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

$$\sigma = (a_{11} a_{12} \dots a_{1 n_1})(a_{21} \dots a_{2 n_2}) \dots (a_{k 1} a_{k 2} \dots a_{k n_k})$$
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 \(S_5\).

Parents:

  • Symmetric group

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