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QUESTION


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\item[(a)]
State Burnside's formula, carefully defining the terms used in the
formula.

\item[(b)]
Describe the elements of the rotation group of the cube, giving
the order of each element, its fixed set, and describing the
orbits of the faces for each rotation.

\item[(c)]
Use your answers to parts (a) and (b) to find the number of
distinct ways there are to label the faces of a cube with six
colours, where each colour may be used more than once. (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.)

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ANSWER


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\item[(a)]
Let $G$ be a finite group acting on a finite set $X$. For each
$g\in G$ let $X_g$ denote the set $\{x\in X|g(x)=x\}$, then the
number $r$ of orbits of the action is given by
$\begin{displaystyle}\left(\sum_{g\in
G}|X_g|\right)/|G|\end{displaystyle}$.

\item[(b)]

Rotation of order 3 about a diagonal preserving two opposite
vertices- 2 orbits each of three mutually adjacent faces.

Rotation of order 2 about a line bisecting opposite edges-3
orbits, each fixed point yields an orbit consisting of two
adjacent faces containing that fixed point, and the remaining two
faces form the third orbit.

Rotation of order 4 about a line joining the midpoints of opposite
faces- each face containing a fixed point is invariant so we
obtain two orbits of length one. The remaining four faces split
into two orbits of length 2, each consisting of a pair of opposite
faces.

The identity element- 6 orbits each of length 1.

\item[(c)]
$r=\begin{displaystyle}\left(\sum_{g\in
G}|X_g|\right)/24\end{displaystyle}$. There are 8 rotations of the
first type and each preserves $6^2$ labellings. There are 6
rotations of the second type and each preserves $6^3$ labellings.
There are 6 rotations of the third type and each preserves $6^3$
labellings, and there are 3 rotations of the fourth type each
preserving $6^4$ labellings. The identity element preserves all
$6^6$ labellings. Hence there are
$\begin{displaystyle}\frac{(8.36+6.192+6.192+3.1296+46.656)}{12}\end{displaystyle}=4428$
distinguishable labellings.

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