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$R$ is a clock-wise twist of a right-hand face and $U$ is a clock-wise twist of the upper face of a $2 \times 2 \times 2$ Rubik's Cube. Find the order of the subgroup generated by $\{R, U\}$

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I'm surprised no one has answered this yet.

Anyway, the answer is (6!/6)(3^5) = 29, 160.

I will cover both the 3x3x3 case and 2x2x2 case for completeness, as the amount of extra work to cover the 3x3x3 case is negligible.

Proof

Lemma 1: Edge Orientation is preserved in < R,U >. (This should be obvious to see.)

Lemma 2: There are 3^5 possible corner orientation states for the 6 corners in < R,U >.

Proof

The move sequence, R U2 R' U R' U2 R U2 R U R' U' R' U R U2, twists the top-front-right (UFR) corner -90 degrees and top-back-right corner (UBR) +90 degrees. Since we can place any two corners in these two slots and since every legal corner orientation twist type can be expressed as a composition of a {+90,-90} corner twist, then we can apply conjugates of this move sequence to twist all 6 corners in any of 3^5 different ways.

Lemma 3: All 7! edge permutations (this is not referring to orientation, only the placement of pieces) are possible to reach in < R,U >.

Proof

Assume that the list {1,2,3,4,5} is the solved state. We can apply the following 3 4-cycles (a face quarter turn does a 4-cycle of edges in its face) to create a 2-cycle.

(2->1->4->3)(5->3->1->4)(1->5->4->3) = {2,1,3,4,5}.

We can place any edges in the face U or R in any order, and therefore we can do any 4-cycle of edges by conjugating the moves U, U' or R, R', respectively. Therefore, we can create a 2-cycle from the above decomposition, for example. Lastly, since any permutation can be expressed as a product of transpositions (2-cycles), and since we can apply conjugates of any 2-cycle maneuver to create any other 2-cycle, we can generate all 7! edge permutations reachable in < R,U >.

Lemma 4: We cannot do a 3-cycle of corners using moves in < R,U > alone.

Proof 1

Consider either the U face or R face. There are (4!)/4 = 6 possible 4-cycles and (4!)/(2^2*2!) = 3 possible 2 2-cycles in 4 objects. Using < R,U > moves alone, only two of the six possible 4-cycle permutations are reachable in < R,U > (one of the three distinct 4-cycles and its inverse) and only one of the three 2 2-cycles is reachable in < R,U >. Since we can write any of the 6 possible 4-cycles as a product of any of one of the other 4-cycles and a 3-cycle, clearly this shows that it's not possible to do a 3-cycle of corners in < R,U >. A similar argument holds true for the 2 2-cycle.

Proof 2

We cannot isolate a corner in either the U face or R face using < R,U > turns. For example, we cannot put the bottom back right corner in the top back right slot preserving the formation of the remaining 3 corners in the U layer. (This can be achieved with the sequence D R D' R', however, but that is obviously outside of < R,U >). Since we cannot isolate a corner in the U or R face using < R,U > turns only, we cannot make a 3-cycle corner commutator, as many simple commutators are based on isolating a single piece into a face with moves X and then turning that face for move Y for the commutator [X,Y].

Corollary of Lemma 4

Since we cannot do a 3-cycle of corners in < R,U >, it follows that we cannot generate a 2-cycle of corners in < R,U > because if we could generate a 3-cycle of corners in < R,U >, then we could apply the proper 4-cycle to the 3 cycled corners and a corner outside of the 3-cycle to create a 2-cycle.

Lemma 5: Only 1/6 of corner permutations are reachable in < R,U >.

Proof

Assume we have a scrambled state generated by some maneuver strictly in < R,U >. Speaking strictly of placing corners (ignoring orientation), we can place the first corner in any of the 6 corner slots with moves in < R,U >. We can then place a second corner in any of the 5 remaining slots. We can then place a third corner in any of the remaining 4 slots.

This gives (6)(5)(4) = 120 = 6!/6 possible permutations.

Since only 3 corners remain to be placed in 3 slots now, and since Lemma 4 and its corollary are true, we cannot choose the arrangement of these three remaining corners because we are unable to do a 3-cycle or a 2-cycle in < R,U >.

Conclusion (End of Proof)

Since lemmas 1-5 and Lemma 4's corollary are true, and since the parity of the corners and edges on the 3x3x3 are dependent, clearly the diameter of the < R,U > subgroup is (6!/6)(3^5)(7!)(1/2) = 73, 483, 200 for the 3x3x3 case.

By lemmas 2, 4, 5 and lemma 4's corollary, the diameter of the < R,U > subgroup for the 2x2x2 case is: (6!/6)(3^5) = 29, 160.

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  • $\begingroup$ Is this calculation for a 2x2x2? There are no edge pieces. $\endgroup$ – Cheerful Parsnip Nov 11 '14 at 23:22
  • $\begingroup$ Oh yeah. I just edited my post for the answer for the 2x2x2. Same answer pretty much. $\endgroup$ – Christopher Mowla Nov 11 '14 at 23:48
  • $\begingroup$ This is actually pretty cool, but there are some things that aren't so clear: 1) How do you formally define the orientation of a piece? 2) What does it mean that a list of numbers is a solved state, like the proof of Lemma 3. 3) Where did you lear all this? $\endgroup$ – hjhjhj57 Nov 12 '14 at 7:06
  • $\begingroup$ You can see how we define corner and edge orientation here. Stefan Pochmann made a nice post regarding edge orientation here. As far as the solved state list, if {1,2,3,4,5} is the solved state, then {2,3,5,1,4}, {1,3,2,4,5}, etc. are all scrambles (permutations) of that solved state. Lastly, I have been studying the cube for a few years now, so I got all of this info from my studies. I wrote up this entire proof in about an hour. $\endgroup$ – Christopher Mowla Nov 12 '14 at 9:16

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