This document summarizes a research paper about error-correcting codes derived from the cyclic group of order three. It describes generating two linear codes by using the regular representation of the cyclic group and bordering the resulting block matrices. The codes are investigated over Galois fields of characteristic p, where p is a prime number. For p not equal to three, the codes are shown to have no error-correction capability. For p equal to three, the generator matrix of the code is determined and its weight distribution is given by a theorem.

1. Journal of Advanced Computing and Communication Technologies (ISSN: 2347 - 2804)
Volume No2 Issue No 2, April 2014
1
Codes from the Cyclic Group of Order Three
By
M. Asifuzzaman, Kamrul Hasan,
Partha Pratim Dey
Grameenphone Ltd., Dhaka, Bangladesh
BdREN, Dhaka, Bangladesh
North South University, Dhaka,Bangladesh
tamal56@yahoo.com, kamrul@bdren.net.bd,
ppd@northsouth.edu
ABSTRACT
In this paper we use cyclic group 3Z and its regular
representation to produce a couple of linear error-correcting
codes. We also discuss their duals.
Keywords
Regular representation, linear code, generator matrix, parity-
check matrix
1. INTRODUCTION
Throughout this paper pF , for some prime ,p will denote the
Galois field )( pGF and
k
pF will be the vector space
comprising of vectors ),...,( 1 kxxx = where pi Fx ∈ for
.,...,1 ki = Let },,{ 321 ggg be an enumeration of the
elements of the cyclic group 3Z of order 3with identity
element 01 =g , ,12 =g 23 =g and let )( igR denote
the regular representation of ig in 3Z using the enumeration
},,{ 321 ggg to index rows and columns of the
representation matrix. Then
=)( 1gR





0
0
1
0
1
0





1
0
0
,





=
1
0
0
)( 2gR
0
0
1





0
1
0
and





=
0
1
0
)( 3gR
1
0
0





0
0
1
.
For each ,mg 3,2,1=m let miw be the
th
i row of
)( mgR and let )( mgR∗
be the block matrix given by:






−
−
=∗
31
21
)(
mm
mm
m
ww
ww
gR .
Consider now the following two block matrices:




= ∗
∗
∗
)(
)(
3
2
]3,2[
gR
gR
B




∗
∗
)(
)(
2
3
gR
gR
and




= ∗
∗
∗
)(
)(
2
3
]2,3[
gR
gR
B




∗
∗
)(
)(
3
2
gR
gR
.
In ]3,2[
∗
B , the first row of blocks is )([ 2gR∗
)]( 3gR∗
and
the second row is the cyclic shift of the first, whereas in ]2,3[
∗
B ,
the first row of blocks is )([ 3gR∗
)]( 2gR∗
and the second
row is the cyclic shift of the first. We now border each of ]3,2[
∗
B
and ]2,3[
∗
B by a row and a column of )([ 1gR∗
)( 1gR∗
)]( 1gR∗
as follows to obtain:
]3,2[M
=





∗
∗
∗
)(
)(
)(
1
1
1
gR
gR
gR
)(
)(
)(
3
2
1
gR
gR
gR
∗
∗
∗





∗
∗
∗
)(
)(
)(
2
3
1
gR
gR
gR










=
1
1
1
1
1
1
0
1
0
1
0
1
−
−
−
1
0
1
0
1
0
−
−
−
0
1
1
0
1
1
−
−
1
0
1
1
0
1
−
−
1
1
0
1
1
0
−
−
1
0
0
1
1
1
−
−
1
1
1
0
0
1
−
−










−
−
0
1
1
1
1
0
and
]2,3[M
=





∗
∗
∗
)(
)(
)(
1
1
1
gR
gR
gR
)(
)(
)(
2
3
1
gR
gR
gR
∗
∗
∗





∗
∗
∗
)(
)(
)(
3
2
1
gR
gR
gR

2. Journal of Advanced Computing and Communication Technologies (ISSN: 2347 - 2804)
Volume No2 Issue No 2, April 2014
2










=
1
1
1
1
1
1
0
1
0
1
0
1
−
−
−
1
0
1
0
1
0
−
−
−
1
0
0
1
1
1
−
−
1
1
1
0
0
1
−
−
0
1
1
1
1
0
−
−
0
1
1
0
1
1
−
−
1
0
1
1
0
1
−
−










−
−
1
1
0
1
1
0
.
Notice that swapping the
rd
3 row with the
th
5 and the
th
4 row
with the
th
6 in ]2,3[M , we obtain the ]3,2[M . Hence both
]3,2[M and ]2,3[M comprise of the same set of rows. We can
view each row of ]3,2[M or ]2,3[M as a row-vector of
9
pF .
Thus the row-vectors of ]3,2[M and ]2,3[M are identical as
set. Hence the linear codes generated i.e. spanned by the row-
vectors of ]3,2[M or ]2,3[M over pF are identical too.
Throughout the rest of the paper we will investigate this linear
code over pF for various 'p s and for convenience, denote
]3,2[M by M only.
2. M over pF for
Various 'p s
We remove the
rd
3 ,
th
6 and
th
9 columns of ]3,2[M to obtain
the following square matrix:










=
1
1
1
1
1
1
Q
0
1
0
1
0
1
−
−
−
0
1
1
0
1
1
−
−
1
0
1
1
0
1
−
−
1
0
0
1
1
1
−
−










−
−
1
1
1
0
0
1
.
Since det
3
3=Q , the inverse of Q exists in pF where
3≠p . We use elementary row operations method to evaluate
the
1−
Q . The steps of the deduction are shown in the
Appendix, whereas here below we produce the result
1−
Q
only:










−
−
−
=−
α
α
α
0
0
0
1
Q
α
α
α
α
α
α
0
0
α
α
α
α
−
−
α
α
α
α
−
−
0
0
α
α
α
α
0
0
−
−










−
−
0
0
α
α
α
α
The entry α in
1−
Q above is the inverse of 3 in pF . Since
3≠p , the inverse of 3 in pF exists. Consider now:
MQ 1−










−
−
−
=
α
α
α
0
0
0
α
α
α
α
α
α
0
0
α
α
α
α
−
−
α
α
α
α
−
−
0
0
α
α
α
α
0
0
−
−










−
−
0
0
α
α
α
α
.










1
1
1
1
1
1
0
1
0
1
0
1
−
−
−
1
0
1
0
1
0
−
−
−
0
1
1
0
1
1
−
−
1
0
1
1
0
1
−
−
1
1
0
1
1
0
−
−
1
0
0
1
1
1
−
−
1
1
1
0
0
1
−
−










−
−
0
1
1
1
1
0
=
=










0
0
0
0
0
1
0
0
0
0
1
0
0
0
0
0
1
1
−
−
0
0
0
1
0
0
0
0
1
0
0
0
0
0
1
1
0
0
−
−
0
1
0
0
0
0
1
0
0
0
0
0










−
−
1
1
0
0
0
0
.
Thus )0,0,0,0,0,0,1,0,1( − is a code-word of weight 2 of the
linear code generated by the row-vectors of M over pF with
3≠p . Hence these codes do not have error-correction
capabilities.
It is thus appropriate that we would like to consider the linear
code generated by the row-vectors of M over pF where 3=p .
Towards that goal, we gaussjord M over 3F to obtain:





0
0
1
0
1
0
0
2
2
1
0
0
1
1
2
1
2
1
2
0
2
2
1
1





2
2
0
which after appropriate permutation of columns becomes





=
0
0
1
G
0
1
0
1
0
0
2
2
0
2
0
2
0
2
2
2
1
1
1
1
2





1
2
1
.
Notice that each row of G above is a vector of
16
3F and the
subspace spanned by its 3rows over 3F is a linear code and G




























3 a 0 −3 a 0 0 0 0 0 0
0 3 a −3 a 0 0 0 0 0 0
0 0 0 3 a 0 −3 a 0 0 0
0 0 0 0 3 a −3 a 0 0 0
0 0 0 0 0 0 3 a 0 −3 a
0 0 0 0 0 0 0 3 a −3 a

3. Journal of Advanced Computing and Communication Technologies (ISSN: 2347 - 2804)
Volume No2 Issue No 2, April 2014
3
is its generator matrix. We will denote this code by )(GC and
explore it throughout the rest of the paper. We will also explore
the dual code
⊥
)(GC . For an understanding of the linear code
at a basic level one may please consult [1] and [2].
3.Weight
Distribution of )(GC
We begin with a theorem.
Theorem (3.1) The code )(GC has the following weight
distribution.
Weight Number of Words
0 1
9 2
6 24
Moreover each code-word of )(GC but for 99 1,0 and 92
contains exactly three zeros, three ones and three twos.
Proof. Given ,3F∈α let 9α denote the row-vector
),...,( αα with 9co-ordinates each of whose co-ordinates is
α . Notice that for any )(GCc ∈ ,
)2,2,2
),(2),(2),(2,,,(
γβαλβαγβα
βαγαγβγβα
++++++
+++== wGc
for some .),,( 3
3Fw ∈= γβα Let γβα == . Then
)4,4,4,22,22,22,,,( ααααααααα ⋅⋅⋅=c
91),...,( ααα == , giving 3 code-words: 99 1,0 and 92 .
Assume now that βα, and γ are all distinct. Without loss,
let 0=α . Then γβ 2= , ,2βγ = 0=+ γβ and
therefore
=+++
++++⋅=
)0
,00,0,2,2,02,,,0(
γββ
γγββγγβc
),0,,,,0,,,0( βγγβγβ . Finally, let us suppose that 2 of
βα, and γ are identical. Let without loss, βα = and
αγ ≠ . Then
).,),(2,),(2),(2,,,( γγγααγαγαγαα +++=c
Suppose .)(2 αγα =+ Then γα = , a contradiction.
Hence .)(2 αγα ≠+ Similarly γγα ≠+ )(2 . Hence
)(2,, γαγα + are distinct elements of 3F and therefore
),),(2,),(2),(2,,,( γγγααγαγαγαα +++=c
contains three zeros, three ones and three twos. ■
Corollary (3.2) )(GC can correct 2 errors.
Proof. Since 2 is the largest integer less than half of minimum
weight 6 of the code, )(GC can correct 2 errors. ■
Next we show that this code )(GC is in fact an −2 error-
correcting extended ]6,3,9[ BCH code.
Let ][1)( 3
8
xFxxf ∈−= and we choose the primitive
polynomial 2)( 2
++= aaap in ][3 aF . Then
))(/(][3 apaF is a finite field of order 9 and
82
,,, aaa  constitute all the non-zero elements in
))(/(][3 apaF . Let C be the code that results from
considering the first four powers of a , namely
32
,, aaa and
4
a . To determine the generator polynomial )(xg for C , we
must find the minimum polynomials
)(,),(),( 421 xmxmxm  for
42
,,, aaa  respectively.
Since
)22)(2)(1)(2)(1(1 2228
+++++++=− xxxxxxxx
we have 2)()( 2
31 ++== xxxmxm ,
1)( 2
2 += xxm and 1)(4 += xxm .Thus
543222
22)1)(1)(2()( xxxxxxxxxg ++++=++++=
. Hence >=< )(xgC and generator matrix J of C is given
by:





=
0
0
2
J
0
2
0
2
0
1
0
1
1
1
1
2
1
2
1
2
1
0





1
0
0
Then =ext
J





0
0
2
0
2
0
2
0
1
0
1
1
1
1
2
1
2
1
2
1
0
1
0
0





2
2
2
.
We gaussjord
ext
J to get





0
0
1
0
1
0
1
0
0
0
2
2
2
2
0
2
1
1
1
2
1
2
0
2





1
1
2
which after appropriate permutation of columns becomes G .
Thus we have the following theorem.
Theorem (3.3) )(GC is the double error-correcting extended
]6,3,9[ BCH code generated by
5432
22)( xxxxxg ++++= .
4 Weight Distribution of the Dual Code ⊥
)(GC
Since [ ]:3 MIG = where
=M





2
2
0
2
0
2
0
2
2
2
1
1
1
1
2





1
2
1
,
we have ]2:[ 6IMH tr
= for the parity check matrix H of
)(GC .
Hence










=
1
2
1
2
2
0
H
2
1
1
2
0
2
1
1
2
0
2
2
0
0
0
0
0
2
0
0
0
0
2
0
0
0
0
2
0
0
0
0
2
0
0
0
0
2
0
0
0
0










2
0
0
0
0
0
.

4. Journal of Advanced Computing and Communication Technologies (ISSN: 2347 - 2804)
Volume No2 Issue No 2, April 2014
4
Notice that each row of H above is a vector of
9
3F and the
subspace spanned by its 6 rows over 3F is a linear code
)(HC and H is its generator matrix. As 0=tr
GH ,
⊥
= )()( GCHC . We will find weight distribution of
)(HC from the weight distribution of )(GC . Below we
state a theorem [3] due to Mac Williams that will help us to
find the weight distribution of the other code-words.
Theorem (4.1) (Mac Williams) Let C be an ],[ kn code over
)(qGF with ,iA the number of vectors of weight i in C and
iB , the number of vectors of weight i in
⊥
C . The following
relations relate the }{ iA and }{ iB :
,
00
j
n
j
k
j
n
j
B
j
jn
qA
jn
∑∑ =
−
=






−
−
=




 −
υυ
υ
where
n,...,0=υ .
Let )(HCC = . Then )()( GCHCC == ⊥⊥
and
10 =B , ,246 =B and 29 =B by Theorem (3.1). Now
taking 8=υ in Mac Williams equation, we obtain:
j
j
j
j
B
j
j
A
j
∑∑ =
−
=






−
−
=




 − 9
0
86
9
0 8
9
3
8
9
or 








+








=





+





2
3
8
9
9
1
8
8
8
9
6010 BBAA
or 109 AA + )39(
9
1
60 BB +=
or =+⋅ 119 A )24319(
9
1
⋅+⋅
01 =∴ A
Inserting 7=υ again in Mac Williams equation,









+








=





+




 −
1
3
7
9
3
7
7
7
9
60
76
20 BBAA
or 21
7
9
A+⋅














⋅+⋅








=
1
3
241
7
9
3
1
02 =∴ A .
Similarly inserting 2,3,4,5,6=υ and1 in the Mac Williams
equation, we obtain:
,243 =A ,1084 =A ,1085 =A 1926 =A ,
54,216 87 == AA .
Thus we have the following theorem.
Theorem )2.4( The dual code =⊥
)(GC )(HC has the
following weight distribution.
Weight Number of Words
0 1
3 24
4 108
5 108
6 192
7 216
8 54
9 26
7. REFERENCES
[1] F. J. MacWilliams, “A theorem on the distribution of weights
in a systematic code”, Bell Syst. Tech. Journal, 42 pp 79-94, 1993.
[2] R.E., Klima, N. Sigmon and E. Stitzinger, “Applications of
Abstract Algebra with MAPLE”, ISBN 0-8493-8170-3, CRC
Press, Boca Raton, 2000.
[3] V. Pless, “Introduction to the Theory of Error Correcting
Codes”, ISBN 9814-12-688-8, Wiley Student Edition, John Wiley
& Sons (Asia) Pte. Ltd., Singapore, 2003.