How to Notate Commutators and Conjugates?

[MrRedHair]

I understand how commutators work. How a commutator is an algorithm, AB, and its inverse, A'B', written as [A,B], and also how a conjugate is CBC', or [C:B]. My question is more directed to the notation. [A:B] [C:B'] is the same as A B A' C B' C' - how can those moves turn into [C: [C' A, B]]? The last part makes no sense to me. If anybody could try and explain how commutator/conjugate notation works, it would be most appreciated. Thank you.

This commutator/conjugate example came from

http://snk.digibase.ca/k4/7.htm

 

[aronpm]

C C' A B A' C B' C' (expanded), cancels to ABA' CB'C'

 

[qqwref]

[A:B] [C:B'] = (A B A B') (B C B' C') = A B A' C B' C'

[C: [C' A, B]] = C (C' A B A' C B') C' = A B A' C B' C'

What's going on here isn't a weird identity, but just a different way of dividing up the expression. The first one works because you can do the two halves of the expression as commutators and have the extra two moves cancel each other out; the second one works because the commutator [C' A, B] ends up being just a setup move away from the expression you want.

Incidentally, [C: [C' A, B]] = [A' C', C B C'], by the property that [O: [P, Q]] = [[O], [O:Q]]. This identity can be pretty useful to know.

 

[MrRedHair]

I thought [C: [C A, B']] was read as do C, do the commutator of CA,B', which is C A B' C' A' B, then do C' to finish the conjugate.

 

[aronpm]

It is, but it is [c:[c' here, not c and c

 

[Stefan]

AB, and its inverse, A'B'

There's your mistake. It's B'A'.

the commutator of CA,B', which is C A B' C' A' B

Same here.

 

[reThinking the Cube]

[A:B] [C:B'] = (A B A B') (B C B' C') = A B A' C B' C'

[C: [C' A, B]] = C (C' A B A' C B') C' = A B A' C B' C'

What's going on here isn't a weird identity, but just a different way of dividing up the expression. The first one works because you can do the two halves of the expression as commutators and have the extra two moves cancel each other out; the second one works because the commutator [C' A, B] ends up being just a setup move away from the expression you want.

Incidentally, [C: [C' A, B]] = [A' C', C B C'], by the property that [O: [P, Q]] = [[O], [O:Q]]. This identity can be pretty useful to know.

cancelling setup move (conjugate) ~ cyclic shift.

Good post, but the intermediate results are incorrect as you have shown them.
[A:B] [C:B'] = (A B A B') (B C B' C') is wrong.
The 2nd A should be a A', and [A:B] [C:B'] is conjugates - not commutators, and so that extra "B') (B" does not belong here.

Incidentally, [C: [C' A, B]] = [A' C', C B C']

This is wrong too, since that A'C' needs to be AC'

@michael, I sent you a courtesy PM a couple of days ago to inform you of these mistakes so that you could correct them yourself. Did you not get it?