Logical Subtlety in Division Algorithm 
Let $\sigma$ be the $m$-cycle $(a_1~a_2...a_m)$. Deduce that $|\sigma| = m$. 

Here is my solution, with questions about the correctness of this solution scattered throughout: 
I already proved that $\sigma^i(a_k) = a_{k+i}$, where $k \in \{1,...,m\}$ and $k+i$ is replaced by the least positive residue when $k+i$ is divided by $m$. Hence, by this composition formula $\sigma^m (a_k) = a_{k+m}$. By the division algorithm, there exist unique integers $q$ and $r$ such that $k+m = qm + r$, where $r \in \{1,...,m-1\}$ is the least positive residue we are seeking. Now clearly $q = 1$ and $r=k$ are choices that satisfy the equation. My question is, does this imply that these are in fact the $q$ are $r$ guaranteed by the division algorithm? It seems that the uniqueness part, in particular, implies that $q = 1$ and $r=k$. 
If this is the case, then I can say $\sigma^m(a_k) = a_k$ for all $k \in \{1,...,m\}$, implying $|\sigma| \le m$. Now, suppose that $|\sigma| = d \le m$. Then $\sigma^d(a_k) = a_k$ for all $k$. This means $k$ is the least positive residue, or, in other words, $k+d = q'm + r'$, where $r' = k$. Hence $d = q'm$. This is a contradiction unless $q' = 1$, because $q' < 0$ would imply $|\sigma| < 0$ and $q' > 0$ would imply $d > m$. 
First of all, does any of this sound correct? Second, is there a better/cleaner way of thinking about this? 

I think you may make your life simpler if you 


*

*write the cycle as $\sigma = (a_{0}, a_{1}, \dots, a_{m-1})$, 

*regard the indices as integers modulo $m$, 

*note that $\sigma$ acts as $a_{i} \mapsto a_{i+1}$, 

*so that $\sigma^{k}$ acts as $a_{i} \mapsto a_{i+k}$.

