Duadic Group Algebra Codes

0.11 MB·English

by  Salah A. Aly

Duadic Group Algebra Codes
Salah A. Aly, Andreas Klappenecker, Pradeep Kiran Sarvepalli
Department of Computer Science, Texas A&M University
College Station, TX 77843-3112,USA
Email: {salah, klappi, pradeep}@cs.tamu.edu
Abstract—Duadic group algebra codes are a generalization II. DUADIC GROUP ALGEBRA CODES
of quadratic residue codes. This paper settles an open problem
raised by Zhu concerning the existence of duadicgroup algebra A. Background
codes.Thesecodescanbeusedtoconstructdegeneratequantum
7
stabilizer codes that have the nice feature that many errors of Let G be a finite group of order n with identity element
0
small weightdonotneederrorcorrection; thisfact isillustrated 1 and let F denote a finite field with q elements such that
0 q
2 by an example. gcd(q,n) = 1. Recall that the group algebra R = Fq[G]
consists of elements of the form a g, with a ∈ F .
n g∈G g g q
a I. INTRODUCTION ThesetRisavectorspaceoverFq inwhichtheelementsofG
P
J formabasis.Furthermore,Risequippedwithamultiplication
Binary cyclic duadic codes were introduced by Leon,
9 defined by the convolutional product
Masley and Pless [5] as a generalization of quadradic residue

] codes. Duadic codes share many properties of quadratic
T residue codes, but are more flexible; for example, they are agg bgg = ahbh−1g g.
.I not restricted to prime lengths. Subsequently, cyclic duadic (cid:18)gX∈G (cid:19)(cid:18)gX∈G (cid:19) gX∈G(cid:18)hX∈G (cid:19)
s
code were generalized to nonbinary fields in [7], [9], [10].
c A left ideal in R is an additive subgroup I of R such that
[ Further progress was made by interpreting duadic codes as
ra∈I for all r in R and a in I. A group algebra code in R,

group algebra codes in the sense of MacWilliams [6]. Using
1 or shortly an R-code, is a left ideal I of R.
this point of view, duadic codes were extended to abelian
v For example, if G = Z/nZ is a cyclic group, then
60 gInrotuhpisspbaypReru,swhaenfaunrt[h7e]rarnedlatxotnhoenraebqeuliiraenmgernotuspabnydZalhluow[1f4o]r. Fq[Z/nZ]∼=Fq[x]/hxn−1i;thus,inthiscaseagroupalgebra
code is simply a cyclic code.
0 a wider class of antiautomorphisms.
1 AnelementeinthegroupalgebraRiscalledanidempotent
An open problem by Zhu [14] asks to find necessary
0 if and only if e2 = e. An idempotent in the center of R is
and sufficient conditions for the existence of duadic group
7 called central. Since gcd(n,q) = 1, any R-code is generated
0 algebra codes with splitting given by µ−1 (the terminologyis by an idempotent, that is, for any left ideal I there exists an
/ explainedin the nextsection).Partialanswersto this question
s idempotentelementeinRsuchthatI =Re.Twoidempotent
c were obtained by Smid [10] in the cyclic case, by Ward
elements e and f are called orthogonal if ef = 0 = fe. An
: and Zhu [12] and Rushanan [7] in the abelian case, and by
v nonzero idempotent e in R is called (centrally) primitive if
i Zhang[13]in the case of (nonabelian)finite groupsandfinite
X and only if it cannot be written as the sum of two nonzero
fields of characteristic 2.
orthogonal (central) idempotents in R.
r
The main result of this paper settles Zhu’s question in the
a The elements 0 and 1 are idempotents of R. If N is a
generalcaseofarbitraryfinitegroupsandarbitraryfinitefields.
subgroupofG,thenN =|N|−1 g isanidempotentel-
We show that if G is a finite group of odd order n and F g∈N
q ementinthegroupalgebraR.IfN isanormalsubgroup,then
is a finite field with q elements such that gcd(q,n)=1, then P
N is a central idempobtent. The central idempotent G, known
there exists a splitting given by µ with central idempotents
−1 as the trivial idempotent, will play a significant role in the
e and f if and only if the order of q is odd modulo n. In our
proof,weestablishakeypropositionthatallowsustotransfer sbubsequentsections.Ifwemultiplyanelementb= bg∈Gbgg
by G, then we obtain bG = b G; in particular, we
the existence question of duadic group algebra codes (for an g∈G g P
arbitrarysplitting)to aquestionaboutso-calledFq-conjugacy have dimRG = 1. An elem(cid:0)enPt b = (cid:1) g∈Gbgg in R is
b b b
classesofthegroup.Furthermore,we giveanexampleforthe called even-like if and only if bG = 0 (viz. b = 0).
(cid:0)P g∈G(cid:1) g
existence of a splitting µ when µ -splitting cannot exist. An element obf R that is not even-like is called odd-like.
−1
P
Oneoftheapplicationsofduadicgroupalgebracodesisthe An antiautomorphismon the grboupalgebra R is a bijective
construction of (degenerate) quantum error-correcting codes; map µ on R that satisfies (i) µ(a)+µ(b) = µ(a +b) and
hereduadicgroupalgebracodescanprovideevenbetterresults (ii) µ(ab) = µ(b)µ(a) for all a,b in R. We say that the
thancyclicduadiccodes,butspaceconstraintswillonlyallow antiautomorphism µ is isometric if it preserves the Hamming
ustosketchanexample.Wederiveafamilyofquantumcodes weight. An important isometric antiautomorphism on F [G]
q
with parameters [[n,1,d]] such that d2−d+1≥n. is µ defined as µ (g)=g−1 for g in G.
q −1 −1
B. Duadic Group Algebra Codes Lemma 3: Supposethateandf areidempotentsthatdeter-
mine a pair of even-like duadic codes in F [G] with splitting
LetGbeafinitegroupofordernandF afinitefieldsuch q
q
given by µ. If the group G has order n, then the minimum
that gcd(q,n) = 1. If e and f are two even-like idempotents
weight d of an odd-like element in the odd-like duadic code
in R=F [G] that satisfy the equations o
q
D =R(1−e) or D =R(1−f) satisfies
A1 e+f =1−G and f e
i) d2 ≥n,
A2 µ(e) = f and µ(f) = e for some isometric antiautomor- o
ii) d2−d +1≥n if µ=µ .
phism µ on R,b o o −1
Proof: i) Suppose that a is an odd-like element in D
then the idempotents e and f generate f
of weight d , so there exists an element b∈ F [G] such that
C1 apairofeven-likeduadiccodesC :=ReandC :=Rf, o q
e f
a=b(1−e).Theelementbisodd-like,since06=aG=b(1−
C2 and a pair of odd-like duadic codes D :=R(1−f) and
e
e)G = bG holds. A splitting satisfies µ(G) = G; thus, µ(b)
D :=R(1−e).
f
is odd-like as well, as 06=bG implies 06=Gµ(b)=bµ(b)G.
TheantiautomorphismµgiveninA2issaidtogiveasplitting.
By a slight abuse we also refer to µ−1 as a splitting. Tbhe prboduct of a and µ(a) has Hammbing wbeight ≤ d2o.
Lemma 1: LetGbeagroupofordernandF afinitefield However, we recall that (1−be)(1−f)=Gb, so b
q
such that gcd(n,q)=1. If e and f are even-like idempotents
aµ(a)=b(1−e)(1−f)µ(b)=bGµ(b)=bµ(b)G6=0.
inF [G]thatsatisfyA1andA2withsplittingµ,thenwenote b
q
that
If bµ(b) = c g, then bµ(b)G = c G; thus,
g∈G g b g∈G gb
i) the idempotents e, f, and G are pairwise orthogonal; the product aµ(a) yields an element of Hamming weight n,
ii) dimCe =dimCf =(n−1)/2anddimDe =dimDf = which provePs the bound n≤d2. b (cid:0)P (cid:1)b
o
(n+1)/2; b For part ii), we note that aµ (a) has at most Hamming
−1
iii) in particular, the order n of the group G must be odd; weight d2 − d + 1 when a has Hamming weight d . By
o o o
iv) the codes satisfy the inclusions C ⊆D and C ⊆D .
e e f f symmetry similar results hold for D .
e
Proof: i) Since e and f = (1 − G − e) are even-
like, it follows that the idempotents e,f, and G are pairwise D. Duals of Duadic Codes
orthogonal; hence, R = Re ⊕ Rf ⊕ RGb. For ii) and iii),
We can define a Euclidean inner product on F [G] by
we observe that dimRG = 1 and dimRe =bdimRµ(e) = q
dimRf, which implies dimRe= dimRfb=(n−1)/2. The
dimensions for the odd-blike duadic codes are an immediate agg bgg = agbg.
consequence,since Ce⊕Df =R and Cf ⊕De =R. For iv), (cid:28)gX∈G (cid:12)(cid:12)gX∈G (cid:29) gX∈G
notice that the orthogonality of e and f yields e(1−f) = e (cid:12)
(cid:12)
andf(1−e)=f.Therefore,Ce =Re⊂R(1−f)=De and Lemma 4: Let G be a finite group and R=Fq[G].
Cf =Rf ⊂R(1−e)=Df. i) The product ab=0 for a,b∈R if and only if µ−1(b)∈
Lemma 2: Let F be a finite field of characteristic p. C⊥, where C =Ra.
q
Suppose that µ is an isometric antiautomorphism of a group ii) If C is an R-code, then C⊥ is also an R-code.
algebra F [G] that satisfies µ(G) = G. Then there exists a iii) If e is an idempotent in R and C = Re, then C⊥ =
q
Galois automorphism σ ∈ Gal(Fq/Fp) and an antiautomor- R(1−µ−1(e)).
phism µ of the group G such tbhat b Proof: i) We note that the product of a and b can be
∗
expressed in the form
µ(αg)=σ(α)µ (g) holds for all α∈F ,g ∈G. (1)
∗ q
Proof: The isometry of the antiautomorphism µ implies ab= aghbh−1 g = hg−1a|µ−1(b)ig,
that the image µ(g) of an element g in G is of the form !
g∈G h∈G g∈G
µ(g)= c g′ for some nonzero constant c ∈F and g′ ∈G. X X X
g g q
from which we can directly deduce the claim.
Since µ(G) = G, we have c = 1 for all g ∈ G. Therefore,
g
µ restricts to a antiautomorphismµ on the groupG. Since µ ii) We note that the inner product satisfies hga|gbi=ha|bi
∗
preservesbadditiobnandmultiplicationofscalarsandµ(Fq1)= for all g in G and a,b in R. If a ∈ C and b ∈ C⊥, then for
F 1, we have µ(α1) = σ(α)1 for some automorphism of eachg ∈G,wehaveha|gbi=hg−1a|bi=0,sinceg−1a∈C.
q
F . The elements of the prime field F remain fixed, so σ Extending linearly shows that C⊥ is a left ideal.
q p
is an element of Gal(F /F ). The claim is an immediate iii)Sincee(1−e)=0,propertyi)showsthattheidempotent
q p
consequence of these observations. 1−µ−1(e)iscontainedinC⊥,soR(1−µ−1(e))⊆C⊥ byii).
Since dimC⊥ = dimR(1 −e) = dimR(1− µ (e)), we
−1
C. Odd-like Weights in Duadic Group Algebra Codes actually must have equality.
Our first result is a slight generalization of a theorem by Corollary 5: If e and f are even-like idempotents that
satisfy A1 and A2, then the following statements hold:
Zhu [14]. We allow isometric antiautomorphisms of F [G],
q
whereas Zhu considers only antiautomorphisms that are in- i) If µ−1(e)=f, then Ce⊥ =De and Cf⊥ =Df;
duced from the group G. ii) if µ−1(e)=e, then Ce⊥ =Df and Cf⊥ =De.
III. EXISTENCEOF SPLITTINGS we need some preparation to state this result. We note that
The goal of this section is to prove the following theorem acentrallyprimitiveidempotenteχ ofFq[G]canbe explicitly
of the existence of duadic group algebra codes. written in the form
Theorem 6: Let G be a finite group of odd order and let e = nχ χ(g)g−1, (2)
χ
F beafinitefieldwithgcd(n,q)=1.Thereexistsasplitting |G|
q g∈G
µ=µ withcentralidempotentseandf suchthatequations X
−1
where χ is an irreducible F -character and n is a positive
A1 and A2 are satisfied if and only if the order of q is odd q χ
integer that depends on χ, see Lemma 14 in the Appendix.
modulo n.
Lemma 8: Let F be a finite field and G a finite group
Proof: (Outline)Ourproofis subdividedintothree parts. q
of order n such that gcd(n,q) = 1. Suppose that µ is an
Part A. We show that a splitting µ with central idempo-
−1
antiautomorphism of F [G] of the form (1). Then the action
tentse andf satisfying A1 andA2 existsif andonlyif no q
of µ on a centrally primitive idempotent (2) is given by
nontrivial centrally primitive idempotent is fixed by µ ,
−1
see Proposition 7. µ(e )= nχ χ(µ−1(gk))g−1,
Part B. We then define an action of µ on so-called F - χ |G| ∗
−1 q g∈G
conjugacyclassesinG.Weprovethatnonontrivalcentrally X
where k is a positive integer determined by the Galois auto-
primitive idempotent is fixed by µ if and only if no
−1 morphismσ and µ−1 is the inverse of the groupantiautomor-
F -conjugacy class is fixed by the action of µ , see ∗
q −1
phism µ . In particular, k is a power of the characteristic of
Proposition 9. ∗
F .
Part C. By [13,Lemma 2.3],K 6=µ (K) for allnontrivial q
−1
Proof: If the exponent of the group G is m, then the
F -conjugacy classes K if and only the order of q is odd
q
values of the character are contained in F ∩F (δ), where
modulo n. q p
δ is a primitive m-th root of unity over the prime field F .
The claim follows by combining the three parts. p
Suppose that σ′ is a Galois automorphismof Gal(F (δ)/F )
Actually, our proofs of Part A and Part B are valid for q p
that restricts to the Galois automorphismσ on F . If σ′(δ)=
arbitrary splittings µ. If one can find necessary and sufficient q
δk, then σ(χ(g))=χ(gk) holdsfor all g in G, and the action
conditions such that µ(K) 6= K holds for all nontrival F -
q
of an antiautomorphism µ is given by
conjugacy classes, then one already obtains an extension of
the theorem to µ-splittings. µ(e )= nχ σ(χ(g))µ (g−1)= nχ χ(µ−1(gk))g−1,
One technical difficulty in our proof is that the counting χ |G| ∗ |G| ∗
g∈G g∈G
argument used in Part B cannot be done in nonzero charac- X X
teristic. We circumventthis problem by using an extension of where the latter equality is obtained by substituting µ−∗1(g)
the field of p-adic integers (a local field) such that the ring of for g.
Let us recall the concept of an F -conjugacy class before
integers in this field reduces to the given finite field modulo q
stating our next result. The F -conjugacy class K (g) of an
its maximal ideal; this proof technique is interesting in itself. q q
Part A. We now supply the details of Part A of the proof. element g in a finite group G is the set
Proposition 7: There exists a splitting µ with even-like K (g)={h−1gqkh|h∈G,k ≥0}.
q
(central) idempotents e and f that satisfy A1 and A2 if and
only if each nontrivial (centrally) primitive idempotent h of It is easy to see that two Fq-conjugacy classes are either
F [G] satisfies µ(h)6=h. disjointor coincide.The followingtwo key facts are essential
q
Proof: Suppose that e and f are even-like (central) for our purpose: (i) An irreducible Fq-character is constant
idempotents that satisfy A1 and A2. These equations imply onFq-conjugacyclasses,and(ii)thenumberofFq-conjugacy
that e+f +G = 1, where the idempotents e, f, and G are coincides with the number of irreducible Fq-characters.
pairwise orthogonal. Suppose that e = h + ···+ h is a We define an action of an antiautomorphismµ of the form
1 m
decompositionbof the idempotente into orthogonal(centbrally) (1) on an Fq-conjugacy class Kq(g) by
primitiveidempotents.Seekingacontradiction,weassumethat
Kµ(g)=K (µ (g)ℓ),
µ(h ) = h for some k in the range 1 ≤ k ≤ m. However, q q ∗
k k
thene andf cannotbeorthogonalidempotents,contradiction. where ℓ is a positive integer such that kℓ ≡ 1modm, k is
Conversely, suppose that h 6= µ(h) for all nontrivial the integer given in Lemma 8, and m is the exponent of the
primitive (central) idempotents h of F [G]. Partition the group G.
q
nontrivial (central) primitive idempotents into disjoint pairs Proposition 9 (Key Proposition): Let F be a finite field
q
{h ,µ(h )},...,{h ,µ(h )}. Let e = h + ··· + h and andletGbeafinitegroupofordernsuchthatgcd(n,q)=1.
1 1 ℓ ℓ 1 ℓ
f = µ(e). Then e+f = 1−G and eG = 0 and fG = 0. If µ is an antiautomorphism of F [G] of the form (1), then
q
Further,µ(e+f+G)=1=e+f+Gimpliesthate=µ(f).So the number of F -conjugacy classes of G that are fixed by µ
q
µ is the desired splitting with (cbentral) ebven-like idembpotents coincides with the number of centrally primitive idempotents
e and f. b b of Fq[G] that are fixed by µ.
Part B. The second partof our argumentis more involved. Proof: Step 1. Suppose that the finite field F has
q
Our goal is to prove the key proposition below. However, characteristic p. There exists a monic irreducible polynomial
f(x) in F [x] such that F [x]/hf(x)i = F . Let Z denote We note that U is nonsingular, since the irreducible K-
p p q p
the ring of p-adic integers and Q its quotient field. Then characters are linearly independent over K. Let A =
p
p = pZ is the unique nonzero prime ideal of Z and (A ) and B = (B ) be permutation ma-
p p χ,ψ χ,ψ∈Y L,K L,K∈Kq
Zp/p∼= Fp. Choose a monic polynomial g(x) in Zp[x] such trices that are respectively defined by
that f(x) ≡ g(x) mod p. Then R = Z [x]/hg(x)i is a
p 1 if χ=ψη 1 if Kη =L
discrete valuation ring with nonzero prime ideal P=pR. Aχ,ψ = 0 otherwise and BL,K = 0 otherwise.
If K denotes the field Qp[x]/hg(x)i, then K/Qp is an (cid:26) (cid:26)
unramified Galois extension. The Galois group Gal(K/Q ) Since χη(Kη)=χ(K), we have
p
canbeidentifiedwiththecyclicgroupGal(F /F ).Thelatter
q p A ψ(K)=χ(Kη) and χ(L)B =χ(Kη),
group is generated by the automorphism x 7→ xp, and the χ,ψ L,K
generator of Gal(K/Q ) can be characterized by ψX∈Y LX∈Kq
p
so AU = UB. Since U is invertible, we have A = UBU−1.
b≡bp modP for all b∈R.
Thus, tr(A) = tr(B). The trace of A counts the number of
All facts stated in this step are proved in [8, Chapter 1]. characters that remain fixed under the action of η, and the
Step 2. The number of centrally primitive idempotents trace of B counts the number of F -conjugacy classes that
q
in K[G] and in F [G] is the same. If Y denotes the set remain fixed under η. These facts imply the claim.
q
of irreducible K-characters of G, then the set of centrally Recall that Part C has been proved in [13]; thus, this
primitive idempotents {e |χ ∈ Y} of K[G] is bijectively concludesourproofofTheorem6.Fortheexistenceofduadic
χ
mapped to the centrally primitive idempotents of F [G] by group algebra codes with splitting µ 6= µ one only needs
q −1
reduction modulo P. to modify Part C.
Letk be the integerdefinedas in Lemma 8. Then thereex-
IV. EXTENSIONS AND APPLICATIONS
istsauniqueautomorphismτ inGal(K/Q )suchthatτ(x)=
p
xk modP. Therefore, we can define an antiautomorphism η The natural question following the previous section is the
on K[G] by η(αg)=τ(α)µ (g) such that existence of duadic group algebra codes when the order of q
∗
is even. The splitting is no longer given by µ , but we will
−1
η(e )modP=µ(e modP) (3)
χ χ show that there exist duadic algebra codes. We will confine
holds for all centrally primitive idempotentse of K[G]. The ourselvestotheabeliancase.Thenwewillgiveanapplication
χ
latter equation guarantees that the number of idempotents in of duadic group algebra codes to quantum error-correction.
K[G] fixed by η is the same as the numberof idempotentsin
A. Extensions
F [G] fixed by µ.
q
Cyclic duadic codes exist if and only if q is a quadratic
Step 3. A centrally primitive idempotent in K[G] is of the
residue modulon. However,this condition is not required for
form
e = nχ χ(g)g−1. group codes. We partially generalize some of the existence
χ
|G| results of [7], where characteristic two is considered.
g∈G
X Lemma 10: LetG=ha,b|ap =bp =1,ab=bai, wherep
It follows from Lemma 8 that η(e ) = e if and only if
χ χ is an odd prime and q be a prime power such that ord (q) is
p
χ(g)=χ(η−1(gk)) holdsforallg in G. Therefore,wedefine
evenandgcd(p,q)=1.TheF -conjugacyclassofanelement
q
the action of η on an irreducible K-character by axby in G is given by C(q) = {axqjbyqj | j ∈ Z}. Then the
x,y
χη(g) = χ(µ−1(gk)), (4) automorphismµ definedas µ(axby)=aqybx doesnotfix any
∗
of the F -conjugacy classes if (x,y) 6= (0,0). Further, µ
q −1
for all g in G. fixes each F -conjugacy class i.e., µ (C(q))=C(q).
An irreducible K-character is constant on K-conjugacy q −1 x,y x,y
Proof: Assume that there is an element axby ∈ G ,
classes. The K-conjugacy classes coincide with the F -
q such that µ(C(q)) = C(q). Then there exists an integer j
conjugacy classes, since the Galois groups are isomorphic. x,y x,y
such that µ(axby) = axqjbyqj. This implies that (qy,x) =
Suppose that m is the exponent of the group G. There exists
(xqj,yqj) mod porqy ≡xqj mod pandx≡yqj mod p.
a positive integer ℓ such that kℓ ≡ 1modm. We define the
If x = 0,y 6= 0, then we have qy ≡ 0 mod p or y = 0;
action of η on F -conjugacy classes by
q
a contradiction. If y = 0,x 6= 0, then it follows x = 0,
Kqη(g) = Kq(µ∗(gℓ)), (5) which leads to a contradiction again. Assuming that both
x,y 6= 0 we get qxy ≡ xyq2j mod p. Since q,x,y are all
for all g in G. The definitions are carefully chosen such that
coprime to p this can be written as 1 ≡ q2j−1 mod p. But
χη(Kqη(g))=χ(Kq(g)) as ordp(q) is even, 1 6≡ q2j−1 mod p. Therefore, none of
the F -conjugacy classes are fixed by µ. Let ord (q) = 2w,
holds for all g in G. q p
then q2w ≡ 1 mod p, which implies that qw ≡ −1 mod p.
Step 4. Let K denote the set of F -conjugacyclasses. We
q q Hence, C(q) =C(q) =µ (C(q)).
have |Y|=|K |. Therefore, we can define the square matrix x,y −x,−y −1 x,y
q Theorem 11: Let G = ha,b|ap = bp = 1,ab = bai with
U =(χ(K)) . p an odd prime, q a prime power, gcd(p,q)=1 and ord (q)
χ∈Y,K∈Kq p
even. Then there exist duadic codes over F [G] with splitting codesoverF [G ]withidempotentse =a +a2+a b +a2b2
q 2 i i i i i i i i
givenbyµwhereµ(axby)=aqybxforanyelementaxby ∈G. andf =b +b2+a b2+a2b satisfyingA1andA2,underthe
i i i i i i i
These codes are fixed by µ . action of µ (axby)=a2ybx. Further e ,f are fixed by µ .
−1 i i i i i i i −1
Proof: By Part A (Proposition 7), Part B (Proposi- We can construct a duadic code over F [G ×G ] using
2 1 2
tion 9), and replacing Part C by Lemma 10, we know that a construction similar to [7, Theorem 5.6]. Embedding G
i
there exist a pair of duadic codes over F [G] with splitting into G × G , we get the idempotents of the new code as
q 1 2
given by µ such that the codes are fixed by µ . e = e +e −e e −f e and f = f +f −f f −e f .
−1 1 2 1 2 1 2 1 2 1 2 1 2
The splitting for this code is given by µ=µ ×µ .
1 2
B. Quantum Error-Correction
Theseidempotentsgiveusduadicgroupalgebracodesover
Quantum codes can be utilized to protect quantum infor- F [G ×G ] that are fixed by µ . As ee = e , wt(C ) =
2 1 2 −1 1 1 e
mation over noisy quantum channels. The CSS construction wt(C )≤wt(e )=4,whilewt(D \C )=wt(C⊥\D⊥)=
f 1 e e e e
isparticularlytransparentmethodtoderivequantumstabilizer wt(D \ C ) ≥ 9 by Corollary 5 ii) and Lemma 3. Thus
f f
codes from a pair of classical codes. by Lemma 12 there exists a [[81,1,≥9]] quantumcode;this
2
Lemma 12 (CSS Construction [2], [11]): Suppose that C codeisdegenerateandmanyerrorsofweightsbetween4and9
and D are linear codes over a finite field F such that (contained in C or D⊥) do not need error-correction.
q e e
C ⊆D. If C is an [n,k ] code and D an [n,k ] code, then
1 q 2 q APPENDIX
there exists an [[n,k −k ,d]] stabilizer code with minimum
2 1 q
distance d=minwt{(D\C)∪(C⊥\D⊥)}. Lemma 14: Let G be a finite group of order n and F a
finite field of characteristic coprime to n. If E ⊇ F is a
Theorem 13: Letnbeanoddpositiveintegerandqapower
splittingfieldforG,W anirreducibleFG-modulaffordingthe
of a prime such that the order of q modulo n is odd. Then
there exists an [[n,1,d]] stabilizer code with d2−d+1≥n. characterχ,andV anirreduciblesubmoduloftheEG-modul
q E⊗W affording the character θ, then χ(x)= θσ(x),
Proof: ThereexistsagroupGofordern.ByTheorem6, σ∈H
there exist idempotents e and f in Fq[G] that satisfy A1 and where H = Gal(Fq(θ)/Fq), and e = θ|G(1|) g∈PGχ(g−1)g is
A2 with splitting given by µ = µ . By Lemma 1, we have a centrally primitive idempotent in FG.
−1 P
C ⊂ D . The CSS construction shows that there exists an Proof: Since F is a finite field, we note that the Schur
e e
[[n,(n+1)/2−(n−1)/2,d]] = [[n,1,d]] stabilizer code indexofthecharacterθis1by[3,Theorem24.10].Therefore,
q q
with minimum distance d=min{(D \C )∪(C⊥\D⊥)}= thecharacterχ isof theclaimedformby[3,Theorem24.14].
min{(De\Ce)∪(De\Ce)}, wherethee lateterequealityfoellows We notethateθ = θ|(G1|) g∈Gθ(g−1)g is a centrallyprimitive
from Corollary 5 i). Since D = RG⊕C , we observe that idempotent of EG by [3, Lemma 24.13]. The form of the
e e P
De \Ce contains precisely the odd-like elements of De. By character χ implies that e = σ∈Heθσ, so e is a central
Lemma 3, the minimum weight of anbodd-like element in De idempotent of FG. The primiPtivity of e is shown in [4,
satisfies d2−d+1≥n. Theorem 19.2.9].
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