不变子群、商群与群同态基本定理

不变子群、商群与群同态基本定理

不变子群

定义:设 H ≤ G H\le G HG,若 ∀ a ∈ G , a ⋅ H = H ⋅ a \forall a\in G, a\cdot H = H\cdot a aG,aH=Ha,则称 H H H G G G不变子群,也称正规子群,记作 H ◃ G H \triangleleft G HG

定理1:设 H ≤ G H\le G HG,则下列条件等价:

(1) H ◃ G H\triangleleft G HG

(2) ∀ a ∈ G , a ⋅ H ⋅ a − 1 = H \forall a \in G, a\cdot H\cdot a^{-1}=H aG,aHa1=H

(3) ∀ a ∈ G , a ⋅ H ⋅ a − 1 ⊆ H \forall a\in G,a\cdot H \cdot a^{-1}\subseteq H aG,aHa1H
(4) ∀ a ∈ G , ∀ h ∈ H , a ⋅ h ⋅ a − 1 ∈ H \forall a \in G,\forall h \in H, a\cdot h\cdot a^{-1}\in H aG,hH,aha1H

证明:

( 1 ) ⇒ ( 2 ) (1)\Rightarrow(2) (1)(2) H ◃ G H\triangleleft G HG,则 ∀ a ∈ G \forall a\in G aG,有 a ⋅ H = H ⋅ a a\cdot H = H\cdot a aH=Ha,所以
a ⋅ H ⋅ a − 1 = ( H ⋅ a ) ⋅ a − 1 = H ⋅ ( a ⋅ a − 1 ) = H ⋅ e = H a\cdot H\cdot a^{-1}=\left(H\cdot a\right)\cdot a^{-1}=H\cdot \left(a\cdot a^{-1}\right)=H\cdot e = H aHa1=(Ha)a1=H(aa1)=He=H
( 2 ) ⇒ ( 3 ) (2)\Rightarrow(3) (2)(3),和 ( 3 ) ⇒ ( 4 ) (3)\Rightarrow(4) (3)(4)显然

( 4 ) ⇒ ( 1 ) (4)\Rightarrow(1) (4)(1) ∀ a ∈ G , ∀ h ∈ H , a ⋅ h ⋅ a − 1 ∈ H \forall a \in G,\forall h \in H, a\cdot h\cdot a^{-1}\in H aG,hH,aha1H

因为 a − 1 ∈ G a^{-1}\in G a1G,所以 a − 1 ⋅ h ⋅ a ∈ H a^{-1}\cdot h\cdot a\in H a1haH,于是 ∃ h 1 , h 2 ∈ H \exists h_1,h_2\in H h1,h2H,使 a ⋅ h ⋅ a − 1 = h 1 , a − 1 ⋅ h ⋅ a = h 2 a\cdot h\cdot a^{-1}=h_1, a^{-1}\cdot h\cdot a=h_2 aha1=h1,a1ha=h2,所以
a ⋅ h = h 1 ⋅ a ∈ H ⋅ a , h ⋅ a = a ⋅ h 2 ∈ a ⋅ H a\cdot h=h_1\cdot a\in H\cdot a,\quad h\cdot a=a\cdot h_2\in a\cdot H ah=h1aHa,ha=ah2aH
所以 a ⋅ H ⊆ H ⋅ a , H ⋅ a ⊆ a ⋅ H a\cdot H \subseteq H\cdot a, H\cdot a\subseteq a\cdot H aHHa,HaaH

a ⋅ H = H ⋅ a a\cdot H=H\cdot a aH=Ha,因此 H ◃ G H\triangleleft G HG

定理2:设 H ◃ G H\triangleleft G HG,则 G G G关于 H H H的陪集关系 R R R G G G上的同余关系

证明: R R R是等价关系

∀ a , b , c , d ∈ G \forall a,b,c,d\in G a,b,c,dG,若 a R b , c R d aRb,cRd aRb,cRd,则 a ⋅ H = H ⋅ b , c ⋅ H = H ⋅ d a\cdot H=H\cdot b,c\cdot H = H\cdot d aH=Hb,cH=Hd,所以

( a ⋅ c ) ⋅ H = a ⋅ ( c ⋅ H ) = a ⋅ ( H ⋅ d ) = ( a ⋅ H ) ⋅ d = ( H ⋅ b ) ⋅ d = H ⋅ ( b ⋅ d ) \begin{aligned} \left(a\cdot c\right)\cdot H &= a\cdot\left(c\cdot H\right)\\ &=a\cdot\left(H\cdot d\right)\\ &=\left(a\cdot H\right)\cdot d\\ &=\left(H\cdot b\right)\cdot d\\ &=H\cdot \left(b\cdot d\right) \end{aligned} (ac)H=a(cH)=a(Hd)=(aH)d=(Hb)d=H(bd)

( a ⋅ c ) R ( b ⋅ d ) \left(a\cdot c\right) R \left(b\cdot d\right) (ac)R(bd)

定理3:设 R R R是群 G G G上的同余关系,则 [ e ] R ◃ G \left[e\right]_R\triangleleft G [e]RG,且 R R R G G G关于 [ e ] R \left[e\right]_R [e]R的陪集关系

证明:

(1) [ e ] R ≤ G \left[e\right]_R\le G [e]RG ∀ a , b ∈ [ e ] R \forall a,b\in \left[e\right]_R a,b[e]R,知 a R e , e R b aRe,eRb aRe,eRb。又因为 b − 1 R b − 1 b^{-1}Rb^{-1} b1Rb1,所以 ( a ⋅ e ⋅ b − 1 ) R ( e ⋅ b ⋅ b − 1 ) \left(a\cdot e\cdot b^{-1}\right)R\left(e\cdot b\cdot b^{-1}\right) (aeb1)R(ebb1),即 ( a ⋅ b − 1 ) R e \left(a\cdot b^{-1}\right)Re (ab1)Re,故 a ⋅ b − 1 ∈ [ e ] R a\cdot b^{-1}\in \left[e\right]_R ab1[e]R
(2) [ e ] R ◃ G \left[e\right]_R\triangleleft G [e]RG ∀ a ∈ G , ∀ h ∈ [ e ] R \forall a\in G, \forall h \in \left[e\right]_R aG,h[e]R,则 a R a , h R e , a − 1 R a − 1 aRa,hRe,a^{-1}Ra^{-1} aRa,hRe,a1Ra1。所以 ( a ⋅ h ⋅ a − 1 ) R ( a ⋅ e ⋅ a − 1 ) \left(a\cdot h\cdot a^{-1}\right)R\left(a\cdot e\cdot a^{-1}\right) (aha1)R(aea1),即 ( a ⋅ h ⋅ a − 1 ) ∈ [ e ] R \left(a\cdot h \cdot a{-1}\right)\in \left[e\right]_R (aha1)[e]R

(3)

1. [ a ] R ⊆ a ⋅ [ e ] R \left[a\right]_R\subseteq a\cdot \left[e\right]_R [a]Ra[e]R ∀ x ∈ [ a ] R \forall x\in \left[a\right]_R x[a]R x R a xR a xRa。又因为 a − 1 R a − 1 a^{-1}Ra^{-1} a1Ra1,所以 ( a − 1 ⋅ x ) R e \left(a^{-1}\cdot x\right)Re (a1x)Re,所以 a − 1 ⋅ x ∈ [ e ] R a^{-1}\cdot x\in\left[e\right]_R a1x[e]R,故 x ∈ a ⋅ [ e ] R x\in a\cdot \left[e\right]_R xa[e]R

2. a ⋅ [ e ] R ⊆ [ a ] R a\cdot \left[e\right]_R\subseteq \left[a\right]_R a[e]R[a]R ∀ x ∈ a ⋅ [ e ] R \forall x\in a\cdot \left[e\right]_R xa[e]R,则 ∃ h ∈ [ e ] R \exists h \in \left[e\right]_R h[e]R,使 x = a ⋅ h x=a\cdot h x=ah

因为 a R a , h R e aRa,hRe aRa,hRe,所以 ( a ⋅ h ) R ( a ⋅ e ) \left(a\cdot h\right) R \left(a\cdot e\right) (ah)R(ae),则 x R a xRa xRa,故 x ∈ [ a ] R x\in \left[a\right]_R x[a]R

商群

定理1:设 H ◃ G H\triangleleft G HG,则 G G G关于 H H H的陪集关系的商代数 < G / H , ⊙ > \left G/H,使群,称为 G G G关于 H H H商群

其中 G / H = { a ⋅ H ∣ a ∈ G } G/H=\left\{a\cdot H|a\in G\right\} G/H={aHaG},并且 ∀ a ⋅ H , b ⋅ H ∈ G / H , ( a ⋅ H ) ⊙ ( b ⋅ H ) = ( a ⋅ b ) ⋅ H \forall a\cdot H, b\cdot H\in G/H, \left(a\cdot H\right)\odot \left(b\cdot H\right)=\left(a\cdot b\right)\cdot H aH,bHG/H,(aH)(bH)=(ab)H

证明:

由于 H ◃ G H\triangleleft G HG,则 ⊙ \odot G / H G/H G/H上的二元运算,因为若 a ⋅ H = a ′ ⋅ H , b ⋅ H = b ′ ⋅ H a\cdot H = a^{\prime}\cdot H, b\cdot H = b^{\prime} \cdot H aH=aH,bH=bH
a = a ′ ⋅ h 1 , b = b ′ ⋅ h 2 ( h 1 , h 2 ∈ H ) a=a^{\prime}\cdot h_1, b = b^{\prime}\cdot h_2\left(h_1,h_2\in H\right) a=ah1,b=bh2(h1,h2H),所以
a ⋅ b = a ′ ⋅ ( h 1 ⋅ b ′ ) ⋅ h 2 = a ′ ⋅ b ′ ⋅ ( h 3 ⋅ h 2 ) ( h 3 ∈ H ) a\cdot b = a^{\prime}\cdot \left(h_1\cdot b^{\prime}\right)\cdot h_2=a^{\prime}\cdot b^{\prime}\cdot \left(h_3\cdot h_2\right)\quad \left(h_3\in H\right) ab=a(h1b)h2=ab(h3h2)(h3H)
( a ′ ⋅ b ′ ) − 1 ⋅ ( a ⋅ b ) ∈ H \left(a^{\prime}\cdot b^{\prime}\right)^{-1}\cdot \left(a\cdot b\right) \in H (ab)1(ab)H,故 ( a ⋅ b ) ⋅ H = ( a ′ ⋅ b ′ ) ⋅ H \left(a\cdot b\right) \cdot H=\left(a^{\prime}\cdot b^{\prime}\right)\cdot H (ab)H=(ab)H,从而
( a ⋅ H ) ⊙ ( b ⋅ H ) = ( a ′ ⋅ H ) ⊙ ( b ′ ⋅ H ) \left(a\cdot H\right)\odot \left(b\cdot H\right) = \left(a^{\prime}\cdot H\right) \odot\left(b^{\prime}\cdot H\right) (aH)(bH)=(aH)(bH)
⋅ \cdot 的结合性,易证 ⊙ \odot 也是可结合的,而 H H H < G / H , ⊙ > \left G/H,的单位元
a ⋅ H a\cdot H aH的关于 ⊙ \odot 的逆元为 a − 1 ⋅ H a^{-1}\cdot H a1H,故 < G / H , ⊙ > \left G/H,是群
H H H是有限群 G G G的不变子群,则由拉格朗日定理由, ∣ G / H ∣ = ∣ G ∣ / ∣ H ∣ \left|G/H\right|=\left|G\right|/\left|H\right| G/H=G/H

群同态基本定理

定义:设 f f f < G , ⋅ > \left G, < G ′ , ∗ > \left G,的群同态, e ′ e^{\prime} e G ′ G^{\prime} G的单位元,则集合 { a ∈ G ∣ f ( a ) = e ′ } \left\{a\in G|f\left(a\right)=e^{\prime}\right\} {aGf(a)=e}称为 f f f同态核,记为 k e r f \mathop{ker} f kerf

群同态基本定理:设 f f f < G , ⋅ > \left G, < G ′ , ∗ > \left G,的群同态,则 k e r f ◃ G \mathop{ker} f\triangleleft G kerfG
并且 < G / k e r f , ⊙ > ≅ < f ( G ) , ∗ > \left\cong \left G/kerf,f(G),

证明:先证明 k e r f ◃ G \mathop{ker} f\triangleleft G kerfG。因为 ∀ k 1 , k 2 ∈ k e r f \forall k_1,k_2\in \mathop{ker} f k1,k2kerf,由 f ( k 1 ) = f ( k 2 ) = e ′ f\left(k_1\right)=f\left(k_2\right)=e^{\prime} f(k1)=f(k2)=e,所以
f ( k 1 ⋅ k 2 − 1 ) = f ( k 1 ) ∗ f ( k 2 ) − 1 = e ′ ∗ e ′ = e ′ f\left(k_1\cdot k_2^{-1}\right) = f\left(k_1\right)*f\left(k_2\right)^{-1}=e^{\prime}*e^{\prime}=e^{\prime} f(k1k21)=f(k1)f(k2)1=ee=e
k 1 ⋅ k 2 − 1 ∈ k e r f k_1\cdot k_2^{-1}\in \mathop{ker} f k1k21kerf.又因为 ∀ g ∈ G , k ∈ k e r f \forall g\in G,k\in \mathop{ker} f gG,kkerf
f ( g ⋅ k ⋅ g − 1 ) = f ( g ) ∗ f ( k ) ∗ f ( g − 1 ) = f ( g ) ∗ e ′ ∗ f ( g ) − 1 = e ′ f\left(g\cdot k \cdot g^{-1}\right)=f\left(g\right) * f\left(k\right) * f\left(g^{-1}\right)=f\left(g\right)*e^{\prime}*f\left(g\right)^{-1}=e^{\prime} f(gkg1)=f(g)f(k)f(g1)=f(g)ef(g)1=e
g ⋅ k ⋅ g − 1 ∈ k e r f g\cdot k \cdot g^{-1}\in \mathop{ker} f gkg1kerf,从而 k e r f ◃ G \mathop{ker} f\triangleleft G kerfG

再证 < G / k e r f , ⊙ > ≅ < f ( G ) , ∗ > \left\cong \left G/kerf,f(G),,作
ϕ : G / k e r f → f ( G ) , ϕ ( g ⋅ k e r f ) = f ( g ) \phi: G/\mathop{ker} f\to f\left(G\right),\phi\left(g\cdot \mathop{ker} f\right)=f\left(g\right) ϕ:G/kerff(G),ϕ(gkerf)=f(g)

1. ϕ \phi ϕ是映射且是单射,因为 ∀ g 1 ⋅ k e r f , g 2 ⋅ k e r f ∈ G / k e r f \forall g_1\cdot \mathop{ker} f,g_2 \cdot \mathop{ker} f\in G / \mathop{ker}f g1kerf,g2kerfG/kerf
ϕ ( g 1 ⋅ k e r f ) = ϕ ( g 2 ⋅ k e r f ) ⇔ f ( g 1 ) = f ( g 2 ) ⇔ f ( g 2 ) − 1 ∗ f ( g 1 ) = e ′ ⇔ f ( g 2 − 1 ⋅ g 1 ) = e ′ ⇔ g 2 − 1 ⋅ g 1 ∈ k e r f ⇔ g 1 ⋅ k e r f = g 2 ⋅ k e r f \begin{aligned} &\phi\left(g_1\cdot \mathop{ker} f\right) = \phi\left(g_2\cdot \mathop{ker} f\right)\\ &\quad\Leftrightarrow f\left(g_1\right) =f\left(g_2\right)\\ &\quad\Leftrightarrow f\left(g_2\right)^{-1}*f\left(g_1\right)=e^{\prime}\\ &\quad\Leftrightarrow f\left(g_2^{-1}\cdot g_1\right)=e^{\prime}\\ &\quad\Leftrightarrow g_2^{-1}\cdot g_1\in\mathop{ker} f\\ &\quad\Leftrightarrow g_1\cdot \mathop{ker}f = g_2\cdot \mathop{ker}f \end{aligned} ϕ(g1kerf)=ϕ(g2kerf)f(g1)=f(g2)f(g2)1f(g1)=ef(g21g1)=eg21g1kerfg1kerf=g2kerf
2. ϕ \phi ϕ是满射。 ∀ f ( g ) ∈ f ( G ) \forall f\left(g\right) \in f\left(G\right) f(g)f(G),由 g ∈ G g\in G gG,所以有 g ⋅ k e r f ∈ G / k e r f g\cdot \mathop{ker}f \in G/ \mathop{ker}f gkerfG/kerf,使 ϕ ( g ⋅ k e r f ) = f ( g ) \phi\left(g\cdot \mathop{ker}f\right)=f\left(g\right) ϕ(gkerf)=f(g)
3. ϕ \phi ϕ是同态映射。 ∀ g 1 ⋅ k e r f , g 2 ⋅ k e r f ∈ G / k e r f \forall g_1\cdot \mathop{ker}f,g_2\cdot \mathop{ker}f\in G/ \mathop{ker}f g1kerf,g2kerfG/kerf
ϕ ( g 1 ⋅ k e r f ⊙ g 2 ⋅ k e r f ) = ϕ ( ( g 1 ⋅ g 2 ) ⋅ k e r f ) = f ( g 1 ⋅ g 2 ) = f ( g 1 ) ⋅ f ( g 2 ) = ϕ ( g 1 ⋅ k e r f ) ∗ ϕ ( g 2 ⋅ k e r f ) \begin{aligned} \phi\left(g_1\cdot \mathop{ker}f \odot g_2\cdot \mathop{ker}f\right)&=\phi\left(\left(g_1\cdot g_2\right)\cdot \mathop{ker}f\right)\\ &=f\left(g_1\cdot g_2\right)\\ &=f\left(g_1\right)\cdot f\left(g_2\right)\\ &=\phi\left(g_1\cdot \mathop{ker}f\right)*\phi\left(g_2\cdot \mathop{ker}f\right)\\ \end{aligned} ϕ(g1kerfg2kerf)=ϕ((g1g2)kerf)=f(g1g2)=f(g1)f(g2)=ϕ(g1kerf)ϕ(g2kerf)
< G / k e r f , ⊙ > ≅ < f ( G ) , ∗ > \left\cong \left G/kerf,f(G),

推论:设 f f f是群 G G G G ′ G^{\prime} G的满同态,则 G / k e r f ≅ G ′ G/\mathop{ker}f\cong G^{\prime} G/kerfG

参考:
离散数学(刘玉珍)

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