valenza-groups

A magma $⟨ S, * ⟩$ is a set $S$ together with an operation $*$
An associative magma is called a semi-group
A semi-group with an identity group is called a monoid
A group $⟨ G, * ⟩$ is a monoid in which every element is invertible, also note
- the operation $*$ is associative
- there exists an element $e \in G$ which is an identity for $G$
- for every $s \in G$ $\exists$ $t \in G$ s.t. $s * t = t * s = e$

A commutative group with operation

+

is called an additive group

Cancellation Laws

Let $G$ be a group. Then for $s, t, u \in G$

\begin{matrix} s t = s u ⟹ t = u \\ s t = u t ⟹ s = u \end{matrix}

Proof:

\begin{aligned} s t & = s u \\ s^{- 1} (s t) & = s^{- 1} (s u) \\ (s^{- 1} s) t & = (s^{- 1} s) u \\ e t & = e u \\ t & = u \end{aligned}

Uniqueness of Identity

Let $G$ be a group. Then the identity element $e \in G$ is unique

Proof: Let $e, e^{'}$ be two identities of $G$

e = e e^{'} = e^{'}

Uniqueness of Inverse

Let $G$ be a group. For every $s \in G$ , there exists a unique $t \in G$ s.t. $s t = e = s u$

Proof: Let $t, t^{'}$ be two inverses of $s$ .

s t = e = s t^{'} ⟹ t = t^{'}

Let

G

be a group. If

s, t \in G

with

s t = e

, then

s = t^{- 1}

and

t = s^{- 1}

Proof:

\begin{aligned} s t & = e \\ s^{- 1} (s t) & = s^{- 1} \\ (s^{- 1} s) t & = s^{- 1} \\ t & = s^{- 1} \end{aligned}

Let

G

be a group

\forall s \in G

(s^{- 1})^{- 1} = s

Proof: By uniqueness of inverses

Let

G

be a group

\forall s, t \in G

(s t)^{- 1} = t^{- 1} s^{- 1}

Proof:

\begin{aligned} (s t) t^{- 1} s^{- 1} = s (t t^{- 1}) s^{- 1} = s s^{- 1} = e \end{aligned}

Thus, $t^{- 1} s^{- 1} = (s t)^{- 1}$

Let

G

be a group. If

s \in G

, then

s s = s

iff

s = e

Proof:

\begin{aligned} s s & = s \\ s & = e \end{aligned}

Let $⟨ G, * ⟩$ be a group. Then a subset $H$ of $G$ is called a subgroup of $G$ if it constitutes a group in its own right w.r.t. $*$ defined on $G$
- $H$ is closed under the operation defined on $G$
- $H$ contains the identity of $G$
- If $s \in H$ , $s^{- 1} \in H$

A non-empty subset

H

of the group

G

is a subgroup of

G

iff

$s, t \in H ⟹ s t^{- 1} \in H$

NOTE: A group $G$ is always a subgroup of itself

Let $G_{0}$ and $G_{1}$ be groups. Then a function $φ : G_{0} \mapsto G_{1}$ is called a homomorphism of groups if it satisfies:

φ (s, t) = φ (s) φ (t) \forall s, t \in G_{0}

* LHS is the operator defined of the domain ( $G_{0}$ )
* RHS is the operator defined of the co-domain ( $G_{1}$ )

The composition of group homomorphisms is also a group homomorphism

Proof: Let $φ_{0} : G_{0} \mapsto G_{1}$ and $φ_{1} : G_{1} \mapsto G_{2}$ , then:

\begin{aligned} φ_{1} \circ φ_{0} (s, t) \\ φ_{1} (φ_{0} (s, t)) \\ φ_{1} (φ_{0} (s) \circ φ_{0} (t)) \\ φ_{1} (φ_{0} (s)) φ_{1} (φ_{0} (t)) \\ φ_{1} \circ φ_{0} (s) φ_{1} \circ φ_{0} (t) \end{aligned}

φ : G_{0} \mapsto G_{1}

is a group homomorphism. Then

φ (e_{0}) = e_{1}

Proof: Note that $e_{0} e_{0} = e_{0}$

φ (e_{0}) φ (e_{0}) = φ (e_{0})

For this to hold $φ (e_{0})$ must be $e_{1}$

φ : G_{0} \mapsto G_{1}

is a group homomorphism, then

φ (s^{- 1}) = (φ (s))^{- 1} \forall s \in G_{0}

Proof:

\begin{aligned} s s^{- 1} & = e_{0} \\ φ (s) φ (s^{- 1}) & = φ (e_{0}) \\ φ (s) φ (s^{- 1}) & = e_{1} \end{aligned}

By previous proposition, $φ (s^{- 1}) = (φ (s))^{- 1}$

A bijective group homomorphism is called an group isomorphism; denoted $G_{0} ≅ G_{1}$
Let $φ : G_{0} \mapsto G_{1}$ be a group homomorphism. Then the kernel of $φ$ denoted $Ker (φ)$ is defined as:

Ker (φ) = {s \in G_{0} : φ (s) = e_{1}}

Ker (φ)

is a subgroup of

G_{0}

Proof: Let $s, t \in Ker (φ)$ , we have

\begin{aligned} φ (s t^{- 1}) & = φ (s) φ (t^{- 1}) \\ = φ (s) [φ (t)]^{- 1} \\ = e e^{- 1} \\ = e \in Ker (φ) \end{aligned}

The image of $φ$ is defined as:

Im (φ) = {t \in G_{1} : \exists s \in G_{0} s.t. φ (s) = t}

Im (φ)

is a subgroup of

G_{1}

Let $φ : G_{0} \mapsto G_{1}$ be a homomorphism of groups and suppose $φ (s) = t$ , then

φ^{- 1} = {s k : k \in Ker (φ)}

A ring (with unity) consists of a non-empty set $A$ together with two operations $+$ and $*$ s.t.:

$⟨ A, + ⟩$ is an additive group
$⟨ A, * ⟩$ is a monoid (no inverse)

\begin{aligned} a * (b + c) & = a * b + b * c \\ (a + b) * c & = a * c + b * c \end{aligned}

A commutative ring $k$ is called a field if $k^{*}$ (set of non-zero elements of $k$ ) forms a group w.r.t ring multiplication (inverse exists for multiplication). Thus every non-zero element of $k$ has a multiplicative inverse

Ring	Field
$⟨ A, * ⟩$ is a monoid	$⟨ A, * ⟩$ is a group