prove that $\operatorname{Hom}_\Bbb{Z}(\Bbb{Z}/n\Bbb{Z},A)\cong A_n.$

Question

Let $A$ be any $\Bbb{Z}$-module, let $a$ be any element of $A$ and let $n\in \Bbb{Z}^+$. Prove that the map $\psi_a\colon\Bbb{Z/nZ}\to A$ given by $\psi_a(\overline{k})=ka$ is a well-defined $\Bbb{Z}$-module homomorphism if and only if $na=0$.

Then, prove that $\text{Hom}_\Bbb{Z}(\Bbb{Z}/n\Bbb{Z},A)\cong A_n$, where $A_n=\{a\in A: na=0\}$.

I am struggling to find a map so that i can prove the isomorphism between $\text{Hom}_\Bbb{Z}(\Bbb{Z}/n\Bbb{Z},A)$ and $A_n$.

Answer 1

Define a map $f:\text{Hom}_\Bbb{Z}(\Bbb{Z/nZ},A)\to A_n$ by $(\psi_a(\overline{k})\mapsto ka)\mapsto a$.

$f$ is a $\Bbb{Z}$-module homomorphism:

• $\psi_a(\overline{x})+\psi_b(\overline{x})=xa+xb=x(a+b)=\psi_{a+b}(\overline{x})$,
• $\psi_a(r\overline{x})=r(xa)=r\psi_a(\overline{x})$ Therefore, $f(\psi_a+\psi_b)=f(a+b)=a+b=f(\psi_a)+f(\psi_b).$

$f(r\psi_a)=ra=rf(\psi_a).$

$f$ is injective: \begin{align}f(\psi_a)=f(\psi_b)\Rightarrow a=b, ka=kb\Rightarrow \psi_a=\psi_b.\end{align}

$f$ is surjective: For any $a\in A_n$, $\psi_a$ is a homomorphism in $\text{Hom}_\Bbb{Z}(\Bbb{Z/nZ},A)$. Therefore $f(\psi_a)=a$.

Answer 2

Another way: $\newcommand{\Z}{\mathbb Z} \newcommand{\Hom}{\operatorname{Hom}}$

In the previous exercise you show that every $a \in A$ such that $na=0$ (that is, $a \in A_n$) determines a $\Z$-module homomorphism $\psi_a : \Z/n\Z \to A$ (that is, $\psi_a \in \Hom_\Z(\Z/n\Z,A)$) given by $\psi_a(m+n\Z) = ma$. Thus, the map $\psi : A_n \to \Hom_\Z(\Z/n\Z,A)$ given by $\psi(a) = \psi_a$ is well-defined and in fact is a $\Z$-module homomorphism (note that $A_n$ and $\Hom_\Z(\Z/n\Z,A)$ are both $\Z$-modules) because if $a,b \in A_n$, then $$(\forall k \in \Z) (\forall m \in \Z) \ m(a+kb) = ma+kmb$$ tells us that $$(\forall k \in \Z) (\forall m \in \Z) \ \psi_{a+kb}(m+n\Z) = (\psi_a+k\psi_b)(m+n\Z)$$ and therefore $(\forall k \in \Z) \ \psi_{a+kb} = \psi_a+k\psi_b$, that is $$(\forall k \in \Z) \ \psi(a+kb) = \psi(a)+k\psi(b).$$ Also, note that $\psi$ is an isomorphism since $\phi : \Hom_\Z(\Z/n\Z,A) \to A_n$ given by $\phi(f) = f(1+n\Z)$ is a well-defined set-map (just check that $f(1+n\Z) \in A_n$ for every $f \in \Hom_\Z(\Z/n\Z,A)$) such that $$\psi \phi = \operatorname{id}_{\Hom_\Z(\Z/n\Z,A)} \quad \textrm{and} \quad \phi \psi = \operatorname{id}_{A_n}$$ (how is that $\phi$ is a $\Z$-module homomorphism follows from the fact that $\psi$ is a $\Z$-module homomorphism?).