## Introduction

Q: 为什么要学群论?

A: 因为好玩.

本帖旨在用小学生也能看懂的语言（英语: ? ）来向您介绍群论的知识. Let’s Start!

## Basic of Groups

Recall that a **binary operation** on a set G is simply a function from G^2 to G. Once we have defined one or more binary operations on a set, we get an **algebraic system**. For reasons that will be clear soon, we are particularly interested in a specific type of algebraic systems, namely **groups**.

回忆一下建立在集合 G 上的**二元运算**, 它就是一个函数 f:G^2 \rightarrow G. 一旦我们在某个集合上定义了一个（或多个）二元运算, 我们就得到了一个**代数系统**. 出于之后你就会懂的原因, 我们特别关注一种特殊的代数系统, 它就叫**群**.

# 以防你之前没学过的说明

**二元运算(Binary Operation)**: 取 a,b,c \in G, f(a,b) =c 是一个二元运算. 例如, +(2, 3) = 5

**代数系统(algebraic system)**: 非空集合 G 和 G 上 的 **封闭的 (\forall a,b \in G, f(a,b) \in G)** 二元运算 f 组成的系统 (G, f) 称为一个代数系统

#### Definition 1: Group

A **group** is an algebraic system (G, \cdot) where G is a set and \cdot is a binary operation on set G, that satisfies the following conditions:

一个**群**是一个代数系统 (G, \cdot), 其中 G 是一个集合, \cdot 是一个建立在 G 上的二元运算, 满足以下条件:

- There exists an identity element 1 \in G such that for any a ∈ G, 1 \cdot a = a \cdot 1 = 1.

存在一个单位元 1 \in G 满足对于任意 a ∈ G, 1 \cdot a = a \cdot 1 = 1. - The operation \cdot is associative, i.e., for any a, b, c ∈ G, (a \cdot b) \cdot c = a \cdot (b \cdot c).

二元运算子 \cdot 满足结合律, 换句话说, 对任意 a, b, c ∈ G, (a \cdot b) \cdot c = a \cdot (b \cdot c). - Every element a ∈ G has an inverse in G, written as a^{-1}, such that a \cdot a^{-1} = a^{-1} \cdot a = 1.

任何 G 中的元素 a ∈ G 都在 G 中存在一个逆元, 记为 a^{-1}, 满足 a \cdot a^{-1} = a^{-1} \cdot a = 1.

For simplicity we often say G, rather than (G, \cdot ), is a group.

为了方便, 我们经常用 G, 而非 (G, \cdot ), 来表示一个群.

# 有关群的性质的说明

事实上, 一般的教材上群有四条性质, 上面没提到的是**封闭性**. 也就是说, \forall a,b \in G, a \cdot b \in G. 这里封闭性已经在代数系统上提及了, 所以没有额外说明.

This definition is pretty abstract—-that is actually why this branch of mathematics is usually called *abstract algebra*. You might prefer concrete examples to such abstract definitions. Here is an example you might like.

这个定义相当抽象——这就是为什么这个数学分支通常被称为 *抽象代数* 的原因. 大家都更喜欢具体的例子而不是这样的抽象定义. 让我们来看一个你可能会喜欢的例子.

#### Example 1: The Klein^{[2]} four-group

The Klein four-group is (\{1, a, b, c\}, \cdot ) where

克莱因四元群是满足以下条件的 (\{1, a, b, c\}, \cdot ) :

- 1 \cdot a = a \cdot 1 = a; \ \ 1 \cdot b = b \cdot 1 = b; \ \ 1 \cdot c = c \cdot 1 = c;
- a \cdot a = b \cdot b = c \cdot c = 1;
- a \cdot b = b \cdot a = c; \ \ b \cdot c = c \cdot b = a; \ \ c \cdot a = a \cdot c = b.

Do you see it is a group? (Why?)

你觉得这是个群吗? 为什么?

思考一下群的四个条件.

The \cdot operation of Klein four-group is commutative *(why?)*, which is not required by the definition of a group. So let us see another example where the group is not commutative.

克莱因四元群上的 \cdot 运算满足交换律 *（为什么? ）* , 这是一个在群的定义中没有要求的性质. 因此, 让我们看一个不满足交换律的群.

#### Example 2: Geometric transformations

Consider a regular triangle ABC. There are altogether six geometric transformations that map it to itself:

考虑一个正三角形 ABC. 总共有6种将其变换到自身的几何变换:

- The identity transformation, which does not do anything.

单位变换（什么也不做） - 120° clockwise rotation.

顺时针旋转120度 - 120° counter-clockwise rotation.

逆时针旋转120度 - reflection.

镜像翻转 - reflection followed by 120° clockwise rotation.

镜像翻转, 然后顺时针旋转120度 - reflection followed by 120° counter-clockwise rotation.

镜像翻转, 然后逆时针旋转120度

Let G be the set of all these six tranformations, and \cdot the composition operation. Do you see (G, \cdot ) is a group? Why? Do you see it is not commutative? Why not?

令 G 为这六种几何变换的集合, 运算 \cdot 是将运算进行复合. 你注意到 (G, \cdot ) 是一个群了吗? 为什么? 它符合交换律吗? 为什么不?

# Solution

We skip why it is a group and jump to why it is not commutative. Observe, for example,

that we can’t switch the order between a reflection and a 120° clockwise rotation.

为什么它是一个群太简单了, 我们跳过, 直接解释为什么它不是可交换的. 注意到, 翻转后120° 顺时针旋转 不等于 120° 顺时针旋转后翻转.

#### Example 3: Matrix group

Let M be the set of n×n non-singular real matrices. Let × be the multiplication of matrices. Then (M, ×) is a group.

设 M 是 n×n 非奇异^{[3]}实矩阵的集合. 令 × 为矩阵乘法. 那么 (M, ×) 是一个群.

After playing with the above examples, you should be now more comfortable with the abstract concept of group. A natural question that may arise is whether the identity element and the inverses of elements, which are required to exist by the definition, are unique or not. The answer is positive.

在和上面的例子做游戏之后, 你现在应该更熟悉群这个抽象概念了.

你可能会自然而然问出一个问题: 单位元和逆元是否唯一?

答案是肯定的.

#### Proposition 1: The identity element is unique

In a group, **the identity element is unique**, and the inverse of each element is also unique.

在一个群中, **单位元是唯一的**, 每一个元素也只有一个逆元.

**Proof:** Suppose that there are two identity elements 1 and 1' in group G. We easily get that 1 \cdot 1' = 1 because 1' is an identity element; similarly we can get that 1 \cdot 1' = 1' because 1 is also an identity element. So, 1 = 1'. Since any two identity elements must be equal, there can be only one identity element.

假设群 G 中有两个单位元 1 和 1'. 我们很容易得到 1 \cdot 1' = 1, 因为 1' 是一个单位元； 同样地, 我们可以得到 1 \cdot 1' = 1', 因为 1 也是一个单位元. 所以, 1 = 1'. 这说明任何两个单位元必须相等, 因此只能有一个单位元.

Suppose that a ∈ G has two inverses a^{-1} and a^*. Clearly a^{-1}=a^{-1} \cdot (a \cdot a^*) = (a^{-1} \cdot a) \cdot a^* = a^*. Since any two inverses of a must be equal, there can be only one inverse.

假设 a ∈ G 有两个逆元 a^{-1} 和 a^*. 显然 a^{-1}=a^{-1} \cdot (a \cdot a^*) = (a^{-1} \cdot a) \cdot a^* = a^*. 由于 a 的任意两个逆元必须相等, 因此 a 只能有一个逆元.

Another natural question is about the operation laws. The most frequently seen laws include the commutative law, the associative law, the distributive law, and the cancellation law. The definition requires every group to be associative, and Example 2 tells us that a group may not be commutative. Since there is no second operation involved, we can’t discuss about the distributive law. So the only remaining question is about the cancellation law.

另一个自然而然的问题是关于运算律的. 最常见的运算律包括交换律、结合律、分配律和消去律. 群的定义已经要求每个群都满足结合律的, Example 2 已经告诉我们群可能不满足交换律. 由于不涉及两种运算, 所以我们不讨论分配律. 所以剩下的唯一问题是: 群是否满足消去律?

#### Proposition 2: Cancellation law

**The cancellation law holds in every group.** For any group G, for any a, b, c ∈ G, a \cdot c = b \cdot c implies that a = b. Similarly, c \cdot a = c \cdot b also implies that a = b.

**每个群都满足消去律**. 对任意一个群 G, 对任意 a, b, c ∈ G, a \cdot c = b \cdot c 可以推出 a=b. 相似的是, c \cdot a = c \cdot b 也能推出 a = b.

**Proof:** a \cdot c = b \cdot c \Rightarrow a \cdot c \cdot c^{-1} = b \cdot c \cdot c^{-1} \Rightarrow a=b. The second half of the proposition is proved similarly.

Closely related to Proposition 2 is the result below about equations.

与命题 2 密切相关的是以下关于方程式的结果.

#### Proposition 3: Equations

In any group G, for any a, b ∈ G, the equations a \cdot x = b and x \cdot a = b both have solutions.

在任意群 G 中, 对于任意 a,b∈G , 方程 a \cdot x = b 和 x \cdot a = b 均有解.

Propositions 1, 2, and 3 are about properties of a group. Please study the groups in Examples 1, 2,and 3 to verify these properties.

Given Propositions 2 and 3, you may be curious about their converses. Does the cancellation law, or the existence of solutions to equations a \cdot x = b and x \cdot a = b, imply that we are working in a group?

Before we answer these questions, we should first introduce a few new definitions.

命题 1、2 和 3 是关于群的性质的. 请研究Example 1、2 和 3 中的群以验证这些属性.

有了命题 2 和命题 3, 你可能会对他们逆命题感到好奇. 如果我们发现某个结构满足消去律, 或方程 a \cdot x = b 和 x \cdot a = b 的解的恒存在, 是否意味着我们研究的对象是一个群?

在回答这些问题之前, 我们应该先介绍几个新的定义.

#### Definition 2 Semigroup

An algebraic systems (G, \cdot ) with the operation \cdot being associative is called a **semigroup**.

拥有在 G 上封闭且满足结合律的二元运算 \cdot 的代数系统 (G, \cdot ) 被称为一个**半群**.

#### Example 4 Semigroup of singular real matrices

The set of n×n singular real matrices is a semigroup (with respect to matrix multiplication). It is not a group.

n \times n 奇异矩阵的集合是一个半群（运算是矩阵乘法）. 它不是一个群.

With Definition 2, we can now state our question more precisely: In a semigroup, if the cancellation law holds, or the equations a \cdot x = b and x \cdot a = b always have solutions, can we say the semigroup is definitely a group? Answering this question requires a technical lemma on left/right identity elements and left/right inverses.

有了定义2, 我们可以更加准确地描述我们的问题: 在一个半群里, 如果满足消去律, 或者方程 a \cdot x = b 和 x \cdot a = b 恒有解, 我们可以说这个半群是一个群吗? 回答这个问题需要一个关于左/右单位元和左/右逆元的引理.

Recall that the definition of identity element requires that, for any a ∈ G, a \cdot 1 = 1 \cdot a = a.

回忆一下单位元的定义. 它要求对于任意 a \in G, a \cdot 1 = 1 \cdot a = a.

#### Definition 3: Left/Right Identity

If for any a ∈ G, a \cdot 1 = a, we say 1 is a **right identity**. If for any a ∈ G, 1 \cdot a = a, we say 1 is a **left identity.**

Hence, an identity is actually an element that is both left identity and right identity.

如果对于任意 a ∈ G, a \cdot 1 = a, 我们说 1 是一个 **右单位元**. 如果对任意 a ∈ G, 1 \cdot a = a, 我们说 1 是一个 **左单位元.**

因此, 单位元事实上既是左单位元又是右单位元.

Similarly, recall that the definition of group requires the existence of a^{-1} such that a \cdot a^{-1}=a^{-1}\cdot a = 1

相似德, 回忆一下群的定义要求存在 a^{-1} 使得 a \cdot a^{-1}=a^{-1}\cdot a = 1

#### Definition 4: Left/Right Inverse

If a^{-1} satisfies a \cdot a^{-1} = 1, we say it is a **right inverse** of a. If a^{-1} satisfies a^{-1} \cdot a = 1, we say it is a **left inverse** of a.

Hence, an inverse of a is actually an element that is both left inverse and right inverse of a.

如果 a^{-1} 满足 a \cdot a^{-1} = 1, 我们称它为 a 的一个**右逆元**. 如果 a^{-1} 满足 a^{-1} \cdot a = 1, 我们称它为 a 的一个**左逆元**

Hence, an inverse of a is actually an element that is both left inverse and right inverse of a.

因此, a 的逆元事实上既是左逆元又是右逆元.

#### Lemma 1

In a semigroup, if there are both a left identity and a right identity, then they must be equal, which means there is an identity.

In a semigroup with an identity element, if an element has both a left inverse and a right inverse, then they must be equal, which means the former element has an inverse.

在半群中, 如果同时存在左单位元和右单位元, 那么它们必须相等, 这意味着存在单位元.

在具有单位元的半群中, 如果一个元素同时具有左逆元和右逆元, 那么它们必须相等, 这意味着该元素具有逆元.

上述引理留作作业, 请读者自行证明.

Using the above lemma, we can easily get the answer to our question: For the cancellation law, the answer is yes for finite semigroups, and no for infinite semigroups; for the equations having solutions, the answer is always yes.

利用上述引理, 我们可以很容易地得到上面的问题的答案: 对于消去律, 有限半群的答案是肯定的, 无限半群的回答是否定的；对于方程恒有解的情况, 答案总是肯定的.

#### Proposition 4

If a finite semigroup satisfies the cancellation law, then it is a group.

如果一个有限半群满足消去律, 那么它就是一个群.

你已经是一个懂群论的小学生了, 该自己阅读英语了

**Proof:** (From now on, we often omit the symbol \cdot for operations.) For a ∈ G, consider the set aG = \{ax|x ∈ G\}. How many elements are there in aG? Since the cancellation law holds, for any b \neq c, ab could not be equal to ac. (Otherwise, by the cancellation law we would get b = c which contradicts b \neq c.) Hence, |aG| = |G|. Since aG ⊆ G, and G is finite, we get that aG = G. Similarly, consider the set Ga = \{xa|x ∈ G\} and we can show that Ga = G.

Since aG = G, there definitely exists e ∈ G such that ae = a. Since G = Ga, for any b ∈ G, there exists c such that b = ca. Hence, be = cae = ca = b. This implies that e is a right identity element. Similarly, we can show there is a left identity element. Consequently, there is an identity element, which we write as 1.

For any a ∈ G, since aG = G, there definitely exists x ∈ G such that ax = 1, which means a has a right inverse. From Ga = G, we get that a also has a left inverse. Combining these two things together, we get that a has an inverse. Consequently, G is a group.

The notations(记号) aG and Ga, which we just used above in the proof of Proposition 4, are quite frequently used.

#### Proposition 5

There is an infinite semigroup that satisfies the cancellation law but is not a group.

**Proof:** One such semigroup is (\mathbb Z_+, +), where \mathbb Z_+ is the set of positive integers and + is the addition operation. Easy to verify that it is a semigroup and that it satisfies the cancellation law. However, it does not have an identity–note that the natrual number 1 is not an identity in this semigroup! Hence, it is not a group.

#### Proposition 6

In a semigroup G, if for any a, b ∈ G, the equations ax = b and xa = b both have solutions, then G is a group.

**Proof:** The proof is analogous to that of Proposition 4.

First, the equation ax = a has a solution. Let’s call it e and we know ae = a. For any b ∈ G, the equation b = xa has a solution. Let’s call it c and thus b = ca. Hence, be = cae = ca = b. This implies that e is a right identity element. Similarly, we can show there is a left identity element. Consequently, there is an identity element, which we write as 1.

Now, for any a ∈ G, ax = 1 and xa = 1 both have solutions, which means a has a right inverse and a left inverse.

So, a must have an inverse.

Consequently, G is a group.

Having studied the basic properties of a group and their converses, we conclude Section 1 with a less straightforward example.

在研究了群的基本属性及其逆命题之后, 我们用一个有点弯弯绕的例子来总结第1节.

#### Example 5

*[5]* Suppose G is a group and x, y ∈ G (x \neq y, x \neq 1, y \neq 1) satisfy that x^3 = y^2 = (xy)^2 = 1. For any non-identity element a ∈ G, there exist a positive integer n and elements a_1, \dots , a_n such that a = a_1 \dots a_n, and that each a_i is either x or y. Find |G|.

考虑 G 是一个群, x, y ∈ G (x \neq y, x \neq 1, y \neq 1) 满足 x^3 = y^2 = (xy)^2 = 1. 对于任意非单位元的元素 a ∈ G, 存在一个正整数 n 和一些元素 a_1, \dots , a_n 满足 a = a_1 \cdot a_2 \cdot \dots {} \cdot a_n, 其中 a_i 要么是 x 要么是 y. 请求出 |G|.

# Solution

For any non-identity element a = a_1 . . . a_n, we easily know that these a_i s are a sequence of y s with x s or x^2 s between each pair of neighboring y s. Of course, there may be x or x^2 before the first y or after the last y. So we view them as a sequence of xy s mixed with x^2y s, with a possible additional head of y, and a possible additional tail of x or x^2.

From y^2=(xy)^2 we get y = xyx. Hence, x^2y=x^2xyx=yx This means we can ignore x^2y s and just consider a sequence of xy s with a possible additional head of y, and a possible additional tail of x or x^2. But (xy)^2=1 tells us that this “sequence” of xy s is actually either a single xy or nothing at all.

If there is a head of y before a single xy, we can use yxy = x^2 (which is obtained from (xy)^2=x^3) to simplify it. If there is a tail of x or x^2 after a single xy, we can use xyx = y to simplify it. After the above simplication, the only remaining case (with both x and y in the sequence) is an x or x^2 going before or after y (but definitely not both). Since x^2y = yx and yx^2 = xy (which is obtained from y = xyx), we only need to consider xy and yx.

If the a_i s are only x or only y, we have three possibilities: x, x^2 , y. Putting everything together, we have G = \{1, x, x^2 , y, xy, yx\} . It is easy to see these listed elements must be distinct, and thus |G| = 6.

简单解释:

x,y 显然是元素. 以他们开头, 分别乘 x,y, 得到 xx, xy, yx, yy

其中 yy 是 1, 忽略. 得到 xx, xy, yx

再乘以 x,y, 得到 xxx, xyx, yxx, xxy, xyy, yxy

xxx = 1

(xy)^2=1\Rightarrow xyx=xyxyy^{-1}=y^{-1} = y

yxx=(xyx)xx=xy

xxy=xx(xyx) = yx

xyy=x

yxy=(xyx)x(xyx)=xyyx=xx

全部重复, 因此 G = \{1,x,y,xx,xy,yx\}. |G|=6.

附录：

## 奇怪的文件

Lec1-4HW1.pdf (384.4 KB)

Lec5-8HW2.pdf (548.7 KB)

Teacher of mine, NJU ↩︎

[Credit: Information fully or partially from Wikipedia] Felix Klein was a German mathematician and educator. Receiving PhD at the age of 19 from Bonn University, he became a professor at 23. The last 27 years of his career were spent at University of Gottingen, where he played a leadership role in establishing it as a world center of mathematics. In particular, he was responsible for hiring David Hilbert. He was the founding president of International Commission on Mathematical Instruction. Besides Klein four-group, his other significant findings includes Klein bottle, a non-orientable surface without boundary. ↩︎

行列式不为0 ↩︎