Rotation Group of a Soccer Ball

Question

I am attempting to show that a soccer ball cannot have a 60 degree rotational symmetry through a line through the centers of two opposite hexagons. My proof so far:

If it had such a symmetry, let's call it Q, then the order of Q is 6. So then the stabilizer of a particular hexagon contains an element of order 6, which in turn implies the stabilizer must be of at least order 6 as well. Now, if I knew that the orbit of any hexagon is of order 12(which i know, but cannot seem to show), then this would mean by orbit-stabilizer theorem that the order of the entire group of rotations would be at least 72, which contradicts that the group is isomorphic to A₅, which is of order 60.

How do I show the orbit of any hexagon is 12?

Answer 1

The symmetry group is $A_5$, which does not contain an element of order $6$.

Answer 2

It seems to be easier just to observe that the six sides of one of the hexagons are not all equal -- three of them border on pentagons and three of them border on hexagons. So a 60° rotation like you describe would take pentagons to hexagons and vice versa -- it wouldn't be a symmetry at all.