If $m \mid n$, show that there is a one-to-one homomorphism $\mathbb{Z}_m \to \mathbb{Z}_n$.

Question

If $m \mid n$, show that there is a one-to-one homomorphism $\mathbb{Z}_m \to \mathbb{Z}_n$. Give an explicit homomorphism $\varphi : \mathbb{Z}_6 \to \mathbb{Z}_{12}$ that is injective, i.e., one-to-one.

Can the map $a+ \mathbb{Z}_m \to a+\mathbb{Z}_n$, where $a$ belong to $\mathbb{Z}$, be an example?

Answer 1

No. What is $a+\mathbb Z_m$ ? Anyway, if $m\mid n$, then $$\varphi (k)=k\frac{n}{m},\quad k\in \mathbb Z_m$$ is the only injective homomorphism.

Answer 2

What does not work

Why doesn't the map $a+m\Bbb Z\mapsto a+n\Bbb Z$ work?

A homomorphism needs to map $0+m\Bbb Z\mapsto 0+n\Bbb Z$. In $\Bbb Z_m$ we have $0\equiv_m m$, but $m+m\Bbb Z\mapsto m+n\Bbb Z$ is not valid, because $m\not\equiv_n 0$. In short, $0$ does not get mapped to $0$, therefore the proposed map is not a homomorphism.

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What does work

Suppose $\phi:\Bbb Z_m\rightarrow\Bbb Z_n$ is a homomorphism, then $\phi(m)=0\iff n\mid m\phi(1)$, therefore $$m\phi(1)=kn\iff \phi(1)=k\left(\frac{n}{m}\right)$$ for some $k\in\Bbb Z$. If $\phi$ is going to be injective we need $k=1$, otherwise we would have $\phi(\frac{m}{k})\equiv_n 0$ and $\frac{m}{k}<m$, which would give $\ker(\phi)\neq (0)$.

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Example

Let $\phi(1)=\frac{12}{6}=2$, such that $\phi(k)=2k$, then $\phi:\Bbb Z_6\rightarrow\Bbb Z_{12}$ is injective.

Answer 3

If $m|n$, then $\Bbb Z_m\le\Bbb Z_n$, because $\Bbb Z_n$ is cyclic. Thus we can consider the inclusion, $i:\Bbb Z_m\hookrightarrow\Bbb Z_n$, an injective homomorphism.