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=4536 =(657306)(30645*6) =(657(963657)*2)((963657)6*(657306*2)) =(6572*963+2*657)(9636576*(6572*(963657))) =3*6572*963(9636576*(6572*963+2*657)) =3*6572*963963+657+6*65712*963+12*657 =22*65715*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 firstorder 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 + an1xn1 +...+ 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)∙ (xpx) + 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 + an1xn1 +... + 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,..., p1 (mod p)
Is the solution of the congruence x^(p1) ≡ 1 (mod p)
Corollary, if p is a prime number, then for any integer x, there is
x^(p1) – 1 ≡ (x – 1) (x – 2) ...(x – (p1)) ( mod p)
The necessary and sufficient condition for integer n to be prime is (n1)+ 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 (p1)/2 square residues and (p1)/2 square non residues.
When finding the square residue of module p, only the following numbers can be calculated:
12,22,...,((p1)/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^((p1)/2) ≡ 1 (mod p)
(2) The necessary and sufficient condition that a is a module p squared non residual is
a^((p1)/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, ap
Legendre sign calculation still uses Euler discriminant method
a^((p1)/2) ≡ 1 (mod p)
a^((p1)/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 as1,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:
 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)

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 a1 be the inverse of a module m, that is, a1 · 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 (p1) primordial roots
Let p be an odd prime number, and all different prime factors of p1 are q1,q2,... qs,
Then g is the primitive root of module p if and only if:
g^((p1)/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^(n1) ≡ 1 (mod n), then n is a prime number
The composite number n satisfying a^(n1) ≡ 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 ≡ p1 (mod p)
Miller Rabin primality test:
Given that n is an odd number, it is easy to calculate the integers r,u so that n1 = 2r,u, where u is an odd number and r ≥ 1 (limiting r ≥ 1 means that n is an odd number).
Fermat test a^(n1) = 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 ^ (r1) * 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 n1 = (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 r1 do if z = n1 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 = (n1), 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 (r1) = 1 times a) x = x^2 mod 13 = 82 mod 13 = 12 b) because x = (n1), 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^(911) = 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 251 = 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,..., n1} 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,..., n1} 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 a1. zn*= ϕ (n)
For example, Z8* = {1,3,5,7},  Z8*= 4
For prime number p, Zp* = Zp\{0} = {1,2,3,..., p1}
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
Onetoone 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 onetoone 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
Integer addition group Z and even addition Group E are isomorphic
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 norder element, then a generates an norder cyclic group, G=ordm (a)
Finite norder cyclic groups can be expressed as
{g0, g1, g2,..., g(n1)}, two different, where g^0 = e,
g^n = e, g is an norder 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 norder finite cyclic group is isomorphic with additive group Z/mZ(norder 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 onetoone 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 onetoone 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+b2, 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