# 部分群と正規部分群の定義

Assume $G$ is a group.
$H \le G$ (is subgroup of $G$) if $H \subset G$ and is a group under the same operarion
Also, $H \vartriangleleft G$ (is normal subgroup of $G$) if $H \le G$ and  $\forall g \in G, \forall h \in H, ghg^{-1} \in H \;$''
-------------------------proposition---------------------
$H \vartriangleleft G$
$\Updownarrow$
(1) - $\forall g \in G, gHg^{-1} \subset H$
$\Updownarrow$
(2) - $\forall g \in G, gHg^{-1} = H$
$\Updownarrow$
(3) - $\{ gH | g \in G \} = \{ Hg | g \in G\}$
$\Updownarrow$
(4) - $\forall g \in G, gH = Hg$
$\Updownarrow$
(5) - $\cup_{h \in H} \{ghg^{-1}| g \in G \} = H$
$\Updownarrow$
(6) - $\exists Z:\mathrm{group}, \exists f:G \to Z, \forall x \in G, \forall y \in G, f(xy) = f(x)f(y) \ \mathrm{and}\ \mathrm{ker} f = H$
-------------------------proof---------------------
(1) $\Rightarrow$ (2)
$x\in G$ then $x^{-1}\in G \Rightarrow x^{-1}H(x^{-1})^{-1} \subset H \Leftrightarrow x^{-1}Hx \subset H$
$H = x(x^{-1}Hx)x^{-1} \subset x H x^{-1}$
$\therefore H = xHx^{-1}$
(2) $\Rightarrow$ (1)

(2) $\Rightarrow$ (4)
Assume $H = gHg^{-1}$
then $Hg = gHg^{-1}g = gH.$
(4) $\Rightarrow$ (2)
Assume $gH = Hg$
then $gHg^{-1} = Hg^{-1}g = H.$
(2) $\Rightarrow$ (5)
(3) $\Rightarrow$ (?)

(4) $\Rightarrow$ (5)
(5) $\Rightarrow$ (6)
(6) $\Rightarrow$ (1)
$\forall g \in G, \forall k \in gHg^{-1} =g\mathrm{Ker}fg^{-1}=\{ghg^{-1}\| h\in\mathrm{Ker}f\}$
$\exists \hat{k} \in \mathrm{Ker}f, k=g\hat{k}g^{-1}$ $f(k)=f(g\hat{k}g^{-1}) =f(g)f(\hat{k})f(g^{-1}) =f(g)e_H f(g^{-1})=f(g)f(g)^{-1}=e_H$ thus, $k\in {\rm{Ker}}f=H$.
(1) $\Rightarrow$ (6)