Let $a$ and $m$ be integers with $0 < a < m$. Let $f:\mathbb{Z}/m\mathbb{Z} \rightarrow \mathbb{Z}/m\mathbb{Z}$ be the function defined by $f(\overline{x}) = \overline{a}\cdot \overline{x}$. Prove that $f$ is injective if and only if $\gcd(a,m) = 1$.
So far, I have two different approaches to this problem, but they don’t get me anywhere.The first way is to assume that f is invective, which means that $f(\overline{x}) =f(\overline{y})$, and thus we have that $\overline{x}=\overline{y}$. Therefore, from this we can say that $x\sim y$, which means that $ x-y \in m\mathbb{Z}$.Hence, m divides $x-y$. Thus, $x-y=km$ for some integer k. After that, I separated this into two possible cases: One in which m is composite, and one in which it is prime. After that, I got stuck. The other approach I had in mind was a proof by contradiction, where I assume that $gcd(a, m)>1$. But I didn't know where to go after that. Any help would be appreciated.