# Are actions in the $3\times 3\times 3$ rubik cube a group?

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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– Derek Allums May 22 '12 at 23:37
A cube, Rubik or otherwise, is not a group. The actions you can perform on a Rubik cube, those form a group. – Gerry Myerson May 22 '12 at 23:37
Of course, it might be a torsor (and here it is). – Zhen Lin May 22 '12 at 23:40
@GerryMyerson I think that the actions in rubik not form a group in case 3x3x3 – Gastón Burrull May 22 '12 at 23:44
@GastónBurrull The first few paragraphs of the answer to the question discuss the $3\times 3 \times 3$ case. – Derek Allums May 22 '12 at 23:50

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.
@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
@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 0:51