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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.

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    $\begingroup$ consider the group generated by $\alpha$. it is cyclic of order $p$ and every non-identity element is a generator. $\endgroup$ – yoyo Dec 16 '11 at 20:58
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    $\begingroup$ 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. $\endgroup$ – Dylan Moreland Dec 16 '11 at 21:04
  • $\begingroup$ @DylanMoreland: I think $\alpha^k$ sends $a_i$ to $a_{ik}$, not $a_{i+k}$. $\endgroup$ – Greg Martin Dec 16 '11 at 22:48
  • $\begingroup$ @DylanMoreland: yeah, I think you're right ... sorry.... $\endgroup$ – Greg Martin Dec 18 '11 at 3:44
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    $\begingroup$ For reference, this is exercise 8.B.8 in Pinter. $\endgroup$ – dharmatech Aug 3 '17 at 2:47
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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.

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