Group structure of the Rubik's Revenge?

[mande]

As I showed, there is no way to construct a group that will really represent the 4x4. Any construction that seems to - like yours - will construct a group that probably won't represent anything like the 4x4 if you look more closely.

Well, that certainly surprises me, your argument seems to imply that there is no way to construct a group that represents any nxnxn cube (n>3)? Are you sure that is right?

 

[JBCM627]

Associativity fails, because, as qq showed, composition is not consistently definable.

Composition of permutations is also easily definable in a consistent manner; that is, it is possible to define it so that it works for any composition for a cube in any state, as opposed to the view of it not working for select compositions in select states. As you said, this may be changing the idea behind the puzzle, but with minimal change, this is still a valid view of the 4x4. In other words, it is ok to distinguish center pieces if you consider multiple states as solved - it is still the same puzzle.

A B = e, but B A ≠e
Breaks inverses.

There are a couple of assumptions initially made in this argument, namely: B is an inverse, and the 4x4 is a group. Since BA≠e, B obviously isn't an inverse element. Since there is still an element that does satisfy BA=AB=e, The 4x4 can still be a group.

 

[Lucas Garron]

A B = e, but B A ≠e
Breaks inverses.

There are a couple of assumptions initially made in this argument, namely: B is an inverse, and the 4x4 is a group. Since BA≠e, B obviously isn't an inverse element. Since there is still an element that does satisfy BA=AB=e, The 4x4 can still be a group.

B is an inverse: Yes, because it satisfies AB=e (this is equivalent to defining an inverse as AB=BA=e).
The 4x4 is a group: You're presuming exactly this fact, and I'm trying to show that it does not satisfy the definition.

It's an elementary property of groups that being a right-inverse/left-inverse to an element is equivalent. Let's consider C the inverse you like:
A = [U r U', l']
B = [l, F r' F']
C = [l', U r U]

C
= C e       [Existence of Identity]
= C (A B)   [AB=e]
= (C A) B   [Associativity]
= e B       [CA=e]
= B         [Identity]

Thus, we get C = B, which you can easily verify is false - it's a contradiction. Our assumption that the 4x4x4 can be a group is false.

The best thing you can do to make it a group is to consider the 4x4x4 a supercube with multiple solved states - but that's still a supercube group, not a 4x4x4 group!

Well, that certainly surprises me, your argument seems to imply that there is no way to construct a group that represents any nxnxn cube (n>3)? Are you sure that is right?

That's right. Unless with "represents" you don't mean "represents exactly."

 

[JBCM627]

A B = e, but B A ≠e
Breaks inverses.

There are a couple of assumptions initially made in this argument, namely: B is an inverse, and the 4x4 is a group. Since BA≠e, B obviously isn't an inverse element. Since there is still an element that does satisfy BA=AB=e, The 4x4 can still be a group.

B is an inverse: Yes, because it satisfies AB=e (this is equivalent to defining an inverse as AB=BA=e).

There are still holes in this though. AB=e is not equivalent to saying AB=BA=e, as demonstrated by your example, since [A,B]≠0. And anyway, AB=e only for a select few cubes.

C
= C e       [Existence of Identity]
= C (A B)   [AB=e]
= (C A) B   [Associativity]
= e B       [CA=e]
= B         [Identity]

As I said above, AB doesn't necessarily = e. Since you have (AB) after C, then in this case, (AB)≠e, and your 3rd line is not true.

If you can explain this further, I'd agree, but there are too many disputable points otherwise.

 

[Lucas Garron]

As I said above, AB doesn't necessarily = e. Since you have (AB) after C, then in this case, (AB)≠e, and your 3rd line is not true.

A = [U r U', l']
B = [l, F r' F']
C = [l', U r U]

Do you deny that AB=e? (left-to-right composition)

If so, please explain how you can have a group where AB=e sometimes and AB≠e at other times.

 

[qqwref]

JBCM, please go look up some basic facts about group theory. The stuff Lucas is talking about is very basic group identities... and the fact that you're trying to dispute (mathematically) obvious points shows that you don't really understand what a group is.

 

[JBCM627]

A = [U r U', l']
B = [l, F r' F']
C = [l', U r U']

Do you deny that AB=e? (left-to-right composition)

If so, please explain how you can have a group where AB=e sometimes and AB≠e at other times.

In some cases, yes, AB=e. But do AB on almost any randomly scrambled cube, and AB will not = e.

I will ask one of my teachers to clarify this, I guess.

 

[Lucas Garron]

A = [U r U', l']
B = [l, F r' F']
C = [l', U r U']

Do you deny that AB=e? (left-to-right composition)

If so, please explain how you can have a group where AB=e sometimes and AB≠e at other times.

In some cases, yes, AB=e. But do AB on almost any randomly scrambled cube, and AB will not = e.

Exactly. This cannot happen if the 4x4x4 is a group.

 

[cuBerBruce]

There are generally considered to be approximately 7.4*(10^45) positions for the non-supercube 4x4x4 (or 24 times that many if you consider the orientation of the cube to matter). qqwref (and Lucas) are correctly claiming that these positions do not map 1-to-1 with the elements of a group in the same way that the positions of the 3x3x3 map to elements of a group.

JBCM627 is looking at the permutation effects that move sequences can have on the 4x4x4 as group elements. Of course, this requires many more group elements than 7.4*(10^45), and really models the 4x4x4 supercube group rather than just the states that the 4x4x4 can have when centers that belong on the same face are considered to indistinguishable from each other.

To have a group theory model of the 4x4x4 (non-supercube), JBCM627's approach needs to be modified to reduce the number of states to the actual number of 4x4x4 states. (As qqwref and Lucas point out, this will not actually be a group.)

You could take the subgroup of the 4x4x4 supercube group that fixes corners, fixes edges, and only permutes centers within the same face. Let's call this subgroup H, and the 4x4x4 supercube group G. You could then model the non-supercube 4x4x4 positions as a coset space formed by G/H. I note that there is a parity constraint between corners and centers in the supercube. Since H has the corners fixed, the centers in H must have even parity, so H only contains (|S_4|^6)/2 (95551488) elements. The number of ways the centers can be permuted (for the supercube) is |S_24| = 24!, but half of these positions require the corners to be in an odd parity configuration. So the number of non-supercube positions of the centers is |A_24|/|H| or (24!/2)/((4!^6)/2) = 24!/(4!^6). So basically the 4x4x4 (non-supercube) is a coset space rather than a group. I have to think this is the concept mande had in mind when saying "quotienting the group."

There is also the issue of positions differing only by cube orientation. One way of handling that is to only consider generators for the supercube group that keep one of the cubies (such as the DBL corner) fixed. So using <U,u,d,Uud',R,r,l,Rrl',F,f,b,Ffb'> for instance rather than <U,u,d,D,R,r,l,L,F,f,b,B> to define the supercube group will avoid redundant positions due to cube orientations.

 

[TMOY]

Since H has the corners fixed, the centers in H must have even parity, so H only contains (|S_4|^6)/2 (95551488) elements. The number of ways the centers can be permuted is |S_24|/(|S_4|^6) but half of these positions require the corners to be in an odd parity configuration.

You're counting the parity constraint twice here. In fact every one of the |S_24|/(|S_4|^6) centers permutations can be reached with the corners being in an even parity configuration.

 

[cuBerBruce]

Since H has the corners fixed, the centers in H must have even parity, so H only contains (|S_4|^6)/2 (95551488) elements. The number of ways the centers can be permuted is |S_24|/(|S_4|^6) but half of these positions require the corners to be in an odd parity configuration.

You're counting the parity constraint twice here. In fact every one of the |S_24|/(|S_4|^6) centers permutations can be reached with the corners being in an even parity configuration.

Oops, thanks. The supercube centers have |S_24| permutations with only half reachable with the corners in an even parity configurations. But the number of configurations of the non-supercube centers is
|S_24|/(|S_4|^6) regardless of the corner parity. Or |A_24|/|H|.

 

[mande]

You could take the subgroup of the 4x4x4 supercube group that fixes centers, fixes edges, and only permutes centers within the same face. Let's call this subgroup H, and the 4x4x4 supercube group G. You could then model the non-supercube 4x4x4 positions as a coset space formed by G/H
So basically the 4x4x4 (non-supercube) is a coset space rather than a group. I have to think this is the concept mande had in mind when saying "quotienting the group."

I guess you meant corners there...anyway, thanks a lot for the explanation . I seemed to assume that H was normal in G, thus assuming G/H to be a group, but as it isn't normal I now realize that G/H will just be a coset space as you mentioned.

 

[MatsBergsten]

Parity on the 4x4

We had a 4x4BLD scramble that at least three of us tried. Mike noted that he had double parity. I got that too.

BUT, and here you theorists may help, CAN YOU SKIP PARITY JUST BY CHOOSING
another orientation of the cube? I did not think so. For corners I have tried and failed (both practically and mentally). (When you solve corners you don't alter the permutation of the centers, or maybe you could?) Corner (OLL) parity in a solve is in the scramble (permutation of the corners), not in the orientation of the cube. Am I not right?

As for edges I am not as sure? When I solve centers I choose commutator from what
feels most convenient (both regarding setup moves and actual execution). So maybe you can fix edge parity by just choosing the right center solving methods. As bld solving centers (and corners and edges) means NOT affecting anything else edge parity is preserved.

 

[qqwref]

Well, the question is: can you do a cube rotation and change the parity of edges or corners? We only need to consider (say) an x rotation since every rotation can be made out of 90-degree rotations around a face.

So what does an x rotation do? Well, for centers it does 6 4-cycles, for edges it does 6 4-cycles, and for corners it does 2 4-cycles. So actually the rotation won't change the parity of any orbit. If you have parity you pretty much just have to deal with it (although of course you could always try a u premove).