Homomorphism between groups of coprime order is trivial

Statement
Suppose $$G,H$$ are finite groups such that their orders are relatively prime. Then, the only homomorphism between $$G$$ and $$H$$ is the trivial map: the map sending every element of $$G$$ to the identity element of $$H$$.

Related facts

 * Subgroup of least prime index is normal: This uses a similar idea -- except that the two groups in question are not of relatively prime orders; rather, the greatest common divisor of their orders is a prime number, which is forced to be the order of the image.
 * Any homomorphism from a simple group to any group is either trivial or injective.

Applications

 * Normal Hall implies order-unique: If $$H$$ is a normal Hall subgroup of a finite group $$G$$, i.e., the order and index of $$H$$ are relatively prime, then it is the unique subgroup of that order.

Facts used

 * 1) uses::Order of quotient group divides order of group: Given a homomorphism $$\varphi:G \to H$$ of groups, the order of the image $$\varphi(G)$$ (which is a subgroup of $$H$$) divides the order of $$G$$.
 * 2) uses::Lagrange's theorem: The order of any subgroup of a group divides the order of the group.

Proof
Given: Groups $$G$$ and $$H$$ of relatively prime order. A homomorphism $$\varphi:G \to H$$.

To prove: $$\varphi$$ is the trivial map.

Proof:


 * 1) The order of $$\varphi(G)$$ divides the order of $$G$$: This follows from fact (1).
 * 2) The order of $$\varphi(G)$$ divides the order of $$H$$: This follows from fact (2).
 * 3) The order of $$\varphi(G)$$ is $$1$$: This follows from the previous two steps, and the fact that the orders of $$G$$ and $$H$$ are relatively prime.
 * 4) $$\varphi(G)$$ is the trivial subgroup and $$\varphi$$ is the trivial map : Any subgroup of order one is trivial, so $$\varphi(G)$$ is the trivial subgroup, i.e., it comprises the identity element of $$H$$. Thus, the map $$\varphi$$ must send every element of $$G$$ to the identity element of $$H$$.