Suppose $A$ is a finite cyclic group of order $n$, and $a$ is an element in $A$ of order $d$. My question is:
Question 1:Can we find a generator $x$ of $A$ such that the element $a$ can be written in the form $x^{\frac{n}{d}}$?
My attempt to solve this question is as follows:
Suppose $x$ is a generator of $A$. Since A has a unique subgroup of order $d$, namely $<x^{\frac{n}{d}}>$. We have $<a>=<x^{\frac{n}{d}}>$. Hence we can write $a$ as: \begin{equation} x^{\frac{mn}{d}}, \end{equation} where $m\in \mathbb{Z}, (m,d)=1$.
Since the generators of $A$ takes the form: $x^l$, where $(l,n)=1$. Our question can be restated as :
Question 1': Suppose $m\in \mathbb{Z}, (m,d)=1$. Can we find an integer $l$, such that $(l,n)=1$ and $x^{\frac{mn}{d}}=x^{\frac{ln}{d}}$?
And this is obviously equivalent to:
Question 1'': Suppose $m\in \mathbb{Z}, (m,d)=1$. Can we find $k\in \mathbb{Z}$, such that $m+kd$ and $n$ are relatively prime?
I fail to prove it. Also I cannot find a counterexample.
Proof of Question 1''. Suppose $n=dn'$. Decompose $n'=d'n''$ such that the prime factors of $d'$ divide $d$ and there is no prime factors of $n'$ that divide $d$. Since $(dd',n')=1$ and $(m,dd')=1$, we only need to prove that we can find $k\in \mathbb{Z}$, such that $(m+kdd',n')=1$. This becomes trivial because the congruence classes represented by: \begin{equation} m,m+dd',m+2dd',\cdots,m+(n'-1)dd' \end{equation} exhaust $\mathbb{Z}/n'\mathbb{Z}$. (Again use the fact $(dd',n')=1$)