Abstract Algebra, Lecture 2

How to induct an equivalent relationship by a trivial relationship

We can induct an equivalent relationship $\hat{R}$ by a trivial relationship $R$:

$$\tag{1} \hat{R}=\bigcap_\limits{S\in\Phi}S .$$

$ \Phi$ is the set of all equivalent relationships containing $R$. If $R$ is represented to a $n \times n$ matrix, then

• Transitive property : $\hat{R}^2\subseteq \hat{R}$
• Symmetric property : $\hat{R}=\hat{R}^T$

Hence,

$$\tag{2} \hat{R}=I+(R+R^T)+(R+R^T)^2+\cdots$$

Maps and equivalent relationship

We can induct an equivalent relationship by a arbitrary map.

$$\begin{equation} \begin{aligned} f: A & \rightarrow B \\ & \Downarrow \\ R_f \in S \times S , R_f=&\{(a_1,a_2)|a_1,a_2 \in S, f(a_1)=f(a_2) \} \end{aligned}\tag{3} \end{equation}$$

Let $Im(f)=C$, then

$$\tag{4} S=\bigcup_\limits{y \in C} f^{-1}(y).$$

There exits an nature Map giving an equivalent relationship $R$.

$$\begin{aligned} \pi: A &\rightarrow A/R \\ a & \mapsto[a] _ R \end{aligned}\tag{5}$$

The relationship inducting by this map $R _ \pi=R$.

Example 1 :

$$\begin{aligned} f: \mathbb{R}^2 &\rightarrow \mathbb{R} \\ (x,y)&\mapsto y-2x \end{aligned}\tag{6}$$

$[(x,y)] _ {R_f}={(x,y)|y=2x+b}.$

Semi-group

Binary operation

A binary operation $(S,\cdot )$ on a set S is a map which sends elements of the Cartesian product $S \times S$ to $S$.

$$\tag{7} f: S \times S \rightarrow S.$$

Semi-group

A semi-group $(S,\cdot)$ is an algebraic structure consisting of a set $S$ together with an associative binary operation. That is for all ${ a,b,c\in S} a,b,c\in S$, the equation ${(a\cdot b)\cdot c=a\cdot (b\cdot c)} (a\cdot b)\cdot c = a\cdot(b\cdot c)$ holds.

Exercise 1 : Proof that $(S, \oplus)$ an arbitrary set $S$ together with symmetric difference is a semi-group.