# §3 群的映射
每当你定义一个想法时，最好能知道这个想法是哪些函数的朋友。

**提问：** 我们要研究哪些类型的**函数**?

**例 3.1**
$$
        \begin{aligned}
            \text{集合}~S,T~&\leftrightarrow~\text{任意函数}~f:S\to T\\
            \text{空间}~X,Y~&\leftrightarrow~\text{连续函数}~f:X\to Y\\
            \text{光滑曲线+曲面}~X,Y~&\leftrightarrow~\text{可微函数}~f:X\to Y\\
            \text{群}~G,H~&\leftrightarrow~\text{群同态}~\phi:G\to H\\
        \end{aligned}
$$



## §3.1 群同态
### §3.1.1 群同态的定义
**定义 3.1** 令$G,H$是群。从$G$到$H$的一个**群同态(group homomorphism)** 是一个函数
$$
\phi:G\to H
$$
使得 $\forall g_{1}, g_{2}\in G$,
$$
\phi(g_{1}g_{2})=\phi(g_{1})\phi(g_{2}).
$$
### §3.1.2 群同态的例子
$\begin{aligned}
        \text{\textbf{\small{例 3.2}}}~\exp:(\mathbb{R},+)&\to \mathbb{R}^{\times}~\text{是一个群同态。}\\
        t&\mapsto \mathrm{e}^{t}
    \end{aligned}$
<details>
<summary>证明.</summary>

$\forall t_{1}$, $t_{2}\in(\mathbb{R},+)$,
    
$$
\mathrm{e}^{t_{1}+t_{2}}=\mathrm{e}^{t_{1}}\cdot \mathrm{e}^{t_{2}}
$$
    
$$
~\tag*{$\square$}
$$
</details>

$\begin{aligned}\text{\textbf{\small{例 3.3}}}~\det: GL_{n}(\mathbb{R})&\to\mathbb{R}^{\times}~\text{是一个群同态。}\\
    M&\mapsto \det(M)
    \end{aligned}$

<details>
<summary>证明.</summary>

$\forall M_{1},M_{2}\in GL_{n}(\mathbb{R})$, 

$$
\det(M_{1}M_{2})=\det(M_{1})\det(M_{2})
$$

$$
~\tag*{$\square$}
$$
</details>

$\begin{aligned}\text{\textbf{\small{例 3.4}}}~(\mathbb{R},+)&\to S^{1}~\text{是一个群同态。}\\
    t&\mapsto \mathrm{e}^{\mathrm{i}t}
    \end{aligned}$

<details>
<summary>证明.</summary>

$\forall t_{1},t_{2}\in(\mathbb{R},+)$,

$$
\mathrm{e}^{\mathrm{i}(t_{1}+t_{2})}=\mathrm{e}^{\mathrm{i}t_{1}}\cdot \mathrm{e}^{\mathrm{i}t_{2}}
$$

$$
~\tag*{$\square$}
$$

</details>


**例 3.5** 如果$V,W$是向量空间，它们在加法$+$下是群。任意线性映射
$$
        \phi: V\to W
$$
是一个群同态。任何线性子空间是一个子群。

<details>
<summary>证明.</summary>
    
(a) 向量空间$V$是群, 因为$V$满足

(1) 结合律: $V$中得加法满足结合律, i.e. 对于$u,v,w\in V$, $(u+v)+w=u+(v+w)$;

(2) 恒元: 存在一个元素$0\in V$, 使得对于任意$v\in V$, 都有$v+0=v$;

(3) 逆元: 对于任意$v\in V$, 存在$-v\in V$, 使得$v+(-v)=0$。

(b) 任何两个向量空间$V$和$W$之间的线性映射$\phi:V\to W$是一个群同态, 因为$\phi$满足

(1) 同态的性质: 对于任意$u, v\in V$, 有
$$
\phi(u+v)=\phi(u)+\phi(v)\tag*{(线性映射的定义)}
$$

(2) 保持恒元: 因为$\phi$是线性的,
$$
\phi(0_{V})=0_{W}
$$
其中, $0_{V}$和$0_{W}$分别是$V$和$W$中的恒元;

(3) 保持逆元: 对于任意$v\in V$,
$$
\phi(-v)=-\phi(v)
$$
由于$\phi$的线性性, 依赖于标量乘法和加法逆元的$\phi$的这条性质成立。

(c) 任意线性子空间是一个子群, 因为

- 一个线性子空间$U\subseteq V$在加法下也是一个群, 因为它继承了$V$的群结构;

- $V$的恒元$0_{V}$也是$U$中的恒元;

- $U$中的逆元也从$V$中继承。

$$
~\tag*{$\square$}
$$
</details>

### §3.1.3 群同态的性质
**命题 3.1** 令$\phi:G\to H$是一个群同态。我们有

(a) $\phi(1_{G})=1_{H}$。

(b) $\phi(g^{-1})=\phi(g)^{-1}$。

<details>
<summary>证明.</summary>
    
(a) 对于任意$g\in G$,
$$    
    \begin{aligned}
        \phi(g)&=\phi(1_{G}\cdot g)\\
        &=\phi(1_{G})\cdot\phi(g)
    \end{aligned}~~~~~~~~~~~~~~~~~~
    \begin{aligned}
        \text{(2)}&\\
        \text{(同态的定义)}&
    \end{aligned}
$$
令$h$是$\phi(g)$的逆。则
$$
    \begin{aligned}
        \phi(g)\cdot h=\phi(1_{G})\cdot\phi(g)\cdot h\quad&\Rightarrow\quad 1_{H}=\phi(1_{G})\cdot 1_{H}\\
        &\Rightarrow\quad 1_{H}=\phi(1_{G})
    \end{aligned}~~~~~~~~~~~~~~~~~~
    \begin{aligned}
        \text{(3)}&\\
        \text{(2)}&
    \end{aligned}
$$    
(b) $1_{H}=\phi(1_{G})=\phi(g\cdot g^{-1})=\phi(g)\cdot\phi(g^{-1})\quad \Rightarrow \phi(g^{-1})=\phi(g)^{-1}$
$$
~\tag*{$\square$}
$$
</details>

**命题 3.2** $\mathrm{id}_{G}: G\to G$是一个同态。

<details>
<summary>证明.</summary>

考虑恒等映射$\mathrm{id}_{G}:G\to G$定义为$\mathrm{id}_{G}(g)=g$, 对于任意$g\in G$。要说明$\mathrm{id}_{G}$是一个同态, 我们需要说明:对于任意$g_{1}$, $g_{2}\in G$, 我们有
$$
\mathrm{id}_{G}(g_{1}\cdot g_{2})=\mathrm{id}_{G}(g_{1})\cdot \mathrm{id}_{G}(g_{2})
$$

对于任意$g_{1}$, $g_{2}\in G$, 我们有
$$
\mathrm{id}_{G}(g_{1}\cdot g_{2})=g_{1}\cdot g_{2}\tag*{($\mathrm{id}_{G}$的定义)}
$$
$$
\mathrm{id}_{G}(g_{1})\cdot \mathrm{id}_{G}(g_{2})=g_{1}\cdot g_{2}\tag*{(因为$\mathrm{id}_{G}(g_{i})=g_{i}$)}
$$
所以, 我们有
$$
\mathrm{id}_{G}(g_{1}\cdot g_{2})=g_{1}\cdot g_{2}=\mathrm{id}_{G}(g_{1})\cdot \mathrm{id}_{G}(g_{2})
$$
因此, $\mathrm{id}_{G}:G\to G$是一个同态。
$$
~\tag*{$\square$}
$$
</details>

**命题 3.3** 如果 $G\stackrel{\phi}{\longrightarrow}H$, $H\stackrel{\psi}{\longrightarrow}K$是同态，那么$\psi\circ\phi$是一个同态。

<details>
<summary>证明.</summary>

为了证明两个同态映射$\phi$和$\psi$的复合映射也是一个同态, 我们需要验证$\psi\circ\phi$满足: 对于任意$g_{1}, g_{2}\in G$, 有
$$
(\psi\circ\phi)(g_{1}\cdot g_{2})=(\psi\circ\phi)(g_{1})\cdot (\psi\circ\phi)(g_{2})
$$
根据复合映射的定义, 我们得到
$$
(\psi\circ\phi)(g_{1}\cdot g_{2})=\psi(\phi(g_{1}\cdot g_{2}))
$$
应用$\phi$的同态性质, 我们得到
$$
\psi(\phi(g_{1}\cdot g_{2}))=\psi(\phi(g_{1})\cdot \phi(g_{2}))
$$
再应用$\psi$的同态性质, 我们得到
$$
\psi(\phi(g_{1})\cdot \phi(g_{2}))=\psi(\phi(g_{1}))\cdot\psi(\phi(g_{2}))
$$
根据复合映射的定义, 我们有
$$
\psi(\phi(g_{1}))\cdot\psi(\phi(g_{2}))=(\psi\circ\phi)(g_{1})\cdot (\psi\circ\phi)(g_{2})
$$
至此, 我们证明了
$$
(\psi\circ\phi)(g_{1}\cdot g_{2})=(\psi\circ\phi)(g_{1})\cdot (\psi\circ\phi)(g_{2})
$$
因此, 如果 $G\stackrel{\phi}{\longrightarrow}H$, $H\stackrel{\psi}{\longrightarrow}K$是同态，那么$\psi\circ\phi$是一个同态。
$$
~\tag*{$\square$}
$$
</details>

**命题 3.4** 如果$H\subset G$是一个子群，那么包含映射$i: H\hookrightarrow G$是一个同态。

<details>
<summary>证明.</summary>

为了证明包含映射$i: H \hookrightarrow G$是一个同态, 其中$H \subset G$是一个子群, 我们需要验证同态性质: 对于所有$h_1, h_2 \in H$,
$$
i(h_1 \cdot h_2) = i(h_1) \cdot i(h_2)
$$
包含映射$i:H\hookrightarrow G$定义为: 对于任意$h\in H$, 有
$$
i(h)=h
$$
根据包含映射的定义, 对于任意$h_{1}, h_{2}\in H$, 有
$$
\begin{aligned}
i(h_{1})=h_{1}\\
i(h_{2})=h_{2}
\end{aligned}
$$
所以
$$
i(h_{1})\cdot i(h_{2})=h_{1}\cdot h_{2}
$$
同时, $h_1 \cdot h_2\in H$, 根据包含映射定义
$$
i(h_1 \cdot h_2) = h_1 \cdot h_2
$$
因此, 对于任意$h_1, h_2 \in H$, 有
$$
i(h_1 \cdot h_2) = i(h_1) \cdot i(h_2)
$$
即: 包含映射$i: H \hookrightarrow G$是一个同态。
$$
~\tag*{$\square$}
$$
</details>

### §3.1.4 群同态的核和像
**定义 3.2** 给定一个群同态
$$
\phi:G\to H
$$
$\phi$的**核(kernel)** 是集合
$$
\ker(\phi)=\{g\in G\mid\phi(g)=1_{H}\}.
$$
$\phi$**像(image)** 是集合
$$
\mathrm{im}(\phi)=\{h\in H\mid h=\phi(g)~\text{对于一些}~g\in G\}.
$$

**命题 3.5** $\ker(\phi)\subset G$, $\mathrm{im}(\phi)\subset H$是子群。

<details>
<summary>证明.</summary>

$\ker(\phi)$是$G$的子群, 因为:

(1) $\phi(g_{1}), \phi(g_{2})=1_{H}$
$$
\begin{aligned}
                \Rightarrow \phi(g_{1}g_{2})&=\phi(g_{1})\cdot\phi(g_{2})\\
                &=1_{H}\cdot 1_{H}\\
                &=1_{H}
            \end{aligned}
$$

(2) $\phi(1_{G})=1_{H}$, 所以$1_{G}\in\ker(\phi)$.

(3) $\phi(g)=1_{H}$
$$
\begin{aligned}
                &\Rightarrow\quad \phi(g^{-1})=1_{H}^{-1}=1_{H}\\
                &\Rightarrow\quad g^{-1}\in\ker(\phi)
        \end{aligned}
$$

$\mathrm{im}(\phi)$是$H$的子群, 因为
 
(1) $h_{i}=\phi(g_{i})$ $\Rightarrow$ $h_{1}h_{2}=\phi(g_{1})\phi(g_{2})=\phi(g_{1}g_{2})$.
 
(2) $\phi(1_{G})=1_{H}$, 所以$1_{H}\in\mathrm{im}(\phi)$.

(3) $h=\phi(g)$ $\Rightarrow$ $h^{-1}=\phi(g^{-1})$.

$$
~\tag*{$\square$}
$$
</details>

## §3.2 群同构
### §3.2.1 群同构的定义
**定义 3.3** 如果一个群同态$\phi$是一个双射，那么$\phi$被称为是一个**群同构(group isomorphism)**。

同构不是等于，就像集合的双射不是等于一样。

**例 3.6** 五个香蕉的集合不等于五个苹果的集合。

但我们是用同构对群的分类，就像我们用双射对集合分类。

### §3.2.2 群同构的例子
**例 3.7** $\exp:(\mathbb{R},+)\to(\mathbb{R}_{>0},\times)$是一个群同构。

<details>
<summary>证明.</summary>

为了证明$\exp: (\mathbb{R}, +) \to (\mathbb{R}_{>0}, \times)$是一个群同构, 我们需要说明两点: 

(1) **同态**: $\exp$保持群作用, i.e., 对于任意$x, y \in \mathbb{R}$, 都有
$$
\exp(x + y) = \exp(x) \cdot \exp(y)
$$
    
(2)**双射**: $\exp$是一个双射, 意味着它同时是一个单射和满射。
    
现在让我们进行验证:    

对于$x, y \in \mathbb{R}$,
$$
\exp(x + y) = e^{x+y}
$$
应用指数的性质,
$$
e^{x+y} = e^x \cdot e^y = \exp(x) \cdot \exp(y)
$$
因此, $\exp(x+y) = \exp(x) \cdot \exp(y)$, 说明了$\exp$是一个同态。
    
为了说明$\exp$是单射, 假设对于一些$x, y \in \mathbb{R}$有$\exp(x) = \exp(y)$。
$$
\exp(x) = \exp(y) \Rightarrow e^x = e^y
$$
在等式两边取自然对数(这都是有效的, 因为对于所有$x$, 都有$e^x > 0$),
$$
x = y
$$

因此, $\exp$是单射。
    
为了说明$\exp$是满射, 我们需要说明对于任意$y \in \mathbb{R}_{>0}$, 存在$x \in \mathbb{R}$使得$\exp(x) = y$。

因为$y > 0$, 存在$x = \ln(y) \in \mathbb{R}$ (其中, $\ln$表示自然对数) 使得$\exp(x) = e^{\ln(y)} = y$。

因此, $\exp$是满射。
    
因为$\exp$既是一个同态又是一个双射, 它在$(\mathbb{R}, +)$和$(\mathbb{R}_{>0}, \times)$之间是一个群同构。
    
因此, $\exp: (\mathbb{R}, +) \to (\mathbb{R}_{>0}, \times)$是一个群同构。  
$$
~\tag*{$\square$}
$$
</details>

**定理 3.6** 每个阶为 $n$ 的有限循环群都同构于 $\mathbb{Z}/n\mathbb{Z}$。

<details>
<summary>证明.</summary>
设 $G=\langle g\rangle$ 是一个阶为 $n$ 的循环群。定义函数 $\phi: G \to \mathbb{Z}/n\mathbb{Z}$ 为
$$
\phi(g^{k})=k \pmod n。
$$
(1) 同态性：
$$
\phi(g^{a}\cdot g^{b})=\phi(g^{a+b})=(a+b) \pmod n。
$$
另一方面，
$$
\phi(g^{a})+\phi(g^{b})=a+b \pmod n。
$$
两边相同，所以 $\phi$ 是一个同态。

(2) 单射性：如果 $\phi(g^{a})=\phi(g^{b})$，则 $a\equiv b \pmod n$，意味着 $g^{a}=g^{b}$。

(3) 满射性：对于任意 $k\in\mathbb{Z}/n\mathbb{Z}$，我们有 $\phi(g^{k})=k \pmod n$。
$$
~\tag*{$\square$}
$$
</details>

### §3.2.3 群同构的性质
**命题 3.7** 如果$\phi:G\to H$是一个群同构, $\phi^{-1}$是一个群同构。

<details>
<summary>证明.</summary>
很明显, 它是一个双射。需要证明$\phi^{-1}$是一个同态: $\forall g_{1},g_{2}\in G$, $h_{1}=\phi(g_{1}),h_{2}=\phi(g_{2})$, 我们有
$$
    \begin{align*}
        \phi^{-1}(h_{1}\cdot h_{2})&=\phi^{-1}(\phi(g_{1})\cdot\phi(g_{2}))\tag{$\phi$满射}\\
        &=\phi^{-1}(\phi(g_{1}\cdot g_{2}))\tag{$\phi$同态}\\
        &=(\phi^{-1}\circ\phi)(g_{1}\cdot g_{2})\tag{记号}\\
        &=g_{1}\cdot g_{2}\tag{$\phi^{-1}$的定义}\\
        &=\phi^{-1}(h_{1})\cdot \phi^{-1}(h_{2})\tag{$g_{1}$, $g_{2}$的定义}
    \end{align*}
$$
$$
~\tag*{$\square$}
$$
</details>

## §3.3 乘积群
乘积群是一种通过组合已知群的元素和运算来构造新群的方法。

**定义 3.4** 给定两个群 $G_{1}$ 和 $G_{2}$，**乘积群(product group)** $G_{1}\times G_{2}$ 是有序对的集合：
$$
G_{1}\times G_{2}=\{(g_{1},g_{2})\mid g_{1}\in G_{1}, g_{2}\in G_{2}\}
$$
其群运算定义为
$$
(g_{1},g_{2})\cdot (h_{1},h_{2})=(g_{1}\cdot h_{1},g_{2}\cdot h_{2})。
$$
**命题 3.8** 给定两个群 $G_{1}$ 和 $G_{2}$，$G_{1}\times G_{2}$ 是一个群。

<details>
<summary>证明.</summary>
(1) 结合性：由 $G_{1}$ 和 $G_{2}$ 的结合性直接得出。

(2) 单位元：$(1_{G_{1}},1_{G_{2}})$ 是单位元。

(3) 逆元：$(g_{1},g_{2})^{-1}=(g_{1}^{-1},g_{2}^{-1})$。
$$
~\tag*{$\square$}
$$
</details>

**例 3.8 (实数的Cartesian积)**
群 $\mathbb{R}\times\mathbb{R}$ 在标准加法下与Euclid平面 $\mathbb{R}^{2}$ 同构。

**例 3.9 (Klein四元群)**
$\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$ 被称为**Klein四元群(Klein four group)**。它是最小的非循环群。

**例 3.10** 如果 $\{z\in T\subset \mathbb{C}: |z|=1\}$，则 $\mathbb{C}^{\times}=\mathbb{R}^{\times}\times T$。
## §3.4 自同构
理解乘积群自然引出了对群内对称性的研究。自同构是一个从群映射到自身的双射同态，它描述了群的结构如何以不同的方式映射到自身。当自同构应用于乘积群时，尤其引人注目，因为它揭示了多个群的内部结构在映射下如何相互作用。

**定义 3.5** 令$X$是一个集合。我们记
$$
\mathrm{Aut}(X):=\mathrm{Aut}_{\text{set}}(X):=\{\text{双射}~X\to X\}
$$
为从$X$到它本身的双射的集合。

**命题 3.9** $\mathrm{Aut}(X)$在复合下构成一个群。 

<details>
<summary>证明.</summary>

$\mathrm{Aut}(X)$在复合下是一个群, 因为

(1) 复合函数是结合的:
$$
(f\circ g)\circ h=f\circ (g\circ h).
$$

$\begin{aligned}
         (2)~\mathrm{id}_{X}:X&\to X\text{是恒元}, \text{因为}f\circ \mathrm{id}_{X}=\mathrm{id}_{X}\circ f=f.\\
            x&\mapsto x
        \end{aligned}$
        
(3) $f^{-1}$是$f$的逆:
$$
f\circ f^{-1}=\mathrm{id}_{X}=f^{-1}\circ f.
$$
$$
~\tag*{$\square$}
$$
</details>

## §3.5 对称群
**定义 3.6** 令
$$
\underline{n}=\{1,\ldots,n\}.
$$
则
$$
\mathrm{Aut}_{\text{set}}(\underline{n})=:S_{n}
$$
是$n$个元素的集合上的**对称群(symmetric group)**。

**例 3.11** 下面是对称群的例子:
- $n=1$: $\mathrm{Aut}(\{1\})$是有一个元素的群:
$$
\begin{aligned}
S_{1}&=\{\text{双射}~\{1\}\to\{1\}\}\\
&=\{\mathrm{id}_{\underline{1}}\}
\end{aligned}       
$$
- $n=2$: $\mathrm{Aut}(\{1,2\})$是由两个元素的群:
$$
S_{2}=\left\{\left(\mathrm{id}:\begin{aligned}
				1\mapsto 1\\
				2\mapsto 2
			\end{aligned}
			\right),\left(
			\sigma:\begin{aligned}
				1\mapsto 2\\
				2\mapsto 1
			\end{aligned}
			\right)\right\}
$$
其满足
$$
\sigma\circ\sigma=\sigma^{2}=\mathrm{id}.
$$
- $S_{3}$有$3!$个元素。我们将很快学到它的结构。

- 一般地, $S_{n}$是有$n!$元素的群。

**定义 3.7** **集合$G$上的对称群**是从$G$到其自身的所有双射构成的群：
$$
\mathrm{Sym}(G) = \mathrm{Aut}_{\text{set}}(G).
$$

该群包含了$G$的所有元素的排列。当$G$是一个包含$n$个元素的有限集合时，$\mathrm{Sym}(G)$本质上与$S_n$相同，但这种推广允许我们考虑任何集合$G$上的排列，不论其基数大小。

对称群在群论的研究中是基本的，因为它们通过涵盖有限集合的所有可能排列，捕捉了对称性的本质。令人感兴趣的是，每个群，无论是有限的还是无限的，都可以表示为排列群的一个子群。这一深刻的联系由Cayley定理形式化了。

## §3.6 Cayley定理
**定理 3.10 (Cayley定理)** 每个群$G$同构于集合$G$上的对称群$\mathrm{Sym}(G)$的一个子群。

<details>
<summary>证明.</summary>

令$G$为任意群。我们的目标是构造一个单射群同态$\phi: G \to \mathrm{Sym}(G)$，从而证明$G$同构于$\mathrm{Sym}(G)$的一个子群。

对于每个元素$g \in G$，定义一个由左乘法确定的函数$L_g: G \to G$：
$$
L_g(h) = g h \quad \text{对于所有 } h \in G。
$$
由于$G$是一个群，每个$L_g$都是一个双射（其逆为$L_{g^{-1}}$），因此$L_g \in \mathrm{Sym}(G)$。

定义映射$\phi: G \to \mathrm{Sym}(G)$为
$$
\phi(g) = L_g.
$$

- 同态性质：对于所有$g_1, g_2 \in G$和$h \in G$，
$$
\phi(g_1 g_2)(h) = L_{g_1 g_2}(h) = (g_1 g_2) h = g_1 (g_2 h) = L_{g_1}(L_{g_2}(h)) = (\phi(g_1) \circ \phi(g_2))(h).
$$
因此，$\phi(g_1 g_2) = \phi(g_1) \circ \phi(g_2)$，所以$\phi$是一个群同态。
- 单射性：假设$\phi(g) = \phi(h)$，对于某些$g, h \in G$。那么对于所有$k \in G$，
$$
L_g(k) = L_h(k) \implies g k = h k。
$$
特别地，当$k$是$G$的单位元$e$时，
$$
g e = h e \implies g = h.
$$
因此，$\phi$是单射的。

由于$\phi$是一个单射同态，$G$同构于$\mathrm{Sym}(G)$的子群$\phi(G)$。
$$
~\tag*{$\square$}
$$
</details>

Cayley定理揭示了每个群都可以看作是排列的群，强调了对称群在理解所有群的结构中的核心作用。

# §3 Maps of Groups

Whenever you define a new idea, it is useful to know what kinds of functions are naturally associated with it.

**Question:** What types of **functions** should we study?

**Example 3.1**

$$
\begin{aligned}
\text{sets}~S,T~&\leftrightarrow~\text{arbitrary functions}~f:S\to T\\
\text{spaces}~X,Y~&\leftrightarrow~\text{continuous functions}~f:X\to Y\\
\text{smooth curves + surfaces}~X,Y~&\leftrightarrow~\text{differentiable functions}~f:X\to Y\\
\text{groups}~G,H~&\leftrightarrow~\text{group homomorphisms}~\phi:G\to H\\
\end{aligned}
$$



## §3.1 Group Homomorphisms

### §3.1.1 Definition of a Group Homomorphism

**Definition 3.1** Let $G,H$ be groups. A **group homomorphism** from $G$ to $H$ is a function

$$
\phi:G\to H
$$

such that for all $g_1,g_2\in G$,

$$
\phi(g_1g_2)=\phi(g_1)\phi(g_2).
$$

### §3.1.2 Examples of Group Homomorphisms

$\begin{aligned}
\text{\textbf{\small{Example 3.2}}}~\exp:(\mathbb{R},+)&\to \mathbb{R}^{\times}~\text{is a group homomorphism.}\
t&\mapsto \mathrm{e}^{t}
\end{aligned}$

<details>
<summary>Proof.</summary>
    
For all $t_{1}$, $t_{2}\in(\mathbb{R},+)$,

$$
\mathrm{e}^{t_1+t_2}=\mathrm{e}^{t_1}\cdot \mathrm{e}^{t_2}.
$$

$$
~\tag*{$\square$}
$$
</details>

$\begin{aligned}
\text{\textbf{\small{Example 3.3}}}~\det: GL_n(\mathbb{R})&\to\mathbb{R}^{\times}~\text{is a group homomorphism.}\
M&\mapsto \det(M)
\end{aligned}$

<details>
<summary>Proof.</summary>

For all $M_1,M_2\in GL_n(\mathbb{R})$,

$$
\det(M_1M_2)=\det(M_1)\det(M_2).
$$

$$
~\tag*{$\square$}
$$
</details>

$\begin{aligned}
\text{\textbf{\small{Example 3.4}}}~(\mathbb{R},+)&\to S^{1}~\text{is a group homomorphism.}\
t&\mapsto \mathrm{e}^{\mathrm{i}t}
\end{aligned}$

<details>
<summary>Proof.</summary>

For all $t_1,t_2\in(\mathbb{R},+)$,

$$
\mathrm{e}^{\mathrm{i}(t_1+t_2)}
=
\mathrm{e}^{\mathrm{i}t_1}\cdot\mathrm{e}^{\mathrm{i}t_2}.
$$

$$
~\tag*{$\square$}
$$
</details>

**Example 3.5** If $V,W$ are vector spaces, then they are groups under addition $+$. Any linear map

$$
\phi:V\to W
$$

is a group homomorphism. Any linear subspace is a subgroup.

<details>
<summary>Proof.</summary>

(a) A vector space $V$ is a group because $V$ satisfies

(1) Associativity: addition in $V$ is associative, i.e. for $u,v,w\in V$,

$$
(u+v)+w=u+(v+w);
$$

(2) Identity: there exists an element $0\in V$ such that for every $v\in V$,

$$
v+0=v;
$$

(3) Inverses: for every $v\in V$, there exists $-v\in V$ such that

$$
v+(-v)=0.
$$

(b) A linear map $\phi:V\to W$ between any two vector spaces $V$ and $W$ is a group homomorphism because $\phi$ satisfies

(1) The homomorphism property: for all $u,v\in V$,

$$
\phi(u+v)=\phi(u)+\phi(v)\tag*{(definition of a linear map)}
$$

(2) Preservation of the identity: since $\phi$ is linear,

$$
\phi(0_V)=0_W,
$$

where $0_V$ and $0_W$ are the identity elements of $V$ and $W$, respectively;

(3) Preservation of inverses: for every $v\in V$,

$$
\phi(-v)=-\phi(v).
$$

This follows from the linearity of $\phi$, which involves scalar multiplication and additive inverses.

(c) Any linear subspace is a subgroup because

* a linear subspace $U\subseteq V$ is also a group under addition, since it inherits the group structure of $V$;

* the identity $0_V$ of $V$ is also the identity in $U$;

* inverses in $U$ are inherited from $V$.

$$
~\tag*{$\square$}
$$
    
</details>


### §3.1.3 Properties of Group Homomorphisms

**Proposition 3.1** Let
$\phi:G\to H$
be a group homomorphism. Then

(a) $\phi(1_G)=1_H$.

(b) $\phi(g^{-1})=\phi(g)^{-1}$.

<details>
<summary>Proof.</summary>

(a) For any $g\in G$,

$$
\begin{aligned}
\phi(g)&=\phi(1_G\cdot g)\\
&=\phi(1_G)\cdot\phi(g)
\end{aligned}
~~~~~~~~~~~~~~~~~~
\begin{aligned}
\text{(2)}&\\
\text{(definition of a homomorphism)}&
\end{aligned}
$$

Let $h$ be the inverse of $\phi(g)$. Then

$$
\begin{aligned}
\phi(g)\cdot h=\phi(1_G)\cdot\phi(g)\cdot h
\quad&\Rightarrow\quad
1_H=\phi(1_G)\cdot1_H\\
&\Rightarrow\quad
1_H=\phi(1_G)
\end{aligned}
~~~~~~~~~~~~~~~~~~
\begin{aligned}
\text{(3)}&\\
\text{(2)}&
\end{aligned}
$$

(b)

$$
1_H=\phi(1_G)=\phi(g\cdot g^{-1})
=\phi(g)\cdot\phi(g^{-1})
\quad\Rightarrow\quad
\phi(g^{-1})=\phi(g)^{-1}.
$$

$$
~\tag*{$\square$}
$$
  
</details>

**Proposition 3.2**
$\mathrm{id}_G:G\to G$
is a homomorphism.

<details>
<summary>Proof.</summary>

Consider the identity map
$\mathrm{id}_G:G\to G$
defined by
$\mathrm{id}_G(g)=g$
for every $g\in G$.
To show that $\mathrm{id}_G$ is a homomorphism, we need to show that for all
$g_1,g_2\in G$,

$$
\mathrm{id}_G(g_1\cdot g_2)
=
\mathrm{id}_G(g_1)\cdot\mathrm{id}_G(g_2).
$$

For any $g_1,g_2\in G$,

$$
\mathrm{id}_G(g_1\cdot g_2)
=
g_1\cdot g_2
\tag*{(definition of $\mathrm{id}_G$)}
$$

and

$$
\mathrm{id}_G(g_1)\cdot\mathrm{id}_G(g_2)
=
g_1\cdot g_2
\tag*{(since $\mathrm{id}_G(g_i)=g_i$)}.
$$

Therefore,

$$
\mathrm{id}_G(g_1\cdot g_2)
=
g_1\cdot g_2
=
\mathrm{id}_G(g_1)\cdot\mathrm{id}_G(g_2).
$$

Hence
$\mathrm{id}_G:G\to G$
is a homomorphism.

$$
~\tag*{$\square$}
$$
    
</details>

**Proposition 3.3** If
$G\stackrel{\phi}{\longrightarrow}H$
and
$H\stackrel{\psi}{\longrightarrow}K$
are homomorphisms, then
$\psi\circ\phi$
is a homomorphism.

<details>
<summary>Proof.</summary>

To prove that the composition of two homomorphisms $\phi$ and $\psi$ is again a homomorphism, we need to verify that for all $g_1,g_2\in G$,

$$
(\psi\circ\phi)(g_1\cdot g_2)
=
(\psi\circ\phi)(g_1)\cdot(\psi\circ\phi)(g_2).
$$

By the definition of composition,

$$
(\psi\circ\phi)(g_1\cdot g_2)
=
\psi(\phi(g_1\cdot g_2)).
$$

Using the homomorphism property of $\phi$,

$$
\psi(\phi(g_1\cdot g_2))
=
\psi(\phi(g_1)\cdot\phi(g_2)).
$$

Then using the homomorphism property of $\psi$,

$$
\psi(\phi(g_1)\cdot\phi(g_2))
=
\psi(\phi(g_1))\cdot\psi(\phi(g_2)).
$$

By the definition of composition,

$$
\psi(\phi(g_1))\cdot\psi(\phi(g_2))
=
(\psi\circ\phi)(g_1)\cdot(\psi\circ\phi)(g_2).
$$

Thus,

$$
(\psi\circ\phi)(g_1\cdot g_2)
=
(\psi\circ\phi)(g_1)\cdot(\psi\circ\phi)(g_2).
$$

Therefore, if
$G\stackrel{\phi}{\longrightarrow}H$
and
$H\stackrel{\psi}{\longrightarrow}K$
are homomorphisms, then
$\psi\circ\phi$
is a homomorphism.

$$
~\tag*{$\square$}
$$
</details>

**Proposition 3.4** If $H\subset G$ is a subgroup, then the inclusion map
$i:H\hookrightarrow G$
is a homomorphism.

<details>
<summary>Proof.</summary>

To prove that the inclusion map
$i:H\hookrightarrow G$
is a homomorphism, where $H\subset G$ is a subgroup, we need to verify that for all $h_1,h_2\in H$,

$$
i(h_1\cdot h_2)=i(h_1)\cdot i(h_2).
$$

The inclusion map
$i:H\hookrightarrow G$
is defined by

$$
i(h)=h
$$

for every $h\in H$.

Thus, for any $h_1,h_2\in H$,

$$
\begin{aligned}
i(h_1)&=h_1\\
i(h_2)&=h_2.
\end{aligned}
$$

Therefore,

$$
i(h_1)\cdot i(h_2)=h_1\cdot h_2.
$$

At the same time,
$h_1\cdot h_2\in H$,
and by the definition of the inclusion map,

$$
i(h_1\cdot h_2)=h_1\cdot h_2.
$$

Hence, for all $h_1,h_2\in H$,

$$
i(h_1\cdot h_2)=i(h_1)\cdot i(h_2).
$$

Therefore, the inclusion map
$i:H\hookrightarrow G$
is a homomorphism.

$$
~\tag*{$\square$}
$$

</details>

### §3.1.4 Kernel and Image of a Group Homomorphism

**Definition 3.2** Given a group homomorphism

$$
\phi:G\to H,
$$

the **kernel** of $\phi$ is the set

$$
\ker(\phi)=\{g\in G\mid\phi(g)=1_H\}.
$$

The **image** of $\phi$ is the set

$$
\mathrm{im}(\phi)
=
\{h\in H\mid h=\phi(g)~\text{for some}~g\in G\}.
$$

**Proposition 3.5**
$\ker(\phi)\subset G$
and
$\mathrm{im}(\phi)\subset H$
are subgroups.

<details>
<summary>Proof.</summary>

$\ker(\phi)$ is a subgroup of $G$ because

(1) If
$\phi(g_1)=\phi(g_2)=1_H$,
then

$$
\begin{aligned}
\phi(g_1g_2)
&=\phi(g_1)\cdot\phi(g_2)\\
&=1_H\cdot1_H\\
&=1_H.
\end{aligned}
$$

(2)
$\phi(1_G)=1_H$,
so
$1_G\in\ker(\phi)$.

(3) If
$\phi(g)=1_H$,
then

$$
\begin{aligned}
&\Rightarrow\quad \phi(g^{-1})=1_H^{-1}=1_H\\
&\Rightarrow\quad g^{-1}\in\ker(\phi).
\end{aligned}
$$

Similarly,
$\mathrm{im}(\phi)$
is a subgroup of $H$ because

(1)
$h_i=\phi(g_i)$
implies

$$
h_1h_2
=
\phi(g_1)\phi(g_2)
=
\phi(g_1g_2).
$$

(2)
$\phi(1_G)=1_H$,
so
$1_H\in\mathrm{im}(\phi)$.

(3)
$h=\phi(g)$
implies

$$
h^{-1}=\phi(g^{-1}).
$$

$$
~\tag*{$\square$}
$$

</details>

## §3.2 Group Isomorphisms

### §3.2.1 Definition of a Group Isomorphism

**Definition 3.3** If a group homomorphism $\phi$ is a bijection, then $\phi$ is called a **group isomorphism**.

Isomorphic does not mean equal, just as two sets being in bijection does not mean that the sets are equal.

**Example 3.6** A set of five bananas is not equal to a set of five apples.

Nevertheless, we classify groups up to isomorphism, just as we classify sets up to bijection.

### §3.2.2 Examples of Group Isomorphisms

**Example 3.7**

$$
\exp:(\mathbb{R},+)\to(\mathbb{R}_{>0},\times)
$$

is a group isomorphism.

<details>
<summary>Proof.</summary>

To prove that
$\exp:(\mathbb{R},+)\to(\mathbb{R}_{>0},\times)$
is a group isomorphism, we need to show two things:

(1) **Homomorphism:** $\exp$ preserves the group operation, i.e. for all
$x,y\in\mathbb{R}$,

$$
\exp(x+y)=\exp(x)\cdot\exp(y).
$$

(2) **Bijection:** $\exp$ is bijective, meaning that it is both injective and surjective.

Let us verify these.

For $x,y\in\mathbb{R}$,

$$
\exp(x+y)=e^{x+y}.
$$

Using the exponential law,

$$
e^{x+y}=e^x\cdot e^y=\exp(x)\cdot\exp(y).
$$

Therefore,

$$
\exp(x+y)=\exp(x)\cdot\exp(y),
$$

so $\exp$ is a homomorphism.

To show that $\exp$ is injective, suppose that for some
$x,y\in\mathbb{R}$,

$$
\exp(x)=\exp(y).
$$

Then

$$
e^x=e^y.
$$

Taking the natural logarithm of both sides (which is valid since
$e^x>0$
for all $x$),

$$
x=y.
$$

Therefore, $\exp$ is injective.

To show that $\exp$ is surjective, we must show that for every
$y\in\mathbb{R}_{>0}$,
there exists an
$x\in\mathbb{R}$
such that
$\exp(x)=y$.

Since $y>0$, we may take
$x=\ln(y)\in\mathbb{R}$,
where $\ln$ denotes the natural logarithm. Then

$$
\exp(x)=e^{\ln(y)}=y.
$$

Therefore, $\exp$ is surjective.

Since $\exp$ is both a homomorphism and a bijection, it is a group isomorphism between
$(\mathbb{R},+)$
and
$(\mathbb{R}_{>0},\times)$.

Hence,

$$
\exp:(\mathbb{R},+)\to(\mathbb{R}_{>0},\times)
$$

is a group isomorphism.

$$
~\tag*{$\square$}
$$

</details>

**Theorem 3.6** Every finite cyclic group of order $n$ is isomorphic to
$\mathbb{Z}/n\mathbb{Z}$.

<details>
<summary>Proof.</summary>

Let
$G=\langle g\rangle$
be a cyclic group of order $n$. Define
$\phi:G\to\mathbb{Z}/n\mathbb{Z}$
by

$$
\phi(g^k)=k\pmod n.
$$

(1) Homomorphism:

$$
\phi(g^a\cdot g^b)
=
\phi(g^{a+b})
=
(a+b)\pmod n.
$$

On the other hand,

$$
\phi(g^a)+\phi(g^b)
=
a+b\pmod n.
$$

The two sides are equal, so $\phi$ is a homomorphism.

(2) Injectivity: If
$\phi(g^a)=\phi(g^b)$,
then
$a\equiv b\pmod n$,
which implies
$g^a=g^b$.

(3) Surjectivity: For every
$k\in\mathbb{Z}/n\mathbb{Z}$,
we have

$$
\phi(g^k)=k\pmod n.
$$

$$
~\tag*{$\square$}
$$

</details>

### §3.2.3 Properties of Group Isomorphisms

**Proposition 3.7** If
$\phi:G\to H$
is a group isomorphism, then
$\phi^{-1}$
is also a group isomorphism.

<details>
<summary>Proof.</summary>

Clearly, $\phi^{-1}$ is a bijection. It remains to prove that
$\phi^{-1}$
is a homomorphism.

Let
$g_1,g_2\in G$,
and let
$h_1=\phi(g_1)$,
$h_2=\phi(g_2)$.
Then

$$
\begin{align*}
\phi^{-1}(h_1\cdot h_2)
&=\phi^{-1}(\phi(g_1)\cdot\phi(g_2))
\tag{$\phi$ is surjective}\\
&=\phi^{-1}(\phi(g_1\cdot g_2))
\tag{$\phi$ is a homomorphism}\\
&=(\phi^{-1}\circ\phi)(g_1\cdot g_2)
\tag{notation}\\
&=g_1\cdot g_2
\tag{definition of $\phi^{-1}$}\\
&=\phi^{-1}(h_1)\cdot\phi^{-1}(h_2)
\tag{definitions of $g_1,g_2$}.
\end{align*}
$$

$$
~\tag*{$\square$}
$$

</details>


## §3.3 Product Groups

Product groups provide a way to construct new groups by combining the elements and operations of groups we already know.

**Definition 3.4** Given two groups $G_1$ and $G_2$, the **product group**
$G_1\times G_2$
is the set of ordered pairs

$$
G_1\times G_2
=
\{(g_1,g_2)\mid g_1\in G_1,\ g_2\in G_2\},
$$

with group operation defined by

$$
(g_1,g_2)\cdot(h_1,h_2)
=
(g_1\cdot h_1,g_2\cdot h_2).
$$

**Proposition 3.8** Given two groups $G_1$ and $G_2$,
$G_1\times G_2$
is a group.

<details>
<summary>Proof.</summary>

(1) Associativity follows directly from associativity in $G_1$ and $G_2$.

(2) Identity:

$$
(1_{G_1},1_{G_2})
$$

is the identity element.

(3) Inverse:

$$
(g_1,g_2)^{-1}
=
(g_1^{-1},g_2^{-1}).
$$

$$
~\tag*{$\square$}
$$
    
</details>

**Example 3.8 (Cartesian Product of the Real Numbers)**

The group
$\mathbb{R}\times\mathbb{R}$
under ordinary addition is isomorphic to the Euclidean plane
$\mathbb{R}^2$.

**Example 3.9 (Klein Four Group)**

$\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$
is called the **Klein four group**. It is the smallest noncyclic group.

**Example 3.10** If
${z\in T\subset\mathbb{C}:|z|=1}$,
then

$$
\mathbb{C}^{\times}
=
\mathbb{R}^{\times}\times T.
$$

## §3.4 Automorphisms

Understanding product groups naturally leads to the study of symmetries within a group. An automorphism is a bijective homomorphism from a group to itself, describing the different ways in which the structure of a group may be mapped onto itself. Automorphisms are especially interesting when applied to product groups, because they reveal how the internal structures of several groups can interact under mappings.

**Definition 3.5** Let $X$ be a set. We write

$$
\mathrm{Aut}(X)
:=
\mathrm{Aut}_{\text{set}}(X)
:=
\{\text{bijections}~X\to X\}
$$

for the set of bijections from $X$ to itself.

**Proposition 3.9**
$\mathrm{Aut}(X)$
forms a group under composition.

<details>
<summary>Proof.</summary>

$\mathrm{Aut}(X)$
is a group under composition because

(1) composition of functions is associative:

$$
(f\circ g)\circ h=f\circ(g\circ h).
$$

$\begin{aligned}
(2)~\mathrm{id}_X:X&\to X
\text{ is the identity, since }
f\circ\mathrm{id}_X=\mathrm{id}_X\circ f=f.\
x&\mapsto x
\end{aligned}$

(3)
$f^{-1}$
is the inverse of $f$:

$$
f\circ f^{-1}
=
\mathrm{id}_X
=
f^{-1}\circ f.
$$

$$
~\tag*{$\square$}
$$

</details>

## §3.5 Symmetric Groups

**Definition 3.6** Let

$$
\underline{n}=\{1,\ldots,n\}.
$$

Then

$$
\mathrm{Aut}_{\text{set}}(\underline{n})=:S_n
$$

is the **symmetric group** on a set of $n$ elements.

**Example 3.11** The following are examples of symmetric groups:

* $n=1$:
  $\mathrm{Aut}({1})$
  is a group with one element:

$$
\begin{aligned}
S_1
&=
\{\text{bijections}~\{1\}\to\{1\}\}\\
&=
\{\mathrm{id}_{\underline{1}}\}.
\end{aligned}
$$

* $n=2$:
  $\mathrm{Aut}({1,2})$
  is a group with two elements:

$$
S_{2}=\left\{\left(\mathrm{id}:\begin{aligned}
1&\mapsto 1\\
2&\mapsto 2
\end{aligned}
\right),\left(
\sigma:\begin{aligned}
1&\mapsto 2\\
2&\mapsto 1
\end{aligned}
\right)\right\}.
$$

It satisfies

$$
\sigma\circ\sigma=\sigma^2=\mathrm{id}.
$$

* $S_3$ has $3!$ elements. We will soon study its structure.

* In general, $S_n$ is a group with $n!$ elements.

**Definition 3.7** The **symmetric group on a set $G$** is the group of all bijections from $G$ to itself:

$$
\mathrm{Sym}(G)
=
\mathrm{Aut}_{\text{set}}(G).
$$

This group contains all permutations of the elements of $G$. When $G$ is a finite set containing $n$ elements,
$\mathrm{Sym}(G)$
is essentially the same as $S_n$, but this generalization allows us to consider permutations of an arbitrary set $G$, regardless of its cardinality.

Symmetric groups are fundamental in group theory because, by containing all possible permutations of a finite set, they capture the essence of symmetry. Remarkably, every group, whether finite or infinite, can be represented as a subgroup of a permutation group. This profound connection is formalized by Cayley's theorem.

## §3.6 Cayley's Theorem

**Theorem 3.10 (Cayley's Theorem)** Every group $G$ is isomorphic to a subgroup of the symmetric group
$\mathrm{Sym}(G)$
on the set $G$.

<details>
<summary>Proof.</summary>

Let $G$ be an arbitrary group. Our goal is to construct an injective group homomorphism

$$
\phi:G\to\mathrm{Sym}(G),
$$

thereby proving that $G$ is isomorphic to a subgroup of
$\mathrm{Sym}(G)$.

For each element
$g\in G$,
define a function
$L_g:G\to G$
by left multiplication:

$$
L_g(h)=gh
\quad
\text{for all }h\in G.
$$

Since $G$ is a group, every $L_g$ is a bijection, with inverse
$L_{g^{-1}}$.
Hence

$$
L_g\in\mathrm{Sym}(G).
$$

Define

$$
\phi:G\to\mathrm{Sym}(G)
$$

by

$$
\phi(g)=L_g.
$$

* **Homomorphism property:** For all
  $g_1,g_2\in G$
  and
  $h\in G$,

$$
\phi(g_1g_2)(h)
=
L_{g_1g_2}(h)
=
(g_1g_2)h
=
g_1(g_2h)
=
L_{g_1}(L_{g_2}(h))
=
(\phi(g_1)\circ\phi(g_2))(h).
$$

Therefore,

$$
\phi(g_1g_2)
=
\phi(g_1)\circ\phi(g_2),
$$

so $\phi$ is a group homomorphism.

* **Injectivity:** Suppose
  $\phi(g)=\phi(h)$
  for some
  $g,h\in G$.
  Then for every
  $k\in G$,

$$
L_g(k)=L_h(k)
\implies
gk=hk.
$$

In particular, taking $k$ to be the identity element $e$ of $G$,

$$
ge=he
\implies
g=h.
$$

Therefore, $\phi$ is injective.

Since $\phi$ is an injective homomorphism,
$G$
is isomorphic to the subgroup
$\phi(G)$
of
$\mathrm{Sym}(G)$.

$$
~\tag*{$\square$}
$$

</details>

Cayley's theorem shows that every group can be regarded as a group of permutations, emphasizing the central role of symmetric groups in understanding the structure of all groups.

§ DS-03 群的映射

§ DS-03 Map of Groups

# §2 子群
## §2.1 子群的定义
**定义 2.1** 令 $G$ 是一个群，并且 $H \subset G$。 $H$ 被称为是 $G$ 的一个**子群(subgroup)**，如果满足

(1) $\forall h_1, h_2 \in H$，$h_1 h_2 \in H$ 。(在乘法下封闭)

(2) $1_G \in H$ 。

(3) 如果 $h \in H$ ，则 $h^{-1} \in H$ 。



## §2.2 子群的例子
**例 2.1** $\mathbb{R}_{>0} \subset \mathbb{R}^{\times}$是一个子群。

<details> <summary>证明.</summary> 
    
$\mathbb{R}_{>0}$ 是一个 $\mathbb{R}^{\times}$的一个子群，因为

(1) $\forall h_1, h_2 \in \mathbb{R}_{>0}$, $h_1 h_2 \in \mathbb{R}_{>0}$ 。

(2) $1 \in \mathbb{R}_{>0}$ 。

(3) 如果 $h \in \mathbb{R}_{>0}$，则 $h^{-1}=\dfrac{1}{h} \in \mathbb{R}_{>0}$ 。

$$
~\tag*{$\square$}
$$

</details>

**例 2.2** $\mathbb{Z}_{\geqslant 0} \subset(\mathbb{Z},+)$ 不是一个子群。

<details> <summary>证明.</summary> 
    
$\mathbb{Z}_{\geqslant 0}$ 不是 $\mathbb{Z}$ 的一个子群，因为不是所有元素 $z \in \mathbb{Z}$ 有一个逆。举例而言， $z=1$ ，它的逆应该是$-1$，但它不在 $\mathbb{Z}_{\geqslant 0}$ 中。

$$
~\tag*{$\square$}
$$

</details>

**例 2.3** $SL_n(\mathbb{R}):=\left\{M \in GL_n(\mathbb{R}) \mid \det M=1\right\} \subset GL_n(\mathbb{R})$ 是一个子群。

<details> <summary>证明.</summary> 
    
$SL_n(\mathbb{R})$ 是 $GL_n(\mathbb{R})$ 的一个子群，因为

(1) $\forall M_1, M_2 \in SL_n(\mathbb{R}) ， M_1 M_2 \in SL_n(\mathbb{R})$ ，因为
$$
\det\left(M_1 M_2\right)=\det\left(M_1\right) \det\left(M_2\right)=1 \times 1=1 .
$$

(2) 单位矩阵 $1_{GL_n(\mathbb{R})} \in SL_n(\mathbb{R})$ ，因为 $\det\left(1_{GL_n(\mathbb{R})}\right)=1$ 。

(3) $\forall M \in SL_n(\mathbb{R})$ ，则 $M^{-1} \in SL_n(\mathbb{R})$ ，因为 $\det\left(M^{-1}\right)=\frac{1}{\det(M)}=1$。

$$
~\tag*{$\square$}
$$

</details>

**例 2.4** 令 $\mathbb{C}^{\times}:=\mathbb{C} \setminus\{0\}$ 。定义 $m$ 为复数乘法: $\forall z_1, z_2 \in \mathbb{C}^{\times}$，
$$
m\left(z_1, z_2\right)=z_1 \times z_2 .
$$
我们有

(a) $\mathbb{C}^{\times}$是一个群。

(b) 令
$$
S^1=\{z\mid | z |=1\} .
$$
则 $S^1 \subset \mathbb{C}^{\times}$是一个子群。

<details> <summary>证明.</summary> 
    
 (a) $\mathbb{C}^{\times}$是一个群，因为

(1) 复数乘法满足结合律。

(2) $\forall z \in \mathbb{C}^{\times}, 1 \times z=z \times 1=z$ 。

(3) $\forall z \in \mathbb{C}^{\times}$， 令 $z^{-1}=\frac{\bar{z}}{\|z\|^2}$ 。

(b) $S^1$ 是 $\mathbb{C}^{\times}$的一个子群，因为

(1) $\left|z_1\right|=\left|z_2\right|=1 \Rightarrow\left|z_1 \times z_2\right|=1$ ，所以 $S^1$ 在 $\times$ 下封闭。

(2) $|1|=1$ ，所以恒元在 $S^1$ 中。

(3) 如果 $|z|=1$ ，则 $|\bar{z}|=1$ ，并且 $z^{-1}=\bar{z}$。

$$
~\tag*{$\square$}
$$

# §2 Subgroups

## §2.1 Definition of a Subgroup

**Definition 2.1** Let $G$ be a group and let $H \subset G$. We say that $H$ is a **subgroup** of $G$ if the following conditions are satisfied:

(1) For all $h_1, h_2 \in H$, we have $h_1 h_2 \in H$. (Closure under multiplication)

(2) $1_G \in H$.

(3) If $h \in H$, then $h^{-1} \in H$.



## §2.2 Examples of Subgroups

**Example 2.1** $\mathbb{R}_{>0} \subset \mathbb{R}^{\times}$ is a subgroup.

<details> <summary>Proof.</summary> 

$\mathbb{R}_{>0}$ is a subgroup of $\mathbb{R}^{\times}$ because

(1) For all $h_1, h_2 \in \mathbb{R}*{>0}$, we have $h_1 h_2 \in \mathbb{R}*{>0}$.

(2) $1 \in \mathbb{R}_{>0}$.

(3) If $h \in \mathbb{R}*{>0}$, then
$h^{-1}=\dfrac{1}{h} \in \mathbb{R}*{>0}$.

$$
~\tag*{$\square$}
$$

</details>

**Example 2.2** $\mathbb{Z}_{\geqslant 0} \subset(\mathbb{Z},+)$ is not a subgroup.

<details> <summary>Proof.</summary> 

$\mathbb{Z}*{\geqslant 0}$ is not a subgroup of $\mathbb{Z}$ because not every element has an inverse in $\mathbb{Z}*{\geqslant 0}$. For example, for $z=1$, its inverse under addition should be $-1$, but
$-1\notin\mathbb{Z}_{\geqslant 0}$.

$$
~\tag*{$\square$}
$$

</details>

**Example 2.3**
 $SL_n(\mathbb{R}):=\left\{M \in GL_n(\mathbb{R}) \mid \det M=1\right\} \subset GL_n(\mathbb{R})$ is a subgroup.

<details> <summary>Proof.</summary> 

$SL_n(\mathbb{R})$ is a subgroup of $GL_n(\mathbb{R})$ because

(1) For all
$M_1, M_2 \in SL_n(\mathbb{R})$,
we have
$M_1 M_2 \in SL_n(\mathbb{R})$,
since

$$
\det\left(M_1 M_2\right)=\det\left(M_1\right) \det\left(M_2\right)=1 \times 1=1 .
$$

(2) The identity matrix
$1_{GL_n(\mathbb{R})} \in SL_n(\mathbb{R})$,
since
$\det\left(1_{GL_n(\mathbb{R})}\right)=1$.

(3) For every
$M \in SL_n(\mathbb{R})$,
we have
$M^{-1} \in SL_n(\mathbb{R})$,
since
$\det\left(M^{-1}\right)=\frac{1}{\det(M)}=1$.

$$
~\tag*{$\square$}
$$

</details>

**Example 2.4** Let
$\mathbb{C}^{\times}:=\mathbb{C} \setminus\{0\}$.
Define $m$ to be complex multiplication: for all
$z_1, z_2 \in \mathbb{C}^{\times}$,

$$
m\left(z_1, z_2\right)=z_1 \times z_2 .
$$

Then:

(a) $\mathbb{C}^{\times}$ is a group.

(b) Let

$$
S^1=\{z\mid | z |=1\}.
$$

Then
$S^1 \subset \mathbb{C}^{\times}$
is a subgroup.

<details> <summary>Proof.</summary> 

(a) $\mathbb{C}^{\times}$ is a group because

(1) Complex multiplication is associative.

(2) For every
$z \in \mathbb{C}^{\times}$,
we have
$1 \times z=z \times 1=z$.

(3) For every
$z \in \mathbb{C}^{\times}$,
let
$z^{-1}=\frac{\bar{z}}{|z|^2}$.

(b) $S^1$ is a subgroup of $\mathbb{C}^{\times}$ because

(1)
$\left|z_1\right|=\left|z_2\right|=1
\Rightarrow
\left|z_1 \times z_2\right|=1$,
so $S^1$ is closed under multiplication.

(2)
$|1|=1$,
so the identity lies in $S^1$.

(3) If
$|z|=1$,
then
$|\bar{z}|=1$,
and
$z^{-1}=\bar{z}$.

$$
~\tag*{$\square$}
$$

§ DS-02 子群

§ DS-02 Subgroup

# §0 简介
<table class="tg">
<thead>
  <tr><th class="tg-0pky">概念(名字)</th>
    <th class="tg-0pky">所有数字</th>
    <th class="tg-0pky">导数</th>
    <th class="tg-0pky">群</th>
    <th class="tg-0pky">环</th>
  </tr>
</thead>
<tbody>
  <tr>
    <td class="tg-0pky">这个概念解释了...?<br> (数学是一种表达思想的语言; <br>这些词语体现了什么思想呢？)</td>
    <td class="tg-0pky">计数,<br>数量</td>
    <td class="tg-0pky">变化率,<br>线性化</td>
    <td class="tg-0pky">对称性</td>
    <td class="tg-0pky">空间上的函数</td>
  </tr>
  <tr>
    <td class="tg-0pky">一些数学结果</td>
    <td class="tg-0pky"></td>
    <td class="tg-0pky"></td>
    <td class="tg-0pky" colspan="2">几何的"代数化" <br>(Descartes至今),<br>代数的"几何化"</td>
  </tr>
  <tr>
    <td class="tg-0pky">一些应用<br>(纯数之外)</td>
    <td class="tg-0pky"></td>
    <td class="tg-0pky"></td>
    <td class="tg-0pky" colspan="2">Nöther定理(物理)<br>RSA算法(密码学)<br>逻辑电路作为"余层"<br>数据集的同调形状, 等等</td>
  </tr>
</tbody>
</table>



# §1 群
## §1.1 群

考虑某个对象 $X$ 。

**什么是 $X$ 的一个对称性呢?**

通常来说，我们对于这个对象 $X$ 会有一些关于**结构**的前置印象 (形状，距离，线性等等，比如说圆、三角、方块儿，再比如说度量，抑或是向量空间$V$等等) 。

<p align="center">
  <img src="https://tex.xumin.net/svgb/fZLNToQwFIX3PkWXkMBNYSgwMTwJsKhtxSoppFMy0cmsTdyoK_fu3BsXJj6OmcxbmAIzjImyPPfnO7cnLS5EJdXGyJu7VjLTabHNa6kEYrTNdNMp7qFeXzdSDYXyrOCaroexteTmKsOQtqZEDoaEeAGExEVMalaLXFMuu1VmG-X5_4shLLCHgWAX-T5yFoAtZ0kOMj3psltWixlWBISM01owQ1VVC-TER-TMKgZsjcLIRQCINcropl4d60HqIqp4r61FkPRzDoagbyfDuQSWJ92JEo-UZKT0Jw1USxllGNkLVcNFftkokxU11ZUoETVjLj6GIHTR5vvrdffwsZ15Tmxzm2JMpmCsTG3mh0T-9us3rB-2fk_P-_vH3dvn_uXdugrFf_2bHw">
</p>

$X$ 的一个对称性应该是一个映射(为什么是一个从 $X$ 到 $X$ 的映射呢? 因为我们是对 $X$ 进行了变换操作，而 $X$ 本身并没变)
$$
\phi: X \rightarrow X
$$

满足
1. 保持结构，比如 $\operatorname{dist}(x, y)=\operatorname{dist}(\phi(x), \phi(y))$ ，再比如 $\phi(x+y)=\phi(x)+\phi(y)$ ，并且
2. 可以撤销这一操作。

现在，我们首先来尝试一般性地瞅瞅这玩意儿咋回事，以给我们一点启发:

令 $G=\{\phi\}$ 是 $X$ 的对称性的集合。
1. 如果 $\phi_1, \phi_2$ 保持结构，则它们的复合 $\phi_1 \circ \phi_2, \phi_2 \circ \phi_1$ 也保持。 $\Rightarrow$ 我们可以分解 $G$ 的元素 $\Rightarrow$ $G \times G \xrightarrow{m} G$ ，结合性；
2. “什么都不做”应该也是 $X$ 的一个对称性。 $\Rightarrow \mathrm{id}_X \in G ， \operatorname{id}_X \circ \phi=\phi \circ \mathrm{id}_X=\phi$;
3. 因为 $\phi$ 可以撤销，我们应该有 $\phi^{-1} \in G$ ，所以 $\phi \circ \phi^{-1}=\mathrm{id}_X$， $\phi^{-1} \circ \phi=\mathrm{id}_X$ 。

这样通过最一般的映射来看是不是群的定义马上就呼之欲出呢?

### §1.1.1 群的定义

**定义 1.1** 一个**群 (group)** 是一个有序对
$$
(G, m)
$$
其中 $G$ 是一个集合，而 $m$ 是一个映射
$$
\begin{aligned}
& G \times G \rightarrow G \\
& \begin{aligned}
\left(g_1, g_2\right) \mapsto m\left(g_1, g_2\right) & =: g_1 \cdot g_2 \\
& =: g_1 g_2
\end{aligned}
\end{aligned}
$$
使得
1. $m$ 满足结合律，i.e.
$$
m\left(m\left(g_1, g_2\right), g_3\right)=m\left(g_1, m\left(g_2, g_3\right)\right)
$$
i.e. $\left(g_1 \cdot g_2\right) \cdot g_3=g_1 \cdot\left(g_2 \cdot g_3\right)$ 或 $\left(g_1 g_2\right) g_3=g_1\left(g_2 g_3\right)$ ；

2. $\exists$一个元素$1_G \in G$ ，称为**恒元(identity)**，使得
$$
m\left(1_G, g\right)=g=m\left(g, 1_G\right)
$$
i.e. $1_G \cdot g=g=g \cdot 1_G$ 或 $1_G g=g=g 1_G$ ；

3. $\forall g \in G ， \exists$ 元素 $h \in G$ 使得
$$
m(g, h)=1_G=m(h, g)
$$
i.e. $g \cdot h=1_G=h \cdot g$ 或 $g h=1_G=h g$ ，我们经常记 $g^{-1}:=h$ ，称为 " $g$ 的**逆 (inverse)**" 。

### §1.1.2 群的例子

**例 1.1** 令
$$
G=\{\ldots,-1,0,1, \ldots\}=: \mathbb{Z}
$$
为整数的集合。定义
$$
G \times G \xrightarrow{m} G
$$
为
$$
m(g, h)=g+h
$$
(i.e. 整数加法)

比如，我们有
$$
m(-2,3)=1
$$
则 $(G, m)$ 是一个群。
<details>
<summary>证明.</summary>

$(\mathbb{Z},+)$ 是一个群，因为

(1) $m$ 满足结合律，即:

$$
(g+h)+k=g+(h+k)
$$

(2) $0=1_G$ 是恒元，即:

$$
\begin{aligned}
& m(0, g)=0+g=g \\
& m(g, 0)=g+0=g
\end{aligned}
$$

(3) 每个元素都有一个逆元素:

$$
m(g,-g)=g+(-g)=0
$$

$$
~\tag*{$\square$}
$$
</details>

**例 1.2** 令
$$
G=\{\ldots,-1,0,1, \ldots\}=: \mathbb{Z}
$$
为整数的集合。并且令
$$
\begin{aligned}
m: G \times G & \rightarrow G \\
(a, b) & \mapsto a \times b
\end{aligned}
$$
举例
$$
(2,3) \rightarrow 6
$$
我们有 $(G, m)$ 不是群。

<details>
<summary>证明.</summary>

$(\mathbb{Z}, \times)$ 不是群，因为不是所有元素 $z \in \mathbb{Z}$ 都有逆元素。举例而言， $z=2$ ，它的逆元素应该是 $\dfrac{1}{2}$ ，但不在 $\mathbb{Z}$ 中。

$$
~\tag*{$\square$}
$$

</details>

上面的两个例子

$(\mathbb{Z},+)$ 是群。

$(\mathbb{Z}, \times)$ 不是群。

说明了知道 $m$ 很重要。无论如何，我们经常会缩写，比如说 "令 $G$ 是一个群"，省略提及 $m$ 。

**例 1.3** 令 $G=\mathbb{R} \backslash\{0\}$ (去掉 0 的实数的集合)。令
$$
\begin{aligned}
m: G \times G & \rightarrow G \\
(a, b) & \mapsto a \times b
\end{aligned}
$$
则 $G$ 是一个群。我们从现在起记为 $\mathbb{R}^{\times}$。

<details>
<summary>证明.</summary>

$\mathbb{R}^{\times}$是群，因为

(1) 实数乘法满足结合律。

(2) 数字 1 是恒元。

(3) $\forall g \in \mathbb{R} \backslash\{0\}$ ，存在一个 $\dfrac{1}{g}$ ，使得 $g \dfrac{1}{g}=\dfrac{1}{g} g=1$ 。

$$
~\tag*{$\square$}
$$

</details>

### §1.1.3 群的性质

**命题 3.1 (消去律)** 令 $G$ 是一个群，并且 $g, h, k \in G$ 。假设
$$
g h=g k \text{.}
$$
则
$$
h=k \text{.}
$$

类似地，我们有
$$
h g=k g \quad \Rightarrow \quad h=k
$$

<details>
<summary>证明.</summary>

$\exists g^{-1}$ ，使得 $g^{-1} g=1_G$

$$
\begin{aligned}
g h=g k & \Rightarrow g^{-1}(g h)=g^{-1}(g k) \\
& \Rightarrow\left(g^{-1} g\right) h=\left(g^{-1} g\right) k \\
& \Rightarrow 1_G h=1_G k \\
& \Rightarrow h=k
\end{aligned}
$$

我们运用了作为一个群的所有公理!

$$
~\tag*{$\square$}
$$

</details>

**评论:** 消去律对于矩阵乘法不成立，除非 $g, h, k$ 是可逆的，比如 $g=0$ ?

**命题 3.2 (恒元的唯一性)** 群 $G$ 的恒元是唯一的。（即：如果两个元素 $1_G$ 和 $1_G^{\prime}$ 满足定义的恒元的性质，则 $1_G=1_G^{\prime}$ 。)

<details>
<summary>证明.</summary>

如果 $1_G$ 是恒元，则它必须满足方程: 对于任何 $g \in G$ ，

$$
1_{G}g=g1_{G}=g
$$

特别地，如果 $1_G^{\prime}=g$ ，我们必须有

$$
1_G 1_G^{\prime}=1_G^{\prime} .
$$

另一方面，如果 $1_G^{\prime}$ 也是恒元，我们必须有

$$
1_G 1_G^{\prime}=1_G .
$$

根据传递性，我们有

$$
1_G=1_G^{\prime} .
$$

$$
~\tag*{$\square$}
$$

</details>

**命题 3.3 (逆元的唯一性)** 对于任意元素 $g \in G$ ，它的逆 $g^{-1}$ 是唯一的。(即：给定元素 $h$ ， $h^{\prime}$ 满足定义的 $g^{-1}$ 的性质，则 $h=h^{\prime}$ 。)

<details>
<summary>证明.</summary>

假设 $h$ 和 $h^{\prime}$ 都是 $g$ 的逆。则

$$
g h^{\prime}=1_G .
$$

通过在等式两边都左乘 $h$ ，我们得到

$$
h\left(g h^{\prime}\right)=h .
$$

但是根据结合律，等号左边变成

$$
(h g) h^{\prime}=1_G h^{\prime}=h^{\prime} .
$$

通过等号的传递性，我们有

$$
h^{\prime}=h .
$$

$$
~\tag*{$\square$}
$$

</details>

## §1.2 Abel群

**例 1.4** 令 $n \geqslant 1$ 和 $n \in \mathbb{Z}$ 。那么
$$
G=GL_n(\mathbb{R}):=\{n \times n \text{实矩阵} M \mid \det M \neq 0\}
$$
是群，其中
$$
m: G \times G \rightarrow G
$$

通过矩阵乘法给出（这也说明一般情况下 $g h \neq h g$ ）。

<details>
<summary>证明.</summary>

$GL_n(\mathbb{R})$ 是群，因为

(1) 矩阵乘法满足结合律。

(2) 单位矩阵是恒元。

(3) $\operatorname{det}(g) \neq 0 \Rightarrow g$ 是可逆的。

$$
~\tag*{$\square$}
$$

</details>

由于矩阵的乘法不满足交换律， 这说明一般情况下，$gh\neq hg$! 那么如果群乘法是可交换的呢？ 我们有下面的定义:

**定义 1.2** 群$G$被称为**Abel群(abelian group)**, 如果对于所有$g_{1}, g_{2}\in G$, 我们有$g_{1}g_{2}=g_{2}g_{1}$。

**定义 1.3** 群乘法对于不是所有元素都满足交换律的群叫**非Abel群(non-abelian groups)**。

## §1.3 循环群
尽管Abel群通过交换性提供了有用的结构，但某些Abel群甚至更为简单。如果我们可以用一个单独的元素生成整个群呢？这就引出了循环群的概念，循环群是Abel群中最基本的例子之一。

循环群的特殊之处在于它将整个群的结构简化为某个元素的幂或倍数，这个元素称为生成元。这种简单性使得循环群在群论中成为一个重要的构建块，同时也是解决更复杂问题的关键工具。

**定义 1.4** 当且仅当存在一个元素 $g\in G$，称为**生成元(generator)**，使得 $G$ 中的每一个元素都可以表示为 $g$ 的幂时，群 $G$ 被称为**循环群(cyclic group)**。
$$
G=\langle g\rangle:=\{g^{n}\mid n\in\mathbb{Z}\}
$$

**例 1.5 (整数在加法下构成的群)**
群 $(\mathbb{Z},+)$ 是一个以 $1$ 为生成元的循环群：
$$
\langle 1\rangle=\{\ldots,-2,-1,0,1,2,\ldots\}
$$

**例 1.6 (模运算群)**
在模 $n$ 的加法下，群 $\mathbb{Z}/n\mathbb{Z}=\{0,1,\ldots,n\}$ 是一个循环群。元素 $1$ 生成整个群：
$$
\langle 1\rangle=\{1,2,\ldots,n-1,0\}
$$

**命题 1.4** 循环群是Abel群。

<details>
<summary>证明.</summary>

设循环群为 $G$，由元素 $g$ 生成。那么 $\forall x,y\in G$，存在 $m, n$ 使得 $g^{m}=x$ 且 $g^{n}=y$。因此，

$$
xy=g^{m}g^{n}=g^{m+n}=g^{n}g^{m}=yx
$$

因此，$G$ 是Abel群。

$$
~\tag*{$\square$}
$$

</details>

## §1.4 群的阶
**定义 1.5** 令$G$是一个群。我们令
$$
|G|\in \mathbb{Z}_{\geqslant 1}\cup \{\infty\}
$$
为$G$中元素的数量。我们称$|G|$为$G$的**阶(order)**。

**定义 2.3** 对于$g\in G$，考虑集合
$$
\{\ldots,\underbrace{g^{-1}\cdot g^{-1}}_{=:g^{-2}}, g^{-1},\mathrm{id}_{G},g,\underbrace{g\cdot g}_{=:g^{2}},\underbrace{g\cdot g\cdot g}_{=:g^{3}},\ldots\}
$$
我们定义 **$g$的阶(order of $g$)** 为
$$
|\langle g\rangle|
$$

**例 1.7**

- $1_{G}\in G$的阶为$1$。

- $n\in\mathbb{Z}$, $n\neq 0$有无穷阶。

- $\begin{pmatrix}
     0 & 1\\
     1 & 0
    \end{pmatrix}\in GL_{2}(\mathbb{R})$的阶为$2$，因为
$$
g^{2}=\begin{pmatrix}
0&1\\
1&0
\end{pmatrix}\begin{pmatrix}
0&1\\
1&0
\end{pmatrix}=\begin{pmatrix}
1&0\\
0&1
\end{pmatrix}=1_{GL_{2}(\mathbb{R})}
$$
所以$\{\ldots,g^{-1},1,g,\ldots\}=\{1,g\}$。

**定理 1.5 (Euler函数)** 阶为 $n$ 的循环群的生成元的数量由Euler函数$\varphi(n)$给出。

## §1.5 群的中心

为了研究非Abel群， 我们经常关注满足交换律的子集。一个重要的概念便是群的中心：

**定义 1.7** 群$G$的**中心(center)**, 记为$Z(G)$, 定义为:
$$
Z(G)=\{z\in G\mid zg=gz~\text{对于任意}~g\in G\}.
$$

**命题 1.6** 一个Abel群的中心是整个群。

<details>
<summary>证明.</summary>

一个群 $G$ 的中心定义为：

$$
Z(G) = \{z\in G \mid zg=gz~\text{对于所有}~g \in G\}.
$$

在一个Abel群中，每一对元素都可交换：

$$
g_{1} \cdot g_{2}=g_{2}\cdot g_{1} \quad \text{对于所有}~g_{1}, g_{2} \in G.
$$

因此，$G$ 中的每个元素都在中心，这意味着：
$$
Z(G)=G.
$$

$$
~\tag*{$\square$}
$$

</details>

群$G$的中心是与群$G$中每个元素可交换的元素的集合。这个集合不仅是特殊元素的收集，它实际上构成$G$的一个结构。我们将在下一节引入这个结构。

# §0 Introduction

<table class="tg">
<thead>
  <tr><th class="tg-0pky">Concept (Name)</th>
    <th class="tg-0pky">All Numbers</th>
    <th class="tg-0pky">Derivatives</th>
    <th class="tg-0pky">Groups</th>
    <th class="tg-0pky">Rings</th>
  </tr>
</thead>
<tbody>
  <tr>
    <td class="tg-0pky">What does this concept explain...?<br> (Mathematics is a language for expressing ideas;<br>what ideas do these words represent?)</td>
    <td class="tg-0pky">Counting,<br>quantity</td>
    <td class="tg-0pky">Rate of change,<br>linearization</td>
    <td class="tg-0pky">Symmetry</td>
    <td class="tg-0pky">Functions on spaces</td>
  </tr>
  <tr>
    <td class="tg-0pky">Some mathematical results</td>
    <td class="tg-0pky"></td>
    <td class="tg-0pky"></td>
    <td class="tg-0pky" colspan="2">The "algebraization" of geometry<br>(from Descartes to the present),<br>and the "geometrization" of algebra</td>
  </tr>
  <tr>
    <td class="tg-0pky">Some applications<br>(outside pure mathematics)</td>
    <td class="tg-0pky"></td>
    <td class="tg-0pky"></td>
    <td class="tg-0pky" colspan="2">Noether's theorem (physics)<br>RSA algorithm (cryptography)<br>logic circuits as "cosheaves"<br>homological shapes of data sets, etc.</td>
  </tr>
</tbody>
</table>



# §1 Groups

## §1.1 Groups

Consider some object $X$.

**What is a symmetry of $X$?**

Usually, when we think about an object $X$, we already have some prior notion of its **structure**—its shape, distance, linear structure, and so on. For example, $X$ may be a circle, a triangle, a square, a metric space, or a vector space $V$, etc.

<p align="center">
  <img src="https://tex.xumin.net/svgb/fZLLboMwEEX3_YpZggQjQ3AgqviKSN0AC9eeJk6RjRxHUYvy75WBPCq1Wd55nDu-cvtOO21Grz-_By39ydGl6bUhkGKonT0ZlcCkD1abudC9tMqJ8zx21srva4bV4DuIGJY8yTDnMUjtZE-NE0qfjnVodK__L-a4YglDzmJIU4hWyAJnw6-yeujKL9nTE1aBnC_TjqQXZtcTROsb8skqQxaM8iIGRJDWeGf7462eVTEIoyYdLLJymosYZlO7nM_luHno3inrhVIulOmkmRooi8yLcKGxipoPa3zd9sLtqAPhl1xShlkew7jdi4GOlyfPWYfc7jGW92CCrELm10T-9ps2gh-LYXwj6a2D7SAkBVcy6te_-QE">
</p>

A symmetry of $X$ should be a map (why a map from $X$ to $X$? Because we are performing a transformation on $X$, while $X$ itself remains the same)

$$
\phi: X \rightarrow X
$$

such that

1. it preserves the structure, for example
   $\operatorname{dist}(x, y)=\operatorname{dist}(\phi(x), \phi(y))$,
   or
   $\phi(x+y)=\phi(x)+\phi(y)$,
   and

2. the operation can be undone.

Now let us first look at this in the most general possible way, in order to gain some intuition.

Let $G={\phi}$ be the set of symmetries of $X$.

1. If $\phi_1,\phi_2$ preserve the structure, then their compositions
   $\phi_1\circ\phi_2$ and $\phi_2\circ\phi_1$
   also preserve the structure.
   $\Rightarrow$ We can compose elements of $G$
   $\Rightarrow$
   $G\times G\xrightarrow{m}G$,
   with associativity;

2. "Doing nothing" should also be a symmetry of $X$.
   $\Rightarrow \mathrm{id}_X\in G$,
   and
   $\operatorname{id}_X\circ\phi=\phi\circ\mathrm{id}_X=\phi$;

3. Since $\phi$ can be undone, we should have
   $\phi^{-1}\in G$,
   so
   $\phi\circ\phi^{-1}=\mathrm{id}_X$,
   and
   $\phi^{-1}\circ\phi=\mathrm{id}_X$.

Seen from this most general point of view, does the definition of a group not now almost suggest itself?

### §1.1.1 Definition of a Group

**Definition 1.1** A **group** is an ordered pair

$$
(G, m)
$$

where $G$ is a set and $m$ is a map

$$
\begin{aligned}
& G \times G \rightarrow G \\
& \begin{aligned}
\left(g_1, g_2\right) \mapsto m\left(g_1, g_2\right) & =: g_1 \cdot g_2 \\
& =: g_1 g_2
\end{aligned}
\end{aligned}
$$

such that

1. $m$ satisfies associativity, i.e.

$$
m\left(m\left(g_1, g_2\right), g_3\right)=m\left(g_1, m\left(g_2, g_3\right)\right)
$$

i.e.
$\left(g_1 \cdot g_2\right) \cdot g_3=g_1 \cdot\left(g_2 \cdot g_3\right)$
or
$\left(g_1 g_2\right) g_3=g_1\left(g_2 g_3\right)$;

2. $\exists$ an element $1_G \in G$, called the **identity**, such that

$$
m\left(1_G, g\right)=g=m\left(g, 1_G\right)
$$

i.e.
$1_G \cdot g=g=g \cdot 1_G$
or
$1_G g=g=g 1_G$;

3. $\forall g \in G$, $\exists$ an element $h\in G$ such that

$$
m(g, h)=1_G=m(h, g)
$$

i.e.
$g\cdot h=1_G=h\cdot g$
or
$gh=1_G=hg$.
We usually write
$g^{-1}:=h$
and call it the **inverse** of $g$.

### §1.1.2 Examples of Groups

**Example 1.1** Let

$$
G=\{\ldots,-1,0,1, \ldots\}=: \mathbb{Z}
$$

be the set of integers. Define

$$
G \times G \xrightarrow{m} G
$$

by

$$
m(g, h)=g+h
$$

(i.e. integer addition).

For example,

$$
m(-2,3)=1.
$$

Then $(G,m)$ is a group.

<details>
<summary>Proof.</summary>

$(\mathbb{Z},+)$ is a group because

(1) $m$ is associative:

$$
(g+h)+k=g+(h+k).
$$

(2) $0=1_G$ is the identity:

$$
\begin{aligned}
& m(0, g)=0+g=g \\
& m(g, 0)=g+0=g
\end{aligned}
$$

(3) Every element has an inverse:

$$
m(g,-g)=g+(-g)=0.
$$

$$
~\tag*{$\square$}
$$

</details>

**Example 1.2** Let

$$
G=\{\ldots,-1,0,1, \ldots\}=: \mathbb{Z}
$$

be the set of integers, and let

$$
\begin{aligned}
m: G \times G & \rightarrow G \\
(a, b) & \mapsto a \times b.
\end{aligned}
$$

For example,

$$
(2,3) \rightarrow 6.
$$

Then $(G,m)$ is not a group.

<details>
<summary>Proof.</summary>

$(\mathbb{Z},\times)$ is not a group because not every element
$z\in\mathbb{Z}$
has an inverse. For example, if $z=2$, its inverse would have to be
$\dfrac{1}{2}$,
which is not in $\mathbb{Z}$.

$$
~\tag*{$\square$}
$$

</details>

The two examples above show that

$(\mathbb{Z},+)$ is a group,

while

$(\mathbb{Z},\times)$ is not a group.

This illustrates that knowing $m$ is important. Nevertheless, we will often abbreviate our language and simply say "let $G$ be a group", omitting explicit mention of $m$.

**Example 1.3** Let
$G=\mathbb{R}\backslash{0}$
(the set of nonzero real numbers). Let

$$
\begin{aligned}
m: G \times G & \rightarrow G \\
(a, b) & \mapsto a \times b.
\end{aligned}
$$

Then $G$ is a group. From now on, we denote it by $\mathbb{R}^{\times}$.

<details>
<summary>Proof.</summary>

$\mathbb{R}^{\times}$ is a group because

(1) multiplication of real numbers is associative;

(2) the number $1$ is the identity;

(3) for every
$g\in\mathbb{R}\backslash{0}$,
there exists
$\dfrac{1}{g}$
such that
$g\dfrac{1}{g}=\dfrac{1}{g}g=1$.

$$
~\tag*{$\square$}
$$

</details>

### §1.1.3 Properties of Groups

**Proposition 3.1 (Cancellation Law)** Let $G$ be a group and let
$g,h,k\in G$.
Suppose

$$
gh=gk.
$$

Then

$$
h=k.
$$

Similarly,

$$
hg=kg \quad \Rightarrow \quad h=k.
$$

<details>
<summary>Proof.</summary>

There exists $g^{-1}$ such that
$g^{-1}g=1_G$.

$$
\begin{aligned}
g h=g k & \Rightarrow g^{-1}(g h)=g^{-1}(g k) \\
& \Rightarrow\left(g^{-1} g\right) h=\left(g^{-1} g\right) k \\
& \Rightarrow 1_G h=1_G k \\
& \Rightarrow h=k.
\end{aligned}
$$

We have used all the axioms of a group!

$$
~\tag*{$\square$}
$$

</details>

**Remark:** The cancellation law does not hold for arbitrary matrix multiplication unless the relevant matrices are invertible—for example, what happens if $g=0$?

**Proposition 3.2 (Uniqueness of the Identity)** The identity element of a group $G$ is unique. That is, if two elements $1_G$ and $1_G^{\prime}$ both satisfy the defining properties of the identity, then
$1_G=1_G^{\prime}$.

<details>
<summary>Proof.</summary>

If $1_G$ is an identity, then for every $g\in G$,

$$
1_Gg=g1_G=g.
$$

In particular, setting $g=1_G^{\prime}$, we must have

$$
1_G1_G^{\prime}=1_G^{\prime}.
$$

On the other hand, if $1_G^{\prime}$ is also an identity, then

$$
1_G1_G^{\prime}=1_G.
$$

By transitivity,

$$
1_G=1_G^{\prime}.
$$

$$
~\tag*{$\square$}
$$

</details>

**Proposition 3.3 (Uniqueness of Inverses)** For every element $g\in G$, its inverse $g^{-1}$ is unique. That is, if elements $h$ and $h^{\prime}$ both satisfy the defining properties of the inverse of $g$, then $h=h^{\prime}$.

<details>
<summary>Proof.</summary>

Suppose $h$ and $h^{\prime}$ are both inverses of $g$. Then

$$
gh^{\prime}=1_G.
$$

Multiplying both sides on the left by $h$, we obtain

$$
h\left(gh^{\prime}\right)=h.
$$

By associativity, the left-hand side becomes

$$
(hg)h^{\prime}=1_Gh^{\prime}=h^{\prime}.
$$

Hence, by transitivity,

$$
h^{\prime}=h.
$$

$$
~\tag*{$\square$}
$$

</details>

## §1.2 Abelian Groups

**Example 1.4** Let $n\geqslant1$ and $n\in\mathbb{Z}$. Then

$$
G=GL_n(\mathbb{R}):=\{n \times n \text{ real matrices } M \mid \det M \neq 0\}
$$

is a group, where

$$
m: G \times G \rightarrow G
$$

is given by matrix multiplication. This also illustrates that in general
$gh\neq hg$.

<details>
<summary>Proof.</summary>

$GL_n(\mathbb{R})$ is a group because

(1) matrix multiplication is associative;

(2) the identity matrix is the identity element;

(3)
$\operatorname{det}(g)\neq0\Rightarrow g$
is invertible.

$$
~\tag*{$\square$}
$$

</details>

Since matrix multiplication is not commutative, this shows that in general
$gh\neq hg$.
But what if the group multiplication is commutative? This leads to the following definition.

**Definition 1.2** A group $G$ is called an **Abelian group** if, for all
$g_1,g_2\in G$,
we have
$g_1g_2=g_2g_1$.

**Definition 1.3** A group for which the group multiplication is not commutative for all elements is called a **non-Abelian group**.

## §1.3 Cyclic Groups

Although Abelian groups already possess useful structure through commutativity, some Abelian groups are even simpler. What if the entire group could be generated by a single element? This leads to the notion of a cyclic group, one of the most fundamental examples of Abelian groups.

What makes cyclic groups special is that the entire group structure reduces to powers, or multiples, of a single element called a generator. This simplicity makes cyclic groups important building blocks in group theory and also useful tools in studying more complicated problems.

**Definition 1.4** A group $G$ is called a **cyclic group** if and only if there exists an element
$g\in G$,
called a **generator**, such that every element of $G$ can be written as a power of $g$:

$$
G=\langle g\rangle:=\{g^{n}\mid n\in\mathbb{Z}\}.
$$

**Example 1.5 (The Integers under Addition)**

The group $(\mathbb{Z},+)$ is cyclic, generated by $1$:

$$
\langle 1\rangle=\{\ldots,-2,-1,0,1,2,\ldots\}.
$$

**Example 1.6 (A Group under Modular Arithmetic)**

Under addition modulo $n$, the group
$\mathbb{Z}/n\mathbb{Z}={0,1,\ldots,n}$
is cyclic. The element $1$ generates the whole group:

$$
\langle 1\rangle=\{1,2,\ldots,n-1,0\}.
$$

**Proposition 1.4** Every cyclic group is Abelian.

<details>
<summary>Proof.</summary>

Let $G$ be a cyclic group generated by $g$. For any
$x,y\in G$,
there exist integers $m,n$ such that
$g^m=x$
and
$g^n=y$.
Therefore,

$$
xy=g^{m}g^{n}=g^{m+n}=g^{n}g^{m}=yx.
$$

Hence $G$ is Abelian.

$$
~\tag*{$\square$}
$$

</details>

## §1.4 Order of a Group

**Definition 1.5** Let $G$ be a group. We define

$$
|G|\in \mathbb{Z}_{\geqslant 1}\cup \{\infty\}
$$

to be the number of elements in $G$. We call $|G|$ the **order** of $G$.

**Definition 2.3** For $g\in G$, consider the set

$$
\{\ldots,\underbrace{g^{-1}\cdot g^{-1}}_{=:g^{-2}}, g^{-1},\mathrm{id}_{G},g,\underbrace{g\cdot g}_{=:g^{2}},\underbrace{g\cdot g\cdot g}_{=:g^{3}},\ldots\}.
$$

We define the **order of $g$** to be

$$
|\langle g\rangle|.
$$

**Example 1.7**

* The order of $1_G\in G$ is $1$.

* Any
  $n\in\mathbb{Z}$,
  $n\neq0$,
  has infinite order.

* The matrix
  $\begin{pmatrix}
  0 & 1\
  1 & 0
  \end{pmatrix}\in GL_{2}(\mathbb{R})$
  has order $2$, because

$$
g^{2}=\begin{pmatrix}
0&1\\
1&0
\end{pmatrix}\begin{pmatrix}
0&1\\
1&0
\end{pmatrix}=\begin{pmatrix}
1&0\\
0&1
\end{pmatrix}=1_{GL_{2}(\mathbb{R})}.
$$

Thus
${\ldots,g^{-1},1,g,\ldots}={1,g}$.

**Theorem 1.5 (Euler's Totient Function)** The number of generators of a cyclic group of order $n$ is given by Euler's totient function $\varphi(n)$.

## §1.5 The Center of a Group

To study non-Abelian groups, we often focus on subsets whose elements commute. One important concept is the center of a group.

**Definition 1.7** The **center** of a group $G$, denoted by $Z(G)$, is defined by

$$
Z(G)=\{z\in G\mid zg=gz~\text{for every}~g\in G\}.
$$

**Proposition 1.6** The center of an Abelian group is the whole group.

<details>
<summary>Proof.</summary>

The center of a group $G$ is defined by

$$
Z(G) = \{z\in G \mid zg=gz~\text{for all}~g \in G\}.
$$

In an Abelian group, every pair of elements commutes:

$$
g_{1} \cdot g_{2}=g_{2}\cdot g_{1} \quad \text{for all}~g_{1}, g_{2} \in G.
$$

Therefore every element of $G$ lies in the center, and hence

$$
Z(G)=G.
$$

$$
~\tag*{$\square$}
$$

</details>

The center of a group $G$ is the set of elements that commute with every element of $G$. This is not merely a collection of special elements: it itself forms an important structure inside $G$. We will introduce this structure in the next section.

Concept (Name)	All Numbers	Derivatives	Groups	Rings
What does this concept explain...? (Mathematics is a language for expressing ideas; what ideas do these words represent?)	Counting, quantity	Rate of change, linearization	Symmetry	Functions on spaces
Some mathematical results			The "algebraization" of geometry (from Descartes to the present), and the "geometrization" of algebra
Some applications (outside pure mathematics)			Noether's theorem (physics) RSA algorithm (cryptography) logic circuits as "cosheaves" homological shapes of data sets, etc.