# 04.4.共役類

## Proposition

### Statement

$G$:Group
$\cdot:G \times G \rightarrow G$
$(g,x) \mapsto gxg^{-1}$
then
$\cdot$ is an action of $G$

### Proof

$x \in G$
$e \in G$ is an identity of $G$
$e \cdot x = exe^{-1} = x$
$x,g,g' \in G$
$g \cdot (g' \cdot x) = g (g' x g'^{-1}) g^{-1} = gg' x (gg')^{-1} = gg' \cdot x$

## Proposition

### Statement

$G$:Group
$x,y \in G$
$x \sim y :\Leftrightarrow \exists g \in G, gxg^{-1} = y$
then
$\sim$ is an equivalence relation

### Proof

(1)
$x = e x e^{-1} \therefore x \sim x$

(2)
$x \sim y$
$\Leftrightarrow \exists g \in G, g x g^{-1} = y$
$\Leftrightarrow g^{-1} (g x g^{-1}) g = g^{-1} (y) g$
$\Leftrightarrow x = g^{-1} y g$
$\Leftrightarrow x = g' y g'^{-1} \quad (g' = g^{-1})$
$\Leftrightarrow y \sim x$

(3)
$x \sim y, y \sim z$
$\Leftrightarrow \exists g,g' \in G, g x g^{-1} = y, g' y g'^{-1} = z$
$\Leftarrow \exists g,g' \in G, g' g x g^{-1} g'^{-1} = z$
$g'' = g'g \Rightarrow g'' g^{-1}g'^{-1} = e \therefore g^{-1}g'^{-1} = g''^{-1}$
$\therefore g'' x g''^{-1} = z \Leftrightarrow x \sim z$

## Proposition

### Statement

$G$:Group
$|G| \lt \infty$
(1)
$Z(x) = \{g \in G|gx = xg\}$
then
$|C(x)| = \frac{|G|}{|Z(x)|}$

(2)
$\{x_i\}_{i \le i \le n}$:完全代表系
then
$G = \coprod_{1 \le i \le n} C(x_i)$
$|G| = \sum_{1 \le i \le n} |C(x_i)|\quad$(類等式)
(2-1)
$\exists |C(x_i)| = 1$
(2-2)
$|C(x_i)| | |G|$
(2-3)
$|\{x_i||C(x_i)| = 1\}| | |G|$

### Proof

(1)
$a \in C(x)$
$\Leftrightarrow \exists g \in G, gxg^{-1} = a$
$\Leftrightarrow \exists g \in G, g \cdot x = a$
$\Leftrightarrow a \in G \cdot x$
$\therefore C(x) = G \cdot x$

$Z(x) = \{g \in G|gx = xg\}$
$= \{g \in G|g x g^{-1} = x\}$
$= \{g \in G|g \cdot x = x\} = G_x$
$\therefore$
$|C(x)| = |G \cdot x| = \frac{|G|}{|G_x|} = \frac{|G|}{|Z(x)|} \quad \because \text{Prop.安定化群による剰余類}$
(2-1)
$\because C(e) = \{x = g e g^{-1}| \forall g \in G\} \therefore |C(e)| = 1$
(2-2)
$\because$ (1)
(2-3)
$|C(x_i)| = 1 \Rightarrow C(x_i) = \{ x_i \}$
$\therefore \forall g \in G, g x_i g^{-1} = x_i$
$\therefore x_i \in Z(G)$
$Z(G)$ is a subgroup of $G$
$\therefore |\{x_i||C(x_i)| = 1\}| | |G|$

## Proposition

$G$:Group
$|G| = p^2 \quad q$ is a prime number.
then
$G$ is a commutative Group