let $\omega=(1 \ 2 \ 3\ 4\ 5\ 6\ 7\ 8\ 9\ 10\ 11\ 12\ 13\ 14)$ be $14$ cycle. For which positive integers $i$ is $\omega ^i$ also a $14$ cycle. let $\omega=(1  \ 2 \ 3\ 4\ 5\ 6\ 7\ 8\ 9\ 10\ 11\ 12\ 13\ 14)$ be $14$ cycle.
For which positive integers $i$ is  $\omega ^i$ also a $14$ cycle.
now if $\omega^i$ is $14$ cycle . then $o(\omega^i)=14$.
Hence only possible choice for $i=\{1,3,5,9,11,13\}$
But is this true that for each of these  choices of $i$ $\omega^i$ is $14$ cycle. if yes how to prove that.
Please provide some hint.
 A: The set of indices you have found is exactly the list of integers less than $14$ that are prime with $14$, whose number is $\varphi(14)=\varphi(7) \times \varphi(2)=7-1=6$ ($\varphi$ is Euler's totient function).
In order to prove that, I refer you to the very interesting proofs and discussions in (Intuition and Tricks - Crafty Short Proof - Generators, Order of a Cyclic Group - Fraleigh p. 64 Theorem 6.14). See as well (Intuition and Tricks - Hard Overcomplex Proof - Order of Subgroup of Cyclic Subgroup - Fraleigh p. 64 Theorem 6.14) where the final argument rests on Bezout formula : $i$ is prime with $14$ if and only if there exists $u,v \in \mathbb{Z}$ such that $u.i+v.14=1.$
A: Hint: Consider the cyclic group $\langle \omega\rangle$ in the symmetric group on 14 letters. The subgroup has order 14. Now consider the possible subgroups of a cyclic group of order 14 and you have it.
A: The possible cycle structures for a permutation of order $14$ are $14^1$ (a $14$-cycle) or $2^m 7^n$ (the product of disjoin transpositions and $7$-cycles).
Now, $2m+7n=14$ implies $n\le 2$.
If $n=0$, then the permutation has order $2$, not $14$.
If $n=1$, then $2m=7$, impossible. So, $n=2$ and $m=0$, but then the permutation has order $7$, not $14$.
Thus, $2^m 7^n$ is impossible and only $14^1$ is left.
