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

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

1. Invertible implies cancellative in monoid
2. Equality of left and right inverses in monoid

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:

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