Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Show that the orbits of $S_n$ under the conjugation action of $S_n$ on itself correspond 1-1 with the cycle types.

So, the orbit of $\sigma \in S_n$ is the set $S_n \sigma = \{ \tau .\sigma : \tau \in S_n \} = \{\tau \sigma \tau^{-1} : \tau \in S_n \}$. I know that $\sigma$ and $\tau \sigma \tau^{-1}$ have the same cycle type. Therefore, $S_n \sigma$ is a subset of the set with all elements that have the same cycle type as $\sigma$, say $H_{\sigma}$.

Now I think I need to show that $H_{\sigma} \subset S_n \sigma$. So, let $\lambda \in H_{\sigma}$. We know that $\sigma$ and $\lambda$ now have the same cycle type. How do I prove that $\lambda$ can be written as $\tau_0 \sigma \tau_0 ^{-1}$ for a certain $\tau_0 \in S_n$?

Or am I completely on the wrong track? Thanks in advance.

share|cite|improve this question
up vote 2 down vote accepted

You are on exactly the right track, what you need to do is indeed find some $\rho \in S_n$ such that $\rho\sigma\rho^{-1} = \lambda$.

To do this, suppose that the disjoint cycle composition of $\sigma$ is $c_1c_2\dots c_n$ where $c_i = (a_{i1}\dots a_{ik_i})$. Since $\lambda$ has the same cycle type, it follows that $\lambda = c_1'c_2'\dots c_n'$ where $c_i' = (b_{i1}\dots b_{ik_i})$.

Now that all the tedious naming is done, the choice of $\rho$ becomes more obvious, let $\rho$ be the element of $S_n$ such that $\rho(a_{ij}) = b_{ij}$. It follows immediately from writing out $\rho\sigma\rho^{-1}$ in disjoint cycle notation that $\rho\sigma\rho^{-1}=\lambda$

Note: This technique of dealing with conjugation in $S_n$ is something that it is really useful to be familiar with, because it is useful in many aspects of dealing with conjugation within $S_n$ and its subgroups, particularly in the proof of the statement that "elements of the same cycle type are not conjugate in $A_n$ iff their disjoint cycle notations contain only odd length cycles of distinct lengths".

share|cite|improve this answer
People on stackexchange are really the kindest. Thanks a bunch! – Jeroen Apr 7 '13 at 21:09
@Jeroen No problem, happy to help! – Tom Oldfield Apr 7 '13 at 21:12

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.