# §12 单群和Hölder计划
## §12.1 单群
有些群不能由其他群构建出来。例如，如果$H$不允许任何（非平凡的）正规子群呢？那么就不可能有短正合列，除非$H \cong K$或$G \cong H$。在这种意义上，没有正规子群的群是最简单的群。

**定义 12.1** 一个群$H$称为**单群(simple group)**，如果它没有非平凡的正规子群。




**例 12.1** 一个循环群当且仅当它是有限的素数阶时是单群。(如果它是素数阶，它是单群，因为它没有子群，除了它本身和$\{1\}$。另一方面，阶为$n$的循环群对于每一个除以$n$的数$n/k$都有一个子群；例如，取任意生成元$x$的集合$\{1, x^{k}, \ldots\}$。因此，循环群在它是素数阶时是单群。)

**例 12.2** $\mathbb{Z}$不是单群。（它有很多子群，并且Abel群的任意子群都是正规子群。）

**例 12.3** $A_{1}, A_{2}, A_{3}$是单群。（$A_{1}$和$A_{2}$都有一个元素。根据第一个同构定理，$A_{3}$是$S_{3}$的指数为$2$的子群——它是阶为$3$的群，属于素数阶的循环群。）

**例 12.4** $A_{4}$不是单群。（我们需要找到一个非平凡的正规子群。这个群唯一的非平凡正规子群由2个循环的乘积元素构成。这个正规子群同构于Klein四元群，而商群是阶为$3$的循环群。）

**定理 12.1** 对于$n\geqslant 5$，$A_{n}$是一个单群。

证明：为了证明$n \geqslant 5$时交错群$A_n$是单群，我们需要证明$A_n$除了平凡子群$\{e\}$和$A_n$本身外没有其他正规子群。

设$N$是$A_n$的一个正规子群，且$N \neq \{e\}$。我们将证明$N = A_n$，从而证明$A_n$是单群。

1. $N$包含一个3-循环：
    
因为$N$是非平凡的，存在一个非恒等元$\sigma \in N$。我们考虑$\sigma$的互不相交的循环分解中的循环类型。

情况1：如果$\sigma$包含一个3-循环，则$N$包含一个3-循环。

情况2：如果$\sigma$不包含3-循环，我们将通过以下步骤证明$N$仍然包含一个3-循环。

2. 共轭与正规性：
    
因为$N$在$A_n$中是正规子群，对于任意$\tau \in A_n$，共轭$\tau \sigma \tau^{-1} \in N$。该性质使我们能够从已有的元素中生成$N$中的新元素。

3. 生成3-循环：
- 将元素表示为3-循环的乘积：
        
任何偶置换都可以表示为3-循环的乘积。例如，长度为$k \geqslant 3$的循环$(a_1\, a_2\, \dots\, a_k)$可以写成：
$$
(a_1\, a_2\, a_3)(a_1\, a_3\, a_4)\dots(a_1\, a_{k-1}\, a_k)
$$
- 使用对易子得到3-循环：
        
考虑$\sigma$和适当的$\tau \in A_n$。换位子$\gamma = \sigma \tau \sigma^{-1} \tau^{-1}$是$N$的一个元素（因为$N$是正规子群）且可以是3-循环。

4. $N$包含所有3-循环：
- 共轭类：

在$A_n$中，3-循环分为两个共轭类，但它们的并集是所有3-循环的集合。因为$N$包含至少一个3-循环且是正规子群，它必须包含该共轭类的所有3-循环。

- 用3-循环生成$A_n$：

当$n \geqslant 5$时，群$A_n$由其3-循环生成。因此，由所有3-循环生成的$A_n$的子群是$A_n$本身。

5. 结论：因为$N$包含所有3-循环，$N$生成$A_n$。因此，$N = A_n$。

因此，当$n \geqslant 5$时，$A_n$是单群。
$$
~\tag*{$\square$}
$$
## §12.2 Hölder计划
现在我们已经看到了很多群的例子，我们想开始对它们进行分类。我们是否可以想出一个一般的策略来帮助我们说：“我知道所有的群”？

**问题**：我们如何分类所有群？

在某种程度上，这个问题没有一个令人满意的答案。我们可以尝试了解所有的单群，然后了解它们的构造方式。19世纪开始的策略被称为Hölder计划。它看起来很自然。问题是，我们不知道如何执行它。我们甚至无法分类所有有限单群。

那么我们是否可以了解所有有限单群及其扩张？这至少会分类所有有限群。我们仍然不知道如何做到这一点。我们可以在1985年左右对所有有限单群进行分类，但我们仍然不知道如何解决分类其扩张的问题。为了让您了解分类的困难程度，请考虑以下定理为Thompson赢得了Fields奖：

**定理 12.2 （Feit-Thompson或奇数阶定理）**
每个有限的非Abel单群都有偶数阶。

因此，例如，如果你给我一个非Abel群，其阶是奇数，我就知道它不是单群。

**定义 12.2** 用于分类群的**Hölder计划(Hölder program)** 为：

(1) 分类所有单群。

(2) 分类所有构造单群扩张的方式。

我们不知道如何完成这个计划。例如，我们不知道如何做第(1)步。我们只知道如何对有限单群进行分类，并且直到1985年才完成这一工作。即使对于有限群，我们也没有完成第(2)步。
## §12.3 可解群
继Hölder计划的重点关注复杂群结构的理解和分解后，我们现在探讨可解群，这些群可以系统地分解为简单的Abel分量。

**定义 12.3** 群$G$称为**可解的（Solvable）**，如果它有一个有限的子群序列
$$
G=G_{0}\triangleright G_{1}\triangleright G_{2}\triangleright\cdots\triangleright G_{n}=\{e\}
$$
使得每个$G_{i}$在$G_{i-1}$中是正规子群且相应的商群$G_{i-1}/G$是Abel的。

**注** 群是“可解的”，因为它可以被分解为更小、更简单的部分，最终达到平凡群。

**例 12.5** 每个循环群都是Abel群，因此它是平凡地可解的。

**例 12.6**
$S_{3}$是可解的。正规序列：$S_{3}\triangleright A_{3}\triangleright\{e\}$。商群：$S_{3}/A_{3}\cong\mathbb{Z}/2\mathbb{Z}$，且$A_{3}/\{e\}\cong\mathbb{Z}/3\mathbb{Z}$。它们都是Abel的，因此$S_{3}$是可解的。

**例 12.7** $S_{5}$不是可解的。缺乏合适的正规序列突显了可解群和更复杂群之间的差异。

**命题 12.3** 可解群的每个子群都是可解的。

证明：一个可解群$G$有一个子群链：
$$
G=G_{0} \triangleright G_{1} \triangleright \dots \triangleright G_{n}=\{e\},
$$
其中每个商群$G_{i-1}/G_{i}$都是Abel的。

设$H \subseteq G$为子群。我们可以将$H$与链中的每个群相交：
$$
H \cap G_{0} \triangleright H \cap G_{1} \triangleright \dots \triangleright H \cap G_{n}=\{e\}。
$$
每个商群$(H\cap G_{i-1})/(H \cap G_{i})$是Abel商群$G_{i-1}/G_{i}$的子群，因此它也是Abel的。

这给出了$H$的一个Abel商群的正规序列，意味着$H$是可解的。
$$
~\tag*{$\square$}
$$

**命题 12.4** 可解群的商群是可解的。

证明：如果群$G$是可解的，则它有一个正规子群链：
$$
G=G_{0}\triangleright G_{1} \triangleright \dots \triangleright G_{n}=\{e\},
$$
其中每个商群$G_{i-1}/G_{i}$都是Abel的。

设$N \triangleleft G$为正规子群，我们想要证明商群$G/N$是可解的。

利用$G$的链在商群中形成一个新链：
$$
G_{0}/N \triangleright G_{1}/N \triangleright \dots \triangleright G_{n}/N= \{e\}.
$$

每个新的商群：
$$
   (G_{i-1}/N)/(G_{i}/N) \cong G_{i-1}/G_{i}
$$
是Abel的，因为原始商群$G_{i-1}/G_{i}$是Abel的。

因此，商群$G / N$是可解的。
$$
~\tag*{$\square$}
$$
关于可解群的讨论展示了如何将群分解为更简单的组成部分。虽然可解群结构清晰且易于处理，但非可解群（如单群）的研究在理解群论的广泛图景中仍然至关重要。

在下一章中，我们将继续探讨群论中的更多专门结果，例如Sylow定理，它为有限群的结构提供了更深刻的见解。

**# §12 Simple Groups and the Hölder Program

## §12.1 Simple Groups

Some groups cannot be built out of other groups. For example, what if $H$ admits no nontrivial normal subgroups? Then there can be no short exact sequence unless $H\cong K$ or $G\cong H$. In this sense, groups with no nontrivial normal subgroups are the simplest groups.

**Definition 12.1** A group $H$ is called a **simple group** if it has no nontrivial normal subgroups.



**Example 12.1** A cyclic group is simple if and only if it is finite of prime order.

If it has prime order, then it is simple because it has no subgroups other than itself and ${1}$.

Conversely, a cyclic group of order $n$ has a subgroup for every number $n/k$ dividing $n$; for example, given any generator $x$, consider the set

$$
\{1,x^{k},\ldots\}.
$$

Therefore, a cyclic group is simple precisely when it has prime order.

**Example 12.2** $\mathbb{Z}$ is not simple.

It has many subgroups, and every subgroup of an Abelian group is normal.

**Example 12.3** $A_1,A_2,A_3$ are simple.

$A_1$ and $A_2$ each have one element. By the First Isomorphism Theorem, $A_3$ is a subgroup of index $2$ in $S_3$—it is a group of order $3$, and hence a cyclic group of prime order.

**Example 12.4** $A_4$ is not simple.

We need to find a nontrivial normal subgroup. The unique nontrivial normal subgroup of this group consists of elements that are products of two cycles. This normal subgroup is isomorphic to the Klein four group, while the quotient group is cyclic of order $3$.

**Theorem 12.1** For $n\geqslant5$, $A_n$ is simple.

Proof: To prove that the alternating group $A_n$ is simple for $n\geqslant5$, we must show that $A_n$ has no normal subgroups other than the trivial subgroup ${e}$ and $A_n$ itself.

Let $N$ be a normal subgroup of $A_n$ with

$$
N\neq\{e\}.
$$

We will prove that

$$
N=A_n,
$$

and hence that $A_n$ is simple.

1. $N$ contains a $3$-cycle.

Since $N$ is nontrivial, there exists a nonidentity element

$$
\sigma\in N.
$$

Consider the cycle types appearing in the disjoint-cycle decomposition of $\sigma$.

Case 1: If $\sigma$ contains a $3$-cycle, then $N$ contains a $3$-cycle.

Case 2: If $\sigma$ contains no $3$-cycle, we will show in the following steps that $N$ nevertheless contains a $3$-cycle.

2. Conjugation and normality.

Since $N$ is normal in $A_n$, for every

$$
\tau\in A_n,
$$

we have

$$
\tau\sigma\tau^{-1}\in N.
$$

This property allows us to generate new elements of $N$ from elements already known to lie in $N$.

3. Producing $3$-cycles.

* Express elements as products of $3$-cycles.

Every even permutation can be expressed as a product of $3$-cycles. For example, a cycle of length $k\geqslant3$,

$$
(a_1\,a_2\,\dots\,a_k),
$$

can be written as

$$
(a_1\,a_2\,a_3)
(a_1\,a_3\,a_4)
\dots
(a_1\,a_{k-1}\,a_k).
$$

* Use commutators to obtain a $3$-cycle.

Consider $\sigma$ and a suitable

$$
\tau\in A_n.
$$

The commutator

$$
\gamma
=
\sigma\tau\sigma^{-1}\tau^{-1}
$$

is an element of $N$, because $N$ is normal, and it can be a $3$-cycle.

4. $N$ contains all $3$-cycles.

* Conjugacy classes.

In $A_n$, the $3$-cycles split into two conjugacy classes, but their union is the set of all $3$-cycles.

Since $N$ contains at least one $3$-cycle and is normal, it must contain every $3$-cycle in the corresponding conjugacy class.

* $A_n$ is generated by $3$-cycles.

For

$$
n\geqslant5,
$$

the group $A_n$ is generated by its $3$-cycles.

Therefore, the subgroup of $A_n$ generated by all $3$-cycles is $A_n$ itself.

5. Conclusion.

Since $N$ contains all $3$-cycles, $N$ generates $A_n$. Hence

$$
N=A_n.
$$

Therefore, for

$$
n\geqslant5,
$$

$A_n$ is simple.

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

## §12.2 The Hölder Program

Now that we have seen many examples of groups, we would like to begin classifying them.

Can we come up with a general strategy that would allow us to say:

“I know all groups”?

**Question:** How do we classify all groups?

In some sense, this question has no satisfactory answer.

We can try to understand all simple groups, and then understand all the ways in which they can be assembled.

The strategy that began in the nineteenth century is called the **Hölder program**.

It looks very natural.

The problem is that we do not know how to carry it out.

We cannot even classify all simple groups.

Can we at least understand all finite simple groups and their extensions?

That would classify all finite groups.

We still do not know how to do this.

By around 1985, we were able to classify all finite simple groups, but we still do not know how to solve the problem of classifying their extensions.

To give you some sense of how difficult the classification problem is, consider the following theorem, which helped earn Thompson a Fields Medal:

**Theorem 12.2 (Feit-Thompson Theorem, or Odd Order Theorem)**

Every finite non-Abelian simple group has even order.

Thus, for example, if you give me a non-Abelian group of odd order, I immediately know that it is not simple.

**Definition 12.2** The **Hölder program** for classifying groups is:

(1) Classify all simple groups.

(2) Classify all ways of constructing extensions of simple groups.

We do not know how to complete this program.

For example, we do not know how to carry out Step (1) in general.

We only know how to classify finite simple groups, and this classification was not completed until around 1985.

Even for finite groups, we have not completed Step (2).

## §12.3 Solvable Groups

Following the Hölder program's emphasis on understanding and decomposing complicated group structures, we now study solvable groups, which can be systematically decomposed into simpler Abelian components.

**Definition 12.3** A group $G$ is called **solvable** if it has a finite sequence of subgroups

$$
G=G_{0}\triangleright G_{1}\triangleright G_{2}\triangleright\cdots\triangleright G_{n}=\{e\}
$$

such that each $G_i$ is normal in $G_{i-1}$ and the corresponding quotient group

$$
G_{i-1}/G
$$

is Abelian.

**Remark** A group is called “solvable” because it can be decomposed into smaller and simpler pieces, eventually reaching the trivial group.

**Example 12.5** Every cyclic group is Abelian, and therefore it is trivially solvable.

**Example 12.6**

$S_3$ is solvable.

A normal series is

$$
S_3\triangleright A_3\triangleright\{e\}.
$$

The quotient groups are

$$
S_3/A_3
\cong
\mathbb{Z}/2\mathbb{Z},
$$

and

$$
A_3/\{e\}
\cong
\mathbb{Z}/3\mathbb{Z}.
$$

Both are Abelian, so $S_3$ is solvable.

**Example 12.7** $S_5$ is not solvable.

The absence of a suitable normal series highlights the difference between solvable groups and more complicated groups.

**Proposition 12.3** Every subgroup of a solvable group is solvable.

Proof: A solvable group $G$ has a chain of subgroups

$$
G=G_{0}
\triangleright
G_{1}
\triangleright
\dots
\triangleright
G_{n}
=
\{e\},
$$

where each quotient group

$$
G_{i-1}/G_i
$$

is Abelian.

Let

$$
H\subseteq G
$$

be a subgroup.

We may intersect $H$ with each group in the chain:

$$
H\cap G_{0}
\triangleright
H\cap G_{1}
\triangleright
\dots
\triangleright
H\cap G_{n}
=
\{e\}.
$$

Each quotient group

$$
(H\cap G_{i-1})/(H\cap G_i)
$$

is a subgroup of the Abelian quotient group

$$
G_{i-1}/G_i,
$$

and is therefore itself Abelian.

Thus we obtain a normal series for $H$ whose quotient groups are Abelian.

Therefore,

$$
H
$$

is solvable.

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

**Proposition 12.4** A quotient group of a solvable group is solvable.

Proof: Suppose $G$ is solvable.

Then it has a chain of normal subgroups

$$
G=G_{0}
\triangleright
G_{1}
\triangleright
\dots
\triangleright
G_{n}
=
\{e\},
$$

where each quotient

$$
G_{i-1}/G_i
$$

is Abelian.

Let

$$
N\triangleleft G
$$

be a normal subgroup. We want to prove that

$$
G/N
$$

is solvable.

Use the series in $G$ to form a new series in the quotient:

$$
G_{0}/N
\triangleright
G_{1}/N
\triangleright
\dots
\triangleright
G_{n}/N
=
\{e\}.
$$

Each new quotient

$$
(G_{i-1}/N)/(G_i/N)
\cong
G_{i-1}/G_i
$$

is Abelian because the original quotient

$$
G_{i-1}/G_i
$$

is Abelian.

Therefore,

$$
G/N
$$

is solvable.

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

Our discussion of solvable groups illustrates how groups can be decomposed into simpler components.

Although solvable groups have a clear and manageable structure, the study of nonsolvable groups, such as simple groups, remains essential for understanding the broader landscape of group theory.

In the next chapter, we will continue to explore more specialized results in group theory, such as the Sylow theorems, which provide deeper insight into the structure of finite groups.
**

§ DS-12 单群和Hölder计划

§ DS-12 Simple Group and Hölder Program

§ DS-11 短正合序列和半直积

§ DS-11 SES and Semi-direct Product

# §10 同构定理
## §10.1 第一同构定理
### §10.1.1 商映射作为群同态
**命题 10.1** 设 $H\subset G$ 是正规子群。映射
$$
\begin{aligned}
q: G&\to G/H\\
g&\mapsto Hg
\end{aligned}
$$
(1) 是一个群同态。

(2) 是满射。

(3) 有核$q$。



证明：(1) 
$$
\begin{aligned}
q(g_{1}g_{2})&=Hg_{1}g_{2}\\
&=Hg_{1}Hg_{2}\\
&=q(g_{1})q(g_{2})
\end{aligned}
$$
(2) 对所有 $Hg\in G/H$，
$$
Hg=q(g)
$$
(3) $q(g)=1_{G/H}$ $\Leftrightarrow$ $\begin{aligned}
q(g)&=H1_{G}\\
&=H
\end{aligned}$

但 $Hg=H1_{G}$ $\Leftrightarrow$ $g$ 和 $1_{G}$ 在相同轨道上

$\Leftrightarrow$ $g=h1_{G}$ 对某个 $h\in H$

$\Leftrightarrow$ $g\in H$

所以 $q(g)=1_{G/H}$ $\Leftrightarrow$ $g\in H$。
$$
~\tag*{$\square$}
$$
### §10.1.2 可视化
图像是什么样的？

将 $G$ 视为某个集合
<div align=center>
    <img src=https://tex.xumin.net/svgb/VY_RCsIgFIbv9xRe7MJgO2hDCvYAe4iMMLUliRtmBMnePZVB7e7__-_jwOFXPRoXg3l8ZiPDy-ul4sqLN8KkIbu2xR2wX6DAciQlIAAkJxf8ZJ9Jh64hcEizcOq_Jqvc6it-M9YiTNO8Mmm8tDpPbA7ZcJPSJ2_GezgjETZurOklDku9rFrhHRwT3mc6FKKd2nzzBQ>
</div>

子群 $H\subset G$ 将 $G$ 划分为轨道
<div align=center>
    <img src=https://tex.xumin.net/svgb/jZRNboMwEIX3nIIFi0QiFpiQEFVdp3cAWhFwCKoLEXEUqYi7d4z5scFpu0Ee8_zN82NEdCJ5UTYpKRmpWyMSJSs-v69Fyu41acNbmlDyimMjyurkYa4c21lvNisPBdPCRT5fOt3CRMhMq5LVFb2BHHm2g_awnZSZXIKqY70Yghye6J3EXIHtDTwWHF_l-DIHzrhoD6wZytejAhUVKChfjwp0KBe5MmoqBSroUUZUVhkxE8YlHne-XZuNFaVZxW5WO48AUtRm4KIAq-3GmvfrjvXex34YLIGt39phhId2_Fh4IrR6xHDgLbda1QHmLSQHUs0ddKTewbmglO9s-XWDtZkWdUoJuLyywWBYF_mFxb3PQdhYuc7jThsJRoeZoYNqaKedDA85z6783kTXuvgi4tnOE_CQpzSUat6wAysJeAj_L4FJCAksXPBABBDow-z_QZSUjeV-HAVlHI1tN62YvzxODaYh3Pni9dMutpCKVoO6sS5LGHxdMfQ6FiVnGTVqgZQvURDTc9Tc1ijmLE2mESkz5Xcndvr_4Q8>
</div>

而 $G/H$ 塌缩到所有这些轨道：
<div align=center>
    <img src=https://tex.xumin.net/svgb/hZBBCoMwEEX3niILFwZENNZaKN33DsaCxqmEplFiRKh498aqFBXazTB_eH_-MDSHksuegdSgBotOUvPHq-ZMtwqGpGGZgAtJLVqorEOO7_oYeR5ildSqEg1yAtf34gijTBZfYQiHmD7awKGZkQVexAgf3ACfLYveuRBJLlpIzTIvMsSRYCSrApIcRNWlqLev9oAYV0yAYWptfCvbnLvxlH9Mn6zwhH9B48FhvF9962mt-BOmOuyTQBart06T-e9v>
</div>

### §10.1.3 单射和群同态的核
顺便说一下，

**命题 10.2** 设 $G\xrightarrow{\phi} G^{\prime}$ 是一个群同态。则 $\phi$ 是单射当且仅当
$$
\ker(\phi)=\{1_{G}\}.
$$
证明：单射 $\Rightarrow$ $\exists ! g$（如果存在）使得 $\phi(g)=1_{G^{\prime}}$。

由于群同态总是将 $1_{G}$ 映射到 $1_{G^{\prime}}$，所以 $g\in 1_{G}$。

假设 $\ker(\phi)=\{1_{G}\}$。那么
$$
\begin{aligned}
\phi(g_{1})=\phi(g_{2}) &\Rightarrow \phi(g_{1})\phi(g_{2})^{-1}=1_{G^{\prime}}\\
&\Rightarrow \phi(g_{1}g_{2}^{-1})=1_{G^{\prime}}\\
&\Rightarrow g_{1}g_{2}^{-1}\in\ker(\phi)\\
&\Rightarrow g_{1}g_{2}^{-1}=1_{G}\\
&\Rightarrow g_{1}=g_{2}
\end{aligned}
$$
$$
~\tag*{$\square$}
$$
因此给定任意的正规子群 $H$，商同态 $q$ 表示 $H$ 作为某个群同态的核。
### §10.1.4 核是正规子群
**问题**：是否每个群同态的核都是正规子群？

**命题 10.3** 设 $\phi:G\to G^{\prime}$ 是一个群同态。那么 $\ker(\phi)$ 是正规子群。

证明：需要证明：$\forall h\in\ker(\phi)$，$\forall g\in G$，$ghg^{-1}\in\ker(\phi)$。

那么，
$$
\begin{aligned}
\phi(ghg^{-1})&=\phi(g)\phi(h)\phi(g^{-1})\\
&=\phi(g)1_{G^{\prime}}\phi(g^{-1})\\
&=\phi(g)\phi(g^{-1})\\
&=\phi(gg^{-1})\\
&=\phi(1_{G})\\
&=1_{G^{\prime}}
\end{aligned}
$$
所以 $ghg^{-1}\in\ker(\phi)$。
$$
~\tag*{$\square$}
$$
在这里我们只证明了
$$
g\ker(\phi)g^{-1}\subset \ker(\phi), \quad\forall g
$$
以证明 $\ker(\phi)$ 是正规子群。但我们如何证明
$$
g\ker(\phi)g^{-1}=\ker(\phi)?
$$
### §10.1.5 共轭的相等性意味着正规性
**命题 10.4** 设 $H\subset G$ 是一个子群。“$\forall g\in G$, $gHg^{-1}\subset H$” 意味着 “$\forall g\in G$, $gHg^{-1}=H$”。

证明：我们需要证明 $\forall g\in G$, $H\subset gHg^{-1}$。

因此固定 $h\in H$。设 $g^{\prime}=g^{-1}$。根据假设，
$$
g^{\prime}H(g^{\prime})^{-1}\subset H,
$$
所以
$$
g^{\prime}h(g^{\prime})^{-1}=h^{\prime}
$$
对于某个 $h^{\prime}\in H$。那么
$$
h=gh^{\prime}h^{-1}
$$
因为
$$
\begin{aligned}
gh^{\prime}g^{-1}&=g(g^{\prime}h(g^{\prime})^{-1})g^{-1}\\
&=gg^{-1}hgg^{-1}\\
&=h
\end{aligned}
$$
这表明 $h\in gHg^{-1}$。
$$
~\tag*{$\square$}
$$
### §10.1.6 正规子群的交集
**推论 10.5** 如果 $H_{1}$, $H_{2}$ 是 $G$ 的正规子群，那么 $H_{1}\cap H_{2}$ 也是正规子群。

证明：设 $h\in H_{1}\cap H_{2}$。则对于所有 $g\in G$，

- $ghg^{-1}\in H_{1}$ 因为 $H_{1}$ 是正规子群。

- $ghg^{-1}\in H_{2}$ 因为 $H_{2}$ 是正规子群。

因此 $ghg^{-1}\in H_{1}\cap H_{2}$。

$\Rightarrow$ 对于所有 $g\in G$，$gH_{1}\cap H_{2}g^{-1}\subset H_{1}\cap H_{2}$。
$$
~\tag*{$\square$}
$$

注意

**命题 10.6** 如果 $H_{1}$, $H_{2}$ 只是 $G$ 的子群，那么 $H_{1}\cap H_{2}$ 也是 $G$ 的一个子群。

证明：$1_{G}\in H_{1}$ 和 $H_{2}$，因为它们都是子群。

因此
$$
1_{G}\in H_{1}\cap H_{2}.
$$
如果 $g$ 和 $g^{\prime}$ 在 $H_{1}\cap H_{2}$ 中，那么

- $gg^{\prime}\in H_{1}$ 因为 $H_{1}$ 是子群

- $gg^{\prime}\in H_{2}$ 因为 $H_{2}$ 是子群

因此
$$
gg^{\prime}\in H_{1}\cap H_{2}.
$$
对逆元也是类似。
$$
~\tag*{$\square$}
$$

### §10.1.7 构造包含一个集合的最小正规子群
现在假设你有一个群 $G$，以及一些任意的集合
$$
I
$$
其元素在 $G$ 中。（$I$ 不一定是一个子群；它只是 $G$ 中的一些元素的随机列表。）

**问题**：你能找到一个包含 $I$ 的 $G$ 的正规子群吗？

那么，$G$ 本身是 $G$ 的一个正规子群。并且它当然包含 $I$。

**问题**： 我们能得到更小的子群吗？

可以。考虑交集
$$
\bigcap H
$$
集合 $\{H\subset G\mid H~\text{是正规子群且}~H~\text{包含}~I\}$ 不是空的，因为 $G$ 在其中。因此，我们通过这个交集得到 $G$ 的一个正规子群（根据前面的推论）。这是一种很好的构造方式。

### §10.1.8 第一同构定理
**命题 10.7** 设 $\phi:G\to G^{\prime}$ 是一个满射群同态。那么 $\exists$ 一个同构 $G/\ker(\phi)\xrightarrow{\cong} G^{\prime}$。

证明：给定一个满射群同态
$$
\phi:G\to G^{\prime},
$$
设 $H=\ker(\phi)$。注意，如果 $g_{2}\in Hg_{1}$，
$$
\phi(g_{2})=g_{1}
$$
因为
$$
\begin{aligned}
\phi(g_{2})&=\phi(hg_{1}), \quad\text{对某些}~h\in H\\
&=\phi(h)\phi(g_{1})\\
&=1_{G^{\prime}}\phi(g_{1})\\
&=\phi(g_{1})
\end{aligned}
$$
所以我们有一个良定义的映射
$$
\begin{aligned}
\psi: G/H&\to G^{\prime}\\
Hg&\mapsto \phi(g)
\end{aligned}
$$
（我们已经证明，如果 $Hg_{1}=Hg_{2}$，那么 $\phi(g_{1})=\phi(g_{2})$。）

这是一个同态，因为
$$
\begin{align*}
\psi(Hg_{1}Hg_{2})&=\psi(Hg_{1}g_{2})\tag*{在 $G/H$ 中}\\
&=\phi(g_{1}g_{2})\tag*{定义 $\psi$}\\
&=\phi(g_{1})\phi(g_{2})\tag*{$\phi$ 是一个同态}\\
&=\psi(Hg_{1})\psi(Hg_{2})\tag*{定义 $\psi$}
\end{align*}
$$
它是单射的，因为
$$
\begin{aligned}
\psi(Hg_{1})=1_{G^{\prime}} & \Leftrightarrow \psi(g_{1})=1_{G^{\prime}}\\
&\Leftrightarrow g_{1}\in H\\
&\Leftrightarrow Hg_{1}=H1_{G}, \text{即}~G/H~\text{的单位元}
\end{aligned}
$$

它是满射的，因为 $\phi$ 是满射：
$$
\forall g^{\prime}\in G^{\prime}, \exists~\text{某个}~g\in G, \text{使得}~\phi(g)=g^{\prime}, \text{因此}~\psi(Hg)=g。
$$
$$
~\tag*{$\square$}
$$
**推论 10.8 (第一同构定理)**
设 $\phi:G\to G^{\prime}$ 是任意的群同态。那么 $\exists$ 群同构
$$
G/\ker(\phi)\cong \mathrm{im}(\phi).
$$

证明：$\mathrm{im}(\phi)\subset G^{\prime}$ 是一个子群。根据像的定义，群同态 $\phi: G\to G^{\prime}$ 如下分解：
<div align=center>
    <img src=https://tex.xumin.net/svgb/i0lKTc_Mqy7JzK5KTqnlcleISSwqyi-PLirSUVCKKcjIVIqFC6WkgMTyy1KLcjLzUqtBsrVK6rEKagpA5B4HFCjKzE2tVYiJ4QIKqEHpmNzEkoyi3OrM3FoNkBZNuHmlpUDzssAmcMWk5qXAXAEA>
</div>

这里，$j$ 是 $\mathrm{im}(\phi)$ 到 $G$ 的嵌入。（这是一个单射的群同态。）$\overline{\phi}$ 是与 $\phi$ 相同的函数，但具有不同的目标/到达域。所以我们看到 $\phi=j\circ \overline{\phi}$。

另外，根据像的定义，$\overline{\phi}$ 是满射。因此，这个命题说明
$$
G/\ker(\overline{\phi})\cong \mathrm{im}(\phi)
$$
但 $\ker(\overline{\phi})=\ker(\phi)$，因为 $j(1_{\mathrm{im}(\phi)})=1_{G^{\prime}}$。
$$
~\tag*{$\square$}
$$
### §10.1.9 第一同构定理的应用：指数
**命题 10.9** 设
$$
O_{n}(\mathbb{R}):=\{n\times n~\text{实矩阵}~A~\text{满足}~A^{\mathrm{T}}A=I\}
$$
和
$$
SO_{n}(\mathbb{R}):=\{A\in O_{n}(\mathbb{R})~\text{使得}~\det(A)=1\}.
$$
那么
$$
[O_{n}(\mathbb{R}):SO_{n}(\mathbb{R})]=2.
$$
即 $SO_{n}(\mathbb{R})$ 是 $O_{n}(\mathbb{R})$ 的指数为 $2$ 的子群。

证明：考虑同态
$$
\begin{aligned}
\det:O_{n}(\mathbb{R})&\to \mathbb{R}^{\times}\\
A&\mapsto \det(A)
\end{aligned}
$$
因为 $\mathbb{R}^{\times}$ 的单位元是 $1\in\mathbb{R}^{\times}$，所以 $\det$ 的核是 $SO_{n}(\mathbb{R})$。

另一方面，
$$
\mathrm{im}(\det)=\{+1,-1\}\subset \mathbb{R}^{\times}.
$$
因此
$$
\begin{align*}
[O_{n}(\mathbb{R}):SO_{n}(\mathbb{R})]&=|O_{n}(\mathbb{R})/SO_{n}(\mathbb{R})|\tag*{根据指数的定义}\\
&=|O_{n}(\mathbb{R})/\ker(\det)|\\
&=|\mathrm{im}(\det)|\tag*{根据第一同构定理}\\
&=|\{+1,-1\}|\\
&=2
\end{align*}
$$
$$
~\tag*{$\square$}
$$

## §10.2 第二同构定理
### §10.2.1 第二同构定理
确定一个群 $G$。设 $S\subset G$ 是一个子群，$N\triangleleft G$ 是一个正规子群。

**命题 10.10** 设 $SN$ 是 $G$ 中所有形如 $sx$ 的元素的集合，其中 $s\in S$ 且 $x\in N$。这是 $G$ 的一个子群。

证明：给定 $s_{1}, s_{2}\in S$ 和 $x_{1}, x_{2}\in N$，我们有
$$
s_{1}x_{1}s_{2}x_{2}=s_{1}s_{2}s_{2}^{-1}x_{1}s_{2}x_{2}=s_{1}s_{2}x^{\prime} x_{2}
$$
对某个 $x^{\prime}\in N$ （因为 $N$ 是正规子群）。并且 $s_{1}s_{2}\in S$ 且 $x^{\prime}x_{2}\in N$ 因为它们都在乘法下封闭。单位元在 $SN$ 中，因为 $1\in S$, $N$ 且 $1\cdot 1=1$。最后，$SN$ 包含逆元，因为
$$
x^{-1}s^{-1}=(s^{-1}x^{\prime}s)s^{-1}=s^{-1}x^{\prime}
$$
其中 $x^{\prime}\in N$ 是满足 $x^{\prime}=sx^{-1}s^{-1}$ 的元素。
$$
~\tag*{$\square$}
$$

**命题 10.11** $N$ 是 $SN$ 的正规子群。

证明：我们知道对于每个 $g\in G$ 和 $x\in N$，$gxg^{-1}\in N$。由于 $SN\subset G$，我们特别有，对于任何 $g\in SN$，$gxg^{-1}\in N$。
$$
~\tag*{$\square$}
$$
**命题 10.12** $S\cap N$ 是 $S$ 的正规子群。

证明：如果 $x\in S\cap N$，那么对于所有 $s\in S$，我们知道 $sxs^{-1}\in N$，因为 $N$ 是 $G$ 的正规子群。另一方面，$S$ 在乘法下封闭，所以 $sxs^{-1}\in S$ 也是成立的。这表明 $sxs^{-1}\in N\cap S$。
$$
~\tag*{$\square$}
$$
**定理 10.13 (第二同构定理)**
存在一个同构
$$
S/(S\cap N)\cong SN/N.
$$
证明：考虑以下同态的复合
$$
S\to SN\to SN/N
$$
其中后者是商映射，前者是简单的包含映射（注意 $S\subset SN$）。这个组合是满射的，因为对于任何 $n\in N$，元素 $[sn]\in SN/N$ 等价于元素 $[s]\in SN/N$。它的核是那些在 $N$ 中的元素——即 $S\cap N$。因此，根据第一同构定理，我们完成了证明。
$$
~\tag*{$\square$}
$$
**问题**：商集 $[s]$ 在 $S/(S\cap N)$ 中是否定义了 $SN/N$ 中的商集 $[sn]$？$[sn]$ 中的 $n$ 重要吗？

这给了我们另一种第二同构定理的证明方法的提示：

证明：给定 $[sn]\in SN/N$，考虑 $[s]\in S/(S\cap N)$。

我们声称赋值 $\phi: [sn]\mapsto [s]$ 是良定义的。如果 $sn=s^{\prime}n^{\prime}x$ 且 $x\in N$，那么
$$
s=s^{\prime}(n^{\prime}xn^{-1})。
$$
我们必须证明元素 $n^{\prime}xn^{-1}$ 在 $S\cap N$ 中。我们可以通过在等式两边左乘 $s^{\prime -1}$ 来看它必须在 $S$ 中。我们知道它在 $N$ 中，因为元素 $n^{\prime}$, $x$, $n^{-1}$ 都在 $N$ 中，且 $N$ 在乘法下封闭。

现在我们证明这是一个群同态：
$$
\begin{aligned}
\phi([s_{1}n_{1}][s_{2}n_{2}])=\phi([s_{1}n_{1}s_{2}n_{2}])&=\phi([s_{1}s_{2}(s_{2}^{-1}n_{1}s_{2}n_{2})])\\
&=\phi([s_{1}s_{2}(n^{\prime}s_{2})])\\
&=[s_{1}s_{2}]\\
&=[s_{1}][s_{2}]\\
&=\phi([s_{1}n_{1}])\phi([s_{2}n_{2}]).
\end{aligned}
$$

为了证明它是单射的，我们必须知道核是平凡的。如果 $\phi([sn])=[x]$ 对于 $x\in S\cap N$，那么 $[sn]$ 有一个代表形式 $xn^{\prime}$；但 $x\in X\cap N$，$n^{\prime}\in N$ 意味着 $xn^{\prime}\in N$，由于 $N$ 在乘法下封闭，因此 $[sn]=[sn^{\prime}]=1\in SN/N$。

为了证明它是满射的，注意对于任何 $s\in S$，我们有 $s=s1_{G}\in SN$。所以 $\phi([s1_{G}])=\phi(s)$。
$$
~\tag*{$\square$}
$$
### §10.2.2 第二同构定理的应用
**例 10.1** 设 $G=S_{4}$，即 $4$ 个元素的对称群。设 $S$ 是由置换 $(12)$ 生成的子群，$N$ 是交错群 $A_{4}$，即 $4$ 个元素的交错群。我们有同构 $S\cong\mathbb{Z}_{2}$。

证明：
- 确定 $S$, $N$ 及其交集：

$S=\langle (12)\rangle$，其阶数为 $2$。

$N=A_{4}$，其阶数为 $12$。

$S\cap N$: 由于 $(12)$ 是一个偶置换，只有当它可以表示为偶数个换位时。但是 $(12)$ 是单个换位，所以 $(12)\notin A_{4}$。因此，$S\cap N=\{1_{G}\}$。

- 计算 $SN$：

$SN$ 由所有可以表示为 $sn$ 的元素组成，其中 $s\in S$ 且 $n\in N$。

由于 $S$ 的阶数为 $2$，$N$ 的阶数为 $12$ 且 $S\cap N=\{1_{G}\}$，因此 $SN$ 的阶数为 $|S||N|/|S\cap N|=(2\times 12)/1=24$。

然而，$G$ 的阶数是 $24$，所以 $SN=G$。

- 应用第二同构定理：
$$
S\cong S/\{1_{G}\}\cong S/(S\cap N)\cong (SN)/N\cong G/N\cong S_{4}/A_{4}\cong\mathbb{Z}_{2}。
$$
$$
~\tag*{$\square$}
$$

## §10.3 第三同构定理
第三同构定理回答了以下问题：假设我有一个嵌套的子群序列 $K\subset N\subset G$。我可以将整个 $N$ 商出去，得到轨道集 $G/N$。（在此过程中，$K$ 也被商掉了，因为 $K$ 包含在 $N$ 中。）或者，我可以尝试逐步商出去：首先取 $G/K$，然后再商去剩余的 $N$。结果是一样的吗？答案是肯定的，如果 $K$ 和 $N$ 都是 $G$ 的正规子群（这样才能谈论商群），最终的结果是相同的群。

### §10.3.1 第三同构定理
**命题 10.14** 存在子群 $K\subset N\subset G$。存在一个单射
$$
f: N/K\to G/K.
$$
证明：对于 $N/K$ 中的任何陪集 $nK$（其中 $n \in N$），定义：
$$
f(nK) = nK \in G/K.
$$
这仅意味着我们将来自 $N/K$ 的陪集 $nK$ 视为 $G/K$ 中的一个元素。

良定义性：如果在 $N/K$ 中 $n_1K = n_2K$，则 $n_1^{-1}n_2 \in K$。因为 $K \subset G$，这也意味着在 $G/K$ 中 $n_1K = n_2K$。因此，$f$ 是良定义的。

如果 $f(n_1K) = f(n_2K)$，那么在 $G/K$ 中 $n_1K = n_2K$，意味着 $n_1^{-1}n_2 \in K$。这表明在 $N/K$ 中 $n_1K = n_2K$。因此，$f$ 是单射。

因此，存在一个单射 $f: N/K \to G/K$。
$$
~\tag*{$\square$}
$$
这些只是集合之间的映射，还不是群。毕竟，我们还没有假设 $K$ 是 $G$ 中的正规子群。

**命题 10.15** 设 $K\subset N\subset G$ 是子群，且 $K$ 是 $G$ 中的正规子群。那么 $K\triangleleft N$。

证明：因为 $K \triangleleft G$，对于任何 $g \in G$ 和 $k \in K$，我们有：
$$
gkg^{-1} \in K.
$$
特别地，这对任何 $n \in N$ 也是成立的，因为 $N \subset G$。因此，对于任何 $n \in N$ 和 $k \in K$：
$$
nkn^{-1} \in K。
$$
这表明 $K$ 是 $N$ 的正规子群，所以 $K \triangleleft N$。
$$
~\tag*{$\square$}
$$
现在我们可以谈论 $G/K$ 和 $N/K$ 作为群。

**命题 10.16** 设 $K\subset N\subset G$ 是子群。单射 $f: N/K\to G/K$ 是一个群同态。

证明：对于 $N/K$ 中的任何 $n_1K, n_2K$，在 $N/K$ 中的乘积是：
$$
(n_1K) \cdot (n_2K) = (n_1n_2)K.
$$
应用 $f$：
$$
f((n_1K) \cdot (n_2K)) = f((n_1n_2)K) = (n_1n_2)K.
$$
另一方面：
$$
f(n_1K) \cdot f(n_2K) = (n_1K) \cdot (n_2K) = (n_1n_2)K.
$$
由于两边相等，$f$ 是一个群同态。
$$
~\tag*{$\square$}
$$
这表明 $N/K$ 作为 $G/K$ 的一个子群。

**命题 10.17** 设 $K\subset N\subset G$ 是子群。存在一个双射
$$
\psi: G/N\to (G/K)/(N/K).
$$

证明：对于 $G/N$ 中的任何陪集 $gN$，定义：
$$
\psi(gN) = gK \in (G/K)/(N/K).
$$
这意味着 $gK$ 被视为 $G/K$ 中的一个陪集，以 $N/K$ 为单位的子群。

良定义性：如果 $g_1N = g_2N$，则 $g_1^{-1}g_2 \in N$，所以在 $(G/K)/(N/K)$ 中 $g_1K$ 和 $g_2K$ 是相同的。因此，$\psi$ 是良定义的。

单射性：如果 $\psi(g_1N) = \psi(g_2N)$，那么在 $(G/K)/(N/K)$ 中 $g_1K$ 和 $g_2K$ 是相同的双陪集，意味着 $g_1N = g_2N$。所以 $\psi$ 是单射的。

满射性：对于 $(G/K)/(N/K)$ 中的任何陪集 $gK$，存在一个对应的 $gN \in G/N$，使得 $\psi(gN) = gK$。因此，$\psi$ 是满射的。

因此，$\psi$ 是一个双射。
$$
~\tag*{$\square$}
$$
这只是两个集合之间的函数。为了明确，在右侧，我们利用了 $N/K$ 对 $G/K$ 的作用，因为 $N/K$ 是一个子群。商集 $(G/K)/(N/K)$ 是这个作用的通常轨道空间。

**命题 10.18** 设 $G$ 是有限群，$K\subset N\subset G$ 是子群，则
$$
|G/N|=|G/K|/|N/K|.
$$
证明：$|G/N|$ 是 $N$ 在 $G$ 中的陪集数量。

$|G/K|$ 是 $K$ 在 $G$ 中的陪集数量。

$|N/K|$ 是 $K$ 在 $N$ 中的陪集数量。

$G$ 中 $N$ 的每个陪集对应于 $G$ 中 $K$ 的 $|N/K|$ 个陪集。因此：
$$
|G/K| = |G/N| \cdot |N/K|.
$$
重新排列得到：
$$
|G/N| = \frac{|G/K|}{|N/K|}.
$$
$$
~\tag*{$\square$}
$$
**定理 10.19 (第三同构定理)**
设 $G$ 是一个群。设 $K\triangleleft G$ 且 $N\triangleleft G$，满足 $K\subset N\subset G$。那么商群 $N/K$ 是 $G/K$ 的正规子群，并且
$$
(G/K)/(N/K)\cong G/N。
$$
证明：
- $N/K$ 是 $G/K$ 的正规子群：

由于 $N$ 是 $G$ 的子群，$N/K$ 是 $G/K$ 的子群。

对于 $G/K$ 中的任何陪集 $gK$ 和 $N/K$ 中的 $nK$，考虑共轭：$(gK)(nK)(gK)^{-1}=gKnKg^{-1}K=gng^{-1}K$。

由于 $N\triangleleft G$，$gng^{-1}\in N$。

因此，$gng^{-1}\in N/K$。

因此，$(gK)(nK)(gK)^{-1}\in N/K$。

所以 $N/K\triangleleft G/K$。

- 定义自然同态 $\phi:G/K\to G/N$，其中 $\phi(gK)=gN$。这是良定义的，即如果 $gK=g^{\prime} K$，则 $\phi(gK)=\phi(g^{\prime}K)$。

如果 $gK=g^{\prime}K$，则 $g^{-1}g^{\prime}\in K\subset N$。

因此，$g^{-1}g^{\prime}\in N$，所以 $gN=g^{\prime}N$。

因此，$\phi(gK)=gN=g^{\prime}N=\phi(g^{\prime}K)$。

- $\phi$ 是一个满射的群同态：

对于 $G/K$ 中的任意 $gK, hK$：
$$
\phi((gK)(hK))=\phi(ghK)=ghN=(gN)(hN)=\phi(gK)\phi(hN).
$$
因此，$\phi$ 是一个同态。

对于 $G/N$ 中的任意 $gN$，存在一个 $gK\in G/K$，使得 $\phi(gK)=gN$。因此，$\phi$ 是满射的。

- 确定 $\phi$ 的核：

$\ker(\phi)=\{gK\in G/K\mid \phi(gK)=N\}$。

$\phi(gK)=gN=N$ 意味着 $g\in N$。

因此，$\ker(\phi)=\{gK\mid g\in N\}=N/K$。

- 应用第一同构定理：

如果 $\phi: G/K\to G/N$ 是一个满射同态，且核为 $N/K$，则
$$
(G/K)/\ker(\phi)\cong\mathrm{im}(\phi)=G/N.
$$
由于 $\ker(\phi)=N/K$，我们有
$$
(G/K)/(N/K)\cong G/N.
$$
$$
~\tag*{$\square$}
$$
### §10.3.2 第三同构定理的应用
**例 10.2** 设 $G=\mathbb{Z}$，$N=4\mathbb{Z}$，$K=12\mathbb{Z}$。

$N/K=4\mathbb{Z}/12\mathbb{Z}\cong\mathbb{Z}_{3}$。

$G/K=\mathbb{Z}/12\mathbb{Z}\cong\mathbb{Z}_{12}$。

$(G/K)/(N/K)=(\mathbb{Z}_{12})/(\mathbb{Z}_{3})\cong\mathbb{Z}_{4}$。

$G/N=\mathbb{Z}/4\mathbb{Z}\cong\mathbb{Z}_{4}$。

$(G/K)/(N/K)\cong G/N\cong\mathbb{Z}_{4}$。

§ DS-10 同构定理

§ DS-10 Isomorphisim Theorems

# §9 商群
## §9.1 商群
设 $H\subset G$ 是一个子群。

**问题**：何时轨道集
$$
G/H
$$
可以被赋予群结构？



**注** $Hg$ 是 $g$ 关于 $H$ 在 $G$ 上的作用的轨道。因此
$$
Hg=\mathcal{O}_{g}
$$

因此，问题是：是否存在一种自然的群运算用于 $H$ 的陪集的集合？

一个候选
$$
\begin{aligned}
    G/H\times G/H&\to G/H\\
    (Hg_{1},Hg_{2})&\mapsto Hg_{1}g_{2}
\end{aligned}
$$
这个映射是良定义的吗？

一般来说，不是。但在特殊情况下是良定义的！

**定理 9.1** 如果 $H\subset G$ 是正规子群，则运算
$$
\begin{aligned}
G/H\times G/H&\to G/H\\
(Hg_{1},Hg_{2})&\mapsto Hg_{1}g_{2}
\end{aligned}
$$
是良定义的，并且使得 $G/H$ 成为一个群。

证明：要证明这个运算是良定义的，我们需要证明：如果存在某些 $g_{1}, g_{1}^{\prime}, g_{2}, g_{2}^{\prime}\in G$ 使得 $Hg_{1}=Hg_{1}^{\prime}$ 且 $Hg_{2}=Hg_{2}^{\prime}$，那么
$$
Hg_{1}g_{2}=Hg_{1}^{\prime}g_{2}^{\prime}。
$$
事实上，
$$
Hg_{1}^{\prime}g_{2}^{\prime}=\{hg_{1}^{\prime}g_{2}^{\prime}\mid h\in H\}
$$
其中
$$
\begin{aligned}
Hg_{1}=Hg_{1}^{\prime} &\Rightarrow \mathcal{O}_{g_{1}}=\mathcal{O}_{g_{1}^{\prime}}\\
&\Rightarrow g_{1}, g_{1}^{\prime}~\text{属于相同轨道}\\
&\Rightarrow g_{1}^{\prime}=h_{1}g_{1}, \text{对于某个}~h_{1}\in H.
\end{aligned}
$$
同样，
$$
Hg_{2}=Hg_{2}^{\prime} \Rightarrow g_{2}^{\prime}=h_{2}g_{2}, \text{对于某个}~h_{2}\in H。
$$
因此
$$
\begin{aligned}
Hg_{1}^{\prime}g_{2}^{\prime}&=\{h\cdot h_{1}g_{1}\cdot h_{2}g_{2}\mid h\in H\}\\
&=\{h\cdot h_{1}g_{1}\cdot h_{2}g_{1}^{-1}g_{1}g_{2}\mid h\in H\}\\
&=\{h\cdot h_{1}h_{3}g_{1}g_{2}\mid h\in H\}\subset Hg_{1}g_{2}
\end{aligned}
$$
由于 $H$ 是正规子群，我们用 $h_{3}:=g_{1}h_{2}g_{1}^{-1}\in H$。

由于 $Hg_{1}g_{2}$ 和 $Hg_{1}^{\prime}g_{2}^{\prime}$ 是轨道/等价类，
$$
Hg_{1}g_{2}\supset Hg_{1}^{\prime}g_{2}^{\prime}
$$
因此，
$$
Hg_{1}g_{2}=Hg_{1}^{\prime}g_{2}^{\prime}
$$
良定义证明完毕。

为什么这是一个群？

(1) 结合性
$$
\begin{aligned}
(Hg_{1}\cdot Hg_{2})\cdot Hg_{3}&=Hg_{1}g_{2}\cdot Hg_{3}\\
&=H(g_{1}g_{2})g_{3}\\
&=Hg_{1}(g_{2}g_{3})\\
&=Hg_{1}\cdot Hg_{2}g_{3}\\
&=Hg_{1}\cdot (Hg_{2}\cdot Hg_{3}).
\end{aligned}
$$
(2) 单位元
$$
H1_{G}\cdot Hg=Hg=Hg\cdot H 1_{G}.
$$
(3) 逆元
$$
\begin{aligned}
Hg\cdot Hg^{-1}&=Hgg^{-1}\\
&=H1_{G}\\
&=Hg^{-1}g\\
&=Hg^{-1}\cdot Hg.
\end{aligned}
$$
$$
~\tag*{$\square$}
$$

来看一些例子。

**例 9.1** $H=\{\mathrm{id}_{G}\}\subset G$。

$H$ 是正规子群，因为 $\forall g\in G$，
$$
\begin{aligned}
gHg^{-1}&=\{ghg^{-1}\mid h\in H\}\\
&=\{g\mathrm{id}_{G}g^{-1}\}\\
&=\{\mathrm{id}_{G}\}\\
&=H
\end{aligned}
$$
但 $G/H$ 不是一个新的特殊群，因为存在一个同构
$$
\begin{aligned}
G/H&\to G\\
Hg&\mapsto g
\end{aligned}
$$

**例 9.2** 设 $H=n\mathbb{Z}=\{\text{n 的倍数}\}=\{\ldots,-2n,-n,0,n,2n,\ldots\}$。则 $Ha=\{a^{\prime}\in\mathbb{Z}, \text{使得}~a^{\prime}=kn+a, \text{对于某个}~k\in\mathbb{Z}\}=\mathcal{O}_{a}$。

**命题 9.2** 存在一个双射 $\mathbb{Z}/n\mathbb{Z}\to \{0,1,\ldots,n-1\}$，对于所有 $n\geqslant 1$。

证明：给定 $\mathcal{O}_{a}\in\mathbb{Z}/n\mathbb{Z}$，设 $r_{a}$ 是唯一的数，使得
$$
a=kn+r_{a}, k\in\mathbb{Z}, r_{a}\in\{0,1,\ldots,n-1\}
$$
即 $a\div k$ 的余数。（从小学学过的概念。）

因此，让这个双射为
$$
\begin{aligned}
\mathbb{Z}/n\mathbb{Z}&\to\{0,1,\ldots,n-1\}\\
\mathcal{O}_{a}&\mapsto r_{a}
\end{aligned}
$$
- 良定义？

如果 $\mathcal{O}_{a}=\mathcal{O}_{a^{\prime}}$，
$$
a^{\prime}=a+k^{\prime}n\tag{根据轨道的定义}
$$
因此
$$
\begin{aligned}
a^{\prime}&=k^{\prime}n+kn+r_{a}\\
&=(k^{\prime}+k)n+r_{a}
\end{aligned}
$$
因此
$$
r_{a^{\prime}}=r_{a}
$$
其中 $a^{\prime}=kn+r_{a^{\prime}}$ 中唯一的数属于 $\{0,1,\ldots,n-1\}$。因此，良定义！

- 单射？

给定 $\mathcal{O}_{a}$，$\mathcal{O}_{b}$，
$$
r_{a}=r_{b}
$$
$\Rightarrow$
$$
\begin{aligned}
a&=kn+r_{a}\\
b&=ln+r_{b}\\
&=ln+r_{a}
\end{aligned}
$$
$\Rightarrow$
$$
a-b=(k-l)n
$$
$\Rightarrow$
$$
a\in\mathcal{O}_{b}
$$
$\Rightarrow$
$$
\mathcal{O}_{a}=\mathcal{O}_{b}
$$
- 满射？
$$
\begin{aligned}
\mathcal{O}_{0}&\mapsto 0\\
\mathcal{O}_{1}&\mapsto 1\\
\mathcal{O}_{2}&\mapsto 2\\
&~\,\vdots\\
\mathcal{O}_{n-1}&\mapsto n-1
\end{aligned}
$$
$$
~\tag*{$\square$}
$$
根据这个定理，这意味着
$$
\mathbb{Z}/n\mathbb{Z}
$$
是一个阶为
$$
n=|\{0,1,\ldots,n-1\}|
$$
的群。这个群的结构是什么？
<div align=center>
    <img src=//tex.xumin.net/svgb/U1HhiklKTc_Mqy7JzK5KTqnlislNLMlISqqOqtXPQ7AVYkoyc1OLFXDJJhYV5ZdHFxXFwpgpKToKSjElqRUl1U_7e55uaKlVUtdRSErNS1EoykzPKLE1M4hVUANCXCbGcFVrgPnJiTnV_rXx1Ym1Oij8pFpNuMVAy3ITC4oVSvJjEW5BEoNbBDdMOwm7ZpC1RWDLilCtwDCvCGYKUEtMtYFOTE5KfkmxTp6uYUwtLLTQhFECCmQGhkZg-AODCB4ZKioA>
</div>

即群结构是：将数相加，然后取除以 $n$ 的余数。

**定义 9.1** 设 $a$, $b\in\mathbb{Z}$。我们写作
$$
a\equiv b \pmod n
$$
或
$$
a=b \pmod n
$$
如果
$$
a-b=kn\quad \text{对于某个}~k\in\mathbb{Z}
$$
等价地，$a\equiv b \pmod n \Leftrightarrow \mathcal{O}_{a}=\mathcal{O}_{b}$。

当我们写作
$$
a \pmod n
$$
我们指的是等价类
$$
\mathcal{O}_{a}=Ha\in G/H
$$

**注** 你可能会觉得记住巨大的等价类 $\mathcal{O}_{a}$, $\mathcal{O}_{b}$ 等很麻烦。因此，可以将 $\mathcal{O}_{a}$ 简单地视为一个数，即将 $a$ 除以 $n$ 时得到的余数 $r$：
$$
a=kn+r
$$
这可以通过双射来证明：
$$
\mathbb{Z}/n\mathbb{Z}\cong \{0,\ldots,n-1\}
$$
因此，当你看到 “$a \pmod n$” 时，你可以简单地将其视为数 $r$。同样地，群运算只是“钟表算术”：
$$
(a,b)\mapsto a+b \pmod n
$$
你可以将其视为 $(a+b)\div n$ 的余数。

## §9.2 子群延伸到商群

设 $G$ 是一个任意的群，且 $H\triangleleft G$。

**命题 9.3** 存在一个从包含 $H$ 的 $G$ 的子群集合到 $G/H$ 中的子群集合的双射。

证明：设 $p: G\to G/H$ 是通过将 $g$ 映射到 $[g]$ 给出的群同态。

给定一个子群 $K\subset G$，注意群同态的组合
$$
K\xhookrightarrow{}G\to G/H.
$$
由于任何群同态的像是一个子群，这表明 $p(K)$ 是 $G/H$ 的一个子群。因此我们有一个从 $\{G\text{的子群}\}\to\{G/H\text{的子群}\}$ 的映射，通过将 $K$ 映射到 $p(K)$ 来给出。

我们证明它是满射：给定 $K^{\prime}\subset G/H$，考虑原像 $p^{-1}(K^{\prime})\subset G$。这是 $G$ 的一个子群，因为如果 $p(x), p(y)\in K^{\prime}$，则 $p(xy)=p(x)p(y)\in K^{\prime}$（因为 $K^{\prime}$ 在乘法下是闭合的）。

现在，只需证明对于所有 $K\subset G$，有 $p^{-1}(p(K))=K$。显然，$K\subset p^{-1}(p(K))$。为了证明另一种包含性，设 $x\in p^{-1}(p(K))$。根据 $p(K)$ 的定义，我们知道存在某个 $y\in K$ 使得 $p(x)=p(y)$。于是 $p(xy^{-1})=1_{G/H}$，因此 $xy^{-1}\in H$。由于 $K$ 包含 $H$，所以 $xy^{-1}\in K$，因此 $x\in K$。
$$
~\tag*{$\square$}
$$

**命题 9.4** 存在一个从包含 $H$ 的 $G$ 中的正规子群集合到 $G/H$ 中的正规子群集合的双射。

证明：我们证明，如果 $K$ 是正规子群，那么 $p(K)$ 也是正规子群。（这证明了我们有一个映射 $\{G\text{中的正规子群}\}\to\{G/H\text{中的正规子群}\}$。）

实际上，如果 $[k]\in p(K)$，则 $[g][k][g]^{-1}=[gkg^{-1}]=[k^{\prime}]$ 对某个 $k^{\prime}\in K$ 成立，因为 $K$ 在 $G$ 中是正规子群。因此，$p(K)\subset G/H$ 是正规子群。（注意我们在这里使用了 $G\to G/H$ 是满射这一事实——否则，我们无法知道 $G/H$ 的每个元素都在 $p(G)$ 的像中。）

满射性：我们证明如果 $p(K)$ 是正规子群，则 $K=p^{-1}(p(K))$ 是正规子群。如果 $k\in K$ 且 $g\in G$，我们有 $[gkg^{-1}]=[g][k][g^{-1}]=[k^{\prime}]$，对于某个 $[k^{\prime}]\in p(K)$——即，对于某个 $k^{\prime}\in K$。因此 $gkg^{-1}\in p^{-1}(p(K))=K$。

我们知道这个赋值是单射的。
$$
~\tag*{$\square$}
$$

## §9.3 交换图
**定义 9.2** 假设我们有群 $K$, $G$, $H$, $X$ 以及这些群之间的映射
<div align=center>
<img src=https://tex.xumin.net/svgb/U1HhiklKTc_Mqy7JzK5KTqnl8laISSwqyi-PLtJRUIopyMhUioWJpIBEEnMKMhKV1GMV1BTcUSQKisFKY7g8UExISi1JVAKpjuCKSc1LgdujogIA>
</div>

如果 $\psi\circ\phi=\beta\circ\alpha: K\to X$，我们称这个图是**交换的(commutative)**。

**注** $1$ 将表示平凡群。这个群在同构意义下是唯一确定的！

现在来看这个图
<div align=center>
<img src=https://tex.xumin.net/svgb/U1HhiklKTc_Mqy7JzK5KTqnl8laISSwqyi-PLtJRUIopyMhUioWJpABFFJXUYxXUFNyRxWIKisGqYrgMkTUrKoFURnDFpOalwI1XUQEA>
</div>
                                                                 
这个图交换是什么意思？这当且仅当 $\psi\circ\phi$ 产生将所有元素映射到恒等元的映射时成立。为什么？因为左边和底部的映射是将 $K$ 中的所有元素映射到 $1_{X}\in X$ 的映射（假设它们是群同态），因此 $\psi\circ\phi$ 是常量映射。

或者，我们有 $\mathrm{im}(\phi)\subset \ker(\psi)$。这与我们之前所谈论的精确性很接近。

**例 9.3** $K=G=X=1$。

**例 9.4** $K=SL_{n}(\mathbb{R})$，$G=GL_{n}(\mathbb{R})$，$X=\mathbb{R}^{\times}$，$\phi$ 是包含映射，$\psi=\det$。

**例 9.5** 设 $K\subset G$ 是一个正规子群，因此 $\phi$ 是嵌入映射。设 $X=G/K$（因为 $K$ 是 $G$ 中的正规子群，所以这是一个群），$\psi$ 是正则的商投影。

## §9.4 商群的泛性质
**定理 9.5** 给定一个交换图
<div align=center>
<img src=https://tex.xumin.net/svgb/U1HhiklKTc_Mqy7JzK5KTqnl8laISSwqyi-PLtJRUIopSa0oqX62dvGL5S21SrEwqZRYBTUFdzgPpLCgOBMkH8NlCNcPUhTBFZOalwI3XEUFAA>
</div>

存在一个唯一的同态 $\tilde{\psi}: G/X\to X$ 使得图
<div align=center>
<img src=https://tex.xumin.net/svgb/U1HhiklKTc_Mqy7JzK5KTqnlcleISSwqyi-PLtJRUIopKM5UioWJpIBFMpXUFWIV1BQiUIRLMnNSUqtB6mtBGmK43PW9UUzKTSzJKMqtzkwByaspAKW5YlLzUuD2qqgAAA>
</div>
交换。

换句话说，如果 $\psi: G\to X$ 是这样的，使得 $K\subset\ker(\psi)$ 对于某个 $K\subset G$ 是正规子群，那么存在一个唯一的映射 $\tilde{\psi}: G/K\to X$ 使得 $\psi=\tilde{\psi}\circ\pi$，如果 $\pi: G\to G/K$ 是正则映射到商。这被称为**商群的泛性质(universal property of quotient group)**。

证明：我们要定义这个映射 $\tilde{\psi}: G/K\to X$。通过取一个陪集 $Kg\mapsto \psi(g)$ 或者将 $G/K$ 中的任何陪集的原像通过正则映射 $\pi$ 返回到 $G$，来定义它。我们需要证明这是良定义的。

假设 $Kg=Kg^{\prime}$，或等价地 $g=k\cdot g^{\prime}$。是否有 $\psi(g)=\psi(g^{\prime})$？实际上
$$
\psi(g)=\psi(k\cdot g^{\prime})=\psi(k)\cdot \psi(g^{\prime})=\psi(g^{\prime})，
$$
所以一切正常！我们现在需要验证 $\tilde{\psi}$ 是唯一的且 $\psi=\tilde{\psi}\circ\pi$。但我们有
$$
\psi(g)=\tilde{\psi}(Kg)=\tilde{\psi}(\pi(g))，
$$
因此我们只需证明唯一性。如果我们有另一个这样的映射 $\gamma: G/K\to X$，那么我们有
$$
\psi(g)=\gamma(\pi(g))=\gamma(Kg)，
$$
因此 $\gamma$ 和 $\tilde{\psi}$ 在所有陪集 $K$ 上必须一致，那么我们就完成了。
$$
~\tag*{$\square$}
$$

**注** 唯一性是根据定理中的图交换性的约束得出的。

## §9.5 商群的推广
泛性质的概念允许我们推广商群的概念。

设 $G\overset{\chi}{\twoheadrightarrow}H$ 是任意的满同态（我们现在用双箭头表示它）。那么 $H$ 是 $G$ 的“商群”，即一个如下的图
<div align=center>
<img src=https://tex.xumin.net/svgb/U1HhiklKTc_Mqy7JzK5KTqnl8laISSwqyi-PLoqFsVJiFdQU3OE8HQWlmOSMTCWgfAyXIZJyNQUPrpjUvBS4WSoqAA>
</div>

具有类似的泛性质。为了回答这个问题，$K$ 是什么？我们只需将它设为 $\chi$ 的核，这样：

**定理 9.6** 给定 $G\overset{\chi}{\twoheadrightarrow}H$，设 $K=\ker(\chi)$。那么如下的图
<div align=center>
<img src=https://tex.xumin.net/svgb/U1HhiklKTc_Mqy7JzK5KTqnl8laISSwqyi-PLtJRUMpUioVxU2IV1BTc4TygZExyBlg-hssQrgekyIMrJjUvBW6gigoA>
</div>

其中 $i: K\to G$ 是包含映射，具有相同的泛性质，即给定任何群同态 $\psi: G\to X$ 其核包含 $K$，则存在一个唯一的映射 $\tilde{\psi}: H\to X$ 使得
$$
\psi=\tilde{\psi}\circ\chi.
$$

证明：我们如何定义 $\tilde{\psi}: H\to X$？由于 $\chi$ 是满射，对于任何 $h\in H$，我们有 $h=\chi(g)$，尽管 $g$ 可能不是唯一的！现在定义
$$
\tilde{\psi}:=\psi(g)。
$$
我们需要验证这是一个良定义的同态。假设 $\chi(g)=\chi(g^{\prime})$。那么 $\chi(g)=\chi(g^{\prime})$ 意味着 $\chi(g\cdot(g^{\prime})^{-1})=1_{H}$，因此 $g\cdot (g^{\prime})^{-1}\in K$，于是 $g=g^{\prime}\cdot k$，对于某个 $k\in K$。现在的论证与先前定理的证明完全相同。
$$
~\tag*{$\square$}
$$

**推论 9.7** 假设 $\chi: G\to H$ 是满射且 $K=\ker(\chi)$。那么 $H\cong G/K$。

**注** 这就是所谓的第一同构定理。让我们用泛性质给出证明。在下一章我们会再次看到它。

证明：定理 9.3 给出了一个从 $G/K$ 到 $H$ 的映射，定理 9.4 给出了一个从 $H$ 到 $G/K$ 的映射。组合这些映射给我们一个从 $G/K$ 到 $H$ 再到 $G/K$ 的映射，这样复合映射满足正确的交换性约束。这种映射是由我们从两个定理中得到的两个映射的组合给出的，但从 $G/K$ 到 $G/K$ 的恒等映射也满足这些约束。由于这些映射是唯一确定的，我们必须有这个复合是恒等映射。同样地，我们得到另一个从 $H$ 到 $G/K$ 再到 $H$ 的映射，这种复合映射在另一方向上也是恒等映射。所以我们有两个从 $G/K$ 到 $H$ 以及从 $H$ 到 $G/K$ 的映射，它们在两个方向上的复合都是各自的恒等映射，因此 $G/K$ 和 $H$ 之间存在双射关系，而这些映射是群同态，所以它们实际上作为群是同构的！
$$
~\tag*{$\square$}
$$

**注** 这种泛性质的概念允许我们唯一地区分某些对象（在同构意义下），就像我们在推论的证明中所做的那样，因为泛性质是存在性和唯一性的陈述。如果你在思考，两对象“在一个范畴中”是同构的，当且仅当它们具有相同的泛性质。在我们的例子中，两个群（商群和 $G$ 的任何被满射到的群）在这个“群的范畴”中满足相同的泛性质，所以它们是同构的！这是范畴论中非常重要的一个结果，称为Yoneda引理的一个推论。

**命题 9.8** 设 $K$, $G$ 是任意的群，$\phi: K\to G$ 是任意的群同态。则存在另一个群同态 $\psi: G\to H$ 使得我们有一个泛性质
<div align=center>
<img src=https://tex.xumin.net/svgb/U1HhiklKTc_Mqy7JzK5KTqnl8laISSwqyi-PLtJRUIopyMhUioWJpIBEUguKM3Py85TUYxXUFNxRpIAyIMUxXIZwM0CKPLhiUvNS4BaoqAAA>
</div>

其中 $\epsilon: K\to 1$ 是将 $K$ 中的每个元素映射到平凡群 $1$ 的恒等元的恒等同态，底部箭头 $1\to H$ 是从平凡群到 $H$ 的唯一同态。

证明：1. 构造 $H$ 和 $\psi$: 定义 $H$ 为商群
$$
H=G/\phi(K)
$$
其中 $\phi(K)$ 是 $K$ 在 $\phi$ 下的像，是 $G$ 的一个子群。$G/\phi(K)$ 由所有 $g\phi(K)$ 的陪集组成，其中 $g\in G$。

定义 $\psi: G\to H$ 为自然的投影映射
$$
\psi(g)=g\phi(K)
$$
其中 $\psi$ 将每个元素 $g$ 映射到 $H$ 中的陪集，且 $\psi$ 是一个满射同态。

2. 验证图的交换性：我们需要证明
$$
\psi\circ\phi=\epsilon
$$
对于任何 $k\in K$，计算 $\psi(\phi(k))$:
$$
\psi(\phi(k))=\phi(k)\phi(K)=\phi(K)
$$
因为 $\phi(k)\in \phi(K)$，所以陪集 $\phi(k)\phi(K)$ 等于 $\phi(K)$，这是 $H$ 中的恒等元。因此，$\psi(\phi(k))$ 将 $K$ 中的每个元素映射到 $H$ 中的恒等元。这与通过 $\epsilon$ 进行的组合相匹配：
$$
(\text{从}~1~\text{到}~H~\text{的唯一映射})\circ\epsilon(k)=1_{H}.
$$
因此，这个图是交换的。

3. 这个构造满足以下泛性质：对于任何群 $L$ 和任何同态 $f:G\to L$，使得 $f\circ \phi=\epsilon$，存在一个唯一的同态 $\overline{f}: H\to L$ 使得
$$
f=\overline{f}\circ \psi.
$$
由于
(1) $f\circ\phi=\epsilon$，对于所有 $k\in K$：
$$
f(\phi(k))=1_{L}.
$$
这意味着 $\phi(k)\in \ker(f)$ 对于所有 $k\in K$。因此 $\phi(K)\subset \ker(f)$。

(2) 定义 $\overline{f}: H\to L$ 为
$$
\overline{f}(g\phi(K))=f(g).
$$
这是良定义的，因为如果 $g\phi(K)=g^{\prime}\phi(K)$，那么 $g^{-1}g^{\prime}\in \phi(K)$，所以 $f(g^{-1}g^{\prime})=1_{L}$，这意味着 $f(g)=f(g^{\prime})$。因此，$\overline{f}$ 是一个良定义的同态。

(3) 唯一性：如果存在另一个同态 $\overline{f}^{\prime}$ 使得 $f=\overline{f}^{\prime}\circ \psi$，那么对于所有 $g\in G$：
$$
\overline{f}(g\phi(K))=f(g)=\overline{f}^{\prime}(g\phi(K)).
$$
因此，$\overline{f}=\overline{f}^{\prime}$。
$$
~\tag*{$\square$}
$$

# §9 Quotient Groups

## §9.1 Quotient Groups

Let $H\subset G$ be a subgroup.

**Question:** When can the orbit set

$$
G/H
$$

be given a group structure?



**Remark** $Hg$ is the orbit of $g$ under the action of $H$ on $G$. Thus

$$
Hg=\mathcal{O}_{g}.
$$

Therefore, the question is: is there a natural group operation on the set of cosets of $H$?

One candidate is

$$
\begin{aligned}
G/H\times G/H&\to G/H\\
(Hg_{1},Hg_{2})&\mapsto Hg_{1}g_{2}.
\end{aligned}
$$

Is this map well defined?

In general, no. But in a special case, it is!

**Theorem 9.1** If $H\subset G$ is a normal subgroup, then the operation

$$
\begin{aligned}
G/H\times G/H&\to G/H\\
(Hg_{1},Hg_{2})&\mapsto Hg_{1}g_{2}
\end{aligned}
$$

is well defined and makes $G/H$ into a group.

Proof: To prove that this operation is well defined, we need to show that if there exist

$$
g_{1},g_{1}^{\prime},g_{2},g_{2}^{\prime}\in G
$$

such that

$$
Hg_{1}=Hg_{1}^{\prime}
$$

and

$$
Hg_{2}=Hg_{2}^{\prime},
$$

then

$$
Hg_{1}g_{2}
=
Hg_{1}^{\prime}g_{2}^{\prime}.
$$

Indeed,

$$
Hg_{1}^{\prime}g_{2}^{\prime}
=
\{hg_{1}^{\prime}g_{2}^{\prime}\mid h\in H\},
$$

where

$$
\begin{aligned}
Hg_{1}=Hg_{1}^{\prime}
&\Rightarrow
\mathcal{O}_{g_{1}}
=
\mathcal{O}_{g_{1}^{\prime}}\\
&\Rightarrow
g_{1},g_{1}^{\prime}
~\text{belong to the same orbit}\\
&\Rightarrow
g_{1}^{\prime}
=
h_{1}g_{1},
\quad
\text{for some }h_{1}\in H.
\end{aligned}
$$

Similarly,

$$
Hg_{2}=Hg_{2}^{\prime}
\Rightarrow
g_{2}^{\prime}=h_{2}g_{2},
\quad
\text{for some }h_{2}\in H.
$$

Therefore,

$$
\begin{aligned}
Hg_{1}^{\prime}g_{2}^{\prime}
&=
\{h\cdot h_{1}g_{1}\cdot h_{2}g_{2}\mid h\in H\}\\
&=
\{h\cdot h_{1}g_{1}\cdot h_{2}g_{1}^{-1}g_{1}g_{2}\mid h\in H\}\\
&=
\{h\cdot h_{1}h_{3}g_{1}g_{2}\mid h\in H\}
\subset
Hg_{1}g_{2}.
\end{aligned}
$$

Since $H$ is normal, we used

$$
h_{3}:=g_{1}h_{2}g_{1}^{-1}\in H.
$$

Since
$Hg_{1}g_{2}$
and
$Hg_{1}^{\prime}g_{2}^{\prime}$
are orbits/equivalence classes,

$$
Hg_{1}g_{2}
\supset
Hg_{1}^{\prime}g_{2}^{\prime}.
$$

Therefore,

$$
Hg_{1}g_{2}
=
Hg_{1}^{\prime}g_{2}^{\prime}.
$$

Thus the operation is well defined.

Why is this a group?

(1) Associativity:

$$
\begin{aligned}
(Hg_{1}\cdot Hg_{2})\cdot Hg_{3}
&=
Hg_{1}g_{2}\cdot Hg_{3}\\
&=
H(g_{1}g_{2})g_{3}\\
&=
Hg_{1}(g_{2}g_{3})\\
&=
Hg_{1}\cdot Hg_{2}g_{3}\\
&=
Hg_{1}\cdot(Hg_{2}\cdot Hg_{3}).
\end{aligned}
$$

(2) Identity:

$$
H1_{G}\cdot Hg
=
Hg
=
Hg\cdot H1_{G}.
$$

(3) Inverses:

$$
\begin{aligned}
Hg\cdot Hg^{-1}
&=
Hgg^{-1}\\
&=
H1_{G}\\
&=
Hg^{-1}g\\
&=
Hg^{-1}\cdot Hg.
\end{aligned}
$$

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

Let us look at some examples.

**Example 9.1**

$$
H=\{\mathrm{id}_{G}\}\subset G.
$$

$H$ is normal because for every $g\in G$,

$$
\begin{aligned}
gHg^{-1}
&=
\{ghg^{-1}\mid h\in H\}\\
&=
\{g\mathrm{id}_{G}g^{-1}\}\\
&=
\{\mathrm{id}_{G}\}\\
&=
H.
\end{aligned}
$$

But $G/H$ is not a new or particularly different group, since there is an isomorphism

$$
\begin{aligned}
G/H&\to G\\
Hg&\mapsto g.
\end{aligned}
$$

**Example 9.2** Let

$$
H=n\mathbb{Z}
=
\{\text{multiples of }n\}
=
\{\ldots,-2n,-n,0,n,2n,\ldots\}.
$$

Then

$$
Ha
=
\{a^{\prime}\in\mathbb{Z}
\mid
a^{\prime}=kn+a,
~\text{for some }k\in\mathbb{Z}\}
=
\mathcal{O}_{a}.
$$

**Proposition 9.2** There is a bijection

$$
\mathbb{Z}/n\mathbb{Z}
\to
\{0,1,\ldots,n-1\}
$$

for every
$n\geqslant1$.

Proof: Given

$$
\mathcal{O}_{a}
\in
\mathbb{Z}/n\mathbb{Z},
$$

let $r_{a}$ be the unique number such that

$$
a=kn+r_{a},
\quad
k\in\mathbb{Z},
\quad
r_{a}\in\{0,1,\ldots,n-1\}.
$$

That is, $r_a$ is the remainder obtained when dividing $a$ by $n$—a concept familiar from elementary school.

Define the bijection by

$$
\begin{aligned}
\mathbb{Z}/n\mathbb{Z}
&\to
\{0,1,\ldots,n-1\}\\
\mathcal{O}_{a}
&\mapsto
r_{a}.
\end{aligned}
$$

* Well defined?

If

$$
\mathcal{O}_{a}
=
\mathcal{O}_{a^{\prime}},
$$

then

$$
a^{\prime}
=
a+k^{\prime}n
\tag{by the definition of an orbit}.
$$

Therefore,

$$
\begin{aligned}
a^{\prime}
&=
k^{\prime}n+kn+r_{a}\\
&=
(k^{\prime}+k)n+r_{a}.
\end{aligned}
$$

Hence

$$
r_{a^{\prime}}
=
r_{a},
$$

where $r_{a^{\prime}}$ is the unique element of
${0,1,\ldots,n-1}$
appearing in

$$
a^{\prime}=kn+r_{a^{\prime}}.
$$

Thus the map is well defined.

* Injective?

Given
$\mathcal{O}*{a}$
and
$\mathcal{O}*{b}$,
suppose

$$
r_{a}=r_{b}.
$$

Then

$$
\begin{aligned}
a&=kn+r_{a}\\
b&=ln+r_{b}\\
&=ln+r_{a}.
\end{aligned}
$$

Therefore,

$$
a-b=(k-l)n,
$$

so

$$
a\in\mathcal{O}_{b},
$$

and hence

$$
\mathcal{O}_{a}
=
\mathcal{O}_{b}.
$$

* Surjective?

$$
\begin{aligned}
\mathcal{O}_{0}&\mapsto0\\
\mathcal{O}_{1}&\mapsto1\\
\mathcal{O}_{2}&\mapsto2\\
&~\,\vdots\\
\mathcal{O}_{n-1}&\mapsto n-1.
\end{aligned}
$$

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

By this proposition,

$$
\mathbb{Z}/n\mathbb{Z}
$$

is a group of order

$$
n
=
|\{0,1,\ldots,n-1\}|.
$$

What is the structure of this group?

<div align=center>
    <img src=//tex.xumin.net/svgb/U1HhiklKTc_Mqy7JzK5KTqnlislNLMlISqqOqtXPQ7AVYkoyc1OLFXDJJhYV5ZdHFxXFwpgpKToKSjElqRUl1U_7e55uaKlVUtdRSErNS1EoykzPKLE1M4hVUANCXCbGcFVrgPnJiTnV_rXx1Ym1Oij8pFpNuMVAy3ITC4oVSvJjEW5BEoNbBDdMOwm7ZpC1RWDLilCtwDCvCGYKUEtMtYFOTE5KfkmxTp6uYUwtLLTQhFECCmQGhkZg-AODCB4ZKioA>
</div>

The group operation is: add the numbers and then take the remainder modulo $n$.

**Definition 9.1** Let
$a,b\in\mathbb{Z}$.
We write

$$
a\equiv b\pmod n
$$

or

$$
a=b\pmod n
$$

if

$$
a-b=kn
\quad
\text{for some }k\in\mathbb{Z}.
$$

Equivalently,

$$
a\equiv b\pmod n
\Longleftrightarrow
\mathcal{O}_{a}
=
\mathcal{O}_{b}.
$$

When we write

$$
a\pmod n,
$$

we mean the equivalence class

$$
\mathcal{O}_{a}
=
Ha
\in
G/H.
$$

**Remark** It may seem inconvenient to keep track of large equivalence classes
$\mathcal{O}*{a}$,
$\mathcal{O}*{b}$,
and so on. Instead, we may simply regard
$\mathcal{O}_{a}$
as a number: namely, the remainder $r$ obtained when $a$ is divided by $n$:

$$
a=kn+r.
$$

This is justified by the bijection

$$
\mathbb{Z}/n\mathbb{Z}
\cong
\{0,\ldots,n-1\}.
$$

Thus, whenever you see
“$a\pmod n$”,
you may simply think of the number $r$.

Similarly, the group operation is just “clock arithmetic”:

$$
(a,b)\mapsto a+b\pmod n.
$$

You can think of this as the remainder of $(a+b)$ upon division by $n$.

## §9.2 Subgroups Descend to Quotient Groups

Let $G$ be an arbitrary group and let

$$
H\triangleleft G.
$$

**Proposition 9.3** There is a bijection between the set of subgroups of $G$ containing $H$ and the set of subgroups of $G/H$.

Proof: Let

$$
p:G\to G/H
$$

be the group homomorphism sending $g$ to $[g]$.

Given a subgroup
$K\subset G$,
consider the composition of group homomorphisms

$$
K\xhookrightarrow{}G\to G/H.
$$

Since the image of any group homomorphism is a subgroup, this shows that
$p(K)$
is a subgroup of
$G/H$.

Thus we obtain a map

$$
\{\text{subgroups of }G\}
\to
\{\text{subgroups of }G/H\}
$$

given by

$$
K\mapsto p(K).
$$

We prove that this map is surjective. Given

$$
K^{\prime}\subset G/H,
$$

consider the preimage

$$
p^{-1}(K^{\prime})\subset G.
$$

This is a subgroup of $G$, because if

$$
p(x),p(y)\in K^{\prime},
$$

then

$$
p(xy)
=
p(x)p(y)
\in K^{\prime},
$$

since $K^{\prime}$ is closed under multiplication.

Now it remains only to prove that for every
$K\subset G$,

$$
p^{-1}(p(K))
=
K.
$$

Clearly,

$$
K\subset p^{-1}(p(K)).
$$

To prove the other inclusion, let

$$
x\in p^{-1}(p(K)).
$$

By the definition of $p(K)$, there exists some
$y\in K$
such that

$$
p(x)=p(y).
$$

Then

$$
p(xy^{-1})
=
1_{G/H},
$$

so

$$
xy^{-1}\in H.
$$

Since $K$ contains $H$,

$$
xy^{-1}\in K,
$$

and therefore

$$
x\in K.
$$

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

**Proposition 9.4** There is a bijection between the set of normal subgroups of $G$ containing $H$ and the set of normal subgroups of $G/H$.

Proof: We first prove that if $K$ is normal, then $p(K)$ is normal. This gives a map

$$
\{\text{normal subgroups of }G\}
\to
\{\text{normal subgroups of }G/H\}.
$$

Indeed, if

$$
[k]\in p(K),
$$

then

$$
[g][k][g]^{-1}
=
[gkg^{-1}]
=
[k^{\prime}]
$$

for some
$k^{\prime}\in K$,
because $K$ is normal in $G$.

Therefore,

$$
p(K)\subset G/H
$$

is a normal subgroup.

Notice that here we used the fact that

$$
G\to G/H
$$

is surjective; otherwise, we would not know that every element of $G/H$ lies in the image $p(G)$.

Surjectivity: We prove that if $p(K)$ is normal, then

$$
K=p^{-1}(p(K))
$$

is normal.

If

$$
k\in K,
\qquad
g\in G,
$$

then

$$
[gkg^{-1}]
=
[g][k][g^{-1}]
=
[k^{\prime}]
$$

for some

$$
[k^{\prime}]\in p(K),
$$

i.e. for some
$k^{\prime}\in K$.

Therefore,

$$
gkg^{-1}
\in
p^{-1}(p(K))
=
K.
$$

We already know that this assignment is injective.

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

## §9.3 Commutative Diagrams

**Definition 9.2** Suppose we have groups
$K,G,H,X$
and maps between these groups:

<div align=center>
<img src=https://tex.xumin.net/svgb/U1HhiklKTc_Mqy7JzK5KTqnl8laISSwqyi-PLtJRUIopyMhUioWJpIBEEnMKMhKV1GMV1BTcUSQKisFKY7g8UExISi1JVAKpjuCKSc1LgdujogIA>
</div>

If

$$
\psi\circ\phi
=
\beta\circ\alpha:
K\to X,
$$

we say that the diagram **commutes**, or that it is **commutative**.

**Remark** The symbol $1$ will denote the trivial group. This group is uniquely determined up to isomorphism.

Now consider the diagram

<div align=center>
<img src=https://tex.xumin.net/svgb/U1HhiklKTc_Mqy7JzK5KTqnl8laISSwqyi-PLtJRUIopyMhUioWJpABFFJXUYxXUFNyRxWIKisGqYrgMkTUrKoFURnDFpOalwI1XUQEA>
</div>

What does it mean for this diagram to commute?

It means precisely that

$$
\psi\circ\phi
$$

is the map sending every element to the identity.

Why?

Because the maps along the left and bottom send every element of $K$ to

$$
1_X\in X
$$

assuming they are group homomorphisms. Thus
$\psi\circ\phi$
is a constant map.

Equivalently,

$$
\mathrm{im}(\phi)
\subset
\ker(\psi).
$$

This is very close to the notion of exactness that we discussed earlier.

**Example 9.3**

$$
K=G=X=1.
$$

**Example 9.4**

$$
K=SL_n(\mathbb{R}),
\qquad
G=GL_n(\mathbb{R}),
\qquad
X=\mathbb{R}^{\times},
$$

where $\phi$ is the inclusion map and

$$
\psi=\det.
$$

**Example 9.5** Let

$$
K\subset G
$$

be a normal subgroup, so $\phi$ is the inclusion map.

Let

$$
X=G/K.
$$

Since $K$ is normal in $G$, this is a group. Let $\psi$ be the canonical quotient projection.

## §9.4 The Universal Property of Quotient Groups

**Theorem 9.5** Given a commutative diagram

<div align=center>
<img src=https://tex.xumin.net/svgb/i0lKTc_Mqy7JzK5KTqnl8laISSwqyi-PLtJRUIopSa0oqc7LL8pNzKlVioVJpcQqqCm4w3kghQXFmSD5GC5DuP5YBTxATSECU5ArJjUvBeYSAA>
</div>

there exists a unique homomorphism

$$
\tilde{\psi}:G/X\to X
$$

such that the diagram

<div align=center>
<img src=https://tex.xumin.net/svgb/U1HhiklKTc_Mqy7JzK5KTqnlcleISSwqyi-PLtJRUIopKM5UioWJpIBFMpXUFWIV1BQiUIRLMnNSUqtB6mtBGmK43PW9UUzKTSzJKMqtzkwByaspAKW5YlLzUuD2qqgAAA>
</div>

commutes.

In other words, if

$$
\psi:G\to X
$$

is such that

$$
K\subset\ker(\psi)
$$

for some normal subgroup
$K\subset G$,
then there exists a unique map

$$
\tilde{\psi}:G/K\to X
$$

such that

$$
\psi
=
\tilde{\psi}\circ\pi,
$$

where

$$
\pi:G\to G/K
$$

is the canonical quotient map.

This is called the **universal property of the quotient group**.

Proof: We want to define the map

$$
\tilde{\psi}:G/K\to X.
$$

Define it by taking a coset

$$
Kg\mapsto\psi(g),
$$

or equivalently by choosing any preimage in $G$ of a coset in $G/K$ under the canonical map $\pi$.

We need to prove that this is well defined.

Suppose

$$
Kg=Kg^{\prime},
$$

or equivalently

$$
g=k\cdot g^{\prime}.
$$

Do we have

$$
\psi(g)=\psi(g^{\prime})?
$$

Indeed,

$$
\psi(g)
=
\psi(k\cdot g^{\prime})
=
\psi(k)\cdot\psi(g^{\prime})
=
\psi(g^{\prime}),
$$

so everything works.

We now need to verify that $\tilde{\psi}$ is unique and that

$$
\psi
=
\tilde{\psi}\circ\pi.
$$

But

$$
\psi(g)
=
\tilde{\psi}(Kg)
=
\tilde{\psi}(\pi(g)).
$$

Thus it remains only to prove uniqueness.

Suppose there is another such map

$$
\gamma:G/K\to X.
$$

Then

$$
\psi(g)
=
\gamma(\pi(g))
=
\gamma(Kg).
$$

Hence $\gamma$ and $\tilde{\psi}$ must agree on every coset $K$, and we are done.

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

**Remark** The uniqueness follows from the requirement in the theorem that the diagram commute.

## §9.5 Generalization of Quotient Groups

The idea of a universal property allows us to generalize the notion of a quotient group.

Let

$$
G\overset{\chi}{\twoheadrightarrow}H
$$

be any surjective homomorphism. We now use a double arrow to indicate surjectivity.

Then $H$ is a “quotient group” of $G$, in the sense that a diagram of the form

<div align=center>
<img src=https://tex.xumin.net/svgb/U1HhiklKTc_Mqy7JzK5KTqnl8laISSwqyi-PLoqFsVJiFdQU3OE8HQWlmOSMTCWgfAyXIZJyNQUPrpjUvBS4WSoqAA>
</div>

has a similar universal property.

To answer the question, what should $K$ be? We simply take it to be the kernel of $\chi$.

Thus:

**Theorem 9.6** Given

$$
G\overset{\chi}{\twoheadrightarrow}H,
$$

let

$$
K=\ker(\chi).
$$

Then the diagram

<div align=center>
<img src=https://tex.xumin.net/svgb/U1HhiklKTc_Mqy7JzK5KTqnl8laISSwqyi-PLtJRUMpUioVxU2IV1BTc4TygZExyBlg-hssQrgekyIMrJjUvBW6gigoA>
</div>

where

$$
i:K\to G
$$

is the inclusion map, has the same universal property:

given any group homomorphism

$$
\psi:G\to X
$$

whose kernel contains $K$, there exists a unique map

$$
\tilde{\psi}:H\to X
$$

such that

$$
\psi
=
\tilde{\psi}\circ\chi.
$$

Proof: How do we define

$$
\tilde{\psi}:H\to X?
$$

Since $\chi$ is surjective, for every
$h\in H$
there exists some
$g\in G$
such that

$$
h=\chi(g),
$$

although $g$ need not be unique.

Now define

$$
\tilde{\psi}:=\psi(g).
$$

We need to check that this is a well-defined homomorphism.

Suppose

$$
\chi(g)=\chi(g^{\prime}).
$$

Then

$$
\chi(g)=\chi(g^{\prime})
$$

implies

$$
\chi(g\cdot(g^{\prime})^{-1})
=
1_H.
$$

Therefore,

$$
g\cdot(g^{\prime})^{-1}
\in K,
$$

and hence

$$
g=g^{\prime}\cdot k
$$

for some
$k\in K$.

The remainder of the argument is exactly the same as in the proof of the previous theorem.

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

**Corollary 9.7** Suppose

$$
\chi:G\to H
$$

is surjective and

$$
K=\ker(\chi).
$$

Then

$$
H\cong G/K.
$$

**Remark** This is the so-called **First Isomorphism Theorem**. Let us prove it using universal properties. We will see it again in the next chapter.

Proof: Theorem 9.3 gives a map from

$$
G/K
$$

to

$$
H,
$$

and Theorem 9.4 gives a map from

$$
H
$$

to

$$
G/K.
$$

Composing these maps gives a map

$$
G/K\to H\to G/K.
$$

This composite satisfies the required commutativity constraints.

The composite is given by the two maps obtained from the two theorems, but the identity map

$$
G/K\to G/K
$$

also satisfies the same constraints.

Since these maps are uniquely determined, the composite must be the identity.

Similarly, in the other direction, we obtain a composite

$$
H\to G/K\to H,
$$

and the same argument shows that this composite is the identity on $H$.

Thus we have maps

$$
G/K\to H
$$

and

$$
H\to G/K
$$

whose composites in both directions are identity maps.

Therefore there is a bijection between
$G/K$
and
$H$.
Since the maps are group homomorphisms, they are in fact isomorphic as groups.

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

**Remark** The notion of a universal property allows us to characterize certain objects uniquely, up to isomorphism, just as we did in the proof of the corollary, because a universal property is a statement of existence and uniqueness.

If you are thinking categorically, two objects are “isomorphic in a category” precisely when they satisfy the same universal property.

In our example, two groups—the quotient group and any group onto which $G$ maps surjectively—satisfy the same universal property in the “category of groups”, so they are isomorphic.

This is a very important idea in category theory and is related to a consequence of the Yoneda lemma.

**Proposition 9.8** Let $K,G$ be arbitrary groups and let

$$
\phi:K\to G
$$

be any group homomorphism.

Then there exists another group homomorphism

$$
\psi:G\to H
$$

such that we have a universal property represented by the diagram

<div align=center>
<img src=https://tex.xumin.net/svgb/U1HhiklKTc_Mqy7JzK5KTqnl8laISSwqyi-PLtJRUIopyMhUioWJpIBEUguKM3Py85TUYxXUFNxRpIAyIMUxXIZwM0CKPLhiUvNS4BaoqAAA>
</div>

where

$$
\epsilon:K\to1
$$

is the trivial homomorphism sending every element of $K$ to the identity of the trivial group $1$, and the bottom arrow

$$
1\to H
$$

is the unique homomorphism from the trivial group to $H$.

Proof:

1. Construct $H$ and $\psi$.

Define

$$
H
=
G/\phi(K),
$$

where $\phi(K)$ is the image of $K$ under $\phi$, and hence is a subgroup of $G$.

The quotient

$$
G/\phi(K)
$$

consists of all cosets

$$
g\phi(K),
\qquad
g\in G.
$$

Define

$$
\psi:G\to H
$$

to be the natural projection

$$
\psi(g)
=
g\phi(K).
$$

Thus $\psi$ sends each element $g$ to its coset in $H$, and $\psi$ is a surjective homomorphism.

2. Verify commutativity of the diagram.

We need to show

$$
\psi\circ\phi
=
\epsilon.
$$

For any
$k\in K$,
compute

$$
\psi(\phi(k))
=
\phi(k)\phi(K)
=
\phi(K).
$$

Since

$$
\phi(k)\in\phi(K),
$$

the coset

$$
\phi(k)\phi(K)
$$

is equal to
$\phi(K)$,
which is the identity element of $H$.

Thus
$\psi(\phi(k))$
sends every element of $K$ to the identity in $H$.

This agrees with the composition through $\epsilon$:

$$
(\text{the unique map from }1\text{ to }H)
\circ
\epsilon(k)
=
1_H.
$$

Therefore, the diagram commutes.

3. This construction satisfies the following universal property:

For any group $L$ and any homomorphism

$$
f:G\to L
$$

such that

$$
f\circ\phi
=
\epsilon,
$$

there exists a unique homomorphism

$$
\overline{f}:H\to L
$$

such that

$$
f
=
\overline{f}\circ\psi.
$$

Indeed:

(1) Since

$$
f\circ\phi
=
\epsilon,
$$

for every
$k\in K$,

$$
f(\phi(k))
=
1_L.
$$

This means

$$
\phi(k)\in\ker(f)
$$

for all
$k\in K$.

Hence

$$
\phi(K)
\subset
\ker(f).
$$

(2) Define

$$
\overline{f}:H\to L
$$

by

$$
\overline{f}(g\phi(K))
=
f(g).
$$

This is well defined because if

$$
g\phi(K)
=
g^{\prime}\phi(K),
$$

then

$$
g^{-1}g^{\prime}
\in
\phi(K).
$$

Therefore,

$$
f(g^{-1}g^{\prime})
=
1_L,
$$

which implies

$$
f(g)
=
f(g^{\prime}).
$$

Hence $\overline{f}$ is a well-defined homomorphism.

(3) Uniqueness:

Suppose there is another homomorphism

$$
\overline{f}^{\prime}
$$

such that

$$
f
=
\overline{f}^{\prime}\circ\psi.
$$

Then for every
$g\in G$,

$$
\overline{f}(g\phi(K))
=
f(g)
=
\overline{f}^{\prime}(g\phi(K)).
$$

Therefore,

$$
\overline{f}
=
\overline{f}^{\prime}.
$$

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

§ DS-09 商群

§ DS-09 Quotient Group

# §8 基本群
**定义 8.1** 设$X\subset\mathbb{R}^{n}$为一个子集，并固定$x_{0}\in X$。以$x_{0}$为基点的$X$中的一个闭**路径(loop)** 是一个连续函数
$$
[0,1]\xrightarrow{\gamma}\mathbb{R}^{n}
$$
满足以下条件：
- 对于任意$t\in[0,1]$，有$\gamma(t)\in X$。
- $\gamma(0)=\gamma(1)=x_{0}$。 



**例 8.1** 设$X=\mathbb{R}^{n}\setminus\{0\}$，$x_{0}=(1,0)$。

<p align="center">
  <img src="https://tex.xumin.net/svgb/lZXLbtswEEX3-QouUkAGHIHDNxNklW3RRdNFAdsoGIm2hSqSStGN0yD_XoryQ3Zcp_FKIkd3Du8lxtMHuyiqF1_8_NMUmV85-zpZ3-JU8saPn3cPbWZKe3sF43X_BLMLFH6f0KrK6sdHW3lUzJGzv1aFs9do2hi_RAkeU4xHN6GstR7lhVk484hKO_eo9cZ5ZDzCY2SqHC1Ni5a2WCw9qucofHaxaXC_NE1Q_Gozb6pFadGkyK_zRgqlBaecaU24BjFDsX6aO_OEElB6zPEIXV2hhGqdgjx-BYFT3q90xfvX7DkLTW627e9W7rdFn4vKtn1nIwRoSTQWVFNGKR12niyctdUMJSToMTIGgJQE3TRFWV15V5dt2JMiEhDVtewOn1DgsZySbXlCGYlVwOLSGSAtBVOhnGngINhJHsog1R0PO4KhmEd35I6FUOhWVDSkI-nOEkkgHYLce2diXkMWDopwwrDQVCh8lArRUVmrrfNEs6iM_0OZYUYk54JKIqlS_FiaHUvDKem37kkFoJXkMkhrIg7jfChXtnOP0E5MqvQ4y01KCvZRcjpo3Ce59_4MSLjPQLGEECGjXKlTIETHqyPkWxAcd8LV2oHgHnrPMTzFuQtOBEggmENwlUn9MRCiovGMHzvC-Q5ExBKJ3wEhmkomZTBGUgKK8pPZDMUOLREqZsP2JIINbvI_s_lm1x59qXN7eMN2ueoRqsIumpgqW9butqpdmHZPtvWzSVFV1oVx12xG5wy9XOLL1_PagSIghBHwQeH1jxf8-p44SBrGG_mw-PeB8GYI3xUu201gIgSTisoQEAgxGMHzoix7Z8V-fmX9pwlp_OjmYmqr_OA_5y8">
</p>

**定义 8.2** 给定两条曲线$\gamma_{1}$和$\gamma_{2}$，如果$\gamma_{1}$可以在不改变$\gamma(0)$和$\gamma(1)$的情况下，通过连续变形变为$\gamma_{2}$，我们称$\gamma_{1}$和$\gamma_{2}$**同伦(homotopic)**。即，如果存在一个连续映射
$$
\begin{aligned}
\Gamma: [0,1]\times[a,b]&\to X\\
(t,s)&\mapsto \Gamma(t,s)
\end{aligned} \tag*{}
$$
使得
- $\Gamma(t,a)=\gamma_{1}(t)$。
- $\Gamma(t,b)=\gamma_{2}(t)$。
- 对任意$s$，有$\Gamma(0,s)=\Gamma(1,s)=x_{0}$。 

其中$[a,b]$是某个区间。

<p align="center">
  <img src="https://tex.xumin.net/svgb/tVjbbtNAEH3nK_ahSKnkWt6bL1R95iOSILmJ1UakTuS4KhBV4g4FoYJASNxEHyruhaogWi4qH0Mdlyd-gfXuOvFGRTRonSd7M7vnzMyZmZVr88FCK-zHrbMXuq1GvBwFq9Vzc5bp0G5snB8-9Bp-O5ibgcY58QTrJ2rNyF8BFQgtEzqGa5l0GszMgAqyLdNmC3B69gRgP9UOonFDiDJLbnoSJLsvDve_pG8fpl9_FHZXG512J5rrRwvzBkCUnoqCpgEcMgsWoiAIDQAJe55vLwfsb2SvGoBv63T9Ris-z9gewVMhNCKw_vH4BKA7YoAwzhlATI5iMArM392_fu3XlVfJzevJ2o7gACqD148O13am_0kGEToKh23nZDA9Ohoe52LlsfAEF0tJmg4g5BWdlq_Cab1AWAXCpQFRJY3ytQwgVS8w1wvSDuSoQE5pQK4K5JYWOs9S5O2VB6SqzitLdQiqzQOW5RFCKhAqSwwIq0C4NCCiApHSQkcV1SF6RIO_e49NmPTx1fT9VrL5afBkR5kz1ZleHPjteDE7zYEGtBhp0wSNThhHnXaPrbquibDBBI3caeCHTbbERAdhtiStKxi6JnTZbmaljLaD_ZfJ5p3Bs8sC-Pf3m-m3H2y6_Lx46fD-M_aQrG8ktzZ-f1_TPPuwB7NYUG_MHYIcvk6kL4RiHjMLCU8I5P9DfJxkTUgJ2wLKHuOEMeVppJITpryobUsGd-iKbkYFZ8cYuYIqQTkljIuWRV9kurXdm7DDj_bGORHI122YZ042Csv7j8xNxOjvYbJlVIZhokThNPJlWBSDpxcP9m4PHn4ZvHqjXvkqbJs4zlYvbNgebT_Y20zffhjcfpes74q9YacZAD9Wt_enqlOrs2MGhfOYQT0zUC6hh-_3ks3HyfZ6-nxLFOvgwX6yfTW99fk4BTpR6yLIzrsv95WwDsLfse4mSTAvHwiHSCKbkOpHoiL7UEXiRaIZievMs3IgITNXPw5XjOPlOEjC6sYRtWxL4RNLNETtWsCe6P6uwMGewNF-A5B1b0shYFfgaNcBphzHkToothu9OARKBwSO7eRC1w1UmJECiebFqz10oi0gJ48dLOk6mJ-MSa4GlCPrRnLlBKCKviG2CvNn40a6tV8cO1KdbNibjCGfFlU_bCwyMmEnihfBStCL69VWGAYR6AVd-Y2mXiSc82VzVZDNpiqnymdqgWR_qrbgLy35Z_q1btRaClaHc0oOwYwKu8hORCWbd8A35utjhyHPYpQmPqt2OmOYnVULwqbyreoP">
</p>

你可以将看作是一段持续$b-a$秒的“路径动画”。

**引理 8.1** 若$\gamma_{1}\sim\gamma_{2}$当且仅当$\gamma_{1}$与$\gamma_{2}$同伦，则$\sim$是一个等价关系。

简单来说：
- 任意路径可以通过“无变形”变回自身。
- 如果$\gamma_{1}$可以变形为$\gamma_{2}$，则$\gamma_{2}$也可以通过反向变形变回$\gamma_{1}$。
- 如果$\gamma_{1}$可以变形为$\gamma_{2}$，而$\gamma_{2}$又可以变形为$\gamma_{3}$，那么可以通过连续执行两次变形来将$\gamma_{1}$变形为$\gamma_{3}$。 

**定义 8.3** 定义
$$
\pi_{1}(X,x_{0})=\{\text{以}~x_{0}~\text{为基点的闭路径}\}/\sim.\tag*{}
$$
称其为以$x_{0}$为基点的$X$的**基本群(fundamental group)**。

**注** 如果$X$是连通的，则对于任意两个基点有$\pi_{1}(X,x_{0})\cong\pi_{1}(X,x_{0}^{\prime})$。

基本群的复合如下：
$$
\begin{aligned}       \pi_{1}(X,x_{0})\times\pi_{1}(X,x_{0})&\to\pi_{1}(X,x_{0})\\       ([\gamma_{b}],[\gamma_{a}])&\mapsto [\gamma_{b}][\gamma_{a}].   \end{aligned} \tag*{}
$$
给定两条路径$\gamma_{a}$和$\gamma_{b}$，考虑路径
$$
\begin{aligned}   \widetilde{\gamma_{b}\circ\gamma_{a}}:[0,2]&\to X\\   t&\mapsto\begin{cases}       \gamma_{a}(t)&\text{若}~t\in[0,1]\\       \gamma_{b}(t-1)&\text{若}~t\in[1,2]   \end{cases}   \end{aligned} \tag*{}
$$
将区间$[0,2]$重缩放到$[0,1]$，得到路径
$$
\begin{aligned}   \gamma_{b}\circ\gamma_{a}:[0,1]&\to X\\   t&\mapsto\begin{cases}       \gamma_{a}(2t)&\text{若}~t\in\left[0,\frac{1}{2}\right]\\       \gamma_{b}(2t-1)&\text{若}~t\in\left[\frac{1}{2},1\right]   \end{cases}   \end{aligned}  \tag*{}
$$
定义$[\gamma_{b}][\gamma_{a}]=[\gamma_{b}\circ\gamma_{a}]$。

**例 8.2** 考虑$X=\mathbb{R}^{2}\setminus\{p,q\}$， $x_{0}$是任意点。

<p align="center">
  <img src="https://tex.xumin.net/svgb/zVjLbttGFN37K7hQAQmgB-Q8SE4N9Ucsw6BoxhYikwotI3YFAkUfTgwUTdHumgYpiiBoF-ki6CJ13eZnrMhd5Rc6D3LIGZGyAwtotTF5Z-6dc--c-6AHw3h_lMymo_ufTkbR9DiL8-2TvgN8Mpnap-rhKArHcX_TtU_kk7uzsWGx30fW4vmv879-Fi-DvSx8aHWpa0OCetbmptWFxNFfEAnkC9tVPken0Ti2tnSL87OvFj98eXXxYv7kfHHx9v3l44ltPbAtAMD7y_P6eS70QQBt6FNp2UUuoDYMIICkt6XtlCtqY6HYsBM6uk3ottksVtRGp90m1W0ip81msaJs0labyDdsBq02A81mqdiwE2PdJiZtNosVtRG32iTGHZHWOyL6HRHjjgqOzH88W3z-h3Wy61S62_dG43F_lu0PGeXIx1m8Zztb-1kcJ-zvcHwc205u8z1WOgmj0fSUs9gSv67rURthB5AeI5gVpck0S8dHhRxR4Pk9K0z2hIC9cBnfqdR9ploK6-q-CxCSC1Ldh8peXR02ny7k3IRSl_YwbDhdCnXwAitfKMErezX1yvflZHz39DXLPms_PDwMd4dWd_Hn2-vz171a1KN0nGZ62F3oFYFn2SFDj1Bub35SHMl_trju5ZsQSBpCiRAjgm9DGjA-FP4gBIHjMbIg4JC6S4hlISI29Fj9wiVrPhRuC0DoE8FIaiKEHlYcF_CgHwgBcf97bIFII-QGZegcCBATQAT0yKnwm7nGl8TCejKOWRKZYHJeyoMq42S6cFkTzsBUx0XGKUcxVPbq6rD5dCwzUWVcaQ-7DadLoQ5eZpxbgVf2auqV7zdkXGh1r79_dnPG-biIu4uxjDyEHs-4VWlGRVSBY9YN3ht8tgZd4KkwUgooc5mTV_MFMyLRgLMe0BvZ3AKzDSPxgMtMe2j5nlghdFnmYw_AEiJHAlmFwJQNLf8DiBgCzHscIxCsyMjZBKkHAmQSqroOxYWy7LJJaP7k5dWbz_754pf547O7leEWVzByZKdFhiOYVVleO5wyI7HnyWKiRzmgIiecNde0dlyQyvJaRhcjKcA6QT1ZE9cPrDJsAAskYlUFcNkIdGC0kHrmhYfLF36XKtCCvxytiFnFCJLAsFfgJ4zeQqCVYEJQ2e_Wm1QVkwxc8sIRVLiwZKbrr8ZV9TBmWhheTw_DQdA4twl5fWpkgoapsfBzadTB1NGnRsw_lpamRiFtOl3Ia1NjaU-fGssom9W_xKqmRumlOTXWfF_uYSWNr958o2oY-5S7uviOEVs2uLUSpsokgzGuoxYkY1woBVhjTJUJyoXrp4_YV-jit1fzF79b3YP0MJ2mk7hOHN5c4yjNwmlcPozSpD87SsL7cV61iKIaoKUywfora7OQQeLTa1HCfEAxcxaal00IYLMJ5ClQgXz306PFq7-1T1wP26Lz9Kwk3Yu3wyQ6YMFN0mx6YD2Mj6Y71qxzsjtz8k6ufxtDNpMGZLXixFCCzi2UHhhKyBXz1Eqlwu1B8d-JcHg8DrN8FuXqC2KQpVMW-WF6MqNOPuv0O_lgoFY7A0m72TAfRKMsssr3MO-UtuNkT1kWstz4uPVFdq6O5QcB3RRIB6NEx7rdDnbntmiRT3knWoG1MHT7bON3VwubcR51bJd_ZjSfZofj0X7SH8f3eJDm3359dfnSoE5wB8DNjboOeCgAb8iw1f659S8">
</p>

右边的两张图都是$[\gamma_{b}\circ\gamma_{a}]$的表示。

**注** 

- 恒等闭路径$\gamma(t)=x_{0}$对任意$t$恒为单位元。 

如果$\gamma$是一条闭路径，而$\gamma_{0}$是恒等路径，那么$\gamma_{0}\circ\gamma$是“快速执行$\gamma$ ，然后静止$\dfrac{1}{2}$秒”的路径。设$\Gamma$为从$\dfrac{1}{2}$秒静止时间逐渐收缩到$0$秒的同伦变换。
- $\gamma$的逆路径是“反向执行$\gamma$”的闭路径。 

<p align="center">
  <img src="https://tex.xumin.net/svgb/vZbfb9JQFMef4a84D1sCSSG391LYD_EfAbKU9gLNSovtJWMSEqOLbg-GLZoYZ0yMMerLotnLdCbzn1nH-C-8vaVdaYpOWXwh99c553s-59xe6k3aNqwhM7Yf9gyN9R06qg2qqFhRekzajQauppq0WpClQTCSG9nMat_S7G6XWgyMFjj0Qd9w6AbUeyrrQA5JpUo5vwmrLmWgG2rbUbtg0hYDl6kOA5UBkkC1dOioLnSo0e4wsFvArbLcN1y_eHt9cHr15nRy_hPaarerZjN13VF3oKbZpu1Uh067KQFWlA2H6hJUSpvQdii1JJBLfNw0-5Rv4_JIElZ2T9UMtluVpXuFbKYBADlcQhKu4DwUi6DZFnNs0w1WCZLzQlswk5E4k8Mynyh8azOQ-PqYi_P2n06PX0bidNXtAEfAqGNVh7YFpSInyDNrBaNRELrMQxPuqVAIJqSCIrci8-mrE-9w__rsi3exNzneu_z2aPrkM491RxRmCOS1NARilcQRiJlAEDLztf4fJTfFEDNZTi3G5Px9ol-gqWrbO6qju79XitFaJBVHQisLdJLUpiFzTUPiTUPiOu9Ah1S47wvJZkA0Uio2PFdAEi8guW0B_w7LYh2xuxQvH5kv3-X5B2984H0d-7_js1CaIC5jCZfXg7tC0HooP3nEX42O-OcDz9MfR5PH32GwhaJ8W4ZpJtNFUbIoTBaNJPCP3iQLDcE9h3kIIX4-W7GKiiUlTBit84m_Fm9ZMfHdxCxlVFRmBwNLWY5cBZZyGDBpKTa4dWQ5c4XjMXGKZSgO36iNXAnLMEkOVdvVTBpvmaURklSEJImQJBCSRQhJEiFJICSLEJIkQpJASBYhJEmEJIGQpCJcBe_s4-XFp9Rux_Fux6ndjuPdjue7_erds8nJRXia72IuQCmiSh4sW6dQUy2twy-7ZTv8ld6hLmvUDMuiDri0N3vu_Qs9XBlsDdFoZRRF5snIS_uaOVvyyQhvIeGpl9C_CKqLF-KPim75DQxbWkHL6YE61BkdsKE3fu4dHglidWrpc3_PfgE">
</p>

[“$\gamma$反向”$\circ \gamma$]=?

<p align="center">
  <img src="https://tex.xumin.net/svgb/vVnNbtxGDD7HTzEFEsAGZGE45PzFdS-5Fr20N9sHZVe2ha53t1q5SWrk3ofog-SF8h6lpJU00kreulACxIhEazgk9fH7OPL1-_QuWz8V2e9_bbNF8Zinn68-XsrY6m0RfWovdotklV6eQ_SxvoKbk1dvHteLzcNDui5Edivy9I_HLE_fiuttUtyLUxmRNWcX4s0uLcQyS-7y5EGs0ttC7IokL0RSCBmJZL0U98lO3KfZ3X0hNreCV52w73eP-Z-p-DlbpztxlS3fLhODliwZhwQeSNONOHl1vcyTD-JqsVlt8sun_O59JJTWb_N0yRfSXYi7PE3XfH0h3q8eU76w4rOIqlWbbbLIik-cihDiVJGMlFVnIo7FYrMu8s1qx1Z0EUqIjT2rQmWDiRBkZeAnTxXICDWciYuTV6NRS4fSawNeWmkVWDVb2NH5T1Xk4MYir6wITdi0v6tiblKditkqTyANKCUNkoR5Kz0VL3GlZVfp2uCnKv1rkScVZMLAEcApi4RkUGvfxV29YVCRMv5MnJ_zjfRhDUa9KQ9Ke2kcgEQEHHFXemjdlb4nYeARvdQgnTIGpTPHamqpLSkQtUVVZgQIP55HNYaRK-swhiGKlYstF9PFpsGDsjGXF0yM2JY31s9D2ZMzjH0nLSH_HEXyC1KogeFhNPzKTjZ2bfRlnntDFXuX91TsSiMq0Gis4VdJ-vvFjjEzhrSxpQbYtQl0THai9r_eJ1tm0ndZvlildQZb5j30pIH5hKTTIYBus9VqmIBs45dN9LKMvXy2i13c1Nit9u5jprLKmHRTdO5FKqOUYT_K4bIylf1T9TKA1k-9DMZ2q628tF2296PC3Q7ouYlJdUG2fqplTW7cpItPZTEnAYKaiHUFkQvt-Hpe0kMcIz1UA3lB1ZcXZsTnexJQsSo6VhnnnHZudnVRdlQXbaguVXKNurSZToXMFZYs5Np7ZlVmlKNU-EIhnwqYixop7wMtt1hRYKswnouNqrp_RhaMBGkRjdcA1vmByqAMVcYfVxnnkZjMtHIsN8bSiLtWZfwRlQFDrH4SPQBxNm5OmmtVBtW4yiAMVQZhqDKlwh9RGS4scv-BQs_taEPZnYeqlRun6soeykyVaCAzQeJTwRurtTGSwABJ1pvvFzsOsO3Km5ZcJuA9JjNkFIu7Rx5WWW1wPpnxozLjhzLj-zJTM-BQZiochTLDifZkpr4_gKgcyEzrR4W7HZCHH8qMH8iM_68yY1GjdfyDlodBcxTcL-M-kiX3HWTtXX-0Rm96ozXafY0neYURYUBqzdCQzFLzUXYjM2hcpHQrfEHwlgmQqAu-ypE61m5zfo6yWXLK0VUDef5_SNnlFsbVHMv0c9SdlvwEaa4HO4ZwFGvdWQjclb4nGYOhwMcBnut4qiN9VMD_F2P7krjUIWO7mrFtx9iuZmzdMbatGfsZdHinrVSkgJWRGCFzk940NhxXWrsA2K58ssNGkPg047FmKuIjnvZa88t0szFe-e5HqKu0ku8Yjw0Vu5APunF_E_ZBzVTlL-plFlo_9TIY2622Boy396Mh7P0DZOxj0tAF2fqpljW5HWU8LyWfNllUeMLm09fM0x5pBjAfZg-wQVpxeXTAG7osazdaE8HR7wnOEo96io9dhoN30G90IuxGPep_VBmnIQJttbWOV2rn1Yi7ZtSr3D036mme8pQh_oempOe5m266sMRqQsGZhYjVBH1Y2OOn2Wr2B1XOSDxWoZ6t6crCHeK5tgZjBht6Y0aNhmH3VKmEYwaR6o0Z9f3hbqo_ZrR-VLjbcMxoY1JdkL0xo81t0HTj38JcOWowo3H_Ge8G5wotA80jf1zzLB_UJGs_6xQ3BZkRd43mVe46zRt7_Xxs9VoBlmdAKa2d7_X7Mc6trCHnku9zrpZjnKvlgHO17HOulmOo2Vu719_4aThXy1GM-gHn7nPpXr8f51zxW_qxEL9slmlTxFOWSz4PyFhyT675F-IqWS_umRXWm7y4Fx_SXXFzla3XaS526Xb_lb_s-6fX17vs4fXni3G_3LTfxC-V3089zO-X5e-bxFt-nN5PHi_we5Wssrv1ZflXkBvx9PXLl6___P0D73Gdrpe9P8H8Cw">
</p>

- 最终，路径复合是结合的。 
<p align="center">
  <img src="https://tex.xumin.net/svgb/hZTPSsMwHMfvPkUOO6zQlq3bVJA9yTYkbWMb1n-kmUNDwZsXDx6EgQ8geBdPQ19Gp3sLky7pn62rt6T5fJrfL1-S6SJFFM9vA2wTSG6Yi5yYQIrjKDUTSH2CkgA6OPKyk-k_aBiTxOeknvowQWm7EEIy56ygbOThiAkwwQ5dEMS_uQQuQben90xLMww-MMQIRLGLJjYK4uWMdXqd7EKRljlSrBg20NMrAh3Wz5iVVbyBeVaI-fioOcjYsGoWWqPTr5C8C87ly9COr5H4pQfDEF7ycjrZrnat3soxQRQvi9b2mjimDIQi-Rwg2PPpDECaN6EB1hl3S3jqYOKU22m1eT8_gbIxY2iey3xGrfn0TWtU0LvJoVBEVDvoPE9p5uMW0aonpLRGqR6RMWrNSJas7TV0VLJUsFVHzttz2gmHQfHvIqm9oLrlhntBaWVSEztYoBnY3SKV19A8VZWpdXlzKhk1QOqWlHk0QCVRrFcAXT4ExYC_CGOWRnCOdADDJMB04aIxv1VOqIMUeSGKKAhQ5FF_PBBfsxk4AaB7qrapHBfb3j3_ft7_rF83Tx-b1ft29fb9-PC1fskELmtFkVt7cP4A">
</p>

**例 8.3** 注意到$L$表示一条直线。

<p align="center">
  <img src="https://tex.xumin.net/svgb/pZNNboJAFMfXcIpZUCONkAGkxjS9QVftro6aAaY6KYIBjKaTWfUC3XTdI7SLbmrS9DT24xadUUCMmn4thveG__-9l_kNII8MaMQy7E1CnHDm-1xVNDSmfWbx-nlPPBuzPoNc14Cq1MRCeQW9uh5TP5skhHc8nJKQRuSE1WHDgKal8y44AB-PN293T4v7-fv8dfHw_PlyqyoKChI8BcIHdeDTxA-J2Jiufiy1SxqGoM6gabvgEFhmy7G5bGi7VTPM3VEckI5HwngKEjoYZl2As33VTFseQ-OiEpEo2DiAeIXESuIMZ8SLZ8xoQ8405MfRQOOgBqRcUkEjnA09j53xHrM5Skk2otEkRQyiNSxR8x0oaDqC0yYSw6jbRbBkgEWAVUKW6TYktH1QKjhKa5VAOdJsL1VXzrBED7FpCm8hNKuCu5ywBe-v6Jw1OnD6O27NXdysIuzktrIWHvNIHsgucyvPRVHVXZFgNS9sxWWItkL-2W0U3u3r6AQ4HZKgK-tb8rt1VoNbYu4_0OfIL7iEq6565L_7Fw">
</p>

这里不证明同构，但粗略地，可以将路径$\gamma$映射到它的“绕数”。

“路径$\gamma$绕过缺口的次数是多少？”

<p align="center">
  <img src="https://tex.xumin.net/svgb/7VfNjtJQFN73Ke4CDJO0Te8ftCI-iFMW_bliM6UlpURG0o0x0a07ExczcefCZOLOMerTMMS38MK9_aE4BHAYYyKbHs4993zfPfd8h2K7bBBEs9RxJ6GTZDMvzBSlCbTaBwyc4dABN-8u51ev5m-ugLERotgyVXD2YhR46SRh2em0Z-gdOkrV88IYe07IehpUp8KCquuMWRhErNfyJknCohS48STyg2jAjanucQ9LTvoKAE2wuPg4__aBm7afOM9BCxGotzsqQsvHCScFWsS4zYUhLlxyY8XlnXshA90VzPzievHySwmDYXuVs90pEmAkYDpo6epuxoqFtVixXUDcvP-8uP6Rbzv14jBOerNk4KqI0ocJ89UO6YJBwlikQsJNN5wwfqp21ucJscCgxgpD14EXR2kSh-NyrS3WnMjnPkKEj-TxLUwt4bLW-O9DRNUe78iFkCoXQ_hgyYXIMHwwl349S40IpWLNqhChkjjcUpQm-Pn1baUXTp8GYdhzQ8c7W0LSTh4PeQovSJY91II6b_X8ni9fLz59r7SH3IKRbvAtUeyz_c45a9grNTayrmKzyF9TnPIANOyhMxqnMTDsIOJ2-sx1Z0-yBrDtnaQN_4q0j6nnY4n4oCaF5iqLRTekIg5gkKI9RS9aqOxNWPTaYYJ9pG2jgCwxJcxcITxEKkpSwIZOEP--fBxegrUstSoQq0CUVRBkKdo-tvaT5b1rEe4sP-2__u5Bf9gwfv9ThQ18PP31y4OZt4oP5fiy7sSS-AjnSvgT8ZdZ6spDusVVSYlOzeOL767Klr_ZmOW1IQEP85kl6BB6MDgfiCKlWb8z05SvNzm47FF8R-BrA3uzZ5El4KxiYEtCskxV5v_MqNTkrBRB8o_JLw">
</p>

**例 8.4** 如果路径$\gamma$可微，可以定义为
$$
\dfrac{1}{2\pi\mathrm{i}}\displaystyle\int_{\gamma}\dfrac{\mathrm{d}z}{z}.\tag*{}
$$

**例 8.5**

<p align="center">
  <img src="https://tex.xumin.net/svgb/tVTLTsJAFN3zFbNoSJvQZgpUYwh_wMotA6SPASaWtrRDQCezUqNx5cKVGzeu3Llj5deg-BnOtJTyiDFK7ObOnJ57zm3uSZGDByRg1HYmvh1z5rq8BMSjoIj0mMlVRII-Pa_MegxyTQFlgFBKQHFIbYqdcMb0E8iZgtwwGCi8YBQSI5sOHYed8i6rcpRgOiLBJEEsqowR35DOhFcTkbOLiLh0EmPeduwE-yTATabCCjRqGu-k3JTvxfYUSFzTdbWaF1MWmBeoNYqGPvF9oJpC6NjSgEti18dCwIDWJisIPdyOyWBIO8CmBZ8p6cAKb2yP0J4lQ9KnTR0adXfUkYInluixLDmCKdTFpS4E8hf1zRfWvreD_XCaeUPjSHKkefSt8T_4mkY99x3nvggH3tZ2DkxErUgEUFtIrgO0ugxFMRlhrv0hHtUf41FLv0rejewIxMnMQXMNmhtgftgJU7oBz06G2OvIhmpFX8vIFErlFKtlfgITA-6n0bB-m8d1x04iD1hQeVUX8-fF_GX58PRxe_9-fbmY3y0frz5vBPK6fHsG5VLmsfppfAE">
</p>

<p align="center">
  <img src="https://tex.xumin.net/svgb/7VzNbhvJET5bT9GH3YUEjIjq-ukfe5XLXoNckpukABQ1lojIpEJRWW8EvUEC5CnyDDnkcTbvEfTPDJucGYlekLZ3wZPF7pnqqp6eqvrqq_HFVX0znT1N6tmyXjwfXaSfy_HV49148fz04clNPjxPVhPTv_z9fjpZPi7q5_OPZzCycr-sfmr_eJiM7-qzU119TH_p6mr8UN9NZ_XZ8eRxsahnS3U1f5xdT2c36mr-cZQWPrk8evPt42wy__AhXDF9rxb1Xx-ni_qturgfL2_VMVRszck79e1DvVTX0_HNYvxB3dXvl-phOV4s1XipoFLj2bW6HT-o23p6c7tU8_eKrTl6c_Tm2z8uF-M49vvprH5Q59Prt9djMEY8aEbLhEKaL9XRm4vrxfhHpZQ6Jm1HxlZowj8n6vRUHRNiHLLhnxP1blCyGE0WrHjvEZ3YfslZTCk5LzYomSw4Y7wj67RlMbIh2XV19tvpbC2BZibthb0V1y-4VNl3Vb4d39dv1Q_TxeSuTnLvPRhnSROLcJBdyD1_P727O3ta3FxVCkXeLurrSsE7dbOo61n86-rusa4UqGelqnCxmt-PJ9PlT2daXapjkqSB0EhO1GikJvPZcjG_eyimkr7hZByT0Ig4jMWLj0l4pOM1nZsljebbpBSV7pThhfMUj4hWEioU2Vi1e2dWT9rbsBSVbi4tPj1Vk5_CToe9_-Fx8bd67YlqzywEpC2hA9DFWTmfzO_mi82tt9zuvWZudh_RxP2PN5b7r9Qxel8hu01L0FTIrRkkFbLfsJ5PBrQ2mhARGNkZK9rsUunq-9OoNpEZEVXo7Ibm6HyYIDZZeWSJV0rWvzW4X3lnAcQ6Z70j78nvfscHVSd2FRrfbHrzq2_T1Z_qj0v1h_l13Wh3jA6iWNIjPlGz-XWtzsezye18cTabL5a36sf6YXl5Pp3N6oV6qO-zy79Uv8iwNZuevhl_8_zu6KKeXa8FmKPv1GcIO30ui5wlYWBvHLBot1OfxS49C919fmkCRrr1WOxGjsPbk_3MMbGPDqAdKf0HjKSZyQ5ElyLTUdADy7cTI7fyW1Gi4Obq7ci6-kHVNNVqX4hMAlbWr3mv_mAnxqMHNs6J0bYMSefX44dbdT9eLuvF7OxpPlM8kvsQ89-nv54vw8sCSV9MyzW_SeNLIRYYnBirLZOzmu2nL-vSsmGZuGz6TUBDfgMdkKawshHtUds9OA5Mgcl2vDVziuPcPjeTBnTfU-83QJxoQ-yMMdpoTTv12qe_y8FGRz38pgXok8K-fW8A0kDjAAvbB6KORRHwDhDAyE4jZdp9aBOzzTenMcs2vhuta81JMWdl94D2LIzCIBo8WrOPsKOTDoa7bz4ldXl1frAd6PqsPk9_cXFAGAeEcUAYB4Txq0cYOVYdEMbXhTD-_HSqn3thhvoMMGN_xa2jrQHMAb_sD79sDV92iF62By87xC5bQpevFblsB1x2XW3aFW7ZCrbseOt3BluOtoMtvyLUcgAtB1pkO1qke_I9emMNgQlUi9GvAnbtVmcfiZqzr4lfykGLTHj97FuqkLEgKFxIp3VLMzCEC1z83W8Ak2YBIY3eMdFuDUgvb6nEmvpMOtfkVxYwhQuD81lxJR7LoX47QFs0oq0Dds4j2p0a8spjMMkOa4onYdOY50-z4wVMs6YS7BKd9KWeVoMzaI1YjUY7sw-6r8ebr3ixLt3Xcmf5OWyep0TPFWRdKWqN7utbOE_JJt1nNlY1Q3Rfs9WtDdKl--LCawXzHg6HISEsNCNLu4BYr5z1DYx1NQivDtW9Q3XvECi_mkC5StHDnQT8ZYJjTtElpE8dMtsELVsg7nRF4LfSmZy1GPpYwFp03tI-SpJrMbGsSlKcCPExY4rkkVs82hjbr7sVi8YZ8tp7xh2TaK-3PoT8oiJqAFHqO6nQZfK3NHuo4QQdInIoQQgZ_qzdD2FnV5UbU6HZbLgxw1nLXhn_wfyvTUgKxl-4KEOtJR89NSsxK34-325LkfnMDS3fTJSMf5IondWbkXX1o6pxqtW-ELlZjn-d8fdeGyBnUJi107jZNxfrJeR9wZdFhM8QKkEvkPqMzoGAF-_AAOtewY2UQnCz1rBgq61GZq-1c0KbnX6uq7HfTmNkJnYIxmtNpA31Sl5T2XdU7gFu7DwbbYwXx4hmP8jTRldBrpP0JixDFppwBNFlks0eknUCIMHpDAQlDD1tVjv0RrMBR_sIqYUafSGVfJO1M0a3yToV945ZRwsZh3s9CLVYFBABzWWj5Q6DqgkvXc_-2xiQOMTP9Mb6NICyrfrhtdTOWY3WG_F-D60q6CKYp66XTwiHQj00M3_BX8XjlJwNtXYPaK-NDq-VA2eJwdq9xFiOkZOx43Ih7W4ITRle59OTvWVp-VBOhqANgZAFIg28j2aPQf1FKg7MQNp8EwMIhTwt7n7x1g8HWvFsUQyKE29AdhpoV0-_A7XjhJSBlkL0Itdg-hipgvt0_Shfmpk2UBYi0_1DL107UQbaJNF2Vu9W2ltV41SrfSFy8-y_0hgszoLRaBxjiAUrB3r-fXLg6phTIQiDn1tTRnR6BQMpFVURSK4ydxQ2d1LgBYYKFaFAYvBTeeBI5MYqw4DUWIsNIeYXCL6KTahD6iaTzSeKPR_fTW9mZ4HKvVRPP__nn__71z9-_u-_f5Ms9IEA3SMByokWdB0CVDIH575OAvT1xs2ILiKcT_rryMq1YL-0_DMToG0uCwkBuc2OafShQlyhawGzy1f6PgL0y_CfSUXb0y1OmVpeYeZ8lhr-U_JR8kMHB8h5doJCzqILWG0PaWyjwgaMSMdCXAEj4pU5EeRM_PJQ1wIZBAGwAgBeG3rt5PwyGJRPSAgc6ygCEvkcsuykv06dDC0B96r-4DxYBIfGe5M-q9r17pukPXWOjs1mtV4n0p9pIB6d_CIEu_u1dwJI1oFxzoLd0-6_0OyTqxrcJDHNSeMGSBS2h1abnkzWOIPIgsAMnL6X23WLVSf9bibKgtFa39JrH2mUH3SsfyJSFozyUDf7bya6n4hIZ_WeglGjalEwak3ttlitF4x22jR--N70t_G96ZdqrLDJW3Q53VSUi24jl1eSo6fMBjd-nr9MY0WlsgWlHuuRKWcBIZBmvihdmWsxbaQNoP8LskZEZbPWuo-J5wbWY1NMMbfTf68MUlsiyhmA6TQj6Ow_WwPyA9i-v_2z0EipPI6hUWSDucstjG2JK73JBE1oLSz__DRSufmE3aQ4letD4avMC1bKr9EhBzbpq2GTesAVETnU3mkduqL28pHH0EsQerViAtxCwyZi5Aw5ILKXP0hkZO-d9sgSLdjLB5WFFuv6p1ZegrY3KfUBk26cEMW3_AUenlkjGAErIuR3-5FNEwNWOmy4IGxDVi5UJztNg0-K-D0QA7QHS9oBa4e7dqHtJ0JDMSwVQVCvXGjyqU2Gjib3LsNgCABi8EKWRahks3d4eMxKifXDk7FT2wWBHAcIm8NTuN99VIjH_f-fgPpO5cH839ukX_n_vvk_" />
</p>

# §8 Fundamental Groups

**Definition 8.1** Let $X\subset\mathbb{R}^{n}$ be a subset, and fix $x_{0}\in X$. A **loop** in $X$ based at $x_{0}$ is a continuous function

$$
[0,1]\xrightarrow{\gamma}\mathbb{R}^{n}
$$

satisfying the following conditions:

* For every $t\in[0,1]$, we have $\gamma(t)\in X$.
* $\gamma(0)=\gamma(1)=x_{0}$.



**Example 8.1** Let

$$
X=\mathbb{R}^{n}\setminus\{0\},
\qquad
x_{0}=(1,0).
$$

<p align="center">
  <img src="https://tex.xumin.net/svgb/lZXLbtswEEX3-QouUkAGHIHDNxNklW3RRdNFAdsoGIm2hSqSStGN0yD_XoryQ3Zcp_FKIkd3Du8lxtMHuyiqF1_8_NMUmV85-zpZ3-JU8saPn3cPbWZKe3sF43X_BLMLFH6f0KrK6sdHW3lUzJGzv1aFs9do2hi_RAkeU4xHN6GstR7lhVk484hKO_eo9cZ5ZDzCY2SqHC1Ni5a2WCw9qucofHaxaXC_NE1Q_Gozb6pFadGkyK_zRgqlBaecaU24BjFDsX6aO_OEElB6zPEIXV2hhGqdgjx-BYFT3q90xfvX7DkLTW627e9W7rdFn4vKtn1nIwRoSTQWVFNGKR12niyctdUMJSToMTIGgJQE3TRFWV15V5dt2JMiEhDVtewOn1DgsZySbXlCGYlVwOLSGSAtBVOhnGngINhJHsog1R0PO4KhmEd35I6FUOhWVDSkI-nOEkkgHYLce2diXkMWDopwwrDQVCh8lArRUVmrrfNEs6iM_0OZYUYk54JKIqlS_FiaHUvDKem37kkFoJXkMkhrIg7jfChXtnOP0E5MqvQ4y01KCvZRcjpo3Ce59_4MSLjPQLGEECGjXKlTIETHqyPkWxAcd8LV2oHgHnrPMTzFuQtOBEggmENwlUn9MRCiovGMHzvC-Q5ExBKJ3wEhmkomZTBGUgKK8pPZDMUOLREqZsP2JIINbvI_s_lm1x59qXN7eMN2ueoRqsIumpgqW9butqpdmHZPtvWzSVFV1oVx12xG5wy9XOLL1_PagSIghBHwQeH1jxf8-p44SBrGG_mw-PeB8GYI3xUu201gIgSTisoQEAgxGMHzoix7Z8V-fmX9pwlp_OjmYmqr_OA_5y8">
</p>

**Definition 8.2** Given two curves $\gamma_{1}$ and $\gamma_{2}$, we say that $\gamma_{1}$ and $\gamma_{2}$ are **homotopic** if $\gamma_{1}$ can be continuously deformed into $\gamma_{2}$ without changing $\gamma(0)$ or $\gamma(1)$.

That is, there exists a continuous map

$$
\begin{aligned}
\Gamma:[0,1]\times[a,b]&\to X\\
(t,s)&\mapsto\Gamma(t,s)
\end{aligned}\tag*{}
$$

such that

* $\Gamma(t,a)=\gamma_{1}(t)$.
* $\Gamma(t,b)=\gamma_{2}(t)$.
* For every $s$,

$$
\Gamma(0,s)=\Gamma(1,s)=x_{0}.
$$

Here $[a,b]$ is some interval.

<p align="center">
  <img src="https://tex.xumin.net/svgb/tVjbbtNAEH3nK_ahSKnkWt6bL1R95iOSILmJ1UakTuS4KhBV4g4FoYJASNxEHyruhaogWi4qH0Mdlyd-gfXuOvFGRTRonSd7M7vnzMyZmZVr88FCK-zHrbMXuq1GvBwFq9Vzc5bp0G5snB8-9Bp-O5ibgcY58QTrJ2rNyF8BFQgtEzqGa5l0GszMgAqyLdNmC3B69gRgP9UOonFDiDJLbnoSJLsvDve_pG8fpl9_FHZXG512J5rrRwvzBkCUnoqCpgEcMgsWoiAIDQAJe55vLwfsb2SvGoBv63T9Ris-z9gewVMhNCKw_vH4BKA7YoAwzhlATI5iMArM392_fu3XlVfJzevJ2o7gACqD148O13am_0kGEToKh23nZDA9Ohoe52LlsfAEF0tJmg4g5BWdlq_Cab1AWAXCpQFRJY3ytQwgVS8w1wvSDuSoQE5pQK4K5JYWOs9S5O2VB6SqzitLdQiqzQOW5RFCKhAqSwwIq0C4NCCiApHSQkcV1SF6RIO_e49NmPTx1fT9VrL5afBkR5kz1ZleHPjteDE7zYEGtBhp0wSNThhHnXaPrbquibDBBI3caeCHTbbERAdhtiStKxi6JnTZbmaljLaD_ZfJ5p3Bs8sC-Pf3m-m3H2y6_Lx46fD-M_aQrG8ktzZ-f1_TPPuwB7NYUG_MHYIcvk6kL4RiHjMLCU8I5P9DfJxkTUgJ2wLKHuOEMeVppJITpryobUsGd-iKbkYFZ8cYuYIqQTkljIuWRV9kurXdm7DDj_bGORHI122YZ042Csv7j8xNxOjvYbJlVIZhokThNPJlWBSDpxcP9m4PHn4ZvHqjXvkqbJs4zlYvbNgebT_Y20zffhjcfpes74q9YacZAD9Wt_enqlOrs2MGhfOYQT0zUC6hh-_3ks3HyfZ6-nxLFOvgwX6yfTW99fk4BTpR6yLIzrsv95WwDsLfse4mSTAvHwiHSCKbkOpHoiL7UEXiRaIZievMs3IgITNXPw5XjOPlOEjC6sYRtWxL4RNLNETtWsCe6P6uwMGewNF-A5B1b0shYFfgaNcBphzHkToothu9OARKBwSO7eRC1w1UmJECiebFqz10oi0gJ48dLOk6mJ-MSa4GlCPrRnLlBKCKviG2CvNn40a6tV8cO1KdbNibjCGfFlU_bCwyMmEnihfBStCL69VWGAYR6AVd-Y2mXiSc82VzVZDNpiqnymdqgWR_qrbgLy35Z_q1btRaClaHc0oOwYwKu8hORCWbd8A35utjhyHPYpQmPqt2OmOYnVULwqbyreoP">
</p>

You may think of this as a “path animation” lasting $b-a$ seconds.

**Lemma 8.1** If

$$
\gamma_{1}\sim\gamma_{2}
$$

if and only if $\gamma_{1}$ and $\gamma_{2}$ are homotopic, then $\sim$ is an equivalence relation.

Informally:

* Every path can be deformed back to itself by “doing no deformation”.
* If $\gamma_{1}$ can be deformed into $\gamma_{2}$, then $\gamma_{2}$ can be deformed back into $\gamma_{1}$ by reversing the deformation.
* If $\gamma_{1}$ can be deformed into $\gamma_{2}$ and $\gamma_{2}$ can be deformed into $\gamma_{3}$, then $\gamma_{1}$ can be deformed into $\gamma_{3}$ by performing the two deformations consecutively.

**Definition 8.3** Define

$$
\pi_{1}(X,x_{0})
=
\{\text{loops based at }x_{0}\}/\sim.\tag*{}
$$

This is called the **fundamental group** of $X$ based at $x_{0}$.

**Remark** If $X$ is connected, then for any two base points,

$$
\pi_{1}(X,x_{0})
\cong
\pi_{1}(X,x_{0}^{\prime}).
$$

Composition in the fundamental group is defined as follows:

$$
\begin{aligned}
\pi_{1}(X,x_{0})\times\pi_{1}(X,x_{0})
&\to
\pi_{1}(X,x_{0})\\
([\gamma_{b}],[\gamma_{a}])
&\mapsto
[\gamma_{b}][\gamma_{a}].
\end{aligned}\tag*{}
$$

Given two paths $\gamma_{a}$ and $\gamma_{b}$, consider the path

$$
\begin{aligned}
\widetilde{\gamma_{b}\circ\gamma_{a}}:[0,2]&\to X\\
t&\mapsto
\begin{cases}
\gamma_{a}(t)&\text{if }t\in[0,1]\\
\gamma_{b}(t-1)&\text{if }t\in[1,2].
\end{cases}
\end{aligned}\tag*{}
$$

Rescale the interval $[0,2]$ to $[0,1]$ to obtain the path

$$
\begin{aligned}
\gamma_{b}\circ\gamma_{a}:[0,1]&\to X\\
t&\mapsto
\begin{cases}
\gamma_{a}(2t)&\text{if }t\in\left[0,\frac{1}{2}\right]\\
\gamma_{b}(2t-1)&\text{if }t\in\left[\frac{1}{2},1\right].
\end{cases}
\end{aligned}\tag*{}
$$

Define

$$
[\gamma_{b}][\gamma_{a}]
=
[\gamma_{b}\circ\gamma_{a}].
$$

**Example 8.2** Consider

$$
X=\mathbb{R}^{2}\setminus\{p,q\},
$$

with $x_{0}$ an arbitrary point.

<p align="center">
  <img src="https://tex.xumin.net/svgb/vVhLb-s2Fl6nv4KLFEgAheB5kIe8RbrpdtBFZ3Z2UCi26gjXkVxZmeaO4f8-kCjJlu04D-TenUiKH7_v8Lyk6dM6q_Ov_1vm91VafdvMs1lZpXVeFmu9SuuHKlst01leLLY_TV959bGsVg95sUjWD-kqW5_f8JhWX_NiceKtVVrXWVWcWinXebM7srnPFnmxadZX-ax-qrLt5PnWaLGrOvk2PKxn6TK7vYHkOT7B3U8XFxc__7uh-EX9kc3qtFgsMzXJ51_mKxfEeRO8CWTYB3unmren8yr9Ryl1FSBBS9fq5kZdoTXjAVkfBwGG59m32TJTv8Qj6yrNFw-1-ldeZOt4YOrYeZGAzohl52B8oLoCFO0xQQkRGwh0SNCjRnt9DtgEdpbEWcFgnGM5Qo5IA3B30OvIyBgIyPuARsSyP0RGM-aM8FbO7D1YCSDGIzKLO4KOUANyd9Lr0JaMC4ienHhmCUfAYcyZzBs5O0TL5Lxp-ELgQ-AOaGDcnfM6sAfvvScmgWC8OzIFyQFj_0bGEtgDExMQWGOOHKMDGnC7c95iChKPHiiQFz5yZeYxYbZvJBy8cRacZXbiPR0R7oAGXH6ziZ3hYEScCRbRHTmyPQg--9bg82BZJAQE7wTDkbd1QAPuydiLGeq3vJrt0pMLxolFF4yAgbElJn_ly-XtplrcJwqt_VJl80SZX9SiyrKifbpfPmWJMmqrVNK8rMpVOsvrb7eg7tQVuJAQG22vldZqVhZ1VS7X3TwF7eRapcW8ndBOmrn2zSsQo-0w3NsooIniQtwoOCDFjXj6xHa-2TxsjEiMoxO74Yhqy6xZ6KkOSO3GncaD7PzbU_XfbHSJjNZwCAYtWCaifWNPZuWyrA6NDegGcyNAb3Ci1uLtxsHiibr5Vd218canzUck2kuCwWvf6yFCbVyCQtpEq12REU02QWe08PWLakQ8BTHOgEEGHz5XTVSCYls_DodS0PEQSK0OlEZUghbeJQIFgguWPXtDwSH9WBW-DVMC39-GQU2YEFLnwvt3-XIUe-e9sIXggMiT-9QoJorRehhTcd7vojgGYjM3pu4PN7bhExfiRsYBKW7E0yc282x2UdwjMYxO7IYjqjGKYUd1QGo37jS-GsXeirPe2iaQHYw7utMeIzxYG5h7eyO6V6M4tObU5jAtNSVUJCEE7QYrhqCDTVrH72SxQR18EyU6nIkBR9YBgyASGIOfK6gTY50Gn6DrHXtfC2hwCbLT2GtpiKMkyEGLfY8Wz-yCkEOwYM04En6EFkbNTUEHTUOGZWycFIPTvne38dW-WC_EeBIhE9iCcfw9chOTiZ0CHUhhwjY3mT682bmYrPoL8aENNvOygNB0LE1bz2DECPxYARhiRegvgilOcB8dLqblcwpACJENCRgrguMG6dMk7BEZS_BR25C3uK9yvYS2lCdo3Zk67UAso-Xg0XkL3yMk-n6zq747BZYiQXadAusoTnRFwloaCvlLCigEh96LZbaA1n8PBXvePBYQvYhwEMAxLkBeEnCyz0b0AAGb5p0Iw6dWaPb-ZNfbzu_32ez9qM_uRB81ihzMuM_m5g_EXp_djk-d2M7v9dk9Ut9n92Y-LGg9s6HPjpp2ffaexlcrdPN1yz44CwElvKXP_oC_7EXe2GHADAvRYaB1GLTcOcxerLxYkgMb8BCAm48Ft591Jje_Jt1vr6x_yMvidrMu0q_Z9q774o4k6CijIDSNAwK27X6XF0UHThBxd7_WasYEG-fvWar_ZM-1-r2cZztrXoHj5tI0X6uinGdqkhazh7K6LcqqflD_ZOv6bpIXRVapdbbqfqG1FDeXz39uzPZyew4cbYLefgR8dRYYzYeB_z4LTNB2ox8A7n8-pvdPy7TabmbbBvbiYlqVdVpn9-XzJpjt5vL2cjudxqXL6SJ9fEz_3Nxv1XSWVzPVT6Tby5ZUVswHxHO02UqbB_D7EL9pmU_zYo_7JHJVJ9nfvY8-SWiK53vJfyQVjLLAZriB9LwjUzAJNF9_77LtJF3mi-J2mf1V36nNQ_lY1uUqO-vX_hMNcb6PecEQ950h4t3t_Un_Pw">
</p>

The two diagrams on the right both represent

$$
[\gamma_{b}\circ\gamma_{a}].
$$

**Remark**

* The constant loop

$$
\gamma(t)=x_{0}
$$

for every $t$ is the identity element.

If $\gamma$ is a loop and $\gamma_{0}$ is the constant path, then
$\gamma_{0}\circ\gamma$
is the path that “runs through $\gamma$ quickly and then remains stationary for $\dfrac{1}{2}$ second”.

Let $\Gamma$ be the homotopy that continuously shrinks this $\dfrac{1}{2}$ second of stationary time to $0$ seconds.

* The inverse path of $\gamma$ is the loop obtained by “running $\gamma$ backwards”.

<p align="center">
  <img src="https://tex.xumin.net/svgb/vZbLbus2EIbX8VPM4gSwAdmYGVK8JHU3Z1t00-7soGAkxhZiSyolN0kNv3shKXYUR0mKg7Q7kZrLP99wKC13la-z-7832W1w4Wmf-qQIrs6KvJqVrl4HX25ckuWrw2j5iem2COU6y1dRtXalrz522Lpwn-WrAavS1bUP-dCbosoa707NrV9l-b55X2ZJvQv-sHic40zHZR09nR6qxG38fErRY_dEN6OLy12eFNutz2vI7iD4P3dZ8FewbIqAMUZSq8k1XFa-hjRzq-C2sPF3NVS1CzW4GjACl6ewdhWsfbZa11DcgdRqdDG6uPy-C395-CXLfQWLLL1KnRJaaqmMkGRJxvIGRhfLNLgHWCTFpgjzfVjdRsBxfBV8GoGW17AK3ucRkJTXcLvZ-QiYFRwAotaxKF2S1U9ziuCn6Q0AjFlixJonMJtBUuR1KDZVtyuQJq3ebkXY2oyZMBIxTeB6dHH5Wx1cW0hfN2m2hDKOEYUhJV50wyJ11RqeWzXfFznIWVw2HO66p8NNo0lhxIImMJ12C6Gxy_cWEhqBNlZkUaNm0vyVlKBFRGYIUbsr-ojaVYvoyHRYsmYrCRUxoxIS6f9U_NLUdkU00NQByBKt1NIwKckqjs1nkhnNSTOfFOt3BYvBU9jsvggW_VMoPhasjVGWtWQUUiopv05vNP25m5tBxPzqUIj-oTiVOKxYaEFolJKCyTB_NeD35fbGvH8ixKdjbgRbq1CxILLG9Bi3HSWOWNluhgXal9oHg1mSSpJBK4UlJhoI1vifgjWR22DNN-MKvmch2fguVomKjdFaacOGUErVv4Duss3mHCaeUOIRJbYkG-NXJMeMtsPymmO7izMZH1Gincm42evPF567Ec7iZ6vOjegUp3OjoWzd7iw-ZTvG4X62N80-auIXkac4rduxtukUkqeG53uErY1jQ0ojWmY2_IWAm-6-LVmcAxZngMUgYHEOWJwB7tZvsp0DFmeAu2xvrqtzwOIMsBgGPDQPgsiwFlJIJeLYvh4H7s8WfzpbbIlji8oQoRAkBoIdZ4v7swW_-8cafi1Sf7QfM3Ik4niGegJ5kXpYuDxZF2GeF6Few4Ov6ptFluc-QOXL57-p5gKC_bfHP_Z4-HZ4J7JA-k8if8kndczCRizxh9QtV267da28DzX9-yu9GZMYf0TQSQ3cuuT-wYW0OlyPlj5PX_0R_wM">
</p>

$[\text{“}\gamma~\text{backwards”}\circ\gamma]?$

<p align="center">
  <img src="https://tex.xumin.net/svgb/vVnNbtxGDD7HTzEFEsAGZGE45PzFdS-5Fr20N9sHZVe2ha53t1q5SWrk3ofog-SF8h6lpJU00kreulACxIhEazgk9fH7OPL1-_QuWz8V2e9_bbNF8Zinn68-XsrY6m0RfWovdotklV6eQ_SxvoKbk1dvHteLzcNDui5Edivy9I_HLE_fiuttUtyLUxmRNWcX4s0uLcQyS-7y5EGs0ttC7IokL0RSCBmJZL0U98lO3KfZ3X0hNreCV52w73eP-Z-p-DlbpztxlS3fLhODliwZhwQeSNONOHl1vcyTD-JqsVlt8sun_O59JJTWb_N0yRfSXYi7PE3XfH0h3q8eU76w4rOIqlWbbbLIik-cihDiVJGMlFVnIo7FYrMu8s1qx1Z0EUqIjT2rQmWDiRBkZeAnTxXICDWciYuTV6NRS4fSawNeWmkVWDVb2NH5T1Xk4MYir6wITdi0v6tiblKditkqTyANKCUNkoR5Kz0VL3GlZVfp2uCnKv1rkScVZMLAEcApi4RkUGvfxV29YVCRMv5MnJ_zjfRhDUa9KQ9Ke2kcgEQEHHFXemjdlb4nYeARvdQgnTIGpTPHamqpLSkQtUVVZgQIP55HNYaRK-swhiGKlYstF9PFpsGDsjGXF0yM2JY31s9D2ZMzjH0nLSH_HEXyC1KogeFhNPzKTjZ2bfRlnntDFXuX91TsSiMq0Gis4VdJ-vvFjjEzhrSxpQbYtQl0THai9r_eJ1tm0ndZvlildQZb5j30pIH5hKTTIYBus9VqmIBs45dN9LKMvXy2i13c1Nit9u5jprLKmHRTdO5FKqOUYT_K4bIylf1T9TKA1k-9DMZ2q628tF2296PC3Q7ouYlJdUG2fqplTW7cpItPZTEnAYKaiHUFkQvt-Hpe0kMcIz1UA3lB1ZcXZsTnexJQsSo6VhnnnHZudnVRdlQXbaguVXKNurSZToXMFZYs5Np7ZlVmlKNU-EIhnwqYixop7wMtt1hRYKswnouNqrp_RhaMBGkRjdcA1vmByqAMVcYfVxnnkZjMtHIsN8bSiLtWZfwRlQFDrH4SPQBxNm5OmmtVBtW4yiAMVQZhqDKlwh9RGS4scv-BQs_taEPZnYeqlRun6soeykyVaCAzQeJTwRurtTGSwABJ1pvvFzsOsO3Km5ZcJuA9JjNkFIu7Rx5WWW1wPpnxozLjhzLj-zJTM-BQZiochTLDifZkpr4_gKgcyEzrR4W7HZCHH8qMH8iM_68yY1GjdfyDlodBcxTcL-M-kiX3HWTtXX-0Rm96ozXafY0neYURYUBqzdCQzFLzUXYjM2hcpHQrfEHwlgmQqAu-ypE61m5zfo6yWXLK0VUDef5_SNnlFsbVHMv0c9SdlvwEaa4HO4ZwFGvdWQjclb4nGYOhwMcBnut4qiN9VMD_F2P7krjUIWO7mrFtx9iuZmzdMbatGfsZdHinrVSkgJWRGCFzk940NhxXWrsA2K58ssNGkPg047FmKuIjnvZa88t0szFe-e5HqKu0ku8Yjw0Vu5APunF_E_ZBzVTlL-plFlo_9TIY2622Boy396Mh7P0DZOxj0tAF2fqpljW5HWU8LyWfNllUeMLm09fM0x5pBjAfZg-wQVpxeXTAG7osazdaE8HR7wnOEo96io9dhoN30G90IuxGPep_VBmnIQJttbWOV2rn1Yi7ZtSr3D036mme8pQh_oempOe5m266sMRqQsGZhYjVBH1Y2OOn2Wr2B1XOSDxWoZ6t6crCHeK5tgZjBht6Y0aNhmH3VKmEYwaR6o0Z9f3hbqo_ZrR-VLjbcMxoY1JdkL0xo81t0HTj38JcOWowo3H_Ge8G5wotA80jf1zzLB_UJGs_6xQ3BZkRd43mVe46zRt7_Xxs9VoBlmdAKa2d7_X7Mc6trCHnku9zrpZjnKvlgHO17HOulmOo2Vu719_4aThXy1GM-gHn7nPpXr8f51zxW_qxEL9slmlTxFOWSz4PyFhyT675F-IqWS_umRXWm7y4Fx_SXXFzla3XaS526Xb_lb_s-6fX17vs4fXni3G_3LTfxC-V3089zO-X5e-bxFt-nN5PHi_we5Wssrv1ZflXkBvx9PXLl6___P0D73Gdrpe9P8H8Cw">
</p>

* Finally, path composition is associative.

<p align="center">
  <img src="https://tex.xumin.net/svgb/jZTdbpwwEIXv9yl8sReLBIifJa1U7ZOEVTTAFKwFGxmTqLH87pXB_KWwzd3Ycw4-40847TuU9PFZ00yA-KMKzLkASTnr_BZkJbCtIaes1Kf0P9KGi7airHS7ClrsnhsaEA_Kyh1VC1KiYHsd3lHjHtNkWFKmTL-luewF6lNaCPggl8AN_MjxvEvgeqYijBf4mmHNP-7qHJz1r0kZ-cmkNeWOOv0tIFehVpFe-WL_x2wc6kNnrNV17Zxtu55wpQzcwPG8oQ0Zf0fzyRKaBt5UqM96zO5sRzkymPA29OKw6wNLbCxWPwgELSt5JyCHIRyizrfLIk5zKvLlOGezNnlXc3lX_6fFkzzFE_pRMqvHRXJMaHPPA07rHOrkm2gX265pS8hLniKykZcrnzaeYRrTbsEeWyym0fAvJy8ZQH3hNHOLttxC7czjvWZ1j_dhyGDGdfVfnC99-9-sEO2IYvuPLDh2RItip-_ax2MuKGc31TF4oEugaWsq-wJvgR_ljUs6LBtkktTISlndYrOr7-TyMh2xuik1vSSQ9TUIrWp9qnjDJW9pTt4ppOlJYJdDTVlJJG3wlCIrZr02X7Vxh_3Vi_QX">
</p>

**Example 8.3** Notice that $L$ denotes a line.

<p align="center">
  <img src="https://tex.xumin.net/svgb/pZNNboJAFMfXcIpZUCONkAGkxjS9QVftro6aAaY6KYIBjKaTWfUC3XTdI7SLbmrS9DT24xadUUCMmn4thveG__-9l_kNII8MaMQy7E1CnHDm-1xVNDSmfWbx-nlPPBuzPoNc14Cq1MRCeQW9uh5TP5skhHc8nJKQRuSE1WHDgKal8y44AB-PN293T4v7-fv8dfHw_PlyqyoKChI8BcIHdeDTxA-J2Jiufiy1SxqGoM6gabvgEFhmy7G5bGi7VTPM3VEckI5HwngKEjoYZl2As33VTFseQ-OiEpEo2DiAeIXESuIMZ8SLZ8xoQ8405MfRQOOgBqRcUkEjnA09j53xHrM5Skk2otEkRQyiNSxR8x0oaDqC0yYSw6jbRbBkgEWAVUKW6TYktH1QKjhKa5VAOdJsL1VXzrBED7FpCm8hNKuCu5ywBe-v6Jw1OnD6O27NXdysIuzktrIWHvNIHsgucyvPRVHVXZFgNS9sxWWItkL-2W0U3u3r6AQ4HZKgK-tb8rt1VoNbYu4_0OfIL7iEq6565L_7Fw">
</p>

We do not prove the isomorphism here, but roughly speaking, we can send a path $\gamma$ to its **winding number**.

“How many times does the path $\gamma$ wind around the hole?”

<p align="center">
  <img src="https://tex.xumin.net/svgb/7VfNjtJQFN73Ke4CDJO0Te8ftCI-iFMW_bliM6UlpURG0o0x0a07ExczcefCZOLOMerTMMS38MK9_aE4BHAYYyKbHs4993zfPfd8h2K7bBBEs9RxJ6GTZDMvzBSlCbTaBwyc4dABN-8u51ev5m-ugLERotgyVXD2YhR46SRh2em0Z-gdOkrV88IYe07IehpUp8KCquuMWRhErNfyJknCohS48STyg2jAjanucQ9LTvoKAE2wuPg4__aBm7afOM9BCxGotzsqQsvHCScFWsS4zYUhLlxyY8XlnXshA90VzPzievHySwmDYXuVs90pEmAkYDpo6epuxoqFtVixXUDcvP-8uP6Rbzv14jBOerNk4KqI0ocJ89UO6YJBwlikQsJNN5wwfqp21ucJscCgxgpD14EXR2kSh-NyrS3WnMjnPkKEj-TxLUwt4bLW-O9DRNUe78iFkCoXQ_hgyYXIMHwwl349S40IpWLNqhChkjjcUpQm-Pn1baUXTp8GYdhzQ8c7W0LSTh4PeQovSJY91II6b_X8ni9fLz59r7SH3IKRbvAtUeyz_c45a9grNTayrmKzyF9TnPIANOyhMxqnMTDsIOJ2-sx1Z0-yBrDtnaQN_4q0j6nnY4n4oCaF5iqLRTekIg5gkKI9RS9aqOxNWPTaYYJ9pG2jgCwxJcxcITxEKkpSwIZOEP--fBxegrUstSoQq0CUVRBkKdo-tvaT5b1rEe4sP-2__u5Bf9gwfv9ThQ18PP31y4OZt4oP5fiy7sSS-AjnSvgT8ZdZ6spDusVVSYlOzeOL767Klr_ZmOW1IQEP85kl6BB6MDgfiCKlWb8z05SvNzm47FF8R-BrA3uzZ5El4KxiYEtCskxV5v_MqNTkrBRB8o_JLw">
</p>

**Example 8.4** If the path $\gamma$ is differentiable, this can be defined by

$$
\dfrac{1}{2\pi\mathrm{i}}
\displaystyle\int_{\gamma}
\dfrac{\mathrm{d}z}{z}.\tag*{}
$$

**Example 8.5**

<p align="center">
  <img src="https://tex.xumin.net/svgb/rVM9T8MwEJ2TX3FDVCVSYjlpA0KoKxMTa12Qk7ipRWqnjqsWLP935Hy0RXQBMZ19X-9875kUrObCaFocGqqsKUvrewFp-ZtJbUi42OiP-PRmsI0C8L0ZEOJ7RElNNSvkySQP2JqAlFLUgYUhfCnfUb0tCvNiX01mScf0jotDR0wb74k9t50BGafg758tL_VBMbsqaMcaLtjShDjGaB7Zte95pFL0CM4TJUmYTSZ1Bk8GR48udcObBsI0xug-j6DkqmwYhBjhfIgLWbGV4vVWr4HqS6YJ-skC-zgBrk7dlm_0MsFoUe7WrslDHmOU5w4wRdhdFnl0DiyuA_k1XsEaeRzwMLpzUQfY3gD7V6wULSasfY9FmKi-LfyvxM4vxEL4TNya4fnVkFbxHbPR71jOepZvkjzv53d3NBwhzJCjbHz-5EyvnNNhlMSw3op2W1atXWYWJ-d6JyLXsvfNB6AsTlHWV45aQm7Ht9T0U0zn1Gs5_WXrT4oxqJU8tB1IAfoooWaCKaql6mAG_tB0_MFf">
</p>

<p align="center">
  <img src="https://tex.xumin.net/svgb/7VxLb9tIEj7Hv6IPk4EN0EJ1PfqRjPcy18XsYedmewFZZmxhHckryzvJGv7vi36QaomkrQRSkhnoFKubrK5qNqvqq6-Yi6v6Zjp7mtSzZb14PrpIP5fjq8e78eL56eOTm3x8nqwmpv_-3_10snxc1M_nn85gZOV-WX1u_3iYjO_qs1NdfUp_6epq_FDfTWf12fHkcbGoZ0t1NX-cXU9nN-pq_mmUFj65PHrz9nE2mX_8GK6YflCL-j-P00X9Tl3cj5e36hgqtubkvXr7UC_V9XR8sxh_VHf1h6V6WI4XSzVeKqjUeHatbscP6rae3twu1fyDYmuO3hy9efvP5WIcx_4-ndUP6nx6_e56DMaIB81omVBI86U6enNxvRj_oZRSx6TtyNgKTfjnRJ2eqmNCjEM2_HOi3g9KFqPJghXvPaIT2y85iykl58UGJZMFZ4x3ZJ22LEY2JLuuzn47na0l0MykvbC34voFlyr7rsq34_v6nfp1upjc1UnuvQfjLGliEQ6yC7nnH6Z3d2dPi5urSqHIu0V9XSl4r24WdT2Lf13dPdaVAvWsVBUuVvP78WS6_Hym1aU6JkkaCI3kRI1GajKfLRfzu4diKukbTsYxCY2Iw1i8-JiERzpe07lZ0mi-TUpR6U4ZXjhP8YhoJaFCkY1Vu3dm9aS9DUtR6ebS4tNTNfkcdjrs_a-Pi__Wa09Ue2YhIG0JHYAuzsr5ZH43X2xuveV27zVzs_uIJu5_vLHcf6WO0fsK2W1agqZCbs0gqZD9hvV8MqC10YSIwMjOWNFml0pXv5xGtYnMiKhCZzc0R-fDBLHJyiNLvFKy_q3B_co7CyDWOesdeU9-9zs-qDqxq9D4ZtObX32brn6vPy3Vb_PrutHuGB1EsaRHfKJm8-tanY9nk9v54mw2Xyxv1R_1w_LyfDqb1Qv1UN9nl3-pvsqwNZuefhr_9Pz-6KKeXa8FmKOf1TcIO30ui5wlYWBvHLBot1OfxS49C919fmkCRrr1WOxGjsPbk_3MMbGPDqAdKf0HjKSZyQ5ElyLTUdADy7cTI7fyW1Gi4Obq7ci6-kHVNNVqX4hMAlbWr3mv_mAnxqMHNs6J0bYMSefX44dbdT9eLuvF7OxpPlM8kvsQ8z-kv54vw8sCSV9MyzW_SeNLIRYYnBirLZOzmu2XL-vSsmGZuGz6TUBDfgMdkKawshHtUds9OA5Mgcl2vDVziuPcPjeTBnTfU-83QJxoQ-yMMdpoTTv12qd_y8FGRz38pgXok8K-fW8A0kDjAAvbB6KORRHwDhDAyE4jZdp9aBOzzTenMcs2vhuta81JMWdl94D2LIzCIBo8WrOPsKOTDoa7bz4ldXl1frAd6PqsPk9_cXFAGAeEcUAYB4Txp0cYOVYdEMaPhTD-9XSqn3thhvoGMGN_xa2jrQHMAb_sD79sDV92iF62By87xC5bQpcfFblsB1x2XW3aFW7ZCrbseOt3BluOtoMtfyLUcgAtB1pkO1qke_I9emMNgQlUi9GvAnbtVmcfiZqzr4lfykGLTHj97FuqkLEgKFxIp3VLMzCEC1z83W8Ak2YBIY3eMdFuDUgvb6nEmvpMOtfkVxYwhQuD81lxJR7LoX47QFs0oq0Dds4j2p0a8spjMMkOa4onYdOY5y-z4wVMs6YS7BKd9KWeVoMzaI1YjUY7sw-6r8ebr3ixLt3Xcmf5OWyep0TPFWRdKWqN7utbOE_JJt1nNlY1Q3Rfs9WtDdKl--LCawXzHg6HISEsNCNLu4BYr5z1DYx1NQivDtW9Q3XvECh_mEC5StHDnQT8fYJjTtElpE8dMtsELVsg7nRF4LfSmZy1GPpYwFp03tI-SpJrMbGsSlKcCPExY4rkkVs82hjbr7sVi8YZ8tp7xh2TaK-3PoT8oiJqAFHqO6nQZfK3NHuo4QQdInIoQQgZ_qbdD2FnV5UbU6HZbLgxw1nLXhn_wfyvTUgKxl-4KEOtJR89NSsxK34-325LkfnMDS3fTJSMf5IondWbkXX1o6pxqtW-ELlZjn-d8fdeGyBnUJi107jZNxfrJeR9wZdFhM8QKkEvkPqMzoGAF-_AAOtewY2UQnCz1rBgq61GZq-1c0KbnX6uq7HfTmNkJnYIxmtNpA31Sl5T2XdU7gFu7DwbbYwXx4hmP8jTRldBrpP0JixDFppwBNFlks0eknUCIMHpDAQlDD1tVjv0RrMBR_sIqYUafSGVfJO1M0a3yToV945ZRwsZh3s9CLVYFBABzWWj5Q6DqgkvXc_-2xiQOMTP9Mb6NICyrfrhtdTOWY3WG_F-D60q6CKYp66XTwiHQj00M3_BX8XjlJwNtXYPaK-NDq-VA2eJwdq9xFiOkZOx43Ih7W4ITRle59OTvWVp-VBOhqANgZAFIg28j2aPQf1FKg7MQNp8EwMIhTwt7n7x1g8HWvFsUQyKE29AdhpoV0-_A7XjhJSBlkL0Itdg-hipgvt0_Shfmpk2UBYi0_1DL107UQbaJNF2Vu9W2ltV41SrfSFy8-y_0hgszoLRaBxjiAUrB3r-S3Lg6phTIQiDn1tTRnR6BQMpFVURSK4ydxQ2d1LgBYYKFaFAYvBLeeBI5MYqw4DUWIsNIeYrBF-NBwVTSnwJzReKPR_fTW9mZ4HKvVRPv_3jd3U7_zhfzu-nk78kE30gQfdIgnKiBl2HBJXMw7kfkwR9vXkzIowI6ZP-OjJzLeAvLf_GJGibz0JCQW6zaxp9qBJX6FrQ7PKVvo8E_T4caFLR9nSMU6aXV7g5n6WGA5V8lPzQwQFynp2gkLPoAl7bQyrbqLABJdKxEFdAiXhlTgY5k7881LlABkEArACA14ZeOzlfB4XyCQnBYx1JQCKgQ6ad9Nepm6El4V7VH5wHi-DQeG_Sp1W73n2TtKfO0bHZrNbrRAo0DcSjk1-EYHe_9k4AyTowzlmwe9r9Fxp-cmWDm0SmOWncgInC9tBu05PNGmcQWRCYgdM3c7tus-qk4M1EWTRa61167UON8qOO9c9EyqJRHuoigGai-5mIdFbvKRo1qhZFo9bUbpvVetFop43jh29O_xrfnH6v5gqbvEWX102Fueg2coklOXrKjHDj5_n7NFdUKltQ6rEemXIWEAJp5ozSlbke00baAPy_I3NEVDZsrfuYeG5gPTbFFHM7_ffKIrVlopwBmE5Dgs7-szUgP4Dte9y_CZWUSuQYmkU22LvcxtiWudKbTNCE1sLyb08llZtP2E2KU8k-FL_KvGCl_BolcmCUfhhGqQdcEZFD7Z3WoTNqLx96DL0EoV8rJsAtNGwiRs6QAyJ7-aNERvbeaY8s0YK9fFRZaLGuf2rnJWj7k1IvMOnGCVF8y1_g4pk1ghGwIkJ-tx_aNDFgpcOGC8I2ZOVidbLTNPikiN8DMUB7sKQdsHa4axfafiY0FMNSEQT1yoUmn9pk6Ghy_zIMhgAgBi9kWYRKRnuHh8eslFg_PBk7tZ0QyHGAsDk8hfvdR5V43P9_CqifVR7M_8VN-pX__5v_Aw" />

§12.1 Simple Groups

§11 Short Exact Sequences and Semidirect Products

§11.1 Extensions—Short Exact Sequences

§10 Isomorphism Theorems

§10.1 The First Isomorphism Theorem

§10.1.1 The Quotient Map as a Group Homomorphism

§9 Quotient Groups

§9.1 Quotient Groups

§8 Fundamental Groups