Order of elements in cyclic groups... Let $G$ be a cyclic group of order $n$. Suppose $x$, $y$ are two elements of order $d$, where $d$ divides $n$. Show that $y = x^m$, where $m$ is an integer coprime to $n$. 
I know $y=x^m$ since the subgroups generated by $x$ and $y$ must be identical. I do not know how to show the coprimeness, however.
 A: This seems surprisingly tricky for such an elementary problem.  For a positive integer $m$, let $U(m) = (\mathbb{Z}/m\mathbb{Z})^{\times}$ be the group of units modulo $m$.  Then for any positive integers $d \mid n$, the quotient map 
$\mathbb{Z}/n\mathbb{Z} \rightarrow \mathbb{Z}/d\mathbb{Z}$ induces a map on unit groups
$U(n) \rightarrow U(d)$, which I claim is always surjective.  Thus, if you start with something which is a unit modulo $d$, then you can always correct by a multiple of $d$ to get something which is a unit modulo $n$: that's what you're trying to prove.  
How do you prove this fact?  Carefully!  It is enough to go from any $d$ to $dp$ for a prime $p$, and you want to treat the cases $p \mid d$ and $\operatorname{gcd}(p,d) = 1$ separately.  I'll bet this question has been asked and answered on this site before, but in case not and you need more help, please ask.
Added: Indeed the surjectivity question has been asked here before: see this.  The answer still leaves something to the reader, so still please feel free to ask for more help...
A: Edit: This is nonsense.
Well, the coprime part is false.  Here's a counter-example:
Take $\mathbb{Z}/n\mathbb{Z}$, where $n=12$, $x=3$, and $y=9$.  Then both $x$ and $y$ are of order $4$, but $y=3x$.  So we've found an $m$ that's not coprime to $n$.
On the other hand, if you change the prompt to "$m$ is an integer coprime to $d$," the statement becomes true (and therefore much easier to prove).  :)
