目錄

Definition 2.7

首先讓我們定義G為一個群，我們說G是solvable/soluble(可解)的話代表存在filtration G = G_{0} \rhd G_{1} \rhd \cdots \rhd G_{n} = \{e\}使得說G_{i}/G_{i+1}是abelian的

照這個定義來看如果是要解多項式f \in Q[x]的話就需要滿足求根式解

Proposition 2.8

如果G是solvable的話，則G的subgroups以及商數皆為solvable

證明

讓G = G_{0} \rhd G_{1} \rhd \cdots \rhd G_{n} = \{e\}使得G_{i}/G_{i+1}屬於abelian

讓H < G屬於subgroup

    \[\Rightarrow H = H \cap G_{0} \rhd H \cap G_{1} \rhd \cdots H \cap G_{n} = \{e\}\]

    \[(H \cap G_{i})/(H \cap G_{i+1}) \hookrightarrow G_{i}/G_{i+1} &\Rightarrow (H \cap G_{i})/(H \cap G_{i+1}) \>\> is \>\> abelian.\]

    \[\Rightarrow H \>\> is \>\> solvable\]

假設H \lhd G，則G_{i}H < G以及G_{i+1}H \lhd G_{i}H.

    \[G_{i}/G_{i+1} \twoheadrightarrow (G_{i}H)/(G_{i+1}H)\]

    \[\Rightarrow (G_{i}H)/(G_{i+1}H) \cong (G_{i}H/H)/(G_{i+1}H/H) \>\> is \>\> abelian.\]

    \[\Rightarrow G/H \>\> is \>\> solvable.\]

範例

(1)D_{n} = <x, y| x^{n} = y^{2} = e, yxy^{-1} = x^{-1}>屬於solvable.

證明：D_{n} \rhd <x> \rhd \{e\}, D_{n}/<x> \cong \mathbb{Z}/2\mathbb{Z}, <x> \cong \mathbb{Z}/n\mathbb{Z}

(2)S_{3} \cong D_{3}屬於solvable

(3)S_{4}屬於solvable

(4)如果n \geq 5則S_{n}不屬於solvable，(\xRightarrow[\text{Galois}]{} f \in Q[x], deg(f) \geq 5,一般來說無法在求根式解解出)

證明：假設我們有一個filtration S_{n} = G_{0} \rhd G_{1} \rhd \cdots \rhd G_{n} = \{e\}使得G_{i}/G_{i+1}是abelian的，G_{i}本身應該會包含在S_{n}上所有i的所有3-cycles組合，但在n \geq 5的時候會矛盾

(5)B_{n}(F) \subset GL_{n}(F)，意指Borel subgroup是solvable的，B_{n}(F)例如像是\{\begin{pmatrix}* & \cdots & *\\0 & * & *\\0 & 0 & *\end{pmatrix} \in GL_{n}(F)\}：一組upper triangular matrices

證明：n = 3: B_{3} \supset U_{3} \supset \mathcal{Z}(U_{3}) = \{\begin{pmatrix}1 & 0 & x\\ & 1 & 0 \\ & & 1\end{pmatrix} | x \in F\} \supset \{1\}，其中U_{3} := \{\begin{pmatrix}1 & x & y\\ & 1 & z \\ & & 1\end{pmatrix}|x, y, z \in F\}，而\{\begin{pmatrix}1 & 0 & x\\ & 1 & 0 \\ & & 1\end{pmatrix} | x \in F\} \supset \{1\} \cong F

U_{3} \rightarrow F \times F, \begin{pmatrix}1 & x & y\\ & 1 & z \\ & & 1\end{pmatrix} \mapsto (x, z) with kernel \mathcal{Z}(U_{3})

B_{3} \rightarrow \{\begin{pmatrix}a_{1} & & \\ & a_{2} & \\ & & a_{3}\end{pmatrix}|a_{1}, a_{2}, a_{3} \in F^{x}\}, \begin{pmatrix}a_{1} & * & * \\0 & a_{2} & * \\ 0 & 0 & a_{3}\end{pmatrix} \mapsto \begin{pmatrix}a_{1} & & \\ & a_{2} & \\ & & a_{3}\end{pmatrix}，其中a_{1}, a_{2}, a_{3} \cong F^{x} \times F^{x} \times F^{x}與kernel

    \[U_{3} \Rightarrow B_{3}/U_{3} \cong F^{x} \times F^{x} \times F^{x}\]

    \[\Rightarrow B_{3} \>\> is \>\> solvable.\]