# 01.2.対照群の可解性

## Proposition

### Statement

$S_n$:n次対称群
$n \le 4 \Rightarrow S_n$:可解群
$5 \le n \Rightarrow S_n$:非可解群

### Proof

(1) $n=1$
$S_1 = (1)$
$\therefore D(S_1) = \{(1)\} = \{e\}$
(2) $n = 2$
$S_2 = \{(1), (1 \ 2)\}$
$\therefore D(S_2) = \{e\}$
(3) $n = 3$
$D(S_3) = A_3 \quad (\because \text{Prop.対称群の交換子群})$
$= \lt (1 \ 2 \ 3) \gt \quad (\because \text{Prop.交代群と3文字の巡回置換が生成する群})$
$\therefore D^2(S_3) = D(A_3) = \{e\} \quad (\because \text{Prop.巡回群の可解性})$
(4) $n = 4$
$D(S_4) = A_4 \quad (\because \text{prop.対称群の交換子群})$
Let $V$ Klein four-group
$V = \{e, (1 \ 2)(3 \ 4), (1 \ 3)(2 \ 4), (1 \ 4)(2 \ 3)\}$
$\forall (i \ j)(k \ l) \in V$
$(i \ j)(k \ l) = (i \ j \ k)(i \ k \ l)(k \ j \ i)(l \ k \ i)$
$= (i \ j \ k)(i \ k \ l)(i \ j \ k)^{-1}(i \ k \ l)^{-1}$
$= [(i \ j \ k), (i \ k \ l)]$
$\in D(A_4)$
$(\because (i \ j \ k) = (i \ k)(i \ j) \in A_4 \quad \therefore [(i \ j \ k), (i \ k \ l)] \in D(A_4))$
$\therefore D(A_4) \supset V$

$A_4 \triangleright V \quad (\because \text{prop.クラインの4元群と正規部分群})$
$|A_4 / V| = 3$
$(\because |S_4| = 24, |A_4| = 12, |V| = 4 \therefore |A_4 / V| = |A_4| / |V| = 3) \quad (\because \text{prop.剰余類の指数})$
$\therefore A_4 / V$ は位数3の巡回群 $\quad (\because \text{prop.素数位数の群})$
したがって可換 $\quad (\because \text{prop.巡回群の可換性})$
$\therefore D(A_4) \subset V \quad (\because \text{prop.剰余群が可換群となる必要十分条件})$

$\therefore D(A_4) = V$
$\forall x \in V, x = x^{-1}$
$\therefore x, y \in V, [x,y] = xyx^{-1}y^{-1} = xyxy$
Let $xy = z \in V$ then $xyxy = zz = e$
$\therefore D(V) = \{e\}$

$\therefore D^3(S_4) = D^2(A_4) = D(V) = \{e\}$
(5) $n \ge 5$
$D^2(S_n) = D(A_n)$ ($\because$ Prop.対称群の交換子群)
$= A_n$ ($\because$ Prop.交代群の交換子群)
$\neq \{e\}$
.

## Proposition

### Statement

$S_n$:n次対称群
$A_n$:n次交代群
then
$D(S_n) = A_n$

### Proof

$\forall x, y \in S_n \Rightarrow [x,y] = xyx^{-1}y^{-1} \in A_n$
$\therefore D(S_n) \subset A_n$

$\text{prop.交代群と3文字の巡回置換が生成する群}$ より
$A_n = \lt \{(i \ j \ k)|1 \le i \lt j \lt k \le n \} \gt$
また $[(i \ j),(i \ k)] = (i \ j)(i \ k)(i \ j)(i \ k) = (i \ j \ k)$ より
$A_n = \lt \{(i \ j \ k)|1 \le i \lt j \lt k \le n \} \gt$
$= \lt \{ [(i \ j),(i \ k)] | 1 \le i \lt j \lt k \le n \} \gt$
$\subset D(S_n)$

$\therefore D(S_n) = A_n$
.

## Proposition

### Statement

n次交代群 $A_n \quad (n \ge 3)$ は3文字の置換 $(i \ j \ k) \quad (1 \le i \lt j \lt k \le n)$ によって生成される

### Proof

$(i \ j \ k) = (i \ k)(i \ j)$
$\therefore (i \ j \ k)(l \ m \ n)\cdots(o \ p \ q) \in A_n$

$(i \ j)(i \ j) = (i \ j \ k)^3$
$(i \ j)(i \ k) = (i \ k \ j)$
$(i \ j)(k \ l) = (i \ k \ j)(i \ k \ l)$
のいずれかである.
ゆえに主張は証明された.

.

## Proposition

### Statement

$G$: 巡回群 $\Rightarrow G$ :可換である

### Proof

Let $G = \lt a \gt$.
Then
$\forall x,y \in G, \exists m, n \in \mathbb{N}, x = a^m, y = a^n$
$\therefore x \cdot y = a^m \cdot a^n = a^{m + n} = a^{n + m} = a^n \cdot a^m = y \cdot x$
.

## Proposition

### Proof

$G$:可換群とする
$\forall x,y \in G$
$xyx^{-1}y^{-1} = xx^{-1}yy^{-1} = e$
$\therefore D(G) = \{e\}$
.

クラインの4元群と正規部分群.

## Statement

クラインの4元群は$S_4$の正規部分群である.
Let $V$ Klein four-group
$V = \{e, (1 \ 2)(3 \ 4), (1 \ 3)(2 \ 4), (1 \ 4)(2 \ 3)\}$
then
$S_4 \triangleright V$

### Proof

$S_4 = \lt \{(1 \ 2),(1 \ 3),(1 \ 4),(2 \ 3),(2 \ 4),(3 \ 4) \} \gt$
6つの生成元をそれぞれ$x_1,\cdots,x_6$とおくと
$\forall x \in S_4, \exists x_i, \cdots , x_j, x = x_i \cdots x_j$ $xVx^{-1} = x_i \cdots x_j V x_j^{-1} \cdots x_i^{-1}$
したがって、$\forall i \in 1 \cdots 6, x_i V x_i^{-1} = V$を示せばよい.
$x_1 e x_1^{-1} = (1 \ 2)e(1 \ 2) = e \in V$
$x_1 (1 \ 2)(3 \ 4) x_1^{-1} = (1 \ 2)(1 \ 2)(3 \ 4)(1 \ 2) = (1 \ 2)(3 \ 4) \in V$
$x_1 (1 \ 3)(2 \ 4) x_1^{-1} = (1 \ 2)(1 \ 3)(2 \ 4)(1 \ 2) = (1 \ 4)(2 \ 3) \in V$
$x_1 (1 \ 4)(2 \ 3) x_1^{-1} = (1 \ 2)(1 \ 4)(2 \ 3)(1 \ 2) = (1 \ 3)(2 \ 4) \in V$
$x_2 \cdots x_6$も同様に計算することで
$\forall x \in S_4, xVx^{-1} = V$
$\therefore S_4 \triangleright V$

## Proposition

### Statement

$G$:Group
$H$:Subgroup
$\forall a,b \in G$
(1) $|aH| = |bH|$
(2) $aH \cap bH \neq \emptyset \Rightarrow aH = bH$
therefore
$|G| = (G:H) \cdot |H|$

### Proof

(1)
Let
$a \in G$
$f:H \rightarrow aH$
$f:x \in H \mapsto ax \in aH$
* $a = e$
$f(x) = x \therefore f$ は全単射
* $a \neq e$
$x,y \in H$
$ax = ay \Rightarrow x = a^{-1}ax = a^{-1}ay = y \therefore f$ は単射

$\therefore \forall a,b \in G, |aH| = |bH|$
(2)
Let
$c \in aH \cap bH$
then
$a^{-1}c \in H$
$c^{-1}a \in H$
$b^{-1}c \in H$
$\forall x \in aH$
$a^{-1}x \in H$
$\therefore (b^{-1}c)(c^{-1}a)(a^{-1}x) = b^{-1} x \in H$
$\therefore x \in bH$
$\therefore aH \subset bH$

$\therefore aH = bH$

## Proposition

### Statement

$G$:群
$|G| = p$:素数 $\Rightarrow G$ :巡回群($\therefore$可換群)

### Proof

Let
$x \in G, x \neq e$
$H = \lt x \gt$
then
$|H| \mid |G| = p$(素数)
$\therefore |H|= p$
$\therefore G = H = \lt x \gt$

## Proposition

### Statement

$G$:Group
$G \triangleright N$
then
$G/N$:可換群 $\Leftrightarrow D(G) \subset N$

### Proof

($\Leftarrow$)
Supporse
$D(G) \subset N$
$f:G/D(G) \rightarrow G/N$
$f:xD(G) \mapsto xN$
then $f$ is well-defined($\because D(G) \subset N$)
$G \triangleright D(G) \therefore G/D(G)$ :剰余群 $(\because \text{Prop.交換子群と正規部分群})$
$f$ is a surjective homomorphism
$(\because f(xD(G)yD(G)) = f(xyD(G)) = xyN = xNyN = f(xD(G))f(yD(G)))$
oviously $f$ is surjective
$G/D(G)$ is a commutative group $(\because \text{Prop.交換子群による剰余群の可換性})$
$\therefore G/N$ is a commutative group.
$(\because xNyN = f(xD(G))f(yD(G)) = f(xD(G)yD(G)) = f(yD(G)xD(G)) = f(yD(G))f(xD(G)) = yNxN)$
($\Rightarrow$)
Supporse
$G/N$ is a commutative group
then
$\forall x,y \in G$
$[x,y]N = xyx^{-1}y^{-1}N = xNyNx^{-1}Ny^{-1}N = xNx^{-1}NyNy^{-1}N = N$ $\therefore [x,y] \in N$
$\therefore D(G) \subset N$
.

## Proposition

### Statement

すなわち
$G$:Group
then
$G \rhd D(G)$

### Proof

$\forall [x,y] \in D(G)$
$\forall z \in G$
$z[x,y]z^{-1} = z(xyx^{-1}y^{-1})z^{-1}$
$= zx(z^{-1}z)y(z^{-1}z)x^{-1}(z^{-1}z)y^{-1}z^{-1}$
$= (zxz^{-1})(zyz^{-1})(zx^{-1}z^{-1})(zy^{-1}z^{-1})$
$= (zxz^{-1})(zyz^{-1})(zxz^{-1})^{-1}(zyz^{-1})^{-1}$
$[zxz^{-1}, zyz^{-1}] \in D(G)$
$\therefore zD(G)z^{-1} \subset D(G)$
$\therefore G \rhd D(G)$

## Proposition

### Statement

$G$:Group
then
$G/D(G)$ は可換群

### Proof

Let
$\bar x, \bar y \in G/D(G)$
$\bar x \bar y \bar x^{-1} \bar y^{-1} = xD(G)yD(G)x^{-1}D(G)y^{-1}D(G) = xyx^{-1}y^{-1}D(G) = [x, y]D(G) = D(G) = \bar e$
$\therefore \bar x \bar y = \bar y \bar x$
.

## Proposition

### Statement

$n \ge 5$
then
$D(A_n) = A_n$

### Proof

Let's define $i,j,k\in \mathbb{N}$ such that
$1 \le i \lt j \lt k \le n$.
Since $n \ge 5$,
$\exists l,m \in \mathbb{N}$ such that
$1 \le l \lt m \le n$ and
$l,m \ne i,j,k$
then
$(i j k) = [(i j m)(i k l)]$
$\because [(i j m)(i k l)]$
$= (i j m)(i k l)(i j m)^{-1}(i k l)^{-1}$
$= (i j m)(i k l)(m j i)(l k i)$
(
$i \rightarrow m \rightarrow m \rightarrow i \rightarrow k$,
$j \rightarrow i \rightarrow l \rightarrow l \rightarrow i$,
$m \rightarrow j \rightarrow j \rightarrow m \rightarrow m$,
$k \rightarrow k \rightarrow i \rightarrow j \rightarrow j$,
$l \rightarrow l \rightarrow k \rightarrow k \rightarrow l$
)
$= (i j k)$
* $A_n$は3文字の巡回置換の積で表せる　$(\because \text{Prop.交代群と3文字の巡回置換が生成する群})$
* 3文字の巡回置換は3文字の巡回置換の交換子で表せる
$\therefore$ $A_n \subset D(A_n)$

$A_n = D(S_n) \supset D(A_n)$
$(\because \text{Prop.対称群の交換子群})$ $\therefore$ $A_n = D(A_n)$
. ----------------------

## Proposition

### Statement

$G$:群
$a \in G$
$Z(a)$ is a subgroup of $G$

### Proof

Let
$x,y \in Z(a) \Leftrightarrow xax^{-1} = a, yay^{-1} = a$
(1)
$\therefore xya(xy)^{-1} = xyay^{-1}x^{-1} = xax^{-1} = x$
$\therefore xy \in Z(a)$
(2)
$xax^{-1} = a$
$\Leftrightarrow x^{-1}xax^{-1}x = x^{-1}ax$
$\Leftrightarrow s = x^{-1}ax$
$\therefore x^{-1} = Z(a)$
(3)
$ese^{-1} = s \therefore e \in Z(a)$

## Proposition

### Statement

$G$:群
$a \in G$
$|K(a)| = (G:Z(a))$

### Proof

Let
$f:G \rightarrow K(a)$
$f:x \mapsto xax^{-1}$
$g:K(a) \rightarrow G/Z(a)$
$g:xax^{-1} \mapsto xZ(a) \quad (x \in G)$
$K(a)$ の定義から $f$ は全射
$\therefore g$ も全射.
when
$f(x) = f(y)$
$\Leftrightarrow xax^{-1} = yay^{-1}$
$y^{-1}xax^{-1}y = a$
$y^{-1}xa(y^{-1}x)^{-1} = a$
$y^{-1}x \in Z(a)$
$y \in xZ(a)$
$y \sim x \quad (G/Z(a))$
$\therefore g$ は単射.
$\therefore g$ は全単射.
$\therefore |K(a)| = (G:Z(a))$.

## Proposition

$p$-群

### Statement

$G$:$p$-群
then
$Z(G) \neq \{e\}$

## Proposition

### Statement

$G$:群
$|G| = p^2$ ($p$:素数)
then
$G$は可換群.