Homomorphism (or Group Morphism)

homomorphism

Let $(G, ⋆)$ and $(H, ⋄)$ be Groups. A map $φ : G \to H$ such that:

φ (x ⋆ y) = φ (x) ⋄ φ (y)

for all $x, y \in G$ is called a homomorphism.

When the group operations for $G, H$ are not explicitly written, the homomorphism condition becomes simply:

φ (x y) = φ (x) φ (y)

but it is important to keep in mind that the product $x y$ on the left is computed in $G$ and the product $φ (x) φ (y)$ on the right is computed in $H$ . Intuitively, a map $φ$ is a homomorphism if it respects the group structures of its domain and codomain.

Intuition

This stack exchange answer clarifies how the idea behind homomorphisms (unlike Isomorphisms) is that some elements may merge to be the same thing. They will have the same structure, but elements may get merged, so the isomorphism extension of bijectivity is important in that case since then elements cannot get merged together like that.

Homomorphism literally means "similar shape", so that gives the idea that squares can become triangles (still n-gons but more or less vertices).

Free Lemma

The identities are preserved, as we'll see in the following lemma:

φ (1) = 1

Suppose $φ : G \to H$ is a homomorphism of groups. Then:

$φ (1_{G}) = 1_{H}$
$φ (g^{- 1}) = φ (g)^{- 1}$

See 16 Homomorphisms, Isomorphisms#1.

Example

Suppose $G, H$ are any two groups. The map $φ : G \to H$ where any $g \in G$ has $φ (g) = 1$ is a trivial homomorphism:

φ (g_{1} g_{2}) = φ (g_{12}) = 1 = φ (g_{1}) φ (g_{2})

Consider the map $φ : R \to R^{+}$ where $R$ is under addition and $R^{+}$ is under multiplication. See Isomorphism#Examples.

Inverses

In class we used this definition to give context to homomorphisms and inverses of those isomorphisms:

inverses

Suppose $φ : G \to H$ and $ψ : H \to G$ are Homomorphism (or Group Morphism)s of groups. We say that $φ, ψ$ are inverses if $ψ \cdot φ = I_{G}$ and $φ \cdot ψ = I_{H}$ .
We also call $φ$ and $ψ$ are Isomorphisms, and say $G$ and $H$ are isomorphic.

The two definitions are the same here (both are saying an invertible isomorphism exists).