Equivalence of definitions of homomorphism of groups

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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

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}