Consider the Ring $\mathbb{Z}$ and the two ideals $(n), (m)$, where $n, m \in \mathbb{N}$, and consider $GCF(m, n)$ (the greatest common factor of $m, n$). Let p: $\mathbb{Z}/(m) \to \mathbb{Z}/GCF(m, n)$ be the natural projection.
Show, that the transformation:
$$\phi: Hom_\mathbb{Z}(\mathbb{Z}/(n), \mathbb{Z}/(m)) \to \mathbb{Z}/(GCF(m, n)), f \mapsto p(f(1 + (n))$$
is an isomorphism of $\mathbb{Z}$-modules.
Thanks in advance! I'm not very familiar with constructions like these.
In order to get a handle on $\hom_{\mathbb{Z}}(\mathbb{Z}/n\mathbb{Z}, \mathbb{Z}/m\mathbb{Z})$, consider that these are exactly the maps induced by those homomorphisms $\mathbb{Z} \to \mathbb{Z}/m\mathbb{Z}$ that vanish on $n\mathbb{Z} \subset \mathbb{Z}$. Such a map is completely determined by $f(1)$, and it vanishes on $n\mathbb{Z}$ exactly when the order of $f(1)$ divides $n$. The order of $f(1)$ is $m/\gcd(f(1),m)$. This divides $n$ exactly when $f(1)$ is divisible all the prime factors that are in $m$ and not in $n$, i.e. exactly when $f(1)$ is divisible by $m/\gcd(m,n)$. Therefore the choices for $f(1)$ are in correspondence with $$\frac{\left(\frac{m}{\gcd(m,n)} \right)\mathbb{Z}}{m\mathbb{Z}}\overset{\text{This is your map $p$}}{\cong} \frac{\mathbb{Z}}{\gcd(m,n)\mathbb{Z}},$$
and the group operations on both sides play nice with each other.
Now this is all conceptual...to actually write your answer, you need to show that it's a homomorphism which is injective and surjective. But the above should guide you.