# 04.3.群の作用

## Proposition

### Proof

$G$:Group
$g,x \in G$
$\cdot:G \times G \rightarrow G$
$(g,x) \mapsto g x \in G$
(1) $e \cdot x = ex = x \quad$ ($e$ is identity of $G$)
(2) $g \cdot (h \cdot x) = ghx = (gh) \cdot x \quad (g,h \in G)$
QED

## Proposition

### Statement

$G$:Group
$S$:Set
for $a,b \in S$
$a \sim b \overset{\text{def}}{\Longleftrightarrow} \exists g \in G, g \cdot a = b$
then
$\sim$ is a equivalent relation.
therefore
Let
$S' = \{s \in \bar s| \bar s \in S / \sim \}$
then
$S = \coprod_{s \in S'} G \cdot s$
$|S| = \sum_{s \in S'} |G \cdot s|$

### Proof

(1)
$a \sim a \because e \in G, e\cdot a = a$
(2)
$a \sim b \Rightarrow \exists g \in G, g \cdot a = b$
$\Leftrightarrow g^{-1} \cdot g \cdot a = g^{-1} \cdot b$
$\Leftrightarrow a = g^{-1} \cdot b$
$\therefore b \sim a$
(3)
$a \sim b, b \sim c$
$\exists g \in G, g \cdot a = b$
$\exists g' \in G, g' \cdot b = c$
$\therefore g' \cdot g \cdot a = g' \cdot b = c$
$\therefore a \sim c$

## Proposition

### Statement

$G$:Group
$X$:Set
$\cdot$:Action of $G$ over $S$
$x \in X$
$G_x = \{g \in G|g \cdot x = x\}$
then
$G_x$ is a Group

### Proof

$g,g' \in G_x$
$g \cdot (g' \cdot x) = g \cdot x = x$
$\therefore g \cdot g' \in G_x$
$g,g_1,g_2,g_3 \in G_x$
(1) $g_1 (g_2 g_3) = (g_1 g_2) g_3 \quad \because g_1,g_2,g_3 \in G$
(2) $e \in G$:identity $\Rightarrow e \cdot x = x \therefore e \in G_x$
(3) $g \cdot x = x \Leftrightarrow g^{-1} \cdot g \cdot x = g^{-1} \cdot x$
$\therefore g^{-1} \cdot x = x \therefore g^{-1} \in G_x$

## Proposition

### Statement

$G$:Group
$X$:Set
$\cdot:G \times X \rightarrow X$:Action of $G$ over $S$
$x \in X$
$G_x = \{g \in G|g \cdot x = x\}$
then
$|G \cdot x| = |G/G_x| = [G:G_x]$
therefore
$|G \cdot x| = |G|/|G_x|$

### Proof

define $\phi$ as
$\phi:G/G_x \Rightarrow G \cdot x$
$gG_x \in G/G_x \mapsto g \cdot x$
$\phi$ is well-defined
$\because$
$h \in g G_x$
then
$\Leftrightarrow \exists g' \in G_x,h = gg'$
$\therefore h \cdot x = (g g') \cdot x = g \cdot (g' \cdot x) = g \cdot x$

$\forall g \cdot x \in G \cdot x$
$\phi(g G_x) = g \cdot x$
$\therefore$ surjective.

$h \cdot x = g \cdot x$
$\Leftrightarrow g^{-1} h \cdot x = g^{-1} \cdot h \cdot x = g^{-1} \cdot g \cdot x = x$
$\therefore g^{-1} h \in G_x$
$\therefore g G_x = h G_x$
$\therefore$ injective.
QED