About $\mathbb{Z}$-module homomorphisms

Question

Let $A$ be any $\mathbb{Z}$-module, and let $a\in A$. Is it true that every $\mathbb{Z}$-module homomorphism $\varphi :\mathbb{Z}/n\mathbb{Z}\to A$ is of the form $\varphi_{a}(\bar{k})=ka$? I'm trying to show this, but I'm unsure if my approach is correct or if I'm overthinking things.

I've shown already that IF we define the map $\varphi_{a} : \mathbb{Z}/n \mathbb{Z} \to A$ by $\varphi_{a}(\bar{k})=ka$, then it is a well-defined $\mathbb{Z}$-module homomorphism iff $na=0$. However, now I need to show that $Hom_{\mathbb{Z}}(\mathbb{Z}/n \mathbb{Z}, A) \cong B$ ($B$ is not important as I would like to show that part on my own), and to do that it seems that I need to show what I mentioned above.

I'm thinking that since for any such homomorphism $\varphi$, we must have $\varphi(\bar{1})=b$ for some $b\in A$, and thus $\varphi(\bar{2})=\varphi(\overline{1+1})=\varphi(\bar{1} + \bar{1})=\varphi(\bar{1})+\varphi(\bar{1})=b+b=2b$. So continuing in this manner, we have $\varphi(\bar{k})=kb$, and hence we can get a different homomorphism $\varphi_{a}$ (of this form) for each $a\in A$, and hence every such homomorphism is of this form. Is my reasoning correct?

Answer 1

The group ($\mathbb{Z}$-module) $\mathbb{Z}/n\mathbb{Z}$ is cyclic, so a homomorphism $\varphi\colon \mathbb{Z}/n\mathbb{Z}\to A$ is determined as soon as we know $a=\varphi(\bar{1})$.

Then, as $\bar{k}=k\bar{1}$, we we'll have $$ \varphi(\bar{k})=\varphi(k\bar{1})=k\varphi(\bar{1})=ka $$ The homomorphism is determined because two of them are the same as soon as they map $\bar{1}$ to the same element in $A$, in view of the argument above.

However, not every $a\in A$ is suitable for being chosen as the image of $\bar{1}$: it has to satisfy $na=0$, because $n\bar{1}=\bar{0}$. This condition is also sufficient, as you can prove.

Consequently, the group of homomorphisms $\mathbb{Z}/n\mathbb{Z}\to A$ is in bijection with the subgroup of elements of $A$ annihilated by $n$, via $\varphi\mapsto\varphi(\bar{1})$. Such a bijection is also a group isomorphism (easy verification)

Answer 2

This is correct.

A different way to phrase the argument is in terms of scalar multiplication rather than addition; $\mathbb{Z}/n\mathbb{Z}$ is a cyclic $\mathbb{Z}$-module, meaning every element is a $\mathbb{Z}$-multiple of a generator (in this case, $\bar{1}$).

So, the fact $\varphi$ is a $\mathbb{Z}$-module homomorphism implies

$$ \varphi(k \cdot \bar{1}) = k \cdot \varphi(\bar{1}) $$

This argument works in greater generality, since it applies to cyclic modules over any ring.

Combined with the addition-based argument, one can use a similar argument any time you have a set of generators for a module. e.g. every $\mathbb{R}$-module homomorphism whose domain is $\mathbb{R}^2$ is determined by its values on $(1, 0)$ and $(0, 1)$.