I'm just brushing up on some proof writing and doing some work on my own. Figured I'd post this on here to get feed back and to make sure everything is correct. Hopefully the formatting comes through properly.
Exercise 1. 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 $m$ . Deduce that $|\sigma|=m .$
Proof. We Proceed by induction. For $i=1$ , $\sigma^{1}(a_{k})=a_{k+1}$ is true by the definition of an $m$- cycle. If $k=m$ , $\sigma^{1}(a_{m})=a_{m+1}$ , however by the definition of an $m$- cycle, $\sigma^{1}(a_{m})=a_{1}$ . Therefore $a_{m+1}=a_{1}$ . Since $(m+1)-1=m$ , $m+1$ is replaced with its least positive reside modulo $m$ . Assume this is true for some $j<m$ .
$\sigma^{j+1}(a_{k})=\sigma(\sigma^{j}(a_{k}))=\sigma(a_{k+j})=a_{k+j+1}.$
Suppose $k=m$ , then
$\sigma^{j+1}(a_{m})=\sigma^{j}(\sigma(m))=\sigma^{j}(a_{m+1})=a_{j+m+1}.$
However, we discovered before that $a_{m+1}=a_{1}.$ Therefore,
$\sigma^{j}(a_{1})=a_{j+1}.$
This means that $a_{j+1}=a_{j+m+1}$ and since $(j+m+1)-(j+1)=m$ , $j+m+1$ is replaced with its least positive residue modulo $m .$
Also note that $\sigma^{m}(a_{k})=a_{k+m}$ and since $k+m=k$ modulo $m$ , $\sigma^{m}(a_{k})=a_{k}$ which means that $|\sigma|=m.$
