Question

Using Burnside's Lemma, determine a formula for the number of orbits under the action of D8 on the set of colourings.

(We colour each side of a square with one of k ≥ 1 colours.)

Answer 1

let's call the set of colored triangles X. The number of elements in X equals $k^3$.

Burnside's lemma gives us:

The number of orbits

  $=\frac{1}{8}\sum_{g \in D_8}|X^g|$,

where Xg is the set of elements that are fixed by $D_8$.

e fixes all the elements of X, so $|X^e|=k^3$.

r also fixes all the elements of X because a rotation doesn't really produce a different triangle.

The same goes for $X^{r^2}$.

s fixes the triangles that have at least two edges with the same color, so $|X^s|=k^2$. And since rotations don't really change anything, the same goes for $X^{sr}$ and $X^{sr^2}$.

So now we have

$|X^r|=|X^{r^2}|=k$

and so the number of orbits is

$|X/G|=\frac{k^3+3k^2+2k}{8}$