# 04.1.群の基本的な命題

## Proposition

カーネルは正規部分群

### Statement

$G,G'$:Group
$f:G \rightarrow G'$:Homomorphism
then
$\text{Ker}f = \{x \in G|f(x) = 0\} \triangleleft G$

### Proof

$\forall k \in \text{Ker}f, \forall g \in G \Rightarrow f(g k g^{-1}) = f(g)f(k)f(g)^{-1} = f(g)ef(g)^{-1} = e$
$\therefore g \text{Ker}f g^{-1} \subset \text{Ker}f$
QED

## Proposition

### Statement

$G,G'$:Group
$f:G \rightarrow G'$:Homomorphism
then
$G/\text{Ker}f \simeq \text{Img}f$

### Proof

Let
$\bar f:G/\text{Ker}f \rightarrow G'$
$\bar f:g \text{Ker}f \in G/\text{Ker}f \mapsto f(x) \in G' \quad (x \in g\text{Ker}f)$
then
$x,y \in g\text{Ker}f \Rightarrow \exists x',y'\in g\text{Ker}f, x = gx', y = gy'$
$f(x) = f(g)f(x') = f(g)$
$f(y) = f(g)f(y') = f(g)$
$\therefore f(x) = f(y)$
$\therefore \bar f$ is well-defined
$\bar f(g\text{Ker}f \cdot h\text{Ker}f) = \bar f(gh\text{Ker}f) = f(gh) = f(g)f(h) = \bar f(g\text{Ker}f) \bar f(h\text{Ker}f)$
$\therefore \bar f$ is homomorphism
$\bar f(g\text{Ker}f) = \bar f(h\text{Ker}f)$
$\Leftrightarrow f(g) = f(h)$
$\Leftrightarrow f(g)f(h)^{-1} = f(gh^{-1}) = e$
$\Leftrightarrow gh^{-1} \in \text{Ker}f$
$\Leftrightarrow g\text{Ker}f = h\text{Ker}f$
$\therefore \bar f$ is injective
$\therefore G/\text{Ker}f \simeq \text{Img}f$

## Proposition

ラグランジェの定理

### Statement

$G$:Group
$H$:Subgroup
$G/H$:Set of Left Coset
then
(1)
$aH,bH \in G/H \Rightarrow |aH| = |bH|$

(2)
$aH \cap bH \neq \emptyset \Rightarrow aH = bH$
(3)
$|G| = (G:H)|H|$

### Proof

(1)
Let
$a \in G$
$f_a:H \rightarrow aH$
$f_a:x \mapsto ax$
then
$f_a$ is injective
($\because f_a(x) = f_a(x') \Leftrightarrow ax = ax' \Leftrightarrow x = x'$)
$f_a$ is surjective
($\because y \in aH \Rightarrow \exists h \in H, y = ah = f_a(h)$)

(2)
Let
$c \in aH \cap bH$
then
$\exists h,h' \in H \text{ s.t. } c = ah = bh'$
$\Rightarrow a = bh'h^{-1} \in bH, b = ahh'^{-1} \in aH$
$\therefore$
$\forall x \in aH, \exists h'' \in H, x = ah'' = bh'h^{-1}h'' \in bH \therefore aH \subset bH$
$\forall y \in bH, \exists h''' \in H, y = bh''' = ahh'^{-1}h''' \in aH \therefore bH \subset aH$
$\therefore aH = bH$
(3)
Let
$\phi:G/H \rightarrow G$
$\phi:aH \mapsto a_\lambda \in aH$
$\Lambda = \{\lambda| a_\lambda = \phi(aH), aH \in G/H\}\quad$(left coset representatives)
then
$G = \cup_{\lambda\in \Lambda} a_\lambda H$
$a_\lambda H \cap a_{\lambda'} H = \emptyset \quad (\lambda \neq \lambda')$
$\forall a_\lambda, |H| = |a_\lambda H|$
$\therefore |G| = |\Lambda||H| = (G:H)|H|$
QED

## Proposition

### Statement

$G$:Group
$G = \langle a\rangle \quad(a \in G)$
$|G| = n \quad(n \in \mathbb{N})$
then
$G \simeq \mathbb{Z}/n\mathbb{Z}$

### Proof

$G = \{1 (= a^0 = a^n), a, \cdots, a^{n-1}\}$
Let
$f:\mathbb{Z} \rightarrow G$
$f:i \in \mathbb{Z} \mapsto a^i \in G$
then
$f$:Homomorphism
$\because f(i+j) = a^{i+j} = a^i a^j = f(i)f(j)$
$\text{Ker}f = n\mathbb{Z}$
$\mathbb{Z}/n\mathbb{Z} = \mathbb{Z}/\text{Ker}f \simeq \text{Img}f = G$
QED

## Proposition

### Statement

$G$:Commutative Group
$H_1,\cdots,H_n$:Subgroup of $G$
$H_1 H_2 \cdots H_n = \{h_1 h_2 \cdots h_n | h_1 \in H_1, h_2 \in H_2, \cdots, h_n \in H_n\}$
$|H_i| = p_i^{a_i}\quad (p_i$:Prime Number)
then
$H_i \cap H_j = \{e\}\quad(i \neq j)$
$\text{gcd}(p_i,p_j) = 1 \quad (i \neq j)$
$\Rightarrow$
$|H_1 H_2 \cdots H_n| = |H_1||H_2| \cdots |H_n|$

### Proof

Let $h_1 h_2 \in H_1 H_2, h'_1 h'_2 \in H_1 H_2$
then
$h_1 h_2 = h'_1 h'_2 \Leftrightarrow (h'_1)^{-1}(h_1 h_2)(h_2)^{-1} = (h')_1^{-1}(h'_1 h'_2)(h_2)^{-1}$
$\Leftrightarrow H_1 \ni (h'_1)^{-1}(h_1) = (h'_2)(h_2)^{-1} \in H_2$
$\therefore (h'_1)^{-1}(h_1) = (h'_2)(h_2)^{-1} = e \in H_1 \cap H_2$
$\therefore h'_1 = h_1, h'_2 = h_2$
$\therefore |H_1 H_2| = |H_1||H_2| = p_1^{a_1} p_2^{a_2}$

Let $K_2 = H_1 H_2, K_i = K_{i-1} H_i\quad(i = 3 \cdots n)$
then
Suppose $K_i = p_1^{a_1}\cdots p_i^{a_i}$
then
$K_i \cap H_{i+1} = \{e\}$
$\because x \in K_i \cap H_{i+1} \Rightarrow |\langle x \rangle| = p_1^{b_1}p_2^{b_2}\cdots p_i^{b_i} = p_{i+1}^{b_{i+1}} = 1\quad(\because\text{Prop.ラグランジェの定理}) \therefore x = e$
$k_i h_{i+1} = k'_i h'_{i+1} \Leftrightarrow (k'_i)^{-1}k_i h_{i+1}(h_{i+1})^{-1} = (k'_i)^{-1}k'_i h'_{i+1}(h_{i+1})^{-1}$
$\Leftrightarrow K_i \ni (k'_i)^{-1}k_i = h'_{i+1}(h_{i+1})^{-1} \in H_{i+1}$
$\therefore (k'_i)^{-1}k_i = h'_{i+1}(h_{i+1})^{-1} = e \in K_i \cap H_{i+1}$
$\therefore k_i = k'_i, h_{i+1} = h'_{i+1}$
$\therefore |K_{i+1}| = |K_i||H_{i+1}| = p_1^{a_1} p_2^{a_2} \cdots p_{i+1}^{a_{i+1}}$

$\therefore |H_1 \cdots H_n| = |K_n| = p_1^{a_1} p_2^{a_2} \cdots p_n^{a_n}$
QED