Homomorphisms

Group Homomorphism

Suppose that

(G, ⋆)

and

(H, \cdot)

are groups, then we say that a group homomorphism from $(G, ⋆)$ to $(H, \cdot)$ is a function

ϕ : G \to H

such that for any

u, v \in G

ϕ (u ⋆ v) = ϕ (u) \cdot ϕ (v)

Note that a homomorphism is an isomorphism without the bijective requirement.

The order of The Mapped Element Divides the Original

Suppose that

| g |

is finite, then

| ϕ (g) |

divides

| g |

Suppose

g

has order

n

, so that

g^{n} = e

, now that is to say that

e = ϕ (e) = ϕ (g^{n}) = ϕ {(g)}^{n}

, then

| ϕ (g) |

divides

n = | g |

as needed.

Kernel of a Homomorphism

The kernel of a homomorphism from the group

G

to the identity

e_{H}

H

is the set

{g \in G : ϕ (g) = e_{H}} = ϕ^{- 1} ({e_{H}})

Right Iterated Binary Operation

Suppose that

(⋆, G, ⋆)

is a group, then we define

right_iter (⋆, g, n) = {\begin{matrix} g & if n = 0 \\ ⋆ (g, right_iter (⋆, g, n - 1)) & if n \in ℕ_{1} \end{matrix}

Left Iterated Binary Operation

Suppose that

(G, ⋆)

is a group, then we define

left_iter (⋆, g, n) = {\begin{matrix} g & if n = 0 \\ ⋆ (left_iter (⋆, g, n - 1), g) & if n \in ℕ_{1} \end{matrix}

The above can be generalized to operate on sequences of elements of $g$ .

Associative Implies Left Equals Right Iteration

Suppose that

⋆

is associative, then for any

g \in G

we have

left_iter (⋆, g, n) = right_iter (⋆, g, n)

Induction using the fact that bracket placement doesn't matter with associative operations

Binary Iteration

Let

⋆

be a binary operation such that

right_iter (⋆, g, n) = left_iter (⋆, g, n)

then we define

iter (⋆, g, n) = right_iter (⋆, g, n)

Left Tuples as Iteration

Let

+

be the standard addition on

ℤ \times ℤ

, then

(m, 0) = iter (+, (1, 0), m)

By induction

Right Tuples as Iteration

Let

+

be the standard addition on

ℤ \times ℤ

, then

(0, m) = iter (+, (0, 1), m)

By induction

Iterated Homomorphism

Let

ϕ

be a homomorphism between

(G, ⋆)

and

(H, \cdot)

ϕ (right_iter (⋆, g, n)) = right_iter (\cdot, ϕ (g), n)

By induction, each step using the property of a homomorphism

Addition Factors Through Homomorphism on Cartesian Product of The Integers

Homomorphism Between a Group and it's Cartesian Product

Let

Suppose that

ϕ : ℤ \times ℤ \to ℤ

is a homomorphism between

(ℤ \times ℤ, \oplus︎)

and

(ℤ, +)

then for any

m, n \in ℤ

we have

ϕ ((m, n)) = m ϕ ((1, 0)) + n ϕ ((0, 1))

\begin{matrix} ϕ ((m, n)) & = ϕ ((m, 0) \oplus︎ (0, n)) \\ = ϕ ((m, 0)) + ϕ ((0, n)) \end{matrix}

We now recall that $ϕ (g^{n}) = ϕ {(g)}^{n}$ so that we obtain $ϕ ((m, 0)) = ϕ ({(1, 0)}^{m}) = ϕ {((1, 0))}^{m}$ here the power symbol is acting with respect to $\oplus︎$ and then $+$ respectively. So in multiplicative notation that would mean $ϕ {((1, 0))}^{m} = m ϕ ((1, 0))$ (note that multiplication here is just syntactic sugar for it being added to itself $m$ times)

By symmetry it follows that $ϕ ((0, n)) = n ϕ ((0, 1))$ so re-connecting with our original chain of equalities we have $ϕ ((m, n)) = m ϕ ((1, 0)) + n ϕ ((0, 1))$

Finding Group Homomorphisms

Find all group homomorphisms

ϕ : ℤ \times ℤ \to ℤ

For every

a, b \in ℤ

such that

ϕ ((1, 0)) = a

and that

ϕ ((0, 1)) = b

, we obtain a unique homomorphism. Therefore all homomorphism are given by the collection:

{ϕ : ℤ \times ℤ \to ℤ : ϕ ((1, 0)) = a and ϕ ((0, 1)) = b}

The Image of a Homomorphism is Abelian iff It's Kernel Contains the Subgroup Generated by

x y x^{- 1} y^{- 1}

Suppose that

ϕ : G \to H

is a group homomorphism. Then

im (ϕ)

is abelian iff

[G, G] \subseteq \ker (ϕ)

$⟹$ Suppose that $im (ϕ)$ is abelian, we want to prove that $[G, G] \subseteq \ker (ϕ)$ , we already know that $[G, G]$ is the smallest subgroup containing the set ${x y x^{- 1} y^{- 1} : x, y \in G}$ since we already know that $\ker (ϕ)$ is a subgroup of $G$ , then we just have to show it contains the set, and it would automatically become a superset as needed.

Let $x y x^{- 1} y^{- 1}$ be in the set, to show that this element is also in $\ker (ϕ)$ we have to show that $ϕ (x y x^{- 1} y^{- 1}) = e_{H}$ , we note that $im (ϕ)$ was abelian and a homomorphism, thus $\begin{matrix} ϕ (x y x^{- 1} y^{- 1}) & = ϕ (x) ϕ (y) ϕ (x^{- 1}) ϕ (y^{- 1}) \\ = ϕ (x) ϕ (y) ϕ {(x)}^{- 1} ϕ {(y)}^{- 1} \\ = ϕ (x) ϕ {(x)}^{- 1} ϕ (y) ϕ {(y)}^{- 1} \\ = e_{H} e_{H} \\ = e_{H} \end{matrix}$ Note that pulling the power out of a homomorphism is justified on line 2

$⟸$ Now we're assuming that $[G, G] \subseteq \ker (ϕ)$ and we want to show that $im (ϕ)$ is abelian so let $a, b \in im (ϕ)$ , so that there is some $x, y \in G$ such that $a = ϕ (x), b = ϕ (y)$ . Let's prove $a b = b a$ :

Firstly by definition $[G, G]$ contains the set ${x y x^{- 1} y^{- 1} : x, y \in G}$ therefore since $\ker (ϕ)$ is a superset of $[G, G]$ it also contains that set, in other words $x y x^{- 1} y^{- 1} \in \ker (ϕ)$ , that is $ϕ (x y x^{- 1} y^{- 1}) = e_{H}$ , well $ϕ$ is a homomorphism: $ϕ (x) ϕ (y) ϕ {(x)}^{- 1} ϕ {(y)}^{- 1} = e_{H}$ , the by multiplying each side on the right we have $ϕ (x) ϕ (y) = ϕ (y) ϕ (x)$ , well that is $a b = b a$ as needed.