# 循環群

${\displaystyle 1^{n}=1+1+1+\cdots +1}$
${\displaystyle n}$咁多次。如果${\displaystyle n}$係負數，
${\displaystyle (-1)+(-1)+\cdots +(-1)}$
${\displaystyle |n|}$咁多次。

## 子群定理

${\displaystyle a^{n},a^{m}\in \langle a\rangle }$

${\displaystyle a^{n}(a^{m})^{-1}=a^{n-m}\in \langle a\rangle }$ ，利用一步子群要求${\displaystyle \langle a\rangle }$ ${\displaystyle G}$ 子群

### 例子一

${\displaystyle U(10)}$ ，係${\displaystyle 10}$ 嘅環單元（Units），${\displaystyle \langle 3\rangle :=\{3,9,7,1\}=U(10)}$

{\displaystyle {\begin{aligned}3^{1}&=3\\3^{2}&=9\\3^{3}&=27=7\\3^{4}&=81=1\\3^{5}&=243=3\\\vdots &=\vdots \end{aligned}}}

${\displaystyle 3\times 7=21=1}$ ，所以${\displaystyle 3^{-1}=7}$

### 例子二

${\displaystyle \mathbb {Z} _{10}}$ ${\displaystyle \langle 2\rangle =\{0,2,4,6,8\}}$

### 例子三

${\displaystyle \mathbb {Z} }$ ${\displaystyle \langle -1\rangle =\mathbb {Z} }$ ${\displaystyle -2(-1),-1(-1),0(-1),1(-1),\cdots }$ ${\displaystyle \mathbb {Z} }$ 入面每一粒嘢。

## 非子群例子

### 例子一

${\displaystyle \mathbb {Z} _{n}=\{0,1,2,3,\cdots ,n\}}$ ，對應${\displaystyle n\geq 1}$ 。係一個循環群。${\displaystyle 1}$ ${\displaystyle -1=n-1}$ 都係整群出嚟嘅嘢。

### 例子二

${\displaystyle \mathbb {Z} _{10}=\{0,1,2,3,4,5,6,7,8,9\}}$ 。可以得知，${\displaystyle \mathbb {Z} _{10}=\langle 1\rangle =\langle 3\rangle =\langle 7\rangle =\langle 9\rangle }$

${\displaystyle \langle 1\rangle =\{0,1,2,3,4,5,6,7,8,9,10=0,\cdots \}}$

${\displaystyle \langle 3\rangle =\{3,6,9,12=2,15,=5,18=8,21=1,24=4,27=7,30=0,33=3,\cdots \}}$

${\displaystyle \langle 7\rangle =\{7,14=4,21=1,28=8,35=5,42=2,49=9,56=6,63=3,70=0,77=7,\cdots \}}$

${\displaystyle \langle 9\rangle =\{9,18=8,27=7,36=6,45=5,54=4,63=3,72=2,81=1,90=0,99=9,\cdots \}}$

## 循環子群定理

${\displaystyle G}$ 係循環群，即係${\displaystyle G=\langle a\rangle }$ ${\displaystyle H}$ 就係${\displaystyle G}$ 嘅子群。

${\displaystyle a^{n}\in H}$ ${\displaystyle n\in \mathbb {Z} }$ 係整數。

${\displaystyle m\in \mathbb {Z} ^{+}}$ 係最細整數，令到${\displaystyle a^{m}\in H}$ ，同埋${\displaystyle b\in H}$ ${\displaystyle H}$ 入面是但一粒嘢。

${\displaystyle a^{n}=a^{mq+r}}$

${\displaystyle a^{r}=(a^{m})^{-q}a^{n}}$

${\displaystyle \implies b\in \langle a^{m}\rangle }$

## 相等元素要求

${\displaystyle a^{m}=a^{nq+r}=(a^{n})^{q}a^{r}=ea^{r}=a^{r}}$

${\displaystyle a^{i-j}=e=a^{qn+r}=e^{q}a^{r}=a^{r}}$

${\displaystyle a^{i-j}=a^{nq}=e^{q}=e}$ ，所以${\displaystyle a^{i}=a^{j}}$

### 推論二

${\displaystyle G}$ 係一個群，${\displaystyle a\in G}$ 同埋佢嘅基數係${\displaystyle k}$

## 最大公因數定理

${\displaystyle G}$ 係一個群，${\displaystyle a\in G}$ 同埋佢嘅基數係${\displaystyle n}$ ${\displaystyle k}$ 係正整數

${\displaystyle \langle a^{k}\rangle =\langle a^{\gcd(n,k)}\rangle }$ ${\displaystyle |a^{k}|={\frac {n}{\gcd(n,k)}}}$

${\displaystyle d=\gcd(n,k)}$ 。同埋，${\displaystyle k=dr}$

${\displaystyle \implies \langle a^{d}\rangle \subseteq \langle a^{k}\rangle }$

（想要證明${\displaystyle |a^{d}|={\frac {n}{d}}}$

${\displaystyle (a^{d})^{\frac {n}{d}}=a^{n}=e}$ ，所以${\displaystyle |a^{d}|\leq {\frac {n}{d}}}$

### 推論二

${\displaystyle |a|=n}$

${\displaystyle \langle a^{i}\rangle =\langle a^{j}\rangle \iff \gcd(n,i)=\gcd(n,j)}$ ，同埋${\displaystyle |a^{i}|=|a^{j}|\iff \gcd(n,i)=\gcd(n,j)}$

### 推論三

${\displaystyle |a|=n}$

${\displaystyle \langle a\rangle =\langle a^{j}\rangle \iff \gcd(n,j)=1}$ ${\displaystyle |a|=|\langle a^{j}\rangle |\iff \gcd(n,j)=1}$

### 推論四

${\displaystyle \mathbb {Z} _{n}}$ 入面嘅整數${\displaystyle k}$ ，佢係整${\displaystyle \mathbb {Z} _{n}}$ 出嚟${\displaystyle \iff \gcd(n,k)=1}$