The proof I do not understand is the proof to statement (2). Below I describe exactly what my problem is.
He shows that $n$ divides $im$, or equivalently $\frac{im}{n} \in \mathbb{Z}$. This is fine, I understand this. The particular part I don't understand is the following: he claims that $\frac{n}{\textrm{hcf}(i,n)}$ divides $m$. It seems to me this doesn't follow, for the following reason:
$$\frac{m}{\left(\frac{n}{\textrm{hcf}(i,n)}\right)} = \frac{im}{\left(\frac{in}{\textrm{hcf}(i,n)}\right)} = \frac{im}{n} \frac{1}{\left( \frac{i}{\textrm{hcf}(i,n)} \right)}$$
The $im/n$ is an integer, $i/\textrm{hcf}(i,n)$ is an integer, but there is no reason to think that their ratio is an integer. So I don't understand why $\frac{n}{\textrm{hcf}(i,n)}$ divides $m$.
We know that $\frac{im}{n} \in \mathbb Z$. Suppose $im = nk$. Let $d = \gcd(n,i)$. We know that $d$ divides both $i$ and $n$, say $i=ad$, $n=bd$.
Furthermore, since $d$ is the greatest common divisor of $i$ and $n$, it follows that $\gcd(a,b) = 1$ otherwise if there was something common between them we could have multiplied that quantity with $d$ to get a larger divisor of both $i,n$.
Then, substituting into the earlier equation gives $adm = bdk \implies am = bk$. Now, $b$ divides $am$. But, $\gcd(a,b) = 1$, so it follows that $b$ must divide $m$, since it doesn't share any factors with $a$, so it can only share all its factors with $m$.
But, $b = \frac nd = \frac n{\gcd(n,i)}$, so clearly this is a divisor of $m$.