Verifying injection of an isomorphism

Question

Suppose $n$ and $r$ are coprime and define a map $\phi: \mathbb{Z}_n \rightarrow \mathbb{Z}_n$ which sends $x \mapsto rx$. Show $\phi$ is an isomorphism.

I verified ontoness and the operation preserving property of $\phi$. I'm getting stuck with 1-1 though. My computation:

$$ \phi(a) = \phi(b) \text{ mod }n \\ ra = rb\text{ mod }n \\ r(a-b) = 0 \text{ mod }n \\ \text{Note: }\gcd(r,n)=1 \text{ so } r \neq 0\text{. Thus} \\ (a-b) = 0 \text{ mod }n \\ a=b \text{ mod }n $$

Apparently this reasoning is not correct.

1. What went wrong?
2. How do I argue correctly?

Answer 1

You have first to check that the map is well-defined, that is, if $x\equiv y\pmod{n}$, then $rx\equiv ry\pmod{n}$. This is essentially obvious, but should be pointed out.

The homomorphism property is obvious as well, so let's do injectivity: if $\phi(x)=0$, then $rx\equiv0\pmod{n}$. As $r$ and $n$ are coprime, there are $a$ and $b$ so that $ar+bn=1$, so from $rx=kn$, we also have $$ arx=akn $$ and so $$ x=1x=(ar+bn)x=arx+bnx=akn+bnx=(ak+bx)n $$ Hence $x\equiv0\pmod{n}$. You don't need to prove surjectivity.

Note that the simple fact that $r\not\equiv0\pmod n$ doesn't allow you to conclude $a-b=0$ from $r(a-b)=0$. For instance, $3\not\equiv0\pmod6$, $3(4-2)\equiv0\pmod{6}$, but $4-2\not\equiv0\pmod{6}$.

The key fact is that $r$ and $n$ being coprime implies there is an inverse of $r$ modulo $n$.

Answer 2

Since $r^{-1}$ exists and $ra\equiv rb\mod n$ is a congruence, this mean that $r^{-1}ra\equiv r^{-1}rb\mod n$, then $a\equiv b\mod n$.