Raising a cycle to a power, cycle decomposition Let $\alpha$ be an m-cycle. Is it true that $\alpha ^k$ can be decomposed into $\gcd(m,k)$ disjoint cycles?
for example $(1 2 3 4 5 6)^2 = (1 3 5)(2 4 6)$, 
$(1 2 3 4 5 6 7 8)^6 =(1 7 5 3)(2 8 6 4)$,
It seems to be true, unless I am missing an easy counterexample.
 A: Identify any $m$-cycle with a canonical indexing of it so that it is represented as
$$
\alpha=(1\ 2\ ...\ m)
$$
just like you did in your examples. Then you can identify $\alpha$ with a map defined on the group of congruences modulo $m$, that is
$$
\begin{align}
f_\alpha:&\newcommand{\Z}{\mathbb{Z}}\Z/m\Z\rightarrow\Z/m\Z\\
f_\alpha:&[x]\mapsto [x+1]
\end{align}
$$
where the congruence class usually represented as $[0]$ is here represented as $[m]$. Then $f$ is simply shifting to the next congrunece class between $[1]$ and $[m]$ starting over with $[1]$ after $[m]$.
With this it should be clear that $f_{\alpha^k}([x])=f_\alpha^k([x])=[x+k]$ and that
$$
\begin{align}
\alpha^k=e&\Leftrightarrow f_\alpha^k([x])=[x+k]=[x]\\
&\Leftrightarrow k\mbox{ is a multiple of }m
\end{align}
$$
where $e$ denotes the neutral permutation or empty cycle. In particular this leads to the conclusion that the order of $\alpha^s$ must be the minimal integer $t$ so that $k=s\cdot t$ is a multiple of $m$.
This shows that $\alpha^s$ has order $\mbox{lcm}(s,m)/s$ and since $\mbox{lcm}(s,m)=s\cdot m/\mbox{gcd}(s,m)$ this shows that
$$
\mbox{ord}(\alpha^s)=\frac{m}{\mbox{gcd}(s,m)}
$$
so indeed you will always have $\mbox{gcd}(s,m)$ disjoint cycles of length $m/\mbox{gcd}(s,m)$ each in the disjoint cycle decomposition of $\alpha^s$. So your intuition was correct after all!
