Let $\tau=(12\cdots n)\in S_n$ and $\sigma\in S_n$. Prove that $\sigma\tau=\tau\sigma\iff\sigma=\tau^i$

Asked Mar 14 '17 at 16:03

Active Mar 14 '17 at 16:03

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Let $\tau=(12\cdots n)\in S_n$ and $\sigma\in S_n$. Prove that $\sigma\tau=\tau\sigma\iff\sigma=\tau^i$ for some $i\in\{1,\dots,n-1\}$.

I can show the $"\impliedby"$ direction. However, I'm struggling with the other one. Say $\sigma\tau=\tau\sigma$. The first thing I thought of is rewriting this as $\sigma=\tau\sigma\tau^{-1}$. We also know that for $i\in\{1,\dots,n-1\}:$ $$ \sigma(\tau(i))=\sigma(i+1)=\tau(\sigma(i))=\sigma(i)+1, $$ except for the case when $\sigma(i)=n.$

However, I wouldn't know how to continue from here on. Could someone give ma a hint?

asked Mar 14 '17 at 16:03

Sha Vuklia

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Hint: Can you show any conjugate of a k-cycle is a k-cycle? – Cheerful Parsnip Mar 14 '17 at 16:14
So if you know $\sigma(1)$ you know all $\sigma(k)$. So there are at most $n$ possible $\sigma$'s. But the powers of $\tau$ already provide them all. – ancient mathematician Mar 14 '17 at 16:15

Let $\tau=(12\cdots n)\in S_n$ and $\sigma\in S_n$. Prove that $\sigma\tau=\tau\sigma\iff\sigma=\tau^i$

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