Journal of Algebraic Combinatorics

, Volume 38, Issue 1, pp 197–208 | Cite as

On the permutation groups of cyclic codes

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Abstract

We classify the permutation groups of cyclic codes over a finite field. As a special case, we find the permutation groups of non-primitive BCH codes of prime length. In addition, the Sylow p-subgroup of the permutation group is given for many cyclic codes of length pm. Several examples are given to illustrate the results.

Keywords

Permutation groups Transitive groups Doubly transitive groups Non-primitive BCH codes 

1 Introduction

The permutation groups of cyclic codes are of great theoretical and practical interest, e.g. the permutation group can be used to find the weight distribution of a code [19], and in decoding [16, 19]. They can also be used for cryptographic purposes such as the McEliece cryptosystem and its variants [20]. Despite the significance of this problem, the permutation groups of cyclic codes are known for only a few subclasses such as the Reed–Solomon codes, Reed–Muller codes and some BCH codes [3, 18]. The other cases remain open. Recently, Bienert and Klopsch [4] studied the permutation groups of cyclic codes in the binary case. They gave the primitive groups which can be the permutation group of a binary cyclic code. Dobson and Witte [13, 14] considered the cyclic codes invariant under some transitive groups. Furthermore in some cases they gave the Sylow p-subgroups of some transitive subgroups of \(S_{p^{2}}\) and \(S_{p^{m}}\).

In this paper we classify the permutation groups of cyclic codes. First we generalize the results of [4] concerning the doubly transitive permutation groups with socle PSL(d,q) to the non-binary case. Then we use the classification of the doubly transitive groups which contain a complete cycle, given by McSorley [21, 22], and our previous results to determine the permutation groups in the doubly transitive cases. This allows us to determine the permutation groups of the BCH codes in the prime length case. Further, we give conditions on the primitivity of the permutation groups which are based on the underlying field and the length of the code. For many cyclic codes, we explicitly give the Sylow p-subgroups of the permutation groups in the primitive and imprimitive cases. This is done using some subgroups of \(S_{p^{m}}\) introduced by Brand [5]. Several examples are given to illustrate the results.

2 Preliminaries

Let \(\mathbb{F}_{q}\) be a finite field. A linear code C of length n over \(\mathbb{F}_{q}\) is a subspace of \(\mathbb{F}_{q}^{n}\). A vector x=(x0,…,xn−1)∈C is said to be a codeword of C. Let I denote the set {0,1,…,n−1}, and let Sn be the symmetric group acting on I. Then Sn acts naturally on a codeword of C as follows. If σ is a permutation of Sn, then
$$\sigma(x)=(x_{\sigma^{-1}(0)}, \ldots, x_{\sigma^{-1}(n-1)}), \quad (x_0, \ldots,x_{n-1}) \in C. $$
The permutation group of C is the subgroup of Sn given by
$$Per(C)= \bigl\{\sigma\in S_n \,|\, \sigma(C)=C \bigr\}. $$
A linear code C over \(\mathbb{F}_{q}\) is cyclic if TPer(C), where T=(0,1,…,n−1) is a complete cycle of length n. If C is cyclic then Per(C) is a transitive group. The group AG(n)={τa,b:a≠0,(a,n)=1,b∈ℤn} is the subgroup of Sn formed by the permutation defined as follows: The group AG(n) is called the group of affine transformations. The affine transformations Ma=τa,0 are called a multiplier. The affine group AGL(1,p) is the group of affine transformations over ℤp. The projective semi-linear group PΓL(d,t) is the semi-direct product of the projective linear group PGL(d,t) and the automorphism group \(Z=Gal(\mathbb{F}_{t}/\mathbb{F}_{p})\) of \(\mathbb{F}_{t}\), where t=ps,p prime, i.e.
$$P\varGamma L(d,t)= PGL(d,t)\rtimes Z. $$

Remark 1

The zero code, the entire space, and the repetition code and its dual are called elementary codes. The permutation group of these codes is Sn [16, p. 1410]. Further, it was proven in [16, p. 1410] that there is no cyclic code with permutation group equal to Alt(n).

3 The permutation groups of cyclic codes

A doubly transitive group G has a unique minimal normal subgroup N which is either regular and elementary abelian, or simple and primitive, and the centralizer of N in G is equal to CG(N)=1 [8, p. 202]. All simple groups which can occur as a minimal normal subgroup of a doubly transitive group are known. This result is due to the classification of finite simple groups [9]. Using this classification, McSorley [21] gave the following result.

Lemma 1

A groupGof degreenwhich is doubly transitive and contains a complete cycle has socleNwithNGAut(N), and is equal to one of the cases in Table 1.
Table 1

The doubly transitive groups that contain a complete cycle

G

n

N

AGL(1,p)

p

Cp

S4

4

C2×C2

Sn,n≥5

n

Alt(n)

Alt(n),n odd and ≥5

n

Alt(n)

PGL(d,t)≤GPΓL(d,t)

\(\frac{t^{d}-1}{t-1}\)

PSL(d,t)

(d,t)≠(2,2),(2,3),(2,4)

  

PSL(2,11)

11

PSL(2,11)

M11(Mathieu)

11

M11(Mathieu)

M23(Mathieu)

23

M23(Mathieu)

The arguments given in the following Lemma are similar to those for the binary case [4, Theorem E, Part 3].

Lemma 2

LetCbe a non-elementary cyclic code of length\(n=\frac{t^{d}-1}{t-1}\)over a finite field\(\mathbb{F}_{q}\), whereq=rαandtis a prime power. If the groupPer(C) satisfies
$$PGL(d,t) \leq Per(C) \leq P\varGamma L(d,t), $$
thent=rafor somea≥1, d≥3, andPer(C)=PΓL(d,t).

Proof

Assume d=2. As the group PGL(2,t) acts 3-transitively on the 1-dimensional projective space \(\mathbb{P}^{1}(\mathbb{F}_{t})\), we deduce from [22, Table 1 and Lemma 2], that the underlying code is elementary, which is a contradiction. Hence d≥3, and from [22, Table 1 and Lemma 2], it must be that since C is non-elementary, t must be equal to ra. Now let V denote the permutation module over \(\mathbb{F}_{r}\) associated with the natural action of PGL(d,t) on the (d−1)-dimensional projective space \(\mathbb{P}^{d-1}( \mathbb{F}_{t})\). Let U1 be a PGL(d,t)-submodule of V. Then U1 is PΓL(d,t)-invariant. This is because, if σ is a generator of the cyclic group \(P\varGamma L(d,t)/PGL(d,t) \simeq Gal( \mathbb{F}_{t}/ \mathbb{F}_{r})\), then \(U_{2}=U_{1}^{\sigma}\), regarded as a PGL(d,t)-module, is simply a twist of U1. Let \(\overline{\mathbb{F}}_{r}\) be the algebraic closure of \(\mathbb{F}_{r}\). Then the composition factors of the \(\overline{\mathbb{F}}_{r}PGL(d,t)\)-modules \(\overline{U}_{1}=\overline{\mathbb{F}}_{r}\otimes U_{1}\) and \(\overline{U}_{2}=\overline{\mathbb{F}}_{r}\otimes U_{2}\) are the same. The submodules of the \(\overline{\mathbb{F}}_{r}PGL(d,t)\)-module \(\overline{V}=\overline{\mathbb{F}}_{r}\otimes V\) are uniquely determined by their composition factors [1]. Then we have \(\overline{U}_{1}=\overline{U}_{2}\), which implies that U1=U2, and therefore Per(C)=PΓL(d,t). □

The following theorem establishes the permutation group of a non-elementary cyclic code of prime length over \(\mathbb{F}_{q}\), where q=rα.

Theorem 3

LetCbe a non-elementary cyclic code of lengthpover\(\mathbb{F}_{q}\). ThenPer(C) is a primitive group, and one of the following holds:
  1. (i)

    Per(C) is a solvable group of orderpmwithma divisor ofp−1 andCpPer(C)≤AGL(1,p), withp≥5. FurthermorePer(C) contains a normal Sylowp-subgroup.

     
  2. (ii)

    Ifp=q, thenPer(C)=AGL(1,p).

     
  3. (iii)

    Per(C)=PSL(2,11) andqis a power of 3. Cis either the [11,6] or [11,5] code that is equivalent to the [11,6,5] ternary Golay code or its dual, respectively.

     
  4. (iv)

    Per(C)=M23andqis a power of 2. Cis either the [23,12] or [23,11] code that is equivalent to the [23,12,7] binary Golay code or its dual, respectively.

     
  5. (v)

    \(Per(C)=P\varGamma L (d,r^{d^{b}})\)whereb∈ℕ, d≥3 is a prime number such that\((d,r^{d^{b}}-1)=1\), and\(p=(r^{d^{b+1}}-1)/(r^{d^{b}}-1)\).

     

Proof

A transitive group of prime degree is a primitive group [23, p. 195]. As a consequence of a result of Burnside [13, Theorem 2], a transitive group of prime degree is either a subgroup of AGL(1,p) or a doubly transitive group. In the first case CpPer(C)≤AGL(1,p), and if p=2 or 3, AGL(1,p)=Sp. In this case, C is elementary by Remark 1, which is a contradiction. Since Cp is normal in AGL(1,p) and AGL(1,p)/Cp is abelian, Per(C) is a normal subgroup. By [11, Example 3.5.1] G is solvable. If q=p, Roth and Seroussi [24] proved that any cyclic code of prime length p over \(\mathbb{F}_{p}\) must be an MDS code equivalent to an extended Reed–Solomon code. Berger [2] proved that the permutation group of such codes is the affine group AGL(1,p). In the doubly transitive cases, as C is non-elementary of prime length p, by Lemma 1, Remark 1 and Lemma 2, we see that Per(C) is one of M11, with p=11, PSL(2,11) with p=11, M23 with p=23, or PΓL(d,t) of degree p=(td−1)/(t−1) and t a prime power. If Per(C)=M11, from [22, Table 1, Lemma 2] C must be elementary, which is a contradiction. If Per(C)=PSL(2,11), from [22, Table 1, Lemma 2 and (J)] q must be a power of 3, and there is a unique non-elementary code over \(\mathbb{F}_{q}\) contained in the dual of the repetition code. The [11,5,6] dual of the ternary Golay code is contained in the repetition code and has permutation group PSL(2,11); its dual, an [11,6,5] code, intersects the dual of the repetition code in this [11,5,6] code and also has permutation group PSL(2,11). Part (ii) then follows. Part (iii) is obtained in an analogous way from [22, Table 1, Lemma 2 and (I)]. For Part (iv), we have from Lemma 2 that Per(C)=PΓL(d,t), t=ra for some a≥1 and d≥3. A number theory argument [12, Lemma 3.1] gives the result that if p is prime, then d must be a prime such that (d,ra−1)=1 and a=bd. The result then follows. □

Remark 2

For p prime, the permutation group of a non-elementary BCH code of length p over \(\mathbb{F}_{q}\) is one of those listed in Theorem 3.

In Table 2, we give examples of permutation groups of BCH codes of length p over \(\mathbb{F}_{q}\). Per(C) (respectively Per(C2) and Per(C3)), denotes the permutation group of the narrow sense (b=1) BCH code with designed distance δ (respectively BCH code with designed distance δ and b=2 and b=3).
Table 2

Permutation groups of some BCH codes of length p

q

p

δ

Per(C)

Per(C2)

Per(C3)

2

17

2

C8C17

S17

S17

2

23

3

M23

M23

M23

2

41

2

C20C41

C20C41

C20C41

2

41

3

C20C41

S41

S41

2

43

5

C14C43

C14C43

C14C43

2

43

7

C14C43

S43

S43

3

13

2

C3C13

C3C13

C3C13

3

13

4

PΓL(3,3)

C3C13

C3C13

3

13

5

C3C13

C3C13

C3C13

3

23

3

C11C23

C11C23

C11C23

3

41

5

C8C41

C8C41

C8C41

4

43

9

C7C43

S43

S43

5

11

5

C5C11

C5C11

C5C11

11

5

3

C5

C2C5

C5

The following result is obtained by considering the permutation groups of cyclic codes of composite length.

Theorem 4

LetCbe a non-elementary cyclic code over\(\mathbb{F}_{r^{\alpha}}\)of composite length. ThenPer(C) is either
  1. (i)

    an imprimitive group (in the case thatn=pm, pa prime, the orbit of the subgroup generated by\(T^{p^{m-1}}\)and its conjugate form a complete block system ofPer(C));

    or

     
  2. (ii)
    Per(C) is a doubly transitive group equal to
    $$P\varGamma L \bigl(d,r^a \bigr),\quad \text{\textit{with} }n = \frac{r^{ad}-1}{r^a-1},\ d\geq3,\ a\ge1. $$
     

Proof

The group Per(C) contains a complete cycle and has composite degree. Hence from a theorem of Burnside and Schur [25, p. 65], Per(C) is either imprimitive or doubly transitive. If it is imprimitive and n=pm, by [7, Chap. XVI, Theorem VIII] Per(C) contains an intransitive normal subgroup generated by \(T^{p^{m-1}}\) and its conjugates. By [25, Proposition 7.1] the orbit of such a subgroup forms a complete block system of Per(C).

In the doubly transitive case, we have from Lemma 1 that the only cases when the socle can be abelian are N=Cp and N=C2×C2. In these cases, Per(C) must be equal to AGL(1,p) or S4, which is impossible. Since the socle is not abelian and the degree is not prime, this leads to the only solution given by row six of Table 1 in Lemma 1. Hence from Lemma 2, Part (ii) follows. □

4 The permutation group of cyclic codes of prime power length

In this section, we consider the permutation groups of cyclic codes of length pm, where p is an odd prime.

Lemma 5

Letqbe a prime power, pan odd prime, andzthe largest integer such thatpz|(qt−1), withtthe order ofqmodulop. Ifz=1 we have
$$\operatorname {ord}_{p^m}(q)=p^{m-1}t. $$

Proof

Let t be the order of q modulo p, and u=qt≡1 mod p. Assume that z=1, or equivalently u≠1 mod p2. It is well known from elementary number theory [10, p. 87] that u mod pm is an element of order pm−1 in the group \((\mathbb{Z}_{p^{m}} )^{\ast}\) if and only if u≠1 mod p2. Hence \(\operatorname {ord}_{p^{m}}(q)=p^{m-1}t\). □

According to Brillhart et al. [6], it is unusual to have z>1.

Proposition 6

Letn=pmandq=rαa prime power with (q,n)=1, andCa cyclic code of lengthnover\(\mathbb{F}_{q}\). LetMqbe the multiplier defined byMq(i)=iq mod pm. Then the groupPer(C) contains the subgroupK=〈T,Mqof order\(p^{m}\operatorname {ord}_{p^{m}}(q)\). Letpl, withlmbe thep-part of the order ofK. Then a Sylowp-subgroupPofPer(C) has orderpssuch that
$$l \le s \le p^{m-1}+ p^{m-2}+ \cdots+1. $$
Ifz=1, thens≥2m−1.

Proof

By the definition of a cyclic code, we have TPer(C). It is obvious that each cyclotomic class modulo n over \(\mathbb{F}_{q}\) is invariant under the permutation Mq. This can be deduced from the fact that the polynomial \(f(x)\in\mathbb{F}_{q}[x]\) satisfies f(xq)=f(x)q. Thus MqPer(C). The order of Mq is equal to \(\operatorname {ord}_{n} (q)\), hence K=〈T,Mq〉 is a subgroup of Per(C) of order \(n \operatorname {ord}_{n}(q)\). Since n=pm, the order of K has p-part pl with lm. Let P be a Sylow p-subgroup of Per(C) which contains T. Then P is a p-subgroup of \(S_{p^{m}}\). From Sylow’s Theorem, P is contained in a Sylow p-subgroup of \(S_{p^{m}}\). It is well known that a Sylow p-subgroup of \(S_{p^{m}}\) is of order \(p^{p^{m-1}+p^{m-2}+ \cdots +1}\) [23, Kalužnin’s Theorem]. Since P also contains the subgroup of K of order pl, then lspm−1+pm−2+⋯+1. If z=1, then by Lemma 5 the order of the group K is \(\operatorname {ord}_{p}(q)p^{2m-1}\). This shows that p2m−1 divides |Per(C)|, so Per(C) contains a p-subgroup of order at least p2m−1. □

Theorem 7

LetCbe a non-elementary cyclic code of lengthpmover\(\mathbb{F}_{r^{\alpha}}\), withm≥1. Then the following hold:
  1. (i)

    Ifpαandp∤(d,ra−1), thenPer(C)=PΓL(d,ra), a≥1, d≥3, if and only if the Sylowp-subgroup ofPer(C) is of orderpm.

     
  2. (ii)

    Ifp≥5, α=1 andr=p, m>1, thenPer(C) is an imprimitive group which admits a complete system formed by the orbit of the subgroup generated by\(T^{p^{m-1}}\)and its conjugate. It also contains a transitive normal Sylowp-subgroup of orderpswithm<spm−1+pm−2+⋯+1.

     
  3. (iii)

    Ifz=1, pαandp∤(d,ra−1), thenPer(C) is an imprimitive group which contains a transitive normal Sylowp-subgroup of orderps, with 2m−1≤spm−1+pm−2+⋯+1. Furthermore, Per(C) admits a complete block system formed by the orbit of the subgroup generated by\(T^{p^{m-1}}\)and its conjugate.

     

Proof

For Part (i), we know that the socle of PΓL(d,ra) is the group PSL(d,ra) of order \(\frac{r^{ad(d-1)/2}}{(d,r^{a}-1)}\prod_{i=2}^{d}(r^{ai}-1)\). From a lemma of Zsigmondy [17, Chap. IX, Theorem 8.3], except for the cases d=2, ra=2b−1 and d=6,ra=2, there exists a prime q0 such that q0 divides rad−1, but does not divide rai−1, for 1≤i<d. From Lemma 2, we cannot have d=2. The case d=6 and ra=2 does not give a prime power. Hence if \(n=p^{m}=\frac{r^{ad}-1}{r^{a}-1}\), there is a q0 which divides (rad−1)=(ra−1)pm. Since q0 does not divide ra−1, then q0 divides pm, and hence q0=p and pm is the p-part of the order of PSL(d,ra). Also, since pra−1, we have p∤(d,ra−1). Hence if (α,p)=1, pm is also in the p-part of the order of PΓL(d,ra), and the result follows.

Conversely, if Per(C) has Sylow p-subgroup P of order pm, we can assume that TP, which gives the equality P=〈T〉. Assume that in this case Per(C) is imprimitive. Then by [13, Theorem 33], P is normal. P is then the minimal normal subgroup which is transitive and abelian. From [25, p. 17] Per(C) is primitive, which is impossible. Thus if P=〈T〉, the group Per(C) is equal to PΓL(d,ra), which is possible only if [PΓL(d,ra):PSL(d,ra)] is prime to p, i.e, (p,α)=1 and p∤(d,ra−1).

For Part (ii), from Theorem 4 if Per(C) is primitive, then it is doubly transitive and equal to PΓL(d,ra) with \(n =\frac{r^{ad}-1}{r^{a}-1}, d\geq3\) and a≥1. From [13, Lemma 22], if Per(C) is doubly transitive with non abelian socle, then Soc(Per(C))=Alt(pm). Hence from Remark 1 the code is elementary, which is a contradiction. Therefore, Per(C) is imprimitive, and then by Part (i) the order of the Sylow p-subgroup is ps with s>m. The second inequality then follows by Proposition 6.

For Part (iii), if z=1 then from Proposition 6, we find that the order of a Sylow p-subgroup of Per(C) is at least p2m−1. If Per(C) is doubly transitive, by Theorem 4 it is equal to PΓL(d,ra), with d≥3. By assuming pα and p∤(d,ra−1), we obtain from Part (i) that a Sylow p-subgroup of Per(C) has order pm, which is impossible. Hence Per(C) is an imprimitive group. From [13, Theorem 33] Per(C) contains a transitive normal Sylow p-subgroup, hence the result follows. □

Example 1

The narrow sense BCH code of length 25 over \(\mathbb{F}_{3}\) with designed distance 3 has a permutation group which is the imprimitive group S5S5. The narrow sense BCH code of length 9 over \(\mathbb{F}_{5}\) with designed distance 2 has a permutation group which is the imprimitive group S3S3. The binary [7,4,3] Hamming code has permutation group PΓL(3,2), which contains a Sylow 7-subgroup of order 7.

We now give the Sylow p-group of Per(C) for several cases. Let p be an odd prime. For n<p−1, we define the following subsets of \(S_{p^{m}}\):

\(Q^{n}=\{f: \mathbb{Z}_{p^{m}} \rightarrow\mathbb{Z}_{p^{m}} | f(x)=\sum_{i=0}^{n}a_{i}x^{i}, a_{i} \in\mathbb{Z}_{p^{m}} \mbox{ for each } i,\ (p,a_{1})=1,\mbox{ and} p^{m-1} \mbox{ divides }a_{i}\mbox{ for }i=2,3,\ldots, n\}\).

\(Q_{1}^{n}=\{f \in Q^{n} | f(x)=\sum_{i=0}^{n}a_{i}x^{i},\mbox{ with } a_{1} \equiv1\ \mathrm {mod}\ p^{m-1} \}\).

The sets Qn and \(Q_{1}^{n}\) are subgroups of \(S_{p^{m}}\) [5, Lemma 2.1]. Note that Q1=AG(pm).

The following lemma will be used later. Note that the proof is similar to that of [15, Lemmas 2.4, 2.5].

Lemma 8

If 1≤n<p−1, then
  1. (i)

    |Qn|=(p−1)p2m+n−2and\(|Q_{1}^{n}|=p^{m+n}\).

     
  2. (ii)

    \(AG(p^{m})=N_{S_{p^{m}}}(\langle T\rangle)\).

     
  3. (iii)

    \(N_{S_{p^{m}}}(Q_{1}^{n})=Q^{n+1}\).

     

Proof

For Part (i), by [5, Lemma 3.2], the map (a0,…,an)⟶f, where \(f(x)= \sum_{i=0}^{n} a_{i} x^{i}\) is injective if n<p−1. Thus in Qn, the coefficients of a0 can take pr different values, and a1 can take pm−1(p−1) values. For 2≤in, ai can take p values. From these results we have |Qn|=p2m+n−2(p−1). For \(Q_{1}^{n}\), the coefficients of a0 can take pm different values, and ai for 1≤in can take p values, hence \(|Q_{1}^{n}|=p^{m+n}\).

Now we prove that \(AG(p^{m})=N_{S_{p^{m}}}(\langle T\rangle)\). Let σ be an element of \(N_{S_{p^{m}}}(\langle T\rangle)\). Then, there is a j∈ℤn∖{0} such that σTσ−1=Tj, or equivalently σT=Tjσ. Hence σT(0)=σ(1)=Tjσ(0)=σ(0)+j and σT(1)=σ(1)+j=σ(0)+2j. Therefore σ(k)=σ(0)+kj for any k∈ℤn. Then (j,n)=1 follows from the fact that the order of T equals the order of Tj.

Now we prove Part (iii).

(⊆ part) Let \(h\in N_{p^{m}}(Q_{1}^{n})\) and g=h−1Th. As \(T\in Q_{1}^{n}\), it must be that \(g\in Q_{1}^{n}\). Since the order of g is equal to the order of T (which is pm), from [5, Lemma 3.6] there exists fQn+1 such that f−1gf=T. Thus f−1h−1Thf=T. The only elements of \(S_{p^{m}}\) which commute with T (a complete cycle of length pm), are the powers of T. Thus hf=Tj for some j. Since Qn+1 is a subgroup of \(S_{p^{m}}\) and 〈T〉≤Qn+1, then hQn+1. Hence \(N_{p^{m}}(Q_{1}^{n})\leq Q^{n+1}\).

(⊇ part) Let \(h \in Q_{1}^{n}\), where \(h(x)=\sum_{i=0}^{n}h_{i}x^{i}\), with h1≡1 mod pm−1 and pm−1|hi, for 2≤in. Let gQn+1 where \(g(x)=\sum_{i=0}^{n+1}g_{i}x^{i}\), with pg1 and pm−1|gi for 2≤in. We have
$$hg(x)=\sum_{i=0}^{n}h_i \Biggl(\sum_{j=0}^{n+1}g_jx^j \Biggr)^i=h_0+h_1 \sum _{i=0}^{n+1}g_jx^j+\sum _{i=2}^{n}h_i \Biggl(\sum _{j=0}^{n+1}g_jx^j \Biggr)^i. $$
Since pm−1|hi, for i≥2 and pm−1|gj for j≥2, any terms in \(\sum_{i=2}^{n}h_{i} (\sum_{j=0}^{n+1}g_{j}x^{j} )^{i}\) involving gj for j≥2 vanish modulo pm. Therefore we have
$$hg(x)=h_0+h_1\sum_{j=0}^{n+1}g_jx^j+ \sum_{i=2}^{n}h_i(g_{0}+g_1x)^i. $$
By [5, Lemma 2.1], we have
$$ g^{-1}(x)=\sum_{i=1}^{n+1}b_ix^i \quad \text{with }b_1 =g_1^{-1} \text{ and }b_i=-g_ig_1^{-(i+1)}\text{ for } 2 \leq j \leq n+1. $$
(2)
We now compute g−1hg in order to prove that it is in \(Q_{1}^{n}\). This is given by As pm−1|gj for j≥2, we have pm−1|bk for k≥2. Furthermore, pm−1|hi for i≥2, and thus Let \(g^{-1}hg(x)=\sum_{m=0}^{n+1}c_{m}x^{m}\), and note that cn+1=b1h1gn+1+bn+1(h1g1)n+1. Then by replacing the bi with their values from (2), we obtain
$$c_{n+1}=g_1^{-1}h_1g_{n+1}-g_{n+1}g_1^{-(n+2)}h_1^{n+1}g_1^{n+1}=g_1^{-1}h_1 \bigl(g_{n+1}-g_{n+1}h_1^n \bigr). $$
As h1≡1 mod pm−1, we have \(h_{1}^{n} \equiv1\ \mathrm {mod}\ p^{m-1}\). In addition, since pm−1|gn+1, we have \(g_{n+1}h_{1}^{n} \equiv g_{n+1}\ \mathrm {mod}\ p^{m}\). Therefore, cn+1=0, and also pm−1|ci for 2≤in. Then we only need to show that c1≡1 mod pm−1. Since gj≡0 mod pm−1 for j≥2, hi≡0 mod pm−1 for i≥2, and bk≡0 mod pm−1 for k≥2, then c1b1h1g1 mod pm−1. Finally, as \(b_{1}=g_{1}^{-1}\), we have c1h1≡1 mod pm−1. □

Lemma 9

Let 1≤n<p−1. IfPis ap-subgroup of\(S_{p^{m}}\)with\(Q_{1}^{n} \lneq P \leq Q^{n+1}\), then\(P=Q_{1}^{n+1}\).

Proof

By Lemma 8 Part (ii), we have \(Q_{1}^{n} \lhd Q^{n+1}\). Hence we can consider \(\overline{Q}=Q^{n+1}/Q_{1}^{n}\), which is of order pm−1(p−1) by Lemma 8. Let N be the number of Sylow p-subgroups of \(\overline{Q}\). Then by Sylow’s Theorem, N≡1 mod p and N divides pm−1(p−1). Hence N=1, so there exists a unique Sylow p-subgroup \(\overline{P'}\) of \(\overline{Q}\) which is normal. From the condition on P above, the image \(\overline{P}\) of P in \(\overline{Q}\) is also a Sylow p-subgroup of \(\overline{Q}\). Since there is a unique Sylow p-subgroup \(\overline{P'}=\overline{P}\), by Lemma 8 the image \(\overline{Q}_{1}^{n+1}\) of \(Q_{1}^{n+1}\) in \(\overline{Q}\) is a Sylow p-subgroup of \(\overline{Q}\). Hence \(\overline{Q}_{1}^{n+1}=\overline{P}=\overline{P'}\). As \(Q_{1}^{n}\lneq P\) and \(Q_{1}^{n} \leq Q_{1}^{n+1}\), the result follows. □

Theorem 10

The group\(Q_{1}^{1}\)is a normal subgroup ofQ1and is the unique subgroup of\(S_{p^{m}}\)of orderpm+1which containsT.

Proof

It is obvious that \(T \in Q_{1}^{1}\). By Lemma 8, \(|Q_{1}^{1}|=p^{m+1}\). Consider now an element g of Q1, g(x)=b0+b1x with \(b_{0},b_{1} \in\mathbb{Z}_{p^{m}}\) and (b1,p)=1. It is not difficult to check that the inverse of g in Q1 is given by \(g^{-1}(x)=-b_{1}^{-1}b_{0}+b_{1}^{-1}x\). Consider \(f\in Q_{1}^{1}\), so that f(x)=a0+a1x with \(a_{0},a_{1} \in\mathbb{Z}_{p^{m}}\), (a1,p)=1 and a1≡1 mod pm. We then have \(g^{-1}fg(x)=g^{-1}(a_{0}+a_{1}(b_{0}+b_{1}x))=(-b_{0}+a_{0}+a_{1}b_{0})b_{1}^{-1}+a_{1}x\). This proves that \(g^{-1}fg(x) \in Q_{1}^{1}\). Hence \(Q_{1}^{1}\) is normal in Q1. Now let S be a subgroup of Q1 of order pm+1 which contains T. Thus 〈T〉 has index p in S, and thus 〈T〉 is maximal in S. Furthermore, 〈T〉◁S, because any subgroup of a p-group of index p must be normal. Therefore we have \(S=N_{S}(T) \leq N_{S_{p^{m}}} (T)\), and by Lemma 8, \(S \leq N_{S_{p^{m}}} (T)=AG(p^{m})=Q^{1}\). Thus, such an S must be a subgroup of Q1. It is clear that \(Q_{1}^{1}\) is not abelian, and S cannot be abelian since it is a transitive group. If this were the case it would have to be a regular group [23, Theorem 1.6.3], and thus |S|=pm, which is impossible. Furthermore, the p-groups which contain a cyclic maximal subgroup are known [23, Theorem 5.3.4]. If these groups are not abelian or p≠2, they have the following special forms:
$$Q_1^1= \bigl\langle x,T\bigm{\vert} x^p=1; \, x^{-1}Tx =T^{1+p^{m-1}} \bigr\rangle, $$
and
$$S= \bigl\langle y,T\bigm{\vert} y^p=1;\ y^{-1}Ty =T^{1+p^{m-1}} \bigr\rangle. $$
However, the conditions on x and y give
$$x^{-1}Tx= y^{-1}Ty \quad \iff\quad Tyx^{-1}=yx^{-1}T, $$
so the only elements of \(S_{p^{m}}\) which commute with T (a complete cycle of length pm), are the powers of T. Thus yx−1=Tj for some j. Since the order of yx−1 is p, the only choices for j are j=pm or j=pm−1. For both choices we get \(S=Q_{1}^{1}\), namely j=pm gives x=y−1 (so \(S=Q_{1}^{1}\)), and j=pm−1 gives \(x=T^{-p^{m-1}}y\). Thus we have x∈〈y,T〉, so that 〈x,T〉=〈y,T〉, and hence \(S=Q_{1}^{1}\). □

Theorem 11

Letpbe an odd prime, q=rαa prime power, Ca cyclic code over\(\mathbb{F}_{q} \)of lengthpm, andPa Sylowp-subgroup ofPer(C) of orderpssuch thatTP. Then the following hold:
  1. (a)

    Ifpαandp∤(d,ra−1), thens=m, andP=〈Tif and only ifPer(C)=PΓL(d,ra), d≥3,

     
  2. (b)

    Ifp≥5, α=1 andr=p, m>1, thenPer(C) is an imprimitve group andPis normal of orderps, s>m. Ifm<sp+m−1, then we have\(P=Q_{1}^{s-m}\).

     
  3. (c)

    Ifz=1, pαandp∤(d,ra−1), thenPer(C) is an imprimitve group andPis normal of orderpsp2m−1. Furthermore, if 2m−1<sp+m−1, then we have\(P=Q_{1}^{s-m}\).

     

Proof

Statement (a) and the first parts of (b) and (c) follow from Theorem 7. We thus only need prove that if s<p+m−1, then \(P=Q_{1}^{s-m}\). Assume sp+m−1, so that P contains a p-subgroup P′ of order pm+1. By Theorem 10, we obtain \(P'=Q_{1}^{1}\). Let j≥1 be the largest integer such that \(Q_{1}^{j} \leq P\). If j=p−1, by Lemma 8 we have \(|Q_{1}^{p-1}|=p^{p+m-1}\). Thus \(Q_{1}^{p-1}\) is a subgroup of P of the same order as P, and hence \(P=Q_{1}^{p-1}\), so we can assume that 1≤j<p−1. If \(Q_{1}^{j} \lneq P\), then \(Q_{1}^{j} \lneq N_{P}(Q_{1}^{j})\) and by Lemma 8, \(N_{P}(Q_{1}^{j})\leq Q_{1}^{j+1}\). Since \(Q_{1}^{j} \lneq N_{P}(Q_{1}^{j})\leq Q_{1}^{j+1}\), by Lemma 8 \(N_{p}(Q_{1}^{j})=Q_{1}^{j+1}\), which contradicts the choice of j. □

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Copyright information

© Springer Science+Business Media, LLC 2012

Authors and Affiliations

  1. 1.Faculty of Mathematics USTHBUniversity of Science and Technology of AlgiersAlgiersAlgeria
  2. 2.Department of Electrical and Computer EngineeringUniversity of VictoriaVictoriaCanada

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