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How many ways can we choose a pair of edges of a dodecahedron, up to rotational symmetry?

Taking $180$ rotation we have $15$ axis of rotations.

Since there are $6$ pairs of opposite faces, we have an axis going through every pair of opposite faces and have 4 non-trivial rotations for every pair of opposite faces. There are $4*6 = 24$ such rotations.

Lastly, at the opposite pairs of vertices and we may rotate the axis which passes through both vertices and have $2$ more rotations for each pair of vertices. Here we have $2*10 = 20$ such rotations.

Finally we have the identity.

Hence there are $15+24+20+1 = 60$ rotational symmetries of a dodecahedron.

After this I am not able to proceed. Need some hints.

Source: Homework.

Marko Riedel
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user5210
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    It sounds like you're getting ready to apply Burnside's Lemma. Is that right? What's preventing you from continuing? – Karl Apr 30 '23 at 13:06
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    This is a near duplicate of this MSE link. Substitute $1+x$ into the cycle index and extract the coefficient on $x^2.$ The post says there are eleven such configurations. (Polya Enumeration Theorem.) – Marko Riedel Apr 30 '23 at 19:18
  • @Karl Can you please explain how to use Burnside's Lemma and get it? Thanks. – user5210 Apr 30 '23 at 22:39
  • Think of the problem as "how many ways are there to paint two edges of a blank dodecahedron blue, if two ways are considered the same if one is just a rotation of the other?". Try following the example of another application of Burnside's Lemma that you've seen, and let us know which part you're unsure about. – Karl Apr 30 '23 at 22:47

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