[I don't know from the Penrose involution, but:] Isn't it a bit easier to see the equivalence of the rotational symmetry group of a dodecahedron with A_5 by embedding the compound of 5 tetrahedra in its 20 vertices? Then its 60 rotations correspond to the even permutations of the tetrahedra. —Dan
On Jul 15, 2015, at 6:17 PM, Adam P. Goucher <apgoucher@gmx.com> wrote:
The rotational symmetry group of a dodecahedron is well known to be A_5.
One way to see this is by noticing that precisely five cubes can be formed by taking 8 of the 20 vertices of the dodecahedron; the rotations of the dodecahedron induce even permutations on those cubes:
https://en.wikipedia.org/wiki/Compound_of_five_cubes <https://en.wikipedia.org/wiki/Compound_of_five_cubes>
Then the Penrose involution gives an odd permutation, upgrading the symmetry group to S_5.