Minimal generating set of Rubik's Cube group 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.
 A: 
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)

It depends on your solution procedure. With the one I use, it turns out that I'm actually only turning the bottom side: (a) while assembling the initial cross (which is easy to do in other ways), and (b) while orienting the top-layer corners at the end (which can be done without turning that side by orienting the cube differently in this phase). So solving after a $B$ quarter-turn without using any $B$s was easy for me.
Keeping two of the sides immovable is not possible, however. If the two sides you won't turn are neighbors, the edge piece they have in common cannot move at all; if they are opposites, the 24 stickers on the edge pieces split into two orbits, so it's not possible to flip an edge anymore.

If we don't require the generators to be single twists, we can get down to two. (Of course we can't get down to one, because a one-generator group is abelian, and the Rubik group isn't). Michael Lugo's comment on Derek's answer gives a set of two somewhat optimized generators, but we can also do:


*

*Let $X$ be a combination that cycles $7$ of the corners, cycles $3$ of the edges, keeps at least one edge unchanged, and flips at least one edge that it doesn't move.

*Let $Y$ be a combination that interchanges $2$ corners, cycles $10$ edges, keeps at least one corner unchanged, and twists at least one of the corners that it doesn't move.


such that exactly one corner and one edge is moved by both $X$ and $Y$.


*

*Using $X$ and $Y$ we can permute the edges any way we like, while messing up the corners.

*Now $X^3$ cycles the 7 corners (even corner partity), leaving all edges in place, and $Y$ transposes 2 corners (odd corner parity), while also cycling $10$ edges. We can achieve any even permutation of the corners by $X^3$s and $Y$s, and since there will be an even number of $Y$s, we can get the 10-cycle of edges back to its starting position without disturbing the corners by doing one or more $Y^2$s afterwards.


Thus $X$ and $Y$ are sufficient to realize any legal permutation of the cubies, ignoring the orientations of each cubie.
However, now $X^{21}$ doesn't move any cubie, so $X^{63}$ keeps all cubies in place and doesn't twist any corners. But $X^{63}$ does flip at least one edge and leaves at least one edge unflipped, so by combining appropriate conjugations of $X^{63}$ we can achieve any legal combination of edge flips.
Likewise, $Y^{10}$ does not move any cubie, and $Y^{20}$ keeps all cubies in place without flipping any edge, but does twist one corner and leaves another corner unchanged. So appropriate conjugations of $Y^{20}$ combine into any legal combination of corner turns.
A: 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.
