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.


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.


This result makes it rather easy to list the conjugacy classes of \(S_5\).


  • Symmetric group

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