# Rubik's Cube Not a Group?

I read online that

although the 3x3x3 is a great example of a mathematical group, larger cubes aren't groups at all.

How can that be true? There is obviously an identity and it is closed, so that must mean that some moves aren't invertible. But this seems unlikely to me.

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The moves on a "Rubik cube" of whatever size are a group action. –  hardmath May 13 '12 at 0:35
I don't know what the writer of that web page thought they meant, but the claim is incorrect. Of course it is a group. –  MJD May 13 '12 at 0:36
The author seems to be implying that because you can have a "solved cube" that is not in its original position (by rotation of centers), then the cube is not "a group" because there are many "identities". But this is a confusion between a group and a group action, or else a confusion as to what constitutes an "identity"; or else you can view the cube as a suitable quotient. In short, the author is confused. –  Arturo Magidin May 13 '12 at 0:45
Don't worry too much about what you read online. A lot of it is wrong. Much of the rest is trivial. –  André Nicolas May 13 '12 at 0:54
@AndréNicolas On the other hand, the entirety of MSE and MathOverflow are online. Not saying your advice is good or bad, just an observation :-) –  treble May 13 '12 at 1:02

The 4x4x4 cube and higher aren't groups in the same sense that the 3x3x3 cube is a group.

The set of reachable positions of a 3x3x3 cube, viewed as functions from a 54-element set (representing the locations of the stickers) to a 6-element set (the colors of the stickers) form a group. The operation here is given by the following: For x, y positions of a 3x3x3 cube, let $a_1a_2 \cdots a_n$ be a sequence of moves which, starting from the identity, puts the cube into position x, and $b_1b_2 \cdots b_m$ the same for y. The product xy is the state the cube is in after the sequence $a_1 \cdots a_n b_1 \cdots b_m$.

The fact that this operation indeed forms a group isn't as trivial as it first seems. The issue isn't with associativity, identity, or invertibility. Instead, it's with well-definedness. How do you know that the choice of sequences $a_1 \cdots a_n$ and $b_1 \cdots b_m$ doesn't make a difference?

For the 3x3x3 cube, the way to solve this is viewing it as a subgroup of the symmetric group on the set of all 54 stickers. This doesn't work for larger cubes, because it's possible to come up with moves that move some of the cubies around, without changing how the stickers. The stickers can move, without their apparent colors changing. To see why this is impossible for a 3x3x3 cube, note that any cubie is uniquely specified by its stickers, so any permutation of 2 stickers of the same color in the cube group must permute the cubies, which then can not be the identity.

But on the 4x4x4 cube, this fails. All 4 of the starred stickers here, for example, are indistinguishable, in the sense that you could permute them and still have a solved state. I don't think a permutation of just these 4 stickers is possible, but there are many permutations of indistinguishable cubies which can be done. According to Dustan Levenstein in the comments, any even permutation of these 4 stickers (or more generally of any center stickers) is possible.

The way to prove formally that the 4x4x4 cube is not a group is to find a sequence of moves that acts as the identity on one configuration, but not on another configuration. This is pretty easy to do, but I don't remember the precise solution, so I'll omit it. (If anyone really wants such an example, comment and I can dig one up.)

It is true that if we labeled all of the stickers somehow, forming a so-called "supercube", and required that the labels also match up, then we would have a group. This group would be constructed in the same way as the 3x3x3 group, as a subgroup of the symmetric group on the 96 stickers of the 4x4x4 cube.

This group acts on the set of positions of the 4x4x4 cube transitively, but not freely. This, in group theoretical language, is why the 4x4x4 cube positions do not form a group. We can still study the larger group, but we need to take into account that the action isn't as nice as in the 3x3x3 case.

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Any even permutation of the middle stickers is possible, so in particular, an even permutation of just those those four stickers is possible. –  Dustan Levenstein May 13 '12 at 6:25
What you are saying is that the action of the group of moves on the set of achievable positions of a cube with edges of length 4 or more has nontrivial stabilizers. Statements like "Rubik's Cube is a group" are so unclear that that it is pointless to discuss whether they are true or not! –  Derek Holt May 13 '12 at 15:44
What is true, I believe, is that in the $3\times3$ case, the stabilizer of an achievable position (which is the stabilizer of any achievable position) is a normal subgroup. So, if we want, we can define the Rubik's cube group for the $3\times3$ cube to be the quotient of the full group by this stabilizer. We can't do this for $4\times 4$ and higher since the stabilizers are not normal. –  Will Orrick May 13 '12 at 16:37
@DerekHolt Having nontrivial stabilizers is, of course, precisely what it means for an action to be non-free. As for the statement, of course it's unclear, that's why the OP wanted someone to interpret it. That doesn't mean it has no content. And this particular abuse of language is common among people who study twisty puzzles, so it's certainly worth knowing. –  Logan Maingi May 13 '12 at 17:20
@WillOrrick You are correct, if you consider the orientation, and not just the color, of the stickers. I ignored this in my answer, but perhaps I should not have. The reason why I did is because the group of all moves which preserve the position, but not necessarily orientation, of each sticker is normal even for larger cubes, so we can mod out by it (and I did without explicitly saying so). For the 3x3x3 cube, this just so happens to be the stabilizer of every element, so the result is a normal action on the quotient. For the 4x4x4 cube, stabilizers are bigger. –  Logan Maingi May 13 '12 at 17:26