What is $\text{Hom}(\mathbb{Z},A)$ for an abelian group $A$? I know that $\text{Hom}(A,B)$ is a  group for abelian groups $A$ and $B$. 

Is there some general statement about $\text{Hom}(\mathbb{Z},A)$ and can $\text{Hom}(\mathbb{Z}/n\mathbb{Z},A)$ be seen as a subgroup of $A$?

 A: $\text{Hom}(\mathbb Z,A) \cong A$ canonically. Moreover, $\text{Hom}(\mathbb{Z}/n\mathbb{Z},A)$ is also canonically isomorphic to element of order $n$ in $A$ : more precisely, the set $T_n(A) := \{x \in A : nx = 0\}$ is a subgroup of $A$, and again we have an isomorphism $\text{Hom}(\mathbb{Z}/n\mathbb{Z},A) \cong T_n(A)$. So the answer to your question is "yes".
Let's see why $1)$ is true  : we have a injection $A \to \text{Hom}(\mathbb Z,A)$, $a \to \phi_a : \phi_a(1) = a$ (recall that a morphism $\phi : \mathbb Z \to A$ is uniquely determined by the image of $1$). By any morphism $\psi$ is on the form $\psi = \phi_{\psi(1)}$ by definition, i.e the map is an isomorphism.
A: *

*$$\text{Hom}(\mathbb{Z},A)$$
This is the group of homomorphisms $\phi:\mathbb{Z}\rightarrow A$. Since $\phi$ is a homomorphism, it is completely determined by the value $\phi(1)$. We can let $\phi(1)=a$, for $a\in A$, and every such element yields a homomorphism. Thus $\text{Hom}(\mathbb{Z},A)\cong A$

*$$\text{Hom}(\mathbb{Z}/n\mathbb{Z},A)$$
Again $\phi:\mathbb{Z}/n\mathbb{Z}\rightarrow A$ is completely determined by $\phi(1)$. So let $\phi(1)=a$. Notice that $0=\phi(n)=n\phi(1)=na$. We can therefore choose any $a\in A$ such that the order of $a$ divides $n$. Letting $N=\{a\in A\mid na=0\}$, we have that $\text{Hom}(\mathbb{Z}/n\mathbb{Z},A)\cong N$


It is true that $N\subseteq A$ is a subgroup of $A$. 


*

*If $a,b\in N$ then $na+nb=n(a+b)=0$ so $a+b\in N$.

*Associativity holds (since $A$ is a group)

*The identity element $0\in N$, since $n0=0$

*Every element has an additive inverse, since if $a\in N\iff an=0\iff (-a)n=0$

