# 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.