Introduction to Chapter 8 proof problem of njupt "Xin'an numeral base"

NJUPT 《 Confidence safe number base 》 Introduction to group theory proof problem

notes ： This article is suitable for students who study the mathematical basis of information security to make a surprise attack before the exam , Not applicable to students majoring in Mathematics . Some of the contents may not be rigorous for the Department of Mathematics , But for students majoring in information security who want to cope with the exam, it is completely enough .

The fundamental reason why many students can't do modern algebra is that they don't understand the basic concepts , For the course of mathematics foundation of information security , It won't be as difficult as the mathematics department （ Engineering is mainly applied ）, Therefore, the focus of review is to understand all the concepts . Instead of memorizing the topic .

This article It's all my own summary , Mainly Sort out the conceptual things 、 Sum up some How to solve the proof problem , Build a complete proof idea , Considering that there are few online resources , I'm bored during the winter vacation , Then write a message for the benefit of younger students and younger students .

First put a picture of Zhenlou . Then get to the point

1 Group

1.1 The concept of group

What is a group ？ textbook P233 It's very messy . In short ： A group is a group with ** Sealing property 、 Can be combined with 、 There is only one yuan 、 An algebraic system in which any element is reversible . ** So by definition , We can easily get the properties of groups ：

The properties of groups ： In groups $<G,\ast >$ in , Yes ：

1. Sealing property ： namely $\forall a,b\in G$ , There are $ab\in G$.

2. associativity ： namely $\forall a,b,c\in G$, There are $a(bc)=(ab)c$ establish

3. There is only one yuan ： That is, there is an element in the Group $e$ , bring $\forall a\in G$ , There are $ae=ea=a$ establish .

4. Any element is reversible ： namely $\forall a\in G$ ,$\exists b\in G$ , bring $ab=ba=e$ establish . here $b$ be called $a$ Inverse element , Write it down as $a^{-1}$

Operations in groups are not necessarily commutative , If it is exchangeable , So this group is a commutative group .

So if we want to prove an algebraic system $<G,\ast >$ It's a group , Then we use the definition method to prove , It's like writing eight part essay .

Example 1： prove $<\mathbb{Z},+>$ It's a group . among $\mathbb{Z}$ It's a set of integers ,$+$ It's ordinary addition .

①： because $\forall a,b\in \mathbb{Z}$ , There are $a+b\in \mathbb{Z}$ Hang up . Therefore, the closeness of operation is established

②： because $\forall a,b,c\in \mathbb{Z}$ , There are $(a+b)+c=a+(b+c)$ establish . Therefore, the operation is associative .

③： because $\forall a\in \mathbb{Z}$ , There are $a+0=0+a=a$ , Therefore, the operation has a unitary $0$ .

④： because $\forall a\in \mathbb{Z}$ , There are $-a\in \mathbb{Z}$ , bring $a+(-a)=(-a)+a=0$ establish . Therefore, any element in the set has an additive inverse .

Sum up , from ①②③④ You know ,$<\mathbb{Z},+>$ It's a group .

Is it simple ？. Of course, if the title requires proof $<\mathbb{Z},+>$ It's an exchange group , Then add one step to prove commutativity .（ Please try to prove , Just add the fifth step to the description of commutativity on the basis of the above steps to prove the Group .）

The operation here is addition , So the additive unit is $0$,$a$ The additive inverse of is $-a$ .

1.2 subgroup

Because a group is an algebraic system , Algebraic systems consist of sets and operations defined on sets . Therefore, the concept of subgroup is also well understood

If $<G,\ast >$ It's a group ,$H\subseteq G$ ,$<H,\ast >$ It's also a group , that $<H,\ast >$ yes $<G,\ast >$ Subgroups , Write it down as $H\le G$. among $\ast$ Operation is the same operation .

Proof subgroup can also be proved by definition .

Example 2： prove $<5\mathbb{Z},+>$ It's a group $<\mathbb{Z},+>$ Subgroups . among $5\mathbb{Z}=\text{{}5k|k\in \mathbb{Z}\text{}}$

①： because $\forall a,b\in 5\mathbb{Z}$ , namely $a,b$ All are $5$ Multiple , Obviously $a+b$ still $5$ Multiple . Therefore, the closeness of operation is established

②： because $\forall a,b,c\in 5\mathbb{Z}$ , There are $(a+b)+c=a+(b+c)$ establish . Therefore, the operation is associative .

③： because $\forall a\in 5\mathbb{Z}$ , There are $a+0=0+a=a$ , Therefore, the operation has a unitary $0$ .

④： because $\forall a\in 5\mathbb{Z}$ , There are $-a\in 5\mathbb{Z}$ , bring $a+(-a)=(-a)+a=0$ establish . Therefore, any element in the set has an additive inverse .

Obviously , For any integer $a$ ,$5a$ It's still an integer . therefore $5\mathbb{Z}\subset \mathbb{Z}$.

therefore $<5\mathbb{Z},+>$ yes $<\mathbb{Z},+>$ Subgroups .

Is it almost the same operation ？

Of course , There is another theorem for judging subgroups ：

Theorem ： set up $H\subseteq G$ ,$<G,\ast >$ It's a group . if $\forall a,b\in H$ ,$a{b}^{-1}\in H$ Hang up . that $<H,\ast >$ yes $<G,\ast >$ Subgroups .

So we can prove the above question ：

because $<\mathbb{Z},+>$ It's a group , set up $a,b\in 5\mathbb{Z}$ , namely $a,b$ All are $5$ Multiple .

Again because $b$ The additive inverse of is $-b$ , because $b$ yes $5$ Multiple , that $-b$ It's also $5$ Multiple , $a+(-b)$ It's also $5$ Multiple .

therefore $a+(-b)\in 5\mathbb{Z}$ , so $<5\mathbb{Z},+>$ yes $<\mathbb{Z},+>$ Subgroups .

Do you suddenly feel that the proof is still very simple ？, Proof group is to write eight part essay , Just add up the four properties ！ Of course , In known $G$ In the case of a group , We should also learn to use these four properties .

Let's look at two more questions to consolidate ：

Example 3： set up $G$ It's a group , And $\forall a\in G$, There are $a^2=e$ establish , prove $G$ It's an exchange group .

set up $a,b \in G $, By group Sealing property , You know $ab\in G,ba\in G$.

So we can get $a^2=b^2=(ab{)}^2=e$, namely $(ab{)}^2=a^2{b}^2$ . （ Be careful ：$(ab{)}^2=(ab)(ab)=abab$, General groups do not have commutativity ）

Expand square , Available $abab=aabb$ , Multiply both sides by one $a$ , Multiply one right $b$ , You can get $a^2(ba)b^2=a^2(ab)b^2$ .

Eliminate square , Yes $ba=ab$ establish , therefore $G$ It's an exchange group .

Let's have a little more difficulty （ This question suggests imitating the above method , Do it yourself and see the answer ）

Example 4： In groups $G$ in ,$\forall a,b$, There are $a^3{b}^3=(ab{)}^3,a^4{b}^4=(ab{)}^4,a^5{b}^5=(ab{)}^5$ , prove $G$ It's an exchange group .

because $a^3{b}^3=(ab{)}^3$, therefore $a(ab{)}^{3}b=a(a^3{b}^3)b=a^4{b}^4=(ab{)}^4=a(ba{)}^{3}b$ . namely $a(ab{)}^{3}b=a(ba{)}^{3}b$. Multiply left on both sides of the equal sign $a^{-1}$ , Take the right $b^{-1}$, available $(ab{)}^3=(ba{)}^3$ （P.S. because $G$ It's a group , therefore $a^{-1}$ You can write directly ）

Again because $a^4{b}^4=(ab{)}^4$, therefore $a(ab{)}^{4}b=a(a^4{b}^4)b=a^5{b}^5=(ab{)}^5=a(ba{)}^{4}b$ . namely $a(ab{)}^{4}b=a(ba{)}^{4}b$. Multiply left on both sides of the equal sign $a^{-1}$ , Take the right $b^{-1}$, available $(ab{)}^4=(ba{)}^4$ . （ This step can be seen directly in the same way ）.

Double right $(ab{)}^3$ Inverse element （ namely $(ab{)}^{-3}$）, obtain $(ab{)}^4(ab{)}^{-3}=(ba{)}^4(ab{)}^{-3}$ , namely $(ab{)}^4(ab{)}^{-3}=(ba{)}^4(ba{)}^{-3}$, That is to say $ab=ba$ , Reason group $G$ It's an exchange group .

Here's another reminder , The power here is the abbreviation of the same operation for consecutive times , Not necessarily multiplication . For example, in addition Group $<\mathbb{R},+>$ in ,$4^3$ Three $4$ Add up , So the result is $12$. But in multiplicative groups $<\mathbb{R}-0,\times >$ in , $4^3$ It means three $4$ Multiply , The result is $64$ .

2 Normal subgroups and quotient groups

2.1 Simply understand coset 、 Quotient set 、 Normal subgroup 、 Quotient group

8.2 section , The textbook is confusing .... Let's first look at how coset is defined in the textbook ：

Similar to the classification of modular congruence , People can pass $G$ Subgroups $H$ To the group $G$ To classify
$$aH=\{c|c\in G,c^{-1}a\in H\}$$

So let's talk about what cosets are . Coset is A collection . $ 5\Z$ yes $\mathbb{Z}$ Subgroups , The algorithm is addition . Then we can export $\mathbb{Z}$ On $5\mathbb{Z}$ The cosets of are .
$$5\mathbb{Z},1+5\mathbb{Z},2+5\mathbb{Z},3+5\mathbb{Z},4+5\mathbb{Z}$$
among $1+5\mathbb{Z}=\{...,-4,1,6,11,...\}$,$2+5\mathbb{Z}=\{...,-3,2,7,12,...\}$ , And so on . exactly , It has something to do with congruence ...

above $5$ In a collection , Only $5\mathbb{Z}$ The addition on forms a group , Additions on other sets do not form groups ！！！（ Because it doesn't satisfy the closeness ）

Then the quotient set is A set of all cosets , Its essence is A collection of collections .$\mathbb{Z}/5\mathbb{Z}$ The result is to add braces to all the previous line ..
$$\mathbb{Z}/5\mathbb{Z}=\{5\mathbb{Z},1+5\mathbb{Z},2+5\mathbb{Z},3+5\mathbb{Z},4+5\mathbb{Z}\}$$
Please try to write $\mathbb{R}/\mathbb{Z}$ Result （ answer ：$ {a+\Z|a\in[0,1) }$）.

And then we'll find out ：$\mathbb{Z}/5\mathbb{Z}$ It's also a collection （ It contains $5$ Elements , Each element is a module $5$ Congruence class of ）. If we define operations $\bigoplus_5 $ For the mould $5$ Add , Then it can be proved that ：$<\mathbb{Z}/5\mathbb{Z},{⨁}_5>$ It's a group . let me put it another way , $\mathbb{Z}/5\mathbb{Z}$ It also forms a group .

Because we came to the conclusion at the beginning , When the operation is addition ,$\mathbb{Z}$ It's a group ,$5\mathbb{Z}$ It's also a group , And quotient set $\mathbb{Z}/5\mathbb{Z}$ It can also form groups . Then we can draw two conclusions ：

Because the quotient set $\mathbb{Z}/5\mathbb{Z}$ Can form a group , therefore

① $5\mathbb{Z}$ yes $\mathbb{Z}$ Normal subgroups of

② $\mathbb{Z}/5\mathbb{Z}$ It's a $\mathbb{Z}$ A quotient group on

Let's make it clear here ：$\mathbb{Z}/5\mathbb{Z}$ Follow ${\mathbb{Z}}_5$ It's a different concept , because ${\mathbb{Z}}_5=\{0,1,2,3,4\}$ , and $\mathbb{Z}/5\mathbb{Z}=\{5\mathbb{Z},5\mathbb{Z}+1,5\mathbb{Z}+2,5\mathbb{Z}+3,5\mathbb{Z}+4\}$ . The elements in the former are numbers , The element in the latter is aggregate .

2.2 Relevant proof methods

1. Prove that two cosets are equal

Properties of cosets ： $H$ yes $G$ Subsets on ,$a,b\in G$ . So there are ：

① $aH=bH\iff b^{-1}a\in H$

② $aH\cap bH=\varphi \iff b^{-1}a\notin H$.

③ Any two cosets are either equal , Or not . That is to say $aH\cap bH=\varphi$ or $aH=bH$, No other relationship

for instance ：$4+5\mathbb{Z}=9+5\mathbb{Z}$, because $-9+4=-5\in 5\mathbb{Z}$ . and $4+5\mathbb{Z}\ne 3+5\mathbb{Z}$ , because $-3+4=1 \not \in 5\Z $.

2. Prove a subgroup $H$ It's a group $G$ Normal subgroups of

Proof method of normal subgroup ：

set up $a\in G$, $H$ yes $G$ Subgroups , If we want to prove $H$ yes $G$ Normal subgroups of , Just prove
$$aH{a}^{-1}\subset H$$

Of course, there is another property , Readers can try to prove , This conclusion is better not to use directly , Because one head can come out

if $G$ It's an exchange group ,$H$ yes $G$ Subgroups , be $H$ It must be $G$ Normal subgroups of

Let's take an example ：

Example 5： set up $G$ It's a group ,$C=\{a|a\in G,\forall b\in G\to ab=ba\}$. prove $C$ yes $G$ Normal subgroups of .

First prove that $C$ yes $G$ Subgroups ：

①： because $G$ It's a group , so $G$ There's a dollar on it $e$ , And $\forall g\in G$ There are $ge=eg=g$ , therefore $e\in C$ .$C$ There's a dollar on it $e$ .

②： because $G$ It's a group , so $G$ The operations in are associative ,$C$ yes $G$ Subset , Therefore, the combination of this operation is $C$ China is still established

③： set up $s,t\in C,g\in G$ that $(st)g=s(tg)=s(gt)=(sg)t=gst=g(st)$ , explain $st\in C$ , It can be seen that the operation is in $C$ It is also closed .

（ notes ： here $stg=sgt$ It's using $C$ Properties of the upper element , have a look $C$ How are the elements defined in , since $t\in C,g\in G$, that $tg=gt$ Is established , The reason for changing brackets at will is $G$ It's a group , $G$ The combination of is established ）

④： set up $s\in C,g\in G$ , from $sg=gs$ , be $s^{-1}(sg)s^{-1}=s^{-1}(gs)s^{-1}$ , That is to say $g{s}^{-1}=s^{-1}g$ （ Please go to the brackets and have a look ） , You know $s^{-1}\in C$.

from ①②③④ You know , $C$ It's a group , And $C$ yes $G$ Subgroups . The following proof $C$ yes $G$ Normal subgroups of

set up $g\in G,s\in C$ , Then there are $gs{g}^{-1}=(gs)g^{-1}=(sg)g^{-1}=s$ , explain $gs{g}^{-1}\in C$ , namely $gC{g}^{-1}\subset C$ . You know $C$ yes $G$ Normal subgroups of .

In the above question , You will find proof $C$ yes $G$ Subgroups of are still defined , Four step proof . And prove $C$ yes $G$ Normal subgroups of , It is proved by definition . So the definition is the most important .

3 Homomorphism and isomorphism

3.1 Definition

The book defines homomorphism and isomorphism in this way ：

set up $f$ It's a group $G$ To the group $G^{\prime }$ Mapping / function , if $\forall a,b\in G$ , There are $f(a)f(b)=f(ab)$ Hang up , be $f$ yes $G$ To $G^{\prime }$ Homomorphism of .

Specifically , if $f$ If it is injective, it is simple homomorphism ,$f$ Is a surjection is a full homomorphism . if $f$ It is both simple homomorphism and full homomorphism , be $f$ It's isomorphic .

P.S： One shot is a one-to-one mapping , Using symbolic language is ： if $f(x_1)=f(x_2)$ , There must be $x_1=x_2$ . A surjection is a mapping / The value range of the function fills the whole $G^{\prime }$ aggregate , That is to say, for any $y\in G^{\prime }$ , There is always $x\in G$ , bring $y=f(x)$ establish .

for instance ： set up $f$ yes $\mathbb{R}$ To $\mathbb{R}$ Mapping , be $f(x)=e^x$ It's a single shot but not a full shot ,$f(x)=x\sin x$ It's a full shot, but not a single shot , $f(x)=x^3$ It's both a single shot and a full shot ,$f(x)=\sin x$ Neither single shot nor full shot .

3.2 Method of proof

So we get the proof of homomorphism and isomorphism （ It's still the definition ）：

Methods of proving homomorphism ： Find one from $G$ To $G^{\prime }$ Function of $f$, To prove to $G$ Arbitrary in $a,b$ , There are $f(ab)=f(a)f(b)$ establish .

Methods of proving isomorphism ： Find one from $G$ To $G^{\prime }$ Function of $f$, Prove that this function has ：① Homomorphism ,② Single shot ,③ Full shot . Three fixed steps , Just like proving that groups have four fixed steps .

Of course , There are several properties that need to be understood , It can be used directly ：

if $f$ Is a group homomorphism , There must be $f(e)=e^{\prime }$ ,$f(a^k)=[f(a){]}^k$ It's all established .

The necessary and sufficient condition of injectivity ：$\ker f=\{e\}$ . The necessary and sufficient condition of surjection ： $f(G)=G^{\prime }$

There are still a few concepts to understand ：

Homomorphic kernel $\mathrm{k}\mathrm{e}\mathrm{r}f=\{x|x\in GAndf(x)=e^{\prime }\}$ （ $e^{\prime }$ It's a group $G^{\prime }$ One yuan of ）. for example ： from $<\mathbb{Z},+>$ To $<{\mathbb{Z}}_5,{⨁}_5>$ Mapping on $f(x)=x\mathrm{m}\mathrm{o}\mathrm{d}5$ . Obviously ,$f(a){⨁}_{5}f(b)=f(a+b)$ Is established . So this is a homomorphism . and ${\mathbb{Z}}_5$ The unit in is $0$ , So we want to have $0=f(x)$ , So we can get $x\in 5\mathbb{Z}=\{5k|k\in Z\}$ . So here , $\mathrm{k}\mathrm{e}\mathrm{r}f=5\mathbb{Z}$ .

Please pay attention to the ${\mathbb{Z}}_5$ and $5\mathbb{Z}$ The difference between （ As mentioned above ）, as well as $\mathbb{Z}$ Upper operation and ${\mathbb{Z}}_5$ The difference of upper operation , above $a,b$ yes $\mathbb{Z}$ Medium element ,$f(a),f(b)$ yes ${\mathbb{Z}}_5$ Medium element .

Fundamental theorem of homomorphism ： if $f$ yes $G$ To $G^{\prime }$ The homomorphism of , be $G/\mathrm{k}\mathrm{e}\mathrm{r}f$ And $G^{\prime }$ isomorphism

Example 6： Prove the basic theorem of homomorphism

set up $F$ yes $G/\mathrm{k}\mathrm{e}\mathrm{r}f$ To $G^{\prime }$ Mapping , $F(a\ker f)=f(a)$

①： prove $F$ It's homomorphism ： because $f$ It's homomorphism , therefore $f(ab)=f(a)f(b)$ , therefore $F(a\ker f)F(b\ker f)=F(ab\ker f)$ establish , so $F$ It's homomorphism .

②： prove $F$ It's a single shot ： if $a\ker F\in \ker F$ , So there are $F(a\ker f)=f(a)=e^{\prime }$ . therefore $a\in \ker f$ , so $a\ker f=\ker f$. So we have $\ker F=\{\ker f\}$ . $\ker F$ There is only one element in , so $F$ It's a single shot .

③： because $f$ It's homomorphism , So for any $y\in G^{\prime }$, There is always $x\in G$ bring $y=f(x)$. So we have $F(x\ker f)=y$ establish , so $F$ It's a volley .

from ①②③ You know ： $F$ It's a volley , so $G/\mathrm{k}\mathrm{e}\mathrm{r}f$ And $G^{\prime }$ isomorphism .

It is worth noting that ： $G/\ker f$ It's a set of sets , therefore $F$ The domain of definition yes $G/\ker f$ The elements in （ The type of this element is set ）, The value range is $G^{\prime }$ The elements in .

Let's take another example to prove isomorphism

Example 7： set up $G$ It's a group ,$a\in G$ , $f(x)=ax{a}^{-1}$ . prove $f$ yes $G$ To $G$ Automorphism of .

①： Prove homomorphism ： set up $s,t\in G$ , be $f(s)f(t)=as{a}^{-1}at{a}^{-1}=as(a{a}^{-1})t{a}^{-1}=ast{a}^{-1}=f(st)$ , therefore $f$ Have homomorphism .

②： Prove that single shot ： set up $f(x_1)=f(x_2)$ , namely $a{x}_1{a}^{-1}=a{x}_2{a}^{-1}$, Multiply both sides by one $a^{-1}$, Multiply one right $a$ , You can get $a^{-1}a{x}_1{a}^{-1}a=a^{-1}a{x}_2{a}^{-1}a$ , namely $x_1=x_2$ establish . so $f$ It's a single shot .

③： Prove the full shot ： set up $y=f(x)=ax{a}^{-1}$ , be $a^{-1}ya=x$ , namely $y=f(a^{-1}ya)$ , so $f$ It's a volley .

Because the operations in the group are closed , So ①②③ You know , $f$ yes $G$ To $G^{\prime }$ Automorphism of .

~~ So proving isomorphism is still writing eight part essay .~~ Directly use the definition of a shuttle .

4. Cyclic groups

4.1 Several definitions

set up $G$ It's a group

$G$ Medium element $a$ The order of $\mathrm{o}\mathrm{r}\mathrm{d}(a)$ ： send $a^k=e$ The smallest positive integer of $k$ , And $k$ It must be $|G|$ The factor of

if $G$ There is an element in $g$ , send $G=\{g,g^2,g^3,...\}$ , be $G$ It's a cyclic group , $g$ yes $G$ The generator of .（ Analog primal root ）

Properties of cyclic groups ：

Cyclic groups must be commutative groups

Infinite order cyclic groups and $<\mathbb{Z},+>$ isomorphism , Finite order （ $n$ rank ） Cyclic groups and $<{\mathbb{Z}}_n,{⨁}_n>$ isomorphism .

if $g$ It's a group $G$ The generator of , if $|G|=+\infty$, be $G$ There are only two generators $g$ and $g^{-1}$ . if $|G|=n$ ,$g^k{|}_{\gcd (k,n)=1}$ yes $G$ The generator of .

Please try to prove it yourself ： $<{\mathbb{Z}}_{41}^{\ast },{⨂}_{41}>$ And $<{\mathbb{Z}}_{40},{⨁}_{40}>$ These two groups are isomorphic , among ${⨂}_{41}$ Is the mould $41$ Multiplication ,${⨁}_{40}$ Is the mould $40$ Add .

Tips ： model $41$ A primitive root of is $6$ .

answer ： set up $f$ yes ${\mathbb{Z}}_{40}$ To ${\mathbb{Z}}_{41}^{\ast }$ The function on , $f(x)=6^x\mathrm{m}\mathrm{o}\mathrm{d}41$

①： because $\phi (41)=40$. set up $a,b\in {\mathbb{Z}}_{40}$. so $f(a+b)=6^{a{⨁}_{40}b}\mathrm{m}\mathrm{o}\mathrm{d}41=6^a{⨂}_{41}{6}^b=f(a){⨂}_{41}f(b)$, so $f$ Have homomorphism .

②： if $f(x_1)=f(x_2)$, namely $6^{x_1}\equiv 6^{x_2}(\mathrm{m}\mathrm{o}\mathrm{d}41)$ . therefore $x_1\equiv x_2(\mathrm{m}\mathrm{o}\mathrm{d}40)$. Again because $x_1,x_2\in {\mathbb{Z}}_{40}$ , So only $x_1=x_2$ establish , so $f$ It's a single shot .

③： because $6$ Is the mould $41$ The original root , so model $41$ A complete remainder system of can be written as $\{6,6^2,6^3,...,6^{40}=6^0=1\}$ , namely $6^i,i\in {\mathbb{Z}}_{40}$ Traversed the module $41$ The complete residue system of , so $f$ It's a volley .

from ①②③ You know , $<{\mathbb{Z}}_{41}^{\ast },{⨂}_{41}>$ And $<{\mathbb{Z}}_{40},{⨁}_{40}>$ These two groups are isomorphic

4.2 Method of proof

Prove that a group is a cyclic group , Is to find it The generator , If it is a finite cyclic group （ The steps are $n$ ）, Find an order of $n$ The elements of .

example 7： Prove that a group of prime order must be a cyclic group

set up $|G|=p$ Prime number , Then for groups $G$ Any element in , The order can only be $1$ or $p$ .

set up $a\ne e$ , be $a$ The order of cannot be $1$ , So only $a$ The first step is $p$ . therefore $a$ It's a group $G $ The generator of , $G$ It's a cyclic group

example 8： Known group $G$,$|G|=pq$ , among $p,q$ Are different primes , prove $G$ It's a cyclic group .

prove ： because $|G|=pq$,$p,q$ Are different primes , So for any element $a$,$a$ The order of can only be $1,p,q,pq$ Four situations . if $a\ne e$, be $a$ The order of can only be $p,q,pq$.

set up $g,h\in G$,$\mathrm{o}\mathrm{r}\mathrm{d}(g)=p,\mathrm{o}\mathrm{r}\mathrm{d}(h)=q$ . Because group $G$ The closeness of operations in , so $gh\in G$ .

Then there are $(gh{)}^1\ne e$,$(gh{)}^p=g^p{h}^p=h^p\ne e$,$(gh{)}^q=g^q{h}^q\ne e$, explain $(gh)$ The order of cannot be $1,p,q$.

so $\mathrm{o}\mathrm{r}\mathrm{d}(gh)=pq=|G|$. $gh$ yes $G$ The generator of , Reason group $G$ It's a cyclic group .

Do you feel a little confused ？ Then I'll take a big move ！.

5 Summary of ideas of proof questions in group theory

$a$ The order of can only be $p,q,pq$.

set up $g,h\in G$,$\mathrm{o}\mathrm{r}\mathrm{d}(g)=p,\mathrm{o}\mathrm{r}\mathrm{d}(h)=q$ . Because group $G$ The closeness of operations in , so $gh\in G$ .

Then there are $(gh{)}^1\ne e$,$(gh{)}^p=g^p{h}^p=h^p\ne e$,$(gh{)}^q=g^q{h}^q\ne e$, explain $(gh)$ The order of cannot be $1,p,q$.

so $\mathrm{o}\mathrm{r}\mathrm{d}(gh)=pq=|G|$. $gh$ yes $G$ The generator of , Reason group $G$ It's a cyclic group .

Do you feel a little confused ？ Then I'll take a big move ！.

5 Summary of ideas of proof questions in group theory