Number of homomorphisms from $(\mathbb Q^+ , \,\cdot\,)$ to $(\mathbb Z_n , +)$ 
How many homomorphism are there from $(\mathbb Q^+ , \,\cdot\,)$ to $(\mathbb Z_n , +)$?

Could anyone give a hint how to proceed?
I couldn't even start
 A: There are definitely non-trivial homomorphisms from $\mathbb Q^+$  to $\mathbb Z_n$. 
To see this, note that every positive rational number has a unique-upto-order "prime decomposition with negative powers". For example, $\frac{24}{35} = 2^{3}3^{1}5^{-1}7^{-1}$. Therefore, as is specified in a comment in the deleted answer above/below, $\mathbb Q^+$ under multiplication is isomorphic to a subgroup of $\displaystyle\oplus_{p \ \mathrm{ prime}} \mathbb Z$, via mapping each rational number, to the infinite tuple (with finitely many entries) of what power of each primes occurs in its decomposition.
Consequently, every such homomorphism is uniquely determined by what it does at the prime numbers. Here's why : let's write $p_1 < p_2 < p_3 < ...$ as the infinite sequence of prime numbers, and suppose you have a number $\frac{p}{q} = p_1^{\alpha_1} p_2^{\alpha_2} ...$ where $\alpha_i$ is a sequence of integers that becomes zero after some time, then by the homomorphism property : 
$$
\phi\left(\frac pq\right) = \phi\left(\prod p_i^{\alpha_i}\right) = \sum \alpha_i\phi(p_i) 
$$
where the sum is finite since $\alpha_i$ is eventually zero.
Consequently, to define a homomorphism, all you need to say is what it does at the prime numbers.
For example, let $\phi$ be the unique homomorphism such that $\phi(2) = \bar 1$ and $\phi(p) = \bar 0$ for all primes $p$ greater than $2$. 
Then, for any rational $\frac pq$, $\phi\left(\frac pq\right)$ is equal to $\bar r$, where $\frac{p}{q} = 2^r3^s....$ You can check that this is a homomorphism.

You may ignore this if you do not know what countability means.
If you want the exact number of homomorphisms, then this is clear as well : each prime number (countably many prime numbers) is mapped to one of $n$ elements freely (where $n$ is finite).
Hence, one concludes that the number of homomorphisms is countable (countable number of finite choices amounts to countable choice), and definitely infinite, hence countably infinite.

On a side note, the deleted answer, which refers to the order of elements, makes sense for homomorphisms in the opposite direction i.e. from $\mathbb Z_n$ to $\mathbb Q^+$. Indeed, if $\psi$ is a homomorphism, then let $\psi(1) = q$ for some $q$. Thus, $\psi(0) = \psi(1 + 1 + ... + 1) = \psi(1) \cdot ... \cdot \psi(1) = q^n$. Hence, $q^n = 1$ for some rational $q$ : this forces $q = 1$. 
Hence, $\psi(l) = 1$ for every $l \in \mathbb Z_n$, hence $\psi$ is trivial.
