# Abstract Algebra: conjugates of cyclical groups

#### epr2008

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

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$$\geq$$3).

2. Relevant equations

3. 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.

Last edited:
Related Calculus and Beyond Homework News on Phys.org

#### 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.

Last edited:

#### 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.

"Abstract Algebra: conjugates of cyclical groups"

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving