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.


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".

  • $\begingroup$ People on stackexchange are really the kindest. Thanks a bunch! $\endgroup$ – Jeroen Apr 7 '13 at 21:09
  • $\begingroup$ @Jeroen No problem, happy to help! $\endgroup$ – Tom Oldfield Apr 7 '13 at 21:12

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