My ultimate commutator challenge
Show that every even permutation on a nxnxn cube can be obtained by a series of commutators. If 2 consecutive comms is a single comm (somehow) then every even permutation is also a direct commutator. No restriction on number of layers used here ...
Per
Well, the first part is pretty trivial to see with some cubing intuition, especially since most of this relies on 3-cycles. The second one is gnarlier, and I never managed to get a satisfactory answer (despite Bruce's reference).Originally Posted by mrCage
Elaborate. I fail to see the triviality...
Per
EP, CP: Generated by 3-cycles. Easy with commutators.
EO, CO: Orient two at a time using commutators. Also easy.
Either I missed somethng or the 2nd part is trivial too, you just have to use the first part and a basic recursion.(if N-1 comms can be replaced by one single comm, than N comms are N-1 comms + 1 comm -> 1 comm + 1 comm = 2 commq -> 1 comm).
The really nontrivial part would be to prove that 2 comms can actually be merged into one.
True if we have even permutation in every orbital effected by the "total permutation". Hmmm .... And where are the centers in your sketchy proof?? I didn't mean solely 3x3x3 cube but any size nxnxn regular cube!!
Per
Hi,
It has already been proved that the alternating group An, which is the group of even permutations on the set {1...n}, consists only of (direct) commutators for n >= 5.
More details, including links to published papers on group theory and examples applied to a 7x7x7 cube using an on-line app, are given in document CommutatorSubgroup.
I hope this may be of help.
If we don't have an even permutation in every orbit, then we aren't in the commutator subgroup. Commutators (and products of commutators) can only do even permutations, even if you restrict to a particular orbit (such as edges or corners on a 3x3). So the commutator subgroup G' is contained in the group H of operations that do even permutations on every orbit. But if we mod out orientations (which, as Lucas pointed out, are easy) then H/{orientations} = H_perm is just a direct product of alternating groups A_n (where n=8, 12, or 24, the sizes of the orbits). Then, as glazik pointed out, every element of A_n is a commutator, so everything in H_perm (and therefore H) should be a single commutator.
There's one catch, though: glazik's article only says that everything in A_n is a commutator of two elements of S_n, not necessarily of A_n. So it could be that some element of H_perm is a commutator of two permutations, but that one or both of these is an odd permutation on some orbit; this may give an impossible position of the NxNxN supercube. The question is: is every element of A_n (at least for n=8, 12, and 24) a commutator of two other elements of A_n?
Perhaps we can use E. Bertram's result from glazik's link (bottom of the first page) to prove this. If we set l = n-1 (which is odd for even n), then we get that every element of A_n can be written as a product of two l-cycles, and therefore as a commutator of one l-cycle (an even permutation) with some tau in S_n. But it may be that tau is an odd permutation--for example, (1234567) and (1234576) are conjugate in S_8, but only by the odd permutation (67). So we must instead use l = n-3. Then we can force tau to be an even permutation, so every element of A_n will be a commutator of two even permutations. The final problem is that in the case n=8, n-3 is less than the bound of 3n/4 required by Bertram. For this case, I went through all the possible cycle structures of permutations in A_8, and with a bit of permutation-matrix-multiplying dirty work from Mathematica, I believe I have found direct commutators for all of them. (Exercise for the reader?) Now let's go back up the chain of reasoning and see what we've proved:
- For n=8 or (from Bertram, using l = n-3) n>=12, every element of A_n is a commutator in A_n.
- Since H_perm is a direct product of groups A_n with n=8, 12, or 24, every element of H_perm is a commutator of elements in H_perm (and therefore in the NxNxN supercube group G).
- Since orientations are easy, every element of H is a commutator in G.
- Since all commutators in G belong to G' by definition of the commutator subgroup, and since G' is contained in H, it follows that {commutators} = G' = H.
Thus the commutator subgroup of the NxNxN supercube group is simply the set of operations that are even permutations on all orbits, and all such operations can be written as single commutators.
It's interesting to note, however, that a similar statement about commutator subgroups of more general groups is not true. Dummit and Foote's Abstract Algebra (page 180 in the third edition) gives a rather convoluted-looking group of order 96 in which one element belongs to the commutator subgroup despite not being a commutator of any two elements in the group.
Hmmm.
reThinking the Cube; are you implying that the Sune produces an odd number of swaps in each orbit?
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