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Say we want to color the faces of an octahedron with 2 colors, say red and blue. How many of these colorings have as many blue as red faces?

I want to use Burnside's lemma, by considering the symmetry group $G=S_4$ of the octahedron acting on the set $X$ of all possible colorings using the 2 colors equally; $|X|={8\choose 4}=70$.

The octahedron has three sorts of rotational symmetries:

  1. Rotation about an axis through 2 opposite vertices by $90^\circ,180^\circ,270^\circ$.

  2. Rotation about an axis through the midpoints of opposite faces by $120^\circ,240^\circ$.

  3. Rotation about an axis through the midpoints of opposite edges by $180^\circ$.

Now, rotation (1) about $90^\circ$ or $270^\circ$ (of which there are $2\cdot 3=6$) permutes the faces in two $4$-cycles, which means that it fixes $2$ configurations.

Rotation (1) about $180^\circ$ (of which there are $1\cdot 3=3$) permutes the top faces in two $2$-cycles and the bottom faces too, therefore there are ${4\choose 2}=6$ fixed configurations.

Rotation (2) (of which there are $4\cdot 2=8$) permutes the faces in two $3$-cycles and fixes two faces. This gives $4$ fixed configurations.

Rotation (3) (of which there are $6$) permutes the faces in four $2$-cycles. This gives again ${4\choose 2}=6$ fixed configurations.

Burnside now gives $|X/G|=\frac{1}{24}\left(70+6\cdot 2+3\cdot 6+8\cdot 4+6\cdot 6 \right)$, which is not correct.

Edit: the mistake was due to a typo. I got the right answer now, which is 7. Thanks.

My question is: where did I go wrong in this calculation? And is there possibly a more efficient way to do this problem?

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  • $\begingroup$ Take another look to rotation 2. I see you edited. $\frac{168}{24}=7$ is the correct answer. $\endgroup$ – Václav Mordvinov May 23 '18 at 20:14
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    $\begingroup$ @Václav and everyone else. I got it. Shokran. $\endgroup$ – rae306 May 23 '18 at 20:19

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