Non-brute-force proof that the maximum order of a cube group element is 1260

Based on my proof and Stefan Pochmann's correction/improvement in this thread, July 2008. Bruce Norskog gives another proof of maximality here. However, this involves examining all cycle types of A_12, dividing them by parity, something best left to Mathematica. The proof below avoids generating all cycle types. If someone finds a proof involving less cycle analysis, please let me know.

Initial observations

(a) An n-cycle of edges has order n if an even number of edges are flipped ("oriented (edge) n-cycle") 2n if an odd number of edges are flipped ("flipped n-cycle") (b) An n-cycle of corners has order n if orientation sums to 0 mod 3 ("oriented (corner) n-cycle") 3n otherwise ("twisted n-cycle").

These can be seen by noting that, after n iterations of the cycle, each piece has been in each position once and so has been flipped/twisted once by each of the n orientations, one at each position.

Example of an element of order 1260

There exists an element of the cube group of order 1260=2^2*3^2*5*7. UR'UF'D2. This is easily confirmed by calculating the least common multiple of the order of the component sticker cycles.

Proof of maximality

Write the permutation as a product of disjoint cycles. The order of any cube group element is the least common multiple (lcm) of the orders of its disjoint cycles. Since n in (a) and (b) is at most 12, the greatest possible prime factor of the order is 11. We first show that any order divisible by 11 is less than 1260.

A cycle with order divisible by 11 must be an 11-cycle of edges, leaving a 1-cycle. The total order for edges is maximized by giving both cycles an odd number of edge flips: lcm(22,2)=22. Since this edge permutation is odd, the restriction on permutation parity forces an even corner permutation. We can easily list all distinct cycle types of A_8:

Only even cycles: 1,1,1,1,1,1,1,1 3,1,1,1,1,1 5,1,1,1 7,1 3,3,1,1 5,3 Two odd cycles: 2,2,1,1,1,1 4,2,1,1 6,2 4,4 Two odd cycles and even cycle(s): 3,2,2,1 Four odd cycles: 2,2,2,2.

Ignoring the restriction on corner orientation, the total order for corners with cycle type l_1,...,l_k is maximized by multiplying each length by 3, according to (b). This at most lcm(3l_1, ..., 3l_k)=3lcm(l_1, ..., l_k), which takes the maximum 45 for cycle type 5,3. Thus these cube group elements have order at most lcm(22,45) = 2*3^2*5*11 = 990 < 1260.

The maximum order must therefore only have prime divisors 2, 3, 5, and 7. By (a) and (b), there can only be at most one factor each of 5 and 7. 1*9 (oriented 9-cycle) or 3*3 (twisted 3-cycle) gives the maximum 3^2, and 2*8 (flipped 8-cycle) gives 2^4. So the maximum order divides 2^4*3^2*5*7. We need both 5 and 7 since 2^4*3^2*5 < 2^4*3^2*7 < 1260. To top 1260=2^2*3^2*5*7, we need 2^3*3^2*5*7, 2^4*3^2*5*7, or 2^4*3*5*7. Each case is impossible:

Case 1: Two factors of 3

It suffices to show that we cannot have more than two factors of 2. The following table shows how we can get these factors:


Look at 7. Corners 7 forces edges 5 and 9 or 10 and 9, which are impossible. So edges 7. Then from 3^2, corners 3*3. Edges 5 would use up the edges and leave five corners, and neither can produce 2^3, so assume corners 5. This uses up the corners and leaves five edges. 2^3 can only come from a flipped 4-cycle, but this violates the permutation parity.

Case 2: One factor of 3

We must show that there cannot be four factors of 2. This means a flipped 8-cycle, which forces corners 7 and 5. The 3 must come from a 3-cycle of the edges, but again this violates the permutation parity. Done.