数学 - 群論 - 部分群と生成系(実数, 置換, 対称群)

代数系入門
(岩波書店)
松坂 和夫 (著)

学習環境

• 数式入力ソフト(TeX, MathML): MathType
• MathML対応ブラウザ: Firefox、Safari
• MathML非対応ブラウザ(Internet Explorer, Google Chrome...)用JavaScript Library: MathJax

代数系入門 (松坂 和夫(著)、岩波書店)の第2章(群論)、3(部分群と生成系)、問4.を解いてみる。

問4.

$\begin{array}{l}y\in ℝ\\ x=\frac{y-b}{a}\\ {\sigma }_{a,b}\left(x\right)=y\\ \\ x_1,x_2\in ℝ\\ {\sigma }_{a,b}\left(x_1\right)={\sigma }_{a,b}\left(x_2\right)\\ a{x}_1+b=a{x}_2+b\\ a{x}_1=a{x}_2\\ x_1=x_2\end{array}$

$\begin{array}{l}e\left(x\right)=I_R\left(x\right)\\ =x\\ =1x+0\\ ={\sigma }_{1,0}\left(x\right)\\ \\ {\sigma }_{a,b}°{\sigma }_{a^{\prime },b^{\prime }}\left(x\right)\\ ={\sigma }_{a,b}\left(a\text{'}x+b\text{'}\right)\\ =a\left(a^{\prime }x+b\text{'}\right)+b\\ =\left(aa\text{'}\right)x+\left(a{b}^{\prime }+b\right)\\ \\ aa\text{'}=1\\ ab\text{'}+b=0\\ b\text{'}=-\frac{b}{a}\\ a\text{'}=\frac{1}{a}\\ {\sigma }_{a,b}°{\sigma }_{\frac{1}{a},-\frac{b}{a}}\left(x\right)\\ =a\left(\frac{1}{a}x-\frac{b}{a}\right)+b\\ =x-b+b\\ =x\\ =e\left(x\right)\\ {\sigma }_{\frac{1}{a},-\frac{b}{a}}°{\sigma }_{ab}\left(x\right)\\ =\frac{1}{a}\left(ax+b\right)-\frac{b}{a}\\ =x\\ =e\left(x\right)\end{array}$

$\begin{array}{l}e\left(x\right)=I_R\left(x\right)\\ =x\\ =1x+0\\ ={\sigma }_{1,0}\left(x\right)\\ \\ {\sigma }_{a,b}°{\sigma }_{a^{\prime },b^{\prime }}\left(x\right)\\ ={\sigma }_{a,b}\left(a\text{'}x+b\text{'}\right)\\ =a\left(a^{\prime }x+b\text{'}\right)+b\\ =\left(aa\text{'}\right)x+\left(a{b}^{\prime }+b\right)\\ \\ aa\text{'}=1\\ ab\text{'}+b=0\\ b\text{'}=-\frac{b}{a}\\ a\text{'}=\frac{1}{a}\\ {\sigma }_{a,b}°{\sigma }_{\frac{1}{a},-\frac{b}{a}}\left(x\right)\\ =a\left(\frac{1}{a}x-\frac{b}{a}\right)+b\\ =x-b+b\\ =x\\ =e\left(x\right)\\ {\sigma }_{\frac{1}{a},-\frac{b}{a}}°{\sigma }_{ab}\left(x\right)\\ =\frac{1}{a}\left(ax+b\right)-\frac{b}{a}\\ =x\\ =e\left(x\right)\end{array}$