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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!

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  • $\begingroup$ 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 $\endgroup$ – J. W. Tanner Oct 27 at 17:42
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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.

b) Visualize a tetrahedron as a flat triangle and a point above it. By rotating about the upper point, the triangle below can be rotated to one of three permutations. The other three would also be symmetries of the tetrahedron, but would require a reflection.

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