# Why rotation of dodecahedron corresponds to an even permutation of inscribed five tetrahedra?

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 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: Let's label the tetrahedra $$12345$$ according to the lowest numbered vertex they contain. A clockwise rotation by 72° about the axis passing through the centers of top and bottom face of the dodecahedron corresponds to a permutation of tetrahedra $$(12345)\to(51234)$$, which is an even permutation. The same goes, of course, for a similar rotation about the axis of another couple of opposite faces. But any rotation mapping the dodecahedron to itself is the composition of such rotations, and thus corresponds to an even permutation of the five tetrahedra.