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Journal of Algebraic Combinatorics

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

On the permutation groups of cyclic codes

  • Kenza Guenda
  • T. Aaron Gulliver
Article
  • 343 Downloads

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 p m . 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=(x 0,…,x n−1)∈C is said to be a codeword of C. Let I denote the set {0,1,…,n−1}, and let S n be the symmetric group acting on I. Then S n acts naturally on a codeword of C as follows. If σ is a permutation of S n , 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 S n 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)={τ a,b :a≠0,(a,n)=1,b∈ℤ n } is the subgroup of S n formed by the permutation defined as follows: The group AG(n) is called the group of affine transformations. The affine transformations M a =τ a,0 are called a multiplier. The affine group AGL(1,p) is the group of affine transformations over ℤ p . 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=p s ,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 S n  [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 C G (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 group G of degree n which is doubly transitive and contains a complete cycle has socle N with NGAut(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

C p

S 4

4

C 2×C 2

S n ,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)

M 11(Mathieu)

11

M 11(Mathieu)

M 23(Mathieu)

23

M 23(Mathieu)

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

Lemma 2

Let C be a non-elementary cyclic code of length \(n=\frac{t^{d}-1}{t-1}\) over a finite field \(\mathbb{F}_{q}\), where q=r α and t is a prime power. If the group Per(C) satisfies
$$PGL(d,t) \leq Per(C) \leq P\varGamma L(d,t), $$
then t=r a for some a≥1, d≥3, and Per(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 r a . 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 U 1 be a PGL(d,t)-submodule of V. Then U 1 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 U 1. 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 U 1=U 2, 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=r α .

Theorem 3

Let C be a non-elementary cyclic code of length p over \(\mathbb{F}_{q}\). Then Per(C) is a primitive group, and one of the following holds:
  1. (i)

    Per(C) is a solvable group of order pm with m a divisor of p−1 and C p Per(C)≤AGL(1,p), with p≥5. Furthermore Per(C) contains a normal Sylow p-subgroup.

     
  2. (ii)

    If p=q, then Per(C)=AGL(1,p).

     
  3. (iii)

    Per(C)=PSL(2,11) and q is a power of 3. C is 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)=M 23 and q is a power of 2. C is 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)

    \(Per(C)=P\varGamma L (d,r^{d^{b}})\) where b∈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 C p Per(C)≤AGL(1,p), and if p=2 or 3, AGL(1,p)=S p . In this case, C is elementary by Remark 1, which is a contradiction. Since C p is normal in AGL(1,p) and AGL(1,p)/C p 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 M 11, with p=11, PSL(2,11) with p=11, M 23 with p=23, or PΓL(d,t) of degree p=(t d −1)/(t−1) and t a prime power. If Per(C)=M 11, 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=r a 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,r a −1)=1 and a=b d . 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(C 2) and Per(C 3)), 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(C 2)

Per(C 3)

2

17

2

C 8C 17

S 17

S 17

2

23

3

M 23

M 23

M 23

2

41

2

C 20C 41

C 20C 41

C 20C 41

2

41

3

C 20C 41

S 41

S 41

2

43

5

C 14C 43

C 14C 43

C 14C 43

2

43

7

C 14C 43

S 43

S 43

3

13

2

C 3C 13

C 3C 13

C 3C 13

3

13

4

PΓL(3,3)

C 3C 13

C 3C 13

3

13

5

C 3C 13

C 3C 13

C 3C 13

3

23

3

C 11C 23

C 11C 23

C 11C 23

3

41

5

C 8C 41

C 8C 41

C 8C 41

4

43

9

C 7C 43

S 43

S 43

5

11

5

C 5C 11

C 5C 11

C 5C 11

11

5

3

C 5

C 2C 5

C 5

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

Theorem 4

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

    an imprimitive group (in the case that n=p m , p a prime, the orbit of the subgroup generated by \(T^{p^{m-1}}\) and its conjugate form a complete block system of Per(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=p m , 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=C p and N=C 2×C 2. In these cases, Per(C) must be equal to AGL(1,p) or S 4, 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 p m , where p is an odd prime.

Lemma 5

Let q be a prime power, p an odd prime, and z the largest integer such that p z |(q t −1), with t the order of q modulo p. If z=1 we have
$$\operatorname {ord}_{p^m}(q)=p^{m-1}t. $$

Proof

Let t be the order of q modulo p, and u=q t ≡1 mod p. Assume that z=1, or equivalently u≠1 mod p 2. It is well known from elementary number theory [10, p. 87] that u mod p m is an element of order p m−1 in the group \((\mathbb{Z}_{p^{m}} )^{\ast}\) if and only if u≠1 mod p 2. 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

Let n=p m and q=r α a prime power with (q,n)=1, and C a cyclic code of length n over \(\mathbb{F}_{q}\). Let M q be the multiplier defined by M q (i)=iq mod p m . Then the group Per(C) contains the subgroup K=〈T,M q of order \(p^{m}\operatorname {ord}_{p^{m}}(q)\). Let p l , with lm be the p-part of the order of K. Then a Sylow p-subgroup P of Per(C) has order p s such that
$$l \le s \le p^{m-1}+ p^{m-2}+ \cdots+1. $$
If z=1, then s≥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 M q . This can be deduced from the fact that the polynomial \(f(x)\in\mathbb{F}_{q}[x]\) satisfies f(x q )=f(x) q . Thus M q Per(C). The order of M q is equal to \(\operatorname {ord}_{n} (q)\), hence K=〈T,M q 〉 is a subgroup of Per(C) of order \(n \operatorname {ord}_{n}(q)\). Since n=p m , the order of K has p-part p l 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 p l , then lsp m−1+p m−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 p 2m−1 divides |Per(C)|, so Per(C) contains a p-subgroup of order at least p 2m−1. □

Theorem 7

Let C be a non-elementary cyclic code of length p m over \(\mathbb{F}_{r^{\alpha}}\), with m≥1. Then the following hold:
  1. (i)

    If pα and p∤(d,r a −1), then Per(C)=PΓL(d,r a ), a≥1, d≥3, if and only if the Sylow p-subgroup of Per(C) is of order p m .

     
  2. (ii)

    If p≥5, α=1 and r=p, m>1, then Per(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 Sylow p-subgroup of order p s with m<sp m−1+p m−2+⋯+1.

     
  3. (iii)

    If z=1, pα and p∤(d,r a −1), then Per(C) is an imprimitive group which contains a transitive normal Sylow p-subgroup of order p s , with 2m−1≤sp m−1+p m−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,r a ) is the group PSL(d,r a ) 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, r a =2 b −1 and d=6,r a =2, there exists a prime q 0 such that q 0 divides r ad −1, but does not divide r ai −1, for 1≤i<d. From Lemma 2, we cannot have d=2. The case d=6 and r a =2 does not give a prime power. Hence if \(n=p^{m}=\frac{r^{ad}-1}{r^{a}-1}\), there is a q 0 which divides (r ad −1)=(r a −1)p m . Since q 0 does not divide r a −1, then q 0 divides p m , and hence q 0=p and p m is the p-part of the order of PSL(d,r a ). Also, since pr a −1, we have p∤(d,r a −1). Hence if (α,p)=1, p m is also in the p-part of the order of PΓL(d,r a ), and the result follows.

Conversely, if Per(C) has Sylow p-subgroup P of order p m , 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,r a ), which is possible only if [PΓL(d,r a ):PSL(d,r a )] is prime to p, i.e, (p,α)=1 and p∤(d,r a −1).

For Part (ii), from Theorem 4 if Per(C) is primitive, then it is doubly transitive and equal to PΓL(d,r a ) 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(p m ). 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 p s 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 p 2m−1. If Per(C) is doubly transitive, by Theorem 4 it is equal to PΓL(d,r a ), with d≥3. By assuming pα and p∤(d,r a −1), we obtain from Part (i) that a Sylow p-subgroup of Per(C) has order p m , 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 S 5S 5. 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 S 3S 3. 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 Q n and \(Q_{1}^{n}\) are subgroups of \(S_{p^{m}}\) [5, Lemma 2.1]. Note that Q 1=AG(p m ).

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)

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

     
  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 (a 0,…,a n )⟶f, where \(f(x)= \sum_{i=0}^{n} a_{i} x^{i}\) is injective if n<p−1. Thus in Q n , the coefficients of a 0 can take p r different values, and a 1 can take p m−1(p−1) values. For 2≤in, a i can take p values. From these results we have |Q n |=p 2m+n−2(p−1). For \(Q_{1}^{n}\), the coefficients of a 0 can take p m different values, and a i 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∈ℤ n ∖{0} such that σTσ −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. Therefore 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 .

Now we prove Part (iii).

(⊆ part) Let \(h\in N_{p^{m}}(Q_{1}^{n})\) and g=h −1 Th. 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 p m ), from [5, Lemma 3.6] there exists fQ n+1 such that f −1 gf=T. Thus f −1 h −1 Thf=T. The only elements of \(S_{p^{m}}\) which commute with T (a complete cycle of length p m ), are the powers of T. Thus hf=T j for some j. Since Q n+1 is a subgroup of \(S_{p^{m}}\) and 〈T〉≤Q n+1, then hQ n+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 h 1≡1 mod p m−1 and p m−1|h i , for 2≤in. Let gQ n+1 where \(g(x)=\sum_{i=0}^{n+1}g_{i}x^{i}\), with pg 1 and p m−1|g i 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 p m−1|h i , for i≥2 and p m−1|g j for j≥2, any terms in \(\sum_{i=2}^{n}h_{i} (\sum_{j=0}^{n+1}g_{j}x^{j} )^{i}\) involving g j for j≥2 vanish modulo p m . 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 −1 hg in order to prove that it is in \(Q_{1}^{n}\). This is given by As p m−1|g j for j≥2, we have p m−1|b k for k≥2. Furthermore, p m−1|h i for i≥2, and thus Let \(g^{-1}hg(x)=\sum_{m=0}^{n+1}c_{m}x^{m}\), and note that c n+1=b 1 h 1 g n+1+b n+1(h 1 g 1)n+1. Then by replacing the b i 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 h 1≡1 mod p m−1, we have \(h_{1}^{n} \equiv1\ \mathrm {mod}\ p^{m-1}\). In addition, since p m−1|g n+1, we have \(g_{n+1}h_{1}^{n} \equiv g_{n+1}\ \mathrm {mod}\ p^{m}\). Therefore, c n+1=0, and also p m−1|c i for 2≤in. Then we only need to show that c 1≡1 mod p m−1. Since g j ≡0 mod p m−1 for j≥2, h i ≡0 mod p m−1 for i≥2, and b k ≡0 mod p m−1 for k≥2, then c 1b 1 h 1 g 1 mod p m−1. Finally, as \(b_{1}=g_{1}^{-1}\), we have c 1h 1≡1 mod p m−1. □

Lemma 9

Let 1≤n<p−1. If P is a p-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 p m−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 p m−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 of Q 1 and is the unique subgroup of \(S_{p^{m}}\) of order p m+1 which contains T.

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 Q 1, g(x)=b 0+b 1 x with \(b_{0},b_{1} \in\mathbb{Z}_{p^{m}}\) and (b 1,p)=1. It is not difficult to check that the inverse of g in Q 1 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)=a 0+a 1 x with \(a_{0},a_{1} \in\mathbb{Z}_{p^{m}}\), (a 1,p)=1 and a 1≡1 mod p m . 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 Q 1. Now let S be a subgroup of Q 1 of order p m+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 Q 1. 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|=p m , 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 p m ), are the powers of T. Thus yx −1=T j for some j. Since the order of yx −1 is p, the only choices for j are j=p m or j=p m−1. For both choices we get \(S=Q_{1}^{1}\), namely j=p m gives x=y −1 (so \(S=Q_{1}^{1}\)), and j=p m−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

Let p be an odd prime, q=r α a prime power, C a cyclic code over \(\mathbb{F}_{q} \) of length p m , and P a Sylow p-subgroup of Per(C) of order p s such that TP. Then the following hold:
  1. (a)

    If pα and p∤(d,r a −1), then s=m, and P=〈Tif and only if Per(C)=PΓL(d,r a ), d≥3,

     
  2. (b)

    If p≥5, α=1 and r=p, m>1, then Per(C) is an imprimitve group and P is normal of order p s , s>m. If m<sp+m−1, then we have \(P=Q_{1}^{s-m}\).

     
  3. (c)

    If z=1, pα and p∤(d,r a −1), then Per(C) is an imprimitve group and P is normal of order p s p 2m−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 p m+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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