# All group homomorphism from $\mathbb{Z} _m$ to $\mathbb{Z}_n$ [duplicate]

All group homomorphism from $\mathbb{Z} _m$ to $\mathbb{Z}_n$

How could I find every group homomorphism?

## marked as duplicate by Mikko Korhonen, The Chaz 2.0, Alexander Gruber♦, rschwieb, Davide GiraudoApr 25 '13 at 14:28

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.)
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.
• Why $d '$ is considered as $d'|d$? $\mathbb{Z}_k$ where $k|n$ can be a subgroup of $\mathbb{Z}_n$. Then I suppose that $d'$ would satisfy the condition of $d'|n$ – user73309 Apr 25 '13 at 14:06
Hint: each homomorpism $f$ is uniquely determined by $f(1)$.