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The Permutation Groups and the Equivalence of Cyclic and Quasi

The Permutation Groups and the Equivalence of Cyclic and Quasi

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F stransitive. is ( C p uhthat such q x S = ) r eascaealinear a associate we YCLIC 1 n ecaatrz the characterize We . scci fi verifies it if cyclic is x . . . , N h atsituation last The . ,j i, C H hc seither is which , } ′ . ( C n { ∈ P S ) C ODES ) n n ∈ { ′ hnits Then . 1 fpermu- of 1 ie by: given H = n , . . . , C n , . . . , n F ′ } . ( σ q . P and , ( C ) s. G ly } is ) d } p e . , simple groups. By using this classification J. P. McSorley [24] by their composition factors [2]. Hence we have U 1 = U 2; gave the following result. this implies U1 = U2. This gives P er(C)= P ΓL(d, t). Lemma 2: A group G of degree n which is doubly tran- We recall that the group AG(n) = {τa,b : a 6= 0, (a,n) = sitive and contains a complete cycle, has as socle N, verifies 1,b ∈ Z/nZ} is the group of the affine transformations defined N ≤ G ≤ Aut(N) and is equal to one of the cases in the as follow following Table. τa,b : Zn −→ Zn G n N (1) AGL(1,p) p Cp x 7−→ (ax + b) mod n. S4 4 C2 × C2 Sn, n ≥ 5 n Alt(n) When n = p a prime number, we have AGL(1,p)= AG(p), Alt(n), n odd and ≥ 5 n Alt(n) and is called the affine group. P GL(d,t) ≤ G ≤ P ΓL(d,t) (td − 1)/t − 1 P SL(d,t) Proposition 4: Let C be a non elementary cyclic code of (d,t) =6 (2, 2), (2, 3), (2, 4) ′s length p over Fq, with q = p . Hence P er(C) is a primitive P SL(2, 11) 11 P SL(2, 11) group, and one of the following holds: M11 11 M11 M23 23 M23 1) P er(C) = PSL(2, 11) of degree 11, q = 3 ( C is the Remark 1: The projective semi-linear group P ΓL(d, t) [11, 6, 5]3 ternary Golay code or its dual), is the semi-direct product of the projective linear group 2) P er(C) = M23 of degree 23, q = 2 (C is the binary P GL(d, t) with the group Z = Gal(Ft/Fp′ ) Golay code [23, 12, 7], or its dual), ′s ′ d of finite field Ft, where t = p ,p a prime, i.e. 3) P er(C) = P ΓL(d, t) of degree p = (t − 1)/(t − 1), where t is a power of the prime p′. P ΓL(d, t)= P GL(d, t) ⋊ Z. 4) Cp ≤ P er(C) ≤ AGL(1,p), with a normal Sylow The order of these groups are subgroup N of order p; and such that p ≥ 5. Proof: A transitive of prime degree is a |P GL(d, t)| = (d, t − 1)|PSL(d, t)|, |Z| = s primitive group [28, p. 195]. From a consequence of a result and P ΓL(d, t)= s|P GL(d, t)|. Hence if t is a prime we have of Burnside [12], we have that a transitive group of prime P ΓL(d, t)= P GL(d, t). degree is either a subgroup of the affine group AGL(1,p) or The zero code, the entire space, the repetition code and its a doubly transitive group. dual are called elementary codes. The permutation group of In the doubly-transitive cases, from Lemma 2, Remark 1 these codes is Sn [21, p. 1410]. Furthermore, there is no cyclic and the Lemma 3 we have that these groups are Sp, M11, codes with permutation group equals to Alt(n). PSL(2, 11) of degree 11, M23 of degree 23, or P ΓL(d, t) of d ′s The following Lemma is a generalization of a part of [8, degree p = (t − 1)/(t − 1) over Fq, with q = p and t is a ′ Theorem E] to the non binary cases. prime power of p . From Remark 1, the group Sp corresponds Lemma 3: Let C be a non elementary cyclic code of length to an elementary code. From [25, Table 1, Lemma 2 and (J)], td−1 ′ M11 rules out, PSL(2, 11) is the permutation group of a n = t−1 over a finite field of characteristic p , d ≥ 2, and s t = p be the power of a prime p, if the group P er(C) verifies: ternary code and M23 is the permutation group of a binary code. From [21] p. 1410, these codes are unique, namely P GL(d, t) ≤ P er(C) ≤ P ΓL(d, t), PSL(2, 11) corresponds to the [11, 6, 5]3 ternary Golay code then we have, or its dual. M23 is the permutation group of the [23, 12, 7] binary Golay code and its dual. ′s d ≥ 3,t = p and P er(C)= P ΓL(d, t). Since |AGL(1,p)| = (p − 1)p, then P er(C) ≤ AGL(1,p) Proof: For d = 2. As the group P GL(2,t) acts 3- admits as order pm with m|(p − 1), hence it contains a 1 Sylow subgroup of order . From Sylow’s Theorem, transitively on 1-dimensional projective space P (Ft), we can N p N deduce from [25, Table 1 and Lemma 2], that the underlying is unique, hence it is normal. For p equals to 2 or 3, we have AGL(1,p) = S , hence the codes are elementary. Then we code is elementary, hence P er(C)= Sn, which is a contradic- p tion. Hence d ≥ 3, and similarly, we deduce from [25, Table assume p ≥ 5. ′s If (q,p)=1, the number of cyclic codes of length p over 1 and Lemma 2] that t = p . Let V denotes the permutation p−1 ord (q) +1 module of Fp′ , associated to the natural action of P GL(d, t) Fq, is equal to 2 p , where ordp(q) is the multiplicative d−1 on (d − 1) dimensional projective space P (Ft). Let U1 order of q modulo p [18]. When ordp(q)= p−1, there are only be a P GL(d, t)-submodule of V . Hence U1 is P ΓL(d, t)- four codes, namely the elementary codes. This is the cases for invariant. That is because, for σ a generator of the cyclic q = 13, p =5, 11, 31, 37, 41 or q = 11 and p = 13, 17, 23, 31. ′ σ group P ΓL(d, t)/PGL(d, t) ≃ Gal(Ft/Fp ). Then U2 = U1 , In the following table we give examples of permutation regarded as P GL(d, t)-module, is simply a twist of U1. Let groups of cyclic codes of length p over Fq in the case Fp′ be the algebraic closure of Fp′ , we conclude that the com- P er(C) ≤ AGL(1,p). The m is such that position factors of the Fp′ P GL(d, t)-modules U1 = Fp′ ⊗ U1 P er(C) = Cm ⋉ Cp. The codes are given in pairs, and U2 = Fp′ ⊗ U2 are the same. The submodules of the corresponding to the code and its dual. Fp′ P GL(d, t)-module V = Fp′ ⊗ V are uniquely determined q p m Codes From [8] the permutation group of the Hamming code 11 5 2 [5, 3, 3], [5, 2, 4] [15, 12, 3]2 is P ΓL(4, 2). 5 1 [5, 2, 4], [5, 3, 3] Now we will deal with the permutation group of MDS 7 3 [7, 4, 4], [7, 3, 5] codes. 19 3 [19, 16, 3], [19, 3, 6] T. P. Berger [4] proved that the only linear MDS codes with 37 6 [37, 31, 5], [37, 6, 27] minimum distance dC equals to 2 or n are the repetitions codes 13 7 2 [7, 5, 3], [7, 2, 6]; and their duals. Hence in the following result we assume that [7, 3, 5], [7, 4, 4] n > dC > 2. 17 4 [17, 13, 4], [17, 4, 12]; Theorem 8: Let C be a non elementary [n,n−dC +1, dC] ′s [17, 12, 4], [17, 5, 11]; MDS cyclic code over Fq, q = p , n > 5, n > dC > 2, [17, 8, 8], [17, 9, 7] and gcd(n − 2, dC − 2) = 1. If the group P er(C) is doubly 17 8 [17, 9, 8], [17, 8, 9] transitive, then 23 11 [23, 12, 9], [23, 11, 10] 1) P er(C)= AGL(1,p) with n = p, or ′ 29 14 [29, 15, 11], [29, 14, 12] ′ p d−1 ′ 2) P er(C)= PSL(d, p ) with n = p′−1 , d ≥ 3, gcd(p − Theorem 5: Let C be a non elementary cyclic code over 1, d)=1. F of length n = pr, where r ≥ 2 and q = p′s. Then P er(C) q Proof: If n = p, hence from Lemma 2 P er(C) is is either: i equal to AGL(1,p), PSL(2, 11), M11, M23 or a subgroup 1) an imprimitive group, and admits a system of p blocks of P ΓL(d, p′). From Proposition 4, we eliminate PSL(2, 11), each of length ps,(si = r) formed by the orbits of i M11, M23 since the [11, 6, 5]3 and [23, 12, 7]2 Golay codes are , a such system of block is complete, or not MDS and M11 does not correspond to any cyclic code. 2) a doubly transitive group equals to Now we need the following Lemma. d t − 1 ′ Lemma 9: ([4]) If C is an MDS linear code with length P ΓL(d, t), with pr = , d ≥ 3,t a power of p . t − 1 n and minimum distance dC > 2 such that P er(C) contains r Proof: A simply transitive subgroup of Sn of degree p a doubly transitive subgroup B and gcd(n − 2, dC − 2)= 1, which contains a permutation of order pr (in our case T ) must then B = P er(C). be imprimitive [16, p. 229]. Hence it admits a system of pi If P er(C) is a doubly transitive not equal to AGL(1,p), then blocks each of length ps,(si = r) formed by the orbits of Theorem 5, Theorem 7, Lemma 9 and Remark 1 implies that i ′ ′ T p [17] p. 67, a such system of block is complete. the only solution is P er(C) = PSL(d, p ) = P GL(d, p ) = ′ ′ From the previous we have that if P er(C) is primitive it P ΓL(d, p ) for gcd(p − 1, d)=1. must be doubly transitive. From Lemma 2, Lemma 3 and Remark 2: When p a prime number any cyclic code of Remark 1 we get the results. length p over Fp is MDS and is equivalent to an extended Corollary 6: Let p ≥ 5 be a prime number, C a cyclic codes Reed–Solomon code [29]. The permutation group of the last r of length p , r ≥ 2 over Fp. Then if C is not an elementary codes is the affine group AGL(1,p) [3]. code, the group P er(C) is imprimitive. III. THE EQUIVALENCE OF CYCLIC CODES Proof: From Theorem 5 if P er(C) is primitive, hence it ′ is doubly transitive. From Lemma 2 the only cases when the Let C and C two cyclic codes of length pr, with p an socle can be abelian are N = Cp and C2 ×C2. In which cases odd prime number. The aim of this section is to find the ′ P er(C) must be equal to AGL(1,p) or S4; which is impos- permutations by which C and C can be equivalent. Even if sible. From [17, Lemma 22], if P er(C) is doubly transitive our results are also true for the cyclic combinatorial objects with non abelian socle, then Soc(P er(C)) = Alt(pm), hence of length pr, we consider only cyclic codes. from Remark 1 the code is elementary. The multiplier Ma is the affine transformation Ma = τa,0. Theorem 7: A non elementary cyclic code C of composed It is obvious that the of a cyclic code by a multiplier ′s length n over Fq, where q = p , admits a permutation group is an equivalent cyclic code. If gcd(n, φ(n)) = 1 or n = 4, P er(C) which is either or n = p.r, p > r are primes, and the Sylow p-subgroup ′ 1) imprimitive, of P er(C) has order p. Then two cyclic codes C and C of 2) or doubly transitive; equals to length n are equivalent if and only if they are equivalent by a multiplier [1],[20],[27]. P er(C)= P ΓL(d, t), When C is a cyclic code of length pr, P a Sylow p-subgroup d− ′ t 1 of P er(C), the following subset of Spr was introduced by N. with n = t−1 , d ≥ 3, t a power of p . Proof: The group P er(C) contains a full cycle and is Brand [9] with composed degree. Hence from Theorem of Burnside −1 H(P )= {σ ∈ Spr |σ T σ ∈ P }. and Shur [30, p. 65], P er(C) is either imprimitive or doubly transitive. In the doubly transitive from Lemma 2 The set H(P ) is well defined, since is a subgroup r and Lemma 3 this happens only if P er(C) is Alt(n), Sn of P er(C) of order p , hence it is a p-group of P er(C). or P ΓL(d, t). From Sylow’s Theorem, there exists a Sylow p-subgroup P of P er(C) such that ≤ P . Furthermore in some cases Now we consider σ ∈ NSpr (), then the set H(P ) is a group. −1 ′ σ σ =< T > . Lemma 10: ([9, Lemma 3.1]) Let C and C two cyclic codes of length pr. Let P be a Sylow p-subgroup of P er(C) Since we assumed ≤ P , hence σ ∈ H(P ). ′ −1 which contains T . Then C and C are equivalent if and only Let σ ∈ NS r (P ), hence σ ∈ NS r (P ) verifies σ P σ = ′ p p if C and C are equivalents by an element of H(P ). P . Since we assumed ≤ P , then σ−1T σ ∈ P , i.e. In the case of length p2, H(P ) has been given explicitly (4) by Huffman et al. [20]. They proved that if P is a Sylow p- NSpr (P ) ⊂ H(P ). subgroup of P er(C) and |P | = p2, then the cyclic codes ′ C and C are equivalent if only if they are equivalent by Lemma 13: ([20, Theorem.5.6]) Let p be an odd prime 2 p+1 ′ a multiplier. If p < |P | ≤ p the codes C and C and let q be a prime power with p ∤ q. Let C be a cyclic code are equivalent by a power of a multiplier times a power r k of length p over Fq and tk be the order of q modulo p and of a generalized multiplier. A generalized multiplier is a k suppose that z =1. For 1 ≤ k ≤ r, P er(C) contains the group (p ) k permutation µa ∈ Spr defined as follows: (p ) k k Gk = {µqi,c | 0 ≤ i 1. b

Now, we are interested on the existence of the multiplier Proof: Let P be a Sylow p-subgroup of P er(C), hence P r Mp+1 in the group P er(C). Because it has as effect to is a p-group of Sp . From Sylow’s Theorem P is contained in a Sylow p-subgroup of Spr . It is well known that a Sylow p- simplify Algorithm 3.1 for the cyclic codes and Algorithm − − pr 1+pr 2+...+1 6.1l for extended cyclic objects [20]. subgroup of Spr is of order p . It is isomorphic Lemma 11: Let z be the largest integer such that pz|(qt − with the Zp ≀...≀Zp [28, Kaluˇznin’s Theorem], − − 1), with t order of q modulo p. Hence if z =1 we have : hence r ≤ s ≤ pr 1 + pr 2 + ... +1. r− If , from Lemma 13 we have that contains r 1 z = 1 P er(C) 1) ordp q = p t. r the group Gr, of order trp . From Lemma 11 we have that 2) The multiplier Mp+1 ∈ P er(C). r−1 2r−1 2r−1 r−1 tr = t1p , then we have |Gr| = t1p . Hence p Proof: The proof of r , follows z = 1 ⇒ ordp q = p t divides |P er(C)|, which means that P er(C) contains a p- from [18, Lemma 3.5.4]. 2r−1 r−1 subgroup of order at least p . Since z =1 ⇒ ordpr q = p t, we choose ak, for r kt For n, and H(P )= AG(pr), H(P ). s−2 s−2 s−1 (b) if s > r, P = Q1 , and H(Q1 )= Q . Proposition 12: Let P be a Sylow subgroup of P er(C) − Proof: From Theorem 14 we have that r ≤ s ≤ pr 1 + which contains T . Hence the group H(P ) verifies : − pr 2 + ... 1. In the case r = s, it is obvious to have P =< r 1) AG(p )= NS r () ⊂ H(P ), p T >, and H() = NSn (). Proposition 12 gives r 2) NSpr (P ) ⊂ H(P ). H()= AG(p ). Furthermore, we remark that the part r r Proof: The fact AG(p ) = NSpr () follows (b) of [20, Theorem 2.1] can be generalized for the case p . from [22, p. 710]. Since it is based essentially on the [20, Lemma. 2.4] and the [9, Lemma. 3.2]( the last Lemma was given for the length pr). IV. THE EQUIVALENCE FOR THE QUASI-CYCLIC CODES s−1 The hypothesis s ≤ p +1 is to assure that Q1 ≤ P er(C). Hence the result follows from [9, Theorem 2.2]. A code C of length n = lm is said to be quasi-cyclic of As noticed Brillhart et al. [10], it is quite unusual to have order l over Fq, if it is invariant by the permutation z > 1. Hence the importance of the following result. l r T : Zn −→ Zn Theorem 16: Let C be a cyclic code of length n = p , (6) over Fq, such that gcd(p, q)=1,z = 1 and P be a Sylow i 7−→ i + l mod n. p-subgroup of P er(C) of order p2r−1, then we have : We consider the cycles σi = (i,i + l,i +2l...,i + (m − 1)l) 1) P is normal in Gr, for 0 ≤ i ≤ l − 1. The cycles σi have order m. Furthermore 2) NSn (P )= Gr. ij (i−1)j (i−2)j we have 3) H(P ) = {τ ∈ Sn|τ : i 7→ q a + q c + q c + r−1 l ... + c , 0 ≤ j p-subgroup in Gr. From the Sylow’s Theorem, N ≡ 1 mod p l 2r−1 the subgroup of Sn generated by the permutation T . Hence and N divide |Gr| = t1p . Assume that N =1+ kp, with l 2r−1 2r−1 the normalizer of in Sn contains the following groups. k> 0 and Nα = t1p , α ≥ 1. Hence (1+kp)α = t1p , 2r−1 ′ but ((1 + kp),p )=1. Then (1 + kp)α = t1, absurd. 1) Q =< σ0,...,σl−1, T > . ′ Because t1|(p − 1) and α ≥ 1, hence N = 1. Since N = 1 2) AG(n). and the Sylow subgroups are conjugate this gives that P is l Proof: This is obvious that T ∈ NSn (). Now we normal in Gr. consider σi ∈ Q, from the relation (7) we have σ0 ...σl−1 = Now, we will prove that NSn (P )= Gr : l (pr) T . Furthermore the cycles σi are disjoints, then commute. c − Let σ ∈ P ≤ Gr, then we can write σ = µqj ,c = T Mqj , 1 l l Hence σi T σi = T . j with T the shift and Mq is a multiplier. Let τ ∈ NSn (P ), ′ ′ ′ We consider the affine transformation τa,b ∈ AG(n) proving c ′ hence there exists σ = T Mqj ∈ P such that τσ = σ τ, l that τa,b ∈ NSn () is equivalent to prove the existence this is equivalent to ∗ of an α ∈ N such that, c c′ j ′ τT Mq = T Mqj τ. l −1 lα τa,bT τa,b = T . For 0 we have :

c c The permutation τa,b can be decomposed as follows, τa,b = τT Mqj (0) = τT (0) = τ(c), τ1,bτa,0. Which gives with (7) the following equality : hence we have: l −1 −1 −1 τa,bT τa,b = τ1,bτa,0σ0 ...σl−1.τa,0 τ1,b . (8) ′ ′ ′ c ′ j τ(c)= T Mqj τ(0) = q τ(0) + c . This is well known [19, Lemma. 5.1] if σ = σ0 ...σl−1 is a ′ ′ ′ Then, j , where j . This product of the disjoint cycles and S is a permutation of Sn. τ(c)= cq + d d = c + q (τ(0) − c) − 1 − implies NS (P ) ≤ Gr. Hence SσS = S(σ0)S(σ1) ...S(σl 1). For ra = a mod l n a we obtains that τa,0(σi)= σ . This gives : Furthermore P is normal in Gr, which gives that NGr (P )= ira G . But N (P ) ≤ N (P ) ≤ N (P ) ≤ G , then r Gr P er(C) Sn r a a a la τa,0(σ0)τa,0(σ1) ...τa,0(σl−1)= σ0 σra ...σra(l−1) = T . NSn (P )= Gr. For rb = b mod l, we obtain From the Lemma 12 we have that NSn (P ) ⊂ H(P ). Hence NS (P )= Gr ⊂ H(P ). −1 n τ1,bσiτ = τi+r , Now, let τ ∈ H(P ), hence there exist c and j such that : 1,b b −1 c c hence τT τ = T Mqj ⇐⇒ τT = T Mqj τ. l−1 l −1 l That is because P ≤ Gr. For 0 we obtains : τ1,bT τ1,b = Y σi+rb = T . τT (0) = τ(1) = qj τ(0) + c. i=0 c j τT (1) = τ(2) = T Mqj (τ(1)) = q τ(1) + c c j Finally we obtains : τT (2) = τ(3) = T Mqj (τ(2)) = q τ(2) + c . l −1 l −1 −1 la −1 la . τa,bT τa,b = τ1,bτa,0T τa,0 τ1,b = τ1,bT τ1,b = T . τ(i)= qij (τ(0)) + q(i−1)j c + q(i−2)j c + ... + c. l Then the elements of H(P ) are τ ∈ Sn such that τ(i) = This gives α = a, hence τa,b ∈ NSn (). ij (i−1)j (i−2)j l q a + q c + q c + ... + c. In [14] C. Chabot gave explicitly the group NSn (. r A. Quasi-Cyclic Codes of length p l 2) or Sn if the code is trivial. In the following, we consider the quasi-cyclic codes of If n is odd then, H′(P ) is either r length n = p l, with p a prime number which verifies 1) imprimitive, l (p,l) = (p, q) =1. In this case < T >≤ P er(C) is a 2) or H′(P )= Alt(n), subgroup of order pr. Hence it is contained in a Sylow p- 3) or Sn if the code is trivial. subgroup P . ′ Proof: From Proposition 17 H (P ) contains the group Lemma 18: Let C and C′ two quasi-cyclic codes of length r n = p l and P a Sylow p-subgroup of P er(C) such that Q =< σ0,...,σl−1, T >, T l ∈ P . Hence C and C′ are equivalent only if they are equivalent by the elements of the set by a similar proof to the Theorem 7 we obtains that the group H′(P ) is either imprimitive or doubly transitive. In the case ′ −1 l H (P )= {σ ∈ Sn|σ T σ ∈ P }. doubly transitive we need the following Lemma. Proof: Since C and C′ are equivalent, hence there exist Lemma 22: ([31]) Let G be a primitive group of degree n ′ which contains a cycle of length m> 1. Hence if m verifies a permutation σ ∈ Sn such that C = σ(C). This gives m< (n − m)!, we have G = Alt(n) or Sn. the relation between the permutation groups P er(C) and ′ P er(C′). If the group H (P ) is doubly transitive, then it is primitive. ′ − r P er(C )= σPer(C)σ 1 (9) From Proposition 19 it contains the cycles σi of length p . Furthermore we have pr < (n − pr)!. Hence from Lemma 22 ′ ′ Let P be a Sylow subgroup of P er(C), hence from the H (P ) is either Alt(n) or Sn. If n is even, since T ∈ H (P ) is − ′′ relation 9 we have σP σ 1 = P is a Sylow p-subgroup of a cycle of length n is odd. Then T/∈ Alt(n), hence H′(P )= P er(C′). From the Sylow Theorem there exists τ ∈ P er(C′) ′ ′′ Sn. If n is odd, H (P ) is the group Alt(n) or Sn. such that τP ′τ −1 = P . We can assume that < T l >≤ P ′, − since < T l > is a p-group. Let γ = τ 1σ, then γ is an V. CONCLUSION between C and C′, because γ(C)= τ −1σ(C)= ′′ −1 ′ ′ −1 l −1 l −1 −1 The aim of this work is to find solutions of the three τ C = C and γ T γ = σ τT τ σ ∈ σ P σ = P following problems: (because τT lτ −1 ∈ τP ′τ −1,) hence γ ∈ H′(P ). Problem I The classification of the permutation groups It is obvious that if P =, then we have • of cyclic codes. l ′ l NSn ()= H (). • Problem II The determination of the permutations by which two cyclic codes of length pr can be equivalent. The following proposition gives other properties of H′(P ). • Problem III The determination of the permutations by which two quasi-cyclic codes can be equivalent. Proposition 19: Let P be a Sylow p-subgroup of P er(C), hence the group H′(P ) verifies the following proprieties: Our contribution to the solution of Problem allows to solve all ′ the primitive cases and some imprimitive cases. Even though, • l NSn () ⊂ H (P ), ′ there is still work and extension to do for the imprimitive • N (P ) ⊂ H (P ). Sn cases. l Proof: We consider NSn (< T >), the normalizer of l l As solution to Problem II we found explicitly the set H(P ) in . Then the permutation r < T > Sn σ ∈ NSn (< T >) of permutations by which two cyclic codes of length can −1 l l p verifies σ σ =⊂ P . Hence, be equivalent. We also proved that is sometimes a group. l ′ We think that our contribution to solve Problem III can NSn () ⊂ H (P ). (10) be refined by proving that the set H′(P ) of permutations by We consider NSn (P ), the normalizer of P in Sn. Then the which two quasi-cylic codes are equivalent is a group. If it is −1 permutation σ ∈ NSn (P ) verifies σ P σ = P , hence for the case, since we proved that it will be an imprimitive group l −1 l T ∈ P we have σ T σ ∈ P . Hence or Alt(n) or Sn, hence it will be interesting to find some of ′ ′ other properties of H (P ) in the imprimitve cases. Also, the NS (P ) ⊂ H (P ). (11) n results can be generalised to other situations of length.

Corollary 20: The set H′(P ) verifies AG(n) ⊂ H′(P ). ACKNOWLEDGMENT

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