I'm new to abstract algebra and I'm trying to derive the general form of any function $f: Z_m \to Z_n$, such that $f$ is an additive (homo-)morphism, and where $n, m \in N$, $N$ being the set of natural numbers, $Z$ the set of integers and $Z_n$ the set of integers modulo $n$.
I am able to prove that a suficient condition is to be of the form $f:Z_m\to Z_n: a\mapsto p_n(a)$ (with $m\leq n$ and $p_n:Z\to Z_n$ the the map to the rest of the division by $n$), or to be of the form $f = 0$. Now I suspect that another sufficient condition is to be a projection $f:Z_m\to Z_n: a\mapsto p_n(a)$ with $m$ a multiple of $n$, and that a necessary condition is that $f$ be a projection or equal to $0$, but I can't prove these facts.
Therefore, I can certainly not find general rules for $f$ to be an additive morphism $Z_m\to Z_n$. I will appreciate any guidance, suggestion or proof.
Best Answer
Let $f\colon \mathbb{Z}_m\to \mathbb{Z}_n$ be an additive homomorphism. Let $\mu=f(1)$. Then for $a\in \mathbb{Z}_m$, we have $f(a)=\mu a$, by the additive property. Thus any such homomorphism is completely described by the integer $\mu$.
Two integers $\mu, \mu'$ will represent the same map if and only if $n|\mu-\mu'$, so we may regard $\mu$ as an element on $\mathbb{Z}_n$. The remaining question is which $\mu\in \mathbb{Z}_n$ actually represent homomorphisms.
If $\mu$ represents a homomorphism, then we know $$\mu m=0 \mod n,$$ as $m=0 \mod m$. In fact this is sufficient to guarantee that multiplication by $\mu$ is a homomorphism.
That is $\mu\in \mathbb{Z}_n$ represents a homomorphism $\mathbb{Z}_m\to \mathbb{Z}_n$ if and only if $n|\mu m$. This is equivalent to $\frac{n}{\gcd(n,m)}|\mu$.
Thus the distinct additive homomorphisms $\mathbb{Z}_m\to \mathbb{Z}_n$ are precisely the maps $$a\mapsto \lambda \frac{n}{\gcd(n,m)} a,\qquad {\rm for}\,\, \lambda=0,1,2,\cdots,\gcd(n,m)-1.$$