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I was trying to use Burnside's formula on a regular hexagon. I believe the answer is $13$, but I am trying to show my work. Here is what I have figured out.

$$\frac{1}{12}(2^6 + 36 + 30 + \cdots?)$$

$2^6$ is $D_{6}$ (at least that is what I think the regular hexagon is called). Then I reflected over the vertices. This gave me $3$ different possible reflections with $12$ different possible ways which is $36$. Then I reflected over the midpoints. This gave me $3$ different possible reflections with $10$ different possible ways which is $30$. Now the rotations are a bit confusing to me.

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  • $\begingroup$ I recommend looking at John B. Fraleigh - A First Course in Abstract Algebra $\endgroup$ – David Simmons Sep 5 '13 at 18:21
  • $\begingroup$ which fix(x) is $2^6$? $\endgroup$ – George Tomlinson Sep 5 '13 at 18:24
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    $\begingroup$ I think we need more detail in what you mean by using Burnside's formula on a regular hexagon, as there's more than one way to use Burnside's formula on a hexagon $\endgroup$ – George Tomlinson Sep 5 '13 at 18:26
  • $\begingroup$ I doubt you're talking about the case of counting hexagonal tiles with three black vertices and three white vertices, but this is one way to use Burnside's formula on a hexagon. In which case are you being told to use it? $\endgroup$ – George Tomlinson Sep 5 '13 at 18:28
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    $\begingroup$ That was the answer, but what was the question?! Burnside's formula always takes the sum of fix(g), but it can be applied to answer various different questions. g can be defined in different ways you see $\endgroup$ – George Tomlinson Sep 5 '13 at 18:54
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The action on the dihedral group on the hexagon is illustrated below:

Dihedral group on the hexagon

The number of assignments of $2$ colors to the vertices that are preserved by a group element $\alpha$ is $$2^{\text{Number of vertex orbits under } \langle \alpha \rangle}$$ since each vertex orbit can be assigned any color, and every vertex in any orbit must be colored the same.

The vertex orbits are highlighted below corresponding to the group elements above (vertices in the same orbit are assigned the same color):

Vertex orbits

Inputting this into Burnside's Lemma gives the number of assignments of $2$ colors (inequivalent under rotations and reflections) as $$\tfrac{1}{12}(2^6 + 2^1 + 2^2 + 2^3 + 2^2 + 2^1 + 2^3 + 2^4 + 2^3 + 2^4 + 2^3 + 2^4)=13.$$

Precisely two of these inequivalent assignments of $2$ colors have all colours the same: when they're all white, and when they're all black. That leaves $11$ inequivalent assignments of $2$ colors to the vertices where both colors are used.

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  • $\begingroup$ Consider rotation through $\frac{2\pi}{3}$. You said there are $2^2$ 'assignments of 2 colors to the vertices [preserved in this case]... since each vertex orbit can be assigned any color, and every vertex in any orbit must be colored the same'. Whilst this is true, this is not the number of inequivalent assignments, since coluring all the vertices in one orbit white is equivalent to colouring all the vertices in the other orbit black, so the number of inequivalent assignments of 2 colours to the vertices preserved by rotation through $\frac{2\pi}{3}$ is actually 2 rather than $2^2 = 4$ $\endgroup$ – George Tomlinson Sep 6 '13 at 22:21
  • $\begingroup$ Actually, I can see now that you do only have 2 in your final answer, as you subtract 2 (the ones with all colours the same) from every case $\endgroup$ – George Tomlinson Sep 6 '13 at 22:37
  • $\begingroup$ Where did you get these diagrams/how did you make them? I found them very helpful. $\endgroup$ – goodcow Jun 11 '16 at 11:16
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I upvoted the first answer but I would like to show how to compute the cycle index $Z(D_6)$ of the dihedral group $D_6$ and apply the Polya Enumeration Theorem to this problem.

We need to enumerate and factor the twelve permutations that contribute to $Z(D_6).$

There is the idenity, which contributes $$a_1^6.$$ The two rotations by a distance of one and five contribute $$2 a_6.$$ The two rotations by a distance of two or four contribute $$2 a_3^2.$$ Finally the rotation by a distance of three contributes $$a_2^3.$$

There are three reflections about an axis passing through opposite vertices, giving $$3 a_1^2 a_2^2$$

and three reflections about an axis passing through the midpoints of opposite edges, giving $$3 a_2^3.$$

This finally yields the cycle index $$Z(D_6) = \frac{1}{12} \left(a_1^6 + 2a_6 + 2a_3^2 + 3a_1^2 a_2^2 + 4a_2^3\right).$$

Coloring the hexagon with at most two colors we get $$Z(D_6)(A+B)_{A=1, B=1}.$$

This yields $$\frac{1}{12} \left(2^6 + 2\times 2 + 2\times 2^2 + 3\times 2^2 \times 2^2 + 4 \times 2^3\right)$$ which evaluates to $$13.$$

There is a list of similar computations at MSE Meta.

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