(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 $N$ ?