# Burnside's counting lemma - Octahedron with $k$ colours and opposite faces coloured identically

I was doing a routine exercise on Burnside's counting lemma, but somehow my solution is wrong. I checked every step many times and I don't see where I made a mistake, and one of my classmates I was working with exactly had the same problem. So we thought of the following.

In how many ways can you colour an octahedron with $k$ colours up to rotational symmetry such that opposite faces are coloured identically?

So obviously you can use Burnside to count the number of orbits under different rotations. The rotational group of an octahedron is $S_4$. Now I found the following fixed points under different rotation types.

• $1$ time the identity $\mathrm{Id}$, cycle type $(\cdot)(\cdot)(\cdot)(\cdot)$, which gives $k^4$ possibilities since after the choosing the colour of four faces on one side, each having $k$ possibilities, the remaining four are fixed.
• $6$ rotations over $90^{\circ}$ through a line through two opposite vertices, cycle type $(\cdot\ \cdot\ \cdot\ \cdot)$. This fixes four faces on one side, and because opposite faces are identically coloured, the choice of one face fixes all faces, thus $k$ possibilities.
• $3$ rotations over $180^{\circ}$ through a line through two opposite vertices, cycle type $(\cdot\ \cdot)(\cdot\ \cdot)$. Two free choices on one side of the octahedron, then the rest fixed by symmetry and the condition that opposite faces are identically coloured. Thus $k^2$ possibilities.
• $8$ rotations over $120^{\circ}$ through a line through the middle of two opposite faces, cycle type $(\cdot\ \cdot\ \cdot)$. The two opposite faces are identically coloured. Three adjacent faces to one of these two are in a cycle, thus have the same colour, and the remaining three also have this colour because the condition that opposite faces are identically coloured. Thus $k^2$ possibilities.
• $6$ rotations over $180^{\circ}$ through a line through the middle of two opposite edges, cycle type $(\cdot\ \cdot)$. Similarly as above, $k^2$ possibilities.

This yields $\frac{1}{24}(k^4+17k^2+6k)$ colourings. However, this is not an integer for $k=3,7,11$ for example and surely many other numbers. (I wouldn't say this is because these numbers are prime). This my solution is just wrong. Can anyone spot the mistake? I don't see it.

The rotations over $180^{\circ}$ through a line through the middle of two opposite edges have $k^3$ possibilities, because the faces not adjacent to that edge get sent to their opposites.

• Thank you, I can see this now. I thought they mapped to the face adjacent to the opposite one. – Václav Mordvinov May 25 '18 at 15:49