On semiabelian groups

Suppose that 𝐺 and 𝐻 are isoclinic finite groups. If 𝐺 is semiabelian, then 𝐻 is semiabelian.

If a finite nilpotent group 𝐻 is isoclinic to a group 𝐺 satisfying the following two conditions:

1. the Sylow 𝑝-subgroups of 𝐺 are modular for all primes p ≥ 3 ;

2. the Sylow 2-subgroup 𝐺 is quaternion-free, but not isomorphic to a Wilkens group of type C,

If a group 𝐺 is semiabelian, then 𝐺 is monomial.

Let 𝒢 be an isoclinism invariant class of finite groups with the property that if G ∈ G , then every subgroup of 𝐺 is contained in 𝒢 and every homomorphic image of 𝐺 is in 𝒢. If 𝒢 consists of semiabelian groups, then every G ∈ G is monomial.

A finite group 𝐺 is called semiabelian if it has a sequence

such that G i + 1 is a quotient group of G i ⋉ A i with a finite abelian group A i for all i = 0 , … , n − 1 .

Lemma 2.2 (Thompson [20] (see also [4, Corollary 2.7]))

We have the following.

1. Nilpotent groups of class 2 are semiabelian.

2. If 𝐺 is a solvable group whose Sylow 𝑝-subgroups are all abelian, then 𝐺 is semiabelian.

Lemma 2.3 ([4, Theorem 2.3])

A non-trivial finite group 𝐺 is semiabelian if and only if there are an abelian normal subgroup 𝐴 and a proper semiabelian subgroup 𝐻 such that G = A ⁢ H .

Lemma 2.4 ([4, Corollary 2.4])

A non-trivial semiabelian group 𝐺 has an abelian normal subgroup which is not contained in the Frattini subgroup Φ ⁢ ( G ) .

Lemma 2.5 ([4, Theorem 2.8])

If 𝐺 and 𝐻 are semiabelian groups, then the standard wreath product G ≀ H is semiabelian. In particular, the direct product G × H is semiabelian.

Lemma 2.6 ([18, Corollary 1 (b)])

If a group 𝐺 is an abelian extension of a cyclic group, then 𝐺 is semiabelian.

Lemma 2.7 ([18, Lemma 2])

Every quotient of a semiabelian group is semiabelian.

Definition 2.8 (Hall [6])

The groups 𝐺 and 𝐻 are isoclinic if an isomorphism φ : G / Z ⁢ ( G ) → H / Z ⁢ ( H ) induces an isomorphism ψ : G ′ → H ′ satisfying

If 𝐺 and 𝐻 are isoclinic, then we write H ∼ G and we call the pair ( φ , ψ ) an isoclinism.

Proof of Theorem 1.1

Assume G ∼ H . Let ( φ , ψ ) be an isoclinism between 𝐺 and 𝐻. If we define

then by [11, Theorem 4.2], the group 𝐺 is a quotient of H ⅄ ( C / C ′ ) . Hence 𝐺 is a quotient of H × ( C / C ′ ) . If 𝐻 is semiabelian, then H × ( C / C ′ ) is semiabelian by Lemma 2.5. It follows from Lemma 2.3 that 𝐺 is semiabelian. Since isoclinism is an equivalence relation, the reverse implication follows if we exchange the roles of 𝐺 and 𝐻. ∎

A finite 𝑝-group 𝐺 is modular if the subgroup lattice of 𝐺 is a modular lattice, namely, ⟨ A , B ⟩ ∩ C = ⟨ A , B ∩ C ⟩ for subgroups A , B , C of 𝐺 such that A ≤ C .

Lemma 3.2 ([7, Satz 2.2])

A finite 𝑝-group 𝐺 is modular if and only if A ⁢ B = B ⁢ A for all subgroups A , B of 𝐺.

Lemma 3.3 ([1, Theorem 9.23])

A non-Hamiltonian 𝑝-group 𝐺 is modular if and only if it contains an abelian normal subgroup 𝑁 with cyclic quotient G / N and there exists some t ∈ G with G = ⟨ t , N ⟩ such that a t = a 1 + p s for all a ∈ N with a positive integer 𝑠. If p = 2 , then s ≥ 2 .

A 𝑝-group is called a Hamiltonian group if it is non-abelian and all its subgroups are normal subgroups.

Lemma 3.5 (Dedekind (see [8, III.7.12]))

A Hamiltonian 2-group is isomorphic to a direct product of Q 8 and an elementary abelian 2-group.

Every modular 𝑝-group and every Hamiltonian 2-group is semiabelian.

Proof

If 𝐺 is a non-Hamiltonian modular group, then it has an abelian normal subgroup 𝑁 such that G / N is cyclic by Lemma 3.2. Thus, the result follows from Lemma 2.6.

If 𝐺 is Hamiltonian, then by Lemma 3.5, we have G ≅ Q 8 × E with 𝐸 elementary abelian. In particular, as Q 8 is nilpotent of class 2 and 𝐸 is abelian, 𝐺 is nilpotent of class 2. Hence 𝐺 is semiabelian by Lemma 2.2 (i). ∎

A 2-group 𝐺 is called quaternion-free (resp. D 4 -free) if all homomorphic images of every subgroup of 𝐺 do not contain a subgroup isomorphic to Q 8 (resp. D 4 ).

Lemma 3.8 (Iwasawa (see [10, Proposition 1.6]))

A 2-group 𝐺 is modular if and only if 𝐺 is D 4 -free.

Lemma 3.9 (Wilkens (see [10, Theorem 1.7]))

A 2-group 𝐺 is non-modular and quaternion-free if and only if it is one of the following three types of groups:

1. 𝐺 has a maximal abelian normal subgroup 𝑁 with exponent larger than 2 and 𝐺 is a semidirect product N ⋊ ⟨ x ⟩ for some x ∈ G , and if 𝑡 is an involution in ⟨ x ⟩ , then 𝑡 acts on 𝑁 by inversion;

2. there exists a maximal elementary abelian normal subgroup 𝑁 of 𝐺 and x ∈ G such that G = N ⁢ ⟨ x ⟩ , where ⟨ x ⟩ is not normal in 𝐺;

3. there exist a maximal elementary abelian normal subgroup 𝑁 of 𝐺, some x ∈ G , and an involution t ∈ G with [ N , t ] = 1 such that G = ⟨ N , x , t ⟩ . If the order of x ⁢ N is 2 k , then G / N ≅ M 2 k + 1 with k ≥ 3 and x 2 k ≠ 1 . Furthermore, we have [ x 2 k − 1 , N ] = 1 and ⟨ t , x 2 k − 2 ⟩ ≅ D 4 .

If a non-modular and quaternion-free group is a Wilkens group of type A or type B, then it is semiabelian.

Proof

In either case, the quotient group G / N is cyclic, where 𝑁 is as in Lemma 3.9. Hence the proposition follows from Lemma 2.6. ∎

Lemma 4.1 (Janko)

Let G = ⟨ N , x , t ⟩ be a Wilkens group of type C as described in Lemma 3.9. The group 𝐺 satisfies the following additional properties:

1. 𝑁 is the unique maximal normal elementary abelian subgroup 𝐺;

2. ⟨ t , x 2 k − 1 ⟩ ≅ D 4 ;

3. if we define Ω 1 ⁢ ( G ) = ⟨ u ∈ G ∣ u 2 = 1 ⟩ , then

   Ω 1 ⁢ ( G ) = ⟨ x 2 k − 1 , t ⟩ ⁢ N ≅ D 4 × E

   holds for some elementary abelian 2-group 𝐸 and G / Ω 1 ⁢ ( G ) is a cyclic group of order greater than or equal to 4;

4. subgroups of ⟨ x ⟩ of order greater than 4 are not normal in 𝐺.

Let G = ⟨ N , x , t ⟩ be a Wilkens group of type C as in Lemma 3.9. Assume that the unique normal elementary abelian subgroup 𝑁 of 𝐺 is of rank 𝑛 with basis a 1 , … , a n . The following relations hold in 𝐺:

Proof

Most of these relations are easy consequences of Lemma 3.9. We show (4.2) and (4.3). Since ⟨ a , t ⟩ ≅ D 4 by Lemma 4.1 ii, relation (4.2) holds. Also, from G / N ≅ M 2 k + 1 in Lemma 3.9, we have ( x ⁢ N ) t ⁢ N = ( x ⁢ N ) 1 + 2 k − 1 by (1.1), and thus, noting that [ a , N ] = 1 , we obtain (4.3).

Since 𝐺 is an extension of M 2 l + 1 by an elementary abelian subgroup 𝑁, it is determined by the M 2 l + 1 -module structure of 𝑁 with conjugate action. By Lemma 3.9, we have [ N , t ] = 1 , and therefore, the module structure is determined by the action of 𝑥 on 𝑁. ∎

Let G = ⟨ N , x , t ⟩ be a Wilkens group of type C as in Lemma 3.9. We fix an integer 𝑘 satisfying (4.1). Let a = x 2 k − 1 as in (4.2). Then the element m ∈ N in (4.3) must satisfy

Proof

We first show that

for every ℓ ≥ 2 by induction on ℓ. Noting that [ a , N ] = 1 and [ a , x ] = 1 , we have, for ℓ = 1 ,

and, for ℓ = 2 , we obtain

since a 4 = 1 . If equation (4.5) holds for ℓ − 1 , then we have

This proves (4.5).

On the other hand, raising both sides of (4.3) to the 2 k − 1 -th power yields

Applying (4.2) and (4.5) to each side of this equation, we get

and thus

Since order of 𝑎 is 4 by (4.1), we thus conclude (4.4). ∎

Let G = ⟨ N , x , t ⟩ be a Wilkens group of type C as in Lemma 3.9. We fix an integer 𝑘 satisfying (4.1). Let Z = ⟨ x ⟩ . If 𝑀 is a non-trivial indecomposable 𝑍-submodule of 𝑁, then the rank of 𝑀 is equal to 2 k − 1 .

Proof

We may assume that 𝑁 itself is indecomposable. We denote the rank of 𝑁 by 𝑛. Since 𝑁 is indecomposable, we can choose a basis ( a 1 , … , a n ) of 𝑁 so that x ∈ Z acts on 𝑁 by the Jordan block

of size 𝑛 over the field F 2 of 2-elements. We define K n = J n − E n , where E n is the identity matrix. We then claim that, for all ℓ ≥ 2 ,

We prove this by induction on ℓ. Here we drop the subscript 𝑛 for simplicity. We start with the case ℓ = 2 :

by the binomial theorem. If the claim holds for ℓ − 1 , then we compute

This completes the proof of our claim (4.6).

Now we use the additive notation on 𝑁 and revive the subscript 𝑛. By (4.4), the element a 2 is contained in 𝑁 and it is left fixed by the action of 𝑥, and therefore, we may assume a n = a 2 . By (4.6), the condition from Lemma 4.3 amounts to

Since K n is nilpotent, it follows that n ≥ 2 k − 1 ; otherwise, the image is 0. On the other hand, K n 2 k − 1 − 1 + K n 2 k − 1 sends all a j with j > n − 2 k − 1 + 1 to 0. Hence the image of the map coincides with a n if and only if n = 2 k − 1 and the a 1 -coordinate of 𝑚 is non-zero. ∎

Let G = ⟨ N , x , t ⟩ be a Wilkens group of type C. If we take a basis ( a 1 , … , a 2 k − 1 ) of 𝑁 so that 𝑥 acts on 𝑁 by the Jordan block J 2 k − 1 , then the a 1 coordinate of 𝑚 in (4.3) must not be zero.

If G = ⟨ N , x , t ⟩ is a Wilkens group of type C with a fixed integer 𝑘, then 𝐺 is not a semiabelian group.

Proof

Let Z = ⟨ x ⟩ as in Lemma 4.4 and s = 2 k − 1 .

We first prove the theorem in the case where 𝑁 is an indecomposable 𝑍-module of rank 𝑠. Let ( a 1 , … , a s ) be a basis of 𝑁 on which 𝑥 acts as the Jordan block J s . Then we have

These equations yield

Since the Frattini subgroup Φ ⁢ ( G ) is generated by G 2 = ⟨ g 2 ∣ g ∈ G ⟩ and G ′ by [8, III.3.14], we have a 2 , … , a s ∈ Φ ⁢ ( G ) . By (4.4) with m = a 1 , we have

Since Φ ⁢ ( G ) is a normal subgroup of 𝐺, we see a 1 x i ∈ Φ ⁢ ( G ) for i ≥ 2 , and thus a 1 x ∈ Φ ⁢ ( G ) . This implies that a 1 is also contained in Φ ⁢ ( G ) . Therefore, we have ⟨ a ⟩ ⁢ N ≤ Φ ⁢ ( G ) . Now we shall show that ⟨ a ⟩ ⁢ N is the unique maximal abelian normal subgroup of 𝐺. Let 𝐴 be a maximal abelian normal subgroup of 𝐺. Since ⟨ a ⟩ ⁢ N is a normal abelian subgroup, we have ⟨ a ⟩ ⁢ N ≤ A . Lemma 3.9 implies that G / N ≅ M 2 k + 1 , and hence G / ⟨ a ⟩ ⁢ N is an abelian group generated by the cosets of 𝑥 and 𝑡. Since [ t , a ] ≠ 1 , we see t ∉ A . Therefore, if A > ⟨ a ⟩ ⁢ N , then 𝐴 must contain a power of 𝑥 whose order is greater than or equal to 8. This is impossible by Lemma 4.1 iv. We thus conclude A = ⟨ a ⟩ ⁢ N ≤ Φ ⁢ ( G ) . It follows from Lemma 2.4 that 𝐺 cannot be semiabelian.

Now we prove the general case. In this case, we have N = M ⊕ L , where 𝑀 is an indecomposable 𝑍-submodule of rank 𝑠 as in Lemma 4.4 and 𝐿 is a 𝑍-module. The quotient group G / L is isomorphic to the group considered above, and hence is not semiabelian. By Lemma 2.7, 𝐺 is not semiabelian, too. ∎

If G = ⟨ N , x , t ⟩ is a Wilkens group of type C, then ⟨ x 2 k − 1 ⟩ ⁢ N is the unique maximal abelian normal subgroup of 𝐺.

The smallest example of a Wilkens group of type C is

of order 256. The GAP ID of this group is ( 256 , 330 ) ; it is non-semiabelian but monomial.

Let G 2 = G × C 2 . The group G 2 is clearly a Wilkens group of type C. We see that Φ ⁢ ( G 2 ) = Φ ⁢ ( G ) × Φ ⁢ ( C 2 ) = Φ ⁢ ( G ) and thus G 2 is not irreducible in the above sense. This shows that the reverse implication of Lemma 2.4 is not true in general.

When k = 4 , the smallest Wilkens group of type C is of order 2 13 .

If a finite nilpotent group 𝐺 satisfies the following two properties:

1. the Sylow 𝑝-subgroups are modular for all p ≥ 3 ;

2. the Sylow 2-subgroup is quaternion-free, but is not isomorphic to a Wilkens group of type C,

Proof

Since 𝐺 is nilpotent, it is the direct product of its Sylow subgroups. For all p ≥ 3 , the Sylow 𝑝-subgroups are semiabelian by Proposition 3.6. For p = 2 , the 2-Sylow group is semiabelian by Propositions 3.6 and 3.10. Therefore, 𝐺 is semiabelian by Lemma 2.5. ∎

As in the case of monomial groups [5, Theorem 2.6], one may expect a solvable group satisfying (P1) and (P2) to be semiabelian. In fact, we have not found any counterexample so far.

Let 𝐺 be a solvable group. A Carter subgroup of 𝐺 is a nilpotent self-normalizing subgroup of 𝐺.

Assume that 𝐺 is a solvable group but not a nilpotent group. Let K ∞ ⁢ ( G ) be the intersection of the groups of the lower central series of 𝐺. If K ∞ ⁢ ( G ) is abelian and a Carter subgroup C ⁢ ( G ) of 𝐺 is semiabelian, then 𝐺 is semiabelian.

Proof

Since 𝐺 is not nilpotent, we have K ∞ ⁢ ( G ) ≠ 1 . By [8, III.2.5 d)], the quotient G / K ∞ ⁢ ( G ) is nilpotent. The subgroup C ⁢ ( G ) ⁢ K ∞ ⁢ ( G ) is self-normalizing [3], and thus C ⁢ ( G ) ⁢ K ∞ ⁢ ( G ) = G by [8, III.2.3 c)]. Since C ⁢ ( G ) is semiabelian by assumption, Lemma 2.3 implies that 𝐺 is semiabelian. ∎

In this example, we apply our results to solvable isoclinism classes whose stem group is of order less than 100 to show how our results work. Here a group 𝐺 is called a stem group if it satisfies Z ⁢ ( G ) ≤ G ′ . Recall that every isoclinism class has a stem group as a class representative.

If | G | has no prime divisor ℓ such that ℓ 3 ∣ | G | , then every Sylow 𝑝-subgroup is abelian, and thus 𝐺 is semiabelian by Lemma 2.2. Hence we skip these groups.

The non-semiabelian stem groups of order up to 100 are

where the numbers are the GAP IDs of the groups. This list is verified independently of our results.

For the rest of the solvable isoclinism classes, we first check conditions (P1) and (P2) in Theorem 1.2 for nilpotent stem groups. If it fails to satisfy them, then we check the conditions for isoclinic groups up to order 128. For non-nilpotent groups, we first check whether K ∞ ⁢ ( G ) is abelian or not, and then the conditions in Proposition 5.4.

Among these classes, the following stem groups are those we cannot prove to be semiabelian from our results:

In general, our results works well for the above cases. Note that verifying the condition for a stem group implies the semiabelian property of the infinitely many groups in the same isoclinism class. On the other hand, the limit of our results lies partly in the fact that we can only deal with semiabelian groups whose subgroups are also semiabelian. For example, the group ( 96 , 204 ) = C 2 3 ⋉ A 4 is semiabelian, but has a subgroup isomorphic to SL ⁢ ( 2 , 3 ) , which is not semiabelian. As we mentioned before, this fact makes it difficult to characterize semiabelian groups more precisely.

The computation of this example is done by Magma [2].

Proof of Theorem 1.4

Assume that G ∈ G is semiabelian. There exists a stem group 𝐻 such that H ∼ G , which is also semiabelian by Theorem 1.1. If 𝐻 is abelian, then 𝐺 is abelian, and therefore 𝐺 is monomial. We may assume that 𝐻 is non-abelian. By Lemma 2.4, there exists an abelian normal subgroup of 𝑁 of 𝐻 such that N ⊄ Φ ⁢ ( H ) . On the other hand, it follows from [8, III.3.12] that H ′ ∩ Z ⁢ ( H ) ≤ Φ ⁢ ( H ) . Since 𝐻 is a stem group, this implies Φ ⁢ ( H ) ≥ Z ⁢ ( H ) . Hence we have N ⊄ Z ⁢ ( H ) . By combining our assumption on 𝒢 and [8, V.18.3], this implies that 𝐻 is monomial. Since G ∼ H , the group 𝐺 is also monomial. ∎