# Conjugacy class is cycle type in symmetric group

https://arbital.com/p/conjugacy_class_is_cycle_type_in_symmetric_group

by Patrick Stevens Jun 14 2016 updated Jun 16 2016

There is a neat characterisation of the conjugacy classes in the symmetric group on a finite set.

In the Symmetric group on a finite set, the Conjugacy class of an element is determined exactly by its cycle type.

More precisely, two elements of $S_n$ are conjugate in $S_n$ if and only if they have the same cycle type.

# Proof

## Same conjugacy class implies same cycle type

Suppose $\sigma$ has the cycle type $n_1, \dots, n_k$; write $$\sigma = (a_{11} a_{12} \dots a_{1 n_1})(a_{21} \dots a_{2 n_2}) \dots (a_{k 1} a_{k 2} \dots a_{k n_k})$$ Let $\tau \in S_n$.

Then $\tau \sigma \tau^{-1}(\tau(a_{ij})) = \tau \sigma(a_{ij}) = a_{i (j+1)}$, where $a_{i (n_i+1)}$ is taken to be $a_{i 1}$.

Therefore $$\tau \sigma \tau^{-1} = (\tau(a_{11}) \tau(a_{12}) \dots \tau(a_{1 n_1}))(\tau(a_{21}) \dots \tau(a_{2 n_2})) \dots (\tau(a_{k 1}) \tau(a_{k 2}) \dots \tau(a_{k n_k}))$$ which has the same cycle type as $\sigma$ did.

## Same cycle type implies same conjugacy class

Suppose $$\pi = (b_{11} b_{12} \dots b_{1 n_1})(b_{21} \dots b_{2 n_2}) \dots (b_{k 1} b_{k 2} \dots b_{k n_k})$$ so that $\pi$ has the same cycle type as the $\sigma$ from the previous direction of the proof.

Then define $\tau(a_{ij}) = b_{ij}$, and insist that $\tau$ does not move any other elements.

Now $\tau \sigma \tau^{-1} = \pi$ by the final displayed equation of the previous direction of the proof, so $\sigma$ and $\pi$ are conjugate.

# Example

This result makes it rather easy to list the conjugacy classes of $S_5$.