Show that the group R of rotational symmetries of a dodecahedron is simple and has order 60.
The Attempt at a Solution
I see how to get order 60 using the orbit stabilizer theorem. Letting R act in the natural way on the set of faces, we find the size of the orbit of a face is 12 and the size of the stabilizer of a face is 5. So |R| = 60.
Also, by letting R act on the set of cubes inscribed inside the dodecahedron, we can show that R is isomorphic to a subgroup of S5, so must be A5, which is the only order 60 subgroup of S5. But I don't think I'm supposed to use this method because the next question is to show that R is isomorphic to A5.
Anyway, how do you determine that R is simple without having to totally classify it?
Thanks in advance.