The following is taken from Modules an approach to linear algebra by Blyth
$\color{Green}{Background:}$
$\textbf{Exercise:}$ Let $R$ be a commutative unitary ring and let $M$ be an $R$-module. For every $r\in R$ let $rM=\{rx\mid x\in M\}$, and $M_r=\{x\in M\mid rx=0\}$. Show that $rM$ and $M_r$ are submodules of $M$. In the case where $R=\mathbb{Z}$ and $M=\mathbb{Z}/n\mathbb{Z}$, suppose that $n=rs$ where $r$ and $s$ are mutually prime. Show that $rM=M_s$.
[Hint: use the fact that there exist $a,b\in \mathbb{Z}$ such that $ra+sb=1$.]
[$\textbf{Solution:}\text{ Last part of the question.}$]
Note that $\mathbb{Z}/n\mathbb{Z}_s\subset r\mathbb{Z}/n\mathbb{Z}$, since the set $\{rx\mid x\in \mathbb{Z}/n\mathbb{Z}\}$, for all $x\in \mathbb{Z}/n\mathbb{Z}$, contains $\{x\in \mathbb{Z}/n\mathbb{Z}\mid sx=0\}$. We only need to show the reverse inclusion $r\mathbb{Z}/n\mathbb{Z}\subset \mathbb{Z}/n\mathbb{Z}_s$. Since $\mathrm{gcd}(r,s)=1$, we can find integers $a,b\in \mathbb{Z}$ so that $$ ra+sb=1_Z. \tag{*} $$
Right multiplying $(*)$ on both sides by an element $w\in \mathbb{Z}/n\mathbb{Z}$ gives $(ra)w+(sb)w=1_Z w=w$. $(sb)w=s(bw)\in \mathbb{Z}/n\mathbb{Z}_s$. Also, $(ra)w\in r\mathbb{Z}/n\mathbb{Z},\, (ra)w=1_{\mathbb{Z}} w-(sb)w,\, 1_{\mathbb{Z}} w-(sb)w\in \mathbb{Z}/n\mathbb{Z}_s$. We also have $(sb)w\in \mathbb{Z}/n\mathbb{Z}_s$ implies that $(sb)w=0$. Hence $(ra)w=1_\mathbb{Z} w-(sb)w=1_\mathbb{Z} w\in \mathbb{Z}/n\mathbb{Z}_s$. Along with $\mathbb{Z}/n\mathbb{Z}_s\subset r\mathbb{Z}/n\mathbb{Z}$, we have $\mathbb{Z}/n\mathbb{Z}_s\subset r\mathbb{Z}/n\mathbb{Z}$.
$\color{Red}{Questions:}$
Can someone checked if my solution to the above exercise is correct. I thought I have to judiciously pick some integers $a,b \in \mathbb{Z}$ given $\mathrm{gcd}(r,s)=1$.
Thank you in advance.