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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 $\Psi$ ?

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Hint: Given $g\in G$, what can be said about the order of $\Psi(g)$ in $H$? – Thomas Andrews Dec 2 '11 at 17:51
You have posted two other question (here and here). May I ask the source of the problems, and why you are contemplating them at this point? – Arturo Magidin Dec 2 '11 at 18:14

If $g \in G,$ then $|g|$ divides the order of $G.$ $\psi(g) \in H,$ so $|\psi(g)|$ divides the order of $H.$ Now $|\psi(g)|$ divides $|g|.$ But $|\psi(g)|$ and $|g|$ must be relatively prime. What does this imply about $|\psi(g)|?$

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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.

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Every quotient of $G$ has order a divisor of $G$.

Every subgroup of $H$ has order a divisor of $H$.

The image of $G$ is a subgroup of $H$, isomorphic to a quotient of $G$. Therefore...

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