For a fixed positive integer $n$, let $C_{n}$ be the group generated by the permutation $r = \begin{pmatrix}1 & 2 & \cdots & n\end{pmatrix}$. We want to determine the cycle type of the powers $r^m$ for $1\leq m\leq n$. To find the cycles of $r^m$, we start at any element $i$ and repeatedly apply $\pi^m$ until we return to $i$. The number of steps before returning to $i$ is exactly the smallest positive integer $k$ such that $i + k m \equiv i \pmod{n}$ which, by cancelling $i$, simplifies to $km \equiv 0 \mod{n}$. Thus $k = n/\gcd(m,n)$ showing that every cycle in $r^m$ has length $n/\gcd(m,n)$ and in fact $r^m$ is made up of $\gcd(m,n)$ cycles of length $n/\gcd(m,n)$. Moreover, an integer $m$ satisfies $1 \leq m \leq n$ and $\gcd(m, n)=d$ if and only if $1\le m/d\leq n/d$ and $\gcd(m/d,l/d)=1$. Hence there are $\phi(n/d)$ such integers $m$, where $\phi$ denotes Euler's totient function, so the cycle index polynomial of $C_{n}$ is given by $Z_{C_{n}}(z_{1}, \dots, z_{n})= \frac{1}{n} \sum_{d\mid n} \phi (n\text{/}d) z^{d}_{n\text{/}d},$or (substituting $n/d$ for $d$) $Z_{C_{n}}(z_{1},\dots,z_{n})=\frac{1}{n} \sum_{d\mid n} \phi (d) z^{n\text{/}d}_{d}.$ For example,