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Can anyone review my work on this problem and tell me if I'm missing anything major? Thanks!

Q: Describe the orbits of poles for the group of rotations of an octahedron.

There are $|G|=N=24$ rotational symmetries for an octahedron. These can be split up into three pole orbits for edges, faces, and vertices respectively. Using the book notation of $r_i$ for the size of the stabilization group and $n_i$ for the size of the orbit we have the following pole orbits

  • Edges: $r_i = 2, n_i = 12$.
  • Vertices: $r_i = 3, n_i = 8$.
  • Faces: $r_i = 4, n_i = 6$.
clay
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There are indeed $24$ elements of the group. You should always ask yourself which one is the identity, however.

Archaick
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  • The identity element exists as a stabilizer in all three orbits. How would that elaborate the answer? – clay Apr 21 '15 at 06:45
  • Stabilizers are not the same thing as an identity. There are 24 elements of this group. 23 of them are rotations. There are only 6 possible rotations which have the center of edges as pivots and there are 9 possible rotations with vertices as pivots. Your count for the number of face pviot rotations is correct at 8. These 23 rotations plus the identity form our 24 elements group. – Archaick Apr 21 '15 at 06:59