homomorphism

introduction

%%visits: 2 What is a homomorphic function of groups? $(g *_G g’) = (g) *_H (g’) :g,g’ G $ ## intuition Let G₁ and G₂ be groups. A map Φ : G₁ → G₂ is a homomorphism if Φ(gh) = Φ(g)(Φ(h)∀g, h,  ∈ G₁ It sat

How a rubiks cube is a group

Every pattern is an element in a group, with actions, being the action of moving around with the cube.

Examples pf G₁  = G₂==

G₁ = {−1, 1} under multiplication

$G_2 = \left\{ \begin{bmatrix} 1 & 2 \\ 1 & 2 \end{bmatrix},\begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} \right\}$

Exam Clinic

Homomorphisms may need to be figured out on the spot, we need to think of the operation rule that defines the homomorphism

If the group is defined as a function, then proving homomorphism is simply using the function.

If the group is defined as a set of elements, then it won’t be possible to use the function shortcut.

1. $\Phi: \mathbb{Z}_{N} \to {The n roots of unity under usual arithmetic multiplication)$

the order of an element is just shorthand for the order of the subgroup generated by an element.

extra Read or try to find. Theorem, Two cyclic_groups of the same order are isomorphic. The trivial homomorphism is

$: G H , g,g’ G (gg’) = e_H $

ϕ(g²) = ϕ(g)²

A function between two rings is homomorphic if

sum of inputs = sum of outputs

product of inputs = product of outputs

1_R maps to 1_S ## rigour homomorphis := A function between two groups is homomorphic if $(g *_G g’) = (g) *_H (g’) :g,g’ G $

Theorems

A homomorphism is an [[isomorphism]] iff it is surjective and the kernel is trivial. {{file:../figures/screenshot_20211025_101142.png}}

The kernel is a normal subgroup Theorem 10.2

The image of a homomorphism is an isomorphism

The image of an homomorphism is a subgroup of H.

Proof

Let g_1^′g_2^′ ∈ Im(phi) ⊂ G₂ Then there exists a g₁, g₂ ∈ G₁ such that ϕ(g₁) = g_1^′, ϕ(g₂) = g_2^′ then g_1^′g_2^′ = ϕ(g₁)ϕ(g₂) = ϕ(g₁g₂) ∈ Im(ϕ) therefore g_1^′g_2^′ ∈ Im(ϕ), hence we have closure

Associativity is inherited from the associative product of G₂

Identity since phi of e is e, we must have the identity.

Inverses: Due to surjectivity ∀g^′ ∈ Im(ϕ)∃ginG₁ such that ϕ(g) = g^′, now g^′−1 = ϕ(g)⁻¹ = ϕ(g⁻¹) therefore g^′−1 ∈ Im(ϕ)

Hence Im(ϕ) ⊂ G₂ is a subgroup ▫

A homomorphism ϕ : G → Im(ϕ) is an isomorphism ⇔ the kernel is just the identity element.

proof: By definition, ϕ is surjective. We only wts that it is injective and so ker is e and vice versa

“⟹”

g ∈ ker (ϕ) ⟹ ϕ(g) = e^′ ∈ Im(ϕ) since ϕ is injective then g is the unique pre-image of e^′

“⟸”

Suppose for contradiction sake if ϕ(g₁) = ϕ(g₂) for g₁ ≠ g₂, g₁, g₂ ∈ G then ϕ(g₁)ϕ(g₂)⁻¹ = e^′ then this implies ϕ(g₁g₂⁻¹) = e^′ therefore g₁g₂⁻¹ ∈ ker(ϕ) = {e} ⟹ g₁g₂⁻¹ = e ⟹ g₁ = g₂ which contradicts our assumption. ▫

Theorem: Let G be a group, then G/Z(G) ≅ Inn(G) Proof Consider he map ψ : G → Inn(G) given by ψ(g) = ϕ_g the inner automorphism, conjugation ψ is a homomorphism:

ψ(g₁g₂)(h) = ϕ_g1g2(h) for some g₁, g₂, h ∈ G

 = (g₁g₂)h(g₁g₂)⁻¹ = g₁g₂hg₂⁻¹g₁⁻¹ = ϕ_g1ϕ̇_g2(h)

 = ψ(g₁)ψ(g₂)(h) therefore it is homomorphic

im(ψ) = Inn(G) by construction (since it is a conjugation)

ker(ψ) = {g ∈ G|ψ(g) =  identity map}

 = {g ∈ G|ϕ_g(h) = id(h) = h∀h ∈ G}

 = {g ∈ G|ϕ_g(h) = hgh⁻¹ = g∀h ∈ G} = Z(G)

Therefore by the group homomorphism theorem we have, G \ Z(G) ≅ Inn(G)

exam clinic

Let G₁ and G₂ be groups. A map Φ : G₁ → G₂ is a homomorphism if Φ(gh) = Φ(g)(Φ(h)∀g, h,  ∈ G₁ ;x Isomorphis := Bijective homomorphism Isomorphism:~ this shows the same underlying structure.

n.b. Φ⁻¹ : G₂ → G₁ while (Φ(g))⁻¹ ∈ G₁ The first is a function inverse, the second is an inverse element.

Proof that Φ(e₁) = e₂ (Identity maps to identity) As e₁ = e₁e₁ then Φ(e₁) = Φ(e₁e₁) = Φ(e₁)Φ(e₁) Now as Φ(e₁) ∈ G₂, a group, then (Φ(e₁))⁻¹ exists in G₂, hence when we left multiply the above by, this, e₂ = (Φ(e₁))⁻¹Φ(e₁)Φ(e₁) = e₂Φ(e₁) = Φ(e₁)

Proof that Φ(g⁻¹) = (Φ(g))⁻¹ As e₁ = gg⁻¹∀g ∈ G Therefore Φ(e₁) = Φ(gg⁻¹) = Φ(g)Φ(g⁻¹) But, Φ(e₁) = e₂ So, e₂ = Φ(g)Φ(g⁻¹) Left Multiply by (Φ(g))⁻¹ Φ(g)⁻¹ = Φ(g⁻¹) ## resources tags :math:introduction_to_abstract_algebra:groups_and_symmetries:

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