On the equivalence of cyclic and quasi-cyclic codes over finite fields

This paper studies the equivalence problem for cyclic codes of length p and quasi-cyclic codes of length pl. In particular, we generalize the results of Huffman, Job, and Pless (J. Combin. Theory. A, 62, 183–215, 1993), who considered the special case p. This is achieved by explicitly giving the permutations by which two cyclic codes of prime power length are equivalent. This allows us to obtain an algorithm which solves the problem of equivalency for cyclic codes of length p in polynomial time. Further, we characterize the set by which two quasi-cyclic codes of length pl can be equivalent, and prove that the affine group is one of its subsets. 2010 MSC: 94B05, 94B15, 94B60


Introduction
The equivalence problem for codes has many practical applications such as code-based cryptography [8,9,12,13].As a consequence, this problem has received considerable attention in the literature [2,3,6,11,12].However, progress in obtaining results has been slow.Brand ([3]) characterized the set of permutations by which two combinatorial cyclic objects on p r elements are equivalent.Using these results, Huffman et al. ( [6]) explicitly gave this set for the case p 2 , and constructed algorithms to find the equivalence between cyclic objects and extended cyclic objects.In [6], a negative answer was given to the generalization of their results to the case p r , r > 2. This is due to the fact that the permutations of Brand that are crucial to the proofs do not generate a Sylow subgroup of S p r .Babai et al. ( [2]) gave an exponential time algorithm for determining the equivalence of codes.Sendrier ([11]) proposed the support splitting algorithm to solve the problem of code equivalence in the binary case.However, in [12] Kenza Guenda (Corresponding Author); Faculty of Mathematics USTHB, University of Science and Technology of Algiers, Algeria (email: ken.guenda@gmail.com).
the authors showed that extending the support splitting algorithm to q ≥ 5 results in an exponential growth in complexity, which makes this approach impractical.
In this paper, we study the equivalence problem for cyclic codes of length p r and quasi-cyclic codes of length p r l over finite fields.We generalize the results of [6] (which are only for the special case p 2 ), by explicitly giving the permutations by which two cyclic codes of prime power length are equivalent.Further, the set of Brand is extended to the class of quasi-cyclic codes of length p r l.
The remainder of this paper is organized as follows.In Section 2, some preliminary results are presented.Section 3 considers the equivalence of cyclic codes, in particular cyclic codes of length p r .Then in Section 4, the equivalence of quasi-cyclic codes of length p r l is investigated.

Preliminaries
Let C be a linear code of length n over the finite field of q elements, F q , and σ a permutation of the symmetric group S n , acting on {0, 1, . . ., n − 1}.For a code C, we associate another linear code σ(C) defined by We say that the codes C and C are permutation equivalent if there exists a permutation σ ∈ S n such that C = σ(C).The automorphism group of C is the subgroup of S n given by A linear code C of length n over F q is called quasi-cyclic of index l or an l-quasi-cyclic code if its automorphism group contains the permutation T l given by where T : i → i + 1 is the cyclic shift.This definition is equivalent to saying that for all c ∈ C we have T l (c) ∈ C. The index l of C is the smallest integer satisfying this property.It can easily be proven that l is a divisor of n.In the case l = 1, the code is called a cyclic code.This is a code with an automorphism group that contains the cyclic shift T .

Equivalence of cyclic codes
In this section, we consider the permutation equivalence of cyclic codes.Later we will show that there is a very close link between the equivalence of some quasi-cyclic codes and cyclic codes.This provides further motivation to study the equivalence of cyclic codes.
We begin with some well known results.Let n be a positive integer.The set of permutations is the subgroup of S n formed by the permutations defined as follows The group AG(n) is called the group of affine transformations.The affine transformations are called multipliers.The affine group AGL(1, p) is the group of affine transformations over Z p .
For d ∈ Z * p , the generalized multiplier µ d ∈ S p 2 was defined in [6] as follows.Let k ∈ Z p 2 and k = i + jp for some 0 ≤ i, j ≤ p − 1, so that kµ d = (id) mod p + pj.Then from Palfy ( [10]) and Alspach and Parson ( [1]), we have the following results.For n = p r , r > 2 the equivalence of cyclic codes of length n is very complex, but in the next section this problem is partially solved.

Equivalence of cyclic codes of length p r
Let C be a cyclic code of length p r , p an odd prime and r > 1.Further let T be the cyclic shift modulo p r and P a p-Sylow subgroup of Aut(C).The following subset of S p r was introduced by Brand ( [3]) The set H(P ) is well defined since T is a subgroup of Aut(C) of order p r , so it is a p-group of Aut(C).From Sylow's Theorem, there exists a p-Sylow subgroup P of Aut(C) such that T ≤ P .Furthermore, in some cases the set H(P ) is a group.Lemma 3.3 shows the importance of having information on the p-Sylow subgroup of Aut(C).The following results provide some of this information.Proposition 3.4.[4, Proposition 9] Let C be a cyclic code of length p r with r > 1, and M q be the multiplier defined by M q (i) = iq mod p r .Then the group Aut(C) contains the subgroup K = T, M q of order p r ord p r (q).Let p l , l ≥ r, be the p-part of the order of K. Then a p-Sylow subgroup P of Aut(C) has order p s such that Now we define the sets of Brand.Let p be an odd prime.For n < p, we define the following subsets of S p r and p r−1 divides a i for i = 2, 3, . . ., n}.
The sets Q n and Q n 1 are subgroups of S p r [3, Lemma 2.1].Note that Q 1 = AG(p r ).Lemma 3.5.Let C be a cyclic code of length p r where p is odd and m > 1.Let P be a p-Sylow subgroup of Aut(C) which contains T .If 1 ≤ n < p, then: Thus in Q n , the coefficient a 0 can take p r different values, and a 1 can take p r−1 (p − 1) values.For 2 ≤ i ≤ n, a i can take p values.From these results we have 1 , the coefficient of a 0 can take p r different values, and a i for 1 ≤ i ≤ n can take p values, so that |Q n 1 | = p r+n .Now we prove that AG(p r ) = N S p r ( T ).Let σ be an element of N S p r ( T ).Then there is a j ∈ Z n \ {0} such that σT σ −1 = T j , or equivalently σT = T j σ.Hence σT (0) = σ(1) = T j σ(0) = σ(0) + j and σT (1) = σ(1) + j = σ(0) + 2j, so that σ(k) = σ(0) + kj for any k ∈ Z n .Then (j, n) = 1 follows from the fact that the order of T equals the order of T j .The last inclusion is obvious.
For part (iv), we begin with the Since the order of g is equal to the order of T (which is p r ), from [3, Lemma 3.6] there exists f ∈ Q n+1 such that f −1 gf = T , so then f −1 h −1 T hf = T .The only elements of S p r which commute with T (a complete cycle of length p r ), are the powers of T .Thus hf = T j for some j.Since Q n+1 is a subgroup of S p r and T ≤ Q n+1 , h ∈ Q n+1 , and hence N S p r (Q n 1 ) ≤ Q n+1 .Now consider the ≥ condition.Let g ∈ Q n+1 where g(x) = n+1 i=0 g i x i with p g 1 and p r−1 |g i for 2 ≤ i ≤ n.Further, let h ∈ Q n 1 , where h(x) = n i=0 h i x i with h 1 ≡ 1 mod p r−1 and p r−1 |h i for 2 ≤ i ≤ n.We have Since p r−1 |h i , for i ≥ 2 and p r−1 |g j for j ≥ 2, any terms in n i=2 h i n+1 j=0 g j x j i involving g j for j ≥ 2 vanish modulo p r , so that By [3, Lemma 2.1] We now determine g −1 hg in order to prove that it is in Q n 1 .This is given by As p r−1 |g j for j ≥ 2, hence p r−1 |b k for k ≥ 2. Furthermore, we have p r−1 |h i for i ≥ 2, and thus Then replacing the b i with their values from (4), we obtain As h 1 ≡ 1 mod p r−1 , we have that h n 1 ≡ 1 mod p r−1 .In addition, as p r−1 |g n+1 , it must be that g n+1 h n 1 ≡ g n+1 mod p r .Therefore, c n+1 = 0, and p r−1 |c i for 2 ≤ i ≤ n.Then we only need to show that c 1 ≡ 1 mod p r−1 .As g j ≡ 0 mod p r−1 for j ≥ 2, h i ≡ 0 mod p r−1 for i ≥ 2, and b k ≡ 0 mod p r−1 for k ≥ 2, so then Next, we require the following theorems which characterize some p subgroups of S p r .
. Further, the group Q 1 1 is a normal subgroup of Q 1 and is the unique p subgroup of S p r of order p r+1 which contains T .Theorem 3.7.[4, Theorem 11] Let G be a subgroup of S p r and P a p-Sylow subgroup of G of order p s such that T ∈ P .Then the following hold: 1 .Corollary 3.8.Let C and C be two cyclic codes of length p r , and let P be a p-Sylow subgroup of Aut(C) such that T ∈ P .If |P | = p s and s ≤ p + r − 1, then C and C can be equivalent only under the action of a permutation of the following subgroups of S p r : Proof.The result follows from Lemmas 3.3 and 3.5, and Theorem 3.7.Remark 3.9.Since each affine transformation can be written as the product of a power of T and a multiplier, and T ∈ Aut(C), we must have τ a,b ∈ C whenever M a ∈ C. Hence from Corollary 3.8, if s = r then two cyclic codes of length p r are equivalent if and only if they are equivalent by a multiplier.
In order to solve the equivalence problem for cyclic codes, we need the p-Sylow subgroup of Aut(C).To determine this, for i ≤ i ≤ p − 1 consider the polynomial f i ∈ Q i 1 defined by Theorem 3.10.Let G be a subgroup of S p r with a p-Sylow subgroup P which contains T .Then the following hold: Proof.If there is no f i ∈ G, then there is no f i in P .If |P | = p r , we can take P = T , but then from Theorem 3.6 any p-Sylow subgroup of p s , s > r, must contain Q 1 1 , which is impossible.Assume that I is the largest i such that f i ∈ G and I ≤ p − 2. Let P be a p-Sylow subgroup of G of order s, and s be such that I + r ≤ s < p + r − 1.From Theorem 3.7, we have that a p-Sylow subgroup of any subgroup of G ≤ S p r which contains T has order p s with m < s ≤ p + r − 1.Then we have The assumption on I gives I = s − r.
Assume now that s > p + r − 1.Since I ≤ p − 2, we have that s > p + r − 1 > r + I.We will prove that this case cannot occur.Further, as is the unique subgroup of S p r of order p r+1 which contains T , so that Q , which gives Q 2 1 ≤ P .Using the same approach for 2 ≤ i ≤ I, we obtain Q I ≤ P .The assumption on s gives that Q I P , so can be considered as it was assumed that I ≤ p − 2).Hence from Theorem 3.6, we obtain that , which contradicts the assumption on I.
This theorem suggests the following algorithm for I ≤ p − 2.
Algorithm A: Let p be an odd prime and C and C be two cyclic codes of length p m .Then the equivalence of C and C can be determined as follows.
Step 1: Find the order of the p-Sylow subgroup of Aut(C) as follows.Find the largest I such that f I ∈ Aut(C), and set s = I + r.
Step 2: Find f ∈ Q I+1 such that C = f C. Remark 3.11.To find the required I in Algorithm A we can use (for example), a binary search which requires checking at most log 2 (p−1) +1 of the f i .Furthermore, the cardinality of Q I+1 is (p−1)p 2r+I−2 .This proves that the algorithm has polynomial time complexity.

Equivalence of quasi-cyclic codes
In this section, we characterize the equivalence problem for quasi-cyclic codes.Consider the cycles σ i = (i, i + l, i + 2l, . . ., i + (m − 1)l) for 0 ≤ i ≤ l − 1.The cycles σ i have order m and satisfy This gives that Proposition 4.1.Let n = lm with (m, l) = 1, and T l be the subgroup of S n generated by the permutation T l .Therefore the normalizer of T l in S n contains the following groups: Proof.It is obvious that T ∈ N Sn ( T l ).As the cycles in (5) are pairwise disjoint, it must be that σ 0 . . .σ l−1 = T l .Furthermore, as the cycles σ i are disjoint, we have that σ −1 i T l σ i = T l .

Quasi-cyclic codes of length p r l
We now consider quasi-cyclic codes of length n = p r l with p a prime number such that (p, l) = (p, q) = 1.In this case, T l ≤ Aut(C) is a subgroup of order p r .Hence it is contained in a p-Sylow subgroup P .
It is obvious that if P = T l , then we have The following proposition gives other properties of H (P ).

Theorem 3 . 1 .
[6,  Theorem 1]  Let C and C be cyclic codes of length n over a finite field.Suppose one of the following holds for n: (i) gcd(n, φ(n)) = 1 or n = 4, or (ii) n = pr, p > r are primes and the p-Sylow subgroup of the automorphism group of C has order p.Then C and C are equivalent by a multiplier.In the case n = p 2 , Huffman et al. ([6]) gave the following result.Theorem 3.2.[6, Theorem 3.1] Let C and C be cyclic codes of length p 2 with p an odd prime, where T ∈ Aut(C) and T ∈ Aut(C ).Then if C and C are equivalent, they are equivalent by a multiplier or a generalized multiplier times a multiplier.

Lemma 3 . 3 .
[3, Lemma 3.1]  Let C and C be cyclic codes of length p r , and P be a p-Sylow subgroup of Aut(C) which contains T .Then C and C are equivalent if and only if C and C are equivalent by an element of H(P ).

Lemma 4 . 2 .
Let C and C be two quasi-cyclic codes of length n = p r l, and P be a p-Sylow subgroup of Aut(C) such that T l ∈ P .Then C and C are equivalent only if they are equivalent by the elements of the setH (P ) = {σ ∈ S n |σ −1 T l σ ∈ P }.Proof.Since C and C are equivalent, there exists a permutation σ ∈ S n such that C = σ(C).This gives the following relationship between the automorphism groups Aut(C) and Aut(C )Aut(C ) = σAut(C)σ −1 .