MHB Surjectivity for permutation representation of a group action

[kalish1]

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?

 

[I like Serena]

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?

It is not.
$T$ is isomorphic with $A_4$, which has 12 elements.
However, there are 24 permutations of the faces.
So f cannot be surjective.
Note that for instance the exchange of exactly 2 faces is not possible.

 

[Deveno]

All you are required to do is show that $T$ is isomorphic to some SUBGROUP of $S_4$. In practical terms, this means finding a suitable set of 4 objects on which $T$ acts injectively. The 4 faces of the tetrahedron work just fine, and your proposed map is injective, so you are finished.

If you MUST prove surjectivity of SOMETHING, you need only show that there is a surjective map of $T$ onto $A_4$, which it appears you have already done, since $S_4$ has PRECISELY ONE subgroup of order 12.