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 are conjugate in $S_{n}$ if and only if they have the same cycle type.

Proof

Same conjugacy class implies same cycle type

Suppose $σ$ has the cycle type $n_{1}, \dots, n_{k}$ ; write $σ = (a_{11} a_{12} \dots a_{1 n_{1}}) (a_{21} \dots a_{2 n_{2}}) \dots (a_{k 1} a_{k 2} \dots a_{k n_{k}})$ Let $τ \in S_{n}$ .

Then $τ σ τ^{- 1} (τ (a_{i j})) = τ σ (a_{i j}) = a_{i (j + 1)}$ , where $a_{i (n_{i} + 1)}$ is taken to be $a_{i 1}$ .

Therefore $τ σ τ^{- 1} = (τ (a_{11}) τ (a_{12}) \dots τ (a_{1 n_{1}})) (τ (a_{21}) \dots τ (a_{2 n_{2}})) \dots (τ (a_{k 1}) τ (a_{k 2}) \dots τ (a_{k n_{k}}))$ which has the same cycle type as $σ$ did.

Same cycle type implies same conjugacy class

Suppose $π = (b_{11} b_{12} \dots b_{1 n_{1}}) (b_{21} \dots b_{2 n_{2}}) \dots (b_{k 1} b_{k 2} \dots b_{k n_{k}})$ so that $π$ has the same cycle type as the $σ$ from the previous direction of the proof.

Then define $τ (a_{i j}) = b_{i j}$ , and insist that $τ$ does not move any other elements.

Now $τ σ τ^{- 1} = π$ by the final displayed equation of the previous direction of the proof, so $σ$ and $π$ are conjugate.

Example

This result makes it rather easy to list the conjugacy classes of $S_{5}$ .