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I am trying to find the subgroup of the rotations of the cube that leave the cube invariant. This corresponds to the order of the group of cube rotations (since they all leave the cube invariant), and the order of this group is 24. I found that by considering the 6 faces of the cube. If I fix a face, there are 4 rotations that keep that face in its position. This gives a stabilizer group of order 4 on the 6 vertices, and 6*4 is the order of the whole group. I'm trying to extend this logic to finding the order of the subgroup that leaves the vertices fixed. I am not sure how to manipulate a cube after picking a vertex to find the rotations that leave in fixed. How do do this?

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    $\begingroup$ What about rotation about the diagonal connecting two opposite vertices? $\endgroup$ – Wright-Moran Sep 16 '14 at 0:00
  • $\begingroup$ I get that this is how I must do this (by analogy to faces) but I can not figure out how to determine the rotations. My problem is with visualization/generalization $\endgroup$ – bsm Sep 16 '14 at 1:03
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Consider a vertex $A$. It has $3$ neighbors, and these three neighbors will permute under a rotation that keeps vertex $A$ fixed. However since you're only considering rotations, their will only be three such permutations (the other three permutations on the neighboring vertices would involve reflections).

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Rotations of the cube are of three sorts:

  1. Around an axis through the centers of two opposite faces.
  2. Around an axis through the centers of two opposite edges.
  3. Around n axis through two opposite vertices.

There are 3 axes of the first type (because 3 pairs of opposite faces), 6 axes of the second type, and 4 axes of the third type.

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  • $\begingroup$ How do I find the actual order of the stabilizer group without the stabilizer/orbit identity? I am having a hard time visualizing the diagonal rotations $\endgroup$ – bsm Sep 16 '14 at 0:26
  • $\begingroup$ Get some dice, then you won't have to visualize it. $\endgroup$ – MJD Sep 16 '14 at 1:01

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