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QUESTION

State Burnside's Formula, carefully defining the terms used in the
formula.

Describe the rotation group of the cube, its fixed set, and
describing the orbits of the faces for each rotation.

Use this to find the number of distinct ways there are to label
the faces of a cube with four colours. (As usual, \lq\lq
distinct'' means that the labellings can be distinguished up to a
rotation of the cube, so you will need to consider the action of
the rotation group of the cube on the set of all possible
labellings.)


ANSWER


Burnside's Formula:

Let $G$ act on a set $X$. For each $g\in G$ let $X_g=\{x\in
X|gx=x\}$ and for each $x\in X$ let the orbit of $x$, denoted
$G_x=\{gx|g\in G\}$. Then the number of orbits is given by

$$\frac{1}{|G|}\sum_{g\in G}|X_g|$$

$G$ has 24 elements

\begin{description}

\item[type 0]

$e$ - order 1 - entire cube is fixed, each face is its own orbit.

\item[type 1]

8 rotations of order 3, each with an invariant diagonal and two
orbits of faces each of length 3.

\item[type 2]

6 rotations of order 2 each fixing a line joining the midpoints of
diagonally opposite edges and having 3 orbits each of length 2.

\item[type 3]

3 rotations of order 2 fixing a line joining the midpoints of
opposite faces and having 2 orbits of length 2 and 2 of length 1.

\item[type 4]

6 rotations of order 4 and 2 of length 1, 1 orbit of length 4 and
2 of length 1.

\end{description}

Each face can have one of 4 colour so there are $4^6$ colourings
available. $G$ acts on the set $X$ of these colourings and number
of orbits of colouringd=number of \lq\lq distinct'' ways to colour
the cube.

By Burnside this is $\frac{1}{|G|}\sum_{g\in G}|X_g|$.

\begin{center}
\begin{tabular}{cccc}
$|X_e|=$&$|X|=$&$4^6=$&4096\\ $|X_g|$=&$4^2$&type 1=&16\\
&$4^3$&type 2=&64\\ &$4^4$&type 3=&256\\ &$4^3$&type 4=&64
\end{tabular}
\end{center}

so number of distinct colourings

\begin{eqnarray*}
&=&\frac{1}{24}\left(4^6+8.4^2+6.4^3+3.4^4+6.4^3\right)\\
&=&\frac{1}{24}(4096+128+384+768+384)\\ &=&240
\end{eqnarray*}




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