Surjectivity for permutation representation of a group action

Question

I am having trouble proving that my function is surjective. Here is the problem statement:

Problem statement: Let T be the tetrahedral rotation group. Use a suitable action of T on some set, and the permutation representation of this action, to show that T is isomorphic to a subgroup of $S_4$.

Outline of my attempt at a proof:

Let $S$ be the set of faces of the tetrahedron. There exists a homomorphism $f:T$ $\rightarrow$ (permutations of the faces). There are $4$ faces, so $|Perm(|4|)| = 24$. We want to show that $T$ is a bijection, and show the faithfulness of $f$. Well, $|Ker(f)|=1$, because no rotation fixes all faces. So $f$ is injective. Also, the class equation for $T$ is $12 = 1+3+4+4$, which admits no "subsums" to $6$, implying that there are no odd permutations because $T$ has no subgroup of index $2$. So $T \approx A_4$, which is a subgroup of $S_4$.

The thing that evades me is how to prove that $f$ is surjective?

Answer 1

The easiest way to do this would be to use the Orbit-Stabilizer Theorem. $$ |G|=|\text{orbit}||\text{stabilizer}| $$ Think about the action of a permutation on the tetrahedron. Draw a particular tetrahedron. How many rotations of the tetrahedron fix one of its faces?

There are 3

Then what is the orbit for that particular tetrahedron?

There are 4

So then the group has order?

The group has order 12.

To show that the group is $A_4$. Since we have already have an action of permutations, we need to show that we can only have even permutations of the "correct" order. I'll leave you to flesh out the details. But first ask why there can be no element of order $12$ in the group

There can be no element of order $12$ as then the group would be cyclic--Clearly not the case!

Finally, why are there no elements of order $6$?

It is a permutation of the FOUR faces of the tetrahedron.

Then finally, ask yourself, why does this only leave the subgroup $A_4$ of $S_4$?