Finite 𝑝-groups all of whose 𝓐₂-subgroups are generated by two elements

Lemma 2.1 ([12, Lemma 2.2]).

Suppose that G is a non-abelian p-group. Then the following conditions are equivalent:

1. G is minimal non-abelian;

2. d⁢(G)=2 and |G′|=p;

3. d⁢(G)=2 and Φ⁢(G)=Z⁢(G).

Lemma 2.2 ([6, Theorem 3.1]).

Suppose that G is a p-group, p is an odd prime, d⁢(G)>2. If d⁢(H)⩽2 for all maximal subgroups H of G, then G has an abelian maximal subgroup.

Lemma 2.3 ([5, Theorem 4]).

Let G be a p-group. If both G and G′ can be generated by two elements, then G′ is abelian.

Lemma 2.4 ([6, Theorem 5.1]).

Let G be a group of order 2n, where n⩾5. Suppose, for some integer r with 5⩽r⩽n, that all subgroups of order 2r-1 and 2r have two generators. Then G is metacyclic.

Lemma 2.5 ([6, Theorem 4.2]).

Suppose that G is a group of order pn, where p is odd and n⩾6, and that all maximal subgroups of G have two generators. Then either G is metacyclic or G/G3 is of order p3 and ℧1⁢(G)=G3.

Lemma 2.6 ([14, Lemma 2.8]).

Let G be a non-abelian p-group. If d⁢(G)=2, then H′<G′ for each H<G.

Lemma 2.7 ([9]).

Let G be a non-abelian p-group. Then the number of abelian subgroups of index p in G is 0, 1 or 1+p.

Lemma 2.8 ([1, Proposition 72.1]).

Let G be a metacyclic p-group. Then G is an At-group if and only if |G′|=pt.

Lemma 3.1.

Let G be a non-abelian p-group all of whose non-abelian proper subgroups are generated by two elements. If d⁢(G)=3, then G∈A2 and c⁢(G)=2.

Proof.

We complete the proof in three steps.

Step 1. If c⁢(G)=2, then exp⁡(G′)=p.

Let G=〈a,b,c〉 be a minimal counterexample. Then we have G′≅Cp2. Assume G′=〈[a,b]〉 without loss of generality. Then [a,c],[b,c]∈〈[a,b]〉. Let [a,c]=[a,b]i and [b,c]=[a,b]j. Then

Let c1=c⁢b-i⁢aj. Then G=〈a,b,c1〉 is the central product of 〈a,b〉 and 〈c1〉. Now 〈ap,b,c1〉 is a non-abelian proper subgroup generated by three elements. This is a contradiction.

Step 2.c⁢(G)=2.

Let G=〈a,b,c〉 be a minimal counterexample. Then we have c⁢(G)=3 and |G3|=p. We will prove this is a contradiction by considering two cases.

Case 1.G has an abelian subgroup of index p.

Without loss of generality, assume 〈a,b,Φ⁢(G)〉 is abelian. Since c⁢(G)=3, [a,c]∉G3 or [b,c]∉G3. Assume [a,c]∉G3 without loss of generality. Since [a,c]∈Φ⁢(G),

It follows that [a,c,c]≠1. Thus G3=〈[a,c,c]〉. Let

Then [b⁢a-i,c,c]=1. Therefore, we may assume without loss of generality that [b,c,c]=1. Since [b,c]∈Φ⁢(G),

It follows that [b,c]∈Z⁢(G). Since cp∈Φ⁢(G), [b,cp]=1. It follows that

If [b,c]∉G3, then G/〈[b,c]〉 is another counterexample, which contradicts the minimality of G. Thus [b,c]∈G3. Let [b,c]=[a,c,c]k and b1=b⁢[a,c]-k. Then [b1,c]=1. Notice that 〈b1,c,Φ⁢(G)〉 is maximal in G. We get that 〈b1,c,Φ⁢(G)〉 is another abelian subgroup of index p of G. It follows that |G′|⩽p, which contradicts c⁢(G)=3.

Case 2.G has no abelian subgroup of index p.

In this case, all subgroups of index p are generated by two elements. By Lemma 2.2, we get p=2. Since c⁢(G)=3, G′ is abelian. Since c⁢(G/G3)=2 and G/G3 satisfies the hypothesis, exp⁡(G2/G3)=2 by step 1. It follows that

If G2/G3≅C2×C2×C2, then d⁢(〈G2,a〉)=d⁢(〈G2,a〉/G3)⩾3, so 〈G2,a〉 is abelian. Similarly, we get that 〈G2,b〉 and 〈G2,c〉 are also abelian. It follows that G2⩽Z⁢(G), which contradicts the fact that c⁢(G)=3.

If G2/G3≅C2×C2, then assume [a,b]∈G3 without loss of generality. If [a,b]=1, then 〈a,b,Φ⁢(G)〉 is an abelian subgroup of index 2 of G, which contradicts the fact that G has no abelian subgroup of index p. Thus G3=〈[a,b]〉≠1. It follows that 〈a,b,Φ⁢(G)〉=〈a,b〉 is minimal non-abelian, and so

Thus

Notice that

Hence one of [a,c,c],[b,c,c] and [a⁢b,c,c] is 1. Without loss of generality, assume [a,c,c]=1. Then [a,c]2=1. If [a,c]∈G3, letting [a,c]=[a,b]j and c1=c⁢b-j, then [a,c1]=1, and so 〈a,c1,Φ⁢(G)〉 is an abelian subgroup of index p of G, which contradicts the fact that G has no abelian subgroup of index p. Thus [a,c]∈G2∖G3. Since [a,c]∈Φ⁢(G)=Z⁢(〈a,b〉), [a,c,a]=[a,c,b]=1. It follows that [a,c]∈Z⁢(G). Now G/〈[a,c]〉 is another counterexample, which contradicts the minimality of G.

Step 3. We prove G∈𝒜2.

For any non-abelian subgroup M of index p of G, we have d⁢(M)=2. Let M=〈x,y〉. By step 1 and step 2, we have c⁢(G)=2 and exp⁡(G′)=p. It follows that M′=〈[x,y]〉 and |M′|=p. By Lemma 2.1, we get M∈𝒜1. Thus G∈𝒜2. ∎

Remark 1.

Notice that non-abelian p-groups all of whose non-abelian proper subgroups are generated by two elements have been classified; see [12, main theorem]. By Lemma 3.1, one of the p-groups listed in [12, main theorem] is either a 𝒫2-group or is an 𝒜2-group with three generators.

Theorem 3.2.

Let G be a non-abelian p-group. Then the following statements are equivalent:

1. all non-abelian subgroups of G are generated by two elements;

2. all subgroups of class 2 of G are generated by two elements;

3. all 𝒜2-subgroups of G are generated by two elements.

Proof.

Obviously, it is enough to show that (2) ⇒ (1) and (3) ⇒ (1).

(2) ⇒ (1): This follows from [8, Theorem 3.1 (1)].

(3) ⇒ (1): Without loss of generality, assume G is an 𝒜t-group with t⩾3. Notice that the condition in (3) is inherited by subgroups. We use induction on t. If t=3, then a non-abelian proper subgroup of G is an 𝒜1- or 𝒜2-subgroup. Now an 𝒜1-subgroup is generated by two elements by Lemma 2.1 and so is an 𝒜2-subgroup by hypothesis. Thus all non-abelian proper subgroups of G are generated by two elements. If d⁢(G)=3, then G∈𝒜2 by Lemma 3.1. This contradicts t=3. It follows that d⁢(G)=2. That is, the conclusion holds for t=3. Assume the conclusion holds for any positive integer k with k<t. Let H be a non-abelian proper subgroup of G. Then there exists a positive integer k<t such that H∈𝒜k by the definition of 𝒜t-groups. By the inductive hypothesis, we have d⁢(H)=2. That is, all non-abelian proper subgroups of G are generated by two elements. Notice t⩾3. It follows by Lemma 3.1 that d⁢(G)=2. So the conclusion holds. ∎

Theorem 3.3.

𝒫1⊆𝒫2.

Proof.

Assume G∈𝒫1. It is enough by Theorem 3.2 to show that all 𝒜2-subgroups of G are generated by two elements. We prove this by contradiction. Assume G has an 𝒜2-subgroup K such that d⁢(K)>2. Then the number of maximal subgroups of K is at least 1+p+p2. On the other hand, the number of abelian maximal subgroups of K is at most 1+p by Lemma 2.7. Thus K has at least two distinct non-abelian maximal subgroups H1 and H2. It follows from K∈𝒜2 that H1,H2∈𝒜1. Now we have |K:H1∩H2|=p2. We will prove |K:I𝒜1(G)|⩾p3. This contradicts the hypothesis.

Since K∈𝒜2, non-abelian maximal subgroups of K are exactly 𝒜1-subgroups of K. Thus Φ⁢(K)⩽I𝒜1⁢(K), and hence K/I𝒜1⁢(K) is elementary abelian. Since d⁢(K)⩾3, the number of 𝒜1-subgroups of K is at least p2. By the correspondence theorem, K/I𝒜1⁢(K) has at least p2 maximal subgroups. If |K:I𝒜1(G)|⩽p2, then K/I𝒜1⁢(K)≅Cp or Cp×Cp. This contradicts the fact that K/I𝒜1⁢(K) has at least p2 maximal subgroups. Therefore, |K:I𝒜1(G)|⩾|K:I𝒜1(K)|⩾p3. ∎

Theorem 4.1.

Let G be a p-group with an abelian subgroup of index p. Then G∈P2 if and only if G∈P1.

Proof.

By Theorem 3.3, it is enough to show that if G∈𝒫2, then G∈𝒫1. Assume M0 is an abelian subgroup of index p of G. Let G=〈M0,x〉, and Hi are two distinct 𝒜1-subgroups of G for i=1,2. Then Hi∩M0 is maximal in Hi, and so Hi=〈Hi∩M0,x⁢ai〉 for some ai∈M0, where i=1,2. Let

Since M0 is abelian,

It follows that K′⩽H1′⁢H2′. Notice that G=〈H1,M0〉=〈H2,M0〉. We have Hi′⩽Z⁢(G). Thus K′⩽Z⁢(G) and exp⁡(K′)=p. Since G∈𝒫2, d⁢(K)=2, and so |K′|=p. It follows by Lemma 2.1 that K is an 𝒜1-subgroup of G. Notice that H1⩽K. We get K=H1, and so H2∩M0⩽K∩M0=H1∩M0. Symmetrically, H1∩M0⩽H2∩M0, and so H1∩M0=H2∩M0. It follows that H1∩M0⩽H1∩H2. By the arbitrariness of H2, we have H1∩M0⩽I𝒜1⁢(G). Notice that H1∩M0 is maximal in H1 and H1∩H2<H1. We get

It follows that

Hence H1∩H2=I𝒜1⁢(G). The conclusion holds. ∎

Corollary 4.2.

Suppose that G is a p-group with an abelian subgroup of index p. If G∈P2, then H′ is characteristic in G for any A1-subgroup H of G.

Theorem 4.3.

Let G be a finite p-group with an A1-subgroup of index p. Then G∈P2 if and only if G∈P1.

Proof.

By Theorem 3.3, it is enough to show that if G∈𝒫2, then G∈𝒫1. Assume M is an 𝒜1-subgroup of index p of G. Since G∈𝒫2, there exist a,b∈G such that M=〈Φ⁢(G),b〉 and G=〈a,b〉. Take two distinct 𝒜1-subgroups H1 and H2 of G such that H1,H2⩽̸M. Then Hi⩽Mi for some maximal subgroup Mi of G, and so Hi∩M⩽Mi∩M=Φ⁢(G). Thus Hi∩M=Hi∩Φ⁢(G). Since M is maximal in G, Hi∩M is maximal in Hi. It follows that Hi∩Φ⁢(G) is maximal in Hi. Thus

where ϕi∈Φ⁢(G) and ti∈{0,1,…,p-1}. Let

Then H1=〈H1∩Φ⁢(G),a⁢bt1⁢ϕ1〉⩽K. Notice that

and

We have K′⩽H1′⁢H2′⁢M′. By Corollary 4.2, we get that Hi′ is characteristic in Mi⊴G. Thus Hi′⊴G, and so Hi′⩽Z⁢(G). Since M is an 𝒜1-subgroup of index p of G, |M′|=p, and so M′⩽Z⁢(G). It follows that K′⩽Z⁢(G) and exp⁡(K′)=p. Since G∈𝒫2, d⁢(K)=2, so |K′|=p. It follows from Lemma 2.1 that K is an 𝒜1-subgroup of G. Thus K=H1, and so

Symmetrically, we get H1∩M⩽H2∩M, and so H1∩M=H2∩M. It follows that H1∩M⩽H1∩H2. By the arbitrariness of H2, we have

Notice that H1∩M is maximal in H1 and H1∩H2<H1. We get

It follows that

Hence H1∩H2=I𝒜1⁢(G). The conclusion holds. ∎

Theorem 4.4.

Let G be a non-abelian p-group with neither A0-subgroups of index p nor A1-subgroups of index p.

1. If G is metacyclic, then G∈𝒫2 and G∉𝒫1.

2. If G is non-metacyclic, then G∈𝒫2 if and only if G∈𝒫1.

Proof.

The proof of (1): Obviously, G∈𝒫2. Let G=〈a,b〉. Assume without loss of generality G′⩽〈a〉. Take two maximal subgroups M=〈ap,b〉, N=〈a,bp〉 of G. Let |M′|=pt1 and |N′|=pt2. By Lemma 2.8, we have M∈𝒜t1 and N∈𝒜t2. By the hypothesis, we get t1,t2⩾2. It follows that

Let M1=〈apt1-1,b〉 and N1=〈apt2-2,bp〉. Notice that G′⩽〈a〉⩽CG⁢(a). We have

Thus M1 and N1 are two 𝒜2-subgroups of G by Lemma 2.8. Now M1 has two distinct 𝒜1-subgroups 〈apt1,b〉 and 〈apt1,b⁢apt1-1〉, and their intersection is

However, N1 has two distinct 𝒜1-subgroups 〈apt2-1,bp〉 and 〈apt2-1,bp⁢apt2-2〉, and their intersection is

Thus G∉𝒫1. The results hold.

The proof of (2): By Theorem 3.3, it is enough to show that if G∈𝒫2, then G∈𝒫1 and |G|=p6.

Assume G∈𝒫2. Since G has no 𝒜0-subgroups of index p, d⁢(G)=d⁢(M)=2 for all maximal subgroups M of G. In the following, we will prove G∈𝒫1 and |G|=p6 in six steps.

Step 1.|G|⩾p6, where p>2.

Notice that every group of order p4 has 𝒜0-subgroup of index p. We get |G|⩾p5. If |G|=p5, then |G/N|=p4 for a normal subgroup N of order p. Then G/N has an 𝒜0-subgroup A/N of index p. It follows that A is maximal in G and |A′|=|N|=p. By Lemma 2.1, we get A is an 𝒜1-subgroup. This contradicts the fact that G has no 𝒜1-subgroup of index p. Hence |G|⩾p6. Moreover, it follows by Lemma 2.4 that p>2.

Step 2.|G/G3|=p3, ℧1⁢(G)=G3=Φ⁢(M) for any maximal subgroup M of G, Φ⁢(G) is abelian, Gi/Gi+1 is elementary abelian for i=2,3,…,c⁢(G).

Since G is not metacyclic, by Lemma 2.5, we get G/G3 is of order p3 and ℧1⁢(G)=G3. Thus G′=Φ⁢(G) and M/G3≅Cp×Cp for any maximal subgroup M of G. It follows from d⁢(M)=2 that Φ⁢(M)=G3. Moreover, G′ is abelian by Lemma 2.3, and so Φ⁢(G) is abelian. For any g∈G and gi∈Gi, we have gp,gi∈Φ⁢(G), and so [gi,gp]=1. It follows that

Hence Gi/Gi+1 is elementary abelian.

Step 3.G3/G4≅Cp×Cp.

Notice that d⁢(G)=2. Let G=〈x,y〉 and [x,y]=c. Then G2=〈c,G3〉 and G3=〈[c,x],[c,y],G4〉. Notice that |G|⩾p6 by step 1 and |G/G3|=p3 by step 2. We have G3>G4, so either [c,x]∉G4 or [c,y]∉G4. Without loss of generality, let [c,x]∉G4.

If |G3/G4|=p, then G3=〈[c,x],G4〉. Let

Then

Thus we may let [c,y]∈G4 without loss of generality. Notice that G′ is abelian by step 2. We have [c,x,y]=[c,y,x]∈G5. It follows that G4=〈[c,x,x],G5〉, and so G4/G5 is cyclic. Notice that Gi/Gi+1 is elementary abelian by step 2. We get |G4/G5|=p. Similarly, we can get

and so |Gi/Gi+1|=p. That is, G is of maximal class. However, G is not of maximal class by [12, Lemma 5.1]. This is a contradiction. Thus

By step 2, G3/G4≅Cp×Cp.

Step 4.|G4/G5|=p.

We recall c=[x,y]. Then we have [c,x,y]=[c,y,x] by step 2. Notice that G3=〈[c,x],[c,y],G4〉. By step 3, we get

If all [c,x,x], [c,y,y] and [c,x⁢y,x⁢y] are in G5, then

Notice that p>2 by step 1. We get [c,x,y]∈G5. It follows that G4=G5, and so G4=1. By step 2 and step 3, we get

On the other hand, |G|⩾p6 by step 1. This is a contradiction. Hence at least one of [c,x,x], [c,y,y] and [c,x⁢y,x⁢y] is not in G5. Without loss of generality, assume [c,x,x]∉G5.

Take M=〈Φ⁢(G),x〉. Then M is maximal in G. Notice that

by steps 2 and 3. We have M=〈c,x,G3〉=〈c,x〉. It follows that

Since G′ is abelian by step 2 and c∈G′, [c,x,c]=1. It follows that

Notice that Φ⁢(G) is abelian by step 2. Since c and xp∈Φ⁢(G), [c,xp]=1. It follows that [c,x]p≡[c,xp]≡1(modM3), so |M′/M3|=p. Thus [M′,G]⩽M3. In particular,

Similarly, we get [c,y,x]∈〈[c,y,y],G5〉 by taking M=〈Φ⁢(G),y〉.

If |G4/G5|≠p, then [c,y,y]∉〈[c,x,x],G5〉, and so

Notice that [c,x,y]=[c,y,x]. We get

It follows that