Permutations: $στ = τσ \iff σ = τ^i$

Question

Hi could someone help me with the following:

Suppose $τ = (123...n) \in Sn$ and $σ \in Sn$
Prove that $στ = τσ$ if and only if $ σ = τ^i$ for an $i \in \mathbb{Z}$ with $0 \leq i < n$.

I dont know where to start. Thank you!

Answer 1

Notice that $\sigma\tau=\tau\sigma\iff\tau=\sigma\tau\sigma^{-1}\iff(1,2,\dots n)=(\sigma(1),\sigma(2),\dots \sigma(n))$. Let's now say that $\sigma(1)=i+1$ for some $0\leq i<n$. Then, $\sigma=\begin{pmatrix}1 & 2 & \dots & n\\ i+1 & i+2 & \dots & i\end{pmatrix}$ which is precisely $\sigma=\tau^i$.

Answer 2

if-part is clear. For the converse, suppose $\sigma\tau=\tau\sigma$ and let $i=\sigma (1)\in \{1,\dots,n\}$. Since $$\sigma(\tau(1))=\tau(\sigma(1)),$$ we have $\sigma(2)=\tau(i)=i+1$ (when $i=n$, $n+1$ is regarded as $1$. More generally, we identify the numbers $n+1,n+2,\dots,$ with $1,2,\dots,$ in what follows). Inductively, we get $\sigma(k)=\tau(\sigma(k-1))=\sigma(k-1)+1$, hence $\sigma$ is the permutation $$\begin{pmatrix} 1&2&3&\cdots&n-i+1&n-i+2&\cdots&n-1&n\\ i&i+1&i+2&\cdots&n&1&\cdots&i-2&i-1\end{pmatrix},$$ that is exactly $\tau^{i-1}$.