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 $\Psi$ ?
What can be said given that $\Psi$ is a Homomorphism?
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1You have posted two other question ([here](http://math.stackexchange.com/questions/87790/sizes-of-conjugacy-classes) and [here](http://math.stackexchange.com/questions/87795/how-does-an-odd-order-group-affect-the-kernel)). May I ask the source of the problems, and why you are contemplating them at this point? – 2011-12-02
3 Answers
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 $|\psi(g)|?$
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.
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...