What powers of an $18$ cycle permutation are also $18$ cycles? Suppose we are given a permutation $A = (1,2,...,18)$ which is an $18$ cycle in $S_{18}$. We want to find $i \in \mathbb{Z}$ such that $A^i$ is also an $18$ cycle. Now I know how to do this by trial and error, but I am sure there must be an easier way to do this.
 A: $\,A \in S_{18}$ has length $18$ and hence has order $18$; so the group generated by $A$ is isomorphic to the cyclic group $\mathbb Z_{18}$, the group of integers under addition modulo $18$. 
You want to find another permutation $\,A^i \in S_{18}\,$ that is also of order $\,18,\;$ and hence, is also a generator of the cyclic group generated by $A$: You can find such an $\,A^i$ by finding any $i$ such that $\,i,$ is relatively prime to $18$, any $i$ such that $\,\gcd(i, 18) = 1, i\neq 1\,$ to obtain a permutation $A^i$ that is also an $18$-cycle.
The smallest $i$ for which $\gcd(i, 18) = 1, \; i\neq 1$ is $\,i = 5$
Compute $A^5$ and see what you get.
A: The cycle notation of a permutation allows one to really easy calculate the order of that permuation. A has order 18 since applying A 18 times fixes everything and so is the identity permutaion. Thus, the set $\{A^i\}$ is the group $Z/18 Z$. The $A^i$ that are 18 cycles must be elements of the group with order 18. Conversely, if an element has order 18, then it must be a generator of the group $Z/18Z$. Since we see that this group gives a transitive action on a set of 18 elements, we see that any such generator must be an 18-cycle. Thus, the 18-cycles are given precisely by the $i$ that are relatively prime to 18.
If any of this is unclear, feel free to ask. 
