TITLE: Any power of a prime-length cycle is a cycle
QUESTION [12 upvotes]: Having some doubts proving exercise statement from Pinter's book. Here's quote:

Let $\alpha $ be a cycle of length $s$, say $\alpha = ( a_1, a_2 ... a_s )$. 
  Prove that, if $s$ is a prime number, every power of $\alpha$ is a cycle.

I know well that power $k$ of cycle of length $k \cdot t$ is a product of $k$ disjoint cycles of length $t$. Prime number length will not let this happen with any power. But I'm not sure whether powering the cycle to a divisor of its length is the only way to break cycle apart.
Thank you for your attention.

REPLY [15 votes]: The claim is slightly incorrect. The correct claim is:

The permutation $\beta = \alpha^k$ is a cycle if and only if $s$ does not divide $k$. 

Clearly, if $s$ divides $k$, then $\alpha^k$ is just the identity permutation, and hence not a cycle.  Now for the other direction, assume $s$ does not divide $k$. It suffices to show that for each $t \in \{ 1, 2, \ldots, s \}$, there exists an $i$ such that $\beta^{i}(a_1) = a_t$; i.e., $\alpha^{ki}(a_1) = a_t$. But this happens if and only if 
$$
ki \equiv t-1 \pmod s.
$$
Assuming $s$ does not divide $k$, since $s$ is a prime, $k$ has a multiplicative inverse $k^{-1}$ modulo $s$. Now picking $i = k^{-1} \cdot (t-1)$ gives what we want.