1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Science Advisor
    Homework Helper

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook