What can be said given that $\Psi$ is a Homomorphism? Let G and H be nontrivial finite groups with relatively prime orders  When $\Psi: G\to H$ be a homomorphism, what can be said about $\Psi$ ?

If $\Psi : G \to H$ is a homomorphism then by the (first) isomorphism theorem you know $G / \ker \Psi \cong \Psi (G)$. This means that $\frac{|G|}{|\ker \Psi|} = |\Psi(G)|$ so you know that $\Psi(G)$ divides $|G|$.
Next you know that $\Psi (G) $ is a subgroup of $H$ and hence by the Lagrange theorem it divides the order of $|H|$.
Putting these two things together you now know that $|\Psi (G)|$ divides $|H|$ and $|\Psi(G)|$ divides $ |G| $. 
But $\gcd(|G|, |H|) = 1$ i.e. their only common divisor is $1$ and so $|\Psi(G)| = 1$ and so $\Psi = 0$ is the trivial map.
