1 Hall 部分群 定義 1.1. π \pi π a a a π \pi π a a a π \pi π pi-number )1 1 1 π \pi π π \pi π π ′ \pi' π ′
有限群G G G H H H ∣ H ∣ |H| ∣ H ∣ π \pi π [ G : H ] [G:H] [ G : H ] π ′ \pi' π ′ H H H Hallπ \pi π (Hall pi-subgroup )π \pi π π \pi π pi-group )π \pi π pi-subgroup )
定義 1.2. V V V M M M V V V §E7.17 定義 4.1 ）を、M M M V V V 補部分群 (complement )H ≤ M H\le M H ≤ M M = V H M=VH M = V H V ∩ H = { e M } V\cap H=\{e_M\} V ∩ H = { e M }
命題 1.4. G G G π \pi π H H H G G G π \pi π N ⊴ G N\trianglelefteq G N ⊴ G H ∩ N H\cap N H ∩ N N N N π \pi π H N / N HN/N H N / N G / N G/N G / N π \pi π
2 最小正規部分群 補題 2.1. G G G M M M G G G M M M r r r M M M r r r M M M v ∈ M v\in M v ∈ M v r = e G v^r=e_G v r = e G M M M F r \mathbb F_r F r
証明. M M M A A A §E7.22 補題 2.2 により、A A A M M M [ A , A ] = { e G } [A,A]=\{e_G\} [ A , A ] = { e G } A A A M ⊴ G M\trianglelefteq G M ⊴ G §E7.15 命題 4.2 (3) によりA ⊴ G A\trianglelefteq G A ⊴ G A ≠ { e G } A\ne\{e_G\} A = { e G } M M M A = M A=M A = M M M M
∣ M ∣ |M| ∣ M ∣ r r r §E7.18 定理 1.3 により、M M M r r r R R R M M M R R R R R R §E7.18 定理 2.2 により Sylowr r r R R R M M M R ⊴ G R\trianglelefteq G R ⊴ G R ≠ { e G } R\ne\{e_G\} R = { e G } M M M R = M R=M R = M M M M r r r
Ω 1 ( M ) = { v ∈ M ∣ v r = e G } \Omega_1(M)=\{v\in M\mid v^r=e_G\} Ω 1 ( M ) = { v ∈ M ∣ v r = e G } と置く。M M M Ω 1 ( M ) \Omega_1(M) Ω 1 ( M ) M M M e G e_G e G v ∈ M v\in M v ∈ M M M M r r r v v v r k r^k r k k ≥ 1 k\ge1 k ≥ 1 v r k − 1 v^{r^{k-1}} v r k − 1 r r r Ω 1 ( M ) ≠ { e G } \Omega_1(M)\ne\{e_G\} Ω 1 ( M ) = { e G } Ω 1 ( M ) ⊴ G \Omega_1(M)\trianglelefteq G Ω 1 ( M ) ⊴ G M M M Ω 1 ( M ) = M \Omega_1(M)=M Ω 1 ( M ) = M
M M M F r \mathbb F_r F r M M M F r \mathbb F_r F r M M M r r r ▨
系 2.2. G G G M M M G G G M M M p p p M M M p p p
系 2.3. G G G V V V G G G r r r V V V r r r V V V v ∈ V v\in V v ∈ V v r = e G v^r=e_G v r = e G
3 初等アーベル核の補部分群 補題 3.1. r r r V V V r r r
1 ⟶ V ⟶ M → ρ Q ⟶ 1 1\longrightarrow V\longrightarrow M\xrightarrow{\rho}Q\longrightarrow1 1 ⟶ V ⟶ M ρ Q ⟶ 1 を考える。r ∤ ∣ Q ∣ r\nmid|Q| r ∤ ∣ Q ∣ ρ ∘ s = id Q \rho\circ s=\id_Q ρ ∘ s = id Q s : Q → M s\colon Q\to M s : Q → M
証明. V V V M M M V V V q ∈ Q q\in Q q ∈ Q ρ ( q ~ ) = q \rho(\widetilde q)=q ρ ( q ) = q q ~ ∈ M \widetilde q\in M q ∈ M
q ⋅ v = q ~ v q ~ − 1 ( v ∈ V ) q\cdot v=\widetilde q\,v\,\widetilde q^{-1}\qquad(v\in V) q ⋅ v = q v q − 1 ( v ∈ V ) と定める。別の持上げはa q ~ a\widetilde q a q a ∈ V a\in V a ∈ V V V V a a a V V V q ⋅ v q\cdot v q ⋅ v
( q t ) ⋅ v = q ⋅ ( t ⋅ v ) (qt)\cdot v=q\cdot(t\cdot v) ( q t ) ⋅ v = q ⋅ ( t ⋅ v ) が成り立つため、Q Q Q V V V
σ ( e Q ) = e M \sigma(e_Q)=e_M σ ( e Q ) = e M σ : Q → M \sigma\colon Q\to M σ : Q → M q , t ∈ Q q,t\in Q q , t ∈ Q f ( q , t ) ∈ V f(q,t)\in V f ( q , t ) ∈ V
σ ( q ) σ ( t ) = f ( q , t ) σ ( q t ) \sigma(q)\sigma(t)=f(q,t)\sigma(qt) σ ( q ) σ ( t ) = f ( q , t ) σ ( q t ) (3.1.1) \text{(3.1.1)} (3.1.1) となる。f f f 式 (3.1.1) を用いると
f ( q , t ) + f ( q t , u ) = q ⋅ f ( t , u ) + f ( q , t u ) f(q,t)+f(qt,u)=q\cdot f(t,u)+f(q,tu) f ( q , t ) + f ( q t , u ) = q ⋅ f ( t , u ) + f ( q , t u ) (3.1.2) \text{(3.1.2)} (3.1.2) を得る。ここでは、σ ( q ) f ( t , u ) σ ( q ) − 1 = q ⋅ f ( t , u ) \sigma(q)f(t,u)\sigma(q)^{-1}=q\cdot f(t,u) σ ( q ) f ( t , u ) σ ( q ) − 1 = q ⋅ f ( t , u )
m = ∣ Q ∣ m=|Q| m = ∣ Q ∣ r ∤ m r\nmid m r ∤ m c m ≡ 1 ( m o d r ) cm\equiv1\pmod r c m ≡ 1 ( mod r ) c c c V V V r r r V V V c m cm c m q ∈ Q q\in Q q ∈ Q
A ( q ) = ∑ u ∈ Q f ( q , u ) , B ( q ) = c A ( q ) A(q)=\sum_{u\in Q}f(q,u),\qquad B(q)=cA(q) A ( q ) = u ∈ Q ∑ f ( q , u ) , B ( q ) = c A ( q ) と置く。式 (3.1.2) をu ∈ Q u\in Q u ∈ Q
m f ( q , t ) + A ( q t ) = q ⋅ A ( t ) + A ( q ) mf(q,t)+A(qt)=q\cdot A(t)+A(q) m f ( q , t ) + A ( q t ) = q ⋅ A ( t ) + A ( q ) となる。右辺の最後の和では、u ↦ t u u\mapsto tu u ↦ t u Q Q Q c c c
f ( q , t ) = B ( q ) + q ⋅ B ( t ) − B ( q t ) f(q,t)=B(q)+q\cdot B(t)-B(qt) f ( q , t ) = B ( q ) + q ⋅ B ( t ) − B ( q t ) (3.1.3) \text{(3.1.3)} (3.1.3) を得る。
s ( q ) = ( − B ( q ) ) σ ( q ) s(q)=(-B(q))\sigma(q) s ( q ) = ( − B ( q )) σ ( q ) と定める。σ ( e Q ) = e M \sigma(e_Q)=e_M σ ( e Q ) = e M 式 (3.1.1) に代入するとf ( e Q , u ) = 0 f(e_Q,u)=0 f ( e Q , u ) = 0 B ( e Q ) = 0 B(e_Q)=0 B ( e Q ) = 0 式 (3.1.1) と式 (3.1.3) から
s ( q ) s ( t ) = ( − B ( q ) ) σ ( q ) ( − B ( t ) ) σ ( t ) = ( − B ( q ) − q ⋅ B ( t ) + f ( q , t ) ) σ ( q t ) = ( − B ( q t ) ) σ ( q t ) = s ( q t ) \begin{aligned}
s(q)s(t)
&=(-B(q))\sigma(q)(-B(t))\sigma(t)\\
&=\bigl(-B(q)-q\cdot B(t)+f(q,t)\bigr)\sigma(qt)\\
&=(-B(qt))\sigma(qt)=s(qt)
\end{aligned} s ( q ) s ( t ) = ( − B ( q )) σ ( q ) ( − B ( t )) σ ( t ) = ( − B ( q ) − q ⋅ B ( t ) + f ( q , t ) ) σ ( q t ) = ( − B ( q t )) σ ( q t ) = s ( q t ) が成り立つ。またs ( e Q ) = e M s(e_Q)=e_M s ( e Q ) = e M s s s ρ ∘ s = id Q \rho\circ s=\operatorname{id}_Q ρ ∘ s = id Q
▨
補題 3.2. r r r V V V r r r
1 ⟶ V ⟶ M → ρ Q ⟶ 1 1\longrightarrow V\longrightarrow M\xrightarrow{\rho}Q\longrightarrow1 1 ⟶ V ⟶ M ρ Q ⟶ 1 を考える。r ∤ ∣ Q ∣ r\nmid|Q| r ∤ ∣ Q ∣ ρ \rho ρ s , t : Q → M s,t\colon Q\to M s , t : Q → M w ∈ V w\in V w ∈ V q ∈ Q q\in Q q ∈ Q t ( q ) = w s ( q ) w − 1 t(q)=ws(q)w^{-1} t ( q ) = w s ( q ) w − 1
証明. V V V M M M q ∈ Q q\in Q q ∈ Q v ∈ V v\in V v ∈ V q ⋅ v = s ( q ) v s ( q ) − 1 q\cdot v=s(q)vs(q)^{-1} q ⋅ v = s ( q ) v s ( q ) − 1 s s s Q Q Q V V V m = ∣ Q ∣ m=|Q| m = ∣ Q ∣ c m ≡ 1 ( m o d r ) cm\equiv1\pmod r c m ≡ 1 ( mod r ) c c c z ( q ) ∈ V z(q)\in V z ( q ) ∈ V
t ( q ) = z ( q ) s ( q ) t(q)=z(q)s(q) t ( q ) = z ( q ) s ( q ) と書くことができる。s s s t t t q , u ∈ Q q,u\in Q q , u ∈ Q
z ( q u ) = z ( q ) + q ⋅ z ( u ) z(qu)=z(q)+q\cdot z(u) z ( q u ) = z ( q ) + q ⋅ z ( u ) (3.2.1) \text{(3.2.1)} (3.2.1) を得る。S = ∑ u ∈ Q z ( u ) S=\sum_{u\in Q}z(u) S = ∑ u ∈ Q z ( u ) q ∈ Q q\in Q q ∈ Q 式 (3.2.1) をu ∈ Q u\in Q u ∈ Q u ↦ q u u\mapsto qu u ↦ q u Q Q Q
S = m z ( q ) + q ⋅ S S=mz(q)+q\cdot S S = m z ( q ) + q ⋅ S となる。w = c S w=cS w = c S
z ( q ) = w − q ⋅ w z(q)=w-q\cdot w z ( q ) = w − q ⋅ w (3.2.2) \text{(3.2.2)} (3.2.2) である。式 (3.2.2) から
w s ( q ) w − 1 = ( w − q ⋅ w ) s ( q ) = z ( q ) s ( q ) = t ( q ) ws(q)w^{-1}=(w-q\cdot w)s(q)=z(q)s(q)=t(q) w s ( q ) w − 1 = ( w − q ⋅ w ) s ( q ) = z ( q ) s ( q ) = t ( q ) を得る。したがってt ( Q ) = w s ( Q ) w − 1 t(Q)=ws(Q)w^{-1} t ( Q ) = w s ( Q ) w − 1 ▨
補題 3.3. r r r V V V r r r
1 ⟶ V ⟶ M → ρ Q ⟶ 1 1\longrightarrow V\longrightarrow M\xrightarrow{\rho}Q\longrightarrow1 1 ⟶ V ⟶ M ρ Q ⟶ 1 を考える。r ∤ ∣ Q ∣ r\nmid |Q| r ∤ ∣ Q ∣ M M M V V V V V V
証明. 補題 3.1 により、ρ \rho ρ s : Q → M s\colon Q\to M s : Q → M H = s ( Q ) H=s(Q) H = s ( Q ) ρ ∣ H \rho|_H ρ ∣ H Q Q Q H ∩ V = { e M } H\cap V=\{e_M\} H ∩ V = { e M } x ∈ M x\in M x ∈ M q = ρ ( x ) q=\rho(x) q = ρ ( x ) x s ( q ) − 1 ∈ V xs(q)^{-1}\in V x s ( q ) − 1 ∈ V M = V H M=VH M = V H H H H V V V
K K K V V V ρ ∣ K \rho|_K ρ ∣ K Q Q Q K ↪ M K\hookrightarrow M K ↪ M 補題 3.2 により、任意の二つの補部分群はV V V ▨
4 Hall の三定理 補題 4.1. G G G N ⊴ G N\trianglelefteq G N ⊴ G π \pi π H ≤ G H\le G H ≤ G π \pi π N ≤ H N\le H N ≤ H
証明. N ⊴ G N\trianglelefteq G N ⊴ G N H NH N H
[ N H : H ] = [ N : N ∩ H ] [NH:H]=[N:N\cap H] [ N H : H ] = [ N : N ∩ H ] は∣ N ∣ |N| ∣ N ∣ π \pi π H ≤ N H ≤ G H\le NH\le G H ≤ N H ≤ G [ N H : H ] [NH:H] [ N H : H ] [ G : H ] [G:H] [ G : H ] π ′ \pi' π ′ π \pi π π ′ \pi' π ′ 1 1 1 [ N H : H ] = 1 [NH:H]=1 [ N H : H ] = 1 N H = H NH=H N H = H N ≤ H N\le H N ≤ H ▨
補題 4.2. G G G π \pi π r r r V ⊴ G V\trianglelefteq G V ⊴ G r r r G / V G/V G / V π \pi π G G G π \pi π
証明. G / V G/V G / V π \pi π H ‾ \overline H H G G G M M M
r ∈ π r\in\pi r ∈ π ∣ M ∣ = ∣ V ∣ ∣ H ‾ ∣ |M|=|V||\overline H| ∣ M ∣ = ∣ V ∣∣ H ∣ π \pi π
[ G : M ] = [ G / V : H ‾ ] [G:M]=[G/V:\overline H] [ G : M ] = [ G / V : H ] はπ ′ \pi' π ′ M M M G G G π \pi π
r ∉ π r\notin\pi r ∈ / π r ∤ ∣ H ‾ ∣ r\nmid|\overline H| r ∤ ∣ H ∣
1 ⟶ V ⟶ M ⟶ H ‾ ⟶ 1 1\longrightarrow V\longrightarrow M\longrightarrow\overline H\longrightarrow1 1 ⟶ V ⟶ M ⟶ H ⟶ 1 へ補題 3.3 を適用し、V V V H ≤ M H\le M H ≤ M ∣ H ∣ = ∣ H ‾ ∣ |H|=|\overline H| ∣ H ∣ = ∣ H ∣ π \pi π
[ G : H ] = [ G : M ] [ M : H ] = [ G / V : H ‾ ] ∣ V ∣ [G:H]=[G:M][M:H]=[G/V:\overline H]|V| [ G : H ] = [ G : M ] [ M : H ] = [ G / V : H ] ∣ V ∣ はπ ′ \pi' π ′ H H H G G G π \pi π ▨
補題 4.3. G G G π \pi π r r r V ⊴ G V\trianglelefteq G V ⊴ G r r r G / V G/V G / V π \pi π G G G π \pi π
証明. H H H K K K G G G π \pi π 命題 1.4 により、H V / V HV/V H V / V K V / V KV/V K V / V G / V G/V G / V π \pi π g ∈ G g\in G g ∈ G K K K g K g − 1 gKg^{-1} g K g − 1
H V / V = K V / V HV/V=KV/V H V / V = K V / V とすることができる。この共通の部分群のG G G M M M M = V H = V K M=VH=VK M = V H = V K
r ∈ π r\in\pi r ∈ π 補題 4.1 によってV ≤ H V\le H V ≤ H V ≤ K V\le K V ≤ K H = V H = M = V K = K H=VH=M=VK=K H = V H = M = V K = K
r ∉ π r\notin\pi r ∈ / π ∣ H ∣ |H| ∣ H ∣ ∣ K ∣ |K| ∣ K ∣ π \pi π ∣ V ∣ |V| ∣ V ∣ r r r H ∩ V = K ∩ V = { e G } H\cap V=K\cap V=\{e_G\} H ∩ V = K ∩ V = { e G } H H H K K K M M M V V V M / V = H V / V M/V=HV/V M / V = H V / V π \pi π r ∤ ∣ M / V ∣ r\nmid|M/V| r ∤ ∣ M / V ∣ 補題 3.3 の仮定が成り立ち、H H H K K K V V V H H H K K K G G G ▨
補題 4.4. G G G π \pi π r r r V ⊴ G V\trianglelefteq G V ⊴ G r r r G / V G/V G / V π \pi π π \pi π G G G π \pi π π \pi π
証明. P ≤ G P\le G P ≤ G π \pi π P V / V PV/V P V / V G / V G/V G / V π \pi π P V / V PV/V P V / V G / V G/V G / V π \pi π L ‾ \overline L L G G G M M M
r ∈ π r\in\pi r ∈ π ∣ M ∣ = ∣ V ∣ ∣ L ‾ ∣ |M|=|V||\overline L| ∣ M ∣ = ∣ V ∣∣ L ∣ [ G : M ] = [ G / V : L ‾ ] [G:M]=[G/V:\overline L] [ G : M ] = [ G / V : L ] M M M G G G π \pi π P ≤ P V ≤ M P\le PV\le M P ≤ P V ≤ M
r ∉ π r\notin\pi r ∈ / π M / V = L ‾ M/V=\overline L M / V = L π \pi π r ∤ ∣ M / V ∣ r\nmid|M/V| r ∤ ∣ M / V ∣ 補題 3.3 により、M M M V V V H H H ∣ H ∣ = ∣ L ‾ ∣ |H|=|\overline L| ∣ H ∣ = ∣ L ∣ [ G : H ] = [ G / V : L ‾ ] ∣ V ∣ [G:H]=[G/V:\overline L]|V| [ G : H ] = [ G / V : L ] ∣ V ∣ H H H G G G π \pi π P ∩ V = { e G } P\cap V=\{e_G\} P ∩ V = { e G } P V = V P PV=VP P V = V P P P P P V PV P V V V V
ρ : M → M / V \rho\colon M\to M/V ρ : M → M / V H ∩ P V H\cap PV H ∩ P V P V PV P V V V V ρ ∣ H : H → M / V \rho|_H\colon H\to M/V ρ ∣ H : H → M / V P V / V ≤ M / V PV/V\le M/V P V / V ≤ M / V x ∈ P V x\in PV x ∈ P V ρ ( h ) = ρ ( x ) \rho(h)=\rho(x) ρ ( h ) = ρ ( x ) h ∈ H h\in H h ∈ H h − 1 x ∈ V ≤ P V h^{-1}x\in V\le PV h − 1 x ∈ V ≤ P V h ∈ H ∩ P V h\in H\cap PV h ∈ H ∩ P V P V = V ( H ∩ P V ) PV=V(H\cap PV) P V = V ( H ∩ P V ) V ∩ ( H ∩ P V ) = { e G } V\cap(H\cap PV)=\{e_G\} V ∩ ( H ∩ P V ) = { e G } P V / V PV/V P V / V π \pi π r ∤ ∣ P V / V ∣ r\nmid|PV/V| r ∤ ∣ P V / V ∣
補題 3.3 を群P V PV P V v ∈ V v\in V v ∈ V
P = v ( H ∩ P V ) v − 1 P=v(H\cap PV)v^{-1} P = v ( H ∩ P V ) v − 1 となる。ゆえにP ≤ v H v − 1 P\le vHv^{-1} P ≤ v H v − 1 v H v − 1 vHv^{-1} v H v − 1 G G G π \pi π ▨
定理 4.5 (Hall の三定理). G G G π \pi π
G G G π \pi π 任意の二つの Hallπ \pi π G G G
任意のπ \pi π P ≤ G P\le G P ≤ G π \pi π
証明. ∣ G ∣ = 1 |G|=1 ∣ G ∣ = 1 { e G } \{e_G\} { e G } π \pi π ∣ G ∣ > 1 |G|>1 ∣ G ∣ > 1 ∣ G ∣ |G| ∣ G ∣ G G G V V V 系 2.3 により、ある素数r r r V V V r r r §E7.22 定理 3.3 (2) によりG / V G/V G / V ∣ G / V ∣ = ∣ G ∣ / ∣ V ∣ < ∣ G ∣ |G/V|=|G|/|V|<|G| ∣ G / V ∣ = ∣ G ∣/∣ V ∣ < ∣ G ∣ G / V G/V G / V 補題 4.2 、補題 4.3 、補題 4.4 により、G G G ▨
5 Sylow の定理との対応 系 5.1. G G G p p p π = { p } \pi=\{p\} π = { p } π \pi π p p p 定理 4.5 の存在、共役、包含の三つの主張は、
Sylow の定理における対応する三つの主張に一致する。
証明. ∣ G ∣ = p n m |G|=p^n m ∣ G ∣ = p n m p ∤ m p\nmid m p ∤ m H H H { p } \{p\} { p } ∣ H ∣ |H| ∣ H ∣ p p p [ G : H ] [G:H] [ G : H ] p p p ∣ G ∣ = ∣ H ∣ [ G : H ] |G|=|H|[G:H] ∣ G ∣ = ∣ H ∣ [ G : H ] ∣ H ∣ = p n |H|=p^n ∣ H ∣ = p n H H H p p p p p p p n p^n p n m m m { p } \{p\} { p }
Hall の三定理により、Sylow 部分群の存在、共役、および任意のp p p §E7.18 定理 2.2 により共役作用からさらに与えられる結論である。▨
6 例と仮定の境界 例 6.1. 正規列
{ e } ⊴ A 3 ⊴ S 3 \{e\}\trianglelefteq A_3\trianglelefteq S_3 { e } ⊴ A 3 ⊴ S 3 の因子はC 3 C_3 C 3 C 2 C_2 C 2 S 3 S_3 S 3
π = { 3 } \pi=\{3\} π = { 3 } A 3 A_3 A 3 3 3 3 2 2 2 π \pi π π = { 2 } \pi=\{2\} π = { 2 } 2 2 2 π \pi π S 3 S_3 S 3 { 2 } \{2\} { 2 } { 2 } \{2\} { 2 }
例 6.2. A 5 A_5 A 5 §E7.22 定理 6.1 ）。π = { 3 , 5 } \pi=\{3,5\} π = { 3 , 5 } A 5 A_5 A 5 π \pi π H H H ∣ H ∣ = 15 |H|=15 ∣ H ∣ = 15
§E7.18 定理 4.2 をp = 3 p=3 p = 3 q = 5 q=5 q = 5 3 ∤ 4 3\nmid4 3 ∤ 4 15 15 15 H H H 15 15 15 S 5 S_5 S 5 15 15 15 3 3 3 5 5 5 8 8 8 S 5 S_5 S 5 A 5 A_5 A 5 15 15 15 15 15 15 A 5 A_5 A 5 { 3 , 5 } \{3,5\} { 3 , 5 }
7 演習 問題 7.1.
補題 3.3 について、次の二つを示せ。第一に、仮定r ∤ ∣ Q ∣ r\nmid|Q| r ∤ ∣ Q ∣ M = Z / 4 Z M=\Z/4\Z M = Z /4 Z V = { 0 , 2 } V=\{0,2\} V = { 0 , 2 } r = 2 r=2 r = 2 M = S 3 M=S_3 M = S 3 V = A 3 V=A_3 V = A 3 r = 3 r=3 r = 3 V V V V V V G = A 4 × C 5 G=A_4\times C_5 G = A 4 × C 5 C 5 = Z / 5 Z C_5=\Z/5\Z C 5 = Z /5 Z π = { 3 , 5 } \pi=\{3,5\} π = { 3 , 5 } G G G G G G π \pi π 60 60 60 A 5 A_5 A 5 { 3 , 5 } \{3,5\} { 3 , 5 } 例 6.2 ）と比べよ。命題 1.4 の証明を完成させよ。A 5 A_5 A 5 { 2 , 5 } \{2,5\} { 2 , 5 } 解答 (演習の解答). (1) では、M = Z / 4 Z M=\Z/4\Z M = Z /4 Z V = { 0 , 2 } V=\{0,2\} V = { 0 , 2 } 2 2 2 2 2 2 Q = M / V Q=M/V Q = M / V 2 2 2 r = 2 r=2 r = 2 ∣ Q ∣ |Q| ∣ Q ∣ V V V H H H ∣ H ∣ = ∣ M ∣ / ∣ V ∣ = 2 |H|=|M|/|V|=2 ∣ H ∣ = ∣ M ∣/∣ V ∣ = 2 Z / 4 Z \Z/4\Z Z /4 Z 2 2 2 2 2 2 2 2 2 V V V H ∩ V = { 0 } H\cap V=\{0\} H ∩ V = { 0 } M = S 3 M=S_3 M = S 3 V = A 3 V=A_3 V = A 3 H H H ∣ H ∣ = 2 |H|=2 ∣ H ∣ = 2 H ∩ A 3 = { e } H\cap A_3=\{e\} H ∩ A 3 = { e } 2 2 2 ⟨ ( 1 2 ) ⟩ \langle(1\,2)\rangle ⟨( 1 2 )⟩ ⟨ ( 1 3 ) ⟩ \langle(1\,3)\rangle ⟨( 1 3 )⟩ ⟨ ( 2 3 ) ⟩ \langle(2\,3)\rangle ⟨( 2 3 )⟩ A 3 A_3 A 3 ∣ A 3 H ∣ = ∣ A 3 ∣ ∣ H ∣ / ∣ A 3 ∩ H ∣ = 6 |A_3H|=|A_3||H|/|A_3\cap H|=6 ∣ A 3 H ∣ = ∣ A 3 ∣∣ H ∣/∣ A 3 ∩ H ∣ = 6 A 3 H = S 3 A_3H=S_3 A 3 H = S 3 σ = ( 1 2 3 ) ∈ V \sigma=(1\,2\,3)\in V σ = ( 1 2 3 ) ∈ V σ ( a b ) σ − 1 = ( σ ( a ) σ ( b ) ) \sigma(a\,b)\sigma^{-1}=(\sigma(a)\,\sigma(b)) σ ( a b ) σ − 1 = ( σ ( a ) σ ( b )) σ ⟨ ( 1 2 ) ⟩ σ − 1 = ⟨ ( 2 3 ) ⟩ \sigma\langle(1\,2)\rangle\sigma^{-1}=\langle(2\,3)\rangle σ ⟨( 1 2 )⟩ σ − 1 = ⟨( 2 3 )⟩ σ 2 ⟨ ( 1 2 ) ⟩ σ − 2 = ⟨ ( 3 1 ) ⟩ = ⟨ ( 1 3 ) ⟩ \sigma^2\langle(1\,2)\rangle\sigma^{-2}=\langle(3\,1)\rangle=\langle(1\,3)\rangle σ 2 ⟨( 1 2 )⟩ σ − 2 = ⟨( 3 1 )⟩ = ⟨( 1 3 )⟩ V V V
(2) では、V 4 = { e , ( 1 2 ) ( 3 4 ) , ( 1 3 ) ( 2 4 ) , ( 1 4 ) ( 2 3 ) } V_4=\{e,(1\,2)(3\,4),(1\,3)(2\,4),(1\,4)(2\,3)\} V 4 = { e , ( 1 2 ) ( 3 4 ) , ( 1 3 ) ( 2 4 ) , ( 1 4 ) ( 2 3 )} A 4 ▹ V 4 ▹ { e } A_4\triangleright V_4\triangleright\{e\} A 4 ▹ V 4 ▹ { e } 3 3 3 4 4 4 §E7.22 定理 2.3 によりA 4 A_4 A 4 G ▹ { e } × C 5 ▹ { e } G\triangleright\{e\}\times C_5\triangleright\{e\} G ▹ { e } × C 5 ▹ { e } G / ( { e } × C 5 ) ≅ A 4 G/(\{e\}\times C_5)\cong A_4 G / ({ e } × C 5 ) ≅ A 4 C 5 C_5 C 5 §E7.22 定理 3.3 によりG G G ∣ G ∣ = 60 = 2 2 ⋅ 3 ⋅ 5 |G|=60=2^2\cdot3\cdot5 ∣ G ∣ = 60 = 2 2 ⋅ 3 ⋅ 5 π \pi π 15 15 15 15 15 15 H H H 5 5 5 G G G 5 5 5 ( a , c ) ∈ G (a,c)\in G ( a , c ) ∈ G 5 5 5 a = e a=e a = e c ≠ 0 c\ne0 c = 0 G G G 5 5 5 { e } × C 5 \{e\}\times C_5 { e } × C 5 { e } × C 5 ≤ H \{e\}\times C_5\le H { e } × C 5 ≤ H ( a , c ) (a,c) ( a , c ) 3 3 3 a a a 3 3 3 c = 0 c=0 c = 0 H H H 3 3 3 T × { 0 } T\times\{0\} T × { 0 } T T T A 4 A_4 A 4 3 3 3 H = T × C 5 H=T\times C_5 H = T × C 5 A 4 A_4 A 4 3 3 3 3 3 3 ⟨ ( 1 2 3 ) ⟩ \langle(1\,2\,3)\rangle ⟨( 1 2 3 )⟩ ⟨ ( 1 2 4 ) ⟩ \langle(1\,2\,4)\rangle ⟨( 1 2 4 )⟩ ⟨ ( 1 3 4 ) ⟩ \langle(1\,3\,4)\rangle ⟨( 1 3 4 )⟩ ⟨ ( 2 3 4 ) ⟩ \langle(2\,3\,4)\rangle ⟨( 2 3 4 )⟩ T × C 5 T\times C_5 T × C 5 15 15 15 4 4 4 π \pi π π \pi π §E7.18 定理 2.2 (2) によりA 4 A_4 A 4 3 3 3 g ∈ A 4 g\in A_4 g ∈ A 4 ( g , 0 ) ( T × C 5 ) ( g , 0 ) − 1 = g T g − 1 × C 5 (g,0)(T\times C_5)(g,0)^{-1}=gTg^{-1}\times C_5 ( g , 0 ) ( T × C 5 ) ( g , 0 ) − 1 = g T g − 1 × C 5 π \pi π A 5 A_5 A 5 { 3 , 5 } \{3,5\} { 3 , 5 } 定理 4.5 (1) の可解性の仮定が外せないことを示す。
(3) では、N ⊴ G N\trianglelefteq G N ⊴ G H N HN H N ∣ H N ∣ = ∣ H ∣ ∣ N ∣ / ∣ H ∩ N ∣ |HN|=|H||N|/|H\cap N| ∣ H N ∣ = ∣ H ∣∣ N ∣/∣ H ∩ N ∣ ∣ H ∩ N ∣ |H\cap N| ∣ H ∩ N ∣ ∣ H ∣ |H| ∣ H ∣ π \pi π [ N : H ∩ N ] = ∣ N ∣ / ∣ H ∩ N ∣ = ∣ H N ∣ / ∣ H ∣ = [ H N : H ] [N:H\cap N]=|N|/|H\cap N|=|HN|/|H|=[HN:H] [ N : H ∩ N ] = ∣ N ∣/∣ H ∩ N ∣ = ∣ H N ∣/∣ H ∣ = [ H N : H ] [ H N : H ] [HN:H] [ H N : H ] [ G : H ] = [ G : H N ] [ H N : H ] [G:H]=[G:HN][HN:H] [ G : H ] = [ G : H N ] [ H N : H ] π ′ \pi' π ′ H ∩ N H\cap N H ∩ N N N N π \pi π ∣ H N / N ∣ = ∣ H N ∣ / ∣ N ∣ = ∣ H ∣ / ∣ H ∩ N ∣ |HN/N|=|HN|/|N|=|H|/|H\cap N| ∣ H N / N ∣ = ∣ H N ∣/∣ N ∣ = ∣ H ∣/∣ H ∩ N ∣ ∣ H ∣ |H| ∣ H ∣ π \pi π [ G / N : H N / N ] = [ G : H N ] [G/N:HN/N]=[G:HN] [ G / N : H N / N ] = [ G : H N ] [ G : H ] [G:H] [ G : H ] π ′ \pi' π ′ H N / N HN/N H N / N G / N G/N G / N π \pi π
(4) で Hall{ 2 , 5 } \{2,5\} { 2 , 5 } 20 20 20 20 20 20 H H H 5 5 5 4 4 4 5 5 5 1 1 1 §E7.18 定理 2.2 (3) ）。したがって個数は1 1 1 P P P H H H H ≤ N A 5 ( P ) H\le N_{A_5}(P) H ≤ N A 5 ( P ) §E7.18 例 3.1 によりA 5 A_5 A 5 5 5 5 6 6 6 §E7.18 式 (2.2.1) より[ A 5 : N A 5 ( P ) ] = 6 [A_5:N_{A_5}(P)]=6 [ A 5 : N A 5 ( P )] = 6 ∣ N A 5 ( P ) ∣ = 60 / 6 = 10 |N_{A_5}(P)|=60/6=10 ∣ N A 5 ( P ) ∣ = 60/6 = 10 H ≤ N A 5 ( P ) H\le N_{A_5}(P) H ≤ N A 5 ( P ) ∣ H ∣ ≤ 10 |H|\le10 ∣ H ∣ ≤ 10 ∣ H ∣ = 20 |H|=20 ∣ H ∣ = 20 ▨