$\sigma^p$ is the product of p cycles of length q

Question

Let $p, q$ be two integers and consider $n = p q$. Let $\sigma$ be a cycle of length $n$ of the symmetric group $S_n$.

Show that $\sigma^p$ is the product of $p$ cycles of length $q$.

I've tried it out with the cycle $\sigma = (1\ 2\ 3\ .... 10\ 11\ 12)$ in $S_{12}$:

$\sigma^2 = (1\ 3\ 5\ 7\ 9\ 11)(2\ 4\ 6\ 8\ 10\ 12)$

$\sigma^3=(1\ 4\ 7\ 10)(2\ 5\ 8\ 11)(12\ 3\ 6\ 9)$

$\sigma^4=(1\ 5\ 9)(2\ 6\ 10)(3\ 7\ 11)(4\ 8\ 12)$

Now I need to show it in the general case.

Answer 1

Hints: First prove the result for the $n$-cycle $\sigma = (1 2 \dots n)$. Notice that $\sigma^p$ sends 1 to $p + 1$, 2 to $p + 2$, and so on. For instance, in your example for $\sigma^3$ we have $1 \mapsto 4 \mapsto 7 \mapsto 10 \mapsto 1$, where the addition of $+ 3$ is being done mod 12. How does this description relate to its cycle decomposition?

Second, can you reduce to this case? Think about conjugation in $S_n$.