循环群和循环子群
设 G G G x x x G G G G G G H H H H H H x x x x x x
定义
如果群 H H H x ∈ H x \in H x ∈ H H = { x n ∣ n ∈ Z } H = \left\{ {{x}^{n} \mid n \in \mathbb{Z}}\right\} H = { x n ∣ n ∈ Z }
在加法表示法中，如果 H H H H = { n x ∣ n ∈ Z } H = \{ {nx} \mid n \in \mathbb{Z}\} H = { n x ∣ n ∈ Z } H = ⟨ x ⟩ H = \langle x\rangle H = ⟨ x ⟩ H H H x x x x x x H H H H = ⟨ x ⟩ H = \langle x\rangle H = ⟨ x ⟩ H = ⟨ x − 1 ⟩ H = \left\langle {x}^{-1}\right\rangle H = ⟨ x − 1 ⟩ ( x − 1 ) n = x − n {\left( {x}^{-1}\right) }^{n} = {x}^{-n} ( x − 1 ) n = x − n n n n − n - n − n
{ x n ∣ n ∈ Z } = { ( x − 1 ) n ∣ n ∈ Z } . \left\{ {{x}^{n} \mid n \in \mathbb{Z}}\right\} = \left\{ {{\left( {x}^{-1}\right) }^{n} \mid n \in \mathbb{Z}}\right\} . { x n ∣ n ∈ Z } = { ( x − 1 ) n ∣ n ∈ Z } . 我们将很快展示如何确定给定循环群 H H H ⟨ x ⟩ \langle x\rangle ⟨ x ⟩ x x x x x x x x x
示例
(1) 设 G = D 2 n = ⟨ r , s ∣ r n = s 2 = 1 , r s = s r − 1 ⟩ , n ≥ 3 G = {D}_{2n} = \left\langle {r,s \mid {r}^{n} = {s}^{2} = 1,{rs} = s{r}^{-1}}\right\rangle ,n \geq 3 G = D 2 n = ⟨ r , s ∣ r n = s 2 = 1 , rs = s r − 1 ⟩ , n ≥ 3 H H H n n n H = ⟨ r ⟩ H = \langle r\rangle H = ⟨ r ⟩ H H H 1 , r , r 2 , … , r n − 1 1,r,{r}^{2},\ldots ,{r}^{n - 1} 1 , r , r 2 , … , r n − 1 r r r ∣ H ∣ = n \left| H\right| = n ∣ H ∣ = n H H H r r r n n n r r r n n n
r n = 1 , r n + 1 = r , r n + 2 = r 2 , … {r}^{n} = 1,{r}^{n + 1} = r,{r}^{n + 2} = {r}^{2},\ldots r n = 1 , r n + 1 = r , r n + 2 = r 2 , … r − 1 = r n − 1 , r − 2 = r n − 2 , … etc. {r}^{-1} = {r}^{n - 1},{r}^{-2} = {r}^{n - 2},\ldots \text{ etc. } r − 1 = r n − 1 , r − 2 = r n − 2 , … etc. 一般地，为了将 r r r r t {r}^{t} r t r k {r}^{k} r k n − 1 n - 1 n − 1 k k k
t = n q + k , where 0 ≤ k < n , t = {nq} + k,\;\text{ where }0 \leq k < n, t = n q + k , where 0 ≤ k < n , 使得
r t = r n q + k = ( r n ) q r k = 1 q r k = r k . {r}^{t} = {r}^{{nq} + k} = {\left( {r}^{n}\right) }^{q}{r}^{k} = {1}^{q}{r}^{k} = {r}^{k}. r t = r n q + k = ( r n ) q r k = 1 q r k = r k . 例如，在 D 8 , r 4 = 1 {D}_{8},{r}^{4} = 1 D 8 , r 4 = 1 r 105 = r 4 ( 26 ) + 1 = r {r}^{105} = {r}^{4\left( {26}\right) + 1} = r r 105 = r 4 ( 26 ) + 1 = r r − 42 = r 4 ( − 11 ) + 2 = r 2 {r}^{-{42}} = {r}^{4\left( {-{11}}\right) + 2} = {r}^{2} r − 42 = r 4 ( − 11 ) + 2 = r 2 D 2 n {D}_{2n} D 2 n
(2) 设 H = Z H = \mathbb{Z} H = Z H = ⟨ 1 ⟩ H = \langle 1\rangle H = ⟨ 1 ⟩ H H H H H H n ⋅ 1 n \cdot 1 n ⋅ 1 n ∈ Z n \in \mathbb{Z} n ∈ Z H H H ∣ H ∣ \left| H\right| ∣ H ∣ ∞ \infty ∞ H = ⟨ − 1 ⟩ H = \langle - 1\rangle H = ⟨ − 1 ⟩ x x x ( − x ) ( − 1 ) \left( {-x}\right) \left( {-1}\right) ( − x ) ( − 1 )
在进一步讨论循环群之前，我们证明在前面两个例子中观察到的有限循环群和无限循环群的性质是通用的。这个命题也验证了关于元素“阶”的术语使用和符号 ∣ ∣ || ∣∣
命题 2
如果 H = ⟨ x ⟩ H = \langle x\rangle H = ⟨ x ⟩ ∣ H ∣ = ∣ x ∣ \left| H\right| = \left| x\right| ∣ H ∣ = ∣ x ∣
(1) 如果 ∣ H ∣ = n < ∞ \left| H\right| = n < \infty ∣ H ∣ = n < ∞ x n = 1 {x}^{n} = 1 x n = 1 1 , x , x 2 , … , x n − 1 1,x,{x}^{2},\ldots ,{x}^{n - 1} 1 , x , x 2 , … , x n − 1 H H H
(2) 如果 ∣ H ∣ = ∞ \left| H\right| = \infty ∣ H ∣ = ∞ n ≠ 0 n \neq 0 n = 0 x n ≠ 1 {x}^{n} \neq 1 x n = 1 a ≠ b a \neq b a = b Z \mathbb{Z} Z
证明：设 ∣ x ∣ = n \left| x\right| = n ∣ x ∣ = n n < ∞ n < \infty n < ∞ 1 , x , x 2 , … , x n − 1 1,x,{x}^{2},\ldots ,{x}^{n - 1} 1 , x , x 2 , … , x n − 1 x a = x b {x}^{a} = {x}^{b} x a = x b 0 ≤ a < b < n 0 \leq a < b < n 0 ≤ a < b < n x b − a = x 0 = 1 {x}^{b - a} = {x}^{0} = 1 x b − a = x 0 = 1 n n n x x x H H H n n n x t {x}^{t} x t x x x t = n q + k t = {nq} + k t = n q + k 0 ≤ k < n 0 \leq k < n 0 ≤ k < n
x t = x n q + k = ( x n ) q x k = 1 q x k = x k ∈ { 1 , x , x 2 , … , x n − 1 } , {x}^{t} = {x}^{{nq} + k} = {\left( {x}^{n}\right) }^{q}{x}^{k} = {1}^{q}{x}^{k} = {x}^{k} \in \left\{ {1,x,{x}^{2},\ldots ,{x}^{n - 1}}\right\} , x t = x n q + k = ( x n ) q x k = 1 q x k = x k ∈ { 1 , x , x 2 , … , x n − 1 } , 如所期望的。
接下来假设 ∣ x ∣ = ∞ \left| x\right| = \infty ∣ x ∣ = ∞ x x x x a = x b {x}^{a} = {x}^{b} x a = x b a a a b b b a < b a < b a < b x b − a = 1 {x}^{b - a} = 1 x b − a = 1 x x x H H H ∣ H ∣ = ∞ \left| H\right| = \infty ∣ H ∣ = ∞
注意，命题的证明给出了将有限循环群中生成元的任意次幂减少到“最小余数”次幂的方法。生成元的一个循环群 n n n Z / n Z \mathbb{Z}/n\mathbb{Z} Z / n Z
首先我们需要一个简单的命题。
命题 3
设 G G G x ∈ G x \in G x ∈ G m , n ∈ Z m,n \in \mathbb{Z} m , n ∈ Z x n = 1 {x}^{n} = 1 x n = 1 x m = 1 {x}^{m} = 1 x m = 1 x d = 1 {x}^{d} = 1 x d = 1 d = ( m , n ) d = \left( {m,n}\right) d = ( m , n ) m ∈ Z m \in \mathbb{Z} m ∈ Z x m = 1 {x}^{m} = 1 x m = 1 ∣ x ∣ \left| x\right| ∣ x ∣ m m m
证明：根据欧几里得算法，存在整数 r r r s s s d = m r + n s d = {mr} + {ns} d = m r + n s d d d m m m n n n
x d = x m r + n s = ( x m ) r ( x n ) s = 1 r 1 s = 1. {x}^{d} = {x}^{{mr} + {ns}} = {\left( {x}^{m}\right) }^{r}{\left( {x}^{n}\right) }^{s} = {1}^{r}{1}^{s} = 1. x d = x m r + n s = ( x m ) r ( x n ) s = 1 r 1 s = 1. 这证明了第一个断言。
如果 x m = 1 {x}^{m} = 1 x m = 1 n = ∣ x ∣ n = \left| x\right| n = ∣ x ∣ m = 0 m = 0 m = 0 n ∣ m n \mid m n ∣ m m ≠ 0 m \neq 0 m = 0 x x x n < ∞ n < \infty n < ∞ d = ( m , n ) d = \left( {m,n}\right) d = ( m , n ) x d = 1 {x}^{d} = 1 x d = 1 0 < d ≤ n 0 < d \leq n 0 < d ≤ n n n n x x x d = n d = n d = n n ∣ m n \mid m n ∣ m
定理4
任何两个相同阶的循环群都是同构的。更具体地，
(1) 如果 n ∈ Z + n \in {\mathbb{Z}}^{ + } n ∈ Z + ⟨ x ⟩ \langle x\rangle ⟨ x ⟩ ⟨ y ⟩ \langle y\rangle ⟨ y ⟩ n n n
φ : ⟨ x ⟩ → ⟨ y ⟩ \varphi : \langle x\rangle \rightarrow \langle y\rangle φ : ⟨ x ⟩ → ⟨ y ⟩ x k ↦ y k {x}^{k} \mapsto {y}^{k} x k ↦ y k 是良定义的且是一个同构
(2) 如果 ⟨ x ⟩ \langle x\rangle ⟨ x ⟩
φ : Z → ⟨ x ⟩ \varphi : \mathbb{Z} \rightarrow \langle x\rangle φ : Z → ⟨ x ⟩ k ↦ x k k \mapsto {x}^{k} k ↦ x k 是良定义的且是一个同构。
证明：假设 ⟨ x ⟩ \langle x\rangle ⟨ x ⟩ ⟨ y ⟩ \langle y\rangle ⟨ y ⟩ n n n φ : ⟨ x ⟩ → ⟨ y ⟩ \varphi : \langle x\rangle \rightarrow \langle y\rangle φ : ⟨ x ⟩ → ⟨ y ⟩ φ ( x k ) = y k \varphi \left( {x}^{k}\right) = {y}^{k} φ ( x k ) = y k φ \varphi φ
if x r = x s ,then φ ( x r ) = φ ( x s ) . \text{if}\ {x}^{r} = {x}^{s}\text{,then}\ \varphi \left( {x}^{r}\right) = \varphi \left( {x}^{s}\right) \text{.} if x r = x s ,then φ ( x r ) = φ ( x s ) . 由于 x r − s = 1 {x}^{r - s} = 1 x r − s = 1 n ∣ r − s n \mid r - s n ∣ r − s r = t n + s r = {tn} + s r = t n + s
φ ( x r ) = φ ( x t n + s ) \varphi \left( {x}^{r}\right) = \varphi \left( {x}^{{tn} + s}\right) φ ( x r ) = φ ( x t n + s ) = y t n + s = {y}^{{tn} + s} = y t n + s = ( y n ) t y s = {\left( {y}^{n}\right) }^{t}{y}^{s} = ( y n ) t y s = y s = φ ( x s ) . = {y}^{s} = \varphi \left( {x}^{s}\right) \text{.} = y s = φ ( x s ) . 这证明了 φ \varphi φ φ ( x a x b ) = \varphi \left( {{x}^{a}{x}^{b}}\right) = φ ( x a x b ) = φ ( x a ) φ ( x b ) \varphi \left( {x}^{a}\right) \varphi \left( {x}^{b}\right) φ ( x a ) φ ( x b ) φ \varphi φ ⟨ y ⟩ \langle y\rangle ⟨ y ⟩ y k {y}^{k} y k x k {x}^{k} x k φ \varphi φ φ \varphi φ φ \varphi φ
如果 ⟨ x ⟩ \langle x\rangle ⟨ x ⟩ φ : Z → ⟨ x ⟩ \varphi : \mathbb{Z} \rightarrow \langle x\rangle φ : Z → ⟨ x ⟩ φ ( k ) = x k \varphi \left( k\right) = {x}^{k} φ ( k ) = x k x a ≠ x b {x}^{a} \neq {x}^{b} x a = x b a , b ∈ Z a,b \in \mathbb{Z} a , b ∈ Z φ \varphi φ φ \varphi φ φ \varphi φ φ \varphi φ
我们选择使用旋转群 ⟨ r ⟩ \langle r\rangle ⟨ r ⟩ n n n Z / n Z \mathbb{Z}/n\mathbb{Z} Z / n Z
记号：对于每个 n ∈ Z + n \in {\mathbb{Z}}^{ + } n ∈ Z + Z n {Z}_{n} Z n n n n
在同构意义下，Z n {Z}_{n} Z n n n n Z n ≅ Z / n Z {Z}_{n} \cong \mathbb{Z}/n\mathbb{Z} Z n ≅ Z / n Z n n n ⟨ x ⟩ \langle x\rangle ⟨ x ⟩ Z \mathbb{Z} Z
如前所述，一个给定的循环群可能有多个生成元。下面两个命题精确地确定了哪些 x x x ⟨ x ⟩ \langle x\rangle ⟨ x ⟩
命题 5
设 G G G x ∈ G x \in G x ∈ G a ∈ Z − { 0 } a \in \mathbb{Z} - \{ 0\} a ∈ Z − { 0 }
(1) 如果 ∣ x ∣ = ∞ \left| x\right| = \infty ∣ x ∣ = ∞ ∣ x a ∣ = ∞ \left| {x}^{a}\right| = \infty ∣ x a ∣ = ∞
(2) 如果 ∣ x ∣ = n < ∞ \left| x\right| = n < \infty ∣ x ∣ = n < ∞ ∣ x a ∣ = n ( n , a ) \left| {x}^{a}\right| = \dfrac{n}{\left( n,a\right) } ∣ x a ∣ = ( n , a ) n
(3) 特别地，如果 ∣ x ∣ = n < ∞ \left| x\right| = n < \infty ∣ x ∣ = n < ∞ a a a n n n
∣ x a ∣ = n a \left| {x}^{a}\right| = \dfrac{n}{a} ∣ x a ∣ = a n
证明：(1) 通过反证法假设 ∣ x ∣ = ∞ \left| x\right| = \infty ∣ x ∣ = ∞ ∣ x a ∣ = m < ∞ \left| {x}^{a}\right| = m < \infty ∣ x a ∣ = m < ∞
1 = ( x a ) m = x a m . 1 = {\left( {x}^{a}\right) }^{m} = {x}^{am}. 1 = ( x a ) m = x am . 同时，
x − a m = ( x a m ) − 1 = 1 − 1 = 1. {x}^{-{am}} = {\left( {x}^{am}\right) }^{-1} = {1}^{-1} = 1. x − am = ( x am ) − 1 = 1 − 1 = 1. 现在在 a m {am} am − a m - {am} − am a a a m m m x x x ∣ x ∣ = ∞ \left| x\right| = \infty ∣ x ∣ = ∞ ∣ x a ∣ < ∞ \left| {x}^{a}\right| < \infty ∣ x a ∣ < ∞
(2) 在 (2) 的记号下，设
y = x a , ( n , a ) = d and write n = d b , a = d c , y = {x}^{a},\;\left( {n,a}\right) = d\text{ and write }n = {db},a = {dc}, y = x a , ( n , a ) = d and write n = d b , a = d c , 对于适合的 b , c ∈ Z b,c \in \mathbb{Z} b , c ∈ Z b > 0 b > 0 b > 0 d d d n n n a a a b b b c c c
( b , c ) = 1 . \left( {b,c}\right) = 1\text{.} ( b , c ) = 1 . 为了证明 (2)，我们必须证明 ∣ y ∣ = b \left| y\right| = b ∣ y ∣ = b
y b = x a b = x d c b = ( x d b ) c = ( x n ) c = 1 c = 1 {y}^{b} = {x}^{ab} = {x}^{dcb} = {\left( {x}^{db}\right) }^{c} = {\left( {x}^{n}\right) }^{c} = {1}^{c} = 1 y b = x ab = x d c b = ( x d b ) c = ( x n ) c = 1 c = 1 因此，根据命题3应用于 ⟨ y ⟩ \langle y\rangle ⟨ y ⟩ ∣ y ∣ \left| y\right| ∣ y ∣ b b b k = ∣ y ∣ k = \left| y\right| k = ∣ y ∣
x a k = y k = 1 {x}^{ak} = {y}^{k} = 1 x ak = y k = 1 因此，根据命题3应用于 ⟨ x ⟩ , n ∣ a k ,i.e., d b ∣ d c k \langle x\rangle ,n\left| {{ak}\text{,i.e.,}{db}}\right| {dck} ⟨ x ⟩ , n ∣ ak ,i.e., d b ∣ d c k b ∣ c k b \mid {ck} b ∣ c k b b b c c c b b b k k k b b b k k k b = k b = k b = k
(3)这是（2）的一个特例，记录下来供将来参考。
命题6
令 H = ⟨ x ⟩ H = \langle x\rangle H = ⟨ x ⟩
（1）假设 ∣ x ∣ = ∞ \left| x\right| = \infty ∣ x ∣ = ∞ H = ⟨ x a ⟩ H = \left\langle {x}^{a}\right\rangle H = ⟨ x a ⟩ a = ± 1 a = \pm 1 a = ± 1
（2）假设 ∣ x ∣ = n < ∞ \left| x\right| = n < \infty ∣ x ∣ = n < ∞ H = ⟨ x a ⟩ H = \left\langle {x}^{a}\right\rangle H = ⟨ x a ⟩ ( a , n ) = 1 \left( {a,n}\right) = 1 ( a , n ) = 1 H H H φ ( n ) \varphi \left( n\right) φ ( n ) φ \varphi φ
证明：我们将（1）留作练习。在（2）中，如果 ∣ x ∣ = n < ∞ \left| x\right| = n < \infty ∣ x ∣ = n < ∞ x a {x}^{a} x a H H H ∣ x a ∣ \left| {x}^{a}\right| ∣ x a ∣ H H H ∣ x a ∣ = ∣ x ∣ \left| {x}^{a}\right| = \left| x\right| ∣ x a ∣ = ∣ x ∣
∣ x a ∣ = ∣ x ∣ if and only if n ( a , n ) = n , i.e. if and only if ( a , n ) = 1 . \left| {x}^{a}\right| = \left| x\right| \text{if and only if}\frac{n}{\left( a,n\right) } = n\text{,}\;\text{i.e. if and only if}\left( {a,n}\right) = 1\text{.} ∣ x a ∣ = ∣ x ∣ if and only if ( a , n ) n = n , i.e. if and only if ( a , n ) = 1 . 由于 φ ( n ) \varphi \left( n\right) φ ( n ) ( a , n ) = 1 \left( {a,n}\right) = 1 ( a , n ) = 1 a ∈ { 1 , 2 , … , n } a \in \{ 1,2,\ldots ,n\} a ∈ { 1 , 2 , … , n } H H H
示例
命题6确切地指出了哪些模 n n n Z / n Z \mathbb{Z}/n\mathbb{Z} Z / n Z ( a , n ) = 1 \left( {a,n}\right) = 1 ( a , n ) = 1 a ˉ \bar{a} a ˉ Z / n Z \mathbb{Z}/n\mathbb{Z} Z / n Z 1 ‾ , 5 ‾ , 7 ‾ \overline{1},\overline{5},\overline{7} 1 , 5 , 7 11 ‾ \overline{11} 11 Z / 12 Z \mathbb{Z}/{12}\mathbb{Z} Z / 12 Z
本节最后的定理给出了循环群完整的子群结构。
定理 7
设 H = ⟨ x ⟩ H = \langle x\rangle H = ⟨ x ⟩
(1) H H H K ≤ H K \leq H K ≤ H K = { 1 } K = \{ 1\} K = { 1 } K = ⟨ x d ⟩ K = \left\langle {x}^{d}\right\rangle K = ⟨ x d ⟩ d d d x d ∈ K {x}^{d} \in K x d ∈ K
(2) 如果 ∣ H ∣ = ∞ \left| H\right| = \infty ∣ H ∣ = ∞ a a a b , b, b , ⟨ x a ⟩ ≠ ⟨ x b ⟩ \left\langle {x}^{a}\right\rangle \neq\left\langle {x}^{b}\right\rangle ⟨ x a ⟩ = ⟨ x b ⟩ m , ⟨ x m ⟩ = ⟨ x ∣ m ∣ ⟩ m,\left\langle {x}^{m}\right\rangle = \left\langle {x}^{\left| m\right| }\right\rangle m , ⟨ x m ⟩ = ⟨ x ∣ m ∣ ⟩ ∣ m ∣ \left| m\right| ∣ m ∣ m m m H H H 1 , 2 , 3 , … 1,2,3,\ldots 1 , 2 , 3 , …
(3) 如果 ∣ H ∣ = n < ∞ \left| H\right| = n < \infty ∣ H ∣ = n < ∞ n n n a a a H H H ⟨ x d ⟩ \left\langle {x}^{d}\right\rangle ⟨ x d ⟩ d = n a d = \dfrac{n}{a} d = a n m , ⟨ x m ⟩ = ⟨ x ( n , m ) ⟩ m,\left\langle {x}^{m}\right\rangle = \left\langle {x}^{\left( n,m\right) }\right\rangle m , ⟨ x m ⟩ = ⟨ x ( n , m ) ⟩ H H H n n n
证明： (1) 设 K ≤ H K \leq H K ≤ H K = { 1 } K = \{ 1\} K = { 1 } K ≠ { 1 } K \neq \{ 1\} K = { 1 } a ≠ 0 a \neq 0 a = 0 x a ∈ K {x}^{a} \in K x a ∈ K a < 0 a < 0 a < 0 K K K x − a = ( x a ) − 1 ∈ K {x}^{-a} = {\left( {x}^{a}\right) }^{-1} \in K x − a = ( x a ) − 1 ∈ K K K K x x x
P = { b ∣ b ∈ Z + and x b ∈ K } . \mathcal{P} = \left\{ {b \mid b \in {\mathbb{Z}}^{ + }\text{and }{x}^{b} \in K}\right\} . P = { b ∣ b ∈ Z + and x b ∈ K } . 由此，P \mathcal{P} P P \mathcal{P} P d d d K K K x d ∈ K {x}^{d} \in K x d ∈ K ⟨ x d ⟩ ≤ K \left\langle {x}^{d}\right\rangle \leq K ⟨ x d ⟩ ≤ K K K K H H H K K K x a {x}^{a} x a a a a
a = q d + r 0 ≤ r < d . a = {qd} + r\;0 \leq r < d. a = q d + r 0 ≤ r < d . 那么 x r = x ( a − q d ) = x a ( x d ) − q {x}^{r} = {x}^{\left( a - qd\right) } = {x}^{a}{\left( {x}^{d}\right) }^{-q} x r = x ( a − q d ) = x a ( x d ) − q K K K x a {x}^{a} x a x d {x}^{d} x d K K K d d d r = 0 r = 0 r = 0 a = q d a = {qd} a = q d x a = ( x d ) q ∈ ⟨ x d ⟩ {x}^{a} = {\left( {x}^{d}\right) }^{q} \in \left\langle {x}^{d}\right\rangle x a = ( x d ) q ∈ ⟨ x d ⟩ K ≤ ⟨ x d ⟩ K \leq \left\langle {x}^{d}\right\rangle K ≤ ⟨ x d ⟩
我们将（2）的证明留作练习（推理过程与（3）的证明相似且更简单）。
（3）假设 ∣ H ∣ = n < ∞ \left| H\right| = n < \infty ∣ H ∣ = n < ∞ a ∣ n a \mid n a ∣ n d = n a d = \frac{n}{a} d = a n ⟨ x d ⟩ \left\langle {x}^{d}\right\rangle ⟨ x d ⟩ a a a a a a K K K H H H a a a
K = ⟨ x b ⟩ K = \left\langle {x}^{b}\right\rangle K = ⟨ x b ⟩ 其中 b b b x b ∈ K {x}^{b} \in K x b ∈ K
n d = a = ∣ K ∣ = ∣ x b ∣ = n ( n , b ) , \frac{n}{d} = a = \left| K\right| = \left| {x}^{b}\right| = \frac{n}{\left( n,b\right) }, d n = a = ∣ K ∣ = x b = ( n , b ) n , 所以 d = ( n , b ) d = \left( {n,b}\right) d = ( n , b ) d ∣ b d \mid b d ∣ b b b b d , x b ∈ ⟨ x d ⟩ d,{x}^{b} \in \left\langle {x}^{d}\right\rangle d , x b ∈ ⟨ x d ⟩
K = ⟨ x b ⟩ ≤ ⟨ x d ⟩ . K = \left\langle {x}^{b}\right\rangle \leq \left\langle {x}^{d}\right\rangle . K = ⟨ x b ⟩ ≤ ⟨ x d ⟩ . 由于 ∣ ⟨ x d ⟩ ∣ = a = ∣ K ∣ \left| \left\langle {x}^{d}\right\rangle \right| = a = \left| K\right| ⟨ x d ⟩ = a = ∣ K ∣ K = ⟨ x d ⟩ K = \left\langle {x}^{d}\right\rangle K = ⟨ x d ⟩
(3)的最终断言来自于观察 ⟨ x m ⟩ \left\langle {x}^{m}\right\rangle ⟨ x m ⟩ ⟨ x ( n , m ) ⟩ \left\langle {x}^{\left( n,m\right) }\right\rangle ⟨ x ( n , m ) ⟩ n n n H H H n n n
示例
(1) 我们可以使用命题6和定理7来列出任意给定 n n n Z / n Z \mathbb{Z}/n\mathbb{Z} Z / n Z Z / 12 Z \mathbb{Z}/{12}\mathbb{Z} Z / 12 Z
(a) Z / 12 Z = ⟨ 1 ˉ ⟩ = ⟨ 5 ˉ ⟩ = ⟨ 7 ˉ ⟩ = ⟨ 11 ‾ ⟩ \mathbb{Z}/{12}\mathbb{Z} = \langle \bar{1}\rangle = \langle \bar{5}\rangle = \langle \bar{7}\rangle = \langle \overline{11}\rangle Z / 12 Z = ⟨ 1 ˉ ⟩ = ⟨ 5 ˉ ⟩ = ⟨ 7 ˉ ⟩ = ⟨ 11 ⟩
(b) ⟨ 2 ˉ ⟩ = ⟨ 10 ‾ ⟩ \langle \bar{2}\rangle = \langle \overline{10}\rangle ⟨ 2 ˉ ⟩ = ⟨ 10 ⟩
(c) ⟨ 3 ‾ ⟩ = ⟨ 9 ‾ ⟩ \langle \overline{3}\rangle = \langle \overline{9}\rangle ⟨ 3 ⟩ = ⟨ 9 ⟩
(d) ⟨ 4 ˉ ⟩ = ⟨ 8 ˉ ⟩ \langle \bar{4}\rangle = \langle \bar{8}\rangle ⟨ 4 ˉ ⟩ = ⟨ 8 ˉ ⟩
(e) ⟨ 6 ‾ ⟩ \langle \overline{6}\rangle ⟨ 6 ⟩
(f) ⟨ 0 ‾ ⟩ \langle \overline{0}\rangle ⟨ 0 ⟩
它们之间的包含关系如下：
⟨ a ˉ ⟩ ≤ ⟨ b ˉ ⟩ if and only if ( b , 12 ) ∣ ( a , 12 ) , 1 ≤ a , b ≤ 12. \langle \bar{a}\rangle \leq \langle \bar{b}\rangle \;\text{ if and only if }\left( {b,{12}}\right) \mid \left( {a,{12}}\right) ,\;1 \leq a,b \leq {12}. ⟨ a ˉ ⟩ ≤ ⟨ b ˉ ⟩ if and only if ( b , 12 ) ∣ ( a , 12 ) , 1 ≤ a , b ≤ 12 . (2) 我们还可以将本节的结果与前一节的结果结合起来。例如，我们可以通过构造 C G ( ⟨ x ⟩ ) {C}_{G}\left( {\langle x\rangle }\right) C G ( ⟨ x ⟩ ) N G ( ⟨ x ⟩ ) {N}_{G}\left( {\langle x\rangle }\right) N G ( ⟨ x ⟩ ) G G G x ∈ G x \in G x ∈ G G G G g g g x x x g g g x x x
C G ( ⟨ x ⟩ ) = C G ( x ) {C}_{G}\left( {\langle x\rangle }\right) = {C}_{G}\left( x\right) C G ( ⟨ x ⟩ ) = C G ( x ) 如第2节练习6所述， ⟨ x ⟩ ≤ N G ( ⟨ x ⟩ ) \langle x\rangle \leq {N}_{G}\left( {\langle x\rangle }\right) ⟨ x ⟩ ≤ N G ( ⟨ x ⟩ ) G = Q 8 G = {Q}_{8} G = Q 8 x = i x = i x = i
C G ( ⟨ i ⟩ ) = { ± 1 , ± i } = ⟨ i ⟩ and N G ( ⟨ i ⟩ ) = Q 8 . {C}_{G}\left( {\langle i\rangle }\right) = \{ \pm 1, \pm i\} = \langle i\rangle \text{ and }{N}_{G}\left( {\langle i\rangle }\right) = {Q}_{8}. C G ( ⟨ i ⟩ ) = { ± 1 , ± i } = ⟨ i ⟩ and N G ( ⟨ i ⟩ ) = Q 8 . 注意，我们已经观察到上述两个等式中的第一个，而第二个等式最容易通过使用随后的练习24的结果来计算。
