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 \(S_5\).