Prove that $\sigma$ is a power of $\tau$ when they commute $\sigma\tau=\tau\sigma$.

Question

Let $\tau = (1, 2, 3, ..., n)$ be a cycle of maximal length $n$ in the symmetric group $S_n$ and let $\sigma \in S_n$ be such that $\sigma\tau=\tau\sigma$.

Prove that $\sigma$ is a power of $\tau$.

I'm not sure how I can use this fact to prove the claim holds (or doesn't).

Answer 1

Think of $\{1,2,\ldots,n\}$ as $\Bbb Z/n\Bbb Z$, the integers modulo $n$. Then $\tau$ acts via $\tau(j)\equiv j+1\pmod n$. If $\sigma\tau=\tau\sigma$ then $$\sigma(j+1)\equiv\sigma(j)+1\pmod n$$ and so $$\sigma(j+k)\equiv\sigma(j)+k\pmod n.$$ Taking $j=n\equiv0\pmod n$ then $$\sigma(k)\equiv\sigma(n)+k\pmod n$$ so $\sigma=\tau^{\sigma(n)}$.