 (𝑓)|
|𝐺| 𝑆
𝑓𝜖𝐶
Proof: By lemma 2 we know that for any function contained within CS
|𝑜𝑟𝑏𝐺 (𝑓)||𝑠𝑡𝑎𝑏𝐺 (𝑓)| = |𝐺|
~To begin we divide the cardinality of the orbit to the right side and then the
cardinality of the group G to the left side to get orbits on one side and everything
else on the other, resulting in the following:
1
1
|𝑠𝑡𝑎𝑏𝐺 (𝑓)|
=
|𝑜𝑟𝑏𝐺 (𝑓)|
|𝐺|
~We now sum both sides over all f є CS.
∑
1
1
|𝑠𝑡𝑎𝑏𝐺 (𝑓)|
=
∑
|𝐺|
𝑓∈ 𝐶 𝑆 |𝑜𝑟𝑏𝐺 (𝑓)|
𝑓∈ 𝐶 𝑆
~We are able to pull the one over the cardinality of G out of the summation
because it is constant for each of the functions we are summing over.
1
~Observation: ∑𝑓∈ 𝐶 𝑆 |𝑜𝑟𝑏
𝐺 (𝑓)|
= |O|
~The orbit of a given function may have more than one function in the set. Recall
that the set of all orbits partition the set of functions thus, we know that the orbits
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are disjoint resulting in the overlap to still only count each distinct orbit once,
making the observation above possible.
~Thus, our equation reduces to exactly what we want it to be, and follows.
|O| =
1
|𝑠𝑡𝑎𝑏𝐺 (𝑓)|
∑
|𝐺|
𝑓∈ 𝐶 𝑆
~That is, the number of orbits is given to us in terms of one over the cardinality of
G times the summation of the stabilizer of each function in our set CS. QED.
The remaining step is now simply adjusting Lemma 3 in order to get the number of orbits in
terms of fixed point sets instead of stabilizers. The number of orbits could be found with Lemma
3 or with the CFB; they yield the same results. It is more or less up to the individual which they
are more comfortable working with, fixed point sets or stabilizers.
We will transform Lemma 3 into the result we are looking for in the CFB using yet another
lemma.
Lemma 4: Let S and C be finite sets and G be a permutation group of S. Then the
summation of the cardinality of the stabilizers over f є CS equals the summation of the
cardinality of the fixed point sets over σ є G. That is,
∑
𝑓∈ 𝐶 𝑆
|𝑠𝑡𝑎𝑏𝐺 (𝑓)| = ∑
𝜎 ∈𝐺
|𝑓𝑖𝑥𝐺 (𝜎)|
Proof: For this, we must look at two separate cases. That is, we are looking for the
number of pairs (σ, f) є G x CS where σ(f) = f.
~Each side counts the same thing and when summed they equal one another.
~First, let’s think about the left side. The cardinality of the stabilizer gives us the
number of elements of G that satisfy σ(f) = f for a given function. When summed
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over all f є CS we receive the total instances where a given group element acting
on a function returns that function.
~Second, move to the right side. The cardinality of the fixed point set of a given
function gives us the number of elements of CS that remain unchanged when the
given group element acts on them. Thus, summing over the elements of the group
returns the total instances of a group element acting on a function and returning
the same function. QED.
The CFB follows immediately by combining the results of Lemma 3 and Lemma 4. Substituting
Lemma 4 into Lemma 3 yields an equation by which the cardinality of the set of orbits can be
found in terms of the summation of the cardinality of the fixed point sets times one over the
cardinality of the group G.
|O| =
1
∑|𝑓𝑖𝑥𝐺 (𝜎)|
|𝐺|
𝜎𝜖𝐺
QED.
The Cauchy-Frobenius-Burnside theorem provides us with a very easy and straightforward way to find the number of nonequivalent colorings, which was our original goal.
However, one drawback to the theorem is that it restricts us to having no restrictions, such as
having exactly x amount of one color and y amount of the other. The square example with the
two colored beads is a very simple but useful example. It allowed us to investigate the different
concepts needed in order to formulate working definitions which lead to the CFB without getting
lost in the numbers. The same principles (orbits, fixed point sets, stabilizers) can be applied to a
wide range of bigger problems allowing one to calculate the number of orbits for a necklace of n
beads with c colors. The more beads and additional colors does create more work but the same
concepts are applicable and follow directly from the work with the square example. Additional
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information about the CFB or Burnside’s Theorem can be found in articles titled Enumeration by
Algebraic Combinatorics and Counting and Coloring under Symmetry. Both are referenced
below to give additional background if needed.
Polya’s enumeration theorem allows for those restrictions that the CFB does not, as it
uses a generating function to look at the different colorings in regards to specific properties and
like the CFB can be used to answer a large number of counting questions all at once. For further
reading on the work of Polya himself, refer to the references below along with Bogart (1990).
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References
Atwood, Carolyn. “Enumeration by Algebraic Combinatorics.”
https://www.whitman.edu/mathematics/SeniorProjectArchive/2009/atwood.pdf.
Bjørge, Amanda Noel. "Counting and Coloring with Symmetry."
http://www.diva-portal.org/smash/get/diva2:348745/FULLTEXT01.pdf
Hungerford, Thomas W. Abstract Algebra: An Introduction. Fort Worth: Saunders College Pub.,
1997. Print.
Mazur, David R. Combinatorics a Guided Tour. Washington, DC: Mathematical Association of
America, 2010. Print.
Tucker, Alan. Applied Combinatorics. New York: Wiley & Sons, 1995. Print.