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If two permutations commute they are disjoint

  1. Dec 26, 2009 #1
    1. The problem statement, all variables and given/known data
    If [tex]\alpha,\beta\in S_n[/tex] and if [tex]\alpha \beta = \beta \alpha[/tex], prove that [tex]\beta[/tex] permutes those integers which are left fixed by [tex]\alpha[/tex]. Show that [tex]\beta[/tex] must be a power of [tex]\alpha[/tex] when [tex]\alpha[/tex] 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?
  2. jcsd
  3. Dec 26, 2009 #2


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    Hi 3029298! :smile:

    (have an alpha: α and a beta: β :wink:)
    But that's obviously not true …

    put α = β: then α does not permute those integers which are left fixed by α. :confused:
  4. Dec 26, 2009 #3
    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...
  5. Feb 17, 2010 #4
    Hi. I'm new in the forum. I'd like to prove (or find counter example) the next statement.

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

    Do you think it's true?? I think it is.
  6. Oct 10, 2011 #5
    Here is an example in [itex]S_5.[/itex] Take [itex]\alpha=(12345),[/itex] and [itex]\beta=(13524)[/itex].

    They commute:

    But they are not disjoint (and neither is one a multiple of the other, re Unviaje).
  7. Oct 16, 2011 #6
    oops I just realized that [itex](12345)^2=(13524)[/itex]. I think you are right, Unviaje. I haven't found a proof yet though.
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