Let D be a regular dodecahedron. It is possible to inscribe a cube on the vertices of D thus:
(a) Prove that one can inscribe exactly 5 such cubes inside D.
(b) Deduce that any rigid motion of D (i.e. an element of SO(3) sending D to itself) permutes the 5 cubes.
(c) Prove that any rigid motion which sends every cube to itself must fix D. Note that “fixing every cube” does not mean that every vertex of the cube is fixed, simply that each cube ends up in the same place but possibly rotated.
(d) Deduce that the symmetry group of the dodecahedron is a subgroup of S5 of order 60.
I think I have a proof for part (a) that relies on the symmetries of each regular pentagon that forms a face of the dodecahedron (i.e. 5 distinct diagonals correspond to 5 distinct orientations of cubes) but I can't seem to form a solid argument for the last three parts. I have a very wordy and unclear explanation for part (b) that also relies on the symmetries of the faces of D but I feel like there must be a more succinct argument. I have yet to come up with anything for (c) and (d).
Pick qany edge of the dodecahedron. From each end, wo other edges start. Take the four other endpoints of these. These four vertices form a square, which can be extended unquely to a cube. Each of the $30$ choices fo rthe fist edge gives us a cube, but each cube is obtained in six ways (once per face).Thus there are 5 such cubes.
Given a rigid motion of $D$, see what happens to the initial edge from above. The cube we construct from the moved edge is what the motion turns the cube constructed from the original edge is moved to. So each of the five cubes is moved to such a cube (possibly itself), i.e., the symmetry group permutes the cubes.
If we label the vertices of $D$ with labels $1,2,3,4,5$ depending on which cube they belong to, you can verify that no two faces of $D$ have the same cyclic label order. Hence a movement that fixes all cubes and thereby fixes the labelling must fix $D$.
From the above, the symmetry group of $D$ has an injective group homomorphism into $S_5$. The order of the group is $60$ because we can pick a face of $D$ as "bottom" in $12$ ways and then turn this face in $5$ ways.