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A question about the commutator subgroup

Submitted by Mike G on Thu, 09/02/2004 - 08:43.
We all know that commutators can be used to generate half of the Cube group G. My first question is: Can all elements of the commutator subgroup themselves be written as commutators? i.e., the problem is to determine whether the set of commutators is closed under multiplication; it need not be in general, but is it true here?

If it is closed in this case, then a natural question to ask is How do we write a given element of the commutator subgroup as a single commutator?

On the other hand, if the set of commutators is NOT closed under multiplication, then how many elements of G can be written in commutator form?

I'm sorry if there was a discusssion of these or similar questions in the old Cube Lovers forum, but I could not find it in the archive.
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commutator subgroups

Submitted by cubex on Fri, 09/10/2004 - 08:02.
I'm not sure how to write a given element of the commutator subgroup as a single commutator but I have some thoughts.

It seems to me that any commutator process is going to be 0 q moves, or 4 q moves or greater. Naturally X Y X' Y' could equal the identity. In the case of a single generator, say the U face, all commutators equal the identity. Even in the < U, D> case we still only get the identity from any commutator process.

In other subgroups like < U, R > the derived subgroup is exactly half the size of the group, same as the full cube group. Any commutator element is even.... need to think more about your last question.
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Re: A question about the commutator subgroup

Submitted by jaap on Tue, 09/07/2004 - 07:33.
Interesting question.

I am fairly certain that every element of An is a commutator of elements in Sn. A sketch of a proof follows:

For any element in An, get its disjoint cycle notation. Any cycle of odd length is an even permutation, and can be written as a commutator. For example:
(01)(23)(45) . (12)(34)(56) = (0246531)
The two permutations on the left are conjugates since they have the same cycle structure, so when combined they form a commutator. Note that no items other than those in the cycle itself need be acted on by the constituent parts of the commutator.
Any two even length cycles can be combined and written as a commutator. For example:
(01)(23)(45)(bc)(de) . (12)(34)(ab)(cd)(ef) = (024531)(abdfec)
Again no other items are acted upon.
The parts of these commutators act upon are disjoint sets, so they can be combined into a single commutator.

The question remains however whether the orientation of the pieces makes any difference.
I have tried some simple cases. For example, this corner permutation:
1234 -> 2+ 1+ 4- 3-
where the numbers 1 to 4 represent any four corners.
If we let f(a,b,c) be the permutation
abc-> b a- c+
then the permutation we want can be written as
f(3,4,1) . f_inv(2,1,4)
which is a commutation of f(3,4,1) and (23)(41)

I'm having trouble finding anything that is not a commutator, and I suspect that may be impossible. If there are any bad cases, they would probably involve all corners, all of them twisted.

Jaap

Jaap's Puzzle Page:
http://www.geocities.com/jaapsch/puzzles/
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