When is an integer power of a $12$-cycle a $12$-cycle?

Question

I'm given a $12$ cycle $\sigma = (1\dots 12)$, and want to find for which $i$ is $\sigma^i$ also a $12$-cycle.

We know that if $G$ is a group and $g \in G$, with $|g| = n$, then $|g^a| = \frac{n}{(n,a)}$, so in this case I want to say that $i$ must be coprime with $12$.

However, how do we eliminate the scenario where we have a permutation of order 12 which isn't a 12-cycle? That is, how do we deal with say $\sigma' = (123)(4567)$ which isn't a $12$-cycle but does possess order $12$?

Thanks in advance for the clarifications.

Answer 1

You have to show that every element of $\{1,\dots, 12\}$ is in the same cycle as $1$ in the permutation $\sigma^k$ when $k$ Is coprime to $12$. To do this you must show for each $a$ there is an $r$ so that $(\sigma^k)^r(1) = a$. This indeed happens because in general $\sigma^m(1) \equiv 1+m \bmod 12$. So you need $kr\equiv a-1 \bmod 12$ which is possible since $k$ is coprime to $12$.

Answer 2

This is obvious if you remember that if $\sigma=(a_1\,a_2\,\dots\,a_i\,\dots a_n)$ is an $n$-cycle, we have $$\sigma^k(a_i)=a_{(i+k)\bmod n},$$ and translations of the additive group $\mathbf Z/n\mathbf Z$ are bijections of the group.