# Equivalence of definitions of homomorphism of groups

This article gives a proof/explanation of the equivalence of multiple definitions for the term homomorphism of groups

View a complete list of pages giving proofs of equivalence of definitions

## Contents

## The definitions that we have to prove are equivalent

### Textbook definition (with symbols)

Let and be groups. Then a map is termed a **homomorphism** of groups if satisfies the following condition:

for all in

### Universal algebraic definition (with symbols)

Let and be groups. Then a map is termed a **homomorphism** of groups if satisfies *all* the following conditions:

- for all in

## Related facts

- Equivalence of definitions of subgroup has essentially the same proof
- Groups form a full subcategory of semigroups is a category-theoretic statement obtained by combining this with the equivalence of definitions of group.

## Facts used

## Proof

We need to prove that the condition:

for all

implies the other two conditions.

For clarity of the proofs, we denote the identity element of by and the identity element of by .

### Proof that it preserves the identity element

**Given**: Groups , a map such that for all

**To prove**:

**Proof**: Pick any (we could pick if we wanted). We get:

On the other hand, we have:

Combining, we get:

Cancel from both sides using Fact (1) (note that is a group, so Fact (1) applies to all elements) and get:

### Proof that it preserves inverses

We will build on the result of the previous proof, that has already shown that the map must preserve the identity element.

**Given**: Groups , a map such that for all and .

**To prove**: For any ,

**Proof**: By definition, we know that:

Applying to both sides, we get that:

We know that , so we get:

We can also write:

Equating the right sides and cancelling , we get: