Abstract Algebra: conjugates of cyclical groups

[epr2008]

Homework Statement

If G is a group with operation * and \alpha,\beta\in G, then \beta\ast\alpha\ast\beta^{-1} is called a conjugate of G. Compute the number of conjugates of each 3-cycle in S_{n} (n\geq3).

Homework Equations

The Attempt at a Solution

For any group S_{n} there must be _{n}P_{m} m-cycles. Each m-cycle has m permutations. Then, the number of m-cycles in S_{n} is \frac{_{n}P_{m}}{m}. 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 S_{n}. Then it must be that for odd m, any m-cycle \beta\neq\beta^{-1}. Therefore, for odd m, there are \frac{_{n}P_{m}}{2m} conjugates of order m in S_{n}. In the case of the 3-cycle, there would be \frac{_{n}P_{3}}{2\cdot3}=\frac{n(n-1)(n-2)}{6} conjugates.

 

[ystael]

You are correct that there are \frac{n!}{m(n - m)!} m-cycles in S_n.

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 \kappa, it's asking for the size of the set \{\pi \kappa \pi^{-1} \mid \pi \in S_n\}, that is, conjugates by the entire group -- not merely conjugates by other cycles. Counting the number of expressions \beta \kappa \beta^{-1} doesn't do this, because you could get the same result for several different \beta.

To figure out what the conjugates of a 3-cycle \kappa can be, suppose \kappa = (abc); what is the cycle notation for \pi \kappa \pi^{-1}, in terms of an arbitrary permutation \pi? What can the result be?

 

[epr2008]

Ok I think I understand what you are saying.

So, I have \kappa = (abc).

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

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

So the answer would be _{n}C_{3}?

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.

 

[ystael]

You're making a mistake that you avoided earlier.

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

In your first post you gave a correct calculation for the number of these 3-cycles, but \binom{n}3 is not the right number.

 

[epr2008]

Thank you. I understand what you were saying now. It helped me out a lot.