Permutation group conjugates

[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.

 

[CellCoree]

for your help!

Hi there,

Thank you for your question about conjugation of permutation groups. Finding the number of conjugation permutations that can also conjugate a and b can be a bit tricky, but I can provide some guidance to help you figure it out.

Firstly, let's define what a conjugation permutation is. A conjugation permutation is a permutation that, when applied to a permutation group, results in a conjugate permutation. In simpler terms, it is a way of rearranging the elements of a permutation group to create a new permutation that has the same cycle structure as the original one.

Now, let's look at the example you provided. You correctly identified that (1 4 2)(3 5) and (1 2 4)(3 5) are conjugate permutations under the conjugation permutation (2 4). However, you also found other arrangements that also result in the same conjugate permutation, such as (2 1 4)(3 5) and (4 1 2)(3 5).

To find the number of unique conjugation permutations, we need to consider the possible arrangements of the two cycles (1 4 2) and (3 5). Since each cycle has 3 elements, there are 3! = 6 possible arrangements for each cycle. Multiplying these together, we get 6x6=36 possible arrangements of the two cycles. However, not all of these arrangements will result in a unique conjugation permutation.

For example, (1 4 2)(3 5) and (2 1 4)(3 5) will result in the same conjugation permutation, since the only difference is the starting point of the first cycle. Similarly, (4 1 2)(3 5) and (1 2 4)(5 3) will also result in the same conjugation permutation.

To avoid counting these duplicates, we can divide the total number of arrangements by the number of possible starting points for each cycle, which is 3 for a 3-element cycle. So, the total number of unique conjugation permutations for (1 4 2)(3 5) and (1 2 4)(3 5) is 36/3=12.

In general, for a permutation group with n elements, the number of unique conjugation permutations will be n!/3 for