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 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
- 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.
We need to prove that the condition:
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
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: