# 循環群基本定理

## 定理

1. 子群要求（Every subgroup of cyclic group is cyclic）；
2. 如果${\displaystyle |\langle a\rangle |=n}$，咁${\displaystyle \langle a\rangle }$嘅子群嘅基數係${\displaystyle n}$嘅因數；
3. 對應每一個${\displaystyle n}$嘅正因數${\displaystyle k}$${\displaystyle \langle a\rangle }$就一定會得一個子群，佢嘅基數係${\displaystyle k}$，即係${\displaystyle \langle a^{\frac {n}{k}}\rangle }$

1. 循環群嘅子群要求。
2. 如果${\displaystyle |\langle a\rangle |=n}$同埋${\displaystyle H}$${\displaystyle \langle a\rangle }$子群。用(1)，得知${\displaystyle H=\langle a^{m}\rangle }$${\displaystyle m\in \mathbb {Z} }$係最細正整數令到${\displaystyle a^{m}\in H}$成立。利用(1)嘅證明，${\displaystyle e=b=a^{m}}$，可以得出${\displaystyle n=mq}$
3. ${\displaystyle k}$任何一個${\displaystyle n}$因數。（想證明${\displaystyle \langle a^{\frac {n}{k}}\rangle }$就係唯一一個${\displaystyle \langle a\rangle }$連基數${\displaystyle k}$）利用最大公因數定理${\displaystyle |\langle a^{\frac {n}{k}}\rangle |={\frac {n}{\gcd(n,{\frac {n}{k}})}}={\frac {n}{\frac {n}{k}}}=k}$。設${\displaystyle H}$${\displaystyle \langle a\rangle }$嘅子群。利用(2)，${\displaystyle H=\langle a^{m}\rangle }$${\displaystyle m}$${\displaystyle n}$嘅因數。咁${\displaystyle m=\gcd(m,n)}$同埋${\displaystyle k=|a^{m}|=|a^{\gcd(n,m)}|={\frac {n}{\gcd(n,m)}}={\frac {n}{m}}}$。所以，${\displaystyle m={\frac {n}{k}},H=\langle a^{\frac {n}{k}}\rangle }$

### 例子

{\displaystyle {\begin{aligned}\langle a\rangle &=\{e,a,a^{2},\cdots ,e^{29}\}&\quad {\text{order }}30\\\langle a^{2}\rangle &=\{e,a^{2},a^{4},\cdots ,a^{28}\}&\quad {\text{order }}15\\\langle a^{3}\rangle &=\{e,a^{3},a^{6},\cdots ,a^{27}\}&\quad {\text{order }}10\\\langle a^{5}\rangle &=\{e,a^{5},a^{10},\cdots ,a^{25}\}&\quad {\text{order }}6\\\langle a^{6}\rangle &=\{e,a^{6},a^{12},a^{18},a^{24}\}&\quad {\text{order }}5\\\langle a^{10}\rangle &=\{e,a^{10},a^{20}\}&\quad {\text{order }}3\\\langle a^{15}\rangle &=\{e,a^{15}\}&\quad {\text{order }}2\\\langle a^{30}\rangle &=\{e\}&\quad {\text{order }}1\\\end{aligned}}}

## 指定基數元素數量

${\displaystyle |\langle a\rangle |=d}$