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Showing any transposition and p-cycle generate S_p

  1. Sep 20, 2015 #1
    I was hoping someone could help me understand the following proof from:

    http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/genset.pdf


    The problem I'm having is the line ``Any p-cycle can be written as ##(1\;2\;\cdots\;p)## by relabeling the objects being permuted by applying an overall conjugation on ##S_p##''.

    An example helps me to understand:

    Suppose I have ##R=(3\;5),Q=(4\;1\;2\;5\;3)\in S_5##. Now, I take this to mean if I conjugate the entire group, ##\sigma S_5 \sigma^{-1}## such that ##\sigma Q \sigma^{-1}=(1\;2\;3\;4\;5)##, then I've effectively ``relabeled'' ##Q## with ##(1\;2\;3\;4\;5)##. That's easy to accomplish since in general for ##\rho=(a\;b\;c\;d\;e)##, ##\sigma\rho\sigma^{-1}=(\sigma(a)\;\sigma(b)\;\sigma(c)\;\sigma(d)\;\sigma(e))##. Then let ##\sigma=(1\;2\;3\;5\;4)## and then we have ##(1\;2\;3\;5\;4)(4\;1\;2\;5\;3)(4\;5\;3\;2\;1)=(1\;2\;3\;4\;5)##. Therefore I assume this all means:

    ##
    \big<(3\;5),(4\;1\;2\;5\;3)\big>=\big<(h\;k),(1\;2\;3\;4\;5)\big>
    ##

    since conjugation is an automorphism (it's a bijection) and where ##(h\;k)## is the conjugation of ##(3\;5)## and just accepting for the moment if I can get it to that form, it generates the group via the second part of the theorem which I will accept for now.

    Am I interpreting this correctly?

    Thanks for reading,
    Jack
     
    Last edited: Sep 20, 2015
  2. jcsd
  3. Sep 20, 2015 #2

    andrewkirk

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    Yes.

    The second part of the theorem is proven as Theorem 2.8 of what you linked.

    Given that, you just need to write out the formulas by which an arbitrary element g of the group can be expressed in terms of a transposition T and a p-cycle C by first using Theorem 2.8 to express the conjugate of g in terms of another transposition (why not choose (1 2) WLOG) and the standard p-cycle.
     
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