Final review of modern algebra at Harbin Institute of Technology

Modern algebra is a branch of abstract algebra and the basis of big data in computer science and artificial intelligence. 

The content of this article is a bit long. You can use the index to jump to the chapter you want to read. The summary of Chapter 10 can be downloaded from my homepage.

1.Algebraic system 

Semigroup: an algebraic system that satisfies the associative law

Commutative semigroup: a semigroup that satisfies the commutative law

Group: There are two determination methods

method1

1. There is unit yuan
2. There is an inverse element
3. The operation satisfies the associative law

method2：

1. The operation satisfies the associative law
2. The operation satisfies the left and right elimination laws

Exchange group (Abel group):

Definition: A group that satisfies the commutative law

Application: When we talk about rings later, we will use the Abel group to determine that an algebraic system (R,+,◦) is a ring:

1. ( R, +) 为一个 Abel group:
2. ( R, ◦)为一个半群； ∀ a, b, c ∈ R( a ◦ b) ◦ c = a ◦ ( b ◦ c）
3. The left and right distribution rule: ∀ a, b, c ∈ R

          a ◦ ( b + c) = ( a ◦ b) + ( a ◦ c)

          ( b + c) ◦ a = ( b ◦ a) + ( c ◦ a)

   Commutative groups can derive many good properties

2. Simple properties of groups

The group satisfies the elimination law

Explanation of answer method for exercise 1

 The order of each element of a finite group does not exceed the order of the finite group.

 Determine the order of elements based on the order of the group

Notes on exercise 6: According to exercises 4 and 5, we get: if elements with order greater than 2 appear in pairs, |G| = 2n= 2k+l+1, we get that l is an odd number,

Exercise 6 means that there must be an element of order 2 in the even-order group 2n, which means that there must be an n-order quotient group (will be mentioned later)

Prove: groups of order 1/2/3/4/5 are commutative groups

How to prove that groups of fifth order and below are commutative groups (use Lagrange’s theorem later to kill the first four)

 How to prove that there must be a third-order subgroup in the sixth-order group

 3. Subgroup Generate subgroup

subgroup

H is a subgroup of G and must satisfy three conditions:

1. H element is not empty
2. H is closed for operations in G
3. H is a subset of G

eg: Find all subgroups of the cubic symmetry group.

解 . (1)， { (1) ,(1 , 2) }， { (1) ,(1 , 3) }， { (1) ,(2 , 3) }， { (1) ,(123) ,(132) }， S3。

Sufficient and necessary conditions for judging subgroups

Similarly, there are similar proofs in subrings and subdomains

Theorem: Read( G 1 , ◦ )sum( G 2 , ∗ )Tozegun, ϕ : G 1 → G 2, ∀ a, b ∈ G 1, ϕ ( a ◦ b ) = ϕ ( a ) ∗ ϕ ( b ), certification:       ϕ − 1 ( e 2)为 G 1 child group, among them e 2为 G 2's lowest element.

The intersection of any multiple subgroups of groupG is still Gsubgroup

No group can be the union of its two proper subgroups

 1. Give an example to illustrate that two subgroups do not need to be subgroups.

Solution . S 6 = { [0] , [1] , [2] , [3] , [4] , [5] } , { [0] , [2] , [4] } sum { [0] , [3] } 为 S 6 pairs of children, however { [0] , [2] , [4] }∪{ [0] , [3] } = { [0] , [2] , [3] , [4] } Not right S 6 children, factor [2] + [3] = [5] ∉ { [0] , [2] , [3] , [4] } .

2.设G1和G2为GroupG's two Mako groups, proof:G1 G1. ⊆ 2 G2someG⊆ 1 Gchild Group sufficient necessary condition isG2为G∪

       ​ ​ fruit G 1 ⊆ G 2some person G 2 ⊆ G 1，则 G 1 ∪ G 2 = G 2or G 1, at this time G 1 ∪ G 2为 G 's child group. result G 1 ∪ G 2为 G 's child group, rebuttal proof for: G 1 ⊆ G 2someone G 2 ⊆ G 1.

        music. G 1 ⊈ G 2并且 G 2 ⊈ G 1, another existence g 2 ∈ G 2, but this g 2∉ G 1, same time existence g 1 ∈ G 1, but this g 1

∉ G 2. You are here G 1为 G 1 ∪ G 2 Mako collection, G 2为 G 1 ∪ G 2 Mako collection, easy to get G 1sum G 2为 G 1 ∪ G 2 Mako group, Yuyu Ren's one group is impossible, both Mako group's two, contradiction.

 GroupGcenterCcorrect's commutative insulator group. G

Center: GroupG elementaname G central element, resultayG for each element interchangeable, immediately∀x ∈ G, ax = xa. GPossessive central element constituent setCname center. G

Don’t forget to first prove that it is a group whose form is a bit like a regular subgroup

 Generate subgroups

read M 为 G 's one non-empty collection, G inclusion M 's possessive group commutation M Generative child group, record( M ).

A bit like cell division

 Corollary: Finitely generated subgroups are still cyclic groups

Prove that every finitely generated subgroup of (Q, ) is a cyclic group? - Zhihu (zhihu.com)

4. Transformation group isomorphism

Isomorphism:

      设(G1, ◦a, b < /span>(< /span>则Name group a>2 pieces are the same. G1到GName为从ϕ2. G1  ≅ G, )b(ϕ ∗ ) a(ϕ ) = b◦ a ϕ, G∈ ∀2, useG∗< /span>1 G: ϕ ) Both groups. As a result, existence is one double injection, 2G), (

The difference and connection between isomorphism and homomorphism:

1. All satisfy that equation
2. Different: Homomorphism does not necessarily require bijection

transformation group

name group: read S a non-empty set, 从 S to S Supper title group, record Sym( S). Current S = { 1 , 2 , · · · , n} hours, Sym( S) = Sn.

Transformation group: Sym ( S ) any subgroup is called A transformation group on S . S n is called any subgroup of

is a permutation group.

Theorem: Any group is isomorphic to a certain transformation group. (Caley’s theorem)

Guide: Any one n finite group same structure n order group S n piece n a group, or immediately any one

A finite group is isomorphic to a permutation group.

practise:

1. Show that a group is a transformation group
2. To prove that it is isomorphism, the key is to find a mapping. First prove that it is a mapping, then prove that it is a surjection, and then prove that the equation conforms to the isomorphism.

 5.Cyclic group

Definition of cyclic group:

As a resultGA certain element in the middleaGenerative, immediateG = (a) = 2 · · · }, , e, a, a1 − , a2 −, a{· · ·

• The integer additive group (Z,+) is a cyclic group, and its generator is 1.
• modelnsame moduleZn = { [0], [1], · · · , [n − 1]}为一个阶为n Finite cyclic group, its generator [1].

Order of cyclic group:

 Isomorphisms of cyclic groups:

1. The infinite cyclic group is isomorphic to the integer addition group (Z, +), that is, if isomorphism is not included, there is only one infinite cyclic group, It’s the additive group of integers;
2. The finite cyclic group of ordern is isomorphic to modulenCongruence plus group (Zn, +) , that is, if isomorphisms are not included, there is only one cyclic group of order n, which is module nCongruent classes add groups.

 This theorem tells us that we can use the additive group of integers and the additive group of integers congruent modulo n as media to prove that two subgroups are isomorphic

example

The order of the subgroup of the cyclic group:

Properties of Cyclic Groups

1. The cyclic group is still a commutative group. The reason is that each element in the cyclic group can be expressed as a power of a, then a^(i+j) = a^(j+i) is obviously a commutative group.
2. A subgroup of a cyclic group is still a cyclic group

3. A subgroup of a cyclic group is still a normal subgroup of the cyclic group

greatest common divisor

d 为 a sum b 's maximum common number display d=( a, b )

two theorems

• 设a, b ∈ Z, use(Z∈ m, n ∃deficiency 0, 则b和a) = ma + nb.
• 设 a, b ∈ Z， b > 0， a = qb + r，0 ≤ r < b，则( a, b) = ( b, r)。

example calculation(266 , 112), m · 266 + n · 112 format

Order and greatest common divisor of a cyclic group

Cosets of subgroups Lagrange’s theorem

Concept: Multiplication of group subsets

• AB = { ab | a ∈ A 且 b ∈ B }
• ∀ g ∈ G, A ∈ , AgSimple copy }g{ A. }A ∈ a |ga{ = gA , immediately gASimple copy A}g{ , G 2

  即 Ag = { ag | a ∈ A }。
• read G one group, one ∀ A, B, C ∈ 2 G ，( AB ) C = A ( BC ).

Concept: Definition and properties of cosets

set义:

read H H group G 's one-piece group, a ∈ G , another set aH nominal group A collection of H , Ha name H 's one-sided collection.

Characteristics:

1. 设H为groupGOne child group , 则∀a ∈ G、aH = H is a sufficient necessary condition a ∈ H.
2. read H H group G 's one group, 则 ∀ a, b ∈ G , aH = bH 's sufficient necessary condition is a ^ − 1  b ∈ H .
3. Child group incompatibility: 设 H H group G 's one group, 则 ∀ a, b ∈ G , aH = bH someone aH ∩ bH = φ .
4. Isocapacitance: H H group G 's one group, 则 ∀ a, b ∈ G , | aH | = | bH | .
5. Dividability: LetH be the groupG A subgroup, then the set composed of all left cosets of H is G

Proof: See handout The second proof is more troublesome↓ ↓ ↓ ↓

Lagrange's theorem

Index definition:

       设 H 为 group g , as a resultH's non-possessive left dependent collective number is a finite numberjNamej为Hexist< a i=13>GIntermediate index, recordj = [ G : H], no other nameHExistingGmedium index 为无穷大.

Lagrange's theorem and corollary:

read G one finite group, H 为 G 's one group, 则 | G | = | H | · [ G : H ].

Corollary*3:

• The order of every element in a finite group divides the order of the finite group.
• If the order of a group is prime, then the group is a cyclic group.
• read G one group, one ∀ a ∈ G , a^ | G | = e.

Corollary*2:

• read H H group G 's one-piece group, S l 为 H 's possessive left collection constitutive set, S r 为 H 's possessive right collection constructive set, 则 | S l | = | S r | .

read p proprime number, integer a given p automeric, 则 a ^  p − 1 ≡ 1 (mod p ).

 备Note:

• [a]p-1 = [1] is because of Corollary 3. This is to prove that an operation is in a group
• Think about whether we can use another method of judging groups to prove it: prove the associative law and the left and right elimination law.

Regular subgroup business group

With the concepts of cosets and group subset multiplication, we introduce normal subgroups and quotient groups.

Simple review

Group operations:

Among them, the third theorem is more difficult for us to understand. 

Determine whether the product of two subgroups is also a subgroup

 normal subgroup

Introduction:

In the previous section we learned about the multiplication of subgroups, but similar to mapping, the multiplication of subgroups does not necessarily satisfy the commutative law, for example

Normal subgroup definition:

read H H group G 's child group, result ∀ a ∈ G , aH =  Ha , other name H 为 G 's correct group.

Equivalence proposition of normal subgroups:

 Proof: 3-4 is relatively troublesome

 business group

Import (you don’t need to read it)

It is to prove that the algebraic system is a group

Definitions and formal examples 

      Definition: the normal subgroup of groupGH The group consisting of all left cosets of , and the group consisting of the multiplication of group subsets is called GpairH The business group of is recorded as G/H.

• The business group is actually the Lagrangian division of G mentioned above.
• Quotient groups operate on families of sets rather than sets.
• The elements divided into each business group are disjoint with each other.

      Formal examples

 practise:

Nothing more to say, just watch it

Fundamental theorem of homomorphism 

The purpose of introducing homomorphism is to weaken the condition of isomorphism. The difference between homomorphism and isomorphism lies in whether it must be a bijection.

• A homomorphism is not necessarily a bijection
• Injective is called monohomomorphism, and surjective is called full homomorphism. Pay attention to how to write the symbol.

Homomorphism definition

nature

two theorems

 Homomorphic substitute and original homomorphic relationship

 See notes for proof method 

homomorphic core

definition

A homomorphic image is a set of elements that can be mapped to an image set. The homomorphic kernel is the set of all unitary-like elements in the domain set. The kernel of homomorphism measures the degree of homomorphism injectivity.

Abstract Algebra Study Notes (4) - Zhihu (zhihu.com)

special homomorphic core

What needs to be noted here is that you must first verify that it is a homomorphism, because it is only said that it is a mapping, and it does not necessarily satisfy that homomorphic expression.

 Fundamental Theorem of Homomorphism

practise 

ring body domain

Some symbols are too troublesome to type. Just look at the latex I wrote.

definition

ring

 body

switching ring/domain

 Sub-ring/sub-domain

Conditions for judging sub-rings 

Subtraction is similar to the inverse of addition 

Sub-ring

 Child body/child area

 Example:

zero factor ring

zero factor 

zero factor ring

 Sufficient and necessary conditions for judging zero factor rings (mathematical logic predicate proof method)

 Sufficient and necessary conditions for judging zero-factor-free rings

The relationship between finite rings and bodies 

practise 

Problems and techniques encountered in solving questions

• Through the associative law and elimination law, it is proved that the algebraic system is a group and must be a necessary and sufficient condition for a finite group.
• The center of group G is a subgroup of G
• A group of even order must have at least one element of order 2
• A finite group of order n is a cyclic group if there is an element whose order is equal to the order n of the finite group.
• If the greatest common divisor of two numbers is d, then there are non-zero integers m and n such that ma+nb=d
• Two isomorphic groups have the same order
• The cyclic group must be a commutative group. The subgroup of the cyclic group is the normal subgroup of the original group.
• If the order of a group is prime, then the group must be a cyclic group (Lagrangian) and in turn a commutative group. The special second-order group is a cyclic group and a commutative group.
• The third-order group is a commutative group and the fourth-order group is also a commutative group. The proof is turned up.
• H is a normal subgroup of G. ah is not necessarily equal to ha ah = ah1
• The subgroup with index 2 is a normal subgroup
• To prove that the multiplication of two sets is still the original subgroup, just prove that AB = BA
• When proving that G\H is a quotient group, it must first be shown that H is a normal subgroup
• The identity element of the quotient group G\H is H, and all elements (elements are sets) are either equal or disjoint.
• When proving homomorphism, sometimes it is necessary to first explain that it is a mapping, that is, to prove that any a=b f(a)=f(b) and the same elements map to the same position.
• The core of homomorphism is only relative to the complete homomorphism. It is used to measure the injective degree of homomorphism mapping.
• {e} is a normal subgroup of any group. This has some applications in the proof of homomorphism and the homomorphic kernel.
• Prove that the first ring is an Abelian group and the second one is a semigroup. Don’t get confused.

1. To prove that it is a business group, you must prove that it is a regular subgroup.

q: It’s not a commutative group. Can you explain that it has an n-order quotient group? [G:H]= n and the order of the quotient group are n are the same thing.

a: The quotient group must be a family of left-coordinate sets of regular subgroups, so it is not the same thing

2.

Use proof by contradiction

3. Use the idea of ​​​​classification discussion to solve questions

 4. Use special methods

Find an example 

5. Use known conditions to perform a "construction" method

 

                                                 The key is in the derivation of 3->2 

6. Although this question is also an after-school question, just understand it and expand your ideas.

.Thinking by analogy. For example, in the solution version of this question, can the theorem be extended to the right coset?

8. Extension of Lagrange’s theorem

The reason for the red pen  Because H is a subset of A, the coset formed by the elements in H and A is A. For details, please see Teacher Ren Shijun’s online course 

9. Existential issues 

There must be a third-order subgroup of the sixth-order group

2n-order commutative group must exist an n-order quotient group