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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?

Davide Gallo
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  • Related/duplicate: https://math.stackexchange.com/questions/487011/showing-that-a-homomorphism-between-groups-of-units-is-surjective – egreg Feb 10 '18 at 21:36

2 Answers2

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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.

Davide Gallo
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  • This is completely right in its substance. There's a slight imprecision in your notation: what are $b,m$? Integers, elements of $\mathbb{Z}_n^\star$, or elements of $\mathbb{Z}_m^\star$? You're using them like they're integers, but you wrote $b\in \mathbb{Z}_m^\star$, and at another point $b\in\mathbb{Z}_n^\star$. – Ben Blum-Smith Feb 10 '18 at 20:04
  • And where is the induction? – awllower Feb 10 '18 at 20:06
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    In the fact that he supposed that the quotient n/m is a prime. Then for an arbitrary n/m, you can compose the various surjective maps of jumping-one-prime-per-time – Andrea Marino Feb 10 '18 at 20:14
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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.

fred goodman
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