I’m reading Arnold’s Abel’s Theorem in Problems and Solutions, where it says:
We now prove that the alternating group $A_5$ is not soluble. One of the possible proofs uses the following construction. We inscribe in the dodecahedron five regular tetrahedra, numbered by the numbers 1, 2, 3, 4 and 5 in such a way that to every rotation of the dodecahedron there corresponds an even permutation of the tetrahedra, and that to different rotations there correspond different permutations. So we have defined an isomorphism between the group of rotations of the dodecahedron and the group $A_5$ of the even permutations of degree 5. The non-solubility of the group $A_5$ will thus follow from the non-solubility of the group of rotations of the dodecahedron.
I’m a bit confused. Why rotation of dodecahedron corresponds to an even permutation of inscribed five tetrahedra?
The answer just mentioned
The tetrahedra are‚ for example‚ those with the vertices chosen in the following way1: (1‚ 8‚ 14‚ 16)‚ (2‚ 9‚ 15‚ 17)‚ (3‚ 10‚ 11‚ 18)‚ (4‚ 6‚ 12‚ 19)‚ (5‚ 7‚ 13‚ 20).
, regarding to the notation
The translator provided an image of the inscribed cube:
, and added some explanation
The inscription of the five Kepler cubes inside the dodecahedron helps us to find the five tetrahedra.
The edges of the cubes are the diagonals of the dodecahedron faces. Every pair of opposite vertices of the dodecahedron is a pair of two opposite vertices of two Kepler cubes. Each cube has thus only one pair of vertices in common with any one of the others. (Two Kepler cubes — black and white — having two opposite vertices in common are shown in the figure). In each cube one can inscribe two tetrahedra (see Problems 126 and 127). Since each tetrahedron is defined by four vertices of the cube‚ and any two Kepler cubes have only 2 vertices in common‚ all tetrahedra inscribed in the Kepler cubes are distinct. So there are in all 10 tetrahedra‚ two for every vertex of the dodecahedron. Any two of such tetrahedra either have no vertices in common‚ or they have only one vertex in common. Indeed‚ if two vertices belonged to two tetrahedra‚ the edge of such tetrahedra joining them should be the diagonal of a face of two different cubes‚ but we know that any two cubes have in common only opposite vertices. There are two possible choices of 5 tetrahedra‚ without common vertices‚ inside the dodecahedron: indeed‚ when we choose one of them‚ we have to reject the four tetrahedra having a vertex in common with the chosen tetrahedron. The remaining tetrahedra are five. Amongst them four tetrahedra have disjoint sets of vertices‚ whereas the remaining tetrahedron has one vertex in common with each of the four disjoint tetrahedra. The choice of the first tetrahedron thus forces the choice of the others‚ so there are in all only two choices. (Translator’s note)
, but I still couldn’t see why.