# Conjugacy class is cycle type in symmetric group

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

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

## Same cycle type implies same conjugacy class

Suppose

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 \(S_5\).

Parents:

- Symmetric group
The symmetric groups form the fundamental link between group theory and the notion of symmetry.