GASSMAN EQUIVALENT SUBGROUPS 1 Introduction

Int. J. Contemp. Math. Sci., Vol. 1, 2006, no. 11, 543 - 556

GASSMAN EQUIVALENT SUBGROUPS Mario Pineda-Ruelas Departamento de Matem´aticas, Universidad Aut´onoma Metropolitana-I, cp 09340 M´exico D.F., M´exico, [email protected] Gabriel D.Villa-Salvador Departamento de Control Autom´atico, Centro de Investigaci´on y de Estudios Avanzados del IPN, cp 07000 M´exico D.F., M´exico, [email protected] Abstract In this paper, we construct one infinite family of groups where Gassman-equivalence does not imply conjugation. This family is a particular case of Theorem 1 in [3], but we get as a bonus of our construction that it is explicit and can be used to construct two extensions K/Q and K  /Q with the same Dedekind zeta function ζK = ζK  with K  K  as in [12].

Mathematics Subject Classification: 11R42, 20B35, 20E45. Keywords: Algebraic number fields, arithmetically equivalent fields, solitary fields, Gassman equivalent subgroups.

1

Introduction

One of the most interesting invariants associated to a number field K is its Dedekind zeta function ζK . If two number fields K, K  are isomorphic, then ζK = ζK  . However, the function ζK does not determine the field K. There exist examples of nonisomorphic number fields K and K  such that ζK = ζK  ([5], [10]). The fields K and K  are called arithmetically equivalent if ζK = ζK  . The field K is called solitary if for any field K  such that ζK = ζK  , imply that K and K  are isomorphic.

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An interesting problem is to study when two nonisomorphic number fields have the same zeta function. Possibly the first approach to this problem is due to Kronecker [9], who characterized number fields K by means of the factorization type of the rational primes in K. An important step was given in 1916 by Bauer [1], who proved that if K/Q and K  /Q are normal extensions and the factorization type of any rational prime in the ring of integers OK and OK  is the same, then K = K  . The first examples of nonsolitary number fields were discovered by F. Gassman [5]. He found two nonisomorphic number fields K and K  of degree 180 over Q such that ζK = ζK  . These fields are the fixed fields of the subgroups H = {id, (1, 2)(3, 4), (1, 3)(2, 4), (1, 4)(2, 3)} and H  = {id, (1, 2)(3, 4), (1, 2)(5, 6), (3, 4)(5, 6)} of a Galois extension N/Q where Gal(N/Q)  S6 . In 1977 Perlis [10] found two infinite families of nonsolitary fields. The first one was built by using the cohomology of split group extensions. For the second family Perlis uses permutation representations and K, K  appear as subfields of an extension N/Q such that Gal(N/Q)  Sn . Perlis also proves that if [K : Q] ≤ 6, then K is solitary and gave two examples of nonsolitary fields, one of degree 7 and the other of degree 8. Let G be a finite group. Two subgroups H, H  of G are called Gassmanequivalent, or just G-equivalent, if every x ∈ G satisfies |xG ∩ H| = |xG ∩ H  |, where xG denotes the conjugacy class of x in G. If H and H  are G-equivalent, then [G : H] = [G : H  ]. On the other hand, if H, H  are conjugate, then they are G-equivalent. There exist examples of groups containing G-equivalent nonconjugate subgroups [7]. Let K, K  be two finite extensions of Q and let N/Q be a common normal extension of K, K  with H = Gal(N/K) and H  = Gal(N/K  ). We know from Galois theory that H and H  are conjugate in G if and only if K and K  are isomorphic. Perlis [10] showed that H and H  are G-equivalent if and only if ζK = ζK  . Therefore we have that if two number fields K, K  satisfy ζK = ζK  , then they are isomorphic if the G-equivalence of H and H  implies that H and H  are conjugate in G. Thus, we have a translation from an analytic problem to a problem in group theory. More precisely, if K, K  are number fields such that ζK = ζK  then, under what specific condition are K and K  isomorphic? or equivalently, if the corresponding subgroups H, H  are G-equivalent then, when are H and H  conjugate? From this point we could establish a more general problem concerning group theory: Let G be a finite group and let H, H  be subgroups of G such that H and H  are G-equivalent. Then, when are H and H  conjugate?

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Part of the development of this problem is as follows. Feit [4] considered the case [G : H] = [G : H  ] = p, p prime. Using the classification of finite simple groups, he described the cases when H, H  are nonconjugate. Guralnick [6] solved the problem in the case [G : H] = [G : H  ] = p2 . He also gave an example of a group G of order p5 containing two subgroups H, H  of order p3 , which are G-equivalent and nonconjugate. Later, Guralnick and Wales [7] studied the case [G : H] = [G : H  ] = pq with p, q different primes and they proved that for 18 = n ≤ 40, there exist groups containing nonconjugate subgroups H, H  of index n if and only if n = 1, 2, 3, 4, 5, 6, 9, 10, 17, 19, 23, 25, 29, 37, 38. The case n = 18 was not settled. A natural way of pursuing this line of investigation, is to study the case [G : H] = [G : H  ] = p2 q, or more generally [G : H] = [G : H  ] = pn q, where p, q are different primes and n ≥ 2. Rzedowski-Villa [13] used a different method from [6] and proved that if a transitive group G is of degree 9 or 10, then G-equivalence implies conjugation. The corresponding result in Galois theory is that if [K : Q] = 9 or 10, then K is solitary. In the search of examples, Rzedowski-Villa proved that if G is the Galois group of some extension K of degree n over Q, where the G-equivalence does not imply conjugation, then for each m there exist examples of degree mn. In [3] B. de Smit and H.J. Lenstra Jr. proved a very smart result settling most of the issue for solvable groups. They consider the set R of n ∈ N such that there exists a finite group G of degree n which contains two Gequivalent and nonconjugate subgroups of index n. Let S = {n ∈ R : the corresponding group G is solvable}. Let P be the set of rational primes. The following sets are contained in R: T = U=

 km(q d − 1) q−1  kq(q 2 − 1) 120

 : q = pr , p ∈ P, d ≥ 2, m | q − 1, m + d ≥ 4, k ≥ 1, r ≥ 1 ,

 : q = pr , p ∈ P, gcd(q, 6) = 1, q ≡ ±1(mod 5), k ≥ 1, r ≥ 1 ,

where T and U are the orders of certain groups constructed in [3]. If n ≤ 2000, then the set R ∩ (S ∪ T ∪ U) contains 1303 elements such that 1083 of them are in S. In particular 18 ∈ S. In [3] are classified all the degrees ≤ 2000 for which there exist examples where G-equivalence does not imply conjugation. Therefore, the smallest integer for which it is not known if the G-equivalence implies conjugation is n = 2001. One of the families that we give in this work is an explicit particular case of Theorem 1 in [3]. Furthermore W. Bosma and B. de Smit [2] have obtained a list of all transitive groups of degree n with n ≤ 15, where G-equivalence does not imply conjugation.

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In this paper we explicitly construct one infinite family of groups where G-equivalence does not imply conjugation.

2

Preliminary Notes

If a group G acts on a set X = ∅, φ : G → SX is the corresponding permutation representation of G and StabG (i) denotes the stabilizer of i ∈ X, then we have Theorem2.1. Let G be a finite group and let H ≤ G be such that [G : H g = {id}, then G is isomorphic to some transitive subgroup H] = n. If g∈G

of Sn . For imprimitive groups we have: Theorem 2.2. Let G be transitive group on X. Then G is imprimitive if and only if for i ∈ X there exists Z < G such that StabG (i) < Z < G. That is, the stabilizers of G are not maximal subgroups of G. Proof. See [15] page 14. Theorem 2.2 is important, among other reasons, because it provides a method to produce blocks of certain size in an imprimitive group. More precisely: Proposition 2.3. Let G be an imprimitive group and let i ∈ X. If Z < G is such that StabG (i) < Z < G, then |iZ | = [Z : StabG (i)], where iZ is the block iZ = {ig : g ∈ Z}. The imprimitive groups are of particular interest to us since they appear as subgroups of wreath products. Theorem 2.4. Let G be an imprimitive group of degree n and let Δ be a block of length m. If n = mk, then G ≤ Sm Sk . Proof. See [8], Satz 1.2, page 146. If G = Sm St , then G is transitive of degree n = mt and G has t-blocks of length m. The description of G that we use here is Sm St = (Sm × Sm × · · · × Sm )  St . The product in G is given by:  (σ0, σ1, . . . , σt−1, )(σ0 , σ1 , . . . , σt−1 , ) = (τ0, τ1 , . . . , τt−1 , ),

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where τi = σi σ −1 (i) . Let X = {0, 1, . . . , n − 1}. If i ∈ X, then there exist unique integers q, s such that i = mq + s,

0 ≤ s ≤ m − 1,

0 ≤ q ≤ t − 1.

If ϕ = (σ0, σ1, . . . , σt−1 , ) ∈ G, then, for i ∈ X, the action of G on X is given by iϕ = m(q) + σq (s). It is clear that iϕ = i if and only if  ∈ StabSt (q) and σq ∈ StabSm (s). Let φ : G −→ St be defined by φ((σ0, . . . , σt−1, )) = . Then φ is a group homomorphism and φ(G) is a transitive subgroup of St . If T = Sm × · · · × Sm    t−f actors and N = ker(φ), then N = G T. A fact that will be very useful in this work is the following: Theorem 2.5. Let G be an imprimitive group contained in Sm St . Let Δi ∈ B, where B = {Δ0 , . . . , Δt−1} is an imprimitivity set of G. If Δi is the i-th block of G and Z = StabG (Δi) = {g ∈ G : Δgi = Δi }, then ker(φ) = N ≤ Z and [G : Z] = [φ(G) : Stabφ(G) (i)] = t. Proof. It is clear that N = {g ∈ G : Δgj = Δj , 0 ≤ j ≤ t − 1} ≤ Z since Z = φ−1 (Stabφ(G) (i)) and φ(G) is a transitive group of degree t. The last statement follows. Let K be a number field and let OK be the ring of integers of K. If I ⊆ OK is a nonzero ideal of OK , then OK /I is a finite set. Let N(I) = |OK /I| be the absolute norm of the ideal I. We define the Dedekind zeta function of K as ζK (s) =



1 . N(I)s I=0

In case that K = Q we have O = Z and {|Z/I| : I = 0} = N. In this case ∞

1 ζ (s) = ns n=1

is the well known Riemann zeta function. Let G be a finite group and let H, H  be subgroups of G. We say that H and H  are G-equivalent in G (or Gassman-equivalent) if for x ∈ G we have |xG ∩ H| = |xG ∩ H  |. We denote the G-equivalence of H and H  by writing G H ∼ H  . The properties about G-equivalence that we will use in this work are summarized in the following

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Theorem 2.6. Let G be a finite group and let H, H  be subgroups of G. G

1. If H and H  are conjugate, then H ∼ H  . G

2. If H ∼ H  , then o(H) = o(H  ). G

G

3. If H ∼ H  and N  G, then H ∩ N ∼ H  ∩ N. G

G

4. If G ≤ G1 and H ∼ H  , then H ∼1 H  . Proof. It follows from the definition of G-equivalence. The relationship between G-equivalence and the zeta function, is given in the following result [10]: Theorem 2.7. Let K, K  be number fields and let L be a common normal extension of K and K . If G = Gal(L/Q), H = Gal(L/K), H  = Gal(L/K  ), then the following conditions are equivalent: 1. ζK (s) = ζK  (s). G

2. H ∼ H  . If any of these conditions holds, then [K : Q] = [K  : Q] = n, the discriminants of K and K  are the same, K and K  have the same normal closure, the unit groups of K and K  are isomorphic and the number of real (complex) embeddings of K and K  is the same. Theorem 2.7 allows one to translate arithmetic equivalence problems to G-equivalence problems. Our interest in conditions (1) and (2) of Theorem 2.7 stems from a number theoretic case. We know from Galois theory that two number fields K and K  are isomorphic if and only if they have the same normal closure N and the subgroups H = Gal(N/K), H  = Gal(N/K  ) are conjugate in G = Gal(N/Q). Hence, by Theorem 2.7 and Galois theory, if two fields K and K  are arithmetically equivalent, they are isomorphic in case that H, H  are conjugate in G = Gal(N/Q). The following are well known examples of solitary fields: 1. K/Q is normal. 2. [K : Q] ≤ 6 [10]. The following result gives several families of solitary fields [10]: Theorem 2.8. Let K be a number field and let N be the normal closure of K/Q. Let G = Gal(N/Q), H = Gal(N/K) and [G : H] = n. Then K is solitary if some of the following statements hold:

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1. H is cyclic. 2. G = Sn . 3. n = p, prime, and G = Ap . 4. n = p + 1, p prime, and G = Ap+1. 5. o(H) = pi , p prime and gcd (p, n) = 1. 6. n = p + 1, p prime, o(H) ≡ 0 mod (p), and o(H) ≡ 0 mod (n). 7. G is solvable and gcd(o(H), n) = 1.

3

Main Results

In this section we construct two infinite families of imprimitive groups in which G-equivalence does not imply conjugation. The first of this families appear in [11] and we only sketch the proofs of its consequences.

3.1

The family N  D8

Let p ≥ 3 be a prime number, N = x, y an elementary abelian group of order p2 , D8 = a, b the dihedral group of order 8, where a2 = b4 = (ab)2 = 1 and ba = b−1 . We define G = N  D8 with the following action of D8 on N: θa(x) = xa = x−1, θa(y) = y a = y −1, θb (x) = xb = y, θb (y) = y b = x. We have the following relations in G: xp = y p = (xy)p = 1, [a, x] = x2 , [a, y] = y 2, where [a, x] = a−1x−1 ax denotes the commutator of a and x. Consider the subgroups H = x, a and H  = y, ax. These groups are of order 2p and they are G-equivalent and nonconjugate in G. Some properties of Z(G), G , CG (x) and CG (a): Theorem 3.1. The group G = N  D8 satisfies: 1. Z(G) = [a, b]. 2. G = N, Z(G). 3. CG (x) = N, Z(G). 4. CG (a) = a, Z(G).

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5. H



H b = {id}.

Proof. See [11]. i

i

Note that H = {xi , ax : 0 ≤ i ≤ p − 1} and H  = {y i , axy : 0 ≤ i ≤ p − 1}. From Theorem 3.1 (3) we have that xi , x−i , y i , y −i are the conjugate elements of xi in G. That is [G : CG (xi) ] = 22 . Corollary 3.2. Let H = x, a and H  = y, ax . Then H and H  are G-equivalent and nonconjugate subgroups.  H g . Each element of H  is conjugate to some eleProof. Choose h ∈ K = g∈G ment of H. Therefore H  ⊆ K and |hG H| = |hG H  | = 0. If h ∈ K, then h = hg1 for some h1 ∈ H and some g ∈ G. We have hG = (hg1 )G = hG 1 . If h1 = xi, then:      i G i −i i −i G H| = |(x H| = |{x H  |. ) , x , y , y } H| = |h |hG H| = |hG 1 1 i

If h1 = ax , then   i i H| = |{ax | 0 ≤ i ≤ p − 1}| = p = |{axy | 0 ≤ i ≤ p − 1}| = |hG H  |. |hG 1 1 G

Therefore H ∼ H  . Suppose that H g = H  for some g ∈ G. That is, xg , ag  = y, ax . Then i i −1 ag = axy for some i. It follows that a = axy g and xy ig −1 ∈ CG (a) = i i {id, a, b2, ab2}. If xy ig −1 = a then g = axy i. Thus H g = xaxy , aaxy  = H  . However, i i i xaxy = (x−1 )xy = (x−1)y = x−1 . Therefore x, y ∈ H  and H  should contain a subgroup of order p2 which is absurd. This proves the result in case that xy ig −1 = a. The other three cases are similar. Corollary 3.3. The group G is imprimitive of degree [G : H] = 22 p. Furthermore, G ≤ Sp S22 . Proof. It follows of Theorems 3.1 (5) and 2.5.

3.2

The family G = P  Cn

We construct another group G containing two subgroups that are G-equivalent and nonconjugate. We use the classification of groups of order p4 and give a semidirect product with a cyclic group of order n, where n is even and relatively prime to p.

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The only p-group P of order p4 such that exp(P ) = p,

o(Z(P )) = p2 ,

[P : F rat(P )] = p3

is P = x × E(p3 ), where E(p3 ) = a, b with ap = bp = [a, b]p = id, [a, b] ∈ Z(E(p3 )) and E(p3 ) is a nonabelian maximal subgroup of P of order p3 ([14], page 98). Lemma 3.4. Let T be a nonabelian group of order p3 . Then T  = Z(T ) = F rat(T ). Corollary 3.5. The group P satisfies Z(P ) = x, [a, b], F rat(P ) = [a, b]. Let Cn = g be the cyclic group of order n with gcd(n, p) = 1 and n even. Consider the group G = P  g. The group G has order np4 and g defines an automorphism φg : P → P given by φg (a) = ag = b, φg (b) = bg = a, φg (x) = xg = x[a, b]−1. It is clear that [a, b]−1 = id since p ≥ 3, φg ([a, b]) = [b, a] and φg is an automorphism of P of order 2. We consider the following subgroups of G: H = x, a, H g = x[a, b]−1, b. We have o(H) = o(H g ) = p2 . Lemma 3.6. The group G = P  g satisfies the following properties: 1. CG (P ) = CP (P ), g 2 . 2. CG (x) = P, g 2 . 3. NG (H) = NP (H), g 2   G. 4. CG (a) = CP (a), g 2 . 5. g 2 ∈ Z(G)  CG (P ). 6. H ∩ H g = {id}. n Proof. (1) First notice that o(g 2 ) = , Z(P ) = CP (P ) and CP (P ), g 2  ⊆ 2 n 2 2 CG (P ). Since o(CP (P ), g ) = p , we have 2 [G : CP (P ), g 2 ] = [G : CG (P )][CG(P ) : CP (P ), g 2 ] = 2p2 .

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On the other hand, since a, b, g ∈ CG (P ), it follows that [G : CG (P )] ≥ 2p2 . Thus 2p2 ≤ [G : CG (P )] ≤ [G : CP (P ), g 2 ] = 2p2 . Hence CG (P ) = CP (P ), g 2 . (2) Since x ∈ Z(P ) and x = id we have P, g 2  ⊆ CG (x). Also o(P, g 2 ) = n 4 p , that is, P, g 2  is a maximal subgroup of G of index 2 and P, g 2  ⊆ 2 CG (x) ⊆ G. Since g ∈ CG (x), we have P, g 2  = CG (x). n (3) It is clear that NP (H) = x, y, a. Therefore o(NP (H), g 2 ) = p3 and 2 NP (H), g 2  ⊆ NG (H). We have [G : NP (H), g 2 ] = [G : NG (H)][NG(H) : NP (H), g 2 ] = 2p. Since b, g ∈ NG (H) it follows that [G : NG (H)] ≥ 2p. Therefore 2p ≤ [G : NG (H)] ≤ [G : NP (H), g 2 ] = 2p. Thus, NG (H) = NP (H), g 2 . (4) We have a ∈ Z(P ). Since Z(P ), a ⊆ CP (a)  P, we have o(CP (a)) = n 3 p . Therefore o(CP (a), g 2) = p3 . Hence 2 [G : CP (a), g 2] = [G : CG (a)][CG(a) : CP (a), g 2] = 2p. Since b, g ∈ CG (a), it follows as before that CG (a) = CP (a), g 2. (5) The statements Z(G) ⊆ CG (P ) and x ∈ CG (P ) are clear. Let y = [a, b] = a−1 b−1 ab. Since xg = xy −1 = x we have x ∈ Z(G). Therefore Z(G)  CG (P ). (6) We have that P = x × a, b = xg  × a, b. If h ∈ H ∩ H g , then h = xi aj = xk al bm , that is, in the expression of h does not appear b. On the other hand, since h ∈ H g , necessarily h is of the form h = (xg )t ∈ H. Therefore xg ∈ H, which is absurd. It follows that H ∩ H g = {id}. We will generalize part (6) of Lemma 3.6. For this end, it is useful to know some of the decompositions of P as direct product of certain subgroups. We have P = x × a, b = xg  × a, b, which are evident. On the other hand, since a ∈ Z(P ), Z(P ), a is a subgroup of order p3 contained in CP (a). Therefore CP (a) = NP (H). We have obtained n in Lemma 3.6(3) that o(NG (H)) = p3 . That is, [G : NG (H)] = 2p. This 2 means that H has 2p conjugates in G. Since o(NP (H)) = p3 , then H has p conjugates in P . Since P  G, we obtain that P contains two conjugacy classes of H. Representatives of each class are H and H g respectively. Let L1 = {H, H ω1 , . . . , H ωp−1 }

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be the conjugacy class of H in P , where ωi are certain elements of P giving the conjugates of H in P . Let us consider the set L2 = {H g , H gω1 , . . . , H gωp−1 }. Note that H gωi = xg , bωi . Corollary 3.7. Let P = x × a, b be as in Corollary 3.5. Then, for α, β ∈ P we have that aαbβ ∈ Z(P ) = x, y. If 1 ≤ i, j ≤ p − 1, then Corollary 3.7 holds substituting a, b by ai , bj respectively. We will see that H ωi and H gωj are nonconjugate in P . Suppose that, for some α ∈ P , we have (H ωi )α = H gωj . Then aωiα = (xs )g (bt)ωj ∈ H gωj . Therefore aωi α (b−t )ωj = (xs )g ∈ Z(P ), which is impossible by Corollary 3.7. It is clear that if H gωi = H gωj , then H ωi = H ωj . Thus we have: Corollary 3.8. The set L2 = {H g , H gω1 , . . . , H gωp−1 } is the conjugacy class of H g in P . Corollary 3.9. We have that o(a, aωi ) = p2 and o(a, agωi ) = p3 for ωi and gωi with 1 ≤ i ≤ p − 1. The second part of Corollary 3.9 gives the decomposition P = xg × a, bω  since a, b = a, bω . Now let us see that H ∩ H gωi = {id}. Let h ∈ H ∩ H gωi . Using the decomposition P = xg  × a, bωi , we have that there exist j, k, l, m, r, s ∈ Z such that h = (xg )j (bωi )k = (xg )l am (bωi )r . The later equality implies that m ≡ 0 (mod p) and therefore a does not appear in the expression of h. For this reason, since h ∈ H we have that h = xs . That is, xs = (xg )l (bωi )r . Thus s ≡ l ≡ r ≡ 0 (mod p). Therefore h = id. The next result will be very useful for the study of H. Lemma 3.10. Let m ∈ Z. Then b−k am bk ∈ H if and only if km ≡ 0 (mod p). Proof. Since am bk = bk amy km we have b−k ambk = b−k bk am y km = am y km ∈ H if and only if km ≡ 0 (mod p). As a consequence of the previous lemma, we have the next result which gives one of the most important properties of H. Corollary 3.11. If h ∈ H, then hG ∩ H = {h}.

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Proof. Let us consider G = x, y, a, b, g. Let t ∈ G, h ∈ H and i, r, j, k, l, m,  ∈ Z be such that t = xi y r aj bk g  , h = xl am . Then i r aj bk g 

ht = (xl)x y

i r aj bk g 

(am )x y









= (xl )g (b−k )g (am )g (bk )g .

k

If  is even, then ht = xl(am )b . In this case, by Lemma 3.10, we have that ht ∈ H if and only if km ≡ 0 (mod p). In any case ht = h. The case  odd is similar. Let us consider the subgroup H  = x, b. The following result describes the most important properties of H  . Lemma 3.12. The subgroup H  satisfies: 1. o(H  ) = p2 . 2. H and H  are nonconjugate. 3. NP (H  ) = NP (H g ). 4. If A = i H ωi , B = i H gωi , C = H H g , then H   A ∪ B ∪ C. Proof. (1) It is immediate. (2) It suffices to show that H  ∈ L1 L2 . First we note that H  = H, H g . If H  ∈ L1 , then, for some i, we have that b ∈ H ωi , which is not possible since o(H ωi ) = p2 . Using the same argument we can see that H  ∈ L2 . −1 (3) If H   P , then yb = ba = aba−1 ∈ H . Therefore y ∈ H  . It follows that o(H  ) = p3 , which is absurd. Therefore H   P. Since H  ⊆ NP (H g ) we obtain (3). (4) From the definition of H  and from (2) we have H  ∩H = H  ∩H ωi = x. Note that |H  ∩ H g | = |H  ∩ H gωi | = p, for 1 ≤ i ≤ p − 1. On the other hand, for i = j, we have (H  ∩H gωi )∩(H  ∩H gωj ) = H  ∩(H gωi ∩H gωj ) = H  ∩xg  = H  ∩xy −1  = {id}. Using the last equality and counting the intersections H  ∩ H ωi and H  ∩ H gωj we obtain (4). Lemma 3.13. Each element of H  is conjugate to one and only one element of H. Proof. First, we show that each element of H  is conjugate to some element of H. Let xi bj be any element of H  . If i = 0 or j = 0 the result is evident. Without loss of generality we may assume that i, j = 0. Using (4) of Lemma

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Gassman equivalent subgroups

3.12 we have that for some ωt , xi bj ∈ H gωt = xg , bωt . Therefore there exist l, k ∈ Z such that xibj = (xg )l (bωt )k . Let h = xlak ∈ H. Therefore hgωt = (xl )gωt (ak )gωt = (xg )l (bωt )k = xi bj . Suppose that some element of H  is conjugate to two different elements of H. −1 Let xi bj = hf = hf11 , where h, h1 ∈ H and h = h1. Then hf f1 = h1 = h. It follows that |hG ∩ H| ≥ 2, which contradicts Corollary 3.11. Corollary 3.14. If h ∈ H, then |hG ∩ H  | = 1.  H g . If h ∈ L, then |hG ∩ H  | = 1. Corollary 3.15. Let L = g∈G

We have proved the following: Theorem 3.16. Let G = P  g, H = x, a, H  = x, b defined as in G Lemmas 3.6 and 3.12. Then H and H  are nonconjugate and H ∼ H  .  From the fact that H H g = {id} it follows that H g = {id}. We have g∈G

that G is a transitive subgroup contained in S[G:H] = Snp2 . Moreover, since G [G:H] acts transitively on X = {Hxi }i=1 and H ⊂ P , the set H P = { Hz : z ∈ P } is a block of G. It is clear that |H P | = [P : H] = p2 . Therefore, G has blocks of length p2 and G ⊆ Sp2 Sn (Theorem 2.4). Thus we have the following result: Corollary 3.17. The group G = P  g is imprimitive of degree np2 contained in Sp2 Sn .

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