free_group_isomorphism.saty

@require: stdjabook
@require: code
@require: itemize
@require: tabular
@require: math
@import: local

let-math \group-gen m1 m2 =
  ${\angle-bracket{#m1 \| #m2}}
let-math \Z = ${\math-style!(MathDoubleStruck){Z}}
let-math \fA = ${\alpha^3}
let-math \fB = ${\beta\alpha^{-2}}
let-math \fC = ${\alpha\beta}
let-math \fD = ${\alpha^2\beta\alpha^{-1}}
in
document (|
  title = {自由群${F_4}から${F_2}への、とある準同型写像};
  author = {hsjoihs};
  show-title = true;
  show-toc = false;
|) '<
  +section{問題}<
    +p{
      自由群${F_4 : \group-gen{A,B,C,D}{} } と ${F_2 : \group-gen{\alpha, \beta}{}} があり、
      準同型写像 ${f : F_4 \arrow F_2} が 
      ${\app{f}{A} = \fA}, 
      ${\app{f}{B} = \fB}, 
      ${\app{f}{C} = \fC}, 
      ${\app{f}{D} = \fD} によって定まるとき、 
    }
    +listing{
      * ${\app{f}{F_4}} は ${F_2} の正規部分群
      * ${F_2 / \app{f}{F_4} \simeq \Z / 3\Z}
    }
    +pn{
      であることを示せ。
    }
  >
  +section{「${\app{f}{F_4}} は ${F_2} の正規部分群である」の略解}<
    +p{
      示すべきことは、${
        \all n \in \app{f}{F_4}, \all g \in F_2, gng^{-1} \in \app{f}{F_4}
      } である。ところで、
      \align[
        ${| \alpha \app{f}{A} \alpha^{-1} |=| \alpha \fA \alpha^{-1}  |=| \fA |=| \app{f}{A}|};
        ${| \alpha \app{f}{B} \alpha^{-1} |=| \alpha \fB \alpha^{-1}  |=| \fC\paren{\fA}^{-1} |=| \app{f}{CA^{-1}} |};
        ${| \alpha \app{f}{C} \alpha^{-1} |=| \alpha \fC \alpha^{-1}  |=| \fD |=| \app{f}{D} |};
        ${| \alpha \app{f}{D} \alpha^{-1} |=| \alpha \fD \alpha^{-1}  |=| \fA\fB |=| \app{f}{AB} |};
        ${| \beta \app{f}{A} \beta^{-1} |=| \beta \fA \beta^{-1}  |=| \fB\fA\paren{\fB}^{-1} |=| \app{f}{BAB^{-1}}|};
        ${| \beta \app{f}{B} \beta^{-1} |=| \beta \fB \beta^{-1}  |=| \fB\fD\paren{\fA}^{-1}\paren{\fB}^{-1} |=| \app{f}{BDA^{-1}B^{-1}} |};
        ${| \beta \app{f}{C} \beta^{-1} |=| \beta \fC \beta^{-1}  |=| \fB\fA |=| \app{f}{BA} |};
        ${| \beta \app{f}{D} \beta^{-1} |=| \beta \fD \beta^{-1}  |=| \fB\fA\fC\paren{\fA}^{-1}\paren{\fB}^{-1}|=| \app{f}{BACA^{-1}B^{-1}} |};
        ${| \alpha^{-1} \app{f}{A} \alpha |=| \alpha^{-1} \fA \alpha  |=| \fA |=| \app{f}{A}|};
        ${| \alpha^{-1} \app{f}{B} \alpha |=| \alpha^{-1} \fB \alpha  |=| \paren{\fA}^{-1}\fD |=| \app{f}{A^{-1}D}|};
        ${| \alpha^{-1} \app{f}{C} \alpha |=| \alpha^{-1} \fC \alpha  |=| \fB\fA |=| \app{f}{BA}|};
        ${| \alpha^{-1} \app{f}{D} \alpha |=| \alpha^{-1} \fD \alpha  |=| \fC |=| \app{f}{C}|};
        ${| \beta^{-1} \app{f}{A} \beta |=| \beta^{-1} \fA \beta  |=| \paren{\fC}^{-1}\fA\fC |=| \app{f}{C^{-1}AC}|};
        ${| \beta^{-1} \app{f}{B} \beta |=| \beta^{-1} \fB \beta  |=| \paren{\fA}^{-1}\fC |=| \app{f}{A^{-1}C}|};
        ${| \beta^{-1} \app{f}{C} \beta |=| \beta^{-1} \fC \beta  |=| \paren{\fC}^{-1}\fD\fC |=| \app{f}{C^{-1}DC}|};
        ${| \beta^{-1} \app{f}{D} \beta |=| \beta^{-1} \fD \beta  |=| \paren{\fC}^{-1}\fA\fB\fC |=| \app{f}{C^{-1}ABC}|};
      ];%
      なので、${F_2}の任意の要素が${\alpha}・${\beta}・${\alpha^{-1}}・${\beta^{-1}}の積で書ける以上、
    }
    +math(${
        \all n \in \app{f}{F_4}, \all g \in F_2, gng^{-1} \in \app{f}{F_4}
      });
    +pn{
      は成り立つ。
    }

  >
  +section{「${F_2 / \app{f}{F_4} \simeq \Z / 3\Z}」の略解}<
    +p{
      まだ解いてない。
    }
  >
>