Product of cycles of equal length I am trying to prove that a permutation $\sigma=(1\ 2\ ...\ m)$ at power $k$ is a product of $d$ cycles of equal lengths where $(k,m)=d$. I know that $\sigma^k$ has length $m$ iff $(k,m)=1$ but I'm clueless when it comes to connecting the information.
 A: See what happens to $1$ when you apply $\sigma^k$ to it several times:$$1\overset{\sigma^k}{\mapsto}1+k\overset{\sigma^k}{\mapsto}1+2k\overset{\sigma^k}{\mapsto}\cdots\overset{\sigma^k}{\mapsto}1+\left(\frac md-1\right)k\overset{\sigma^k}{\mapsto}1.$$So, after $\frac md$ steps you'll get $1$ once again. And that will not happen before, because if $r\in\left\{0,1,\ldots,\frac md-1\right\}$, then$$(\sigma^k)^r(1)=1\iff1+rk\equiv1\pmod m\iff m\mid rk,$$which is impossible, since $(m,k)=d$ and the smallest natural $n$ such that $m\mid nk$ is $n=\frac md$.
So, $\sigma^k$ has this cycle of length $\frac{m}{d}$ that I described. But there is nothing peculiar about $1$ here. So, unless the elements that appear on this cycle are all numbers of $\{1,2,\ldots,m\}$, you take one that is missing and you start all over again.
For instance, if $m=10$ and $k=4$, you will have $2$ cycles:$$1\mapsto5\mapsto9\mapsto3\mapsto7\mapsto1$$and$$2\mapsto6\mapsto10\mapsto4\mapsto8\mapsto2.$$
Note that, since all cycles have length $\frac md$, the number of cycles will be $\frac m{m/d}=d$.
