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(pn), 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, ...,
k1 are elements in the fieldF
q, then the sequencea
0, a
1, ...
is called non homogeneous linear recurring sequence of order kiff :or
, , 0 , 1 ,..., 1
...
0
2 2 1 1
k i
F a
a a
a
q i n
k n k k n k k n
or
1 1 k
i
i n i k
n
a
a
(1) The elements
a
0, a
1, ..., a
k1 are called the initial values (or the vector( a
0, a
1, ..., a
k1)
is called the initial vector).If
0 then the sequencea
0, a
1, ...
is called homogeneous linear recurring sequence (H. L. R. S. ), except the zero initial vector, and the polynomial0 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. THE IMPORTANCE OF THIS RESEARCH AND ITS OBJECTIVES
The elements of the fields GF(pn), and their subgroups, can be represented as square matrices by m – sequences, which satisfies the multiplicative properties as a cyclic group, that is it will be useful in many other scientific branches.
Most of the existing communication devices (such as coders channels and decoders) for example, orthogonal sets in the Manuscript received on July, 2014.
Ahmad Hamza Al Cheikha, Department of Mathematical Science, Ahlia University, Kingdom of Bahrain.
forward and the inverse link of communication channels in the CDMA systems especially in the second (IS-95-CDMA), the third…. (CDMA200,…), the pilot channels, the Sync channels, and the Traffic channels uses computational Binary SystemF2 due to ease of manufacture and affordability which shows how the significance of this research. This study contributes notions for making modern communication devices to be efficient, confidential and safe, although the cost may not be low, using computational Fpsystems where p> 2, in the present or in many other scientific branches in the future.
III. RESEARCH METHODS AND MATERIALS Basic Definitions and Theorems
Definition 1.Let S be a nonempty set and
a
0, a
1, ....
is sequence from S and if r0 such that:0
;
00
;
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 nonnegative
n
0such that:0
;
00
;
a n n n
a
n r n ,is called Pre-Period, [1][ 4]
Definition 2.The Ultimately Periodic sequence
a
0, a
1, ....
with the smallest Period r is called a periodic iff:
...
, 1 ,
;
0
a n
a
n r n [1]-[ 4]Definition 3.The complement of the vector:
) ,..., ,
( x
1x
2x
nX
, whenx
i F
p, is the vectorX ( x
1, x
2,..., x
n)
,wheni
i p x
x 1 , and
X ( x
1, x
2,..., x
n)
whenp
x p
x
i
imod
. [1]-[ 4]Definition 4. (Euler function
). (n )
is the number of the natural numbers that are relatively prime with n.[5]-[ 8]Definition 5.AnyPeriodic Sequence
a
0, a
1, ....
overF
p, when p is prime, with prime characteristic polynomial is an orthogonal cyclic code and ideal auto correlation [1]-[ 10].Definition 6.The binary periodic sequence
( a
i)
iN, with the period r has the property of “ Ideal Auto Correlation” if and only if its periodic auto correlationR
a
of the form:
otherwise r for
Ra r
; 1
mod 0
;
Matrix Representation of Groups In the Finite Fields GF(p n )
Ahmad Hamza Al Cheikha
When:
1
0
1 )
(
r
t
t a t a
Ra [1],[2]
Theorem1.
i. If
a
0, a
1, ....
is a homogeneous linear recurring sequence of order kinF
p , satisfies (1) then this sequence is periodic.ii. If this sequence is homogeneous linear recurring sequence, periodic with the period r,and its characteristic polynomial
f (x )
thenr ord f (x )
. [6]iii. If the polynomial
f (x )
is primitive then the period of the sequence isp
k 1 ,
and this sequence is called m –sequence.
Lemma 2.( Fermat’s theorem ). If F is a finite field and has q elements then every element a of F satisfies the equation:
x
x
q
. [6],[9]Theorem 3.For any primitive element p and any positiveInteger n there is a field F, which has
p
nelements and any two fields havingq p
n elements, are isomorphic.[6],[9],[11]
Theorem 4.
i.
( q
m 1 ) ( q
n 1 ) m n
(4) ii. IfF
qis a field of orderq p
nthen any subfield of them of the orderp
m andm n
and by inverse ifm n
then in the fieldF
qthere is a subfield of order mp
. [6],[9],[11]Theorem 5.The number of irreducible polynomials in
)
(x
F
q of degree m and order e is ( e / ) m
,ife2, When m is the order of q by mod e, and equal to 2. Also, if m = e = 1, and equal to zero else where. [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) (x
g
in any splitting field of F2then the general bound of the sequence is:p n k
i i n
C i
a
1
1
. [11]-[13].
* The study here, is limited to the fields Galois
GF ( p
n)
, andp 2
, then the periodr p
k 1
is even.IV. RESULTS AND DISCUSSION A. First step
Theorem 7: Suppose
a
0, a
1, ....
is a non zero homogeneous linear recurring sequence of order k over} 1 ,..., 1 , 0
{
p
F
p andf (x )
is its prime characteristic polynomial then the firstr p
k 1
bounds with all its cyclic shifts forming an additive group.Proof: This sequence is periodic with period r pk 1. We suppose
$ S
1, S
2,..., S
r
whereS
1 a
1a
2... a
r
is the sequence of the firstr p
k 1
bounds, and
1....
1 , ...,
2
a
ra a
rS S
r a
2a
3... a a
r 1
are all its cyclic shifts, and we supposeO 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 , :
) (
k
j j k
j
k j j
i i
k b i
P GF
And the function:
h : GF ( p
k) S
as following:
[ ]) ( )
( 0 1 1 0 1 1 k2
k p k k
i hi hb b b b b b b b
h
Then h is one-to-one corresponding and:
p i
i
j i
j i
F m h
m m
h
h h
h
), ( . ) . (
) ( ) ( ) (
And h is Linear Transformation and isomorphism from the additive group
GF ( p
k),
to the additive group S ,
, but$
is not closed under the addition as F n2 because:
for k
p i
F
then: h(
i)0&h(
i)h(
i)0 and: h(
i)h(
i)0$B. Second Step
Theorem 8: Suppose
a
1, a
2,...
is a non zero homogeneouslinear recurring sequence of order k inF
pand f(x) is their primitive characteristic polynomial, S1 is the initial bounds wherer p
k 1
and$ S
1, S
2,..., S
r
are the all cyclic shifts. Let A is a matrix which its rows are elements $ respectively, then by A
i, i 1 ,..., r
, or by powers of its permutations of Awe can represent all subgroups inF
pk relatively to product and addition of matrices, having the period ofS
1( x )
and rows ofA
iare the shifts to rows of A.Proof: Suppose
Sr
S S A ...
2 1
and we will compute
A
2 A . A
,and the first row
1 in the matrix A then:O S
I i
i
i
1 When I theset of all columns in A which does not start by zero and the of
i
th isa
i 0
then$ S
l
1
, becausemultiplying any element of
$
by any element ofGF ( p )
is an element of $, the sum of any two elements of $ is in $ and
1 0
.The second row
2inA
2 is a result of shift i by 1digit to theright, then: 2 1 1 1
lI i
i
i
S S
, and respectively wehave 1 1 1
l rI i
r 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. In other hand we suppose that x S x , S x ,..., S
r x
$
1 2 then:) ( )
1
( x S x
I i
i
i
;
)
(
1 1 12
I i
i i I
i
i
i
S x x S x
x
;….;
...,
1 1
I ir i r i
r
x S x
And :
1
x S12
x
I i
i I
i
i
x x S x
S
1 1
1When:
S
12 x $ x
, and the calculations are done by
1 mod
x
pk 1 , And we have:
2 x xS
12 x
; …;
r x x
r1S
12 x
Suppose
f
i x
denotes the row of coefficients off
i x
,respectively to the increasing exponents of x, and which has length r, then:
x S x
x xS
x S A x S x
x xS
x S
x S
x S
x S A
r r
r 2
1 1 12 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
1
When:
S
1i x $ x ; i 1 , ...., r
,then:
S
xx x A x S x x
x S
x S
x S A
r r r
2 1 1 2
1 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 toord ( S
1( x ))
and dividesp
k 1
.Result 2: If
ord ( S
1( x )) p
k 1
then S is representation to the fieldGF ( p
k)
.Result 3: If
ord ( S
1( x )) l and l p
k 1
then: 1
p
ml
and A
is a representation to one subgroup of order l in the fieldGF ( p
k)
Result 4:If
ord ( S
1( x )) 1 p
land l k
then :}
{O A
is representation to the SubfieldGF ( p
l)
. Result 5: Ifp
k 1
is prime then all elements ofS
i(x )
are of the orderp
k 1
except only one of them which is of the order one.C. Third Step
Example 1:If
is a root of the prime polynomial 2)
(x x2x
f and generates GF(32) then the
then the representation of the elements of
GF(32)in F is:
3Where ( i ), i 0 , 1 , 2 , ..., 8 is the symbol of the sequence i.
The divisors of the number 8 are 1, 2, 4, and 8, consequently, GF(32) contains four multiplicative subgroups are: <1>, GF(3), fourth order multiplicative subgroup =
2
, and GF(32), and the divisors of 9 are: 1, 3, and 9 .Then GF(32)contains three additive subgroups are: one first order additive subgroups, one third order additive subgroups, and one ninth order additive subgroups, consequently the field GF(32) contains two subfields are :GF ( 3 )
and the same GF(32).. Suppose the
Linear Recurring Sequence be:
) 5 ( 2
0
2
2 11
2 n n n n n
n
a a or a a a
a
Figure(1): Linear feedback register of degree2 generates sequence (5)
With the characteristic equation
x
2 x 2 0
and thecharacteristic polynomial f(x)x2x2 ,which is a prime and generates 2
F3 and if x
GF
32 is a root of)
(x
f
then the solutions of characteristic equation are },
{
n
3n .The general solution of equation (1) is given by
n n n
a 2
(1
)
3 ,and the sequence is periodic with the period
3
2 1 8
. For the initial position:a
1 0 , a
2 1
, then) 1 1 2 0 2 2 1 0
1(
S and by the cyclic permutationsonS1
we have
$ S
1, S
2, S
3, S
4, S
5, S
6, S
7, S
8
where:) 0 1 1 2 0 2 2 1 (
) 1 0 1 1 2 0 2 2 (
; ) 2 1 0 1 1 2 0 2 (
; ) 2 2 1 0 1 1 2 0 (
) 0 2 2 1 0 1 1 2 (
; ) 2 0 2 2 1 0 1 1 (
; ) 1 2 0 2 2 1 0 1 (
8
7 6
5
4 3
2
S
S S
S
S S
S
The first two digits in each sequence are the initial position of the feedback register.
In this example the resulting sequences is:
0 1 2 2 0 2 1 1 0 1 2 2 0 2 …
The matrix A of the cyclic permutations of S1 is:
0 1 1 2 0 2 2 1
1 0 1 1 2 0 2 2
2 1 0 1 1 2 0 2
2 2 1 0 1 1 2 0
0 2 2 1 0 1 1 2
2 0 2 2 1 0 1 1
1 2 0 2 2 1 0 1
1 1 2 0 2 2 1 0
1 A B
Or briefly:
0 1 1 2 0 2 2 1
1 2 0 2 2 1 0 1
1 1 2 0 2 2 1 0 B1
The first row in this matrix is called The head row.
If the function h:GF(32)$where:
i h i hi
h(
) ( ) = [The row of the matrix A corresponding of the initial position i].Thenh
i is isomorphism from the group(GF(32),) on the group($, )
.I- In this matrix the head row is: [01220211]
1
S and
the head polynomial is:
7 6 5 3
1( ) 2 2 2 2
) 1
( S x x x x x x x
h . Thus
0 1 0 1 1 2 0 2
0 1 1 2 2 0 2 2
1 0 1 1 2 0 2 2
12
2 B
B
In this matrix the head row is: (3) [22021101]
7
S h
and the corresponding head polynomial is:
2
2 2 )
( 3 4 5 7
7 x x x x x x
S , Thus
0 2 2 1 0 1 1 2
2 1 0 1 1 2 0 2
2 2 1 0 1 1 2 0 3 3 B1 B
In this matrix the head row is: h(5)S5 [02110122] and the corresponding head polynomial is:
2 2 2
)
( 2 3 5 6 7
5 x x x x x x x
S , Thus
1 2 0 2 1 1 0 1
0 2 2 1 0 1 1 2
2 0 2 2 1 0 1 1 14
4 B
B
In this matrix the head row is: (7) [11012202]
3
S h
and the corresponding head polynomial is:
2 2 2 1
)
( 3 4 5 7
3 x x x x x x
S , Thus
5 1
1 B
B and the set B* =
{ B
1, B
2, B
3, B
4}
is fourth order multiplicative group and the setB=
{ O , B
1, B
2, B
3, B
4}
is not additive group as is shown by the table 1.Then B is not a field, and we see that:
4 ) ( )
( B
1 ord B
3 ord
II- We suppose that;
2 1 0 1 1 2 0 2
0 1 1 2 0 1 2 1
1 0 1 1 2 0 2 2
C1
In this matrix the head row is: (3) [22021101]
7
S h
and the corresponding head polynomial is:
2
2 2 )
( 3 4 5 7
7 x x x x x x
S , Thus
1 2 0 2 1 1 0 1
0 2 2 1 0 1 1 2
2 0 2 2 1 0 1 1 2 1 2 C C
In this matrix the head row is: h(7)S3 [11012202] and the corresponding head polynomial is:
7 5 4
3(x) 1 x x3 2x 2x 2x
S , Thus C13C1.
And the
C
* { C
1, C
2}
is second order multiplicative group andC { O , C
1, C
2}
is an additive group as showing in the table2.We see that C is a representation of the subfield
)
3 (
23
GF
F
and:ord ( C
1) 2
andord ( C
2) 1
. III. We suppose that:
1 1 2 0 2 2 1 0
2 0 2 2 1 0 1 1
1 2 0 2 2 1 0 1 D1
In this matrix the head row is: (8) [10122021]
2
S h
and the corresponding head polynomial is:
2
2 2 1
)
( 2 3 4 6 7
2 x x x x x x
S , Thus
0 1 1 2 0 2 2 1
1 2 0 2 2 1 0 1
1 1 2 0 2 2 1 0
2 2D
1D
In this matrix the head row is:
] 1 1 2 0 2 2 1 0 ( )
1 ( S1 h
and the corresponding head polynomial is:
2
2 2 )
( 2 3 5 6 7
1 x x x x x x x
S , Thus
1 0 1 1 2 0 2 2
1 1 2 0 2 2 1 0
0 1 1 2 0 2 2 1 3 3 D1 D
In this matrix the head row is: h(2)S8 [12202110] and the corresponding head polynomial is:
2
2 2 1 )
( 2 4 5 6
8 x x x x x x
S , Thus
2 1 0 1 1 2 0 2
0 1 1 2 0 2 2 1
1 0 1 1 2 0 2 2 4
4 D1
D
In this matrix the head row is: (3) [22021101]
7
S h
and the corresponding head polynomial is:
2
2 2 )
( 3 4 5 7
7 x x x x x x
S , Thus
0
2 2 1 0 1 1 2 0
1 1 1 1 2 0 2 2
2 1 0 1 1 2 0 2
5 5
D
1D
In this matrix the head row is:
] 2 1 0 1 1 2 0 2 [ )
4
( S6 h
and the corresponding head polynomial is:
2 2
2 )
( 2 3 4 6 7
6 x x x x x x
S , Thus
0 2 2 1 0 1 1 2
2 1 0 1 1 2 0 2
2 2 1 0 1 1 2 0 6 D16 D
In this matrix the head row is:
] 2 2 1 0 1 1 2 0 [ ) 5 ( S5 h
and the corresponding head polynomial is:
2 2 2
)
( 2 3 5 6 7
5 x x x x x x x
S , Thus
2 0 2 1 1 0 1 1
2 2 1 0 1 1 2 0
0 2 2 1 0 1 1 2
17
7
D
D
In this matrix the head row is: h(6)S4[21101220] and the corresponding head polynomial is:
2 2 2
)
( 2 4 5 6
4 x x x x x x
S , Thus
1 9
18
8 ;
1 2 0 2 1 1 0 1
0 2 2 1 0 1 1 2
2 0 2 2 1 0 1 1
D D
D D
In this matrix the head row is: h(7)S3[11012202] and the corresponding head polynomial is:
7 5 4
3(x) 1 x x3 2x 2x 2x
S
.
And D*
= { D
1, D
2, D
3, D
4, D
5, D
6, D
7, D
8}
is a multiplicative group of order 8, andD=
{ O , D
1, D
2, D
3, D
4, D
5, D
6, D
7, D
8}
is additive group as shown in table 3.Consequently, D is a field and representation of the field )
3 ( 2 GF .
The field GF(32) contains
321 22 third degree irreducible polynomials are f(x)x2x2 and2 2 )
(x x2 x
g .
V. RESULTS AND RECOMMENDATIONS 1. The fields GF(pn) and their subfield can be
represented by square matrices.
2. The multiplicative group GF(pn) and their subgroups can be represented by square matrices.
3. The equations of the degree less than or equal to n on
)
( p
GF
can also be solved by square matrices.4. Building encoders on the field
F
q whenq n
are recommended for further study.APPENDIX Table 1: The Addition in B*
B4
B3
B2
B1
+
B B
O
B3
B1
B
O
B4
B
B2
B
B1
B
O B3
B2
B B
O
B4
Table 2: The Addition in C*
C2
C1
+
O C2
C1
C1
O C2
Table 3: Addition in D*
D8
D7
D6
D5
D4
D3
D2
D1
+
D8
D7
D6
D5
D4
D3
D2
D1
D2
D6
D4
D7
D8
D3
D5
D1
D1
D7
D5
D8
D1
D4
D6
D3
D2
D2
D6
D1
D2
D5
D7
D4
D8
D3
D3
D2
D3
D6
D8
D5
D1
D7
D4
D4
D3
D4
D7
D1
D6
D2
D8
D5
D5
D5
D8
D2
D7
D3
D1
D4
D6
D6
D1
D3
D8
D4
D2
D5
D6
D7
D7
D4
D1
D5
D3
D6
D7
D2
D8
D8
ACKNOWLEDGMENT The author express his gratitude
to Prof. Abdulla Yousuf AlHawaj, President of Ahlia University for all the support
provided.
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AUTHORS PROFILE
Dr. Ahmad Hamza Al Cheikha. His Research interests are Design Orthogonal sequences with variable length, Finite Fields, Linear and Non Linear codes, Co-positive Matrices and Fuzzy Sets.