Final review of fundamentals of information security Mathematics (cryptography)

Final review of fundamentals of information security Mathematics:

Chapter 1:

1. Find all prime numbers within 100

```100
2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97
```

2. Solve GCD (45100)

```gcd(45,100)=5
100=45*2+10
45=10*4+5
10=5*2+0
```

3. Find the maximum common divisor of 963657 and express it as the linear combination of integral coefficients of 963657

```gcd(963,657)=9
963=657*1+306
657=306*2+45
306=45*6+36
45=36*1+9
36=9*4+0
9=45-36
=(657-306)-(306-45*6)
=(657-(963-657)*2)-((963-657)-6*(657-306*2))
=(657-2*963+2*657)-(963-657-6*(657-2*(963-657)))
=3*657-2*963-(963-657-6*(657-2*963+2*657))
=3*657-2*963-963+657+6*657-12*963+12*657
=22*657-15*963
//It's very difficult to calculate. It's easy to make mistakes
```

Chapter 2:

Knowledge points:

Modular operation properties:

```1.a (mod n) **±** b (mod n) (mod n)=a **±** b (mod n)
2.a (mod n) ∙ b (mod n) (mod n)=a ∙ b (mod n)
```

Nature of congruence:

```(1) Reflexivity to any integer a，a ≡ a   (mod m)
(2) Symmetry if a ≡b (mod m)，be b ≡a (mod m)
(3) Transitivity if a ≡b (mod m)， b ≡c (mod m)，be a ≡c (mod m)
If a ≡ b (mod m)，c ≡ d (mod m)，be
(4) a±c ≡ b±d (mod m)
(5) ac ≡ bd   (mod m)
(6) ax+ cy ≡ bx+ dy (mod m)，among x，y Is any integer
(7) if ac ≡bc (mod m) ，And gcd(c,m)=1，be a ≡b  (mod m)
Promotion: if ac ≡bc (mod m) ，And c≠0，be a ≡b  (mod m/gcd(c,m))
(8)if a ≡b  (mod mi), i=1,2,...,n ，be a≡b (mod (lcm(m1,m2,...,mn))
if a ≡ b  (mod m), And n Is a positive integer, be
(9)  a∙n ≡ b∙n (mod nm)
(10) an ≡ bn (mod m)
(11) gcd(a,m)=gcd(b,m)
(12) if n| m ，be a ≡ b   (mod n)
```

Complete residual system:

Definition: if M is a positive integer and one residue is taken from the residue class of module m, the set of these residues is called a complete residue system (or simply the complete system) of module M.

Simplified residual system:

Definition: a residue class of module m is called a simplified residue class if there is a residue that is coprime with m. At this time, the residue in the simplified residue class is called simplified residue.

Euler function:

Theorem:

Let n have a standard factor, and the decomposition formula is

n =p1^(k1)· p2^(k2)... · pm^(km)

be φ (n)=n · (1 - 1/p1)…… ·(1 - 1/pm)

Fermat theorem:

p is a prime number, a is a positive integer, and gcd(a,p)=1, then

a^(p−1) ≡ 1(mod p)

The equivalent form is:

a*a^(p − 2) ≡ 1(mod p) or a ^ (− 1) ≡ a^(p − 2) (mod p)

If p is a prime number and a is any positive integer, then

a^p ≡ a (mod p) (mutual prime of a and P is not required)

Euler's theorem:

´ let n be a positive integer, if gcd(a,n)=1, then

a^φ(n) ≡ 1 (mod n)

1. Write a positive integer no more than 100 in a congruence class of 6 of module 15.

```15
6 21 36 51 66 81 96
```

2. Write out two complete residue systems of module 9. It is required that each number in one perfect system is odd and each number in the other is even

```9 Normal complete residual system
{0,1,2,3,4,5,6,7,8}
Odd complete residue system
{1,3,5,7,-9,-7,-5,-3,-1}
Even complete residue system
{0,2,4,6,8,-8,-6,-4,-2}
```

3.a=7, p=19, using Fermat's small theorem to calculate ap mod p

```7^19 mod 19
gcd(7,19) = 1
7^18 ≡ 1,7^19 = 7^18 * 7 = 1 * 7
7^19 mod 19 = 7
```

4. Calculate the Euler function of 120.

```120=2^3 * 3 * 5
φ(120)=120 * (1 - 1/2) * (1 - 1/3) * (1- 1/5)
=120 * 1/2 * 2/3 * 4/5
=32
```

Chapter 3:

Knowledge points:

Primary congruence:

Theorem: let m be a positive integer and a be an integer satisfying m | a, then the first-order congruence

​ ax ≡ b (mod m)

The necessary and sufficient condition for a solution is (a,m)|b, and the number of solutions is (a,m)

When the congruence has a solution, the solution is

x ≡ (( a / (a,m) )^(-1) (mod m / (a,m) ) * b / (a,m) + t * m / (a,m) ) (mod m)

t = 0,1,..., (a,m) - 1

Chinese remainder theorem:

Theorem: let M1 and M2 be mutually prime, then the congruence formula group

x ≡ b1 (mod m1)

x ≡ b2 (mod m2)

...

There must be a solution and the solution is unique if m = m1 ∙ m2

Then the solution of the congruence group is:

x ≡b1 ∙ M1 ∙ M1^(-1) + b2 ∙ M2 ∙ M2^(-1)+... (mod m)

Where Mi^(-1) ∙ mi ≡ 1 (mod mi), i = 1, 2

Modulo prime congruence:

f(x) = anxn + an-1xn-1 +...+ a1x + a0 ≡ 0 (mod p)

The number of solutions does not exceed n

Polynomial Euclidean division / polynomial division with remainder

f(x)=q(x)∙ g(x)+r(x) f(x) = q(x)∙ (xp-x) + r(x)

f(x) ≡ 0 (mod p) is equivalent to r(x) º 0 (mod p)

f(x) ≡ (x –b1) (x –b2)...(x –bk) fk(x) (mod p),x º bi (mod p)

The first congruence f(x) = xn + an-1xn-1 +... + a1x + a0 ≡ 0 (mod p) has n solutions, which is equivalent to f(x)| (x^p – x) under module P

Fermat's theorem:

From Fermat's small theorem, x ≡ 1, 2,..., p-1 (mod p)

Is the solution of the congruence x^(p-1) ≡ 1 (mod p)

Corollary, if p is a prime number, then for any integer x, there is

x^(p-1) – 1 ≡ (x – 1) (x – 2) ...(x – (p-1)) ( mod p)

The necessary and sufficient condition for integer n to be prime is (n-1)+ 1 ≡ 0 ( mod n)

1. Find the solution of the following congruence

x^3 -2x +3 ≡ 0 (mod 5)

```0^3 - 2*0 + 3 ≡ 3 mod 5
1^3 - 2*1 + 3 ≡ 2 mod 5
2^3 - 2*2 + 3 ≡ 2 mod 5
3^3 - 2*3 + 3 ≡ 4 mod 5
4^3 - 2*4 + 3 ≡ 4 mod 5
//Congruence no solution
```

2. Solve the congruence equation 10x ≡ 25 (mod 65)

```gcd(10,65)=5
10x ≡ 25 (mod 65)
2x ≡ 5 (mod 13)
2^(-1)=7
x ≡ (5*7 (mod 13) + 13 * t)mod 65
t=0,1,2,3,4
x ≡ 9,22,35,48,61
```

3. Han Xin's second call of arms: there is a team of soldiers. If they are arranged in a five element column, the last line is one; In six columns, the last five; In seven columns, the last four; If you form an eleven line column, you will end up with ten people. Number of soldiers

```//Subject analysis available
x ≡ 1 (mod 5)
x ≡ 5 (mod 6)
x ≡ 4 (mod 7)
x ≡ 10 (mod 11)
//Chinese remainder theorem
x ≡ 1*6*7*11*3 + 5*5*7*11*1 + 4*5*6*11*1 + 10*5*6*7*1 (mod 5*6*7*11)
x ≡ 2111
```

4. Solve the congruence 21x^18 + 2x^15 - x^10 + 4x -3 ≡ 0 (mod 7)

```//Using Fermat's small theorem
21x^18 + 2x^15 - x^10 + 4x -3 ≡ 0 (mod 7)
x^18 + 2x^15 - x^10 + 4x - 3 ≡ 0 (mod 7)
2x^3 - x^4 + 4x -3 ≡ 0 (mod 7)
Verify 0,1,2,3,4,5,6 unsolvable
```

Chapter 4:

Knowledge points:

Square residue:

Definition: let a be an integer, p be a prime, and (a,p)=1

If the quadratic congruence, x^2 ≡ a (mod p), has a solution, then a is the quadratic residue or square residue of module P, otherwise a is the quadratic non residue or square non residue of module P

Theorem: let p be an odd prime number. In the simplified residue system of module P, there are (p-1)/2 square residues and (p-1)/2 square non residues.

When finding the square residue of module p, only the following numbers can be calculated:

12,22,...,((p-1)/2 )^2 (mod ⁡p)

Euler discriminant method:

Let a be an integer, p be an odd prime, (a,p)=1

(1) The necessary and sufficient condition for a to be the square residue of module p is

a^((p-1)/2) ≡ 1 (mod p)

(2) The necessary and sufficient condition that a is a module p squared non residual is

a^((p-1)/2) ≡ -1 (mod p)

Corollary: let p be an odd prime number, (a1,p)=1, (a2,p) =1, then

(1) If a1 and a2 are the square residue of module P, a1 ∙ a2 is the square residue of module p;

(2) If a1 and a2 are the square residue of module P, then a1 ∙ a2 is the square residue of module p; (similar to negative positive)

(3) If a1 is the square residue of module p and a2 is the square non residue of module p, a1 ∙ a2 is the square non residue of module p.

Legendre symbol:

(a/p) = 1, a is the square residue of module p

(a/p) = -1, a is the square non residue of module p

(a/p) = 0, a|p

Legendre sign calculation still uses Euler discriminant method

a^((p-1)/2) ≡ 1 (mod p)

a^((p-1)/2) ≡ -1 (mod p)

nature:

```//The nature of rational use is particularly simple, but remember
p Odd prime
1) (1/𝑝) = 1
2) ((−1)/𝑝) = (−1)^((𝑝−1)/2)
3) (𝒂/𝒑) = ((𝒂+𝒑)/𝒑)  (Periodicity)
4) ((𝒂∙𝒃)/𝒑) = (𝒂/𝒑)(𝒃/𝒑)  (Complete multiplicity)
5) if (a，p) = 1，be(𝑎^2/𝑝)=1
Inference: if a ≡ b (mod p)，be(𝒂/𝒑) = (𝒃/𝒑)
6) p It's an odd prime,(2/𝑝) = (−1)^((𝑝^2−1)/8)
7) if p，q Is a different odd prime number, then (𝑞/𝑝) = (−1)^((𝑝−1)/2  ∙ (𝑞−1)/2) (𝑝/𝑞)
8) (2/𝑝)= -1 If p ≡ ±3 (mod 8)
(2/𝑝)= 1 If p ≡ ±1 (mod 8)
```

Jacobi symbol:

Definition: let m = p1p2... pr be the product of odd prime number pi. For any integer a, the Jacobian symbol is defined as follows:

(a/m) = (a/p1)*(a/p2)...(a/pr)

Where (a/pi) is the Legendre symbol

nature:

```set up m Is an odd number, then
1)(1/m) = 1
2) ((−1)/𝑚) = (−1)^((𝑚−1)/2)
3) (2/𝑚) = (−1)^((𝑚^2−1)/8)
set up m Is a positive odd number, then
4) (𝒂/𝒎) = ((𝒂+𝒎)/𝒎)
5)((𝒂∙𝒃)/𝒎) = (𝒂/𝒎)(𝒃/𝒎)
6) if (a，m) = 1，be(𝑎^2/𝑚) = (𝑚/𝑎^2 ) = 1
Inference: if a ≡ b (mod m)，be(𝒂/𝒎) = (𝒃/𝒎)
7)if m，n Are odd numbers, then(𝑛/𝑚) = (−1)^((𝑚−1)/2  (𝑛−1)/2) (𝑚/𝑛)
```

1. Find the square residue and square non residue of module p=11,23

```p = 11
1^2 = 1 (mod 11)
2^2 = 4 (mod 11)
3^2 = 9 (mod 11)
4^2 = 5 (mod 11)
5^2 = 3 (mod 11)
Square surplus:1,3,4,5,9
Square non residue:2,6,7,8,10
p = 23
1^2 = 1 (mod 23)
2^2 = 4 (mod 23)
3^2 = 9 (mod 23)
4^2 = 16 (mod 23)
5^2 = 2 (mod 23)
6^2 = 13 (mod 23)
7^2 = 3 (mod 23)
8^2 = 18 (mod 23)
9^2 = 12 (mod 23)
10^2 = 8 (mod 23)
11^2 = 6 (mod 23)
Square surplus:1,2,3,4,6,8,9,12,13,16,18
Square non residue:5,7,10,11,14,15,17,19,20,21,22
```

2. Calculate Legendre symbol: (17 / 37), (911 / 2003)

```(17/37) = (-1)^((16/2)*(36/2))(37/17)
= (3/17)
= (-1)^((2/2)*(16/2))(17/3)
= (2/3)
= -1
(911/2003) = (-1)^((910/2)*(2002/2))(2003/911)
= -(181/911)
= (-1)*(-1)^((180/2)*(910/2))(911/181)
= (-1)*(2/181)*(3/181)
= (-1)*(-1)*(-1)^((2/2)*(180/2))(181/3)
= (1/3)
= 1
```

3. Judge whether the congruence has a solution

(1) x^2 ≡ 7 (mod 227)

```//227 is a prime number, using Legendre symbols
(7/227) = (-1)^((6/2)*(226/2))(227/7)
= (-1)*(3/7)
= (-1)*(-1)^((2/2)*(6/2))(7/3)
= (-1)*(-1)*(1/3)
= 1
Legendre sign is judged as 1,Have solution
```

(2)x^2 =11 (mod 511)

```//511 is not a prime number and uses Jacobian symbols
(11/511) = (-1)^((10/2)*(510/2))(511/11)
= (-1)*(5/11)
= (-1)*(-1)^((4/2)*(10/2))(11/5)
= (-1)*(1/5)
= -1
Jacobian symbol is judged as-1,unsolvable
//PS: the Jacobian sign is judged as 1, and there may not be a solution
```

Chapter 5:

Knowledge points:

Index and root:

Definition: if M > 1 is an integer and a is a positive integer coprime with m, then

a^e ≡ 1 (mod m)

The established minimum positive integer e is called the exponent (or order) of a to module M and is recorded as ordm(a) if the exponent of a to module M is ϕ (m) , then a is called the primitive root of module M

Primitives are also called generators, or primitives

nature:

1. Let m > 1 be an integer, (a,m)=1, then the integer d makes

​ a^d ≡ 1 (mod m)

The necessary and sufficient condition is ordm (a) |d

Corollary: let m > 1 be an integer, (a,m)=1, then ordm(a)| ϕ (m)

1. Let m > 1 be an integer and a be an integer coprime with M

(1) If B ≡ a (mod m), then ordm(b) = ordm(a)

(2) Let a-1 be the inverse of a module m, that is, a-1 · a ≡ 1 (mod m), then

​ ordm(a^(-1)) = ordm(a)

(3) The necessary and sufficient condition for a ^ D ≡ a^k (mod m) is d ≡ k (mod ordm(a))

3. Let m > 1 be an integer and a be an integer coprime with m, then

1 = a0, a1, a2, ····, a(ordm(a) -1), module m is not congruent.

In particular, when a is the primitive root of module m, that is, ordm(a)= ϕ (m) When,

This ϕ (m) Number 1=a0,a1,a2 ···, a ϕ (m) Simplified residue system of module M composed of - 1

4. Let m > 1 be an integer, a be an integer coprime with m, and d be a nonnegative integer, then

​ ordm(a^d) = ordm(a) / (d,ordm(a))

Inference** 😗* Let m > 1 be an integer * *, * * g be the original root of module m * *. * * let d ≥ 0 be an integer,

Then g^d is the primitive root of module m if and only if (d, ϕ (m) )= 1

5. Let m > 1 be an integer, if module M has an original root g,

Then module M has ϕ ( ϕ (m)) different primordial roots

Original root existence:

Let p be an odd prime number, then the primitive root of module P exists and has ϕ (p-1) primordial roots

Let p be an odd prime number, and all different prime factors of p-1 are q1,q2,... qs,

Then g is the primitive root of module p if and only if:

g^((p-1)/qi ) != 1(mod p), i = 1,2,⋯,s

1. Calculate the index of 2, 5, 10 and 13

```ϕ(13) = 12
12 The factor is 1,2,3,4,6,12
2^1 = 2 (mod 13)
2^2 = 4 (mod 13)
2^3 = 8 (mod 13)
2^4 = 3 (mod 13)
2^6 = 12 (mod 13)
2^12 = 1 (mod 13)
ord13(2) = 12
5^1 = 5 (mod 13)
5^2 = 12 (mod 13)
5^3 = 8 (mod 13)
5^4 = 1 (mod 13)
ord13(5) = 4
10^1 = 10 (mod 13)
10^2 = 9 (mod 13)
10^3 = 12 (mod 13)
10^4 = 3 (mod 13)
10^6 = 1 (mod 13)
ord13(10) = 6
```

2. Seek ord41(10)

```ϕ(41)=40
40 The factor is 1,2,4,5,8,10,20,40
10^1 = 10 (mod 41)
10^2 = 18 (mod 41)
10^4 = 37 (mod 41)
10^5 = 1 (mod 41)
ord41(10) = 5
```

3. Find all primitive roots of module 11

```ϕ(11) = 10
10 The factor is 1,2,5,10
1^1 = 1 (mod 11)
2^1 = 2 (mod 11)
2^2 = 4 (mod 11)
2^5 = 10 (mod 11)
2^10 = 1 (mod 11)
2 Original root
(d,10) = 1
d Is 1,3,7,9
2^1 = 2 (mod 11)
2^3 = 9 (mod 11)
2^7 = 7 (mod 11)
2^9 = 6 (mod 11)
11 The original root of is 2,6,7,9
```

4. (optional) solve the congruence x^22 ≡ 5 (mod 41)

```//I won't do what I choose to do for the time being
```

Chapter 6:

Knowledge points:

Pseudoprime:

Inverse proposition of Fermat's theorem:

If the integer a satisfies (a, n)=1, a^(n-1) ≡ 1 (mod n), then n is a prime number

The composite number n satisfying a^(n-1) ≡ 1 (mod n) is called the pseudo prime of base a, or the pseudo prime to base a

Miller Rabin primality test:

If p is an odd prime number and x is a positive integer less than p, the equation

The solutions of x^2 ≡ 1 (mod p) are only x ≡ 1 (mod p) and X ≡ - 1 ≡ p-1 (mod p)

Miller Rabin primality test:

Given that n is an odd number, it is easy to calculate the integers r,u so that n-1 = 2r,u, where u is an odd number and r ≥ 1 (limiting r ≥ 1 means that n is an odd number).

Fermat test a^(n-1) = a(2r * u) ≡ 1 (mod n)

A more refined algorithm is used to investigate the sequence of r terms (au,a(2u),a(4u),..., a (2 ^ (r-1) * U) (all modulo n). Each term in the sequence is the square of the previous term. Therefore, if the value of one term is equal to 1 or - 1, all the subsequent values in the sequence are equal to 1

Miller Rabin (n) algorithm:

```Find integer r, u, r > 0, u Is an odd number, bring n-1 = (2^r * u)
Select a random positive integer a,a < n
z <- a^u (mod n)
if z = 1 then return( n It could be prime)
for i = 0 to r-1 do
if z = n-1 then return( n It could be prime)
else z <- z^2 (mod n)
return (n It's a composite number)
```
```//Because the Miller Rabin Prime test is difficult to understand, the following examples are listed
Example, if n = 13 ，be n − 1 = 2^2 * 3, r = 2, u = 3,
First iteration:
1)Positive random number a, a < n
set up a = 4，calculation a^u (mod n)
2)  x = 4^3 mod 13 = 12
3) because x = (n-1), return "n May be prime "
Second iteration:
1)Positive random number a, a < n
set up a = 5，calculation a^u (mod n)
2)x = 5^3 mod 13 = 8
3) x It is neither equal to 1 nor 12
4) Do following (r-1) = 1 times
a) x = x^2 mod 13 = 82 mod 13 = 12
b) because x = (n-1), return "n May be prime "
Because both iterations return“ n May be prime "“ n May be prime "
```

1. Prove that 91 is a pseudo prime of base 3

```//Inverse proposition of Fermat theorem
(3,91) = 1
3^(91-1) = 3^90 (mod 91) = (3^6)^15 (mod 91) = (729 (mod 91))^15 (mod 91)
= 1 (mod 91)
```

2. Prove that 25 is a strong pseudo prime of base 7

```//The method of proof is fixed, although it is difficult
25 It's an odd number
25-1 = 24 = 2^3 * 3, r = 3, u = 3
take a = 7, 7^3 = 18 (mod 25), 18^2 = 24 (mod 25)
Return 25 may be prime
Then 25 is a strong pseudo prime of base 7
```

Chapter 7:

Knowledge points:

Group:

Definition: Let G be a non empty set. If an algebraic operation "·" is defined on G, the following conditions are met:

① (closure) ∀ a, b ∈ G, with a · b ∈ G;

② (binding) ∀ a, B, c ∈ G, with a · (B · c) = (a · b) · c;

③ (unit element) there is an element E in G. for ∀ a ∈ g, there is e · a = a · e = a. element E is called unit element, also known as unitary element;

④ (inverse element) each element a in G has an inverse element, that is, there is element a ', so that

A ∙ a ′ = a ′ ∙ a = e, a ′ is called the inverse element of a, and if it is recorded as a^(-1), it is called (G, ·) as a group

Group properties:

1. The unit element in G is unique

2. The inverse element of each element in G is unique

3. Let * be a binary operation in set G,a ∈ G,a ≠ e. if a satisfies, for any a,b ∈ g, there are:

If a * b = a * c, then b = c; if b * a = c * a, then b = c

The element a pair * is said to be reducible (erasable)

Properties of inverse element derivation

1.(a^(-1) )^(-1)=a

2. If a and B are reversible, ab is also reversible, and (ab)(-1)=b(-1) a^(-1)

3. If a is reversible, an is also reversible, and (a ^ n) (- 1) = (a (- 1)) n = a (- n)

```//Property 1
prove:
(a^(-1))^(-1) = (a^(-1))^(-1) * e = (a^(-1))^(-1) * a^(-1) * a = e * a = a
//Property 2
prove:
b^(-1)a^(-1) * a*b = b^(-1) * e * b = e
b^(-1)a^(-1) = (ab)^(-1)
```

Order of group:

Definition: if the number of elements of group G is finite, G is called a finite group, otherwise, G is called an infinite group. The number of elements in G is called the order of the group and is recorded as | g| or # g. the order of an infinite group is recorded as infinity

All integers Z are infinite groups for the group (Z, +) formed by addition

We only care about finite groups

Finite addition group Zn: for any positive integer n(n ≥ 1), a completely nonnegative residue system Zn={0,1,..., n-1} of module n constitutes an addition group with n elements. The unit element is 0. Any element a in the group has an inverse element, | Zn|= n.

Multiplication group Zn * is an important group used in many cryptography schemes.

Zn = {0,1,..., n-1} is a minimum nonnegative complete residue system of module n

Definition: Zn * = {x ∈ Zn: (x, n) = 1}

Zn means that the module n multiplication operation is adopted on Zn to form a finite module n multiplication group. The unit element e=1. For any element a in the group, there is a-1. |zn*|= ϕ (n)

For example, Z8* = {1,3,5,7}, | Z8*|= 4

For prime number p, Zp* = Zp\{0} = {1,2,3,..., p-1}

For example, Z7* = {1,2,3,4,5,6}, | Z7*| = 6

Abel group:

Definition: if the operations in group G satisfy the commutative law, the group is called a commutative group or an Abel group

That is to meet the closure, associativity, unit element, inverse element, commutative law

Judgment theorem of subgroups:

Theorem: if h is a nonempty subset of group < g, · >, then < h, · > is a subgroup if and only if a,b ∈ h, then a ⋅ b^(-1) ∈ H

Mapping:

Definition: the mapping f from one set a to another set B is ∀ a ∈ a, and there is a definite b = f(a) ∈ B corresponding to it

Injective: a,c ∈ a, if a! = C, then f (a)! = F ©

Full shot: B ∈ B, there is always A ∈ A, so that f(a) = b

One-to-one mapping: a mapping that is both surjective and injective

If A = B, the mapping f is also called transformation, that is, the mapping from a set to itself is called transformation.

If the mapping f of A set A to itself is defined as:

For ∀ A ∈ A, there is f(a) = a,

Then the mapping f is called identity mapping, unit mapping or identity transformation, and is recorded as I

Homomorphism:

The definition assumes that G and G 'are two groups, if there is a mapping f: G → G',

If ∀ a,b ∈ G, all f(a ∙ b)= f(a) · f(b)

Then f is called a homomorphism of group G to G '

If f is injective, then f is said to be monomorphism;

If f is a surjection, then f is said to be a full homomorphism;

If f is a one-to-one mapping, then f is isomorphic;

If G = G ', homomorphism f is called endomorphism, and isomorphism f is called automorphism

Isomorphism:

Definition: Let G and G 'be two groups. If there is an isomorphic mapping from G to G', G and G 'are isomorphic, which is recorded as G ≅ G'

If G ≅ G ', it is called G automorphism

The real number addition Group R and the positive real number multiplication group R + are isomorphic, and the isomorphic mapping is f(a) = e^a

nature:

The isomorphism of a group has reflexivity, symmetry and transitivity, that is, it is an equivalent relationship

1)G ≅ G；

2) G '≅ G' can be deduced from G '≅ G';

3) From G ≅ G 'and G ^ ≅ G ^ ′, G ≅ G ^ ′' can be deduced

Cyclic group:

Definition: if there is an element g ∈ g, ∀ a ∈ G can be obtained by the power operation of G,

Then group g is called a cyclic group. g is the generator or primitive of group g,

G is said to generate a cyclic group G, which is written as < g >

Any element a of group G can generate a cyclic group, which is a subgroup of group G

If a is an element of infinite order, a generates an infinite cyclic group,

The generators of G are a and a^(-1)

If a is an n-order element, then a generates an n-order cyclic group, |G|=ordm (a)

Finite n-order cyclic groups can be expressed as

{g0, g1, g2,..., g(n-1)}, two different, where g^0 = e,

g^n = e, g is an n-order element

All powers of g are not equal, so the cyclic group with g as generator

{..., g(-2), g(-1), g0, g1, g^2,...}, is an infinite cyclic group, where g^0 = e, g is an element of infinite order,

Obviously, the elements of an infinite cyclic group are elements of infinite order

The order of the generator of a finite cyclic group is the order of the group

Theorem: each infinite cyclic group is isomorphic with integer additive group Z, and each n-order finite cyclic group is isomorphic with additive group Z/mZ(n-order residual class additive group)

A transformation is a mapping of a set to itself

The multiplication (composition of transformations) of two transformations F and G on the specified set a is as follows: ∀ a ∈ R, fg(a) = f(g(a))

Transformation group:

Definition: let set M be a nonempty set. G be a set composed of all one-to-one mappings from m to itself. G forms a group for the compound operation of mapping, which is called transformation group

Theorem (Cayley theorem) any group is isomorphic to a transformation group

Permutation group:

Definition: the one-to-one transformation of a finite set is called permutation

Let a finite set a have n elements, A = {a1, a2, a3... an},

Then a permutation p can be expressed as:

αi →αki，i = 1，2，3，...，n

It can also be expressed as: (A1, A2... an)

​ ( ak1 ak2 ... akn)

If the specific content of the element is removed, the replacement p can also be expressed as: (1, 2... n)

​ (k1 k2 ... kn)

A permutation is actually an arrangement of elements A. in fact, any arrangement of elements in the first row is a representation, but it is generally expressed in the order of (1, 2, 3,..., n)

1. Define the operation "◦" in the integer set Z: ∀ a,b ∈ Z, a ◦ b=a+b-2, and verify that Z constitutes a group with respect to the operation "◦"

```//Proof of associativity
∀a, b, c∈Z, (a ◦ b) ◦ c = (a + b - 2) ◦ c = a + b - 2 + c - 2 = a + b + c -4
a ◦ (b ◦ c) = a ◦ (b ◦ c) = a ◦ (b + c - 2) = a + b + c - 2 -2 = a + b + c -4
(a ◦ b) ◦ c = a ◦ (b ◦ c)
//Proof unit element
a ◦ b = a = a + b - 2 -> b = 2
//Prove the existence of inverse element
∀a∈z, a ◦ a' = e = 2 = a + a' - 2 = 2
a' = (4 - a)∈z
```

2. Let G be a group and prove that ∀ a, b ∈ G has (1) (a(-1))(-1) = a (2) (ab)^(-1) = b(-1)a(-1)

```prove:
(a^(-1))^(-1) = (a^(-1))^(-1) * e = (a^(-1))^(-1) * a^(-1) * a = e * a = a
prove:
b^(-1)a^(-1) * a*b = b^(-1) * e * b = e
b^(-1)a^(-1) = (ab)^(-1)
```

3. Prove that the intersection of two subgroups of group G is also a subgroup of G

```//Judgement theorem of subgroups
set up G1,G2 yes G Subgroup of.
Then for any a,b∈G1∩G2, have a,b∈G1 And a,b∈G2.
because G1,G2 Is a group, therefore a^(-1)b ∈G1 And a^(-1)b∈G2
therefore a^(-1)b∈G1∩G2.
also G1∩G2 Obviously not empty (All have unit yuan e)
therefore G1∩G2 yes G Subgroup of.
```

4. Find the order of each element of order 13 and 16 cyclic groups respectively, and point out its generator

```//The order is ordm(a)
ϕ(13) = 12
12 The factor is 1,2,3,4,6,12
ord13(1) = 1
ord13(2) = 12
ord13(3) = 3
ord13(4) = 6
ord13(5) = 4
ord13(6) = 12
ord13(7) = 12
ord13(8) = 4
ord13(9) = 3
ord13(10) = 6
ord13(11) = 12
ord13(12) = 2
The generator is 2,6,7,11
ϕ(16) = 8//Just find the order of Coprime with 16
8 The factor is 1,2,4,8
ord16(1) = 1
ord16(3) = 4
ord16(5) = 4
ord16(7) = 2
ord16(9) = 2
ord16(11) = 4
ord16(13) = 4
ord16(15) = 2
16 No generator
```

Code implementation order:

```#include<iostream>
#include<cmath>

using namespace std;

bool judge_prinme(int n)
{
if (n <= 3)
{
return n > 1;//If n < 3, it can only be 1.2.3, 2.3 is prime, and 1 is neither prime nor composite
}
if (n % 6 != 1 && n % 6 != 5)
return false;
int t = static_cast<int>(sqrt(n)) + 1;
for (int i = 5; i <= t; i += 6)
{
if (n % i == 0 || n % (i + 2) == 0)
{
return false;
}
}
return true;
}

int gcd(int a, int b)
{
int temp;
if (a == 0 || b == 0)
return 0;
if (a < b)
{
temp = a;
a = b;
b = temp;
}
while (b != 0)
{
temp = b;
b = a % b;
a = temp;
}
return temp;
}

int Find_Euler_function(int num)
{
if (judge_prinme(num))
{
return num - 1;
}
else
{
int count = 0;
for (int i = 1; i < num; i++)
{
if (gcd(i, num) == 1)
count++;
}
return count;
}
}

void Find_order(int num)
{
for (int i = 1; i < num; i++)
{
for (int j = 1; j < num; j++)
{
if (static_cast<unsigned long long int>(pow(i, j)) % num == 1)
{
cout << i << "The order of is:" << j;
if (j == Find_Euler_function(num))
cout << ",This element is a generator." ;
cout << endl;
break;
}
}
}
}

int main()
{
cout << "Please enter the order of the cyclic group of the element:" << endl;
int num;
cin >> num;
int Eul_fun = Find_Euler_function(num);
cout <<"φ(" << num << ")=" << Eul_fun << endl;
cout << "φ(" << Eul_fun << ")=" << Find_Euler_function(Eul_fun) << endl;
Find_order(num);
return 0;
}
```

Chapter 8:

There are no homework questions in Chapter 8. No blind guess

Posted on Wed, 17 Nov 2021 23:17:48 -0500 by eightFX