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 p^{m}. Several examples are given to illustrate the results.
Keywords
Permutation groups Transitive groups Doubly transitive groups Non-primitive BCH codes1 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
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
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)≤G≤PΓL(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
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
- (i)
Per(C) is a solvable group of orderpmwithma divisor ofp−1 andC_{p}≤Per(C)≤AGL(1,p), withp≥5. FurthermorePer(C) contains a normal Sylowp-subgroup.
- (ii)
Ifp=q, thenPer(C)=AGL(1,p).
- (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.
- (iv)
Per(C)=M_{23}andqis 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.
- (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 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.
Permutation groups of some BCH codes of length p
q | p | δ | Per(C) | Per(C_{2}) | Per(C_{3}) |
---|---|---|---|---|---|
2 | 17 | 2 | C_{8}⋉C_{17} | S_{17} | S_{17} |
2 | 23 | 3 | M_{23} | M_{23} | M_{23} |
2 | 41 | 2 | C_{20}⋉C_{41} | C_{20}⋉C_{41} | C_{20}⋉C_{41} |
2 | 41 | 3 | C_{20}⋉C_{41} | S_{41} | S_{41} |
2 | 43 | 5 | C_{14}⋉C_{43} | C_{14}⋉C_{43} | C_{14}⋉C_{43} |
2 | 43 | 7 | C_{14}⋉C_{43} | S_{43} | S_{43} |
3 | 13 | 2 | C_{3}⋉C_{13} | C_{3}⋉C_{13} | C_{3}⋉C_{13} |
3 | 13 | 4 | PΓL(3,3) | C_{3}⋉C_{13} | C_{3}⋉C_{13} |
3 | 13 | 5 | C_{3}⋉C_{13} | C_{3}⋉C_{13} | C_{3}⋉C_{13} |
3 | 23 | 3 | C_{11}⋉C_{23} | C_{11}⋉C_{23} | C_{11}⋉C_{23} |
3 | 41 | 5 | C_{8}⋉C_{41} | C_{8}⋉C_{41} | C_{8}⋉C_{41} |
4 | 43 | 9 | C_{7}⋉C_{43} | S_{43} | S_{43} |
5 | 11 | 5 | C_{5}⋉C_{11} | C_{5}⋉C_{11} | C_{5}⋉C_{11} |
11 | 5 | 3 | C_{5} | C_{2}⋉C_{5} | C_{5} |
The following result is obtained by considering the permutation groups of cyclic codes of composite length.
Theorem 4
- (i)
an imprimitive group (in the case thatn=p^{m}, pa prime, the orbit of the subgroup generated by\(T^{p^{m-1}}\)and its conjugate form a complete block system ofPer(C));
or
- (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
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
Proof
By the definition of a cyclic code, we have T∈Per(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 l≤m. 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 l≤s≤p^{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
- (i)
Ifp∤αandp∤(d,r^{a}−1), thenPer(C)=PΓL(d,r^{a}), a≥1, d≥3, if and only if the Sylowp-subgroup ofPer(C) is of orderp^{m}.
- (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 orderp^{s}withm<s≤p^{m−1}+p^{m−2}+⋯+1.
- (iii)
Ifz=1, p∤αandp∤(d,r^{a}−1), thenPer(C) is an imprimitive group which contains a transitive normal Sylowp-subgroup of orderp^{s}, with 2m−1≤s≤p^{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 p∤r^{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 T∈P, 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 p∤α 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_{5}≀S_{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_{3}≀S_{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
- (i)
|Q^{n}|=(p−1)p^{2m+n−2}and\(|Q_{1}^{n}|=p^{m+n}\).
- (ii)
\(AG(p^{m})=N_{S_{p^{m}}}(\langle T\rangle)\).
- (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≤i≤n, 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≤i≤n 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 f∈Q^{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 h∈Q^{n+1}. Hence \(N_{p^{m}}(Q_{1}^{n})\leq Q^{n+1}\).
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 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 ofQ^{1}and is the unique subgroup of\(S_{p^{m}}\)of orderp^{m+1}which containsT.
Proof
Theorem 11
- (a)
Ifp∤αandp∤(d,r^{a}−1), thens=m, andP=〈T〉 if and only ifPer(C)=PΓL(d,r^{a}), d≥3,
- (b)
Ifp≥5, α=1 andr=p, m>1, thenPer(C) is an imprimitve group andPis normal of orderp^{s}, s>m. Ifm<s≤p+m−1, then we have\(P=Q_{1}^{s-m}\).
- (c)
Ifz=1, p∤αandp∤(d,r^{a}−1), thenPer(C) is an imprimitve group andPis normal of orderp^{s}≥p^{2m−1}. Furthermore, if 2m−1<s≤p+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 s≤p+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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