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Proving one element in the symmetric group (s>=3) commutes with all element

  1. May 17, 2012 #1
    I am really stuck with how to prove that the only element in Sn (with n>=3) commuting with all the other elements of this group is the identity permutation id.

    I have no idea what im supposed to do with it, i know why S3 has only one element that commutes but i dont know how to prove it for all.
  2. jcsd
  3. May 17, 2012 #2
    You could start by proving the following:

    If f and g are in Sn, then the disjoint cycle representation of fgf-1 is obtained by taking the disjoint cycle representation of g and by changing every number n in the representation by g(n).

    Think how that would prove your claim.
  4. May 17, 2012 #3


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    Gold Member

    I think micromass's solution is very good. If you have trouble with his solution, then you can also just do the following:

    If [itex]\sigma \in S_n \setminus \{\mathrm{id}\}[/itex], then there exists an [itex]i \in \{1,\dots,n\}[/itex] such that [itex]\sigma(i) = j[/itex] and [itex]i \neq j[/itex]. Now let [itex]k \in \{1,\dots,n\}[/itex] be such that [itex]k \neq j,\sigma(j)[/itex] and let [itex]\tau \in S_n[/itex] be the transposition which switches [itex]j[/itex] and [itex]k[/itex]. The rest of the proof is trivial from here.
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