We define essential idempotents in group algebras and use them to prove that every mininmal abelian non-cyclic code is a repetition code. Also we use them to prove that every minimal abelian code is equivalent to a minimal cyclic code of the same length. Finally, we show that a binary cyclic code is simplex if and only if is of length of the form $n=2^k-1$ and is generated by an essential idempotent

Chalom, Gladys

Ferraz, Raul A.

Milies, C. Polcino

Journal of Algebra Combinatorics Discrete Structures and Applications

English

Journal of Algebra Combinatorics Discrete Structures and Applications (JACODESMATH, Yildiz Technical University - YTU)

ISSN 2148-838Xhttp://dx.doi.org/10.13069/jacodesmath.284931J. Algebra Comb. Discrete Appl.4(2) • 181–188Received: 15 June 2015Accepted: 22 February 2016Journal of Algebra Combinatorics Discrete Structures and ApplicationsEssential idempotents and simplex codes∗Research ArticleGladys Chalom, Raul A. Ferraz, C. Polcino MiliesAbstract: We define essential idempotents in group algebras and use them to prove that every mininmal abeliannon-cyclic code is a repetition code. Also we use them to prove that every minimal abelian code isequivalent to a minimal cyclic code of the same length. Finally, we show that a binary cyclic code issimplex if and only if is of length of the form n = 2k− 1 and is generated by an essential idempotent.2010 MSC: 16S34, 20C05, 94B15Keywords: Group code, Essential idempotent, Simplex code1. IntroductionLet Fq be a finite field with q elements and m a positive integer. We recall that a linear code oflength n over Fq is any proper subspace C of Fnq . Given a vector v = (a0, a1, . . . , an−2, an−1) ∈ C, its shiftis the vector v1 = (an−1, a0, a1, . . . , an−2). A linear code C is cyclic if, for every vector v ∈ C, its shiftalso belongs to C. Notice that the definition implies that if a vector v1 = (a0, a1, . . . , an−2, an−1) is in C,then every vector obtained as a circular permutation of v is also in C.The map ψ : Fnq → Fq[X]/〈Xn−1〉 given by ψ(a0, a1, . . . , am−2, am−1) = a0+a1X+· · · , am−2Xm−2+am−1Xm−1 is an isomorphism of linear spaces and it is easy to see that a code C of length n over Fq iscyclic if and only its image ψ(C) is an ideal of the ring Fq[X]/〈Xn − 1〉.Since this ring is isomorphic to the group algebra of a cyclic group C, of order n, over Fq, we canthink of cyclic codes as ideals in the group algebra FqC.More generally, an abelian code over Fq is any ideal in the group algebra FqA of a finite abeliangroup A. These codes were introduced independently by S.D. Berman [1] and MacWilliams [8].∗ This work was supported by FAPESP proc. 2009/52665-0 and CNPq 300243/79-0.Gladys Chalom, Raul A. Ferraz; Universidade de Sao Paulo, Brazil (email: agchalom@ime.usp.br,raul@ime.usp.br).C. Polcino Milies (Corresponding Author); Universidade de Sao Paulo and Univeersidade Federal do ABC, Brazil(email: polcino@ime.usp.br).181G. Chalom et al. / J. Algebra Comb. Discrete Appl. 4(2) (2017) 181–188Since in the case when char(F) |6 |A| the group algebra FA is semisimple and all ideals are directsums of the minimal ones, it is only natural to study minimal abelian code - or - equivalently, primitiveidempotents - and these has been done by several authors (see, for example [6], [5], [4] [9] [12] [14]). Also,Sabin and Lomonaco [15] have shown that central codes in metacyclic group algebras are equivalent toabelian codes.In what follows, we shall show that these are not better than minimal cyclic codes. First, we provein Corollary 2.5 that every minimal abelian non-cyclic code is a repetition code. Then, we show, inTheorem 3.3, that every minimal abelian code is equivalent to a minimal cyclic code of the same length.Finally, in section § 4, we show that essential idempotents can be used to characterize simplex codes.2. Basic factsThroughout this paper all groups will be finite, and we shall always assume that the fields F consid-ered are such that char(F) |6 |G|.For an element α in the group algebra FG, the Hamming weight of α is the number of elements inits support; i.e., if α =∑g∈G αgg, thenω(α) = |{g ∈ G | αg 6= 0}|.Given an ideal I ⊂ FG the weight distribution of I is the map which assigns, to each possible weightt, the number of elements of I having weight t.Let H 6= {1} be a normal subgroup of a group G. ThenĤ =1|H|∑h∈Hh.is a central idempotent of FG andFG = FG · Ĥ ⊕ FG · (1− Ĥ).Remark 2.1. As shown in [11, Proposition 3.6.7] we have that FG · (1 − Ĥ) = ∆(G,H), the kernelof the natural projection pi : FG → F[G/H] and it is easy to see that FG · Ĥ ∼= F[G/H] via the mapψ : FG · Ĥ → F[G/H] defined by g · Ĥ 7→ gH ∈ G/H.Also, if α ∈ FA · Ĥ, taking a transversal T of H in A we can rewrite α asα =∑t∈T∑h∈Hαthth.As α ∈ FA · Ĥ, we have thatα = αĤ =∑t∈T∑h∈HαththĤ =∑t∈T∑h∈HαthtĤ.So αth = αth′ for every h, h′ ∈ H and, setting αt = αth,∀h ∈ H we can writeα = |H|∑t∈TαttĤ. (1)Since Ĥ is central, it is a sum of primitive central idempotents called its constituents. Given aprimitive idempotent e we have that either eĤ = e or eĤ = 0, depending on wheather e is, or is not, aconstituent of H.182G. Chalom et al. / J. Algebra Comb. Discrete Appl. 4(2) (2017) 181–188In the first case, e is an element in the ideal FGĤ, and thus an element α in FGe is also in FGĤ.So, it can be written as in eqution 1.Writing T = {t1, t2, . . . , td} and H = {h1, h2, . . . , hm}, the explicit expression of α isα = α1t1h1 + α1t1h2 + · · ·+ α1t1hm + · · ·+ αdtdh1 + αdtdh2 + · · ·+ αdtdhm.In terms of coding theory, this means that the code given by the minimal ideal FGe is a repetitioncode. We shall be interested in idempotents that are not of this type.Definition 2.2. A primitive idempotent e in the group algebra FG, is an essential idempotent ife · Ĥ = 0, for every subgroup H 6= (1) in G.A minimal ideal of FG is called an essential ideal if it is generated by an essential idempotent andnon essential otherwise.Notice that, if e is a central idempotent, then the map pi : G→ Ge, given by pi(g) = g · e is a groupepimorphism. We can use this map to characterize essential idempotents.Proposition 2.3. Let e ∈ FG be a primitive central idempotent. Then e is essential if and only if themap pi : G→ Ge, is a group isomorphism.Proof. Let e be essential and assume, by way of contradiction, that pi is not a monomorphism. Then,there exists 1 6= g ∈ G such that pi(g) = ge = e, hence also gie = e for every positive integer i. Thus〈̂g〉 · e = e, a contradiction.Conversely, assume that e is not essential. Then, there exists H 6= (1) such that eĤ = e. For everyh ∈ H, we have that h · e = h · eĤ = Ĥ · e = e. Hence H ⊂ Ker(pi) and thus pi is not injective.Consequently, if pi is an isomorphism, e is essential.Corollary 2.4. If G is abelian and FG contains an essential idempotent, then G is cyclic.Proof. If e ∈ FG is essential, then G ∼= Ge ⊂ FGe. As G is abelian, a simple component FGe of FG isa field and Ge, being a finite subgroup contained in it, is cyclic.In terms of coding theory, the result above gives the following.Corollary 2.5. Let A be an abelian non-cyclic group. Then, for any finite field Fq, all the minimal codesof FqA are repetition codes.We wish to show, on the other hand, that if G is a cyclic group, then FG always contains an essentialidempotent.To do so, assume that G is cyclic of order n = pn11 · · · pntt . Then, G can be written as a direct productG = C1 × · · · ×Ct, where Ci is cyclic, of order pnii , 1 ≤ i ≤ t. Let Ki be the minimal subgroup of Ci; i.e.the unique subgroup of order pi in Ci and denote by ai a generator of this subgroup, 1 ≤ i ≤ t. Sete0 = (1− K̂1) · · · (1− K̂t)=1− (1 + a1 + · · ·+ apn1−111 )p1 · · ·1− (1 + at + · · ·+ apnt−1tt )pt .Then e0 is a central idempotent and we claim that it is non zero. In fact, it is easy to see that thecoefficient of a1 · · · at in this expression is (−1)t(1/p1) · · · (1/pt) so, e0 6= 0.Theorem 2.6. Let G be a cyclic group. Then, a primitive idempotent e ∈ FG is essential if and only ife · e0 = e.183G. Chalom et al. / J. Algebra Comb. Discrete Appl. 4(2) (2017) 181–188Proof. Let e ∈ FG be essential. Then, in particular, eK̂i = 0, so e(1− K̂i) = e, 1 ≤ i ≤ t. Hencee · e0 = e(1− K̂1)(1− K̂2) · · · (1− K̂t) = e · (1− K̂2) · · · (1− K̂t) = · · · = e.Conversely, if e is not essential then there exists a subgroup H of G such that e · Ĥ = e. Thereexists a minimal subgroup Ki ⊂ H, and for this subgroup we have that Ĥ(1 − K̂i) = Ĥ − ĤK̂i = 0.Consequently Ĥ · e0 = 0. Hence:e · e0 = (e · Ĥ) · e0 = e · (Ĥ · e0) = 0.Remark 2.7. Notice that the previous theorem actually shows that e0 is the sum of all the essentialidempotents of FG and, consequently, the simple components of the ideal FGe0 are precisely the essentialideals.Since e0 is non zero, we have the following.Corollary 2.8. Let G be a cyclic group and F a field such that char(F) |6 |G|. Then, FG always containsan essencial idempotent.3. The equivalenceLet F be a field, A be a finite abelian group such that char(F) |6 |A| and e 6= Â an idempotent in FA.SetHe = {g ∈ G | ge = e}. (2)Clearly, He is the unique maximal subgroup of G is such that Hee = e and it can be shown easilythat He = G if and only if e = Ĝ, the principal idempotent of FG. Actually, He is the kernel of theirreducible representation associated to the simple component FG · e.Theorem 3.1. Let e 6= Â be a primitive idempotent of FA and ψ the natural projection defined inRemark 2.1. Then, the element ψ(e) is an essential idempotent of F[A/He].Proof. Let K 6= 1 be a non-trivial subgroup of A/He. Then, it is of the form K = K/He where K 6= Heis a subgroup of A containing He.Let T be a transversal of He in K. ThenK̂ =1|K|∑k∈Kk =|He||K|∑t∈TtĤe =1|K|∑t∈TtĤe.As ψ(Ĥe)=1, we haveψ(K̂) =1|K|∑t∈Tψ(t) =1|K|∑t∈TtHe = K̂.Then ψ(e) · K = ψ(e) · ψ(K̂) = ψ(e · K̂) = 0, as K 6⊂ He.Corollary 3.2. Let e 6= Â be a primitive idempotent of FA. Then, the factor group A/He is cyclic.184G. Chalom et al. / J. Algebra Comb. Discrete Appl. 4(2) (2017) 181–188Proof. This fact follows immediately from Proposition 3.1 above and Corollary 2.4.Let G1 and G2 denote two finite groups of the same order, F a field, and let γ : G1 → G2 be abiijection. Denote by γ : FG1 → FG2 its linear extension to the corresponding group algebras.Clearly, γ is a Hamming isometry; i.e., elements corresponding under this map have the same Ham-ming weight. Ideals I1 ⊂ FG1 and I2 ⊂ FG2 such that γ(I1) = I2 are thus equivalent, in the sense thatthey have the same dimension and the same weight distribution. In this case, the codes I1 and I2 aresaid to be permutation equivalent and were called combinatorially equivalent in [15].In what follows, we will show that every minimal ideal of an abelian group algebra FA is permutationequivalent to a minimal ideal of the group algebra of a cyclic group of the same length.Let A be an abelian group of order n, F a field, I a minimal ideal of FA and e be the primitiveidempotent generating I. Let He be the subgroup defined in (2) and let C be a cyclic group of the sameorder n. If He = A then e = Â and any bijection σ : A → C is such that I = FAe = Fe is mapped toFCĈ, so I is equivalent to FCĈ.So, assume that He 6= A. Since the order d of the factor group A/He is a divisor of n, there existsa unique subgroup K of C such that |A/He| = |C/K| = d and, as they are both cyclic groups, we havethatA/He ∼= C/K. (3)So, alsoF[A/He] ∼= F[C/K] andFA · Ĥe ∼= F[A/He] ∼= F[C/K] ∼= FC · K̂.Denote by µ : F[A/He]→ F[C/K] and θ : FA · Ĥe → FC · K̂ realizing these isomorphisms.Let a ∈ A be an element such that a = aHe is a generator of A/He. Then {1, a, a2, . . . , ad−1} is atransversal of He in A and we can writeA = {aih | 0 ≤ i ≤ d− 1, h ∈ He}.Similarly, if t ∈ C is such that t = tK = µ(aHe), it is a generator of C/K and we can writeC = {tik | 0 ≤ i ≤ d− 1, k ∈ K}.As He andK have the same order, we can choose a bijection f : He → K and define a map η : A→ Cby η(aih) = tif(h), for all aih ∈ A.Given an element α ∈ FA · Ĥe, using formula (1) we can write it in the formα = |He|d−1∑i=0αiai · Ĥe,and, taking into account that |He| = |K|, we computeθ(α) = µ(|He|d−1∑i=0αia¯i)= |K|d−1∑i=0αitiK̂ =d−1∑i=0αiti(∑h∈Hef(h)).Comparing the expressions of α and θ(α) it is clear that the linear extension of η : A → C to FAcoincides with θ on FA · Ĥe, and it is such that θ(FA · Ĥe) = FC · K̂.Notice that FA · e ⊂ FA · Ĥe and thus e is primitive also in FA · Ĥe, so e′ = θ(e) is primitive inFC · K̂.185G. Chalom et al. / J. Algebra Comb. Discrete Appl. 4(2) (2017) 181–188We claim that e′ is also primitive in FC. In fact, assume that e′ = f1 + f2, with f1, f2 orthogonalidempotents in FC. Then e′ · K̂ = f1K̂ + f2K̂ is a decomposition of e′ in FC · K̂. Hence, either f1K̂ = 0and f2K̂ = e′ or vice-versa. So either f1 = f1e′ = f1e′K̂ = 0 or, in a similar way, f2 = 0.Hence, we have shown the following.Theorem 3.3. Every minimal ideal in the semisimple group algebra FA of a finite abelian group A ispermutation equivalent to a minimal ideal in the group algebra FC of a cyclic group C of the same order.It should be noted that non minimal abelian codes can actually be more convenient than the cyclicones (see, for example, [7], [10] and [13]).4. Binary simplex codesSet r = dimFqC and let B = {v1, v2, . . . , vr} be a basis of C over Fq. We shall denote by GB thegenerating matrix of C; i.e. the matrix whose rows are the componentes of the vectors of B, when writtenin the basis B of FqC.We start with a very simple remark.Lemma 4.1. The matrix GB contains no zero columns.Proof. In fact, assume that the jth column of the given matrix is zero. Then, the jth component ofevery vector in C is equal to 0. Take any non-zero vector v ∈ C. So it has at least one component whichis equal to 1. Since every shift of v is in C, there exists a vector in C whose jth component is equal to 1,a contradiction.Notice also that, if a matrix GB has two columns, in positions i and j, say, that are equal to oneanother, then the ith component of every vector in C is equal to its jth component. Hence, we have shownthe following.Lemma 4.2. The columns of a matrix GB are pairwise different for a given basis B of C, if and only ifthe columns of generating matrices are pairwise different, for every basis of C.Set F = Fq · e. Then F is an isomorphic copy of Fq contained in FqC · e. Notice that, since g is agenerator of C, we have that F[ge] = FqC · e. Then, the evaluation mapping ϕ : F[X]→ FqC · e given byϕ(f) = f(ge), for all f ∈ F[X], is an epimorphism and we have thatFqC · e ∼= F[X]Ker(ϕ).Notice that Ker(ϕ) = 〈h〉, for some h ∈ F[X], which is a polinomial of minimal degree having ge asa root. Since dimFFqC · e = dimFqFqC · e = r we have that the degree of h is precisely r and there existcoeficients b0, . . . , br−1 in F such thatgre = br−1gr−1e+ · · ·+ b0.This clearly implies that B0 = {e, ge, . . . , gr−1e} is a basis of FqC · e over F and also over Fq.Moreover, we have the following.Theorem 4.3. For every basis B of C the generating matrix GB has pairwise different columns if andonly if the mapping ψ : C → C · e is an isomorphism.186G. Chalom et al. / J. Algebra Comb. Discrete Appl. 4(2) (2017) 181–188Proof. Notice that FqC · e is of dimension r over Fq, it contains qr elements.Suppose, by way of contradiction, the ψ is not injective. Then, there exists an index j, 0 < j < n,such that e = gje.Write e = a0 + a1g + · · ·+ an−1gn−1. Then gje = a0gj + a1gj+1 + · · ·+ an−1gj+n−1.Since e = gje, we have that ai = ai+j , 0 ≤ i ≤ n − 1, where the subindexes are taken modulo n.This shows that if G denotes the generating matrix with respect to the basis B then, the first column ofG is equal to its jth column.Conversely, assume that ψ is an isomorphism and, by way of contradiction, that there exists a basiswhose corresponding generating matrix has two equal columns, in positions i and j, say. In view ofLemma 4.2 we can assume, without loss of generality, that this basis is precisely the basis B0. Thismeans that ai+t = aj+t or, equivalently, that at = at+j−i, for all t, where indexes are taken, again,modulo n. This readily implies that e = gj−ie.Recall that a binary linear code of dimension k and length n is called simplex if a generating matrixfor the code contains all possible non zero columns of length k. Since these are 2k − 1 in number, thismatrix must be of size k × (2k − 1) so, we must have n = 2k − 1.Theorem 4.4. Let C be a binary linear code of dimension k and length n = 2k − 1. Then C is a simplexcode if and only if it is essential.Proof. Assume that C is simplex. Since all its columns are different, it follows from Theorem 4.3 thatthe mapping ψ : C → C · e is an isomorphism. To prove thar e is an essential idempotent we are onlyleft to prove that it is primitive.Notice that C · e ⊂ F2C · e \ {0}. Since F2C · e = C is of dimension k over F2, it contains 2k elements.Hence |F2C · e \ {0}| = n = |C|, showing that actually C · e = F2C · e \ {0}. This means that every nonzero element in FC · e is invertible and thus, it is a field. Consequently, e is primitive.As a consequence, and taking the results in [3] into account, we can state the following.Corollary 4.5. Every binary linear code of constant weight is a repetition of a code generated by anessential idempotent.References[1] S. D. Berman, Semisimple cyclic and abelian codes. II, Kibernetika 3(3) (1967) 21–30.[2] S. D. Berman, On the theory of group codes, Kibernetika 3(1) (1967) 31–39.[3] A. Bonisoli, Every equidistant linear code is a sequence of dual Hamming codes, Ars Combin. 18(1984) 181–186.[4] R. A. Ferraz, M. Guerreiro, C. P. Milies, G−equivalence in group algebras and minimal abeliancodes, IEEE Trans. Inform. Theory 60(1) (2014) 252–260.[5] R. A. Ferraz, C. P. Milies, Idempotents in group algebras and minimal abelian codes, Finite FieldsAppl. 13(2) (2007) 382–393.[6] P. Grover, A. K. Bhandari, Explicit determination of certain minimal abelian codes and their mini-mum distance, Asian–European J. Math. 5(1) (2012) 1–24.[7] J. Jensen, The concatenated structure of cyclic and abelian codes, IEEE Trans. Inform. Theory 31(6)(1985) 788–793.[8] F. J. Mac Williams, Binary codes which are ideals in the group algebra of an abelian group, BellSystem Tech. J. 49(6) (1970) 987–1011.187G. Chalom et al. / J. Algebra Comb. Discrete Appl. 4(2) (2017) 181–188[9] R. L. Miller, Minimal codes in abelian group algebras, J. Combinatorial Theory Ser A 26(2) (1979)166–178.[10] C. P. Milies, F. D. de Melo, On cyclic and abelian codes, IEEE Trans. Information Theory 59(11)(2013) 7314–7319.[11] C. Polcino Milies, S. K. Sehgal, An Introduction to Group Rings, Algebras and Applications, KluwerAcademic Publishers, Dortrecht, 2002.[12] A. Poli, Construction of primitive idempotents for a variable codes, Applied Algebra, Algorithmicsand Error–Correcting Codes: 2nd International Conference, AAECC–2 Toulouse, France, October1–5, 1984 Proceedings (1986) 25–35.[13] R. E. Sabin, On minimum distance bounds for abelian codes, Appl. Algebra Engrg. Comm. Comput.3(3) (1992) 183–197.[14] R. E. Sabin, On determining all codes in semi–simple group rings, Applied Algebra, Algebraic Algo-rithms and Error–Correcting Codes: 10th International Symposium,AAECC-10 San Juan de PuertoRico, Puerto Rico, May 10–14, 1993 Proceedings (1993) 279–290.[15] R. E. Sabin, S. J. Lomonaco, Metacyclic error–correcting codes, Appl. Algebra Engrg. Comm. Com-put. 6(3) (1995) 191–210.188

Essential idempotents and simplex codes

Abstract

Similar works

Full text

Available Versions

Journal of Algebra Combinatorics Discrete Structures and Applications (JACODESMATH, Yildiz Technical University - YTU)

Directory of Open Access Journals