# 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. 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. 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 $H$.