Find all $Hom(\mathbb Z/m ,A) $, A- finite Abelian group My attempt: 
I want to prove that $Hom(\mathbb Z/m, A) \simeq A$
Let's build map $k$: $f \to f(1)$. It's a group homomorphism:
$fg \to g(f(1)) = g(a) = g(1+1+..+1) = g(1)a=ba=f(1)g(1) $
$k$ is injective and surjective hence bijective.  
 A: Your conjecture is clearly false, I'm afraid: for $m=2$ and $A=\mathbb{Z}/3\mathbb{Z}$, the only homomorphism is the zero one.
Consider instead $A(m)=\{x\in A:ma=0\}$
If you use multiplicative notation on $A$, $A(m)=\{x\in A:x^m=1\}$
A: A less confusing way to put the question is as follows:


*

*Question: Let $m \in \Bbb Z$ and  $C_m$ the cyclic group of order $m$, $A$ a finite abelian group, describe the group of homomorphisms $f : C_m \rightarrow A$ with group composition defined by $(f*g)(x) = f(x)+g(x)$.

*Answer: Let $a \in C_m$ be a generator of $C_m$ and $f$ any homomorphism, then $f$ is completely determined by the value $f(a)$. But since $a^m = e$ and $f(e)=0$
one must have that the (additive) order $f(a)$ is a divisor of $m$ so $\operatorname{Hom}(C_m,A) \cong G$, The subgroup of A consisting of elements with an order a divisor of $m$. 

*Examples: With $A = \Bbb Z_2 \times  \Bbb Z_2 \times \Bbb Z_6 \times \Bbb Z_{30}$ and $m = 6$ we have $G = \Bbb Z_2 \times  \Bbb Z_2 \times \Bbb Z_6 \times \Bbb Z_6$. On the other hand for $m = 5$ we have $G = \Bbb Z_5$.

*Exercise: Calculate $G$ in the cases $m = 2,3,4$.

