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 H≤G，若 ∀ a ∈ G , a ⋅ H = H ⋅ a \forall a\in G, a\cdot H = H\cdot a ∀a∈G,a⋅H=H⋅a,则称 H H H是 G G G的不变子群，也称正规子群，记作 H ◃ G H \triangleleft G H◃G
定理1：设 H ≤ G H\le G H≤G，则下列条件等价：
（1） H ◃ G H\triangleleft G H◃G
（2） ∀ a ∈ G , a ⋅ H ⋅ a − 1 = H \forall a \in G, a\cdot H\cdot a^{-1}=H ∀a∈G,a⋅H⋅a−1=H
（3） ∀ a ∈ G , a ⋅ H ⋅ a − 1 ⊆ H \forall a\in G,a\cdot H \cdot a^{-1}\subseteq H ∀a∈G,a⋅H⋅a−1⊆H
（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 ∀a∈G,∀h∈H,a⋅h⋅a−1∈H
证明:
( 1 ) ⇒ ( 2 ) (1)\Rightarrow(2) (1)⇒(2) 设 H ◃ G H\triangleleft G H◃G,则 ∀ a ∈ G \forall a\in G ∀a∈G,有 a ⋅ H = H ⋅ a a\cdot H = H\cdot a a⋅H=H⋅a,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 a⋅H⋅a−1=(H⋅a)⋅a−1=H⋅(a⋅a−1)=H⋅e=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) 设 ∀ 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 ∀a∈G,∀h∈H,a⋅h⋅a−1∈H
因为 a − 1 ∈ G a^{-1}\in G a−1∈G，所以 a − 1 ⋅ h ⋅ a ∈ H a^{-1}\cdot h\cdot a\in H a−1⋅h⋅a∈H，于是 ∃ h 1 , h 2 ∈ H \exists h_1,h_2\in H ∃h1,h2∈H，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 a⋅h⋅a−1=h1,a−1⋅h⋅a=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 a⋅h=h1⋅a∈H⋅a,h⋅a=a⋅h2∈a⋅H
所以 a ⋅ H ⊆ H ⋅ a , H ⋅ a ⊆ a ⋅ H a\cdot H \subseteq H\cdot a, H\cdot a\subseteq a\cdot H a⋅H⊆H⋅a,H⋅a⊆a⋅H
故 a ⋅ H = H ⋅ a a\cdot H=H\cdot a a⋅H=H⋅a，因此 H ◃ G H\triangleleft G H◃G
定理2：设 H ◃ G H\triangleleft G H◃G，则 G G G关于 H H H的陪集关系 R R R是 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,d∈G，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 a⋅H=H⋅b,c⋅H=H⋅d,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} (a⋅c)⋅H=a⋅(c⋅H)=a⋅(H⋅d)=(a⋅H)⋅d=(H⋅b)⋅d=H⋅(b⋅d)
故 ( a ⋅ c ) R ( b ⋅ d ) \left(a\cdot c\right) R \left(b\cdot d\right) (a⋅c)R(b⋅d)
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]R◃G，且 R R R是 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]R≤G。 ∀ 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} b−1Rb−1，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) (a⋅e⋅b−1)R(e⋅b⋅b−1)，即 ( a ⋅ b − 1 ) R e \left(a\cdot b^{-1}\right)Re (a⋅b−1)Re，故 a ⋅ b − 1 ∈ [ e ] R a\cdot b^{-1}\in \left[e\right]_R a⋅b−1∈[e]R
（2） [ e ] R ◃ G \left[e\right]_R\triangleleft G [e]R◃G。 ∀ a ∈ G , ∀ h ∈ [ e ] R \forall a\in G, \forall h \in \left[e\right]_R ∀a∈G,∀h∈[e]R,则 a R a , h R e , a − 1 R a − 1 aRa,hRe,a^{-1}Ra^{-1} aRa,hRe,a−1Ra−1。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) (a⋅h⋅a−1)R(a⋅e⋅a−1)，即 ( a ⋅ h ⋅ a − 1 ) ∈ [ e ] R \left(a\cdot h \cdot a{-1}\right)\in \left[e\right]_R (a⋅h⋅a−1)∈[e]R
（3）
1. [ a ] R ⊆ a ⋅ [ e ] R \left[a\right]_R\subseteq a\cdot \left[e\right]_R [a]R⊆a⋅[e]R。 ∀ x ∈ [ a ] R \forall x\in \left[a\right]_R ∀x∈[a]R则 x R a xR a xRa。又因为 a − 1 R a − 1 a^{-1}Ra^{-1} a−1Ra−1,所以 ( a − 1 ⋅ x ) R e \left(a^{-1}\cdot x\right)Re (a−1⋅x)Re,所以 a − 1 ⋅ x ∈ [ e ] R a^{-1}\cdot x\in\left[e\right]_R a−1⋅x∈[e]R，故 x ∈ a ⋅ [ e ] R x\in a\cdot \left[e\right]_R x∈a⋅[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 ∀x∈a⋅[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=a⋅h。
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) (a⋅h)R(a⋅e)，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 H◃G，则 G G G关于 H H H的陪集关系的商代数 < G / H , ⊙ > \left
其中 G / H = { a ⋅ H ∣ a ∈ G } G/H=\left\{a\cdot H|a\in G\right\} G/H={a⋅H∣a∈G},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 ∀a⋅H,b⋅H∈G/H,(a⋅H)⊙(b⋅H)=(a⋅b)⋅H
证明：
由于 H ◃ G H\triangleleft G H◃G,则 ⊙ \odot ⊙是 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 a⋅H=a′⋅H,b⋅H=b′⋅H，
则 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=a′⋅h1,b=b′⋅h2(h1,h2∈H)，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) a⋅b=a′⋅(h1⋅b′)⋅h2=a′⋅b′⋅(h3⋅h2)(h3∈H)
即 ( a ′ ⋅ b ′ ) − 1 ⋅ ( a ⋅ b ) ∈ H \left(a^{\prime}\cdot b^{\prime}\right)^{-1}\cdot \left(a\cdot b\right) \in H (a′⋅b′)−1⋅(a⋅b)∈H,故 ( a ⋅ b ) ⋅ H = ( a ′ ⋅ b ′ ) ⋅ H \left(a\cdot b\right) \cdot H=\left(a^{\prime}\cdot b^{\prime}\right)\cdot H (a⋅b)⋅H=(a′⋅b′)⋅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) (a⋅H)⊙(b⋅H)=(a′⋅H)⊙(b′⋅H)
由 ⋅ \cdot ⋅的结合性，易证 ⊙ \odot ⊙也是可结合的，而 H H H是 < G / H , ⊙ > \left
a ⋅ H a\cdot H a⋅H的关于 ⊙ \odot ⊙的逆元为 a − 1 ⋅ H a^{-1}\cdot H a−1⋅H，故 < G / H , ⊙ > \left
设 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是 < G , ⋅ > \left
群同态基本定理：设 f f f是 < G , ⋅ > \left
并且 < G / k e r f , ⊙ > ≅ < f ( G ) , ∗ > \left
证明：先证明 k e r f ◃ G \mathop{ker} f\triangleleft G kerf◃G。因为 ∀ k 1 , k 2 ∈ k e r f \forall k_1,k_2\in \mathop{ker} f ∀k1,k2∈kerf，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(k1⋅k2−1)=f(k1)∗f(k2)−1=e′∗e′=e′
故 k 1 ⋅ k 2 − 1 ∈ k e r f k_1\cdot k_2^{-1}\in \mathop{ker} f k1⋅k2−1∈kerf.又因为 ∀ g ∈ G , k ∈ k e r f \forall g\in G,k\in \mathop{ker} f ∀g∈G,k∈kerf
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(g⋅k⋅g−1)=f(g)∗f(k)∗f(g−1)=f(g)∗e′∗f(g)−1=e′
故 g ⋅ k ⋅ g − 1 ∈ k e r f g\cdot k \cdot g^{-1}\in \mathop{ker} f g⋅k⋅g−1∈kerf,从而 k e r f ◃ G \mathop{ker} f\triangleleft G kerf◃G
再证 < G / k e r f , ⊙ > ≅ < f ( G ) , ∗ > \left
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/kerf→f(G),ϕ(g⋅kerf)=f(g)
则
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 ∀g1⋅kerf,g2⋅kerf∈G/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(g1⋅kerf)=ϕ(g2⋅kerf)⇔f(g1)=f(g2)⇔f(g2)−1∗f(g1)=e′⇔f(g2−1⋅g1)=e′⇔g2−1⋅g1∈kerf⇔g1⋅kerf=g2⋅kerf
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 g∈G，所以有 g ⋅ k e r f ∈ G / k e r f g\cdot \mathop{ker}f \in G/ \mathop{ker}f g⋅kerf∈G/kerf,使 ϕ ( g ⋅ k e r f ) = f ( g ) \phi\left(g\cdot \mathop{ker}f\right)=f\left(g\right) ϕ(g⋅kerf)=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 ∀g1⋅kerf,g2⋅kerf∈G/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(g1⋅kerf⊙g2⋅kerf)=ϕ((g1⋅g2)⋅kerf)=f(g1⋅g2)=f(g1)⋅f(g2)=ϕ(g1⋅kerf)∗ϕ(g2⋅kerf)
故 < G / k e r f , ⊙ > ≅ < f ( G ) , ∗ > \left
推论：设 f f f是群 G G G到 G ′ G^{\prime} G′的满同态，则 G / k e r f ≅ G ′ G/\mathop{ker}f\cong G^{\prime} G/kerf≅G′
参考：
离散数学（刘玉珍）