Presentation of Rubik's Cube group

Question

The Rubik's Cube group is the group of permutations of the 20 cubes at the edges and vertices of a Rubik's group (taking into account their specific rotation) which are attainable by succesive rotations of its sides (the cubes in the middle of the sides are considered as fixed). Wikipedia says that this group is given as $(\mathbb Z_3^7 \times \mathbb Z_2^{11}) \rtimes \,((A_8 \times A_{12}) \rtimes \mathbb Z_2)$.

Is there also a nice presentation of this group in terms of generators and relations (i. e. at most as many relations as one can write down on a single sheet of paper)? In particular, are 6 generators (e. g. the six standard rotations) the minimum number needed for a presentation or is it even possible to obtain a rotation around one side as a composition of rotations around the other five faces?

Answer 1

To flesh my comment out into a light answer: while I don't know if I'd really characterize them as 'nice', the page at https://web.archive.org/web/19990202074648/http://www.math.niu.edu/~rusin/known-math/95/rubik includes an explicit characterization via 44 relations of total length about 600 symbols. The page is a bit scattershot, but hopefully it'll offer at least a starting point. I don't know if any explicit characterization using the five (or six) face moves as generators, and I'm not sure if anyone does.

Answer 2

You can certainly achieve a rotation of side as a composition of moves of the other sides only. The most instructive way to see this is to take your favorite algorithm for solving the cube from an arbitrary position and figure out how to modify it such that it never turns the first side you solve. (Whether this is easy depends on what your favorite algorithm is, of course. My own usual one is almost there except for the initial-cross stage and the final turning of the top corners, both of which are easily replaced). Then turn your chosen side by one quarter, and use the modified algorithm to solve from that position. Whatever moves you make during the solve will add up to a "turn this side without turning it" combination.

Also, 5 is the minimum number of single-side turns that generate the group, because restricting movement to any set of 4 sides will either leave one edge unmovable or be unable to flip an edge to the opposite orientation.

A full set of relations between 5 basic quarter-turns would be absolute horrible, though.

If the generators are not required to be simple moves, we don't need as many as 5 of them. It's easy to get down to 3 quite complex operations:

• $\alpha$, a cyclic permutation of 11 of the 12 edges simultaneously with 7 of the 8 corners.
• $\beta$, swaps the edge that $\alpha$ doesn't touch with another one, and swaps the corner that $\alpha$ doesn't touch with another one.
• $\gamma$, turns two corners and flips two edges.

Here there's some hope of being able to write down some reasonably natural relations, but I haven't tried to see whether the details work out.

I'm not sure if we can get down to two generators -- could the role of $\gamma$ be folded into $\beta$, perhaps? Edit: The next time the question came up (and I had forgotten this one existed) I tried anew and came up with a blueprint for a two-generator solution: Minimal generating set of Rubik's Cube group

One generator is of course not possible, since the group is not abelian.