# Any power of a prime-length cycle is a cycle

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.

• consider the group generated by $\alpha$. it is cyclic of order $p$ and every non-identity element is a generator. – yoyo Dec 16 '11 at 20:58
• Here's one way of looking at it. Let's label the cycle as $(a_0, \ldots, a_{s - 1})$ instead. Then we can view the subscripts as being the distinct elements of the group $\mathbf Z/s\mathbf Z$. Then $\alpha^k$ sends $a_i$ to $a_{i + k}$. Note that every non-zero element of $\mathbf Z/s\mathbf Z$ generates that group. – Dylan Moreland Dec 16 '11 at 21:04
• @DylanMoreland: I think $\alpha^k$ sends $a_i$ to $a_{ik}$, not $a_{i+k}$. – Greg Martin Dec 16 '11 at 22:48
• @DylanMoreland: yeah, I think you're right ... sorry.... – Greg Martin Dec 18 '11 at 3:44
• For reference, this is exercise 8.B.8 in Pinter. – dharmatech Aug 3 '17 at 2:47

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.