In a finite cyclic group of order n, number of solutions to $x^m = e$ - Fraleigh p. 68 6.53,54 (53.) Show that in a finite cyclic group G of order n, written multiplicatively, the equation $x^m = e$ has exactly m solutions $x$ in G for each $m \in \mathbb{N}$ that divides n. 
(54.) With reference to Exercise 53, what is the situation if $1 < m < n$ and $m \not| n$? Tried this.
Question 5 on http://www.math.drexel.edu/~rboyer/courses/math533_03/hw2_soln.pdf 
(53.) By  means of Fraleigh p. 63 Theorem 6.10, group of order n $\cong <\mathbb{Z_n},+_n >, $ hence can work in the group $\mathbb{Z_n}$. 
We can write $x^m = e$ as $mx \equiv 0 (\mod n).$
This has at least m solutions: $0, n/m, 2n/m,...,(m - 1)n/m$.   

(1.) Where do these $m$ solutions loom from?  

If x is any solution of
$mx \equiv 0 (\mod n).$, then $n|mx \iff \exists \; q \in \mathbb{N} \; : \;nq = mx$.
 Hence $\frac{nq}{m} = x$ are the solutions. We next find $x = \dfrac{qn}{m} < n  \qquad (♥)$   

(2.) How's this inequality true?
   (3.) How does this induce $q = 0, 1, ..., m - 1$?     } }$  

In other words, $x$ must have the form of the solution already given.  

(4.) Did we use the presupposition $m|n$ anywhere? If not, what's it for?    

Note: Typically, the class tried to show this problem by using the existence of subgroups of order m in a
cyclic group of order n. But no one cited the key fact that such subgroups are unique so the equation has
exactly m solutions.
(54.) Let $d = gcd(m,n), \color{blue}{M = m/d}, \color{magenta}{N = n/d}$ such that $gcd(M,N) = 1$.
Working in $\mathbb{Z_n}$ again, we see that $0, n/d, 2n/d, · · · , (d − 1)n/d$ are solutions of $mx \equiv 0 (\mod n).$ 

(5.) I don't see this. Can someone please flesh out how?   

By means of (♥), $x = q\dfrac{\color{magenta}n}{\color{blue}m} = q\dfrac{\color{magenta}{dN}}{\color{blue}{dM}} = q\dfrac{N}{M} \qquad (☼).$
Since $gcd(M,N) = 1$, hence by means of Euclid's Lemma, $M|q \iff$ $Ms = q$ for some integer s. 
Then by dint of (☼), $x = q\dfrac{N}{M} =(Ms)\dfrac{N}{M} = \color{magenta}Ns = \color{magenta}{\dfrac{n}{d}}s$.
By means of (♥) and the overhead line, $x = \color{magenta}{\dfrac{n}{d}}s \quad < n \iff s < d.$
Consequently, $s$ must be one of the numbers $0, 1, 2, · · · , d − 1$
and we see that the solutions exhibited above are indeed all the solutions.

(6.) What's the proof blueprint for (54.)? All the algebra perplexes me. 
  I understand we define $M, N$ to induce $gcd(M,N) = 1$, and this gcd is to induce Euclid's Lemma. 
  But how do you preordain all this? And in the end, we actually sub back $n/d$ after defining $N$ ? 

 A: The formatting of your question makes it very difficult to read. Your choice of words is often odd and sometimes plain wrong. Please take to heart the suggestions in the comments to your question.


*

*If $m$ divides $n$ then $x=\tfrac{n}{m}$ is an integer that is a solution to the congruence relation
$$mx\equiv0\pmod n.$$
Moreover any multiple of $\tfrac{n}{m}$ is also
a solution to the congruence, yielding the solutions
$$0,\ \frac{n}{m},\ 2\cdot\frac{n}{m}\,\ldots,\ (m-2)\cdot\frac{n}{m},\ (m-1)\cdot\frac{n}{m}.$$
Every element of $\Bbb{Z}_n$ has unique representative $x\in\Bbb{Z}$ satisfying $0\leq x<n$, so for solutions in $\Bbb{Z}_n$ we find precisely the representatives above. After all, the next integer multiple of $\tfrac{n}{m}$ is $m\cdot\tfrac{n}{m}=n$, which is too big.
To see that we have found all solutions, let $x$ be a solution to $mx\equiv0\pmod n$. Then there exists $q\in\Bbb{N}$ such that $mx=qn$. Because $m$ divides $n$ it follows that $x=q\cdot\tfrac{n}{m}$, so $x$ is an integer multiple of $\tfrac{n}{m}$. This shows that we have listed all solutions to the congruence relation, and hence there are precisely $m$ solutions.
Because any cyclic group $G$ of order $n$ is isomorphic to $\Bbb{Z}_n$, it follows that the equation $x^m=e$ has precisely $m$ solutions if $m$ divides $n$.

*We are looking for solutions in $\Bbb{Z}_n$. Every element of $\Bbb{Z}_n$ has a unique representative $x\in\Bbb{Z}$ satisfying $0\leq x<n$.

*If $q$ is an integer such that $0\leq q\cdot\tfrac{n}{m}<n$, where $m$ is a divisor of $n$, then $0\leq q<m$. So
$$q=0,\ 1,\ldots,\ m-2,\ m-1.$$

*In 1 we have used the assumption that $m$ divides $n$ first by stating that $\tfrac{n}{m}$ is an integer, and later by stating that $x=q\cdot\tfrac{n}{m}$ is an integer.

*It should be clear, also by the arguments above, that 
$$0,\ \frac{n}{d},\ 2\cdot\frac{n}{d},\ \ldots,\ (d-2)\cdot\frac{n}{d},\ (d-1)\cdot\frac{n}{d},$$
are solutions to the congruence relation $mx\equiv0\pmod n$. Again any larger multiple of $\tfrac{n}{d}$ would be too big.
To see that these are all solutions, suppose $x$ is a solution. Because $mx\equiv0\pmod n$ there exists $q\in\Bbb{N}$ such that $mx=qn$. In particular $m$ divides $qn$. Because $\gcd(m,n)=d$ it follows that $\tfrac{m}{d}$ divides $q$. Hence there exists $k\in\Bbb{N}$ such that $q=k\cdot\tfrac{m}{d}$, or equivalently $dq=km$. It follows that
$$mx=qn=q\cdot\left(d\cdot\frac{n}{d}\right)=dq\cdot\frac{n}{d}=km\cdot\frac{n}{d}=m\cdot\left(k\cdot\frac{n}{d}\right),$$
which shows that $x=k\cdot\tfrac{n}{d}$, so $x$ is an integer multiple of $\tfrac{n}{d}$. This shows that we have listed all solutions to the congruence relation above.

*I am not sure that I understand your question, but I think I might have answered it in 5.
