Homomorphism from Rubik's Cube group to Symmetric Group.

Question

$R^2$ is the rotation of the right side by 180, $F^2$ is the rotation of the front by 180. Other than this I am completely lost and this blowing my mind. I wish I had more to build off of, but I am super confused.

Answer 1

Here is a picture with the relevent seven edge cubies labeled.

If you think about the action of $G = \langle \phi, \rho \rangle$ on that set $\Omega$ of cubies, it shouldn't be too hard to believe that the orbits are $\{1, 3\},~\{2, 4, 6\}$, and $\{5, 7\}$, which explains the $S_3 \times \Bbb Z_2^2$.

Now that we know our group $G$ lives inside $S_3 \times \Bbb Z_2^2$, we can indeed project onto $S_3$ denoting by $\pi$ the homomorphism $\pi : G \to S_3$.

To think about this map's kernel, note that $\phi \rho \phi = (2~6)(5~7)$ while $\rho \phi \rho$ permutes analogously, so that $(\phi \rho)^3 = (\rho \phi)^3 \in \ker \pi$.

Could anything else be in the kernel? No, the kernel is $\langle (\phi \rho)^3 \rangle \cong \Bbb Z_2$ and I'll let you think about why. Truthfully, I haven't quite got a "good" reason, only a messy one. If I manage to articulate something, I'll update this answer accordingly.

This should be enough guidance for you to get your bearings, at least.