最后更新于3年前
定义
设HHH为GGG的子群，aaa为GGG中的任意元，则aH={ah∣h∈H}aH=\left\{ah\mid h\in H\right\}aH={ah∣h∈H}为GGG中的左陪集，Ha={ha∣h∈H}Ha=\left\{ha\mid h\in H\right\}Ha={ha∣h∈H}为右陪集
aHaHaH中的元素叫aHaHaH的代表元
若aH=HaaH=HaaH=Ha，则aHaHaH为GGG中HHH的陪集
定理
∀a∈G,aH={c∣c∈G,a−1c∈H},Ha={c∣c∈G,ca−1∈H}\forall a\in G,aH=\left\{c\mid c\in G,a^{-1}c\in H\right\},Ha=\left\{c\mid c\in G,ca^{-1}\in H\right\}∀a∈G,aH={c∣c∈G,a−1c∈H},Ha={c∣c∈G,ca−1∈H}
∀a,b∈G,aH=bH⟺b−1a∈H\forall a,b\in G,aH=bH\Longleftrightarrow b^{-1}a\in H∀a,b∈G,aH=bH⟺b−1a∈H
∀a,b∈G,aH∩bH=∅⟺b−1a∉H\forall a,b\in G,aH\cap bH=\varnothing\Longleftrightarrow b^{-1}a\not\in H∀a,b∈G,aH∩bH=∅⟺b−1a∈H
∀a∈H,aH=H=Ha\forall a\in H,aH=H=Ha∀a∈H,aH=H=Ha
群(Zha,+)\left(Z_{ha},+\right)(Zha,+)子群<a><a><a>的所有陪集STEP 1: <a><a><a>生成子群{0,a,2a,⋯ ,(h−1)a}\left\{0,a,2a,\cdots,\left(h-1\right)a\right\}{0,a,2a,⋯,(h−1)a}STEP 2: 陪集为{m+0,m+a,m+2a,⋯ ,m+(h−1)a},m=0,⋯ ,a−1\left\{m+0,m+a,m+2a,\cdots,m+\left(h-1\right)a\right\},m=0,\cdots,a-1{m+0,m+a,m+2a,⋯,m+(h−1)a},m=0,⋯,a−1
群(Zha,+)\left(Z_{ha},+\right)(Zha,+)子群<a><a><a>的所有陪集
STEP 1: <a><a><a>生成子群{0,a,2a,⋯ ,(h−1)a}\left\{0,a,2a,\cdots,\left(h-1\right)a\right\}{0,a,2a,⋯,(h−1)a}
STEP 2: 陪集为{m+0,m+a,m+2a,⋯ ,m+(h−1)a},m=0,⋯ ,a−1\left\{m+0,m+a,m+2a,\cdots,m+\left(h-1\right)a\right\},m=0,\cdots,a-1{m+0,m+a,m+2a,⋯,m+(h−1)a},m=0,⋯,a−1
G/H={aH∣a∈G}G/H=\left\{aH\mid a\in G\right\}G/H={aH∣a∈G}
G/HG/HG/H中左（右）陪集的个数叫做HHH在GGG中的指标，记作[G:H]\left[G:H\right][G:H]
拉格朗日定理
H≤G⟹∣G∣=[G:H]∣H∣H\leq G\Longrightarrow \left|G\right|=\left[G:H\right]\left|H\right|H≤G⟹∣G∣=[G:H]∣H∣
K,H≤G,K≤H⟹[G:K]=[G:H][H:K]K,H\leq G,K\leq H\Longrightarrow \left[G:K\right]=\left[G:H\right]\left[H:K\right]K,H≤G,K≤H⟹[G:K]=[G:H][H:K]
H≤G,HH\leq G,HH≤G,H满足∀a∈G,aH=Ha\forall a\in G,aH=Ha∀a∈G,aH=Ha
NNN为GGG的正规子群，(aN)(bN)=(ab)N\left(aN\right)\left(bN\right)=\left(ab\right)N(aN)(bN)=(ab)N，G/NG/NG/N构成一个商群
m+<a>m+<a>m+<a>在Zka/<a>Z_{ka}/<a>Zka/<a>里的阶
写出<a><a><a>
(m+<a>)⋅ord(m+<a>)=<a>{\left(m+<a>\right)}\cdot{\mathrm{ord}\left(m+<a>\right)}=<a>(m+<a>)⋅ord(m+<a>)=<a>