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Are actions in the $3\times 3\times 3$ rubik cube a group? You can see here Rubik's Cube Not a Group? that $4\times 4\times 4$ rubik cubes or higher arent groups. But what about $3\times 3\times 3$?

Added: Each state of cube like an element, where solved cube is identity.

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Gaston Burrull
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1 Answers1

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It is most certainly a group. In fact, it is isomorphic to a subgroup of the symmetric group $S_{54}$ generated by $6$ elements, corresponding to rotating each row to the right and each column up. To see this, one need only label the $54$ colored stickers on a cube with the numbers $1$ through $54$. Since each of the motions of the cube are permutations of the $54$ stickers, numbering the stickers gives us a subgroup of $S_{54}$. One such numbering gives
$$ (1\; 10)(2\; 11)(3\; 12)(10\; 30)(11\; 29)(12\; 28)(30\; 39)(29\; 38)(28\; 37)(39\; 1)(38\; 2)(37\; 1) \cong \text{ rotating the top row right}$$ and similarly the other $5$ elements are products of $12$ $2$-cycles. It is worth noting that each of these permutations is even, which gives an easy proof of the unsolvability of certain configurations, such as any configuration in which two stickers have been switched.

Edit: Some people consider the cube to be the same after being rotated as a whole. We can modify our group to deal with this by omitting the last row and column rotations, giving us a subgroup of $S_{54}$ generated by $4$ elements. We can do this because rotating the last row to the right is the same as rotating the cube to the right and the first two rows to the left, and a similar procedure works for columns.

Alex Becker
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  • are you sure that actions in 3x3x3 rubik cube form a group? I really think that there are more than one identity element like 4x4x4 case. Not all combinations in a rubik cube can be made I dont think that is isomorphic to a simmetrycx group. – Gaston Burrull May 22 '12 at 23:52
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    @Gastón: depending on what you mean by "actions," this is true for all $n \times n \times n$ cubes. What is not true for higher $n$ is that this group acts freely and transitively on the set of states of the cube. – Qiaochu Yuan May 22 '12 at 23:59
  • @QiaochuYuan I think an element of rubik cube like a "order" of colors in each part. – Gaston Burrull May 23 '12 at 00:02
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    @QiaochuYuan The action is always transitive, simply because that's how one defines a state of the cube to be accessible. If you consider all states of the cube (accessible and inaccessible) then even in the 3x3x3 case the action is not transitive, and there are 12 orbits. If you want to include center orientations, it gets more complicated. Of course, the fact that it fails to act freely is what is really important. – Logan M May 23 '12 at 00:08
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    @GastónBurrull This answer is correct. The cubies on the 3x3x3 cube can be uniquely identified by their collection of stickers, which is what fails in the 4x4x4 case. If you get a cube take it apart, you'll see that no two cubies are identical. – Logan M May 23 '12 at 00:13
  • @LoganMaingi ¿what about the central stickers? you can rotate central stikers in relation with color fixed sides without changing colors of rest of cube, is a complex move but exist, is the same thing that in 4x4x4 rubik fails... there are more than 1 identity (in this case rotating center equals to rotare 4 center stickers) – Gaston Burrull May 23 '12 at 00:50
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    @GastónBurrull "downvote: for example some pieces of cube has two or three stickers, you cant permute corners like S_3 only in anticlockwise or clockwise form" I'm not talking about permutations of the pieces (of which there are only 26) but rather of the stickers, so trivially each motion is well-defined. This in fact works perfectly well for $n\times n\times n$ cubes as well (as noted in the question you linked to, these do in fact form a group, just not under permutations of the pieces). As for "more than one identity" it depends on what you consider a solved group. – Alex Becker May 23 '12 at 00:51
  • @GastónBurrull con't: If you consider several other positions to be solved, you can quotient the group I gave by the permutations generating those partitions, provided this is in fact a normal subgroup. – Alex Becker May 23 '12 at 00:52
  • @AlexBecker can you edit answer with detail of "what group do you refering" for remove down vote? – Gaston Burrull May 23 '12 at 00:54
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    A few of us had this discussion on the original question previously. You can define a larger group, a sort of super-cube group, which pays attention to the orientations of the center pieces. The subgroup of all permutations which fix all stickers in position, but not necessarily orientation, is normal, and if you quotient out by it you get back the ordinary cube group. This group would be important if you were trying to solve a colored version of the cube here: http://twistypuzzles.com/museum/large/00226-01.jpg – Logan M May 23 '12 at 00:57
  • @LoganMaingi my question was answer with your last comment. Thanks for clarification if we cuocient by subsemigroups that are not groups we can loss the group structure. But we can reciprocally considering a larger set that it is a group. – Gaston Burrull May 23 '12 at 00:59
  • @GastónBurrull I've edited the question. Hopefully I've dealt with your concerns. – Alex Becker May 23 '12 at 01:06
  • Thank you for clarification. – Gaston Burrull May 23 '12 at 01:12