# The Primitive Idempotents in FC - II

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```					       International Journal of Algebra, Vol. 4, 2010, no. 25, 1243 - 1254

The Primitive Idempotents in FC 2n - II

Kulvir Singh

Department of Mathematics, SGHS College
Sri Jiwan Nagar (Sirsa)-125075, India
kulvirsheoran@yahoo.com

S. K. Arora

Department of Mathematics, Maharshi Dayanand University
Rohtak-124001, India

Abstract. Let FG be the semi simple group algebra of the cyclic group G of
order 2n (n ≥ 4) over the finite field F( = GF(q) ) of prime power order q
(odd), where q is quadratic residue modulo 2n. Then q = 8k + 1 (k is even or odd).
In case k is even and is of the form k = 2t (t-odd), then FG has 8(n−2) primitive
idempotents, the explicit expressions for these primitive idempotents are
obtained.

Mathematics Subject Classification (2000) : 94B15, 16S34, 20C05

Keywords : Group algebra; primitive idempotents ; cyclotomic cosets

1. Introduction
Let G = Cm = < g > be a finite cyclic group of order m and F( = GF (q)) be a
field of order q, a power of its prime (odd) characteristic ρ, (say). Let t be the
φ(2n )
multiplicative order of q modulo m. The case when q = 8k ± 3 and t =                       is
2
discussed by Batra et al.[1], [2] . Now suppose that q is quadratic residue modulo
2n. By Theorem 9.12 [3, p. 204], q ≡ 1 (mod 8) and conversely i.e. q = 8k + 1
(k = 2α0 p1 1 ........ pr r ; αi ≥ 0 ). In case α 0 = 0 , i.e. k is odd, then FG has 4(n−1)
α             α

primitive idempotents and the explicit expressions for these primitive
1244                                                                        K. Singh and S. K. Arora

idempotents are obtained by Singh and Arora in [4]. In this paper we
consider the case when α 0 = 1 i.e. k = 2t ( t-odd ). By Lemma 2.1, order of q
φ(2n )
modulo 2n ( n ≥ 4 ) is              .   Let S be the set { 0, 1, …, 2n −1 }. For a , b ∈ S,
8
the relation a ≡ bqi (mod 2n), partitions S into 8( n−2 ) disjoint q-cyclotomic
cosets, as obtained in Theorem 2.2. Therefore, FC2n has 8(n−2) primitive
idempotents. The explicit expressions                     for    these primitive idempotents are
obtained in Theorem 4.1.

2. Cyclotomic Cosets

φ(2n )
Lemma 2.1. Let q = 8k + 1 , k = 2t ( t-odd ). Then,                                      is the order of q
8
n
modulo2 .

Proof. Trivial.

Theorem 2.2. Suppose q = 8k + 1 , k = 2t ( t-odd ). Then, 8( n−2 ) q-
cyclotomic cosets modulo 2n are given by :
Ω* = {0} ; Ω* n = {2n−1} , Ω* n−1 = {2n−2 } , Ω* n−1 = {3.2n−2 }
0               (1),               (1),        (2),

Ω* n−2 = {2n−3} , Ω* n−2 = {3.2n−3} , Ω* n−2 = {5.2n−3} , Ω* n−2 = {7.2n−3}
(1),              (2),                (3),                (4),
for 1 ≤ i ≤ n − 3 and 1 ≤ β ≤ 8,
Ω*β), i = {2i −1 (16λ + 2β − 1) : 0 ≤ λ ≤ 2n−(i +3) − 1} .
(

Proof. Follows by Lemma 2.1.

Notations 2.3.

(i) For 1 ≤ i ≤ n and 1 ≤ β ≤ 8, denote by S(*β), i the elements                                    ∑         gs
s∈Ω*β ),i
(

of FC2n .
1
(ii) For 1 ≤ i ≤ n − 3,       G*(l ,m ), i =   n −i +1
( S(*l ), i − S(*m ), i ) , where 1 ≤ l, m ≤ 8
2
and l ≠ m.
(iii) We follow the notation 2.3 (i),(ii) [4] and observe that, for 1 ≤ i ≤ n −3,
∗           ∗                       ∗           ∗
G(1,2), i = G(1,2), i + G(5,6), i , G(1,3), i = G(1,3), i + G(5,7), i
Primitive idempotents                                                                                             1245

∗           ∗                       ∗           ∗
G(2,4), i = G(2,4), i + G(6,8), i , G(3,4), i = G(3,4), i + G(7,8), i
1                  i       i     i               n −i −1).2i
(iv) For 0 ≤ i ≤ n , let X i =                         ( 1 + x 2 + x 2.2 + x3.2 + .... + x (2                 )
2 n −i
2n −i −1
1
∑
i
=                       xl 2
2 n −i     l =0
Set X i = X i ( g ) . Further for 1 ≤ i ≤ n , let Yi = X i − X i −1 and Y0 = X 0 .
Then, for 0 ≤ i ≤ n,

1 ⎡ ⎛ n −3 8 * ⎞                                                                           *     ⎤
Xi =        ⎢1 + ⎜ ∑ ∑ S(β), j ⎟ + ( S *                                            *
+ ... + S * n− 2 ) + ( S * n−1 + S(2), n−1 ) + S(1), n ⎥
2n−i ⎢ ⎜ j = i +1 β =1
⎣ ⎝
⎟
⎠
(1), n − 2           (4),           (1),
⎥
⎦

and for 1 ≤ i ≤ n,
1 ⎡ ⎧ ⎛ n−3 8
⎪                   ⎞
Yi = n−i +1 ⎢ ⎨1 + ⎜ ∑ ∑ S(*β), j ⎟ + ( S(1), n−2 + ... + S(4), n−2 )
*                 *
2       ⎢ ⎪ ⎜ j =i +1 β=1     ⎟
⎣⎩ ⎝                  ⎠
⎫ 8 * ⎤
⎪
+ ( S(1), n−1 + S(2), n−1 ) + S(1), n ⎬ − ∑ S(β), i ⎥
*           *             *
⎪ β=1
⎭             ⎥
⎦
(v) The suffix δ > 8 in S(*δ ), i is considered reduced modulo 8.

3. Some families in the Group Algebra and their properties

Lemma 3.1.
I. Suppose j ≤ i − 3 and k ≥ 0 be any integer. Then for 4 ≤ i ≤ n,
g k 2 S(*β ), j = S(*β ), j
i

II.  If j = i − 2, then for 3 ≤ i ≤ n − 1,
⎧ S(*β), j if k = 2, 4, 6,...
k 2i *      ⎪
g S(β), j = ⎨
*
⎪ S(β+ 4), j if k = 1,3,5,...
⎩
III. If j = i − 1, then for 2 ≤ i ≤ n − 2,
⎧ S(*β+ 2), j if k = 1,5,9,...
k 2i *      ⎪
g S(β), j = ⎨
*
⎪ S(β+6), j if k = 3, 7,11,...
⎩
IV. If j = i, then for 1 ≤ i ≤ n − 3,
1246                                                                                    K. Singh and S. K. Arora

⎧ S(*β+1), j if k = 1,9,17,...
⎪
⎪ S(*β+3), j if k = 3,11,19,...
k2i *        ⎪
g S(β), j = ⎨
*
⎪ S(β+5), j if k = 5,13, 21,...
⎪ *
⎪ S(β+7), j if k = 7,15, 23,...
⎩
Proof. I. Since, for j ≤ i − 3,
k 2i + Ω∗β ), j ≡ Ω∗β ), j ( mod 2n ), therefore,
(           (

∑           g k 2 +s =     ∑
i                               i
*
g k 2 S(β), j =                                              g s = S(*β), j .
s∈Ω*β ),
(       j                s∈Ω*β ),
(       j

II. Let k = 2t for some positive integer t. Then,
k 2i + Ω*β ), i − 2 = { 2t.2i + 2i−3r : r ≡ 2β -1 (mod 16) }
(

= { 2i−3r : r ≡ 2β -1 (mod 16) } = Ω*β ), i − 2 .
(

Now, let k = 2t + 1, for some positive integer t. Then,
k 2i + Ω*β ), i − 2 = { (2t + 1).2i + 2i−3r : r ≡ 2β -1 (mod 16) }
(

={ 2i−3r : r ≡ 2β +7 (mod 16) } = Ω*β+ 4), i − 2 , proving II.
(

III and IV follows on similar lines as above.
Lemma 3.2. Suppose ξβ ∈ Ω*β ), i . Then for 1 ≤ i ≤ n − 3,
(

I.           ξβ + Ω*β ), i = Ω*β ), i +1 ∪ Ω*β+ 4), i +1 .
(          (             (

II.(i)       ξ1 + Ω* i = ξ3 + Ω* i = ξ4 + Ω* i = ξ5 + Ω*
(2),        (8),        (7),        (6),i

= Ω * i + 2 ∪ Ω * i + 2 ∪ Ω* i + 2 ∪ Ω * i + 2
(1),        (3),       (5),        (7),

(ii)    ξ1 + Ω* i = ξ4 + Ω* i = ξ5 + Ω* i = Ω* i +1 ∪ Ω* i +1
(3),        (8),        (7),   (2),      (6),

(iii)   ξ1 + Ω*4), i = ξ2 + Ω* i = ξ5 + Ω* i = ξ6 + Ω* i
(              (3),        (8),        (7),

= Ω* i+3 ∪ … ∪ Ω* i+3
(1),         (8),

(iv)    ξ1 + Ω* i = ξ2 + Ω* i = ξ6 + Ω* i = Ω* i+1 ∪ Ω* i+1
(5),        (4),        (8),   (3),     (7),

(v)     ξ1 + Ω * i = ξ 2 + Ω * i = ξ 3 + Ω * i = ξ 7 + Ω * i
(6),          (5),          (4),          (8),

= Ω* i+ 2 ∪ Ω* i+ 2 ∪ Ω* i+ 2 ∪ Ω* i+ 2
(2),      (4),      (6),      (8),

(vi)    ξ1 + Ω* i = ξ2 + Ω* i = ξ3 + Ω* i = Ω* i+1 ∪ Ω* i+1
(7),        (6),        (5),   (4),     (8),

(vii) ξ1 + Ω* i = ξ2 + Ω* i = ξ3 + Ω* i = ξ4 + Ω* i
(8),        (7),        (6),        (5),

= ( Ω* i + 4 ∪ ... ∪ Ω* i + 4 ) ∪ … ∪ ( Ω* n−3 ∪…. ∪ Ω* n−3 )
(1),             (8),               (1),         (8),

∪ ( Ω* n− 2 ∪ … Ω* n− 2 ) ∪ ( Ω* n−1 ∪ Ω* n−1 )
(1),        (4),          (1),     (2),

∪ Ω* ∪ Ω *
(1),n 0
Primitive idempotents                                                                                                   1247

(viii) ξ2 + Ω* i = ξ3 + Ω* i = ξ4 + Ω* i = Ω* i+1 ∪ Ω* i+1 .
(8),        (7),        (6),   (1),     (5),

Proof. I. By definition,
Ω* i = { 2i−1r : r ≡ 1 (mod 16) }. Then
(1),

ξ1 + Ω* i = { 2i−1 (r + s) : r = 16m + 1, s = 16n + 1 ; m, n ≥ 0 }
(1),

= {2i r : r ≡ 1, 9 (mod 16)}= Ω* i +1 ∪ Ω* i +1
(1),      (5),

Similarly, for 2 ≤ β ≤ 8,
ξ β + Ω*β ), i = Ω*β ), i +1 ∪ Ω*β+ 4), i +1 .
(         (             (

II. Follows on similar lines as above.

Lemma 3.3. Suppose 1 ≤ i ≤ n − 3. Then
I.      S(*β), i S(*β), i = 2n −i −3 ( S(*β), i +1 + S(*β+ 4), i +1 )
*       *         *       *         *       *         *       *
II.(i)        S(1), i S(2), i = S(3), i S(8), i = S(4), i S(7), i = S(5), i S(6), i
= 2n −i −3 ( S(1), i + 2 + S(3), i + 2 + S(5), i + 2 + S(7), i + 2 )
*             *             *             *

(ii)      S(1), i S(3), i = S(4), i S(8), i = S(5), i S(7), i = 2n −i −3 ( S(2), i +1 + S(6), i +1 )
*       *         *       *         *       *                    *            *

*       *         *       *        *       *
(iii)     S(1), i S(4), i = S(2), i S(3),i = S(5), i S(8),i
= S(6), i S(7), i = 2n −i −3 ( S(1), i +3 + ... + S(8), i +3 )
*       *                    *                  *

(iv)      S(1), i S(5), i = S(2), i S(4), i = S(6), i S(8), i = 2n −i −3 ( S(3), i +1 + S(7), i +1 )
*       *         *       *         *       *                    *            *

*       *         *       *         *       *         *       *
(v)       S(1), i S(6), i = S(2), i S(5), i = S(3), i S(4), i = S(7), i S(8), i
= 2n −i −3 ( S(2), i + 2 + S(4), i + 2 + S(6), i + 2 + S(8), i + 2 )
*             *             *             *

(vi) S(1), i S(7), i = S(2), i S(6), i = S(3), i S(5), i = 2n −i −3 ( S(4), i +1 + S(8), i +1 )
*       *         *       *         *       *                    *            *

(vii) S(1), i S(8), i = S(2), i S(7), i = S(3), i S(6), i = S(4), i S(5), i
*       *         *       *         *       *         *       *

= 2n −i −3 [( S(1), i + 4 + ... + S(8), i + 4 ) + ... + ( S(1), n −3 + ... + S(8), n −3 )
*                   *                       *                  *

+ ( S(1), n − 2 + ... + S(4), n − 2 ) + ( S(1), n −1 + S(2), n −1 ) + S(1), n + 1]
*                   *                 *            *              *

(viii)      S(2), i S(8), i = S(3), i S(7), i = S(4), i S(6), i
*       *         *       *         *       *

= 2n −i −3 ( S(1), i +1 + S(5), i +1 ).
*            *

Proof. The result follows by Lemma 3.2.

Lemma 3.4. For j < i,
1248                                                                                   K. Singh and S. K. Arora

⎧ S(*γ ), j     if j ≤ i − 4; ∀β                      (i)
⎪
⎪ S(*γ+ 4), j   if j = i − 3, ∀β                     (ii)
⎪ *
⎪ S( γ+ 2), j   if j = i − 2, β = 1,3,5, 7           (iii)
⎪ *
⎪ S( γ+ 6), j
⎪               if j = i − 2, β = 2, 4, 6,8          (iv)
S(*β), i S(*γ ), j    = 2 n −i −3 ⎨ *
⎪ S( γ+1), j    if j = i − 1, β = 1,5                (v)
⎪ *
⎪ S( γ+3), j    if j = i − 1, β = 2, 6               (vi)
⎪ *
⎪ S( γ+5), j    if j = i − 1, β = 3, 7              (vii)
⎪ *
⎪ S( γ+ 7), j
⎩               if j = i − 1, β = 4,8               (viii)

Proof. Since |Ω∗(β), i | = 2n −i −3 , therefore result follows by lemma 3.1.

Lemma 3.5. Suppose j ≥ i. Then
*                                                      *
(i)     X i G(1, 5), j = 0                                (ii) X i G(2, 6), j = 0
*                                        *
(iii)            X i G(3, 7), j = 0                  (iv) X i G(4,8), j = 0.

Proof. (i) Assume that j = i. Then, by definition of X i , G( l, m ), i and Lemma 3.4,
we get that :
1
*
X i G(1, 5), i =       2 n − 2 i +1
[2.2n−i−4{ ( S(2), i − S(6), i ) + ( S(4), i − S(8), i ) + + ( S(8), i − S(4), i )}
*         *             *         *               *         *

2
+ 22.2n −i −5{( S(3), i − S(7), i ) + ( S(7), i − S(3), i )} + 23.2n −i −6 ( S(5), i − S(1), i )
*         *             *         *                          *         *

+ 2n−i−3 ( S(1), i − S(5), i ) ] = 0.
*         *

Next assume j > i. Then,
*             1          *         *
X i G(1, 5), j = n − j +1 ( S(1), j − S(5), j ) X i
2
1 ⎡                          j −1                j −1 ⎤
= n − j +1 ⎢ ∑              gs 2 −         ∑ gt 2 ⎥ Xi
2        ⎢ s ≡ 1( mod16 )
⎣                          t ≡ 9( mod16 )      ⎥
⎦
Since i ≤ j − 1 and | Ω* j | = | Ω* j | , it follows from Lemma 2.4 [4] that
(1),       (5),

| Ω* j |
(1),
*
X i G(1,5), j =            ( X i − X i ) = 0.
2n − j +1
(ii), (iii) and (iv) follows on similar lines as above.
Primitive idempotents                                                                                      1249

Lemma 3.6.
⎧ *
⎪G                 if j = i,
(i)      Yi +3 G(1,5), j = ⎨ (1, 5), i
*

⎪0
⎩                otherwise.
⎧ *
⎪G                  if j = i,
(ii)     Yi +3 G(2, 6), j = ⎨ (2, 6), i
*

⎪0
⎩                otherwise.
⎧ *
⎪G                  if j = i,
(iii)    Yi +3 G(3, 7), j = ⎨ (1, 5), i
*

⎪0
⎩                 otherwise.
⎧ *
⎪G                  if j = i,
(iv)      Yi +3 G(4,8), j = ⎨ (4,8), i
*

⎪0
⎩                 otherwise.

Proof. Follows by Lemmas 3.1, 3.3, 3.4 and 3.5.

*         *            *         *            *          *
Lemma 3.7. (i)               G(1,5), i G(1,5), j = −G(3,7), i G(3,7), j = −G(2, 6), i G(4,8),   j

⎧1 ∗               ∗
⎪ (G            + G(5,7), i +1 ) if j = i,
= ⎨16 (1,3), i +1
⎪ 0
⎩                               otherwise.
*         *           *          *            *         *
(ii)    G(2,6), i G(2,6), j = G(1, 5), i G(3,7), j = −G(4,8), i G(4,8),   j

⎧1       ∗             ∗
⎪     (G(2,4), i +1 + G(6,8), i +1 ) if j = i,
= ⎨16
⎪ 0
⎩                                   otherwise.
*          *                *        *
(iii) G(1, 5), i G(2,6),   j = −G(3, 7), i G(4,8), j

⎧1          ∗         ∗               ∗               ∗
⎪ ( G(1,2), i + 2 + G(3,4)), i + 2 + G(5,6), i + 2 + G(7,8), i + 2 ) if j = i,
= ⎨ 32
⎪0
⎩                                                                 otherwise.
*          *               *          *
(iv) G(1, 5), i G(4,8),    j = G(2, 6), i G(3,7), j

⎧ 1
⎪− Y                     if j = i,
= ⎨ 64 i +3
⎪ 0
⎩   otherwise.
Proof. Follows by Lemmas3.3 and 3.4.

Lemma 3.8.
⎧ 1 *
⎪− G                           if j = i − 1,
(i)     G(1,3), i G(4,8), j = ⎨ 4 (1,5), i −1
*

⎪
⎩   0                          otherwise.
1250                                                                               K. Singh and S. K. Arora

⎧1 *
⎪ G                             if j = i − 1,
(ii)                 *
G(1,3), i G(3,7), j= ⎨ 4 (4,8), i −1
⎪
⎩       0                        otherwise.
⎧1 *
*          ⎪ G(3,7), i −1                 if j = i − 1,
(iii)     G(1,3), i G(2,6), j = ⎨ 4
⎪
⎩      0                       otherwise.
⎧1 *
⎪ G                           if j = i − 1,
(iv)                *
G(1,3), i G(1,5), j   = ⎨ 4 (2,6), i −1
⎪
⎩       0                     otherwise.
⎧1 *
⎪ G                          if j = i − 1,
(v)                 *
G(2,4), i G(1,5),   j = ⎨ 4 (4,8), i −1
⎪0
⎩                            otherwise.
⎧ 1 *
⎪− G                         if j = i − 1,
(vi)                *
G(2,4), i G(2,6), j   = ⎨ 4 (1,5), i −1
⎪
⎩    0                        otherwise.
⎧ 1 *
*               ⎪ − G(2,6), i −1              if j = i − 1,
(vii) G(2,4), i G(3,7),      j =⎨   4
⎪
⎩    0                        otherwise.
⎧ 1 *
⎪− G                          if j = i − 1,
(viii)              *
G(2,4), i G(4,8), j    = ⎨ 4 (3,7), i −1
⎪
⎩   0                         otherwise.

Proof. (i) First assume that j = i. Then by Lemma 3.3 II ( (i), (iii), (v), (vii) )
*        *          1    *               *                *               *                 1
G(1,3), i G(4,8), i =    (G(1,2), i + 2 + G(3,4), i + 2 + G(5,6), i + 2 + G(7,8), i + 2 ) −     Yi +3
64                                                                    128
1                                       1
=    (G(1,2), i + 2 + G(3,4), i + 2 ) −         Yi +3 .
64                                    128

Next assume that j ≠ i.
If j = i − 1, then by Lemma 3.4(v), (vii)
*        *               −1          *             *                1 *
G(1,3), i G(4,8), i −1 = n −i +5 ( S(1), i −1 − S(5), i −1 ) = − G(1,5), i −1 .
2                                            8
If j = i − 2, then by Lemma 3.4(iii)
*        *               2 n −i −3      *             *             *             *
G(1,3), i G(4,8), i − 2 = 2 n − 2i + 4 ( S(6), i − 2 − S(2), i − 2 − S(6), i − 2 + S(2), i − 2 ) = 0.
2
If j = i − 3, then by lemma 3.4(ii)
Primitive idempotents                                                                                 1251

*           *           2 n −i −3     *            *            *            *
G(1,3), i             = 2 n − 2i +5 ( S(8), i −3 − S(4), i −3 − S(8), i −3 + S(4), i −3 ) = 0.
G(4,8), i −3
2
If j ≤ i −4, then by Lemma 3.4(i)
*             *            *           *
G(1,3), i G(4,8), j = (G(1,3), i + G(5,7), i ) G(4,8), j = 0.
If j = i + 1, then by Lemma 3.4(viii)
*         *              2 n −i − 4     *         *         *          *
G(1,3), i G(4,8), i +1 = 2 n − 2i +1 ( S(8), i − S(8), i − S(2), i + S(2), i ) = 0.
2
If j = i + 2, then by Lemma 3.4(iv)
*         *              2 n −i −5 *            *         *         *
G(1,3), i G(4,8), i + 2 = 2 n − 2i ( S(7), i − S(7), i − S(1), i + S(1), i ) = 0
2
If j = i + 3, then by Lemma 3.4(ii)
2 n −i − 6    *          *         *          *
G(1,3), i G(4,8), i +3 = 2 n − 2i −1 ( S(5), i − S(5), i − S(7), i + S(7), i ) = 0
*        *

2
Finally suppose j ≥ i + 4. Then by Lemma 3.4(i)
*         *
G(1,3), i G(4,8), j = 0.
Similarly,
⎧ −1                                     1
⎪ 64 (G(1,2), i + 2 + G(3,4), i + 2 ) + 128 Yi +3       if j = i,
⎪
*          *          ⎪ 1 *
G(5,7), i G(4,8), j = ⎨ − G(1,5), i −1                                     if j = i − 1,
⎪ 8
⎪ 0                                                  otherwise.
⎪
⎩
The required form of result follows by using notation 2.3.(iii).
Proof of (ii) to (viii) follows on similar lines as above.
Lemma 3.9.
⎧1 *                    *               *             *
⎪16 (G(2,4), i −1 + G(2,6), i −1 − G(4,8), i −1 − G(6,8), i −1 ) if j = i − 1,
⎪
*        ⎪1 *
(i) G(1,2), i G(1,5), j = ⎨ G(3,7), i − 2                                                    if j = i − 2,
⎪4
⎪0                                                                otherwise.
⎪
⎩
⎧1 *                  *
⎪ 8 (G(3,7), i −1 + G(1,5), i −1 )                 if j = i − 1,
⎪
*        ⎪1 *
(ii) G(1,2), i G(2,6), j = ⎨ G(4,8), i − 2                                   if j = i − 2,
⎪4
⎪0                                                otherwise.
⎪
⎩
1252                                                                           K. Singh and S. K. Arora

⎧ 1 *
⎪ 8 (G(4,8), i −1 + G(2,6), i −1 )                 if j = i − 1,
*

⎪
⎪ 1
(iii) G(1,2), i G(3,7), j = ⎨ − G ∗(1,5), i − 2
*
if j = i − 2,
⎪ 4
⎪0                                                  otherwise.
⎪
⎩
⎧ 1 *
⎪− 16 (G(1,3), i −1 + G(1,5), i −1 − G(3,7), i −1 − G(5,7), i −1 ) if j = i − 1,
*             *               *

⎪
⎪ 1 *
(iv) G(1,2), i G(4,8), j = ⎨− G(2,6), i − 2
*
if j = i − 2,
⎪ 4
⎪0                                                                  otherwise.
⎪
⎩
⎧ 1 *                  *              *              *
⎪− 16 (G(2,4), i −1 + G(2,6), i −1 − G(4,8), i −1 − G(6,8), i −1 ) if j = i − 1,
⎪
*              ⎪1 *
(v) G(3,4), i G(1,5),     j = ⎨ G(3,7), i − 2                                                    if j = i − 2,
⎪4
⎪0                                                                 otherwise.
⎪
⎩
⎧ 1 *                 *
⎪− 8 (G(1,5), i −1 + G(3,7), i −1 ) if j = i − 1,
⎪
*             ⎪1 *
(vi) G(3,4), i G(2,6),    j = ⎨ G(4,8), i − 2                     if j = i − 2,
⎪4
⎪0                                  otherwise.
⎪
⎩

⎧ 1 *                 *
⎪− 8 (G(2,6), i −1 + G(4,8), i −1 ) if j = i − 1,
⎪
*            ⎪ 1 *
(vii) G(3,4), i G(3,7),   j = ⎨− G(1,5), i − 2                   if j = i − 2,
⎪  4
⎪0                                otherwise.
⎪
⎩
Primitive idempotents                                                                                         1253

⎧1 *                 *              *              *
⎪16 (G(1,3), i −1 + G(1,5), i −1 − G(5,7), i −1 − G(3,7), i −1 ) if j = i − 1,
⎪
*          ⎪ 1 *
(viii) G(3,4), i G(4,8), j = ⎨− G(2,6), i − 2                                                 if j = i − 2,
⎪ 4
⎪0                                                               otherwise.
⎪
⎩
Proof. Follows by Lemmas 3.3 and 3.4 and by using notation 2.3(iii).

4. Main Result

Theorem 4.1. FC 2n has 8(n − 2) primitive idempotents given by :
e0 = Y0
*
;                  e(1), n = Y1
*

1                                                     1
e(1), n −1 = [Y2 − 2θ2 (G(1,2), 1 + G(3,4), 1 )] , e(2), n −1 = [Y2 + 2θ2 (G(1,2), 1 + G(3,4), 1 )] ;
*                                                         *

2                                                      2
1
e(1), n − 2 = [Y3 − 2θ (G(1,2), 2 + G(3,4), 2 ) − 4θ (G(2,4),1 + θ2G(1,3),1 )] ,
*                      2

4
1
e(2), n − 2 = [Y3 + 2θ2 (G(1,2), 2 + G(3,4), 2 ) − 4θ (θ2G(2,4),1 + G(1,3),1 )] ,
*

4
1
e(3), n − 2 = [Y3 − 2θ2 (G(1,2), 2 + G(3,4), 2 ) + 4θ (G(2,4),1 + θ2G(1,3),1 )] ,
*

4
1
e(4), n − 2 = [Y3 + 2θ2 (G(1,2), 2 + G(3,4), 2 ) + 4θ (θ2G(2,4),1 + G(1,3),1 )] ;
*

4
for 1 ≤ i ≤ n − 3
1
e(1), i = [Yi + 3 − 2θ2 (G(1,2), i + 2 + G(3,4), i + 2 ) −4θ (G(2,4), i +1 + θ2G(1,3), i +1 )
*

8
−8 θ (G(4,8), i + θ G(3,7), i + θ2G(2,6), i + θ3G(1,5), i )] ,
*              *                *               *

1
e(2), i = [Yi +3 + 2θ2 (G(1,2), i + 2 + G(3,4), i + 2 ) − 4θ (θ2G(2,4), i +1 + G(1,3), i +1 )
*

8
+ 8 θ (G(2,6), i − θ G(1,5),i − θ2G(4,8), i + θ3G(3,7), i )] ,
*                *               *                *

1
e(3), i = [Yi +3 − 2θ2 (G(1,2), i + 2 + G(3,4), i + 2 ) + 4θ (G(2,4), i +1 + θ 2G(1,3), i +1 )
*

8
+ 8 θ (G(3,7), i −θ G(4,8),i −θ2G(1,5), i + θ3G(2,6), i )] ,
*            *               *               *

1
e(4), i = [Yi +3 + 2θ2 (G(1,2), i + 2 + G(3,4), i + 2 ) + 4θ(θ2G(2,4), i +1 + G(1,3), i +1 )
*

8
1254                                                                       K. Singh and S. K. Arora

− 8 θ (G(1,5), i + θ G(2,6), i + θ2G(3,7), i + θ3G(4,8), i )] ,
*             *             *             *

1
e(5), i = [Yi +3 − 2θ2 (G(1,2), i + 2 + G(3,4), i + 2 ) − 4θ (G(2,4), i +1 + θ2G(1,3), i +1 )
*

8
+ 8 θ (G(4,8), i + θ G(3,7), i + θ2G(2,6), i + θ3G(1,5), i )] ,
*               *                *               *

1
e(6), i = [Yi + 3 + 2θ2 (G(1,2), i + 2 + G(3,4), i + 2 ) − 4θ (θ2G(2,4), i +1 + G(1,3), i +1 )
*

8
− 8 θ (G(2,6), i − θ G(1,5),i − θ2G(4,8), i + θ3G(3,7), i )] ,
*                 *               *                *

1
e(7), i = [Yi + 3 − 2θ2 (G(1,2), i + 2 + G(3,4), i + 2 ) + 4θ (G(2,4), i +1 + θ2G(1,3), i +1 )
*

8
− 8 θ (G(3,7), i −θ G(4,8),i −θ2G(1,5), i + θ3G(2,6), i )] ,
*           *               *              *

1
e(8), i = [Yi +3 + 2θ2 (G(1,2), i + 2 + G(3,4), i + 2 ) + 4θ(θ 2G(2,4), i +1 + G(1,3), i +1 )
*

8
+ 8 θ (G(1,5), i + θ G(2,6), i + θ2G(3,7), i + θ3G(4,8), i )] ,
*                *               *                *

where θ 2 = − 1 and             θ belongs to GF(ρ) or in some extension of GF(ρ).

Proof. This can be proved by means of Lemmas 2.4[4], 3.6[4] to 3.9[4] and
Lemmas 3.6 to 3.9 as proved above.

Acknowledgements. Research supported by : C.S.I.R.,( F.No.9/382(101), 2k3-
EMR-I), New Delhi, India.

References
[1]      Arora, S.K., Batra, S. and Cohen, S. D. ; The primitive idempotents of a
cyclic group Algebra II, Southeast Asian Bulletin of Mathematics, 29, 197-
208(2005).

[2]      Batra, S. and Arora, S.K. ; Minimal quadratic residue cyclic codes of
length 2n, J. App. Math. & Computing, 18(2005), No. 1-2, 25-43.

[3]      Burton, David M. ; Elementary Number Theory, McGraw Hill Book
Company, New York, (Fifth Edition), 2002.

[4]      Singh, K. and Arora, S.K. ; The Primitive Idempotents in                                  FC2n − I ,
International Journal of Algebra, Vol. 4 (2010), No. 25, 1231-1241.