Let $c$ be an $r$-cycle in $S_n$, where $r$ > 1. Let $d$ be an integer not divisible by $r$. Then the cycle decomposition of $c^d$ is the product of $k$ disjoint $s$-cycles.
I know that $k$ equals the greatest common divisor of $r$ and $d$, but what is $s$?
How can I have express $s$ in terms of $r,n,d$, $gcd(r,d)$?
For convenience, think of the $r$-cycle $\sigma=(0,1,2,\ldots,r-1).$
We give two modes of argument.
First:
Notice that $\sigma^d$ sends $x$ to $x+d \pmod{r}.$ Hence, $s$ is the least positive integer such that $$x+sd = x \pmod{r}$$ for all $x\in \lbrace 0,1,\ldots,r-1\rbrace$.
Simplifying, we want the least positive integer $s$ such that $$sd = 0 \pmod{r},$$ i.e., that $sd$ is a multiple of $r$.
Since $sd$ is also a multiple of $d$, we desire $sd=lcm(r,d)$. Hence, $s=\frac{lcm(r,d)}{d}$.
Second:
Notice that if $\sigma^d(x)=x$ for any $x\in \lbrace 0,1,\ldots, r-1 \rbrace$, then $\sigma: x \mapsto x + d\pmod{r}$ means that $d$ must be divisible by $r.$ This contradicts the assumption, so that each $x\in \lbrace0,1,2\,\ldots,r-1\rbrace$ belongs to one of the $k$ cycles in the cycle decomposition of $\sigma^d$. By a symmetry argument, each of the $k$ cycles have the same length $s$. Thus, we must have $$s = \frac{r}{k}=\frac{r}{gcd(r,d)}=\frac{lcm(r,d)}{d}.$$