Let $G$ be the group of rotational symmetries of a regular tetrahedron. I'm trying to think of an argument proving that since $G\cong H$, where $H\leq S_4$, $H=A_4$.

There are 12 rotational symmetries in $G$ and thus $|H|=12$ (since the homomorphism induced by $G$ acting of the vertices of a tetrahedron is injective). But the only group of index $2$ in $S_4$ is $A_4$. Hence the isomorphism.

However, the difficult part for me here is to prove that the above action is faithful. I've used a geometric argument, but I don't think it's rigorous enough. I've also thought about the orbits of this action, but I don't seem to come up with the idea of how an orbit of $x\in X$ (the set of the vertices) can be of size 12 (to conclude that the stabilizer has size 1). Some hints would be appreciated.

  • $\begingroup$ Orientation preserving is the clue. $\endgroup$ – Melquíades Ochoa Dec 12 '15 at 17:50
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    $\begingroup$ How could an orbit have 12 elements when there are only 4 vertices? You should review the definition of faithful. Points can have nontrivial stabilizers in a faithful action. $\endgroup$ – Matt Samuel Dec 12 '15 at 17:55
  • $\begingroup$ I'm confused about exactly what you know and the specific route you are trying to take. Why don't you conclude that $G \cong A_4$ after the second paragraph? $\endgroup$ – pjs36 Dec 12 '15 at 18:07

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