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Permutation group conjugates

  1. Mar 12, 2012 #1

    I just have a small question regarding the conjugation of permutation groups.

    Two permutations are conjugates iff they have the same cycle structure.

    However the conjugation permutation, which i'll call s can be any cycle structure. (s-1 a s = b) where a, b and conjugate permutations by s

    My question is, how can you find out how many conjugation permutations (s) are within a group which also conjugate a and b.

    So for example (1 4 2)(3 5) conjugates to (1 2 4)(3 5) under s = (2 4), how could you find the number of alternate s's in the group of permutations with 5 objects?

    Would it be like

    (1 4 2) (3 5) is the same as (2 1 4) (35) which gives a different conjugation permutation,
    another is

    (4 1 2)(3 5), then these two with (5 3) instead of ( 3 5),

    so that gives 6 different arrangements, and similarly (1 2 4) (35) has 6 different arrangements,

    and each arrangement would produce a different conjugation permutation (s)

    so altogether there would be 6x6=36 permutations have the property that
    s-1 a s = b ?

    Would each of the arrangements produce a unique conjugation permutation (s) ?
    I went through about 6 and I got no overlapping conjugation permutations but I find it a little hard to a imagine there would be unique conjugation permutations for each of the 36 arrangements.

    Thanks in advance
  2. jcsd
  3. Mar 12, 2012 #2


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    I'm really confused by your question. Every single s will produce a conjugate of a, namely ##sas^{-1}##. Of course, different s and t might give the same conjugate ##sas^{-1}=tat^{-1}##.

    But surely that's not what you're asking about... Did you intend to say that you have a fixed a and b in S_5, and you want to count the number of elements s such that ##sas^{-1}=b##?
  4. Mar 12, 2012 #3
    Yea thats right I want to count the number of s for fixed a and b,

    Sorry for not explaining it well,

    Is the way I wrote correct for a = (1 4 2)(3 5) and b = (1 2 4)(3 5)?

    The first s would be (2 4),

    Then rewritting a as (2 1 4)(3 5) the next would be (1 2)

    Then rewritting as (4 2 1) (3 5) to get another (1 4)

    Then (1 4 2)(5 3) gives (3 5)

    and so on

    I checked and each of these, (2 4), (1 2), (1 4) and (3 5) correctly conjugate (1 4 2)(35) to b

    so would that suggest there are 6 different possible s for a and b?

    Since there are 3 arrangements of (1 4 2) and 2 arrangements of (3 5) which give the same permutation.

    Thanks for answering =]
  5. Mar 12, 2012 #4


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    Yes, that's correct.

    You can get a formula for general a and b (of the same cycle type) in S_n as follows. Begin by noting that $$ |\{s\in S_n \mid sas^{-1}=b\}| = |\{s\in S_n \mid sas^{-1}=a\}|. $$ But the RHS is simply the order ##|C_{S_n}(a)|## of the centralizer of a in S_n, and this is the number you want. Now recall that the order of the centralizer of a is equal to the order of S_n divided by the size of the conjugacy class of a (this follows, for example, from the orbit-stabilizer formula), and there is a general formula for the latter - see e.g. here.

    Let's work this out for a=(142)(35) in S_5. The size of the conjugacy class of a is (5*4*3)/3=20, so the order of the centralizer of a is 5!/20=6, confirming your answer.
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