If two permutations commute they are disjoint

[3029298]

Homework Statement

If \alpha,\beta\in S_n and if \alpha \beta = \beta \alpha, prove that \beta permutes those integers which are left fixed by \alpha. Show that \beta must be a power of \alpha when \alpha is a n-cycle.

The other way round is easy to see, since if two cycles are disjoint they do not do anything with the numbers permuted by the other cycle, hence they commute. But I don't know how to start when I want to prove the statement above... can anyone hint me?

 

[tiny-tim]

Hi 3029298!

(have an alpha: α and a beta: β )

If \alpha,\beta\in S_n and if \alpha \beta = \beta \alpha, prove that \beta permutes those integers which are left fixed by \alpha.

But that's obviously not true …

put α = β: then α does not permute those integers which are left fixed by α.

 

[3029298]

Yes, I saw that as well, but I think (for the question to make sense) alpha must permute different elements than beta. But this makes the second part of the question questionable...

 

[Unviaje]

Hi. I'm new in the forum. I'd like to prove (or find counter example) the next statement.

If two permutations \alpha,\beta\in S_n such that \alpha\neq\beta^i for every i\in\mathbb{Z} commute then they're disjoint.

Do you think it's true?? I think it is.

 

[postylem]

Here is an example in S_5. Take \alpha=(12345), and \beta=(13524).

They commute:
(12345)(13524)=(14253),
(13524)(12345)=(14253).

But they are not disjoint (and neither is one a multiple of the other, re Unviaje).

 

[postylem]

oops I just realized that (12345)^2=(13524). I think you are right, Unviaje. I haven't found a proof yet though.