Show that the map $\mathbb{Z}_n^*\to \mathbb{Z}_m^*$ is surjective

Question

Let $m,n \in \mathbb{Z}$ such that $m|n$. Show that the map $$f: \mathbb{Z}_n^*\to \mathbb{Z}_m^*$$ $$f({a \pmod n}) = (a \pmod m)$$ is surjective. I am not able to figure out any simple way to tackle this... Any hints?

Answer 1

Got it. Will prove by induction. We just have to show that the result holds when $n=mp$. Let $b \in \mathbb{Z}_m^*$.

If $b \not\equiv 0 \pmod p$ then $(b,m)=1 \land (b,p)=1 \implies (b,n)=1 \implies b \in \mathbb{Z}_n^*$. Now we have $f(b)=b$.

If $b \equiv 0 \pmod p$ then $m \not\equiv 0 \pmod p \implies b+m \not\equiv 0 \pmod p$. Therefore $(b+m,p)=1 \land (b+m,m)=1 \implies (b+m,n)=1 \implies b+m \in \mathbb{Z}_n^*$. Now we have $f(b+m)=b$.

The inductive step is straightforward.

Answer 2

This should be a comment, but I had trouble with formatting in the comments.

Evidently, the same proof will work for $R/I \to R/J$ where $R$ is a PID. The more general problem was considered here and here.

A nice generalization is that a surjection $f: R \to S$, where the rings are commutative Artinian, maps units onto units.