On Thu, Aug 23, 2012 at 3:19 AM, Chris Smith <[email protected]> wrote: > 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
Let a, b be your two elements determined by rotation. Is this your question? Q: Does the group have the property that every element has the form a^mb^n, for some m,n? I don't know. What I am saying is that I am willing to believe (because it is very very often true for most permutation groups) that all elements of G can be written in the form a^m1*b^n1*a^m2*b^n2*...*a^mk*b^nk, for some m1, ..., mk, n1, ..., nk. You'd have to check this using Sage or Gap or, now Sympy. ( I'm not sure if Aleksander ever finished correctly implementing the disjoint cycle notation. If not, Sage or Gap would probably be easist.) You enter a, b as generators of the group, compute its order and see if the order if what you think it should be (60 for A_5, for example). Incidently, Jesse Douglas, who solved Plateau's problem (a problem in differential geometry which arose in the study of how soap film forms a minimal surface when the boundary is some closed wire loop) and got one of the two first Fields Medals for his work, turned to researching finite groups as he got older. He apparently classified the finite groups which do have the property mentioned in your question above. I think they are all published in the PNAS and on the web, but I have not read them. So, I think the answer to your question might follow from what is known theoretically from Douglas' work. I'll try to look up those papers as soon as I get the time and let you know. > 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. > -- 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.
