The Rubik's Cube group is generated by the six moves $\{F,B,U,D,L,R\}$. However, is this the minimal generating set for the group? In other words, can I simulate the move $F$ just by making the moves $B,U,D,L,R$? If I try this out on an actual Rubik's cube, it doesn't seem quite simple (starting from a solved state twist the front face and try to solve the cube without turning it again), but I don't see any reason why it would be impossible either.
[Math] Minimal generating set of Rubik’s Cube group
group-theorypermutationsrecreational-mathematicsrubiks-cube
Best Answer
Yes, you can achieve one of the generators from the other $5$. I have some old "Notes on Rubik's Magic Cube" by David Singmaster, according to which $D^{-1}$ is equal to:
$$R^2L^2UF^2B^2UF^2R^2F^2B^2U^2L^2U^2L^2R^2U^2R^2U^2R^2F^2U^{-1}R^2B^2R^2L^2F^2L^2UB^2F^2U $$
I have tried this out now, and it works. It is attributed to Roger Penrose.
You cannot dispense with any more of the standard generators - you need at least $5$ of them.
But the group is a $2$-generator group. Two randomly chosen elements have a reasonably high probability of generating the whole group.