01.3.対照群の可解性

可換群の可解性

Statement

可換群は可解群である.

Proof

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

巡回群の可解性

Statement

巡回群は可解群である.

Proof

巡回群は可換群である $(\because \text{巡回群の可換性})$
可換群は可解群である $(\because \text{可換群の可解性})$
QED

交換子群と正規部分群

Statement

$G$:Group
then
$G \triangleright D(G)$

Proof

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

交換子群による剰余群の可換性

Statement

$G$:Group
then
$G/D(G)$:Abelian Group

Proof

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

剰余群が可換群となる必要十分条件

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
$\forall x,y \in G$
$[x,y]N = xyx^{-1}y^{-1}N = xNyNx^{-1}Ny^{-1}N = xNx^{-1}NyNy^{-1}N = N$ $\therefore [x,y] \in N$
$\therefore D(G) \subset N$
QED

対称群の交換子群

Statement

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

Proof

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

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

$\therefore D(S_n) = A_n$
QED

交代群の交換子群

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
$\therefore$ $A_n \subset D(A_n)$
一方
$A_n = D(S_n) \supset D(A_n)$
$(\because \text{Prop.対称群の交換子群})$ $\therefore$ $A_n = D(A_n)$
QED