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Mathematics of Cryptography Algebraic Structures

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Groups

A group (G) is a set of elements with a binary operation

(•)  that satisfies four properties (or axioms). A

commutative group satisfies an extra property,

commutativity:

Closure: If a and b are elements of G, then c=a•b isalso an element of G. 

Associativity: a•(b•c)=(a•b)•c 

Commutativity: a•b=b•a 

Existence of identity: For all element a in G, there

exists and element e, identity element, s. t. e•a=a•e=a 

Existence of inverse: for each a there exists an a’, 

inverse of a, s.t. a•a’=a’•a=e. 

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Figure Group

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The set Zn * with the multiplication operator, G = , is also

an abelian group.

Identity element is 1.

Example 

Let us define a set G = < {a , b , c , d }, •> and the operation as shown

below.

Example 

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A very interesting group is the permutation group. The set is the

set of all permutations, and the operation is composition: applying

one permutation after another.

Example 

Figure Composition of permutation

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Example

Table Operation table for permutation group

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In the previous example, we showed that a set of permutations

with the composition operation is a group. This implies that usingtwo permutations one after another cannot strengthen the security

of a cipher, because we can always find a permutation that can do

the same job because of the closure property.

Example 

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Is the group H = a subgroup of the group G = ?

Example 

SolutionThe answer is no. Although H is a subset of G, the operations

defined for these two groups are different. The operation in H is

addition modulo 10; the operation in G is addition modulo 12.

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Cyclic Subgroups

If a subgroup of a group can be generated using thepower of an element, the subgroup is called the cyclic

subgroup.

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Four cyclic subgroups can be made from the group G = .

They are H1 = , H2 = , H3 = , and

H4 = G.

Example 

H1H4

H2

H3

H2

H4

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Three cyclic subgroups can be made from the group

G = . G has only four elements: 1, 3, 7, and 9. The cyclicsubgroups are H1 = , H2 = , and H3 = G.

Example 

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Cyclic Groups

A cyclic group is a group that is its own cyclic subgroup.

Generator: The element that generates the cyclicsubgroup can also generate the group itself. The

elements of a finite cyclic group are:

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Three cyclic subgroups can be made from the group

G = . G has only four elements: 1, 3, 7, and 9. The cyclicsubgroups are H1 = , H2 = , and H3 = G.

Example 

a. The group G = is a cyclic group with two generators,

g  = 1 and g  = 5.

b. The group G = is a cyclic group with two generators,g  = 3 and g  = 7.

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Lagrange’s Theorem: relates the order of a group to the

Order of its subgroup.Assume that G is a group, and H is a subgroup of G. If

the order of G and H are |G| and |H|, respectively, then,

based on this theorem, |H| divides |G|.

Order of an Element

The order of an element is the order of the cyclicsubgroup it generates.

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Example 

a. In the group G = , the orders of the elements are:ord(0) = 1, ord(1) = 6, ord(2) = 3, ord(3) = 2, ord(4) = 3,

ord(5) = 6.

b. In the group G = , the orders of the elements are:ord(1) = 1, ord(3) = 4, ord(7) = 4, ord(9) = 2.

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Ring

A ring, R = , is an algebraic structure with two

operations.

Ring

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Example 

The set Z with two operations, addition and multiplication, is acommutative ring. We show it by R = . Addition satisfies

all of the five properties; multiplication satisfies only three

properties.

Fi ld

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Field

A field, denoted by F = is a commutative ring

in which the second operation satisfies all five properties

defined for the first operation except that the identity of

the first operation has no inverse.

Field

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Galois showed that for a field to be finite, the number ofelements should be p n , where p   is a prime and n   is a

positive integer.

Finite Fields: are normally used in cryptography

A Galois field, GF(p n ), is a finite field

with p n  elements.

Note

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When n  = 1, we have GF(p ) field. This field can be the setZp , {0, 1, …, p − 1}, with two arithmetic operations.

GF(p ) Fields

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Example 

A very common field in this category is GF(2) with the set {0, 1}

and two operations, addition and multiplication, as shown below.

GF(2) f ield

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Example 

A very common field in this category is GF(2) with the set {0, 1}

and two operations, addition and multiplication, as shown below.GF(2) f ield

1. The set has only two elements –  binary digits or bits 0 and 1.

2. Addition = XOR operation. Multiplication = AND

3. Addition and multiplication operations are same opn. XOR.

4. Multiplication and division are same AND operation.

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Example 

We can define GF(5) on the set Z5 (5 is a prime) with addition and

multiplication operators as shown below.

GF(5) f ield

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Summary

Summary

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GF(2n) FIELDS

I n cryptography, we often need to use four operations(addition, subtraction, mul tiplication, and division). I n

other words, we need to use fields. We can work in

GF(2 n  ) and uses a set of 2 n  elements. The elements in

this set are n-bi t words.

1 Polynomials

2  Using A Generator

3  Summary

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Example 

Let us define a GF(22) field in which the set has four 2-bit words:

{00, 01, 10, 11}. We can redefine addition and multiplication for

this field in such a way that all properties of these operations are

satisfied, as shown below.

An example of GF (2 2  ) f ield

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Example 

Let us define a GF(22) field in which the set has four 2-bit words:

{00, 01, 10, 11}. We can redefine addition and multiplication for

this field in such a way that all properties of these operations are

satisfied, as shown below.

An example of GF (2 2  ) fieldEach word is a additive

inverse of itself.

Every word except 00 has a

mul tiplicative inverse.

Addition and multiplication are

defined in terms of polynomials.

Polynomials

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Polynomials

A polynomial of degree n  − 1 is an expression

of the form

where x i  is called the ith term and a i  is called coefficient

of the i th term.

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Example 

Figure below shows how we can represent the 8-bit word(10011001) using a polynomials.

Representation of an 8-bi t word by a polynomial

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Example 

To find the 8-bit word related to the polynomial x 5  + x 2  + x , we

first supply the omitted terms. Since n   = 8, it means the

polynomial is of degree 7. The expanded polynomial is

This is related to the 8-bit word 00100110.

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GF(2 n ) Fields

Any operation on polynomials actually

involves two operations: on coefficients,

on two polynomials.Coefficients 0 and 1 belong to GF(2).

The polynomials belong to GF(2n ).

Note

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GF(2 n ) Fields

Polynomials representing n -bit words

use two fields: GF(2) and GF(2n ).

Note

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Addition of two polynomials never creates a polynomial out of the

set.

Multiplication of two polynomials may create a polynomial with

degree more than n-1.

Need to divide the result by a modulus and keep the remainder.

For the sets of polynomials in GF(2n ), a group of polynomials of

degree n  is defined as the modulus.

Modulus

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The modulus in this case acts as the prime polynomial, that is no

polynomial in this set can divide this polynomial.

A prime polynomial cannot be factored into a polynomial with

degree less than n.

Such polynomials are referred to as irreducible polynomials. 

Modulus

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For the sets of polynomials in GF(2n ), a group of

polynomials of degree n  is defined as the modulus. Suchpolynomials are referred to as irreducible polynomials. 

Modulus

L ist of irreducible polynomials

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Addition

Add the coefficients in GF(2).

Adding two polynomials of degree n-1

always creates a polynomial of degree n-1.

Additive identity: zero polynomial  – all

coeffn set to zero.

Additive inverse: of a polynomial in GF(2) isthe polynomial itself.

Addition and subtraction operations are the

same operation.

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Example 

Let us do (x5 + x2 + x) (x3 + x2 + 1) in GF(28). We use the symbol

to show that we mean polynomial addition. The following shows

the procedure:

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1. The coefficient multiplication is done in GF(2).

Multiplication

Sum of multiplication of each term of the first polynomial

with each term of the second polynomial.

3. The multiplication may create terms with degree more

than n  − 1, which means the result needs to be reducedusing a modulus polynomial.

2. The multiplying x i  by x  j  results in x i + j .

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Multiplicative Identity: The multiplicative identity is

always 1, in GF(2^8) it is 00000001.

Multiplication

Multiplicative Inverse:  It can be found using the

extended Euclidean algorithm.

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Example 

Find the result of (x 5 + x 2 + x ) (x 7 + x 4 + x 3 + x 2 + x ) in GF(28)

with irreducible polynomial (x 8 + x 4 + x 3 + x  + 1). Note that we use

the symbol to show the multiplication of two polynomials.

Solution

To find the final result, divide the polynomial of degree 12 by the

polynomial of degree 8 (the modulus) and keep only the

remainder.

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Polynomial division with coeff icients in GF (2)

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Example 

In GF (2

4

), find the inverse of (x 2

 + 1) modulo (x 4

 + x  + 1).

Using the Extended Euclidean Algorithm

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Example 

Find the multiplicative inverse of 11 in Z26.

Solution

The gcd (26, 11) is 1; the inverse of 11 is 7 or 19.

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Example 

In GF (24), find the inverse of (x 2 + 1) modulo (x 4 + x  + 1).

Solution

Using Extended Euclidean Algorithm 

The answer is (x 3 + x  + 1) as shown below.

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Example 

In GF(28), find the inverse of (x5) modulo (x 8 + x 4 + x 3 + x + 1).

Solution

Euclidean algorithm for above example

The answer is (x 5 + x 4 + x 3 + x ) as shown below.

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The computer implementation uses a better algorithm,

repeatedly multiplying a reduced polynomial by x .

Multliplication Using Computer

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Example 

Find the result of multiplying P1 = (x 5 + x 2 + x ) by P2 = (x 

7 + x 4 + x 3 

+ x 2 + x ) in GF(28) with irreducible polynomial (x 8 + x 4 + x 3 + x  +

1)

Solution

We first find the partial result of multiplying x 0, x 1, x 2, x 3, x 4, and

x 5  by P2. Note that although only three terms are needed, the

product of x m   P2  for m   from 0 to 5 because each calculation

depends on the previous result.

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Example 

An eff icient algor ithm

Continued

Find the result of multiplying P1 = (x 5 + x 2 + x ) by P2 = (x 

7 + x 4 + x 3 

+ x 2 + x ) in GF(28) with irreducible polynomial (x 8 + x 4 + x 3 + x  +

1)

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Example 

Repeat above Example using bit patterns of size 8.

Solution

We have P1 = 000100110, P2 = 10011110, modulus = 100011010

(nine bits). We show the exclusive or operation by .

An eff icient algor ithm for multipli cation using n-bit words

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Example 

Addition table for GF(2^3)

Continued

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Example 

Multiplication table for GF(2 3  )

Continued

Using a Generator

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g

Sometimes it is easier to define the elements of the

GF(2n ) field using a generator.

In this Field with the irreducible polynomial f(x), an

element in the field, a, must satisfy the relation f(a) = 0.

In particular, if g is a generator of the Field, then f(g)=0.

It can be proved that the elements can be generated as:

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Example 

Generate the elements of the field GF(24) using the irreducible

polynomial ƒ(x ) = x 4 + x  + 1.

The elements 0, g 0

, g 1

, g 2, and g 3

 can be easily generated, becausethey are the 4-bit representations of 0, 1, x 2, and x 3. Elements g 4 

through g 14, which represent x 4  though x 14 need to be divided by

the irreducible polynomial. To avoid the polynomial division, the

relation ƒ(g ) = g 4 + g  + 1 = 0 can be used.

Solution

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Example  Continued

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Inverses 

Additive inverses:

Additive inverse of each element is the element itselfbecause addition and subtraction in this f ield are same.

 – g^ 3 = g^ 3

Mul tipli cative inverses:To find the multiplicative inverses g3

(g 3  ) -1 = g-3 = g12 = g3+g2+g+1

Note the exponents are calculated mudulo 2^n-1=15.

See g3 x g12 = g15 = g0 = 1

So g3 and g12 are multiplicative inverses of each other.

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Example 

The following show the results of addition and subtractionoperations:

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Example 

The following show the result of multiplication and division

operations:

Summary

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The finite field GF(2

n 

) can be used to define fouroperations of addition, subtraction, multiplication and

division over n -bit words. The only restriction is that

division by zero is not defined.