# Homomorphism between groups of coprime order is trivial

From Groupprops

## Statement

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

## 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 is a normal Hall subgroup of a finite group , i.e., the order and index of are relatively prime, then it is the unique subgroup of that order.

## Facts used

- Order of quotient group divides order of group: Given a homomorphism of groups, the order of the image (which is a subgroup of ) divides the order of .
- Lagrange's theorem: The order of any subgroup of a group divides the order of the group.

## Proof

**Given**: Groups and of relatively prime order. A homomorphism .

**To prove**: is the trivial map.

**Proof**:

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