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Let $G$ be Rubik's cube group. It is generated by the rotations by 90 degrees $L,R,D,U,F,B$ (left, right, down, up, front, behind), but what relations beyond $L^4=R^4=...=B^4=1$ do they satisfy? Thus I would like to know a presentation of the group as

$G = \langle L,R,D,U,F,B ~:~ ?\rangle$.

After playing aroumd I'have also found the relations $LR=RL$, $(LU)^{105}=1$, $(LRFB)^{12}=1$, $(LRFBFB)^4=1$, $(LRLRFBFB)^2=1$ (of course together with the symmetric relations).

From "The Mathematics of Rubik's cube" by W. D. Joyner I know that $G$ is generated by two elements and presentations are known, but I have not found one. Besides, I'm only interested in the standard generating set above. Remark that there is a well-known abstract group-theoretic description of $G$, it is the kernel of the homomorphism $(S_{12} \ltimes (\mathbb{Z}/2)^{12}) \times (S_{8} \ltimes (\mathbb{Z}/3)^{8}) \to \mathbb{Z}/2 \times \mathbb{Z}/2 \times \mathbb{Z}/3$ which maps $(a,x,b,y) \mapsto (\text{sign}(a) \text{sign}(b),\sum_i x_i,\sum_j y_j)$.

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If you have a group which you know a presentation for, and a finite-index subgroup, then you may obtain a presentation via the Reidemeister-Schreier method. Also, it may be easier (and more natural) to obtain a presentation for the slightly larger group where you allow generators which are rotations of the Rubiks cube, not just of its faces (this will have 3 generators). – Ian Agol Aug 9 '11 at 18:37
@Agol: I think the Reidemeister-Schreier method will give me some more complicated generators. – Martin Brandenburg Aug 10 '11 at 11:07
Once you've found a presentation for the kernel $(S_{12} \ltimes (\mathbb{Z}/2)^{11}) \times (S_{8} \ltimes (\mathbb{Z}/3)^{7}) \to \mathbb{Z}/2$, you may then express $L,R,D,U,F,B$ in terms of the generators (adding in 6 relations expressing this), and then express the generators in terms of $L,R,D,U,F,B$ (which will be a consequence of the relators). Then eliminate all of the original generators to get a presentation in terms of $L,R,D,U,F,B$. Expressing $L,R,D,U,F,B$ in terms of the other generators shouldn't be hard, but the other direction might be. – Ian Agol Aug 10 '11 at 15:23

This discussion: seems to culminate in a presentation (due to Dan Hoey). I did not read it carefully, I must admit. The presentation is quite complicated. For the 2x2x2 group there is this:

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Thanks! It seems to me that Dan Hoey only had heuristic arguments for his presentation and GAP was not able to prove it. But this was back in 1995 ... – Martin Brandenburg Aug 9 '11 at 18:37

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