# Finding the order of the rotational symmetry group of a polyhedron

In my math class we were introduced to the symmetries of polyhedra and then to the Orbit-Stabilizer formula that helps us compute the order of rotational symmetry groups of the polyhedra.

I find it extremely hard to picture the symmetries of 3-dimensional solids so I'm struggling with this concept. There are a couple of things I don't understand:

1. We were told that the rotational symmetry group of the tetrahedron acts transitively on the set $$\{1,2,3,4\}$$ - why is this so? I understand what a group action is but don't see what the group action is in this case.
2. Using the above action, we were told that if we picked a vertex, say $$1$$, then we could find $$3$$ rotations that fix it. How is this so? How do we know there are only $$3$$?

I think once I understand these I would be able to extend this example to find the order of the rotational symmetry group of the octahedron, which is on my homework. Thank you for your help!

• The tetrahedron has $4$ vertices, which are permuted by the group; if you keep one fixed and where one of the remaining three is taken (it has to be to one of the remaining three), then you have determined the group element – J. W. Tanner Oct 27 at 17:42

a) Label the vertices of a regular tetrahedron as $$1$$, $$2$$, $$3$$, $$4$$. Then, any symmetry of the tetrahedron will have the effect of permuting these labels, and therefore, these numbers. That’s why you can build such an equivalence.