Matrix Representation of Groups in the Finite Fields GF (2n)

(1)



Abstract— The representation of mathematical fields can be accomplished by binary rows (or columns) of a binary triangular matrix as the Hamming’s matrices, but this representation don’t show the basic product properties of the fields, that is the nonzero elements of the fields forms a cyclic multiplicative group.

In this paper we show that the elements of the fields GF(2ⁿ), and their subgroups, can represent as square matrices by m – sequences, which satisfies the product properties as a cyclic group.

Index Term - Galois fields, m-sequences,cyclic groups, Orthogonal sequences.

I. INTRODUCTION

m- Linear Recurring Sequences

Let k be a positive integer and

 , 

₀

, 

₁

, ..., 

_k_₁ are elements in the field

F

q, then the sequence

a

₀

, a

₁

, ...

^is

called non homogeneous linear recurring sequence of order kiff :

) 1 ( 1 ,..., 1 , 0 , ,

...

1

1 0

2 2 1 1



















^

 















k

i

i n i k

n

q i n

k n k k n k k n

a a

or

k i

F a

a a

a

The elements

a

₀

, a

₁

, ..., a

_k_₁ are called the initial values (or the vector

( a

₀

, a

₁

, ..., a

_k_₁

)

is called the initial vector).

If

  0

then the sequence

a

₀

, a

₁

, ...

is called homogeneous linear recurring sequence (H. L. R.

S. ), except the zero initial vector, and the polynomial 0

1 1

1

...

)

( x  x  

_

x

^

   x  

f

^k _k ^k (2)

is called the characteristic polynomial. In this study, we are limited to



₀

 1

. [1]-[3]

II. THEIMPORTANCEOFTHISRESEARCHAND ITSOBJECTIVES

The elements of the fields GF(2ⁿ), and their subgroups, can be represented assquare matrices by m – sequences, which satisfies the multiplicative properties as a cyclic group, that is it will be useful in many other scientific branches. For example orthogonal sets in the forward and the inverse link of communications channels in the CDMA systems especially in

Manuscript Received May, 2014

Ahmad Hamza Al Cheikha, Department of Mathematical Science, Ahlia University, Kingdom of Bahrain.

the second (IS-95-CDMA), the third…. (CDMA200,…), the pilot channels, the Sync channels, and the Traffics channelsin the present or in many other scientific branches in the future.

III. RESEARCHMETHODSANDMATERIALS Basic Definitions and Theorems

Definition 1.Let S be a nonempty set and

a

₀

, a

₁

, ....

^is sequence from S and if there is

r  0

such that:

0 ;

₀

0

;

 



 a n n n

a

_n _r _n (3)

Then this sequence is called Ultimately Periodic Sequence, and r is called a period of this sequence, the smallest positive integer between these r’s is called the period of this sequence, and the smallest non negative

n

₀ such that:

0 ;

₀

0

;

 



 a n n n

a

_n _r _n ,

is called Pre-Period, [1][ 4]

Definition 2.The Ultimately Periodic Sequence

....

, ,

₁

0

a

with the smallest period r is called a periodic iff:

a

_n__r

 a

_n _;

n  0 , 1 , ...

[1]-[ 4]

Definition 3.The complement of the binary vector

)

,..., ,

( x

₁

x

₂

x

_n

X 

is the vector

X  ( x

₁

, x

₂

,..., x

_n

)

, when

 





 

1 0

0 1

i i

i

if x

x

x if

. [1]-[ 4]

Definition 4. (Euler function



).

 (n )

is the number of the natural numbers that are relatively prime

withn.[5]-[ 8]

Definition 5.AnyPeriodic Sequence

a

₀

, a

₁

, ....

over

F

_q with prime characteristicpolynomial is an orthogonal cyclic code and ideal auto correlation [1]-[ 10].

Definition 6.The binary periodic sequence

( a

_i

)

_i__N, with the period r has the property of “ Ideal Auto

Correlation” if and only if its periodic auto correlation

  

R

a of the form:

    



 

otherwise

r for

R

_a

r

; 1

) mod(

0 ;



When:



^

  ^{   }











¹

0

1 )

(

r

t

t a t a

R

a



^ [1],[2]

Theorem1.

i. If

a

₀

, a

₁

, ....

is a homogeneous linear recurring sequence of order k in

F

_q , satisfies (1) then this sequence is periodic.

Matrix Representation of Groups in the Finite Fields GF (2 ⁿ )

AhmadHamza Al Cheikha

(2)

ii. If this sequence is homogeneous linear recurring sequence, periodic with the period r, and its characteristic polynomial

f (x )

then

r ord f (x )

. [6]

iii. If the polynomial

f (x )

is primitive then the period of the sequence is

q

^k

 1

, and this sequence is called m – sequence.

Lemma 2.( Fermat’s theorem ). IfF is a finite field and has q elements then every elementa of F satisfies the

equation:

x

^q

 x

. [6],[9]

Theorem 3.For any primitive element p and any positive integern there is a field F, which has

p

ⁿ^elements

and any two fields having

q  p

ⁿ Elements are isomorphic. [6],[9],[11]

Theorem 4.

i.

( q

^m

 1 ) ( q

ⁿ

 1 )  m n

(4)

ii.If

F

_qis a field of order

q  p

ⁿthen any subfield of them of the order

p

^m and

m n

and by inverse if

n

m

then in the field

F

_qthere is a subfield of order

p

m . [6],[9],[11]

Theorem 5.The number of irreducible polynomials in

)

(x

F

_q of degree m and order e is

 ( e / ) m

,if

e  2

, whenm is the order of q by mod e, and equal to 2 if m = e = 1, and equal to zeroelsewhere. [6]-[9]

Theorem 6. If

g (x )

is a characteristic prime polynomial of the (H. L. R. S.)

a

₀

, a

₁

, ....

of degree k, and



is a root of

g (x )

in any splitting field of F₂ then the general bound of the sequence is:

k n

i i n

i

C

a 



 



  

^

 2 1

1



.

[11],[12].

* The study here, is limited to the fields Galois

GF ( 2

ⁿ

)

IV. RESULTSAND DISCUSSION

A. First step

Theorem 7

:

^Suppose

^a

₀

^, ^a

₁

^, ^....

is a non zero homogeneous linear recurring sequence of order k over

F

₂

 { 0 , 1 }

and

) (x

f

is its prime characteristic polynomial then the first

1 2 



^k

r

bounds with all its cyclic shifts forming an additive group.

Proof: This sequence is periodic with period

r  2

^k

 1

. We suppose

$   S

₁

, S

₂

,..., S

_r



where

S

₁

  a

₁

a

₂

... a

_r



is the sequence of the first

r  2

^k

 1

bounds, and



₁

....

₁

 , ...,

2

 a

_r

a a

_r

S S

_r

  a

₂

a

₃

... a a

_r ₁



are allits cyclic shifts, and we suppose

O  S

₀

  0 ... 0 

^,

 

₀

$ S

S  

and if



is a root of the prime polynomial

)

(x f

and

:















 

 





 

  



1

0 1

0

1

0 0 }, 0 {

2 ..., , 2 , 1 , 0 , :

) 2 (

k

j j k

j

k j

j i

i

k

b i

GF



And the function:

h : GF ( 2

^k

)  S

as following:



0 1 1

 

0 1 1 ₂ ₂



) ( )

(  h i  h b b b

_

 b b b

_

b b

k_

h 

ⁱ



_k _k _k

Then h is one-to-one corresponding and:



 











1 0 ),

( . ) . (

) ( ) ( ) (

or m h

m m

h

h h

h

i i

j i



And

h

is Linear Transformation and isomorphism from the additive group

 ^GF ⁽ ²

^k

^), ^ 

to the additive group

  S ,  .

B. Second Step

Theorem 8: Suppose

a

₁

, a

₂

,...

is a non zerohomogeneous linear recurring sequence of order k in F₂ and f(x) is their primitive characteristic polynomial, S₁ is the initial bounds where

r  2

^k

 1

and

$   S

₁

, S

₂

,..., S

_r



^are

the all cyclic shifts. Let A is a matrix which its rows are elements $ respectively, then

 ^A

ⁱ

^, ⁱ ^ ¹ ^,..., ^r 

is a cyclic multiplicative group, relatively to product of matrices, having the period of

S

₁

( x )

and rows ofAⁱare the shifts to rows of A.

Proof:

Suppose

 





 







S

r

S S A ...

2 1

and we will compute

A

²

 A  A

.

Suppose the first row



₁ in the matrix

A

²then:







I i

S

i



1

whenI the set of allcolumns in A which does not start by zero, and we see that:

$

&

$

, Y  X  Y  X  Y  X

Then



₁

 S

_l

 $

The second row



₂in

A

² is a result of shift i by 1digit to

the right, then: ₂ ₁ _₁

 



 

^l

I i

i

S

 S

, and respectively we

have ₁ _ _₁

  



 

^l ^r

I i

r i

r

S S



, when the

indexescomputed by mod r, then the rows of the matrix

A

2are shifts to rows of A, On other hand we suppose that

  x  S     x , S x ,..., S

_r

  x 

$ 

₁ ₂ then:

(3)

 

)

1

( 





I i

i

x S

 x ;

    )

(

₁

2

 



 



I i

i I

i

x xS x

S

 x ;…

  )

( 

¹





 I i

i r

r

x x S x



And:

  x x S   x   x S   x S

I i

i I

i

i 2

1 1 1

1

     









When:

^S

₁²

    ^x ^ ^$ ^x

, and the calculations are done by

 

 



  

 





^

 1 mod

x

²^k ¹ , And we have:

  x xS

₁²

  x

2





; … ;

^

_r

  ^x ^ ^x

^r^¹

^S

₁²

  ^x

Suppose

 ^f

_i

  ^x 

denotes the row of coefficients of

^f

_i

  ^x

^,

respectively to increasing exponents of x, and which has the length r, then:

   

 

   

 

  ^

 





 









 







 









 





 







 

x S x

x xS

x S A x

S x

x xS

x S

x S A

r r

r 2

1 1 2 1 2 1 2

1 1 1 1 2

1

.

; .

. ; ... ;

 

r i

x S x

x xS

x S

A

i r

i i

i

, 1 , 2 ,...., .

1 1

1



 





 









When:

S

₁ⁱ

    x  $ x ; i  1 , ...., r

, then:

   

 

  S   x

x A x x S x x

x S

x S A

r r r

12

1 1 2

1 2

1

. 1 , .

1 .

 







 









 







 









 





 









,…

…,   , .

1

1 1

1,2,...,r i

x S x x

A

ⁱ

r

i



 







 











Result 1: The period of the sequence

A , A

²

, A

³

,....

is equal to

ord ( S

₁

( x ))

and divides

2

^k

 1

.

Result 2: If

ord ( S

₁

( x ))  2

^k

 1

then S is representation to the field

GF ( 2

^k

)

.

Result 3: If

ord ( S

₁

( x ))  l and l 2

^k

 1

then

1 2 



^m

l

and

 A 

is a representation to one subgroup of order l in the field

GF ( 2

^k

)

Result 4: If

ord ( S

₁

( x ))  1  2

^l

and l k

then :

}

{O A  



is representation to the Subfield

GF ( 2

^l

)

. Result 5: If

2

^k

 1

is prime then all elements of

S

_i

(x )

are of the order

2 ^k  1

except only one of them that is of the order one

C. Third Step

Example 1: If



1 )

( x  x

³

 x 

f

and generates

GF ( 2

³

)

then the Binary representation of the elements of

GF ( 2

³

)

^is:

Where

( i ), i  0 , 1 , 2 , ..., 7

is the symbol of the sequence i.

The field

GF ( 2

³

)

contains two subfields :

GF ( 2 )

and the same

GF ( 2

³

)

and the divisorsof the number 7 are 1 and 7 consequently

GF ( 2

³

)

contains twomultiplicativesubgroups are:

GF

^

( 2 )

^and

GF

^

( 2

³

)

^.

Suppose the Linear Recurring Sequence be n

n

a a

a

_₃



_₁



or

a

_n_₃

 a

_n_₁

 a

_n

 0

(1)

Figure(1): Linear feedback register of degree3 generates sequence (1)

With the characteristic equation

x

³

 x  1  0

and the characteristic polynomial

f ( x )  x

³

 x  1

,which is a prime and generates 3

F

2 and if

^x ^ ^ ^ ^GF   ²

³ is a root of

)

(x

f

then the solutions of characteristic equation are

}

, ,

{ 

ⁿ



²ⁿ



⁴ⁿ . The general solution of equation (1) isgiven by

a

_n

 

²

 

ⁿ

 

⁴

 

²ⁿ

   

⁴ⁿ, and thesequence is periodic with the period

2

³

 1  7 .

For the initial position:

a

₁

 1 , a

₂

 0 , a

₃

 0

, then S1 = (1 0 0 1 0 1 1

)

and by the cyclic permutationsonS1 we have

$   S

₁

, S

₂

, S

₃

, S

₄

, S

₅

, S

₆

, S

₇



where:

The first three digits in each sequence are the initial position of the feedback register.

(4)

The matrix A of the cyclic permutations of S₁ is:

The first row in this matrix is called The head row.

If the function

h : GF ( 2

³

)  $

^where:

i

h i h

i

h (  )  ( ) 

= [The row of the matrix A corresponding of the initial position I ].

Then

h

_i is isomorphism from the group

( GF ( 2

³

),  )

^{on the} group

($, )

.

I- In this matrix the head row is: S₁



1 0 0 1 0 1 1



and the head polynomial is:

6 5 3 1

( ) 1 )

1 ( S x x x x

h     

.

We see that:

A

₂

 A

²

 A

and

 O, A 

is a representation of the field

GF ( 2 )

, where O is zero matrix

.

II- We suppose that;















 



1 1 0 1 0 0 1

0 1 0 0 1 1 1

1 0 1 0 0 1 1 B1 B

In this matrix the head row is:



1 1 0 0 1 0 1



) 3 ( S₂

h and the corresponding

head polynomial is:

S

₂

( x )  1  x  x

⁴

 x

⁶, Thus















 



1 0 1 0 0 1 1

1 0 0 1 1 1 0

0 1 0 0 1 1 1 2

2 B

B

1 1 1 0 0 1 0

) 5 ( S₃

head polynomial is: ²

( ) 1

² ⁵

2

x x x x

S    

, Thus















 



0 1 0 0 1 1 1

0 0 1 1 1 0 1

1 0 0 1 1 1 0 3 3 B B

In this matrix the head row is: h(4)S₄0 1 1 1 0 0 1 and the corresponding

head polynomial is: ³

( )

² ³ ⁶

2

x x x x x

S    

, Thus















 



1 0 0 1 1 1 0

0 1 1 1 0 1 0

0 0 1 1 1 0 1 4 4 B B

1 0 1 1 1 0 0

) 6 ( S₅

head polynomial is: ⁴

( ) 1

² ³ ⁴

2

x x x x

S    

, Thus















 



0 0 1 1 1 0 1

1 1 1 0 1 0 0

0 1 1 1 0 1 0

5 5 B B



0 1 0 1 1 1 0



) 1 ( S₆

head polynomial is: ⁵

( )

³ ⁴ ⁵

2

x x x x x

S    

, Thus















 



0 1 1 1 0 1 0

1 1 0 1 0 0 1

1 1 1 0 1 0 0 6 B6 B



0 0 1 0 1 1 1



) 2 ( S₇

head polynomial is: ⁶

( )

² ⁴ ⁵ ⁶

2

x x x x x

S    

, Thus















 



1 1 1 0 1 0 0

1 0 1 0 0 1 1

1 1 0 1 0 0 1

7

7 B

B



1 0 0 1 0 1 1



) 7 ( S₁

head polynomial is: ⁷

( ) 1

³ ⁵ ⁶

2

x x x x

S    

, and

B B B

₈



⁸



.

We see that

 O , B

₁

, B

₂

, B

₃

, B

₄

, B

₅

, B

₆

, B

₇



is a representation of

GF ( 2

³

)

and

7 ) ( ) ( ) ( ) ( ) ( )

( B

₁

 ord B

₂

 ord B

₃

 ord B

₄

 ord B

₅

 ord B

₆

 ord

The field 3

F

2 contains

^   ²

^k

^ ¹ ^k ^ ^   ²

³

^ ¹ ³ ^ ²

third degree irreducible polynomials that are:

1 )

( x  x

³

 x 

f

and its conjugate

g ( x )  x

³

 x

²

 1

, and

g (x )

is a prime polynomial then we can represent

23

F

by two different ways.

D. Fourth step

Example 2:If



1 )

( x  x

⁴

 x 

f

and generates

GF ( 2

⁴

)

then the binary representation of the elements of

GF ( 2

⁴

)

is:

Where

( i ), i  0 , 1 , 2 , ..., 15

is the symbol of the sequence

i.

The field

GF ( 2

⁴

)

contains three subfields are:

GF ( 2 )

,

(5)

) 2 (

²

GF

and the same

GF ( 2

⁴

)

and the divisors of the number 15 are 1, 3, 5 and 15 consequently

GF ( 2

⁴

)

contains four multiplicative subgroups :

GF

^

( 2 )

,

GF

^

( 2

²

)

, one group of the order 5 and

GF

^

( 2

⁴

)

.Suppose the Linear Recurring Sequence:

n n

n

a a

a

_₄



_₁



^or

a

_n_₄

 a

_n_₁

 a

_n

 0

(2)

Figure(2): Linear feedback register of degree 4 generates sequence (2)

With the characteristic equation

x

⁴

 x  1  0

and the characteristic polynomial

f ( x )  x

⁴

 x  1

, which is prime and generates ₄

F

2 ^{and if}

^x ^ ^ ^ ^GF   ²

⁴ is a root of

)

(x

f

then the solutions of characteristic equation are

 ^ ^, ^

²

^, ^

⁴

^, ^

⁸

^ ^

²

^ ¹ 

and the general solution of equation (2) is given by:

n n

n

c c a c c

a 

₁

 

₂ ²



₃



⁴



₄



⁸

For

a

₁

 1

,

a

₂

 0

,

a

₃

 1

,

a

₄

 0

we have:

n n

n

a

a  

⁹

  

¹² ²

 

⁹



⁴

 

³



⁸

and the sequence is periodic with the period

2

⁴

 1  15

.

   

 ⁰ ^, ¹ ^, ⁰ ^, ¹ ^, ¹ ^, ¹ ^, ¹ ^, ⁰ ^, ⁰ ^, ⁰ ^, ¹ ^, ⁰ ^, ⁰ ^, ¹ ^, ¹ 

...;

;..

0 , 0 , 1 , 0 , 0 , 0 , 1 , 1 , 1 , 1 , 0 , 1 , 0 , 1 , 1

; 1 , 0 , 0 , 1 , 0 , 0 , 0 , 1 , 1 , 1 , 1 , 0 , 1 , 0 , 1

15

2 1



s

s s

The first four digits in each sequence are the initial position of the feedback register.

The matrix of the cyclic permutations of S₁ is the following matrix:















1 1 0 0 1 0 0 0 1 1 1 1 0 1 0

0 1 1 0 0 1 0 0 0 1 1 1 1 0 1

1 0 1 1 0 0 1 0 0 0 1 1 1 1 0

0 1 0 1 1 0 0 1 0 0 0 1 1 1 1

1 0 1 0 1 1 0 0 1 0 0 0 1 1 1

1 1 0 1 0 1 1 0 0 1 0 0 0 1 1

1 1 1 0 1 0 1 1 0 0 1 0 0 0 1

1 1 1 1 0 1 0 1 1 0 0 1 0 0 0

0 1 1 1 1 0 1 0 1 1 0 0 1 0 0

0 0 1 1 1 1 0 1 0 1 1 0 0 1 0

0 0 0 1 1 1 1 0 1 0 1 1 0 0 1

1 0 0 0 1 1 1 1 0 1 0 1 1 0 0

0 1 0 0 0 1 1 1 1 0 1 0 1 1 0

0 0 1 0 0 0 1 1 1 1 0 1 0 1 1

1 0 0 1 0 0 0 1 1 1 1 0 1 0 1

A

Or briefly:















 



1 1 0 0 1 0 0 0 1 1 1 1 0 1 0

0 0 1 0 0 0 1 1 1 1 0 1 0 1 1

1 0 0 1 0 0 0 1 1 1 1 0 1 0 1

A1

A

The first row in this matrix is called The head row.

If the function

h : GF ( 2

⁴

)  $

^where:

i

h i h

i

h (  )  ( ) 

= [The row of the matrix A corresponding of the initial position i], then

h

_i is isomorphism from the additive group

( GF ( 2

⁴

),  )

on the additive group

($, )

_. In this matrix the head row is:



1 0 1 0 1 1 1 1 0 0 0 1 0 0 1



) 8 ( S₁ h

And the corresponding head polynomial is:

14 11 7 6 5 4

1

( x ) 1 x

2

x x x x x x

S        

, Thus:















 



1 1 1 1 0 1 0 1 1 0 0 1 0 0 0

1 1 0 1 0 1 1 0 0 1 0 0 0 1 1

1 1 1 0 1 0 1 1 0 0 1 0 0 0 1

2 A2

A



1 0 0 0 1 0 0 1 1 0 1 0 1 1 1



) 15 ( S₉ h

and the corresponding head polynomial is:

12 8 7 6 5 3 3

( ) 1

1

x x x x x x x x

S        

Thus:















 



1 0 0 1 0 0 0 1 1 1 1 0 1 0 1

0 1 0 0 0 1 1 1 1 0 1 0 1 1 0

0 0 1 0 0 0 1 1 1 1 0 1 0 1 1 3

3 A

A



1 0 0 0 1 0 0 1 1 0 1 0 1 1 1



) 15 ( S₉ h

12 8 7 6 5 3 3

( ) 1

1

x x x x x x x x

S        

Thus:















 



1 1 1 0 1 0 1 1 0 0 1 0 0 0 1

1 0 1 0 1 1 0 0 1 0 0 0 1 1 1

1 1 0 1 0 1 1 0 0 1 0 0 0 1 1 4

4 A

A

1 1 0 0 0 1 0 0 1 1 0 1 0 1 1

) 4

( S₁₀ h

14 13 11 9 8 5 4

( ) 1

1

x x x x x x x x

S        

Thus:















 



0 0 1 0 0 0 1 1 1 1 0 1 0 1 1

1 0 0 0 1 1 1 1 0 1 0 1 1 0 0

0 1 0 0 0 1 1 1 1 0 1 0 1 1 0 5

5 A

A



0 1 1 0 1 0 1 1 1 1 0 0 0 1 0



) 5 ( S₃ h

and the corresponding head polynomial

is: ⁵

( )

² ⁴ ⁶ ⁷ ⁸ ⁹ ¹³

1

x x x x x x x x x

S        

Thus:















 



1 1 0 1 0 1 1 0 0 1 0 0 0 1 1

0 1 0 1 1 0 0 1 0 0 0 1 1 1 1

1 0 1 0 1 1 0 0 1 0 0 0 1 1 1 6 A6 A



1 1 1 0 0 0 1 0 0 1 1 0 1 0 1



) 10 ( S₁₁ h

and the corresponding head polynomial

is: ⁶

( ) 1

² ⁶ ⁹ ¹⁰ ¹² ¹⁴

1

x x x x x x x x

S        

Thus:















 



0 1 0 0 0 1 1 1 1 0 1 0 1 1 0

0 0 0 1 1 1 1 0 1 0 1 1 0 0 1

1 0 0 0 1 1 1 1 0 1 0 1 1 0 0 7 A7 A

1 0

0 1