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I want to find the conjugacy classes of the permutation group $S_n$

To start with I think I have to prove that $\pi(\sigma_1\dots \sigma_m)\pi^{-1} = (\pi(\sigma_1)\dots \pi(\sigma_m))$. Where $\pi$ is an arbitrary element and $\sigma$ is a simple cycle.

I have a proof where I basically write out the lhs. I won't post it here as I understand "check my work" question are frowned upon. I was just wondering if there is a more elegant (shorter) proof.

john
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  • A related question: http://math.stackexchange.com/questions/1382105/what-does-it-means-to-multiply-a-permutation-by-a-cycle-pix-1-cdots-x-n-pi – coldnumber Aug 03 '15 at 19:27

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The easiest way to see this is to ask yourself: what happens to $\pi(\sigma_{1})$ when we act on it by $\pi (\sigma_{1} \ldots \sigma_{n})\pi^{-1}$? Well the $\pi^{-1}$ sends us to $\sigma_{1}$, the cycle sends us to $\sigma_{2}$ and then the $\pi$ sends us to $\pi(\sigma_{2})$. This is the definition of the RHS.

This implies that conjugation preserves cycle type in $S_{n}$, so you can find the conjugacy classes by writing down all the permutations with the same cycle type.

preferred_anon
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