---
title: Group(群) 定義
tags: crypto
lang: zh_tw
---

* [筆記總覽](https://hackmd.io/@LJP/rkerFdnqS)

[TOC]

# 符號定義
- $Z = \{ … , −3, −2, −1, 0, 1, 2, 3, … \}$ of integers
- $Z_n = \{0, 1, . . . , n – 1\}$
- $Z_n^∗ = \{1,2,3, … , n − 1\}$
- $N = \{0, 1, 2, 3, … \}$ of non-negative integers
- $P = \{1, 2, 3, … \}$ of positive integers
- Group(群)
- Ring(環)
- Field(體)

# Group(群) 定義
A group $(G,\ \bullet)$ is a set $G$ with an operation $\bullet$, such that the following conditions are satisfied:
- Closure
- $a\ \bullet\ b\in G$ for all $a,\ b\in G$
- Associativity
- $a\ \bullet(b\ \bullet\ c)\ =\ (a\ \bullet\ b)\ \bullet\ c$ for all $a,\ b,\ c\in G$
- Identity
- there is an element $e\in G$ such that $a\ =\ a\ \bullet\ e\ =\ e\ \bullet\ a$
- Inverse
- for each $a\in G$, there is an element $b\in G$ such that $a\ \bullet\ b\ =\ b\ \bullet\ a\ =\ e$

$\star$ If commutative $a\ \bullet\ b\ =\ b\ \bullet\ a$
- then forms an **abelian group**

## 例子
- $Z,\ Q,\ R,\ C$ with $+$
- 符合 Closure, 怎麼加都在 set 裡面
- 符合 Associativity, 先結合哪一個沒差
- 符合 Identity, 加 0 都還是自己
- 符合 Inverse, 乘上負號就能找到反元素

- $Q^*,\ R^*,\ C^*$ with $\times$
- 符合 Closure, 怎麼乘都在 set 裡面
- 符合 Associativity, 先結合哪一個沒差
- 符合 Identity, 乘 1 都還是自己
- 符合 Inverse, 自己的倒數就是反元素

- $\{1, -1\}$ with $\times$
- $Z_6$ with $+\ mod\ 6$

## Proposition
- $(a\ \times\ b)^{-1}=\ b^{-1}\ \times\ a^{-1}$
- Proof:
$\begin{split}(a\ \times\ b)\times(a\ \times\ b)^{-1}&=1\\
a^{-1}\times a\ \times\ b\times(a\ \times\ b)^{-1}&=a^{-1}\\
b\times(a\ \times\ b)^{-1}&=a^{-1}\\
b^{-1}\times b\times(a\ \times\ b)^{-1}&=b^{-1}\times a^{-1}\\
(a\ \times\ b)^{-1}&=b^{-1}\times a^{-1}
\end{split}$

- $(a^{-1})^{-1}=a$
- Proof:
$\begin{split}(a^{-1})\times(a^{-1})^{-1}&=1\\
a\times(a^{-1})\times(a^{-1})^{-1}&=a\\
(a^{-1})^{-1}&=a\\
\end{split}$