01.4.対称群の次数と可解性

対称群の次数と可解群

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) $n = 2$
$S_2 = \{e, (1 \ 2)\}$
$\therefore D(S_2) = \{e\}$
(3) $n = 3$
$D(S_3) = A_3 \quad (\because \text{Prop.対称群の交換子群})$
$= \langle (1 \ 2 \ 3) \rangle \quad (\because \text{Prop.交代群と3文字の巡回置換が生成する群})$
$\therefore D^2(S_3) = D(A_3) = \{e\} \quad (\because \text{Prop.巡回群の可解性})$
(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
$V = \{e, (1 \ 2)(3 \ 4), (1 \ 3)(2 \ 4), (1 \ 4)(2 \ 3)\}$
$\forall (i \ j)(k \ l) \in V$
$(i \ j)(k \ l) = (i \ j \ k)(i \ k \ l)(k \ j \ i)(l \ k \ i)$
$= (i \ j \ k)(i \ k \ l)(i \ j \ k)^{-1}(i \ k \ l)^{-1}$
$= [(i \ j \ k), (i \ k \ l)]$
$\in D(A_4)$
$(\because (i \ j \ k) = (i \ k)(i \ j) \in A_4 \quad \therefore [(i \ j \ k), (i \ k \ l)] \in D(A_4))$
$\therefore D(A_4) \supset V$
一方
$A_4 \triangleright V \quad (\because \text{prop.クラインの4元群と正規部分群})$
$|A_4 / V| = 3$
$(\because |S_4| = 24, |A_4| = 12, |V| = 4 \therefore |A_4 / V| = |A_4| / |V| = 3) \quad (\because \text{prop.剰余類の指数})$
$\therefore A_4 / V$ 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)
$\neq \{e\}$
QED