Abstract Algebra: conjugates of cyclical groups

1. The problem statement, all variables and given/known data

If G is a group with operation * and [tex]\alpha,\beta\in G[/tex], then [tex]\beta\ast\alpha\ast\beta^{-1}[/tex] is called a conjugate of G. Compute the number of conjugates of each 3-cycle in S[tex]_{n}[/tex] (n[tex]\geq[/tex]3).

2. Relevant equations

3. The attempt at a solution
For any group [tex]S_{n}[/tex] there must be [tex]_{n}P_{m}[/tex] m-cycles. Each m-cycle has m permutations. Then, the number of m-cycles in [tex]S_{n}[/tex] is [tex]\frac{_{n}P_{m}}{m}[/tex]. Since an m-cycle exchanges m objects, it's inverse must also exchange m-objects. So, the inverse must also be an m-cycle.

I have an idea of how to go on from here, but I am not sure about it since some m-cycles are their own inverse...

If m is odd then there are an even number of m-cycles in [tex]S_{n}[/tex]. Then it must be that for odd m, any m-cycle [tex]\beta\neq\beta^{-1}[/tex]. Therefore, for odd m, there are [tex]\frac{_{n}P_{m}}{2m}[/tex] conjugates of order m in [tex]S_{n}[/tex]. In the case of the 3-cycle, there would be [tex]\frac{_{n}P_{3}}{2\cdot3}=\frac{n(n-1)(n-2)}{6}[/tex] conjugates.
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You are correct that there are [tex]\frac{n!}{m(n - m)!}[/tex] [tex]m[/tex]-cycles in [tex]S_n[/tex].

To think about the conjugates, you don't need to worry about whether a cycle is its own inverse or not. When the problem asks for the number of conjugates of a cycle [tex]\kappa[/tex], it's asking for the size of the set [tex]\{\pi \kappa \pi^{-1} \mid \pi \in S_n\}[/tex], that is, conjugates by the entire group -- not merely conjugates by other cycles. Counting the number of expressions [tex]\beta \kappa \beta^{-1}[/tex] doesn't do this, because you could get the same result for several different [tex]\beta[/tex].

To figure out what the conjugates of a 3-cycle [tex]\kappa[/tex] can be, suppose [tex]\kappa = (abc)[/tex]; what is the cycle notation for [tex]\pi \kappa \pi^{-1}[/tex], in terms of an arbitrary permutation [tex]\pi[/tex]? What can the result be?
Ok I think I understand what you are saying.

So, I have [tex]\kappa = (abc)[/tex].

Then, [tex]\pi\kappa\pi^{-1} = \pi(abc)\pi^{-1} = (\pi(a)\pi(b)\pi(c))[/tex].

Alright then, I think what I'm looking for are all the 3-cycles in [tex]S_{n}[/tex]?

So the answer would be [tex]_{n}C_{3}[/tex]?

EDIT: Wait that's exactly what I said before... Well I guess if the conjugate stays closed to a, b, and c, then it would just be the permutations of (abc). Then the answer would be 6... But I have no clue if this is right or not. Any thoughts? Anyone? I just really don't understand what a conjugate is.
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You're making a mistake that you avoided earlier.

You are correct that the conjugates [tex]\pi\kappa\pi^{-1}[/tex] of any 3-cycle [tex]\kappa = (abc)[/tex], where [tex]\pi[/tex] ranges over all of [tex]S_n[/tex], are all the 3-cycles [tex](\pi(a)\pi(b)\pi(c))[/tex] -- that is, all the 3-cycles in [tex]S_n[/tex].

In your first post you gave a correct calculation for the number of these 3-cycles, but [tex]\binom{n}3[/tex] is not the right number.
Thank you. I understand what you were saying now. It helped me out a lot.

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