# 04.2.群の直積

## Proposition

### Statement

$G_1,\cdots,G_n$:Group
then
$G_1 \times \cdots \times G_n$ is a Group.

### Proof

$(g_1,\cdots,g_n) \in G_1 \times \cdots \times G_n$
$(h_1,\cdots,h_n) \in G_1 \times \cdots \times G_n$
$(k_1,\cdots,k_n) \in G_1 \times \cdots \times G_n$
then
(1)
$((g_1,\cdots,g_n)(h_1,\cdots,h_n))(k_1,\cdots,k_n)$
$= (g_1 h_1,\cdots,g_n h_n) (k_1,\cdots,k_n)$
$= ((g_1 h_1)k_1,\cdots,(g_n h_n)k_n)$
$= (g_1 (h_1 k_1),\cdots,g_n (h_n k_n))$
$= (g_1,\cdots,g_n)(h_1 k_1,\cdots,h_n k_n)$
$= (g_1,\cdots,g_n)((h_1,\cdots,h_n)(k_1,\cdots,k_n))$
(2)
$(g_1,\cdots,g_n)(e_1,\cdots,e_n) = (g_1,\cdots,g_n)$
(3)
$(g_1,\cdots,g_n)(g_1^{-1},\cdots,g_n^{-1}) = (e_1,\cdots,e_n)$
QED

## Proposition

### Statement

$G_1,\cdots,G_n$:Group
$Prj_i:G_1 \times \cdots \times G_i \times \cdots \times G_n \rightarrow G_i$
$Prj_i:(g_1,\cdots,g_i,\cdots,g_n) \mapsto g_i \in G_i$
$Inj_i:G_i \rightarrow G_1 \times \cdots \times G_i \times \cdots \times G_n$
$Inj_i:g_i \mapsto (e_1,\cdots,g_i,\cdots,e_n)$
then
I
$Prj_i$:Homomorphism
II
$Inj_i$:Homomorphism
III
(a.)$\text{Ker }Prj_i = \text{Img }Inj_i \triangleleft G_1 \times \cdots \times G_n$
(b.)$\text{Img }Inj_1\cdots \text{Img }Inj_n = \{g_1 \cdots g_n|g_1 \in \text{Img }Inj_1,\cdots,g_n \in \text{Img }Inj_n\} = G_1 \times \cdots \times G_n$
(c.)$\text{Img }Inj_1 \cap \cdots \cap \text{Img }Inj_n = e = (e_1,\cdots,e_n) \in G_1 \times \cdots \times G_n$

### Proof

$(g_1,\cdots,g_n) \in G_1 \times \cdots \times G_n$
$(h_1,\cdots,h_n) \in G_1 \times \cdots \times G_n$
$(k_1,\cdots,k_n) \in G_1 \times \cdots \times G_n$
I
$Prj_i((g_1,\cdots,g_n)(h_1,\cdots,h_n))$
$= Prj_i((g_1 h_1,\cdots,g_n h_n))$
$=g_i h_i$
$=Prj_i((g_1,\cdots,g_n))Prj_i((h_1,\cdots,h_n))$
II
$Inj_i(g_i h_i)$
$= (e_1,\cdots,g_i h_i,\cdots,e_n)$
$= (e_1,\cdots,g_i,\cdots,e_n)(e_1,\cdots, h_i,\cdots,e_n)$
$= Inj_i(g_i)Inj_i(h_i)$
III
(a.)
$\text{Img}Inj_i = \{(x_1,\cdots,e_i,\cdots,x_n)|x_1 \in G_1,\cdots,x_n \in G_n\}$
$\text{Ker}Prj_i = \{(x_1,\cdots,x_n)|\text{Prj}_i(x_1,\cdots,x_n) = x_i = e_i \in G_i\}$
$= \{(x_1,\cdots,e_i,\cdots,x_n)|x_1 \in G_1,\cdots,x_n \in G_n\}$
$\therefore \text{Img}Inj_i = \text{Ker}Prj_i$
$\therefore \text{Img }Inj_i = \text{Ker }Prj_i \triangleleft G_1 \times \cdots \times G_n \quad(\text{Prop.カーネルは正規部分群})$
(b.)
$\text{Img}Inj_1 \cdots \text{Img}Inj_n = \{a_1 \cdots a_n|a_1 \in \text{Img}Inj_1, \cdots, a_n \in \text{Img}Inj_n\}$
$\forall i \in \{1 \cdots n\},\text{Img}Inj_i \subset G_1 \times \cdots \times G_n$
$\therefore a_1 \cdots a_n \in G_1 \times \cdots \times G_n$
$\therefore \text{Img}Inj_1 \cdots \text{Img}Inj_n \subset G_1 \times \cdots \times G_n$

$\forall (x_1,\cdots,x_n) \in G_1 \times \cdots \times G_n$
$(x_1,\cdots,x_n) = (e_1,x_2,\cdots,x_n) \cdots (e_1,\cdots,x_i,\cdots,e_n) \cdots (e_1,\cdots,e_{n-1},x_n)$
$\therefore (x_1,\cdots,x_n) \in \text{Img}Inj_1 \cdots \text{Img}Inj_n$
$\therefore \text{Img}Inj_1 \cdots \text{Img}Inj_n \supset G_1 \times \cdots \times G_n$

$\therefore \text{Img}Inj_1 \cdots \text{Img}Inj_n = G_1 \times \cdots \times G_n$
(c.)
$(x_1,\cdots,x_n) \in \text{Img}Inj_1 \cap \text{Img}Inj_2$
$(x_1,\cdots,x_n) = (x_1,e_2,e_3\cdots,e_n) \in \text{Img}Inj_1$
$(x_1,\cdots,x_n) = (e_1,x_2,e_3\cdots,e_n) \in \text{Img}Inj_2$
$\therefore \forall i \in \{1,\cdots,n\},x_i = e_i \in G_i$
$\therefore (x_1,\cdots,x_n) = (e_1,\cdots,e_n) \in \text{Img}Inj_1 \cap \cdots \cap \text{Img}Inj_n$
$\therefore \text{Img}Inj_1 \cap \cdots \cap \text{Img}Inj_n = \{e\}$
QED

## Proposition

### Statement

$G$:Group
$H,K$:Subgroup of $G$
then
$[H,K] = \{e\}$
$\Leftrightarrow$
$hk = kh \quad (\forall h \in H, \forall k \in K)$

### Proof

$[H,K] = \{e\}$
$hkh^{-1}k^{-1} = e \quad (\forall h \in H, \forall k \in K)$
$hk = kh \quad (\forall h \in H, \forall k \in K)$

## Proposition

### Statement

$G$:Group
$H,K$:Subgroup of $G$
$\phi:H \times K \rightarrow G$
$H \times K \ni (h,k) \mapsto hk \in G$
then
$H \times K \stackrel{\phi}{\simeq} G$
$\Leftrightarrow$
(1) $H,K \triangleleft G$
(2) $H \cap K = \{e\}$
(3) $G = HK \quad (= \{hk|h \in H, k \in K\})$

### Proof

$\Rightarrow$
(1)
$\forall x = \phi(h_x,k_x) \in G$
$\forall h \in H$
$\phi^{-1}(x) = (h_x,k_x)$
$\phi^{-1}(x^{-1}) = (h_x^{-1},k_x^{-1})$
$(\because x \cdot \phi(h_x^{-1}, k_x^{-1}) = \phi(h_x,k_x) \cdot \phi(h_x^{-1}, k_x^{-1}) = \phi(h_x h_x^{-1}, k_x k_x^{-1}) = e)$
$\phi^{-1}(h) = (h,e)$
$\therefore xhx^{-1} = \phi\phi^{-1}(xhx^{-1}) = \phi(\phi^{-1}(x)\phi^{-1}(h)\phi^{-1}(x^{-1})) = \phi((h_x,k_x )(h,e)(h_x^{-1},k_x^{-1})) = \phi(h_x h h_x^{-1}, e) = h_x h h_x^{-1} \in H$
$\therefore H \triangleleft G$
Similarly
$K \triangleleft G$
(2)
$x \in H \cap K$
$x^{-1} \in H \cap K \subset K$
$(x,x^{-1}) \in H \times K$
$\phi(x,x^{-1}) = x x^{-1} = e \therefore (x,x^{-1}) \in \text{Ker }\phi$
$\text{Ker }\phi = \{e\} \because \phi$ is injection.
$\therefore (x,x^{-1}) = (e,e)$ $\therefore x = e$
$\therefore H \cap K = \{e\}$
(3)
$\text{Img }\phi = G \quad \because \phi$ is surjection.
$\text{Img }\phi = HK \quad \because$ definition of $\phi$
$\therefore G = HK$
$\Leftarrow$
$\forall h \in H, \forall k \in K$
$kh^{-1}k^{-1} \in H \because H \triangleleft G$
$\therefore [h,k] = h(kh^{-1}k^{-1}) \in H$
$hkh^{-1} \in K \because K \triangleleft G$
$\therefore [h,k] = (hkh^{-1})k^{-1} \in K$
$\therefore [h,k] = hkh^{-1}k^{-1} \in H \cap K = \{e\}$
$\therefore hk = kh$
$\phi(h_1,k_1) \phi(h_2,k_2) = (h_1 k_1)(h_2 k_2)$
$\phi((h_1,k_1) \circ (h_2,k_2)) = \phi(h_1 h_2, k_1 k_2) = (h_1 h_2)(k_1 k_2)$
$k_1 h_2 = h_2 k_1$
$\therefore \phi(h_1,k_1) \phi(h_2,k_2) = \phi((h_1,k_1) \circ (h_2,k_2))$

$\phi(H \times K) = HK = G$
$\phi$ is a surjection.

$\phi(h,k) = e$
$\Rightarrow hk = e$
$\Rightarrow h = k^{-1} \in H \cap K = \{e\}$
$\therefore (h,k) = (e,e) = e \in H \times K$
$\therefore \text{Ker }\phi = \{e\}$
$\phi$ is a injection.
$\therefore H \times K \simeq G$

## Proposition

### Statement

(I)
$m,n \in \mathbb{Z}$
$\text{gcd}(m,n) = 1$
then
$\mathbb{Z}/\langle mn \rangle \simeq \mathbb{Z}/\langle m\rangle \times \mathbb{Z}/\langle n\rangle$
($\mathbb{Z}/mn\mathbb{Z} \simeq \mathbb{Z}/m\mathbb{Z} \times \mathbb{Z}/n\mathbb{Z}$)

(II)
$m \in \mathbb{Z}$
$m = p_1^{e_1} \cdots p_n^{e_n} \quad(p_1,\cdots,p_n:\text{Prime Number},e_1,\cdots,e_n \in \mathbb{N})$
then
$\mathbb{Z}/\langle m \rangle = \mathbb{Z}/\langle p_1^{e_1}\rangle \times \cdots \times \mathbb{Z}/\langle p_n^{e_n}\rangle$

### Proof

(I)
Let
$f:\mathbb{Z} \rightarrow \mathbb{Z}/m\mathbb{Z} \times \mathbb{Z}/n\mathbb{Z}$
$f:x \in \mathbb{Z} \mapsto (xm\mathbb{Z}, xn\mathbb{Z})$
then
$f$:Homomorphism
$\because f(x + y) = ((x + y)m\mathbb{Z}, (x + y)n\mathbb{Z}) = (xm\mathbb{Z}, xn\mathbb{Z}) + (ym\mathbb{Z}, yn\mathbb{Z}) = f(x) + f(y)$
$x \in \text{Ker}f \Leftrightarrow x \in m\mathbb{Z}, x \in n\mathbb{Z} \Leftrightarrow x \in mn\mathbb{Z}$
$\therefore \text{Ker}f = mn\mathbb{Z}$
$\therefore \mathbb{Z}/\text{Ker}f = \mathbb{Z}/mn\mathbb{Z} \simeq \text{Img}f = \mathbb{Z}/m\mathbb{Z} \times \mathbb{Z}/n\mathbb{Z}$
(II)
$\text{gcd}(p_1^{e_1}, p_2^{e_2}\cdots p_n^{e_n}) = 1$
$\therefore \mathbb{Z}/\langle m \rangle \simeq \mathbb{Z}/\langle p_1^{e_1}\rangle \times \mathbb{Z}/\langle p_2^{e_2}\cdots p_n^{e_n}\rangle$
$\text{gcd}\langle p_2^{e_2}, p_3^{e_3}\cdots p_n^{e_n}\rangle = 1$
$\therefore \mathbb{Z}/\langle m \rangle \simeq \mathbb{Z}/\langle p_1^{e_1}\rangle \times \mathbb{Z}/\langle p_2^{e_2}\cdots p_n^{e_n}\rangle \simeq \mathbb{Z}/\langle p_1^{e_1}\rangle \times \mathbb{Z}/\langle p_2^{e_2}\rangle \times \mathbb{Z}/\langle p_3^{e_3}\cdots p_n^{e_n}\rangle$
then
$\therefore \mathbb{Z}/\langle m \rangle \simeq \mathbb{Z}/\langle p_1^{e_1}\rangle \times \cdots \times \mathbb{Z}/\langle p_n^{e_n}\rangle$
QED