Show that the action of the symmetry group of a cube on pairs of opposite faces of the cube defines a surjective homomorphism from the symmetric group $S_4$ to $S_3$.
My attempt/logic:
I know that the rigid symmetries of a cube is isomorphic to $S_4$. I began by identifying the vertices of the cube with points {($\pm 1$,$\pm 1$,$\pm 1$) $\in$ $\mathbb{R}^3$}. Then, it should be clear that rotations of angle $\pi$ about the axises $x,y,z$ fix any pair of opposite faces. The number of elements in the set {$\pi_1$,$\pi_2$,$\pi_3$} of pairs of opposite faces is 3. Thus, the action defines a homomorphism $f:$ $S_4$ $\rightarrow$ $S_3$.
I am not sure if my proof is clear enough. I feel like I jumped a step or two since the proof seems a little too short/simple. How can I improve my proof? Any help or suggestion is appreciated. Thank you.
Best Answer
Your proof starts out well: you have correctly shown that the map is a homomorphism from $S_4$ to $S_3$. (I might add a little more detail justifying why the map respects composition, but what you have seems all right.)
However, you have not yet shown that the map is surjective, i.e., that it maps to every element of $S_3$.
As a hint for how to show this, recall that the symmetric group $S_n$ is generated by the transpositions on $n$ elements - if you can find a way to locate each transposition in $S_3$ as an image of some rotation of the cube, that will suffice.