Burnside's Theorem will allow us to count the orbits, that is, the different colorings, in a variety of problems. We first need some lemmas.

If $c$ is a coloring, $[c]$ is the orbit of $c$, that is, the equivalence class of $c$. Let $G(c)$ be the set of permutations in $G$ that fix $c$, that is, those $\phi$ such that $\phi(c)=c$; the permutation in figure 6.1.4 fixes the coloring in the bottom row, for example.

Lemma 6.2.1 $G(c)$ is a group of permutations.

**Proof.**
We check the properties of a group from definition 6.1.1.

Suppose $\phi$ and $\sigma$ both fix $c$; then $\phi(\sigma(c))=\phi(c)=c$, so $\phi\circ\sigma$ fixes $c$ and $\phi\circ\sigma\in G(c)$.

The identity permutation fixes all colorings, so $\id\in G(c)$.

If $\phi(c)=c$ then $\phi^{-1}(c)=\phi^{-1}(\phi(c))=\id(c)=c$, so $\phi^{-1}\in G(c)$. $\qed$

Lemma 6.2.2 $|G|=|[c]|\cdot|G(c)|$.

**Proof.**
For $\phi$ and $\sigma$ in $G$, define $\phi\sim\sigma$ if
$\sigma^{-1}\circ\phi\in G(c)$. This is an equivalence relation:

1. $\sigma^{-1}\circ\sigma$ is the identity function, which is in $G(c)$. Thus $\sigma\sim\sigma$, so the relation is reflexive.

2. If $\phi\sim\sigma$, $\sigma^{-1}\circ\phi\in G(c)$. Then $(\sigma^{-1}\circ\phi)^{-1}\in G(c)$, and $(\sigma^{-1}\circ\phi)^{-1}=\phi^{-1}\circ\sigma\in G(c)$, so $\sigma\sim\phi$ and $\sim$ is symmetric.

3. If $\phi\sim\sigma$ and $\sigma\sim\tau$, then $\sigma^{-1}\circ\phi\in G(c)$ and $\tau^{-1}\circ\sigma\in G(c)$, so $(\tau^{-1}\circ\sigma)\circ(\sigma^{-1}\circ\phi)\in G(c)$. Since $(\tau^{-1}\circ\sigma)\circ(\sigma^{-1}\circ\phi)= \tau^{-1}\circ\phi$, $\phi\sim\tau$, and $\sim$ is transitive.

Now we claim that the equivalence class of $\phi$ is $A=\{\phi\circ\sigma\mid\sigma\in G(c)\}$. First, suppose that $\sigma\in G(c)$; then $\phi^{-1}\circ\phi\circ\sigma=\sigma\in G(c)$, so $\phi\circ\sigma\sim\phi$ and $A\subseteq[\phi]$. Next, suppose $\phi\sim\tau$, so $\tau^{-1}\circ\phi=\gamma\in G(c)$. Then $\phi\circ\gamma^{-1}=\tau$, so $\tau\in A$ and $[\phi]\subseteq A$.

Now we show that each equivalence class is the same size as $G(c)$. Define $f\colon G(c)\to \{\phi\circ\sigma\mid\sigma\in G(c)\}$ by $f(\gamma)=\phi\circ\gamma$. If $f(\gamma_1)=f(\gamma_2)$, then $$\eqalign{ \phi\circ\gamma_1&=\phi\circ\gamma_2\cr \phi^{-1}\circ\phi\circ\gamma_1&=\phi^{-1}\circ\phi\circ\gamma_2\cr \gamma_1&=\gamma_2\cr }$$ so $f$ is 1–1. Since every $\phi\circ\gamma\in\{\phi\circ\sigma\mid\sigma\in G(c)\}$ is $f(\gamma)$, $f$ is onto.

Thus the number of equivalence classes is $|G|/|G(c)|$.

Finally, we show that the number of equivalence classes is $|[c]|$. Let the set of equivalence classes in $G$ be $E$, that is, $E=\{[\phi]\mid \phi\in G\}$. We define $g\colon [c]\to E$ and show that $g$ is a bijection. Suppose $d\in[c]$, so $d=\sigma(c)$ for some $\sigma\in G$. Let $g(d)=[\sigma]$.

First, we show $g$ is well-defined. If $d=\sigma_1(c)=\sigma_2(c)$, then $\sigma_2^{-1}\circ\sigma_1(c)=c$, so $\sigma_1\sim\sigma_2$ and $[\sigma_1]=[\sigma_2]$, that is, $g(\sigma_1(c))=g(\sigma_2(c))$.

Next, suppose $g(d_1)=g(d_2)$. This means that $d_1=\sigma_1(c)$, $d_2=\sigma_2(c)$, and $[\sigma_1]=[\sigma_2]$. Hence $\sigma_2^{-1}\circ\sigma_1(c)=c$, so $\sigma_1(c)=\sigma_2(c)$ and thus $d_1=d_2$, so $g$ is 1–1.

Suppose that $[\sigma]\in E$. Then $g(\sigma(c))=[\sigma]$, so $g$ is onto.

Thus we have $$|[c]|=|E|={|G|\over|G(c)|}$$ and $$|G(c)|\cdot|[c]|=|G|$$ as desired. $\qed$

Corollary 6.2.3 If $c\sim d$ then $|G(c)|=|G(d)|$.

**Proof.**
Since $c\sim d$, $[c]=[d]$, and
$$\eqalign{
{|G|\over|G(c)|} = |[c]|&=|[d]|= {|G|\over|G(d)|}\cr
|G(c)|&=|G(d)|\cr
}$$
$\qed$

Definition 6.2.4 If group $G$ acts on the colorings of an object and $\sigma\in G$, $\fix(\sigma)$ is the set of colorings that are fixed by $\sigma$. $\square$

Theorem 6.2.5 (Burnside's Theorem) If group $G$ acts on the colorings of an object, the number of distinct colorings modulo $G$ is $${1\over|G|}\sum_{\sigma\in G} |\fix(\sigma)|.$$

**Proof.**
Let $C$ be the set of all colorings, and let $\cal O$ be the set of
orbits. Let $c_1,c_2,\ldots,c_k$ be a list of colorings, one in each
orbit, so that the orbits are $[c_1],[c_2],\ldots,[c_k]$.
Consider the sum $\ds\sum_{c\in C}|G(c)|$:
$$\eqalign{
\sum_{c\in C}|G(c)|&=\sum_{i=1}^k\sum_{c\in[c_i]}|G(c)|\cr
&=\sum_{i=1}^k\sum_{c\in[c_i]}|G(c_i)|\cr
&=\sum_{i=1}^k\sum_{c\in[c_i]}{|G|\over|[c_i]|}\cr
&=\sum_{i=1}^k |[c_i]|{|G|\over|[c_i]|}\cr
&=\sum_{i=1}^k |G|=|G|\sum_{i=1}^k 1 = |G||{\cal O}|.\cr
}$$
Then
$$|{\cal O}| = {1\over|G|}\sum_{c\in C}|G(c)|.$$
This already gives an interesting formula for $|{\cal O}|$, but it is
unwieldy, since the number of colorings is typically quite large;
indeed, since we typically want to compute the number of orbits for
$k$ colors, the number of colorings is not a fixed number.

With just a little more work we can fix this problem: $$\eqalign{ \sum_{c\in C}|G(c)|&=\sum_{c\in C}\sum_{\sigma\in G(c)} 1\cr &=\sum_{\sigma\in G}\sum_{c\in \fix(\sigma)} 1\cr &=\sum_{\sigma\in G}|\fix(\sigma)|.\cr }$$ Now $$|{\cal O}| = {1\over|G|}\sum_{\sigma\in G}|\fix(\sigma)|$$ as desired. $\qed$

Since the group of permutations in a typical problem is fairly small, the sum in Burnside's Theorem is usually manageable. Moreover, we can make the task of computing $|\fix(\sigma)|$ fairly straightforward. Let's consider a particular example, the permutation of figure 6.1.4, shown again in figure 6.2.1. If we are using $k$ colors, how many colorings of the pentagon are fixed by this permutation? Since the permutation swaps vertices 2 and 5, they must be the same color if $\phi$ is to fix the coloring. Likewise vertices 3 and 4 must be the same color; vertex 1 can be any color. Thus, the number of colorings fixed by $\phi$ is $k^3$. It is easy to see that every "flip'' permutation of the pentagon is essentially the same, so for each of the five flip permutations, the size of $\fix(\sigma)$ is $k^3$.

$\phi\colon$ | $\longmapsto$ |

Every permutation can be written in
**cycle form**:
The permutation in figure 6.2.1, for
example, is $(1)(2,5)(3,4)$. A cycle in this context is
a sequence $(x_1,x_2,\ldots,x_k)$, meaning that
$\phi(x_1)=x_2$, $\phi(x_2)=x_3$, and so on until
$\phi(x_k)=x_1$. Following our reasoning above, the vertices in each
cycle must be colored the same color, and the total number of colors
fixed by $\phi$ is $k^m$, where $m$ is the number of cycles.

Let's apply this to another permutation, shown in figure 6.2.2. This permutation consists of a single cycle, so every vertex must have the same color, and the number of colorings fixed by $\phi$ is $k^1$. All rotations of the pentagon consist of a single cycle except the trivial rotation, that is, the identity permutation. In cycle form, the identity permutation is $(1)(2)(3)(4)(5)$, so the number of colorings fixed by the identity is $k^5$. Putting everything together, we thus have $$|{\cal O}|={1\over 10}(k^5+k+k+k+k+k^3+k^3+k^3+k^3+k^3)= {1\over 10}(k^5+5k^3+4k).$$ For example, the number of different 3-colorings is $(3^5+5\cdot3^3+4\cdot 3)/10=39$.

$\phi\colon$ | $\longmapsto$ |

Example 6.2.6 We find the number of distinct colorings of the vertices of a square with $k$ colors, modulo $D_4$, the dihedral group of size 8. The elements of $D_4$ are the four rotations $r_0$, $r_{90}$, $r_{180}$, $r_{270}$, where $r_i$ is the counterclockwise rotation by $i$ degrees, and the four reflections $f_H$, $f_V$, $f_D$, $f_A$, as indicated in figure 6.2.3.

In cycle notation these permutations are: $$\eqalign{ r_0&=(1)(2)(3)(4)\cr r_{90}&=(1,4,3,2)\cr r_{180}&=(1,3)(2,4)\cr r_{270}&=(1,2,3,4)\cr f_H&=(1,4)(2,3)\cr f_V&=(1,2)(3,4)\cr f_D&=(1)(2,4)(3)\cr f_A&=(1,3)(2)(4).\cr }$$ so the number of colorings is $$f(k)={1\over 8}(k^4+k+k^2+k+k^2+k^2+k^3+k^3)={1\over8}(k^4+2k^3+3k^2+2k).$$ For example, $f(2)=6$; the six colorings are shown in figure 6.2.4. $\square$

Example 6.2.7
Here is a more complicated example: how many different graphs
are there on four vertices? In this case, of course, "different''
means "non-isomorphic''. We can interpret this as a coloring problem:
Color the *edges* of the complete graph $K_4$ with two colors,
say black and white. The black edges form a graph; the white edges are
the ones left out of the graph. The group $G$ we need to consider is
all permutations of the six edges of $K_4$ induced by a permutation of the
vertices, so $|G|=4!=24$. All we need to know is the number of cycles
in each permutation; we consider a number of cases.

**Case 1.** The identity permutation on the vertices
induces the identity permutation on the edges, with 6 cycles, so the
contribution to the sum is $2^6$.

**Case 2.** A 4-cycle on the vertices induces a
permutation of the edges consisting of one 4-cycle and one 2-cycle,
that is, two cycles. There are $3!=6$ 4-cycles on the vertices, so the
contribution of all of these is $6\cdot2^2$.

**Case 3.** A permutation of the vertices consisting
of a 3-cycle and a 1-cycle induces a permutation of the edges
consisting of two 3-cycles. There are $4\cdot2=8$ such permutations of
the vertices, so the contribution of all is $8\cdot2^2$.

**Case 4.** A permutation of the vertices consisting
of two 2-cycles induces a permutation of the edges consisting of
two 1-cycles and two 2-cycles. There are ${1\over2}{4\choose2}=3$ such
permutations, so the contribution is $3\cdot 2^4$.

**Case 5.** A permutation of the vertices consisting
of a 2-cycle and two 1-cycles induces a permutation of the edges
consisting of two 1-cycles and two 2-cycles. There are
${4\choose2}=6$ such permutations, so the contribution is $6\cdot
2^4$.

The number of distinct colorings, that is, the number of distinct graphs on four vertices, is $${1\over24}(2^6+6\cdot2^2+8\cdot2^2+3\cdot 2^4+6\cdot2^4)= {1\over24}(264)=11.$$ $\square$

It is possible, though a bit difficult, to see that for $n$ vertices the result is $$\eqalignno{ f(n)&=\sum_{\bf j}\prod_{k=1}^n{1\over k^{j_k} j_k!} \prod_{k=1}^{\lfloor n/2\rfloor}2^{k j_{2k}} \!\!\prod_{k=1}^{\lfloor (n-1)/2\rfloor}\!\!2^{kj_{2k+1}} \prod_{k=1}^{\lfloor n/2\rfloor} 2^{kC(j_k,2)} \!\!\!\!\prod_{1\le r< s\le n-1}\!\!\!\!\!\!\!\! 2^{\gcd(r,s)j_rj_s}\qquad (6.2.1)\cr} $$ where the sum is over all partitions ${\bf j}=(j_1,j_2,\ldots,j_n)$ of $n$, that is, over all $\bf j$ such that $j_1+2j_2+3j_3+\cdots+nj_n=n$, and $C(m,2)={m\choose2}$. With this formula and a computer it is easy to compute the number of graphs when $n$ is not too large; for example, $f(5)=34$, so there are 34 different five-vertex graphs.

In light of the forgoing discussion, we can restate theorem 6.2.5. If $\sigma$ is a permutation, let $\#\sigma$ denote the number of cycles when $\sigma$ is written in cycle form.

Corollary 6.2.8 If group $G$ acts on the colorings of an object, the number of distinct colorings modulo $G$ with $k$ colors is $${1\over|G|}\sum_{\sigma\in G} k^{\#\sigma}.$$ $\qed$

## Exercises 6.2

**Ex 6.2.1**
Write the 12 permutations of the vertices of the regular
tetrahedron corresponding to the 12 rigid motions of the regular
tetrahedron in cycle form. Use the labeling below.

**Ex 6.2.2**
Find the number of different colorings of the
vertices of a regular tetrahedron with $k$ colors, modulo the rigid
motions.

**Ex 6.2.3**
Write the 12 permutations of the edges of the regular
tetrahedron corresponding to the 12 rigid motions of the regular
tetrahedron in cycle form. Use the labeling below.

**Ex 6.2.4**
Find the number of different colorings of the
edges of a regular tetrahedron with $k$ colors, modulo the rigid
motions.

**Ex 6.2.5**
Find the number of non-isomorphic graphs on 5
vertices "by hand'', that is, using the method of
example 6.2.7.