Consider $\text{Hom}(\Bbb Z/n\Bbb Z,\Bbb Q/\Bbb Z)$, and $\text{Hom}(\Bbb Q/\Bbb Z,\Bbb Z/n\Bbb Z)$. $\text{Hom}(A,B)$ is the set of all homomorphisms from $A$ to $B$, where $A,B$ are groups. Find $|\text{Hom}(\Bbb Z/n\Bbb Z,\Bbb Q/\Bbb Z)|$, and $|\text{Hom}(\Bbb Q/\Bbb Z,\Bbb Z/n\Bbb Z)|$.
So, here are my thoughts and questions. First, $\text{Hom}(\Bbb Z/n\Bbb Z,\Bbb Q/\Bbb Z)$. Consider some $f\in \text{Hom}(\Bbb Z/n\Bbb Z,\Bbb Q/\Bbb Z)$. $f$ is completely determined by $f(1) = a$, since $f$ is a homomorphism. What do elements in $\Bbb Q/\Bbb Z$ look like? I am not too comfortable with quotient groups yet. They'd be of the form $q + \mathbb Z$ where $q\in \Bbb Q$, right? Assuming this holds, I feel that $|\text{Hom}(\Bbb Z/n\Bbb Z,\Bbb Q/\Bbb Z)| = \infty$, because there are infinitely many choices of $q\in \mathbb Q$ for $q + \mathbb Z$. I know that if $q_1-q_2\in \mathbb Z$, then $q_1 + \mathbb Z = q_2 + \Bbb Z$ in $\Bbb Q/\Bbb Z$, but even then, the choices are infinite indeed - the representatives are rational numbers in $[0,1)$.
About $|\text{Hom}(\Bbb Q/\Bbb Z,\Bbb Z/n\Bbb Z)|$, consider $f\in \text{Hom}(\Bbb Q/\Bbb Z,\Bbb Z/n\Bbb Z)$. This seems tougher! For $q\in [0,1)$, I must assign values to $f(q + \mathbb Z)$ from $\{0,1,...,n-1\} \in \Bbb Z/n\Bbb Z$. I'm unable to figure out the number of ways of doing this.
Could you please drop any hints to point me in the right direction? Solutions are good too. Thank you!
Update:
Could it be the case that $\text{Hom}(\Bbb Z/n\Bbb Z,\Bbb Q/\Bbb Z) \cong \text{Hom}(\Bbb Z/n\Bbb Z,\Bbb Z/n\Bbb Z)$ in general, or at least for some $n$?
Also, on second thoughts, feels like $\text{Hom}(\Bbb Q/\Bbb Z,\Bbb Z/n\Bbb Z)$ is a singleton, i.e. $|\text{Hom}(\Bbb Q/\Bbb Z,\Bbb Z/n\Bbb Z)| = 1$ for all $n$ (consisting of only the homomorphism which maps all elements of $\Bbb Q/\Bbb Z$ to $0\in \Bbb Z/n\Bbb Z$. Is that right, and how do I show it?
It is much simpler than that. We have $|Hom(\mathbb{Z}/n\mathbb{Z},\mathbb{Q}/\mathbb{Z})|=n$ and $|Hom(\mathbb{Q}/\mathbb{Z},\mathbb{Z}/n\mathbb{Z})|=1$. Indeed, in first case $f_k(1+n\mathbb{Z})=\frac{k}{n}+\mathbb{Z}$ is homomorphism for all $0\leq k\leq n$. Conversely, if $f$ is homomorphism and $f(1+nZ)=\frac{a}{b}+\mathbb{Z}$, where $a$ relatively prime to $b$, then $n\frac{a}{b}\in\mathbb{Z}$ and $b$ is divisor of $n$ and $\frac{a}{b}+\mathbb{Z}=\frac{k}{n}+\mathbb{Z}$, where $k\equiv a\frac{n}{b} (\rm mod\, n)$.
In second case, if $f:\mathbb{Q}/\mathbb{Z}\rightarrow \mathbb{Z}/n\mathbb{Z}$ is homomorphism and $f(\frac{1}{mn})=k+n\mathbb{Z}$, then $f(\frac{1}{m})=f(n\cdot\frac{1}{mn})=nk+n\mathbb{Z}=n\mathbb{Z}$ for all $m\in\mathbb{Z}$. So, $f(\mathbb{Q}/\mathbb{Z})=0$, where $0=n\mathbb{Z}$. That is, in this case there is a single homomorphism equal to zero