Let $G$ be a finite group with prime order $p$ and $H$ a finite group of prime order $q$ such that $p \ne q$. How many homomorphisms $f: G \to H$ are there?
If $|G|=p$, then for $g \in G$ we have that $g^p=e_G$. If $f$ is a homomorphism, then $f(e_G)=e_H$ which means that $f(g^p)=f(g)^p=e_H$, thus $p$ must divide the order of $H$. That is we have $p \mid q$. I think that this implies that $p=q$ which contradicts the fact that $p \ne q$ so there are no homomorphisms?
More generally, let $G,H$ be finite groups of relatively prime order. Then there is only the trivial homomorphism from $G$ to $H$.
The argument is just the same: for any $g\in G$, and any homomorphism $h:G\to H$, we have that the order of $h(g)$ divides both the order of $G$ and of $H$. Thus $h(g)=e$.