On Mon, Aug 20, 2012 at 8:20 PM, smichr <[email protected]> wrote:
> In a docstring the permutation group for the tetrahedron is given. I would
> like to inlclude the same for the cube and dodecahedron (and hence their
> doubles, octahedron and icosahedron). I've googled a bit without finding the
> explicit form that I need. This is outside my field and perhaps someone else
The cube has 8 vertices, so the permutation group representation
of the symmetry group of the cube is a copy of S_4 sitting inside S_8 in a
particular way.
The dodecahedron has 20 vertices, so the permutation group representation
of the symmetry group of the dodecahedron is a copy of A_5 sitting
inside S_{20} in a
particular way.
It's not pretty.
If you really want these, I would just take two randomish symmetries,
compute their
permutations by labeling a cube/dodecahedron's vertices in some way, and then
computing the group generated by just those two. My bet is that will
be the whole thing.
It's a bit messy. I did stuff like that in my book Adventures in Group Theory
(and there is a free but very errata-filled copy on the web at
http://www.permutationpuzzles.org/rubik/webnotes/).
> is a little more savvy at coming up with this? I need something like this
> which is defined for the tetrahedron:
>
> [Permutation([[0,1,2], [3]]),\
> Permutation([[0,1,3], [2]]),\
> Permutation([[0,2,3], [1]]),\
> Permutation([[1,2,3], [0]]),\
> Permutation([[0,1], [2,3]]),\
> Permutation([[0,2], [1,3]]),\
> Permutation([[0,3], [1,2]]),\
> Permutation([0,1,2,3])]
>
> Any help would be appreciated.
>
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