# 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

## The definitions that we have to prove are equivalent

### Textbook definition (with symbols)

Let $G$ and $H$ be groups. Then a map $\varphi: G \to H$ is termed a homomorphism of groups if $\varphi$ satisfies the following condition:

$\varphi(ab) = \varphi(a) \varphi(b)$ for all $a, b$ in $G$

### Universal algebraic definition (with symbols)

Let $G$ and $H$ be groups. Then a map $\varphi:G \to H$ is termed a homomorphism of groups if $\varphi$ satisfies all the following conditions:

• $\varphi(ab) = \varphi(a) \varphi(b)$ for all $a, b$ in $G$
• $\varphi(e) = e$
• $\varphi(a^{-1}) = (\varphi(a))^{-1}$

## Proof

We need to prove that the condition:

$\varphi(ab) = \varphi(a) \varphi(b)$ for all $a,b \in G$

implies the other two conditions.

For clarity of the proofs, we denote the identity element of $G$ by $e_G$ and the identity element of $H$ by $e_H$.

### Proof that it preserves the identity element

Given: Groups $G,H$, a map $\varphi:G \to H$ such that $\varphi(ab) = \varphi(a) \varphi(b)$ for all $a,b \in G$

To prove: $\varphi(e_G) = e_H$

Proof: Pick any $a \in G$ (we could pick $a = e_G$ if we wanted). We get:

$\varphi(a) = \varphi(ae_G) = \varphi(a)\varphi(e_G)$

On the other hand, we have:

$\varphi(a) = \varphi(a) e_H$

Combining, we get:

$\varphi(a)\varphi(e_G) = \varphi(a)e_H$

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

$\varphi(e_G) = e_H$

### 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 $G,H$, a map $\varphi:G \to H$ such that $\varphi(ab) = \varphi(a)\varphi(b)$ for all $a,b \in G$ and $\varphi(e_G) = e_H$.

To prove: For any $a \in G$, $\varphi(a^{-1}) = (\varphi(a))^{-1}$

Proof: By definition, we know that:

$e_G = aa^{-1}$

Applying $\varphi$ to both sides, we get that:

$\varphi(e_G) = \varphi(aa^{-1}) = \varphi(a)\varphi(a^{-1})$

We know that $\varphi(e_G) = e_H$, so we get:

$e_H = \varphi(a)\varphi(a^{-1})$

We can also write:

$e_H = \varphi(a)(\varphi(a))^{-1}$

Equating the right sides and cancelling $\varphi(a)$, we get:

$\varphi(a^{-1}) = (\varphi(a))^{-1}$