0 - Magma, Semigroups, Monoids, Groups

magma

A set $S$ with a map $S \times S \overset{\circ}{⟶} S$ is called a magma. The map $\circ$ is called a binary operator (or just operator), and for $(x, y) \in S \times S$ , we write $\circ (x, y)$ as $x \circ y$ , and often shorten it to just $x y$ . Note that $x y$ is called the product of $x$ and $y$ .

A binary operator of a magma is associative if $x (y z) = (x y) z$ . In this case we can use induction to show that the product of $n$ elements can simply be written $x_{1} x_{2}, \dots x_{n}$ , and any parenthesizing of this product is equivalent.

A binary operator of a magma is commutative if $x y = y x$ . It is customary to use additive notation in such cases, that is we often denote $\circ (x, y)$ as $x + y$ , when $\circ$ is commutative.

identity element

let $S$ be a magma. an element $e \in S$ is called the identity element (or just identity) if for every $s \in S$ , $s e = e s = s$ . Note that the identity of a magma is unique. Otherwise, suppose $e^{'}$ is also an identity of $S$ , then $e^{'} e = e = e^{'}$ .

Semigroup

A semigroup $S$ is a magma whose operator is associative.

Monoid

A monoid $S$ is a semigroup with a (unqiue) identity.

Group

A monoid $G$ where each element $g \in G$ has an inverse $g^{- 1} \in G$ such that $g g^{- 1} = g^{- 1} g = e$ is a group. A group is abelian if it's operator is commutative.

The inverse of an element in a group is unique

Let $G$ be a group, then for each $g \in G$ , there is a unique $g^{- 1} \in G$ such that $g g^{- 1} = g^{- 1} g = e$ .

\begin{proof}
Suppose both $g_{}^{- 1}, g_{-}^{- 1}$ are inverses of an arbitrary element $g \in G$ . Then, $e = g g^{- 1}$ multiplying $g_{-}^{- 1}$ on boh sides, we get $g_{-}^{- 1} e = (g_{-}^{- 1} g) g^{- 1}$ . Therefore, $g_{-}^{- 1} = g^{- 1}$ .

\end{proof}

The unique property of the identity of a group

let $G$ be a group, then the identity $e \in G$ is the only element in $G$ such that $e^{2} = e$

\begin{proof}
Suppose for some $g \in G$ , we have that $g g = g$ . Then, multiplying by $g^{- 1}$ on both sides, we have $g = e$ .

\end{proof}

A one way condition to guarantee abelian groups

let $G$ be a group, such that for each $g \in G$ , $g^{2} = e$ , then $G$ is abelian.

\begin{proof}
let $x, y$ be arbitrary elements in $G$ . Then, $x y \in G$ .
Hence there is a unique $(x y)^{- 1} \in G$ , such that $(x y) (x y)^{- 1} = e$ . Using the algebraic properties of groups, it is not hard to see that $(x y)^{- 1} = y^{- 1} x^{- 1}$ . However, its given to us that $(x y) (x y) = e$ . which means that $(x y) = (x y)^{- 1}$ (they are "both" the unique inverse elements of $x y$ ). Equivalently, $x y = y^{- 1} x^{- 1}$ . Again, in $G$ , $x x = e = y y$ . Hence, $x = x^{- 1}$ and $y = y^{- 1}$ . Therefore, $x y = y x$ , hence $G$ is abelian.

To show that this condition is only one way, consider the group of integers with addition as the operator, $Z, +$ . This group is abelian, however, it is not true that $x + x = 0$ for each $x \in Z$ .

\end{proof}

A semigroup with left identity and left inverses is a group

let $G$ be a semigroup, such that there exists $e \in G$ such that $e g = g$ for all $g \in G$ , and for each $g \in G$ , there exists $g^{- 1} \in G$ such that $g^{- 1} g = e$ . Then $G$ is a group.

\begin{proof}
we play some algebraic tricks here:
Let $g \in G$ , then $e g = g$ . Consider the equation $g e = e g e$ . Notice that $g^{- 1} g = e = {g^{- 1}}^{- 1} g^{- 1}$ (every element has a left inverse). Now, $g e = ({g^{- 1}}^{- 1} g^{- 1}) (g) (g^{- 1} g)$ . Hence, upon simplifying, we have $g e = g$ . Hence $e$ is also a right identity, therefore, $e$ is the identity.
Similarly, $g g^{- 1} = e g g^{- 1}$ . Hence, $g g^{- 1} = ({g^{- 1}}^{- 1} g^{- 1}) g g^{- 1}$ . Upon simplifying, we have $g g^{- 1} = e$ . Therefore $g^{- 1}$ is also the right inverse and hence the inverse of $g$ .

\end{proof}

A semigroup with right identity and right inverses is a group

Proof Outline

Give a proof, symmetrical to the proof of [[#^e5f52d]]

A finite semigroup with two sided cancellation law is a group

Let $G$ be a finite semigroup, $| G | = n$ , such that $a b = a c$ implies $b = c$ and $b a = c a$ implies $b = c$ . Then $G$ is a group.

\begin{proof} for an arbitrary element $g \in G$ , consider the map $G \overset{f_{g}}{⟶} G$ , given by $f_{g} (x) = g x$ for all $x \in G$ . This map is injective, as $g x_{1} = g x_{2}$ implies $x_{1} = x_{2}$ . Hence, each $x \in G$ goes to a unique $g x \in G$ . therefore, the image of $G$ under $f_{g}$ has exactly $n$ elements, Hence $f g (G) = G$ , therefore, the map is bijective. hence, there exists a unqiue $e \in G$ such that $f_{g} (e) = g$ that is $g e = g$ . Now consider an arbitrary element $h \in G$ . notice that $g h = (g e) h$ . Hence $h = e h$ . therefore $e$ is the left identity. moreover, $h g = h (e g)$ (e is the left identity) Hence, $h = h e$ therefore, $e$ is also the right identity. Now, for an arbitrary element $g$ the map $f_{g}$ as described above is bijective, so there is a unique $g^{- 1}$ such that $f_{g} (g^{- 1}) = e$ . Equivalently every element $g$ has a unique right inverse $g^{- 1}$ such that $g g^{- 1} = e$ . We finish the proof by applying [[#^c54aa6]].

\end{proof}

Product of groups:

The direct product of groups

Let $G_{1}, G_{2}, \dots G_{n}$ be a collection of groups. Then, the set $\prod_{i = 1}^{n} G_{i}$ with the product of two tuples $g^{*} = (g_{1}, g_{2}, \dots g_{n})$ and $h^{*} = (h_{1}, h_{2}, \dots h_{n})$ , where $g_{i}, h_{i} \in G_{i}$ . Is given by $g^{*} h^{*} = (g_{1} h_{1}, g_{2} h_{2}, \dots g_{n} h_{n})$ is called the direct product of a collection of groups.

The direct product of a collection of groups, is a group

\begin{proof} (outline)

Associativity of the product defined above, follows directly from the associativity of the products of each group $G_{i}$ . The tuple of identity elements from each group $e^{*} = (e_{1}, e_{2}, \dots e_{n})$ acts as the identity element, and for the arbitrary element $g^{*} = (g_{1}, g_{2}, \dots g_{n})$ , the inverse element is given by ${g^{*}}^{- 1} = (g_{1}^{- 1}, g_{2}^{- 1}, \dots g_{n}^{- 1})$ .
\end{proof}