MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $(a,x,b,y) \mapsto (\text{sign}(a) \text{sign}(b),\sum_i x_i,\sum_j y_j)$.

share|cite|improve this question
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:

share|cite|improve this answer
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
Your fist link is broken! – Yoyontzin Apr 22 at 19:35
@Yoyontzin remember that just fine – Vadym Fedyukovych Jun 6 at 21:34

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.