All group homomorphism from $ \mathbb{Z} _m $ to $\mathbb{Z}_n $ All group homomorphism from $ \mathbb{Z} _m $ to $ \mathbb{Z}_n $
How could I find every group homomorphism?
 A: A homomorphism from the cyclic group $\mathbb{Z}_m$ into any other group is determined by where it sends a generator.  The generator must be sent to an element whose order divides $m$.
In the case of this problem, let $d=\operatorname{gcd}(m,n)$.  For every $d'|d$, there exists a unique subgroup $\mathbb{Z}_{d'}\le\mathbb{Z}_n$.  Map a generator of $\mathbb{Z}_m$ to each generator of $\mathbb{Z}_{d'}$ for all $d'|d$ to obtain all possible homomorphisms.  There are $\varphi(d')$ generators for $\mathbb{Z}_{d'}$, so the total number group homomorphisms is
$$\sum_{d'\mid d}\varphi(d')=d$$
Here $\varphi$ is Euler's totient function.  The above formula is a standard property of Euler's function.
A: Hint: each homomorpism $f$ is uniquely determined by $f(1)$.
A: The map $\phi : Z_m \to Z_n$ given by $\phi(r\mod m) = k r\mod n $ is a well-defined homomorphism if and only if $n$ divides $km$. Every homomorphism from $Z_m \to Z_n$  is of this form.
A homomorphism  is determined by $\phi(1)$ in this case. (Of course, if the group has more
generators, you should check all the generators.)
