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How many ways are there to 2-color 64 squares of an $8 \times 8$ chess board with free rotation (no mirroring)?

I'm not understanding Burnside's theorem really. I have $\frac{1}{4} (1 \times 2^{64} ...)$ but I don't know how to figure out the other numbers...

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closed as off-topic by colormegone, choco_addicted, Ian Miller, Shailesh, Claude Leibovici Apr 8 '16 at 6:29

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The group $\mathbb Z_4$ acts on the chess board by rotation. Then we know that:

  • $\bar 0$ fixes everything, so we have $|X_e| = 2^{64}$
  • $\bar 1$ fixes colorings determined by one of the $4x4$ subboards, so we have $|X_1| = 2^{16}$
  • $\bar 2$ is similar, but now the $4x8$ rectangle determines the coloring, so $|X_2| = 2^{32}$
  • $\bar 3$ is identical to $\bar 1$, so we have $|X_3| = 2^{16}$

Apply the theorem, $$|X/G| = \frac 1 {|G|} \sum_{g\in G} |X_g| = \frac 1 4 (|X_0| + |X_1| + |X_2| + |X_3|)$$ $$= \frac{1}{4} (2^{64} + 2\times2^{16} + 2^{32}) = 2^{62} + 2^{15} + 2^{30}$$

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