# I suggest to rename [a:b] other than the term conjugate



## uiqoo (Mar 12, 2016)

[a,b] = a b a' b' = commutator
[a:b] = a b a' = conjugate (a = setup move)

In group theory, we use the word 'commutator' for (a b a' b') or (a' b' a b), and the sign '[a,b]' is also used. 
However, we use 'b^a' for (a b a') or (a' b a) and we call (a b a') a conjugate of b by a. The sign '[a:b]' is not used. 

∀g,h∈G, if ∃a such that a g a' = h, g and h are called to be a _conjugate of each other_. 
And every h that is conjugate to g forms a conjugacy class Cl(g). (∀g,h∈Cl(g), ∃a s.t. a g a' = h)
A good example of a conjugacy class is {every corner 3-cycle}. 


But, here's the point. 
When we form a commuatator [a,b] = a b a' b', we can say a and b are in some kind of relationship. 
For example, if [a,b]=I then we call a and b commutate to each other. 
However when we form a conjugate [a:b] = a b a', we do NOT call that a and b are in conjugate. a and b are not in any relationship. 
When c := [a:b], b and c are a conjugate of each other, not a and b. This is not intuitive at all. 
Also the word 'commutator' and 'conjugate' is in different word class. (conjugator seems better?)



To be clear, in group theory, [a,b] is called a commuatator of a and b. And, a and b form some kind of relationship. 
But [a:b], which is b^a, is NOT called a _conjugate of a and b_. It is called a _conjugate of b by a_. And, a and b do NOT form some kind of relationship. 
So if we want to call [a:b] something, we need a term for this: [a,b] is called a (_________) of a and b. 
The problem is the terms 'commutator' and 'conjugator' are not symmetric. 


Any suggestions?


----------



## irontwig (Mar 12, 2016)

Firstly, what is "correct" in languages is dictated by usage, not logic or committee. Secondly, I for one find the the use of conjugate rather intuitive. In the same way that "eat" and "ate" (i.e. a conjugation of the verb "eat") carry the same basic idea of consuming food, a conjugated T-perm still switches two corners and two edges. Thirdly, they do share a word class, "commutator" being a noun and "conjugate" being both a noun and a verb.


----------



## uiqoo (Mar 13, 2016)

irontwig said:


> Firstly, what is "correct" in languages is dictated by usage, not logic or committee. Secondly, I for one find the the use of conjugate rather intuitive. In the same way that "eat" and "ate" (i.e. a conjugation of the verb "eat") carry the same basic idea of consuming food, a conjugated T-perm still switches two corners and two edges. Thirdly, they do share a word class, "commutator" being a noun and "conjugate" being both a noun and a verb.



But the word commutator and conjugate have a different usage? We say [setup move: T-perm] a 'conjugated T-perm' but [something, T-perm] is not a 'commutated T-perm'. What I'm saying is the word usage are different for commutator and conjugate but we still say [a:b] as a conjugate and [a,b] as a commutator.


----------



## cuBerBruce (Mar 13, 2016)

I would call *[noparse][a : b][/noparse]* a "conjugation," not a "conjugate."

Unfortunately, the speedcubing community tends to mangle mathematical terminology.

Of course, unlike the commutator notation, the notation *[noparse][a : b][/noparse]* is not used in mathematics, at least not for this purpose. It is purely a "cubing" notation convention.


----------



## coldsun0630 (Mar 14, 2016)

How about using "*conjugator*"?

Well the expression [a:b] doesn't mean that 'a' and 'b' are in conjugate, but it gives us an information that the output 'c' will be a 'b' which conjugated by 'a'.
So I think calling *[a:b]* as *a conjugator which conjugates 'b' with setup move 'a'* is better.

In addition, The word "conjugator" is not a symmetric with "commutator", but it preserves the term "conjugate" as well.


----------

