Q: 为什么要学群论?
A: 因为好玩.

# 小学生真的能看懂吗?

Sorry, life is hard.

—— Professor Z[1]

## 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.

# 以防你之前没学过的说明

#### 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:

• 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.

# 有关群的性质的说明

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.

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

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

• 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?)

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.

#### Example 2: Geometric transformations

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

• 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?
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.

#### 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.
Mn×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.

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

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

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.

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.

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.

#### 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.

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.

#### 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.

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.

#### Definition 2 Semigroup

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

#### 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.

Recall that the definition of identity element requires that, for any a ∈ 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.

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

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

#### 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.

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.

#### Proposition 4

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

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.

#### 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|.

# 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.

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

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

Lec1-4HW1.pdf (384.4 KB)
Lec5-8HW2.pdf (548.7 KB)

1. Teacher of mine, NJU ↩︎

2. [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. ↩︎

3. 行列式不为0 ↩︎

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