06.1.可解群の性質

第３同型定理

Statement

$G$:Group
$N \triangleleft G$
$H \triangleleft G$
$N \subset H$
then
$N \triangleleft H$
$H/N \triangleleft G/N$
$(G/N)/(H/N) \simeq G/H$

Proof

$\forall x \in G, xNx^{-1} = N \quad(\because N \triangleleft G)$
$\therefore \forall x \in H \subset G, xNx^{-1} = N$
$\therefore N \triangleleft H$

$\forall hN \in H/N, \forall gN \in G/N$
$(gN)(hN)(gN)^{-1} = ghg^{-1}N \in H/N \quad\because ghg^{-1} \in H$
$\therefore (gN)(H/N)(gN)^{-1} \subset H/N$
$\therefore H/N \triangleleft G/N$

let
$\phi:G/N \rightarrow G/H$
$gN \in G/N \mapsto gH \in G/H$
then
$\phi$ is well-defined
( $\because$
If $g_1 N = g_2 N \quad(g_1 N, g_2 N \in G/N)$
then
$g_1 n_1 = g_2 n_2 \quad(\exists n_1, n_2 \in N)$
$\Rightarrow g_1 = g_2 n_2 n_1^{-1}$
$\therefore g_1 H = g_2 n_2 n_1^{-1}H = g_2 H \quad(\because n_2 n_1^{-1} \in N \subset H)$
)
$\text{Ker}\phi = H/N, \text{Im}\phi = G/H$
$\therefore (G/N)/(H/N) = (G/N)/\text{Ker}\phi \simeq \text{Im}\phi = G/H$
QED

部分群の対応定理

Statement

$G$:Group
$N \triangleleft G$
$\mathbb{H} = \{H:\text{SubGroup of }G|N \subset H \subset G\}$
$\mathbb{S} = \{S:\text{SubGroup of }G/N\}$
then
$\exists \phi:\mathbb{H} \rightarrow \mathbb{S} \text{ s.t. surjective}$
$\exists \psi:\mathbb{S} \rightarrow \mathbb{H} \text{ s.t. surjective}$

Proof

let
define $f,f^{-1}$ as
$f:G \rightarrow G/N$
$f:g \mapsto gN$
$f^{-1}:G/N \rightarrow \{H | H \subset G\}$
$f^{-1}:gN \mapsto gN = \{x \in G| f(x) = gN\} \subset G$
$\phi:H \in \mathbb{H} \mapsto H/N \in \mathbb{S} \quad (= \{hN \in G/N|h \in H\} = f(H) \subset G/N)$
$\psi:S \in \mathbb{S} \mapsto f^{-1}(S) = \{x \in G| f(x) = xN \in S\} \in \mathbb{H}$
then
$\therefore \psi\cdot\phi(H) = \psi(H/N) = f^{-1}(H/N) = H$
$(\because a \in f^{-1}(H/N) \Leftrightarrow f(a) \in H/N \Leftrightarrow a \in H)$
$\therefore \psi \cdot \phi = id$

$\phi \cdot \psi(S) = \phi(f^{-1}(S)) = f^{-1}(S)/N$

let $gN \in f^{-1}(S)/N$ then
$\Rightarrow g \in f^{-1}(S) = \{x \in G|f(x) = S\} \subset G$
$\Leftrightarrow f(g) \in S$
$f(g) = gN \therefore gN \in S$
$\therefore f^{-1}(S)/N \subset S$

let $gN \in S$ then
$\Leftrightarrow f(g) \in S$
$\therefore g \in f^{-1}(S)$
$\therefore gN \in f^{-1}(S)/N$
$\therefore f^{-1}(S)/N \supset S$
$\therefore f^{-1}(S)/N = S$
$\therefore \phi \cdot \psi(S) = S$
$\therefore \phi \cdot \psi = id$

QED

交換子群の基本性質

Statement

let
$G$:Group
$H$:SubGroup
$N \triangleleft G$

define $D(G)=\langle \{[x,y]|x,y \in G\}\rangle$
define $G$:Solvable Group $\Leftrightarrow \exists n \text{ s.t. } D^n(G) = \{e\}$

then
(i) $G$:Solvable Group $\Rightarrow H$:Solvable Group
(ii) $G$:Solvable Group $\Leftrightarrow N, G/N$:Solvable Group

Proof

(i)
$G$:Solvable $\therefore \exists n, \text{ s.t. } D^n(G) = \{e\}$
$H \subset G, D(H) \subset D(G), D^2(H) \subset D^2(G), \cdots, D^n(H) \subset D^n(G) = \{e\}$
$\therefore D^n(H) = \{e\}, \therefore H$:Solvable

(ii)
$\Rightarrow$
$N \subset G \therefore N$:Solvable $\quad(\because \text{Prop.交換子群の基本性質 (i)})$
for $aN, bN \in G/N$
$[aN,bN] = aN bN a^{-1}N b^{-1}N = aba^{-1}b^{-1}N = [a,b]N$
$D(G/N) = \{[aN,bN]|a,b \in G\} = \{[a,b]N|a,b \in G\} = D(G)N/N$
$\therefore D^n(G/N) = D^n(G)N/N = N/N = \{\bar e\}$
$\therefore G/N$:Solvable
$\Leftarrow$
$\exists l \text{ s.t. } D^l(N) = \{e\}, \exists m \text{ s.t. } D^m(G/N) = \{\bar e\}$
$D^m(G/N) = D^m(G)N/N = \{\bar e\} \therefore D^m(G) \subset N$
$\therefore D^{l+m}(G) = D^l(D^m(G)) \subset D^l(N) = \{e\}$
$\therefore D^{l+m}(G) = \{e\}$
$G$:Solvable
QED

可換群と正規列

Statement

$G:\text{Abelian Group}$
then
$\exists (G_i) \text{ s.t. }$
$G = G_0 \supset G_1 \supset \cdots \supset G_t = \{e\}$
$G_{i+1} \triangleleft G_i$
$G_i/G_{i+1} \simeq \mathbb{Z}/a_i\mathbb{Z}$:Cyclic Group

Proof

$G\simeq (\mathbb{Z}/a_1\mathbb{Z}) \times \cdots \times (\mathbb{Z}/a_t\mathbb{Z})\quad(\because\text{Prop.巡回群の直積分解})$
let
$G_1\simeq (\mathbb{Z}/a_2\mathbb{Z}) \times \cdots \times (\mathbb{Z}/a_t\mathbb{Z})$
$G_2\simeq (\mathbb{Z}/a_3\mathbb{Z}) \times \cdots \times (\mathbb{Z}/a_t\mathbb{Z})$
$\vdots$
$G_{t-1}\simeq (\mathbb{Z}/a_t\mathbb{Z})$
$G_t \simeq \{e\}$
then
$G = G_0 \supset G_1 \supset \cdots \supset G_t = \{e\}$
$g_i = (b_{i+1}, b_{i+2}, \cdots, b_t) \in (\mathbb{Z}/a_{i+1}\mathbb{Z}) \times (\mathbb{Z}/a_{i+2}\mathbb{Z}) \times \cdots (\mathbb{Z}/a_t\mathbb{Z}) \simeq G_i$
$g_{i+1} = (id, c_{i+2}, \cdots, c_t) \in \{id\} \times (\mathbb{Z}/a_{i+2}\mathbb{Z}) \times \cdots (\mathbb{Z}/a_t\mathbb{Z}) \simeq G_{i+1}$
$g_i g_{i+1} g_i^{-1} = (b_{i+1}b_{i+1}^{-1}, b_{i+2}c_{i+2}b_{i+2}^{-1}, \cdots, b_t c_t b_t^{-1}) \in \{id\} \times (\mathbb{Z}/a_{i+2}\mathbb{Z}) \times \cdots (\mathbb{Z}/a_t\mathbb{Z})$
$\therefore \forall g \in G_i, g G_{i+1} g^{-1} \subset G_{i+1}$
$\therefore G_{i+1} \triangleleft G_i$
$G_i/G_{i+1} \simeq \mathbb{Z}/a_{i+1}\mathbb{Z}$
$\mathbb{Z}/a_{i+1}\mathbb{Z}$:Cyclic Group
$\therefore G_i/G_{i+1}$:Cyclic Group
QED

有限可解群の同値条件

Statement

$G$:Finite Solvable Group
$\Leftrightarrow$
$\exists (G_n) = (G = G_0, \supset \cdots \supset G_n = \{e\})$
$\text{ s.t. } G_{i+1} \triangleleft G_i, G_i / G_{i+1}:\text{Cyclic Group}$

Proof

$\Rightarrow$
$G$:Solvable $\Rightarrow \exists n \text{ s.t. } D^n(G) = \{e\}$
(i)$n = 1$
$D(G) = \{e\} \therefore G/D(G) \simeq G$
$G/D(G):\text{Abelian Group}\quad(\because\text{Prop.交換子群による剰余群の可換性})$
$\therefore G$:Abelian Group
then
$G = G_0 \supset G_1 \supset \cdots \supset G_t = \{e\}$
$G_{i+1} \triangleleft G_i$
$G_i/G_{i+1} \simeq \mathbb{Z}/a_i\mathbb{Z}$:Cyclic Group
$(\because\text{Prop.可換群と正規列})$ (ii)$n \ge 2$
$D^n(G) = D^{n-1}(D(G)) = \{e\}$
$\therefore D(G)$:Solvable
$\therefore \exists (H_i) \text{ s.t. }$
$D(G) = H_0 \supset H_1 \supset \cdots \supset H_s = \{e\}$
$H_{i+1} \triangleleft H_i, H_i/H_{i+1}$:Cyclic Group
($\because$:the statement is true when $n-1$ Hypothetically)
$D(G) \triangleleft G, G/D(G)\text{:Finite Abelian Group}\quad(\because\text{Prop.交換子群による剰余群の可換性})$
$\exists (S_i)_{S_i \subset G/D(G)} \quad(\because\text{Prop.可換群と正規列})$
$\text{ s.t. }$
$G/D(G) = S_0 \supset S_1 \supset \cdots \supset S_t = \{\bar e\} = D(G)/D(G)$
$S_i \triangleright S_{i+1}$
$S_i/S_{i+1}$:Cyclic Group
$\exists (G_i)_{G_i/D(G) = S_i, G_i \subset G} \quad(\because\text{Prop.部分群の対応定理})$
$G_i/D(G) \triangleright G_{i+1}/D(G)$
$\Leftrightarrow (g_i D(G)) G_{i+1}/D(G) (g_i D(G))^{-1} \subset G_{i+1}/D(G) \quad(\forall g_i D(G) \in G_i/D(G))$
$\Leftrightarrow (g_i D(G)) G_{i+1}/D(G) (g_i^{-1} D(G)) \subset G_{i+1}/D(G)$
$\Leftrightarrow (g_i D(G)) g_{i+1} D(G) (g_i^{-1} D(G)) \in G_{i+1}/D(G) \quad(\forall g_{i+1}D(G) \in G_{i+1}/D(G))$
$\Leftrightarrow g_i g_{i+1} g_i^{-1} D(G) \in G_{i+1}/D(G)\quad(\forall g_{i+1} \in G_{i+1})$
$\Leftrightarrow g_i g_{i+1} g_i^{-1} \in G_{i+1}\quad(\forall g_{i+1} \in G_{i+1}, \forall g_i \in G_i)$
$\Leftrightarrow g_i G_{i+1} g_i^{-1} \subset G_{i+1}\quad(\forall g_i \in G_i)$
$\Leftrightarrow G_{i+1} \triangleleft G_i$
$(G_i/D(G))/(G_{i+1}/D(G)) = S_i/S_{i+1}$:Cyclic
$G_i/G_{i+1} \simeq (G_i/D(G))/(G_{i+1}/D(G)) \quad(\because\text{Prop.第３同型定理})$
$\therefore G_i/G_{i+1}$:Cyclic
$G = G_0 \supset G_1 \supset \cdots \supset G_t = D(G) = H_0 \supset \cdots H_s = \{e\}$
$G_i \triangleright G_{i+1}, G_i/G_{i+1}$:Cyclic Group
$H_i \triangleright H_{i+1}, H_i/H_{i+1}$:Cyclic Group
$\Leftarrow$
(i) $n = 1$
$G = G_0 \subset G_1 = \{e\}, G_1 \triangleleft G_0 = G$
$G \simeq G/G_1 = G_0/G_1$:Cyclic
$\therefore G:\text{Solvable}\quad(\because\text{Prop.可換群の可解性})$

(ii) $n \ge 2$
Suppose the statement is true when (n-1)
then $G_1$:Solvable ($\because G_1 \supset \cdots \supset G_n = \{e\} \Rightarrow G_1$:Solvable Hypothetically)
$G_1 \triangleleft G_0,G_0/G_1:\text{Cyclic}\quad(\because\text{Given Conditions})$
$\therefore G_0/G_1:\text{Solvable}\quad(\because\text{Prop.可換群の可解性})$
$G_0/G_1$:Solvable $\Rightarrow G_0=G:\text{Solvable}\quad(\because\text{Prop.交換子群の基本性質(ii)})$
QED