On Tue, Aug 21, 2012 at 6:49 AM, David Joyner <[email protected]> wrote: > 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.
OK, I've convinced myself that all permutations can be generated from two fixed axes of rotation -- the axes stay fixed as the polyhedron moves. What is special about the "group"? Is it that all the permutations can be generated with only powers of those permutations, not from mixtures of the same? So it's a bit like a group of prime factors (the two axes) as opposed to general divisors (members of the group that just get applied once). /c -- You received this message because you are subscribed to the Google Groups "sympy" group. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/sympy?hl=en.
