# 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.