# Homomorphisms $\mathbb{Z}/m\mathbb{Z} \to \mathbb{Z}/n\mathbb{Z}$

please provide an explicit description of $Hom_{\mathbb{Z}}(\mathbb{Z}/m\mathbb{Z} ,\mathbb{Z}/n\mathbb{Z})$ and also $\mathbb{Z}/m\mathbb{Z} \otimes_{\mathbb{Z}} \mathbb{Z}/n\mathbb{Z}$.

Thank you

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What's the context? What have you tried? – Julian Kuelshammer Oct 30 '12 at 12:02
I solved for $m,n$ coprime. Applying Chinese remainder we get a sort of decomposition. Am I on the right track? – user17090 Oct 30 '12 at 12:05
Yes, yes. What have you got exactly? (You could include it in the text.) For non coprime $n,m$ you will have to consider the $\gcd$ and $\lcm$ of them. – Berci Oct 30 '12 at 12:13
Try specific examples such as ${\mathbb Z}/3$ and ${\mathbb Z}/12$. Write down everything explicitly. – Scott Carter Oct 30 '12 at 12:26
Any element $\psi\in Hom_{\mathbb{Z}}(\mathbb{Z}/m\mathbb{Z} ,\mathbb{Z}/n\mathbb{Z})$ is uniquely determined by where it sends $1$, and is well defined only if $m\cdot\psi(1) = 0$. – Arthur Oct 30 '12 at 13:07

Here is a start: You have the exact sequence

$$0 \rightarrow \Bbb{Z} \stackrel{f}{\longrightarrow} \Bbb{Z} \longrightarrow \Bbb{Z}/m \longrightarrow 0.$$

where $f$ is multiplication by $m$. Now apply $\textrm{Hom}(-,\Bbb{Z}/n)$ to get that

$$0 \rightarrow \textrm{Hom}(\Bbb{Z}/m,\Bbb{Z}/n) \rightarrow \textrm{Hom}(\Bbb{Z},\Bbb{Z}/n) \stackrel{f_\ast}{\rightarrow} \textrm{Hom}(\Bbb{Z},\Bbb{Z}/n)$$

is exact. This tells you that the number of homomorphisms from $\Bbb{Z}/m\rightarrow \Bbb{Z}/n$ is at most $n$ because the middle term is the cyclic group of order $n$. Now the map $f_\ast$ is defined by $f_\ast(\phi) = \phi \circ f$. What are those $\phi$ such that $f_\ast(\phi) = 0$?

As for the second problem, can you use the chinese remainder theorem to give you a map $\phi : \Bbb{Z}/m \times \Bbb{Z}/n \longrightarrow \Bbb{Z}/(\textrm{gcd}(m,n))$? If you can then you have a unique map out of the tensor product which you can show is an isomorphism.

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Thank you for the nice answer. – user17090 Oct 31 '12 at 10:45