# Abstract Algebra: conjugates of cyclical groups

• epr2008
In summary, the number of conjugates of each 3-cycle in S_n (n≥3) is equal to the number of 3-cycles in S_n, which can be calculated as n(n-1)(n-2)/6. This is because the conjugates of a 3-cycle are all the 3-cycles in S_n, and not just permutations of the original 3-cycle.
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$$\geq$$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.

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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?

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.

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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.

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

## 1. What is a cyclical group in abstract algebra?

A cyclical group is a type of mathematical structure that is formed by a single element, called the generator, and its powers. It is denoted by ⟨g⟩ where g is the generator. The group operation is defined as multiplication of the generator with itself, and its inverse is the reciprocal of the generator.

## 2. What are conjugates of cyclical groups?

Conjugates of cyclical groups are groups that are isomorphic to the original cyclical group but are formed by different elements. This means that the group operation and structure are the same, but the elements are rearranged or expressed in a different way.

## 3. How do you find conjugates of a cyclical group?

To find conjugates of a cyclical group, you can use the group's defining rule or operation to generate new elements that have the same structure. For example, if the cyclical group is formed by the generator g, then its conjugates can be found by raising g to different powers or multiplying it by other elements in the group.

## 4. Why are conjugates important in abstract algebra?

Conjugates are important in abstract algebra because they allow for the study of different representations of the same mathematical structure. This can provide insight into the properties and behaviors of the group, and can also be used in applications such as cryptography and coding theory.

## 5. Can all cyclical groups have conjugates?

Yes, all cyclical groups have conjugates. This is because the group operation and structure remain the same, only the elements are changed. Therefore, any rearrangement or transformation of the elements in a cyclical group can be considered a conjugate of the original group.

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