Timeline for A Presentation for Rubik's cube group?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Aug 10, 2011 at 15:23 | comment | added | Ian Agol | 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. | |
| Aug 10, 2011 at 11:07 | comment | added | Martin Brandenburg | @Agol: I think the Reidemeister-Schreier method will give me some more complicated generators. | |
| Aug 10, 2011 at 8:56 | history | edited | Martin Brandenburg | CC BY-SA 3.0 |
added 164 characters in body; added 18 characters in body
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| Aug 9, 2011 at 18:37 | comment | added | Ian Agol | 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). | |
| Aug 9, 2011 at 18:30 | history | edited | Martin Brandenburg | CC BY-SA 3.0 |
edited title
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| Aug 9, 2011 at 16:02 | answer | added | Igor Rivin | timeline score: 10 | |
| Aug 9, 2011 at 10:01 | history | asked | Martin Brandenburg | CC BY-SA 3.0 |