How Can You Find the Number of Conjugation Permutations in a Group?

[physicsjock]

Hey,

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

 

[morphism]

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 $sas^{-1}=b$?

 

[physicsjock]

Yea that's 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 =]

 

[morphism]

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.