Consider the (regular) dodecahedron and the cube. The latter can be inscribed in the former.
I am trying to deduce from this that the symmetry group of the cube is a subgroup of the symmetry group of the dodecahedron. I know that there are three axis of symmetry of the cube, and that it suffices to come up with one reflection of the cube and then compose it with the rotation group to get all the 24 reflections.
If we consider the axis of rotation given by the center of the faces then I guess we can clearly see that rotating around it also gives a symmetry of the dodecahedron. Taking the vertices of the cube and again doing the same thing also gives us new symmetries of the dodecahedron. And I guess to do the same thing but now taking the midpoints of the edges also works.
As you see I'm struggling to base my proof on purely geometric arguments. Can you give comments or hints on what I have done or how to proceed? Thank you.
Your view is essentially concrete, and that can be helpful, but note that there the edges of the dodecahedron which do not touch the cube come in two orientations relative to the edges of the squares they sit above - they are perpendicular to one pair of edges, and parallel to the other. Some of the symmetries of the cube invert the relative orientation and don't may the dodecahedron to itself. So I don't think you can do it quite this way.