I am trying to, for self-study, prove that the group $G$ of rotations of a cube is isomorphic to the symmetric group $S_4$. There are other questions on this, but I am trying to work through the details on my own without consulting previous attempts. I used this visual tool from Geogebra to inform the geometry of my answer.
Every rotation of the cube corresponds to a permutation of the four body diagonals joining opposite vertices. By mapping each element of $G$ to the permutation it induces on the diagonals, we get a homomorphism $f: G \to S_4$. By considering the rotation through the axis joining the midpoints of two opposite sides of the cube, we find that $\text{Im}(f)$ contains each of the $6$ transpositions in $S_4$. These transpositions generate $S_4$, so $f(G) \supset S_4$. That is, $f$ is surjective. The order of $G$ is $24$, however. The cube has six faces, any of which can be facing upward, and given a fixed such square face, there are $4$ possible rotations of this square that preserve the cube. So $|G| = 6 \times 4 = 24$. As $|G| = |S_4|$ and $f$ is surjective, it is bijective. So $G \cong S_4$.
Is my reasoning here sound? I have been struggling with this question for some time, so I would appreciate any clarifications on the steps of the argument for which my reasoning may be fuzzy.
