Presentation of Rubik's Cube group












14












$begingroup$


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?






group-theory finite-groups group-presentation rubiks-cube







share|cite|improve this question











$endgroup$












  • $begingroup$
    I don't know what relations you'll need, but you definitely only need five generators. This is why Rubik's cubes with pictures on them are (slightly) harder to solve than solid-color Rubik's cubes. :)
    $endgroup$
    – Micah
    Dec 2 '12 at 21:28






  • 2




    $begingroup$
    Right now I found a paper(129.81.170.14/~erowland/courses/2009-2/projects/Cordell.pdf) asserting that even $L2BRD^{−1}L^{−1}, UFRUR^{−1}U^{−1}F^{−1}$ serve as a set of generators of the group and we can express $D^{-1}$ as $R2L2UF2B2UF2R2R2B2U2L2U2L2R2U2R2U2R2F2U−1R2B2R2L2F2L2UB2F2U$.
    $endgroup$
    – Dominik
    Dec 2 '12 at 22:15






  • 2




    $begingroup$
    'Nice' is probably an overstatement, but the page at 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.
    $endgroup$
    – Steven Stadnicki
    Dec 3 '12 at 6:16










  • $begingroup$
    @StevenStadnicki: Could you pleas transform this into an answer so that I have got something to accept?
    $endgroup$
    – Dominik
    Dec 8 '12 at 23:14


















14












$begingroup$


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?






group-theory finite-groups group-presentation rubiks-cube







share|cite|improve this question











$endgroup$












  • $begingroup$
    I don't know what relations you'll need, but you definitely only need five generators. This is why Rubik's cubes with pictures on them are (slightly) harder to solve than solid-color Rubik's cubes. :)
    $endgroup$
    – Micah
    Dec 2 '12 at 21:28






  • 2




    $begingroup$
    Right now I found a paper(129.81.170.14/~erowland/courses/2009-2/projects/Cordell.pdf) asserting that even $L2BRD^{−1}L^{−1}, UFRUR^{−1}U^{−1}F^{−1}$ serve as a set of generators of the group and we can express $D^{-1}$ as $R2L2UF2B2UF2R2R2B2U2L2U2L2R2U2R2U2R2F2U−1R2B2R2L2F2L2UB2F2U$.
    $endgroup$
    – Dominik
    Dec 2 '12 at 22:15






  • 2




    $begingroup$
    'Nice' is probably an overstatement, but the page at 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.
    $endgroup$
    – Steven Stadnicki
    Dec 3 '12 at 6:16










  • $begingroup$
    @StevenStadnicki: Could you pleas transform this into an answer so that I have got something to accept?
    $endgroup$
    – Dominik
    Dec 8 '12 at 23:14
















14












14








14


6



$begingroup$


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?






group-theory finite-groups group-presentation rubiks-cube







share|cite|improve this question











$endgroup$




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?






group-theory finite-groups group-presentation rubiks-cube




group-theory finite-groups group-presentation rubiks-cube






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 20 '18 at 15:39









Shaun

9,310113684




9,310113684










asked Dec 2 '12 at 21:14









DominikDominik

7,35742675




7,35742675












  • $begingroup$
    I don't know what relations you'll need, but you definitely only need five generators. This is why Rubik's cubes with pictures on them are (slightly) harder to solve than solid-color Rubik's cubes. :)
    $endgroup$
    – Micah
    Dec 2 '12 at 21:28






  • 2




    $begingroup$
    Right now I found a paper(129.81.170.14/~erowland/courses/2009-2/projects/Cordell.pdf) asserting that even $L2BRD^{−1}L^{−1}, UFRUR^{−1}U^{−1}F^{−1}$ serve as a set of generators of the group and we can express $D^{-1}$ as $R2L2UF2B2UF2R2R2B2U2L2U2L2R2U2R2U2R2F2U−1R2B2R2L2F2L2UB2F2U$.
    $endgroup$
    – Dominik
    Dec 2 '12 at 22:15






  • 2




    $begingroup$
    'Nice' is probably an overstatement, but the page at 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.
    $endgroup$
    – Steven Stadnicki
    Dec 3 '12 at 6:16










  • $begingroup$
    @StevenStadnicki: Could you pleas transform this into an answer so that I have got something to accept?
    $endgroup$
    – Dominik
    Dec 8 '12 at 23:14




















  • $begingroup$
    I don't know what relations you'll need, but you definitely only need five generators. This is why Rubik's cubes with pictures on them are (slightly) harder to solve than solid-color Rubik's cubes. :)
    $endgroup$
    – Micah
    Dec 2 '12 at 21:28






  • 2




    $begingroup$
    Right now I found a paper(129.81.170.14/~erowland/courses/2009-2/projects/Cordell.pdf) asserting that even $L2BRD^{−1}L^{−1}, UFRUR^{−1}U^{−1}F^{−1}$ serve as a set of generators of the group and we can express $D^{-1}$ as $R2L2UF2B2UF2R2R2B2U2L2U2L2R2U2R2U2R2F2U−1R2B2R2L2F2L2UB2F2U$.
    $endgroup$
    – Dominik
    Dec 2 '12 at 22:15






  • 2




    $begingroup$
    'Nice' is probably an overstatement, but the page at 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.
    $endgroup$
    – Steven Stadnicki
    Dec 3 '12 at 6:16










  • $begingroup$
    @StevenStadnicki: Could you pleas transform this into an answer so that I have got something to accept?
    $endgroup$
    – Dominik
    Dec 8 '12 at 23:14


















$begingroup$
I don't know what relations you'll need, but you definitely only need five generators. This is why Rubik's cubes with pictures on them are (slightly) harder to solve than solid-color Rubik's cubes. :)
$endgroup$
– Micah
Dec 2 '12 at 21:28




$begingroup$
I don't know what relations you'll need, but you definitely only need five generators. This is why Rubik's cubes with pictures on them are (slightly) harder to solve than solid-color Rubik's cubes. :)
$endgroup$
– Micah
Dec 2 '12 at 21:28




2




2




$begingroup$
Right now I found a paper(129.81.170.14/~erowland/courses/2009-2/projects/Cordell.pdf) asserting that even $L2BRD^{−1}L^{−1}, UFRUR^{−1}U^{−1}F^{−1}$ serve as a set of generators of the group and we can express $D^{-1}$ as $R2L2UF2B2UF2R2R2B2U2L2U2L2R2U2R2U2R2F2U−1R2B2R2L2F2L2UB2F2U$.
$endgroup$
– Dominik
Dec 2 '12 at 22:15




$begingroup$
Right now I found a paper(129.81.170.14/~erowland/courses/2009-2/projects/Cordell.pdf) asserting that even $L2BRD^{−1}L^{−1}, UFRUR^{−1}U^{−1}F^{−1}$ serve as a set of generators of the group and we can express $D^{-1}$ as $R2L2UF2B2UF2R2R2B2U2L2U2L2R2U2R2U2R2F2U−1R2B2R2L2F2L2UB2F2U$.
$endgroup$
– Dominik
Dec 2 '12 at 22:15




2




2




$begingroup$
'Nice' is probably an overstatement, but the page at 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.
$endgroup$
– Steven Stadnicki
Dec 3 '12 at 6:16




$begingroup$
'Nice' is probably an overstatement, but the page at 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.
$endgroup$
– Steven Stadnicki
Dec 3 '12 at 6:16












$begingroup$
@StevenStadnicki: Could you pleas transform this into an answer so that I have got something to accept?
$endgroup$
– Dominik
Dec 8 '12 at 23:14






$begingroup$
@StevenStadnicki: Could you pleas transform this into an answer so that I have got something to accept?
$endgroup$
– Dominik
Dec 8 '12 at 23:14












3 Answers
3






active

oldest

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2












$begingroup$

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.






share|cite|improve this answer











$endgroup$









  • 4




    $begingroup$
    Thanks for the link, Steven. I've noticed that your reference contains much older version of an example of analyzing Rubik's Cube with GAP which has been updated for the GAP 4.4 release. Generating the group cube as shown there, one can then try G:=Image(IsomorphismFpGroup(cube)); RelatorsOfFpGroup(G); and then H:=SimplifiedFpGroup(G); RelatorsOfFpGroup(H); to get two finite presentations, with 6 and 23 generators and with 282 and 180 relations respectively.
    $endgroup$
    – Alexander Konovalov
    Apr 29 '13 at 0:13










  • $begingroup$
    It appears that the link in this answer is broken; I can't open it, at least.
    $endgroup$
    – Shaun
    Dec 20 '18 at 15:36



















6












$begingroup$

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.






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$endgroup$













  • $begingroup$
    I love the fact that Rubik's group is not cyclic. It is easy to prove, but it also means that there is no mysterious twist which will get you back to the solved cube from any unsolved one. Which is...pretty cool (for want of a better word)!
    $endgroup$
    – user1729
    Dec 3 '12 at 10:08






  • 1




    $begingroup$
    @user1729 even if the rubik's cube was cyclic, would you really want to apply the same algorithm up to 43 quintillion times? lol
    $endgroup$
    – timidpueo
    Jan 13 '13 at 19:38










  • $begingroup$
    @timidpueo No, but it would be a cool if it were true! Just as it is cool that it is not true...
    $endgroup$
    – user1729
    Jan 14 '13 at 10:20



















4












$begingroup$

This belongs as a comment, but due to being new here I can't comment yet. The link Dominik mentioned (129.81.170.14/~erowland/courses/2009-2/projects/Cordell.pdf) is indeed an interesting paper, but I noticed several things wrong, and would advise skepticism. For example, I don't know if the two elements mentioned are indeed generators, but I do know that the orders he claims they have (253 and 60) are wrong; they are in fact 77 and 12. In another section, he claims that "Any permutation of two adjacent sides (for example, F R) has order 105." This is a false generalization. F R does in fact have order 105, but F R' has order 63.



Nevertheless I was quite impressed with the paper; I just felt like mentioning that not everything there is accurate. I have very little understanding of the group theory of the Rubik's Cube, and it's possible there are many more errors I didn't recognize.






share|cite|improve this answer









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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    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.






    share|cite|improve this answer











    $endgroup$









    • 4




      $begingroup$
      Thanks for the link, Steven. I've noticed that your reference contains much older version of an example of analyzing Rubik's Cube with GAP which has been updated for the GAP 4.4 release. Generating the group cube as shown there, one can then try G:=Image(IsomorphismFpGroup(cube)); RelatorsOfFpGroup(G); and then H:=SimplifiedFpGroup(G); RelatorsOfFpGroup(H); to get two finite presentations, with 6 and 23 generators and with 282 and 180 relations respectively.
      $endgroup$
      – Alexander Konovalov
      Apr 29 '13 at 0:13










    • $begingroup$
      It appears that the link in this answer is broken; I can't open it, at least.
      $endgroup$
      – Shaun
      Dec 20 '18 at 15:36
















    2












    $begingroup$

    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.






    share|cite|improve this answer











    $endgroup$









    • 4




      $begingroup$
      Thanks for the link, Steven. I've noticed that your reference contains much older version of an example of analyzing Rubik's Cube with GAP which has been updated for the GAP 4.4 release. Generating the group cube as shown there, one can then try G:=Image(IsomorphismFpGroup(cube)); RelatorsOfFpGroup(G); and then H:=SimplifiedFpGroup(G); RelatorsOfFpGroup(H); to get two finite presentations, with 6 and 23 generators and with 282 and 180 relations respectively.
      $endgroup$
      – Alexander Konovalov
      Apr 29 '13 at 0:13










    • $begingroup$
      It appears that the link in this answer is broken; I can't open it, at least.
      $endgroup$
      – Shaun
      Dec 20 '18 at 15:36














    2












    2








    2





    $begingroup$

    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.






    share|cite|improve this answer











    $endgroup$



    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.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jun 13 '17 at 12:02









    Community♦

    1




    1










    answered Dec 9 '12 at 3:27









    Steven StadnickiSteven Stadnicki

    41.3k868122




    41.3k868122








    • 4




      $begingroup$
      Thanks for the link, Steven. I've noticed that your reference contains much older version of an example of analyzing Rubik's Cube with GAP which has been updated for the GAP 4.4 release. Generating the group cube as shown there, one can then try G:=Image(IsomorphismFpGroup(cube)); RelatorsOfFpGroup(G); and then H:=SimplifiedFpGroup(G); RelatorsOfFpGroup(H); to get two finite presentations, with 6 and 23 generators and with 282 and 180 relations respectively.
      $endgroup$
      – Alexander Konovalov
      Apr 29 '13 at 0:13










    • $begingroup$
      It appears that the link in this answer is broken; I can't open it, at least.
      $endgroup$
      – Shaun
      Dec 20 '18 at 15:36














    • 4




      $begingroup$
      Thanks for the link, Steven. I've noticed that your reference contains much older version of an example of analyzing Rubik's Cube with GAP which has been updated for the GAP 4.4 release. Generating the group cube as shown there, one can then try G:=Image(IsomorphismFpGroup(cube)); RelatorsOfFpGroup(G); and then H:=SimplifiedFpGroup(G); RelatorsOfFpGroup(H); to get two finite presentations, with 6 and 23 generators and with 282 and 180 relations respectively.
      $endgroup$
      – Alexander Konovalov
      Apr 29 '13 at 0:13










    • $begingroup$
      It appears that the link in this answer is broken; I can't open it, at least.
      $endgroup$
      – Shaun
      Dec 20 '18 at 15:36








    4




    4




    $begingroup$
    Thanks for the link, Steven. I've noticed that your reference contains much older version of an example of analyzing Rubik's Cube with GAP which has been updated for the GAP 4.4 release. Generating the group cube as shown there, one can then try G:=Image(IsomorphismFpGroup(cube)); RelatorsOfFpGroup(G); and then H:=SimplifiedFpGroup(G); RelatorsOfFpGroup(H); to get two finite presentations, with 6 and 23 generators and with 282 and 180 relations respectively.
    $endgroup$
    – Alexander Konovalov
    Apr 29 '13 at 0:13




    $begingroup$
    Thanks for the link, Steven. I've noticed that your reference contains much older version of an example of analyzing Rubik's Cube with GAP which has been updated for the GAP 4.4 release. Generating the group cube as shown there, one can then try G:=Image(IsomorphismFpGroup(cube)); RelatorsOfFpGroup(G); and then H:=SimplifiedFpGroup(G); RelatorsOfFpGroup(H); to get two finite presentations, with 6 and 23 generators and with 282 and 180 relations respectively.
    $endgroup$
    – Alexander Konovalov
    Apr 29 '13 at 0:13












    $begingroup$
    It appears that the link in this answer is broken; I can't open it, at least.
    $endgroup$
    – Shaun
    Dec 20 '18 at 15:36




    $begingroup$
    It appears that the link in this answer is broken; I can't open it, at least.
    $endgroup$
    – Shaun
    Dec 20 '18 at 15:36











    6












    $begingroup$

    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.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I love the fact that Rubik's group is not cyclic. It is easy to prove, but it also means that there is no mysterious twist which will get you back to the solved cube from any unsolved one. Which is...pretty cool (for want of a better word)!
      $endgroup$
      – user1729
      Dec 3 '12 at 10:08






    • 1




      $begingroup$
      @user1729 even if the rubik's cube was cyclic, would you really want to apply the same algorithm up to 43 quintillion times? lol
      $endgroup$
      – timidpueo
      Jan 13 '13 at 19:38










    • $begingroup$
      @timidpueo No, but it would be a cool if it were true! Just as it is cool that it is not true...
      $endgroup$
      – user1729
      Jan 14 '13 at 10:20
















    6












    $begingroup$

    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.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I love the fact that Rubik's group is not cyclic. It is easy to prove, but it also means that there is no mysterious twist which will get you back to the solved cube from any unsolved one. Which is...pretty cool (for want of a better word)!
      $endgroup$
      – user1729
      Dec 3 '12 at 10:08






    • 1




      $begingroup$
      @user1729 even if the rubik's cube was cyclic, would you really want to apply the same algorithm up to 43 quintillion times? lol
      $endgroup$
      – timidpueo
      Jan 13 '13 at 19:38










    • $begingroup$
      @timidpueo No, but it would be a cool if it were true! Just as it is cool that it is not true...
      $endgroup$
      – user1729
      Jan 14 '13 at 10:20














    6












    6








    6





    $begingroup$

    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.






    share|cite|improve this answer











    $endgroup$



    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.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Apr 13 '17 at 12:20









    Community♦

    1




    1










    answered Dec 2 '12 at 22:26









    Henning MakholmHenning Makholm

    241k17308546




    241k17308546












    • $begingroup$
      I love the fact that Rubik's group is not cyclic. It is easy to prove, but it also means that there is no mysterious twist which will get you back to the solved cube from any unsolved one. Which is...pretty cool (for want of a better word)!
      $endgroup$
      – user1729
      Dec 3 '12 at 10:08






    • 1




      $begingroup$
      @user1729 even if the rubik's cube was cyclic, would you really want to apply the same algorithm up to 43 quintillion times? lol
      $endgroup$
      – timidpueo
      Jan 13 '13 at 19:38










    • $begingroup$
      @timidpueo No, but it would be a cool if it were true! Just as it is cool that it is not true...
      $endgroup$
      – user1729
      Jan 14 '13 at 10:20


















    • $begingroup$
      I love the fact that Rubik's group is not cyclic. It is easy to prove, but it also means that there is no mysterious twist which will get you back to the solved cube from any unsolved one. Which is...pretty cool (for want of a better word)!
      $endgroup$
      – user1729
      Dec 3 '12 at 10:08






    • 1




      $begingroup$
      @user1729 even if the rubik's cube was cyclic, would you really want to apply the same algorithm up to 43 quintillion times? lol
      $endgroup$
      – timidpueo
      Jan 13 '13 at 19:38










    • $begingroup$
      @timidpueo No, but it would be a cool if it were true! Just as it is cool that it is not true...
      $endgroup$
      – user1729
      Jan 14 '13 at 10:20
















    $begingroup$
    I love the fact that Rubik's group is not cyclic. It is easy to prove, but it also means that there is no mysterious twist which will get you back to the solved cube from any unsolved one. Which is...pretty cool (for want of a better word)!
    $endgroup$
    – user1729
    Dec 3 '12 at 10:08




    $begingroup$
    I love the fact that Rubik's group is not cyclic. It is easy to prove, but it also means that there is no mysterious twist which will get you back to the solved cube from any unsolved one. Which is...pretty cool (for want of a better word)!
    $endgroup$
    – user1729
    Dec 3 '12 at 10:08




    1




    1




    $begingroup$
    @user1729 even if the rubik's cube was cyclic, would you really want to apply the same algorithm up to 43 quintillion times? lol
    $endgroup$
    – timidpueo
    Jan 13 '13 at 19:38




    $begingroup$
    @user1729 even if the rubik's cube was cyclic, would you really want to apply the same algorithm up to 43 quintillion times? lol
    $endgroup$
    – timidpueo
    Jan 13 '13 at 19:38












    $begingroup$
    @timidpueo No, but it would be a cool if it were true! Just as it is cool that it is not true...
    $endgroup$
    – user1729
    Jan 14 '13 at 10:20




    $begingroup$
    @timidpueo No, but it would be a cool if it were true! Just as it is cool that it is not true...
    $endgroup$
    – user1729
    Jan 14 '13 at 10:20











    4












    $begingroup$

    This belongs as a comment, but due to being new here I can't comment yet. The link Dominik mentioned (129.81.170.14/~erowland/courses/2009-2/projects/Cordell.pdf) is indeed an interesting paper, but I noticed several things wrong, and would advise skepticism. For example, I don't know if the two elements mentioned are indeed generators, but I do know that the orders he claims they have (253 and 60) are wrong; they are in fact 77 and 12. In another section, he claims that "Any permutation of two adjacent sides (for example, F R) has order 105." This is a false generalization. F R does in fact have order 105, but F R' has order 63.



    Nevertheless I was quite impressed with the paper; I just felt like mentioning that not everything there is accurate. I have very little understanding of the group theory of the Rubik's Cube, and it's possible there are many more errors I didn't recognize.






    share|cite|improve this answer









    $endgroup$


















      4












      $begingroup$

      This belongs as a comment, but due to being new here I can't comment yet. The link Dominik mentioned (129.81.170.14/~erowland/courses/2009-2/projects/Cordell.pdf) is indeed an interesting paper, but I noticed several things wrong, and would advise skepticism. For example, I don't know if the two elements mentioned are indeed generators, but I do know that the orders he claims they have (253 and 60) are wrong; they are in fact 77 and 12. In another section, he claims that "Any permutation of two adjacent sides (for example, F R) has order 105." This is a false generalization. F R does in fact have order 105, but F R' has order 63.



      Nevertheless I was quite impressed with the paper; I just felt like mentioning that not everything there is accurate. I have very little understanding of the group theory of the Rubik's Cube, and it's possible there are many more errors I didn't recognize.






      share|cite|improve this answer









      $endgroup$
















        4












        4








        4





        $begingroup$

        This belongs as a comment, but due to being new here I can't comment yet. The link Dominik mentioned (129.81.170.14/~erowland/courses/2009-2/projects/Cordell.pdf) is indeed an interesting paper, but I noticed several things wrong, and would advise skepticism. For example, I don't know if the two elements mentioned are indeed generators, but I do know that the orders he claims they have (253 and 60) are wrong; they are in fact 77 and 12. In another section, he claims that "Any permutation of two adjacent sides (for example, F R) has order 105." This is a false generalization. F R does in fact have order 105, but F R' has order 63.



        Nevertheless I was quite impressed with the paper; I just felt like mentioning that not everything there is accurate. I have very little understanding of the group theory of the Rubik's Cube, and it's possible there are many more errors I didn't recognize.






        share|cite|improve this answer









        $endgroup$



        This belongs as a comment, but due to being new here I can't comment yet. The link Dominik mentioned (129.81.170.14/~erowland/courses/2009-2/projects/Cordell.pdf) is indeed an interesting paper, but I noticed several things wrong, and would advise skepticism. For example, I don't know if the two elements mentioned are indeed generators, but I do know that the orders he claims they have (253 and 60) are wrong; they are in fact 77 and 12. In another section, he claims that "Any permutation of two adjacent sides (for example, F R) has order 105." This is a false generalization. F R does in fact have order 105, but F R' has order 63.



        Nevertheless I was quite impressed with the paper; I just felt like mentioning that not everything there is accurate. I have very little understanding of the group theory of the Rubik's Cube, and it's possible there are many more errors I didn't recognize.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered May 21 '13 at 6:28









        user78729user78729

        432




        432






























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