On a conjecture by Haipeng Qu

Theorem 1.1.

Let G be a finite non-elementary abelian 2-group of order 2n, where n≥3. Then

Lemma 1.2.

Let G be a finite 2-group.

1. If G is extraspecial, then |G|=22⁢r+1 for some positive integer r and either G≅D8*r or G≅Q8*D8*(r-1).

2. If G is almost extraspecial, then |G|=22⁢r+2 for some positive integer r and G≅D8*r*C4.

3. If G is generalized extraspecial, then either G≅E×A or G≅(E*C4)×A, where E is an extraspecial 2 -group and A is an elementary abelian 2 -group.

Lemma 1.3.

Let G be a finite 2-group of order 2n and let M be a normal subgroup of order 2 of G. Then

Lemma 1.4.

Let G be a finite non-elementary abelian 2-group of order 2n, n≥3, and let L1⁢(G) be the set of cyclic subgroups of G. Then

Theorem 2.1.

Let A and B be two finite groups. Then every subgroup H of the direct product A×B is completely determined by a quintuple (A1,A2,B1,B2,φ), where A1⊴A2≤A, B1⊴B2≤B and φ:A2/A1→B2/B1 is an isomorphism, more precisely

Moreover, we have |H|=|A1|⁢|B2|=|A2|⁢|B1|.

Lemma 2.2.

Let G be a finite 2-group of order 2n and let M be a normal subgroup of order 2r of G. Then

Proof.

Take a chief series of G containing M and use induction on r. ∎

Lemma 2.3.

Let G be a finite abelian 2-group of order 2n, n≥3. If G is non-elementary abelian, then

Proof.

Since G is not elementary abelian, we infer that it has a subgroup M of order 2n-2 such that G/M≅C4. Then Lemma 2.2 leads to

and so it suffices to prove that

where G1=C4×C2. By Theorem 2.1, we know that a subgroup of order 2k of G1×C2n-3 is completely determined by a quintuple (A1,A2,B1,B2,φ), where A1⊴A2≤G1, B1⊴B2≤C2n-3, φ:A2/A1→B2/B1 is an isomorphism, and |A2|⁢|B1|=2k. Note that φ can be chosen in a unique way for the elementary abelian sections of orders 1 and 2 of G1, and in 6=|Aut⁢(C22)| ways for the elementary abelian sections of order 4 of G1. We distinguish the following cases:

Case (a): A2=1. Then A1=1 and B1=B2 is one of the (n-3k)2 subgroups of order 2k of C2n-3. Clearly, these determine (n-3k)2 distinct subgroups of the group G1×C2n-3.

Case (b): A2 is one of the three subgroups of order 2 of G1. Then B1 is of order 2k-1 and can be chosen in (n-3k-1)2 ways. If A1=A2, then we have B2=B1, while if A1=1, then B2 is one of the 2n-k-2-1 subgroups of order 2k of C2n-3 containing B1. So, in this case we have

distinct subgroups of G1×C2n-3.

Case (c): A2 is one of the two cyclic subgroups of order 4 of G1. Then B1 is of order 2k-2 and can be chosen in (n-3k-2)2 ways. If A1=A2, then we have B2=B1, while if |A1|=2, then B2 is one of the 2n-k-1-1 subgroups of order 2k-1 of C2n-3 containing B1. So, in this case we have

distinct subgroups of G1×C2n-3.

Case (d): A2 is the unique subgroup isomorphic to C22 of G1. Again, B1 is of order 2k-2 and can be chosen in (n-3k-2)2 ways. If A1=A2, then we have B2=B1; if |A1|=2, then B2 is one of the 2n-k-1-1 subgroups of order 2k-1 of C2n-3 containing B1; if A1=1, then B2 is one of the (n-k-12)2 subgroups of order 2k of C2n-3 containing B1. One obtains

distinct subgroups of G1×C2n-3.

Case (e): A2=G1. In this case B1 is of order 2k-3 and can be chosen in (n-3k-3)2 ways. If A1=A2, then B2=B1; if |A1|=4, then B2 is one of the 2n-k-1 subgroups of order 2k-2 of C2n-3 containing B1; if A1 is the unique subgroup of order 2 of G1 such that A2/A1≅C22, then B2 is one of the (n-k2)2 subgroups of order 2k-1 of C2n-3 containing B1. One obtains

distinct subgroups of G1×C2n-3.

By summing all the above quantities, we get

A similar computation leads to

It is now clear that the inequalities in (2.1) are true, completing the proof. ∎

Lemma 2.4.

If G is an (almost) extraspecial 2-group, then L⁢(G) consists of:

1. the trivial subgroup.

2. all subgroups containing Φ⁢(G); moreover, these are the normal non-trivial subgroups of G.

3. all complements of Φ⁢(G) in the elementary abelian subgroups of order ≥4 containing Φ⁢(G); moreover,

   1. Φ⁢(G) has 2i complements in an elementary abelian subgroup of order 2i+1 containing Φ⁢(G),

   2. two non-normal subgroups H and K of G are conjugate if and only if H⁢Φ⁢(G)=K⁢Φ⁢(G),

   3. given two non-normal subgroups H and K of G, if Hx⊆K for some x∈[G/NG⁢(H)], then x is the unique element of [G/NG⁢(H)] with this property.

Corollary 2.5.

If G is an extraspecial 2-group of order 22⁢r+1, then

while if G is an almost extraspecial 2-group of order 22⁢r+2, then

Examples.

(a) For G=D8*D8 we obtain e2⁢(G)=9 and e3⁢(G)=6, implying that

(b) For G=Q8*D8 we obtain e2⁢(G)=5 and e3⁢(G)=0, implying that

(c) For G=D8*C4 we obtain e2⁢(G)=3, and so

Remark.

As we have seen above, the subgroup structure of an (almost) extraspecial 2-group G depends on its elementary abelian subgroups containing the subgroup Φ⁢(G)=〈x〉. We remark that these are the totally singular subspaces of G/Φ⁢(G) with respect to the quadratic form q:G/Φ⁢(G)→𝔽2, where q⁢(v¯) is the element a∈𝔽2 such that v2=xa for all v¯∈G/Φ⁢(G).

Lemma 2.6.

Under the above notation, we have e2⁢(G)≤e2⁢(D8×C2n-3).

Proof.

By Lemma 1.4, we know that

Let ci⁢(G) be the number of cyclic subgroups of order 2i of G. Since both G and D8×C2n-3 are of exponent 4, the above inequality can be rewritten as

On the other hand, we have

and consequently

i.e.

It is clear that the cyclic subgroups of order 4 of these groups contain the Frattini subgroup. Thus (2.4) leads to e2⁢(G)≤e2⁢(D8×C2n-3), as desired. ∎

Corollary 2.7.

If G is an (almost) extraspecial 2-group of order 2n, n≥3, then

Corollary 2.8.

Given a 2-group G, we denote by S(α,β)⁢(G) the set of all elementary abelian sections H2/H1≅C2α of G with |H1|=2β. If G is (almost) extraspecial of order 2n, then

Proof.

Write

where

Since G/Φ⁢(G)≅D8×C2n-3/Φ⁢(D8×C2n-3), we have

Clearly,

is the number of elementary abelian subgroups of order 2α of G, and so

implying that

Furthermore, we observe that every section H2/H1∈𝒮(α,β)3⁢(G) determines a section H2⁢Φ⁢(G)/H1⁢Φ⁢(G)∈𝒮(α,β+1)1⁢(G) with

Conversely, every section A/B∈𝒮(α,β+1)1⁢(G) with

determines 2β sections H2/H1∈𝒮(α,β)3⁢(G). As there are eα+β+1⁢(G)⁢(α+β+1β+1)2 such sections A/B∈𝒮(α,β+1)1⁢(G), we infer that

Let H2/H1∈𝒮(α,β)4⁢(G). Then Φ⁢(H2)⊆H1. On the other hand, we have

because H2 is normal in G. Thus Φ⁢(H2)=1, i.e. H2 is elementary abelian, and it can be chosen in eα+β⁢(G) ways. Also, H1 is a complement of Φ⁢(G) in one of the (α+ββ+1)2 subgroups of order 2β+1 of H2, and it can be chosen in (α+ββ+1)2⁢2β ways. Hence

Obviously, relations (2.5)–(2.8) lead to

as desired. ∎

Proof of Theorem 1.1.

Let |G′|=2m. If m=0, then G is abelian and the conclusion follows from Lemma 2.3. Assume that m≥1.

If G/G′ is not elementary abelian, then

where the first inequality is obtained by Lemma 2.2, while the second one by Lemma 2.3.

If G/G′ is elementary abelian, then G′=Φ⁢(G). Let M be a normal subgroup of G such that M⊆G′ and [G′:M]=2. Then

where G1=G/M satisfies the conditions

i.e. it is a generalized extraspecial 2-group. Then either

where E is an extraspecial 2-group and A is an elementary abelian 2-group, by Lemma 1.2. In other words, G1 is a direct product of an (almost) extraspecial 2-group and an elementary abelian 2-group. So, it suffices to prove that if G2 is an (almost) extraspecial 2-group of order 2q, then

Let A be one of the groups G2 or D8×C2q-3. From Theorem 2.1 it follows that a subgroup H≤A×C2n-q of order 2k is completely determined by a quintuple (A1,A2,B1,B2,φ), where A1⊴A2≤A, B1⊴B2≤C2n-q, φ:A2/A1→B2/B1 is an isomorphism, and |A2|⁢|B1|=2k. By fixing the section B2/B1 of C2n-q and φ∈Aut⁢(B2/B1), we infer that H depends only on the choice of the section A2/A1∈𝒮(α,β)⁢(A), where (α,β)∈{(i,j)∣i=0,1,…,k,j=0,1,…,k-i}. So, to prove the inequalities in (2.9), it suffices to compare the numbers of elementary abelian sections A2/A1 of the two groups G2 and D8×C2q-3, where |A1| and |A2| are arbitrary. It is now clear that the conclusion follows from Corollary 2.8, completing the proof. ∎

Corollary 2.9.

Let G be a finite non-elementary abelian 2-group of order 2n, n≥3. Then