In course of my attempt of the following Dummit-Foote exercise of Symmetric Group:

- Prove that if $\sigma$ is the $m$-cycle $(a_{1}a_{2}...\, a_{m})$, then for all $i\in\{1,2,...m\}$ , $\sigma^{i}(a_{k})=a_{k+i}$, where $k+i$ is replaced by its least positive residue modulo $m$.
I've found myself that the result is true for all positive integer $i.$ Even though the same problem was posted for verification (Working on proofs from Dummit and Foote 3rd edition 3rd-edition) I still want to verify it to get sure whether saying $i\in\{1,2,...,m\}$ is a overstatement. Here's my attempt:

Without loss of generality let $m\ge2.$

For $i=1,$ the result holds trivially.

Let the result be true for some $i\ge1.$ Then $\sigma^{i+1}(a_k)=\sigma(\sigma^i(a_k)).$ For further calculation let's split it into cases.

Case I: $k+i+1\le m.$ Then $k+i< m$ whence $\sigma^i(a_k)=a_{k+i}$$\implies \sigma^{i+1}(a_k)$$=\sigma(\sigma^i(a_k))$$=\sigma^1(a_{k+i})$$=a_{k+(i+1)}.$

Case 2: $k+i+1>m.$

Subcase 2a: $k+i=m.$ Then $\sigma^{i+1}(a_k)=\sigma(\sigma^i(a_k))=\sigma(a_{k+i})=a_{k+i+1~(\mod m)}$

Subcase 2a: $k+i>m.$ Then $\sigma^{i+1}(a_k)=\sigma(\sigma^i(a_k))=\sigma(a_{k+i~(\mod m)})=a_{k+i+1~(\mod m)}$ since $k+i~(\mod m)+1=k+i+1~(\mod m)$ follows from $m\ge2.$

Hence the result is true for $i+1.$ Consequently by mathematical induction, $\forall~i\in\mathbb Z^+,~\sigma^{i}(a_{k})=a_{k+i}$, where $k+i$ is replaced by its least positive residue modulo $m$.

*Please comment on the validity of my attmpt.*

Added: **PLEASE HELP:** I can see due to the result, $σ^m(a_m)=a_0$ since $m+m>m$ but $a_0$ is undefined. Where did I get it wrong?