I am told that the (orientation preserving) isometries of the cube, taken as a group, are isomorphic to $S_4$. I don't understand anything "4" about the cube. At least, nothing "4" about it was obvious. So, instead of looking for "4", I took a cube and drew a picture with this labeling:
And I decided that the three rotations are probably going to generate what I want, so I codified them (in parity with this picture) by letting:
$r_1=(1 \ 3 \ 6 \ 4)$ $r_2=(1 \ 2 \ 6 \ 5)$ $r_3=(2 \ 4 \ 5 \ 3)$
Am I heading down a bad alley? Please no exhaustive answers, I am just looking for helpful hints (including "change your course of action, this won't work")
thank you!
The group $G$ of $24$ rotations acts on the set $D$ of $4$ diagonals of the cube. It can be shown that the corresponding group homomorphism from $G$ into $Sym(D)$ is injective. Hence, the image is $Sym(D) \cong S_4$.