Group Theory: Rotational Symmetries

[upsidedowntop]

Homework Statement

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.

 

[mighty2000]

Thank you for your question. To show that the group R of rotational symmetries of a dodecahedron is simple, we can use the same approach as you did in finding the order of the group. By letting R act on the set of faces, we can see that the orbit of a face is 12 and the stabilizer of a face is 5. This means that R is a subgroup of S12, the symmetric group on 12 elements. Since |R| = 60, R must be isomorphic to a subgroup of S5, which is isomorphic to A5.

To show that R is simple, we need to prove that it has no non-trivial normal subgroups. One way to do this is by using the fact that A5 is simple. Since R is isomorphic to A5, any normal subgroup of R must also be a normal subgroup of A5. But since A5 is simple, it has no non-trivial normal subgroups, which means that R also has no non-trivial normal subgroups. Therefore, R is simple.

I hope this helps. Let me know if you have any further questions.