TITLE: Burnside's Lemma applied to grids with interchanging rows and columns
QUESTION [7 upvotes]: I've recently learned about Burnside's Lemma (https://en.wikipedia.org/wiki/Burnside%27s_lemma) and its applications to rotating necklaces, coloring cubes and such, but I fear my understanding of it isn't mature enough and am unable to apply it to the following situation:
Suppose you have a 2x3 matrix, and a set of 3 distinct colors $R, Y, B$. How many non-equivalent ways are there to color the matrix? Note that two matrices $m_1$ and $m_2$ are considered equivalent if you can turn the former into the latter by swapping any rows and/or columns as many times as you want.
So I got started by setting up the Burnside's equation as follows:
\begin{align*}\frac{1}{|G|}\sum_{g \in G} |X^g| &= \frac{1}{H! W!} \sum_{\sigma \in S_H}\sum_{\tau \in S_W} |X^{(\sigma, \tau)}|.
\end{align*}
where $|G|$ is the total no. of elements that act to permute the set $X$ of matrices with height and width $H\times W$. Since there are $H!$ ways to permute the rows and $W!$ ways to permute the columns, the total no. of ways to permute the matrix is $|G| = H!W!$. So we can iterate through every way to permute the rows and every way to permute the columns (hence the double summation above), and for each permutation, find out the value of |$X^{(\sigma, \tau)}$| -- which is the number of matrices that are fixed (a.k.a. not changed) when applying permutation $\sigma$ to the rows and $\tau$ to the columns.
For example, say we have a 2x3 grid with the following indices for each cell:
(1,1) (1,2) (1,3)
(2,1) (2,2) (2,3)
There would be 2! = 2 ways to permute the rows and 3! = 6 ways to permute the columns, so the Burnside's equation becomes:
\begin{align*}\frac{1}{2! 3!} \sum_{\sigma \in S_2}\sum_{\tau \in S_3} |X^{(\sigma, \tau)}|
\end{align*}
It is here that I hit a stumbling block. Given we have 3 distinct colors to work with, it isn't clear to me how to count up |$X^{(\sigma, \tau)}$| for each permutation. If someone could show me a step-by-step way to compute the answer for this specific example, I feel I could probably learn to apply it to a more general situation, with arbitrary values of $W$, $H$, and number of colors. Thanks!

REPLY [9 votes]: Let me  just observe that  while rigorous notation  is a must  for any
serious mathematician it can sometimes block the path of the beginning
reader. What is happening here is very simple. We need the cycle index
of the  permutations from $S_2\times S_3$  acting on the slots  of the
matrix. While  the present case can  be computed by inspection  I will
explain the general method. 
We require the  cycle index $Z(M_{2,3})$ of  the pairs $(\sigma,\tau)$
of row and  column permutations acting simultaneously on  the slots of
the matrix  ($6$ of  them, and with  $12$ permutations  total).  Start
from the two cycle indices for the row and column permutations
$$Z(S_2) = \frac{1}{2} a_1^2 + \frac{1}{2} a_2$$
and
$$Z(S_3) =\frac{1}{6} b_1^3 + \frac{1}{2} b_1 b_2 + \frac{1}{3} b_3.$$
The method here is as follows. We draw a diagram of the cycle types of
$\sigma$ and $\tau,$ perhaps one beneath  the other. Now for the cycle
index of the  cartesian product of $S_2$ and $S_3$  we must factor the
combined action of the two permutations  on the slots into cycles.  We
represent row-column pairs  i.e. slots by marking say the  row and the
column and  connecting them with a  edge, not to be  confused with the
directed edges of the two cycles.  This edge travels in parallel along
the two  cycles it  is on  and returns to  its initial  position after
$\mathrm{lcm}(l_1, l_2)$ steps, where $l_{1,2}$ are the lengths of the
cycles.  As  the  pair  of cycles  contributes  $l_1\times  l_2$  slot
identifiers we  get for the  contribition to the combined  cycle index
the term
$$a_{\mathrm{lcm}(l_1, l_2)}^{l_1 l_2/\mathrm{lcm}(l_1, l_2)}.$$
We now do the computation. There are thus six possible combinations of
cycle types  that combine to  form $Z(M_{2,3})$.  We process  these in
turn, leaving the most difficult part for last.
First, combining $a_1^2$  and $b_1^3.$ This fixes all six  slots for a
contribution of
$$\frac{1}{12} a_1^6.$$
Second, combining  $a_1^2$ and $b_3.$  This partitions the  pairs into
three-cycles for a contribution of
$$\frac{1}{6} a_3^2.$$
Third,  combining  $a_2$  and  $b_1^3.$ Here  we  have  everything  on
two-cycles to get
$$\frac{1}{12} a_2^3.$$
Fourth, combining $a_2$ and $b_3$. This will produce a six-cycle to get
$$\frac{1}{6} a_6.$$
Now for the tricky part. Fifth, combining $a_1^2$ and $b_1 b_2.$ There
are two  pairs that are fixed  by these, and two  pairs on two-cycles
and we obtain
$$\frac{1}{4} a_1^2 a_2^2.$$
Finally, sixth, combining  $a_2$ and $b_1 b_2.$ We  have everything on
two-cycles and obtain 
$$\frac{1}{4} a_2^3.$$
Adding everything we now have the  cycle index
$$Z(M_{2,3})
= \frac{1}{12} a_1^6 
+ \frac{1}{3} a_2^3
+ \frac{1}{6} a_3^2
+ \frac{1}{4} a_1^2 a_2^2
+ \frac{1}{6} a_6.$$
In order  to apply Burnside and  ask about colorings with  at most $N$
colors we have  that the assignment of the colors  must be constant on
each cycle and we obtain
$$\frac{1}{12} N^6 
+ \frac{1}{3} N^3
+ \frac{1}{6} N^2
+ \frac{1}{4} N^4
+ \frac{1}{6} N.$$
This yields the sequence
$$M_n = 1, 13, 92, 430, 1505, 4291, 10528, 23052, \ldots$$
which  is OEIS  A027670. In  particular we
find there are $92$ colorings using at most three distinct colors.  We
could apply PET at this point since  we have the cycle index. E.g. for
three colors we obtain
$$1/12\, \left( R+G+B \right) ^{6}
\\ +1/4\, \left( R+G+B \right) ^{2} \left( {B}^{2}+{G}^{2}+{R
}^{2} \right) ^{2}+1/6\, \left( {B}^{3}+{G}^{3}+{R}^{3} \right) ^{2}
\\ +1/3\, \left( {B}^{2}+
{G}^{2}+{R}^{2} \right) ^{3}+1/6\,{B}^{6}
+1/6\,{G}^{6}+1/6\,{R}^{6}$$
which expands to
$${B}^{6}+{B}^{5}G+{B}^{5}R+3\,{B}^{4}{G}^{2}+3\,{B}^{4}GR
\\ +3\,{B}^{4}{R}^{2}+3\,{B}^{3}{G}^{3}+6\,{B}^{3}{G}^{2}R
\\ +6\,{B}^{3}G{R}^{2}+3\,{B}^{3}{R}^{3}+3\,{B}^{2}{G}^{4}
\\ +6\,{B}^{2}{G}^{3}R+11\,{B}^{2}{G}^{2}{R}^{2}+6\,{B}^{2}G{R}^{3}
\\ +3\,{B}^{2}{R}^{4}+B{G}^{5}+3\,B{G}^{4}R+6\,B{G}^{3}{R}^{2}
\\ +6\,B{G}^{2}{R}^{3}+3\,BG{R}^{4}+B{R}^{5}+{G}^{6}+{G}^{5}R
\\ +3\,{G}^{4}{R}^{2}+3\,{G}^{3}{R}^{3}+3\,{G}^{2}{R}^{4}
\\+G{R}^{5}+{R}^{6}.$$
The reader might want to verify some of these with pen and paper.
We may also apply inclusion-exclusion to obtain the count of colorings
of  the matrix  with  exactly  $N$ colors.   The  nodes  of the  poset
correspond to sets $P\subseteq [N]$  of colors which include colorings
that use some  subset of these colors, with the  top node representing
at most $N$ colors, which is the only $P$ that includes colorings with
exactly  $N$ colors.  Colorings with  exactly $p  \lt N$  colors where
$p\ge  1$ are  included at  all nodes  that are  supersets of  the $p$
colors. We thus obtain for the total weight
$$\sum_{q=p}^N {N-p\choose q-p} (-1)^{N-q}
= (-1)^{N-p} \sum_{q=0}^{N-p} {N-p\choose q} (-1)^q = 0$$
since  $N-p\ge 1.$  Colorings with  less  than $N$  colors have  total
weight of zero in the poset. We thus obtain
$$\sum_{q=1}^N {N\choose q} (-1)^{N-q} M_q.$$
We get a  finite sequence since with  six slots it is  not possible to
have a coloring with more than six distinct colors:
$$1, 11, 56, 136, 150, 60$$
The last term represents colorings with exactly six colors. This means
all slots  in the matrix are  distinct. Therefore all orbits  have the
same size, the  number of permutations in the group,  which is twelve,
and indeed  $6!/12 = 60.$ The  term for two colors  indicates that the
two monochrome colorings have been excluded.
This  MSE  link
has the Maple code for the general case.
Addendum. Here is what we mean  when we say in the introduction
that the cycle index can be computed by inspection. This refers to the
isomorphism between $M_{2,3} = S_2\times  S_3$ and $D_6$, the dihedral
group (reflections  and rotations of  regular polygons) acting  on six
slots in this case.  The cycle indices $Z(D_p)$ are tabulated and have
simple               closed               forms,               consult
e.g. Wikipedia. Label the
vertices  of a  hexagon in  clockwise  order with  the labels  $(0,0),
(1,1), (0,2), (1,0),  (0,1)$ and $(1,2).$ Then it is  not difficult to
see that  the rotations  of the  hexagon are in  a bijection  with the
pairs of cycles from $C_2 \times  C_3$ embedded in $M_{2,3}.$ E.g. the
rotation  that  takes  the  top   vertex  to  its  clockwise  neighbor
corresponds to the  two cycles $(0,1)$ and  $(0,1,2)$. The reflections
in  an   axis  passing  through  opposite  vertices   preserve  parity
(permutation $(0)(1)$  from $S_2$)  and fix  one element  from $0,1,2$
while permuting  the other two  in two-cycles.  The reflections  in an
axis passing  through opposite edges flip  parity (permutation $(0,1)$
from  $S_2$)  and  fix  one  of  three  elements  from  $0,1,2$  while
exchanging the other two. In this way we have bjectively accounted for
all permutations and the proof of the isomorphism is complete.