设 < G , ⋅ > \left
易证 ( a ) ≤ G \left(a\right) \le G (a)≤G, ( a ) \left(a\right) (a)称为由 a a a生成的子群
特别地,当 G = ( a ) G = \left(a\right) G=(a)时, G G G为一个具有特殊结构的群
设 < G , ⋅ > \left
例子:
(1) < I m , + m > \left<\mathbb{I}_m, +_m\right> ⟨Im,+m⟩是 m m m阶循环群,生成元是 1 1 1
(2) < I , + > \left<\mathbb{I},+\right> ⟨I,+⟩是无限阶循环群,生成元是 1 1 1和 − 1 -1 −1
定理1:每个循环群都是可交换的
定理2:设 G = ( a ) G=\left(a\right) G=(a)
(1)若 ∣ a ∣ \left|a\right| ∣a∣无限,则 G ≅ < I , + > G\cong \left<\mathbb{I}, +\right> G≅⟨I,+⟩
(2)若 ∣ a ∣ = n ∈ I + \left|a\right| = n \in \mathbb{I}_+ ∣a∣=n∈I+,则 G ≅ < N n , + n > G\cong \left<\mathbb{N}_n, +_n\right> G≅⟨Nn,+n⟩
证明:(1)设 ∣ a ∣ \left|a\right| ∣a∣无限,作 h : G → I , a i ↦ i h:G\to \mathbb{I}, a^i \mapsto i h:G→I,ai↦i
1. h h h是良定的,若 a i = a j a^i=a^j ai=aj,则 a i − j = e a^{i-j}=e ai−j=e,因为 ∣ a ∣ \left|a\right| ∣a∣无限,所以 i − j = 0 , i = j i-j=0,i=j i−j=0,i=j
2.显然 h h h双射
3. h h h同态, ∀ a i , a j ∈ G , h ( a i ⋅ a j ) = h ( a i + j ) = i + j = h ( a i ) + h ( a j ) \forall a^i,a^j\in G, h\left(a^i\cdot a^j\right) = h\left(a^{i+j}\right) = i+j=h\left(a^i\right) + h\left(a^j\right) ∀ai,aj∈G,h(ai⋅aj)=h(ai+j)=i+j=h(ai)+h(aj)
综上所述, h h h是同构, G ≅ < I , + > G\cong \left<\mathbb{I}, +\right> G≅⟨I,+⟩
(2)设 ∣ a ∣ = n \left|a\right|=n ∣a∣=n,作 h : G → N n , a i ↦ i m o d n h:G\to \mathbb{N}_n, a^i\mapsto i\mod n h:G→Nn,ai↦imodn
1. h h h是良定的,若 a I = a j a^I=a^j aI=aj则 a i − j = e , n ∣ ( i − j ) a^{i-j}=e, n\mid \left(i-j\right) ai−j=e,n∣(i−j),所以 i m o d n = j m o d n i \mod n = j \mod n imodn=jmodn,即 h ( a i ) = h ( a j ) h\left(a^i\right)=h\left(a^j\right) h(ai)=h(aj)
2. h h h是单射,若 h ( a I ) = h ( a j ) h\left(a^I\right) = h\left(a^j\right) h(aI)=h(aj)即 i m o d n = j m o d n i\mod n=j\mod n imodn=jmodn,则 n ∣ ( i − j ) n\mid \left(i-j\right) n∣(i−j),所以 a i − j = e a^{i-j}=e ai−j=e,故 a i = a j a^i=a^j ai=aj
3. h h h是满射, ∀ i ∈ N n , a i ∈ G , h ( a I ) = i m o d n = i \forall i \in\mathbb{N}_n, a^i\in G, h\left(a^I\right) = i\mod n=i ∀i∈Nn,ai∈G,h(aI)=imodn=i
4. h h h是同态, ∀ a i , a j ∈ G \forall a^i,a^j\in G ∀ai,aj∈G
h ( a i ⋅ a j ) = h ( a i + j ) = ( i + j ) m o d n = ( i m o d n ) + n ( j m o d n ) = h ( a i ) + n h ( a j ) \begin{aligned} h\left(a^i\cdot a^j\right) &= h\left(a^{i+j}\right)\\ &=\left(i+j\right)\mod n\\ &=\left(i\mod n\right) +_n \left(j\mod n\right)\\ &=h\left(a^{i}\right) +_nh\left(a^j\right) \end{aligned} h(ai⋅aj)=h(ai+j)=(i+j)modn=(imodn)+n(jmodn)=h(ai)+nh(aj)
综上所述, h h h是同构, G ≅ < N n , + n > G\cong \left<\mathbb{N}_n,+_n\right> G≅⟨Nn,+n⟩
由本定理,同阶的循环群必同构,因此常把 n n n阶循环群记为 C n C_n Cn
推论1:设 G = ( a ) G=\left(a\right) G=(a)
(1)若 G G G为无限群,则 ∣ a ∣ \left|a\right| ∣a∣无限,且 G = { ⋯ , a − 2 , a − 1 , e , a , a 2 , ⋯ } G=\left\{\cdots,a^{-2},a^{-1},e,a,a^2,\cdots\right\} G={⋯,a−2,a−1,e,a,a2,⋯}
(2)若 ∣ G ∣ = n ∈ I + \left|G\right|=n\in\mathbb{I}_+ ∣G∣=n∈I+,则 ∣ a ∣ = n \left|a\right|=n ∣a∣=n,且 G = { e , a , a 2 , ⋯ , a n − 1 } G=\left\{e,a,a^2,\cdots,a^{n-1}\right\} G={e,a,a2,⋯,an−1}
推论2:设 G G G为 n n n阶有限群, a ∈ G a\in G a∈G,则 G = ( a ) G=\left(a\right) G=(a)当且仅当 ∣ a ∣ = n \left|a\right|=n ∣a∣=n
定理3:设群 G = ( a ) G=\left(a\right) G=(a)
(1)若 G G G为无限群,则 G G G只有两个生成元 a a a和 a − 1 a^{-1} a−1
(2)若 ∣ G ∣ = n ∈ I + \left|G\right| = n\in \mathbb{I}_+ ∣G∣=n∈I+,则 G = ( a r ) G=\left(a^r\right) G=(ar)当且仅当 ( r , n ) = 1 \left(r,n\right)=1 (r,n)=1,即生成元有 ϕ ( n ) \phi\left(n\right) ϕ(n)个
其中 ϕ ( n ) = ∣ { r < n ∣ ( r , n ) = 1 } ∣ \phi\left(n\right) = \left|\left\{r
证明:
(1)容易验证 a − 1 a^{-1} a−1是生成元
设 a m ∈ G a^m\in G am∈G是生成元,即 G = ( a m ) G=\left(a^m\right) G=(am)
因为 a ∈ G a\in G a∈G,所以 ∃ t ∈ I \exists t \in \mathbb{I} ∃t∈I使得 a = ( a m ) t a=\left(a^{m}\right)^t a=(am)t,所以 a m t − 1 = e a^{mt-1}=e amt−1=e
因为 G G G为无限群,所以 ∣ a ∣ \left|a\right| ∣a∣无限,所以 m t − 1 = 0 mt-1=0 mt−1=0,故 m = t = 1 m=t=1 m=t=1或 m = t = − 1 m=t=-1 m=t=−1
(2)必要性,设 G = ( a r ) G=\left(a^r\right) G=(ar),则有 ∣ a r ∣ = n \left|a^r\right|=n ∣ar∣=n
另外由元素的阶-定理2, ∣ a r ∣ = n ( r , n ) \left|a^r\right| = \frac{n}{\left(r,n\right)} ∣ar∣=(r,n)n,故 ( r , n ) = 1 \left(r,n\right)=1 (r,n)=1
充分性:设 ( r , n ) = 1 \left(r,n\right)=1 (r,n)=1,则由扩展欧几里得, ∃ s , t ∈ I \exists s,t\in\mathbb{I} ∃s,t∈I, 使 r s + n t = 1 rs+nt=1 rs+nt=1
于是 a = a r s + n t = ( a r ) s ⋅ ( a n ) t a=a^{rs+nt}=\left(a^{r}\right)^{s}\cdot \left(a^n\right)^t a=ars+nt=(ar)s⋅(an)t
因为 ∣ a ∣ = ∣ G ∣ = n \left|a\right| = \left|G\right|=n ∣a∣=∣G∣=n,所以 a = ( a r ) s a=\left(a^r\right)^s a=(ar)s,故 G = ( a r ) G=\left(a^r\right) G=(ar)
因此 G = ( a r ) G=\left(a^r\right) G=(ar)当且仅当 ( r , n ) = 1 \left(r,n\right)=1 (r,n)=1,又 G = { e , a , a 2 , ⋯ , a n − 1 } G=\left\{e,a,a^2,\cdots, a^{n-1}\right\} G={e,a,a2,⋯,an−1}
故 G G G的生成元又 ϕ ( n ) \phi\left(n\right) ϕ(n)个
定理4:设群 G = ( a ) , { e } ≠ H ≤ G G=\left(a\right), \left\{e\right\} \neq H \le G G=(a),{e}=H≤G, a m a^m am是 H H H中 a a a的最小正幂
(1) H = ( a m ) H=\left(a^m\right) H=(am)
(2)若 G G G为无限群,则 H H H为无限群
(3)若 ∣ G ∣ = n ∈ I + \left|G\right|=n\in \mathbb{I}_+ ∣G∣=n∈I+,则 m ∣ n m\mid n m∣n且 ∣ H ∣ = n m \left|H\right|=\frac{n}{m} ∣H∣=mn
证明:
(1) ∀ a i ∈ H \forall a^i \in H ∀ai∈H,设 i = q m + r , ( 0 ≤ r < m ) i=qm + r,\left(0\le r < m\right) i=qm+r,(0≤r<m),则 a r = a i ⋅ ( a m ) − q a^r = a^{i}\cdot \left(a^m\right)^{-q} ar=ai⋅(am)−q
因为 a i , a m ∈ H a^i,a^m\in H ai,am∈H,所以 a r ∈ H a^r \in H ar∈H,但是 a m a^m am是 H H H中 a a a的最小正幂,所以 r = 0 r=0 r=0
a i = ( a m ) q a^i=\left(a^m\right)^q ai=(am)q,故 H = ( a m ) H=\left(a^m\right) H=(am)
(2)若 G G G为无限群,则 ∣ a ∣ \left|a\right| ∣a∣无限,所以 ∣ a m ∣ \left|a^m\right| ∣am∣无限,又因为 H = ( a m ) H=\left(a^m\right) H=(am),故 H H H为无限群
(3)若 ∣ G ∣ = n \left|G\right|=n ∣G∣=n,则 ∣ a ∣ = n \left|a\right|=n ∣a∣=n,所以 a n = e ∈ H a^n=e\in H an=e∈H
因为 H = ( a m ) H=\left(a^m\right) H=(am),故 m ∣ n m\mid n m∣n
且 ∣ H ∣ = ∣ a m ∣ = n ( m , n ) = n m \left|H\right|=\left|a^m\right|=\frac{n}{\left(m,n\right)}=\frac{n}{m} ∣H∣=∣am∣=(m,n)n=mn
定理5:设群 G = ( a ) , ∣ G ∣ = n G=\left(a\right),\left|G\right|=n G=(a),∣G∣=n,则对于 n n n的每个正因子 d d d,有且仅有一 d d d阶子群
因此, n n n阶循环群的子群个数恰为 n n n的正因子的个数
证明: H = ( a n d ) H=\left(a^{\frac{n}{d}}\right) H=(adn)是一 d d d阶子群
设 H ′ = ( a m ) H^{\prime}=\left(a^m\right) H′=(am)也是 d d d阶子群,则 ∣ a m ∣ = d , a m d = e \left|a^m\right|=d,a^{md}=e ∣am∣=d,amd=e,所以
n ∣ m d , n d ∣ m n\mid md,\quad \frac{n}{d} \mid m n∣md,dn∣m
所以 a m ∈ H a^{m}\in H am∈H,从而 H ′ ⊆ H H^{\prime}\subseteq H H′⊆H,又 ∣ H ′ ∣ = ∣ H ∣ = d \left|H^{\prime}\right|=\left|H\right|=d ∣H′∣=∣H∣=d,故 H ′ = H H^{\prime}=H H′=H
给定一个结合 A A A, < A A , ∘ > \left ⟨AA,∘⟩是独异点,其中 ∘ \circ ∘是函数合成运算
令 P A P_A PA为 A A A到 A A A的所有双射的集合,则 < P A , ∘ > \left
定义:设 A A A为集合,群 < P A , ∘ > \left
Cayley定理:任意一个群都与某个变换群同构
证明:设 < G , ∗ > \left
(1) f a f_a fa是双射
∀ x , y ∈ G \forall x,y\in G ∀x,y∈G,若 f a ( x ) = f a ( y ) f_a\left(x\right) = f_a\left(y\right) fa(x)=fa(y),即 a ∗ x = a ∗ y a*x=a*y a∗x=a∗y,由消去律 x = y x=y x=y
∀ y ∈ G , f a ( a − 1 ∗ y ) = a ∗ a − 1 ∗ y = y \forall y \in G, f_a\left(a^{-1} * y\right) = a*a^{-1} * y = y ∀y∈G,fa(a−1∗y)=a∗a−1∗y=y,故 f a f_a fa是满射
(2)令 G ′ = { f a ∣ a ∈ G } G^{\prime}=\left\{f_a|a\in G\right\} G′={fa∣a∈G},则 ∅ ≠ G ′ ⊆ P G \empty \neq G^{\prime} \subseteq P_G ∅=G′⊆PG, ∀ f a , f b ∈ G ′ , ∀ x ∈ G \forall f_a, f_b \in G^{\prime},\forall x \in G ∀fa,fb∈G′,∀x∈G
f a ∘ f b ( x ) = f a ( f b ( x ) ) = a ∗ ( b ∗ x ) = ( a ∗ b ) ∗ x f_a \circ f_b\left(x\right) = f_a\left(f_b\left(x\right)\right) = a * \left(b*x\right) = \left(a*b\right) * x fa∘fb(x)=fa(fb(x))=a∗(b∗x)=(a∗b)∗x
故 f a ∘ f b = f a ∗ b ∈ G ′ f_a\circ f_b = f_{a*b} \in G^{\prime} fa∘fb=fa∗b∈G′。 ∀ f a ∈ G ′ , ( f a ) − 1 = f a − 1 ∈ G ′ \forall f_a\in G^{\prime}, \left(f_a\right)^{-1} = f_{a^{-1}} \in G^{\prime} ∀fa∈G′,(fa)−1=fa−1∈G′
故 G ′ ≤ P G G^{\prime}\le P_G G′≤PG, < G ′ , ∘ > \left
(3)作 ϕ : G → G p r i m e , a ↦ f a \phi:G\to G^{prime}, a\mapsto f_a ϕ:G→Gprime,a↦fa
1. ϕ \phi ϕ是单射, ∀ a , b ∈ G \forall a,b\in G ∀a,b∈G,若 ϕ ( a ) = ϕ ( b ) \phi\left(a\right) = \phi\left(b\right) ϕ(a)=ϕ(b),即 f a = f b f_a=f_b fa=fb,则 f a ( e ) = f b ( e ) f_a\left(e\right) = f_b\left(e\right) fa(e)=fb(e),即 a ∗ e = b ∗ e a*e = b*e a∗e=b∗e,所以 a = b a=b a=b
2.显然 ϕ \phi ϕ满射
3. ϕ \phi ϕ是同态。 ∀ a , b ∈ G , ϕ ( a ∗ b ) = f a ∗ b = f a ∘ f b = ϕ ( a ) ∘ ϕ ( b ) \forall a,b\in G, \phi\left(a*b\right) = f_{a*b}=f_a\circ f_b =\phi\left(a\right)\circ \phi\left(b\right) ∀a,b∈G,ϕ(a∗b)=fa∗b=fa∘fb=ϕ(a)∘ϕ(b)
因此 ϕ \phi ϕ是 G G G到 G ′ G^{\prime} G′的同构,故 < G , ∗ > ≅ < G ′ , ∘ > \left
定义:有限集合 S S S到自身的双射称为 S S S上的置换, ∣ S ∣ \left|S\right| ∣S∣称为置换的阶
S S S中的元素有时亦称为文字
定义:一个包含 n n n个元素的集合上的所有置换在合成运算下构成的群称为 n n n次对称群,记作 S n S_n Sn
S n S_n Sn的含有 n n n个元素的子群称为 n n n次置换群
容易证明 ∣ S n ∣ = n ! \left|S_n\right|=n! ∣Sn∣=n!
设 S = { 1 , 2 , ⋯ , n } S=\left\{1,2,\cdots, n\right\} S={1,2,⋯,n},则 π ∈ S n \pi \in S_n π∈Sn常表示为
π = ( 1 2 ⋯ n π ( 1 ) π ( 2 ) ⋯ π ( n ) ) \pi = \begin{pmatrix} 1& 2&\cdots&n\\ \pi\left(1\right)&\pi\left(2\right)&\cdots &\pi\left(n\right) \end{pmatrix} π=(1π(1)2π(2)⋯⋯nπ(n))
Cayley定理-推论:任意一个有限群都与某个置换群同构
定义:把 S S S中的元素 i 1 i_1 i1变为 i 2 i_2 i2, i 2 i_2 i2变为 i 3 i_3 i3… i k i_k ik变为 i 1 i_1 i1,并使 S S S中的其余元素保持不变的置换称为循环,也成为轮换,记为 ( i 1 i 2 ⋯ i k ) \left(i_1\ i_2\ \cdots\ i_k\right) (i1 i2 ⋯ ik), k k k称为循环长度,特别地,长度为2的循环称为对换
定理:(1)任一置换可表示成若干个无公共元素的循环之积
(2)任一置换可表示成若干对换之积,且对换个数的奇偶性不变
定义:若置换 π \pi π可表示为奇数个对换的积,则称 π \pi π为奇置换,否则称为偶置换
定理:令 A n A_n An为 S n S_n Sn中所有偶置换的集合,则 A n ≤ S n A_n\le S_n An≤Sn,称为 n n n次交代群,且 ∣ A n ∣ = n ! 2 \left|A_n\right|=\frac{n!}{2} ∣An∣=2n!
证明:群的有限子群的判定只需要考虑封闭性,由于偶置换的合成仍然是偶置换,故 A n ≤ S n A_n\le S_n An≤Sn,作
f : A n → S n − A n , π ↦ ( 1 2 ) π ; g : S n − A n → A n , π ↦ ( 1 2 ) π f:A_n\to S_n - A_n,\quad \pi \mapsto\left(1\ 2\right)\pi;\\ g:S_n-A_n\to A_n,\quad \pi \mapsto \left(1\ 2\right)\pi f:An→Sn−An,π↦(1 2)π;g:Sn−An→An,π↦(1 2)π
显然 f f f和 g g g都是单射,故 ∣ A n ∣ = ∣ S n − A n ∣ = n ! 2 \left|A_n\right|=\left|S_n-A_n\right|=\frac{n!}{2} ∣An∣=∣Sn−An∣=2n!
参考:
离散数学(刘玉珍)