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
\[U_{3} \Rightarrow B_{3}/U_{3} \cong F^{x} \times F^{x} \times F^{x}\]
\[\Rightarrow B_{3} \>\> is \>\> solvable.\]
