On extension to "other" iteration roots

[Leo.W]

Not to disappoint you either, Leo.W, but *you* missed one thing!
Your explanations might be well and right, however did not address what I was pointing to.
For me Abelian property was just \(f^s\circ f^t=f^t\circ f^s\), because everywhere else Abelian refers to commutativity.
And I just wanted to point out to you that there was more to a continuous iteration - which in turn you pointed out to me! LoL.

Only now when I look back at at one of your answers, I see you meant already \(f^s\circ f^s = f^{s+t}\) by "Abelian property". However I never encountered that naming in some literature. Even when I googled it now, I didn't see clear definitions (On Wikipedia e.g. there is a chapter "Abelian Property" but it is not explicitly defined and they rather use the name "translation functional equation" for non-integers. I always used the name iteration (semi)group - which though is also quite cumbersome to use, because it refers to more than just the equation. So from which literature did you obtain this definition?

I'm sorry if I abused this term *I just thought Abelian can be used as an adjective word* to replace commutative.
The point is that the iterations are defined in the way \(f^s\circ f^s = f^{s+t}\) when you say that f^s is an iteration, I don't see why would iterations not guarantee \(f^s\circ f^s = f^{s+t}\)...? And if you're not referring to iterations, then I personally think \(f_s\) is a lil better than \(f^s\) as notation.

I'm sorry I pointed out to you, I think there's some misunderstanding. The solutions g(z) to the functional equation \(f(g(z))=g(f(z))\) given f(z) is denoted \([f,f]\) as MphLee, but not all solutions are iterations, or continuous iterations, your \(f^s\) can be interpreted as elements in \([f,f]\), I barely think that it is iteration only after \(f^sf^t=f^{s+t}\) because that will make sense, otherwise we can give such solutions in any forms but doesn't provide us with any useful info, like, what about replacing multiplication with another bunch of solutions of \(f(z+1)=f(z)+1\)
But, yeah, you're right, there's more to continuous iteration  I had no intention to disrupt your research or something else, I just love communication with you fantastic guys
I consider that my ability to arrange what things I intended to say is just sh*t, and without termilogies, lol, maybe my English is so pretentiously spoken, plz forgive me for that, I'm still learning and struggling with GRE and TOEFL and all these stuffs

But it does not mean the semi-group property is followed; at best it means the super function property is defined. I know this might not mean much, but if you read Mphlee's work this is very much his disagreement with some of the things Leo has said.

I'd love to know what the disagreement is against, James
And I wonder where can I browse his works? I want to learn about how much has he did and study more