{%hackmd @RintarouTW/About %}

# Abstract Algebra Theory and Applications

https://open.umn.edu/opentextbooks/textbooks/abstract-algebra-theory-and-applications

## Preliminaries

### Equivalence Relation
- Rational Number Equivalence
- Function Equivalence
- Equivalence Classes (form a partition) of a Set
- Matrix Equivalence
:::info
Matrix Equivalence (similar relation)
$$
A,B\in Matrix\\
A \sim B \iff \exists P, PAP^{-1} = B
$$
$A$ and $B$ are **smiliar**.
:::
- Points on the same circle
$\cases{p_1=(x_1,y_1)\\p_2=(x_2,y_2)},{x_1}^2+{y_1}^2={x_2}^2+{y_2}^2$
- Integer Modulo $n$
$r\equiv s\pmod n\iff r\equiv_n s\iff r-s=kn\mid k\in \mathbb{Z}$
$r$ is **congruent to** $s$ **modulo** $n$ (r 與 s 同餘)

### The Division Algorithm

:::info
$$
\cases{
a,b\in\mathbb{Z}\\
S=\{am+bn\mid m,n\in\mathbb{Z}, am+bn\gt 0\}\implies am+bn\in\mathbb{N}\implies S\text{ is well-ordered}\\
d\text{ is the least element of S}, \exists x,y\in\mathbb{Z}, d=ax+by,d\gt 0
}\\
a=q_1d+r_1,0\le r_1\lt d\implies
\begin{array}l
r_1&=a-q_1d\\
&=a-q_1(ax+by)\\
&=a(1-q_1x)+b(-q_1y)\\
\end{array}\\
r_1\gt 0\implies r_1 \in S\land r_1\lt d, \text{contradiction to $d$ is the least element of } S\\
\therefore r_1=0,a=q_1d\implies d\mid a\\
b=q_2d+r_2,0\le r_2\lt d\implies
\begin{array}l
r_2&=b-q_2d\\
&=b-q_2(ax+by)\\
&=a(-q_2x)+b(1-q_2y)\\
\end{array}\\
r_2\gt 0\implies r_2 \in S\land r_2\lt d, \text{contradiction to $d$ is the least element of } S\\
\therefore r_2=0,b=q_2d\implies d\mid b\\
d\mid a\land d\mid b\implies d\text{ is the common divisor of $a$ and $b$}\\
assume\ \exists d'\text{ is the other common divisor of $a$ and $b$}\\
\cases{
a=k_ad',k_a\in\mathbb{Z}\\
b=k_bd',k_b\in\mathbb{Z}
}\\
\begin{array}l
d&=ax+by\\
&=k_ad'x+k_bd'y\\
&=d'(k_ax+k_by)
\end{array}\\
\therefore d'\mid d\implies d=gcd(a,b)\\
gcd(a,b)=\text{the least element of } \{am+bn \mid m,n\in\mathbb{Z}, am+bn\in\mathbb{N}\}
$$
:::

## Integers

### Prime Number

$$
p\in prime\implies\cases{
p\in\mathbb{N}\\
p\not\mid 1\\
\forall a\in\mathbb{N},a\lt p\implies gcd(a,p)=1\\
\forall p_1,p_2\in prime, gcd(p_1,p_2)=1
}
$$

### Euclid

Let a and b be integers and p be a prime number. If p | ab, then either p|a or p|b.
:::info
$$
a,b\in\mathbb{Z},p\in prime, p\mid ab\implies p\mid a\lor p\mid b
$$

Proof:

$$
if\ p\not\mid a\implies gcd(a,p)=1\implies\exists x,y\in\mathbb{Z}, ax+py=1\\
b=b\cdot 1=b(ax+py)= (ab)x+p(by)\\
p\mid ab\land p\mid p\implies p\mid b\\
\text{for the same reason},if\ p\not\mid b\implies p\mid a
$$
:::

### Infinite Prime Numbers

:::info
Assume there're only finite prime numbers $p_1, p_2, \ldots, p_n$
Let $P = p_1p_2\cdot p_n + 1$, then $\exists p_i, 1\le i\le n, p_i\mid P$
$$
\cases{
p_i\mid (p_1p_2\cdot p_n)\\
p_i\mid P\\
}\implies p_i\mid (P-p_1p_2\cdot p_n)\implies p_i\mid 1
$$
Since all prime numbers can't divide 1, contradiction to $p_i\in prime$.
Therefore exists infinite prime numbers.
:::

### $ab \equiv 1 \pmod n$

Let a be a nonzero integer. Then $gcd(a, n) = 1$ if and only if there exists a multiplicative inverse $b$ for $a \pmod n$; that is, a nonzero integer $b$ such that $ab \equiv 1 \pmod n$

:::info
$$
\mathbb{Z^*}=\mathbb{Z}\setminus\{0\}\\
\forall a,n\in\mathbb{Z^*},
gcd(a,n)=1
\iff\exists b\in\mathbb{Z^*},
ab \equiv 1 \pmod n
$$
Proof:
$\implies)$
$$
gcd(a,n)=1\implies \exists r,s\in\mathbb{Z^*}, ar+ns=1\\
ar=1-ns\implies ar\equiv 1\pmod n\\
\therefore b=r\pmod n\in\mathbb{Z_n}\\
$$
$\impliedby)$
$$
ab\equiv 1\pmod n\implies n\mid ab-1\\
\exists k\in\mathbb{Z^*},ab-1=kn\implies ab-kn=1\\
let\ d=gcd(a,n)\implies d\mid a\land d\mid n\\
\therefore d\mid (ab-kn)\implies d\mid 1\implies d=1\\
gcd(a,n)=1
$$
:::

## Groups

### Group of Units of $\mathbb{Z_n}$

Unit of $\mathbb{Z_n}$
:::info
$$
\forall a\in\mathbb{Z_n}, gcd(a,n)=1\text{ aka. $a,n$ are relatively prime or coprime},\\
a\text{ is a unit of }\mathbb{Z_n}
$$
:::

U(n) : the set of all units of $\mathbb{Z_n}$ form a group called **the group of units of $\mathbb{Z_n}$** under $\times_n$
:::info
$$
U(n)=\{u\mid u\in unit\ of\ \mathbb{Z_n}\}
$$
:::

Cayley Table of $(\mathbb{Z_8}, \times_8)$

$$
U(8)=\{1,3,5,7\}\\
\begin{array}{c|ccc}
\times_8 & 1 & 3 & 5 & 7\\\hline
1 & 1 & 3 & 5 & 7\\
3 & 3 & 1 & 7 & 5\\
5 & 5 & 7 & 1 & 3\\
7 & 7 & 5 & 3 & 1\\
\end{array}
$$

### General Linear Group $GL_n(\mathbb{R})$

:::info
$$
\cases{
M_n(\mathbb{R}) \iff n\times n\ matrix\\
I_n \iff n\times n\ identity\ matrix
}\\
GL_n(\mathbb{R})\iff\forall a\in M_n(\mathbb{R}),\exists b\in M_n(\mathbb{R}), ab=ba=I_n
$$
:::

### Quaternion Group

:::info
$$
1=\pmatrix{1&0\\0&1},I=\pmatrix{0&1\\-1&0}\\
J=\pmatrix{0&i\\i&0},K=\pmatrix{i&0\\0&-i}\\
\begin{array}{l|cccc}
\cdot & 1 & I & J & K\\\hline
1 & 1 & I & J & K\\
I & I & -1 & K & -J\\
J & J & -K & -1 & I\\
K & K & J & -I & -1\\
\end{array}
\iff\cases{
i^2=-1\\
I^2=J^2=K^2=-1\\
IJ=K,JI=-K\\JK=I,KJ=-I\\KI=J,IK=-J\\
}\\
Q_8=\{\pm 1, \pm I, \pm J,\pm K\}
$$
:::

Complex Plane Example:

$$
\mathbb{C^*}=\mathbb{C}\setminus\{0\}\\
\forall z\in\mathbb{C^*},z=a+bi\implies z^{-1}=\dfrac{a-bi}{a^2+b^2}
$$

### Heisenberg Group

:::info
$$
a=\pmatrix{1&x&y\\0&1&z\\0&0&1},
b=\pmatrix{1&x'&y'\\0&1&z'\\0&0&1}
\implies ab=\pmatrix{1&x+x'&y+y'+xz'\\0&1&z+z'\\0&0&1}\\
Identity=\pmatrix{1&0&0\\0&1&0\\0&0&1}\\
Inverse\ of\ a=\pmatrix{1&-x&xz-y\\0&1&-z\\0&0&1}
$$
:::

## Subgroups

### Subgroup of Complex Number under Multiplication
$$
H=\{1,-1,i,-i\}\subset\mathbb{C^*}\\
\begin{array}{c|rrrr}
\cdot & 1 & -1 & i & -i\\\hline
1 & 1 & -1 & i & -i\\
-1 & -1 & 1 & -i & i\\
i & i & -i & -1 & 1\\
-i & -i & i & 1 & -1\\
\end{array}\\
\therefore H\lhd \mathbb{C^*}
$$

### Special Linear Subgroup $SL_n(\mathbb{R})$ of $GL_n(\mathbb{R})$

:::info
$$
SL_n(\mathbb{R})\iff\{x\in GL_n(\mathbb{R})\mid det(x)=1\}
$$
:::

## Cyclic Groups

## Permutation Groups

## Coset and Lagrange's Theorem

### Coset

### Lagrange's Theorem

### Euler's Theorem

:::info
Euler's $\phi$-function
$$
\phi(n):|\{m\mid 1\le m\lt n,gcd(m,n)=1\}|\\
$$
Example
$$
|U(12)| = |\{1,5,7,11\}| = \phi(12) = 4\\
\forall p\in prime, \phi(p) = p-1
$$
:::

###### tags: `math`