homomorphism_theorem

introduction

%%visits: 2 Also known as noether 1st homomorphism_theorem

Simply put it is the equation $\mathbb{R}^{*} / \ker \phi \cong \im \phi$

The Right [[quotient]], $^{*} $ mod ker ϕ is isomorphic ([[isomorphism]]) to the image of phi ## intuition $Im() = {(g) | g G} $ $ker() = {g G | (g) =e g H} $

ker(ϕ) ⊲ G

{{file:../figures/screenshot_20211108_110556.png}}

The figure above shows how the homomorphisms split up, and the trivial maps get destroyed then we find a isomorphism (The bottom arrow then diagonal arrow). ## rigour If ϕ : G → H is a group [[homomorphism]] then im

The homomorphism_theorem

Isomorphism between a [[coset]] and another group.

Let ℝ^* = ℝ {0} be a group under multiplication

Let ℝ⁺ = {x ∈ ℝ|x > 0} be a group under multiplication.

Consider the map ℝ^* → ℝ⁺ given by ϕ(x) = |x| (make a negative or positive number into a positive?)

wts, ϕ is a homomorphism:

Let x, x^′ ∈ ℝ^* then ϕ(xx^′) = |xx^′| = |x||x^′| = ϕ(x)ϕ(x^′)

It’s kernel is, ker ϕ = {x ∈ ℝ^*|ϕ(x) = 1} = {1, −1}

So ker ϕ ≅ ℤ₂, ϕ is 2 to 1

We know that ℤ₂ ⊲ ℝ^* since it is the kernel of ϕ

We may form the quotient group ℝ^*/ℤ₂ = {xℤ₂|x ∈ ℝ^*}, note that the left_coset is equal to the right_coset. A quotient group is a set of sets

 = {{−x, x}|x ∈ ℝ^*}

Each coset {−x, x} can be labelled by |x|, the positive element in the set.

N.B. The quotient group action is the multiplication on sets: AB = {a_ib_j|∀a_i ∈ A, b_j ∈ B} think of an cartesian_product

Therefore {−x, x}{−y, y} = {xy, −xy}

There is an isomorphism between ℝ^*/ℤ₂ and ℝ⁺ given by ψ({−x, x}) = |x|

This is evidently onto ℝ⁺ and is injective

ψ is a homomorphism

ψ({−x, x}{−y, y}) = ψ({−xy, xy}) = |xy| = |x||y| = ψ({−x, x})ψ({−y, y})

ψ is a bijective homomorphism $$ is a homomorphism. Therefore ℝ^*/ℤ₂ ≅ ℝ⁺

Let G and G^′ be groups and let ϕ : G → G^′ be a homomorphism then G/kerϕ ≅ Im(ϕ)

Proof. Let K = ker ϕ an Ĝ = G/K = G/ker ϕ and let us find an isomorphism from Ĝ onto Im(ϕ)

ϕ̂ : Ĝ → Im(ϕ) given by

ϕ̂(gK) = ϕ(gK) = {ϕ(gk)|k ∈ ker ϕ}

We may be concerned that this is not well-defined: it should map a whole coset, an element of Ĝ = G/K to a single element in Imϕ but it initially appears as if it maps onto many elements in Imϕ however since K = ker ϕ we have ϕ̂(gk) = {ϕ(gk₁), ϕ(gk₂), …, ϕ(gk_n)} = ϕ(g)since the elements in the kernel maps the identity

this hold for any k.

ϕ̂ is a homomorphism:ϕ̂(g₁Kg₂K) = ϕ̂(g₁g₂K) = ϕ̂(g₁g₂)

ϕ̂ is injective.

Suppose that ∃g₁K ≠ g₂K:

θ̂(g₁K) = θ̂(g₂K)

 ⟹ ϕ(g₁) = ϕ(g₂)

 ⟹ e^′ = (ϕ(g₁))⁻¹ϕ(g₂)

 = ϕ(g₁⁻¹)ϕ(g₂)

 ⟹ g₁⁻¹g₂ ∈ K

 ⟹ g₁K = g₁(g₁⁻¹g₂L) = g₂K

Which is our contradiction

Examples and Non-Examples

ϕ : D₃ → ℤ₂

given by ϕ(aⁿb^m) = b^m by the homomorphism_theorem: D₃ \ ℤ₃ ≅ ℤ₂ :todo: make more homomorphisms yourself.

$\frac{O(n)}{SO(n)} \cong \mathbb{Z}_2$

exam clinic

resources

Section 10.1 of the lecture notes tags :math:

backlinks

• index_of_files.pd