Finite p-groups with abelian maximal subgroups generated by two elements

Abstract

Assume that G is a finite non-abelian p-group. If G has an abelian maximal subgroup whose number of Generators is at least n, then G is called an M_n-group. For p = 2, M₂-groups have been classified. For odd prime p, this paper provides the isomorphism classification of M₂-groups, thereby achieving a complete classification of M₂-groups.

Keywords

Finite p-group / regular p-group / abelian maximal subgroupnumber of Generators

1 Introduction

Let p be a prime number. A group whose order is a power of a primeis called a finite p-group, abbreviated as p-group. The subgroup of a p-group whose index is a power of p is also called a maximal abelian subgroup. p-group is an important component of group theory, and the classification problem is a crucial aspect of p-group studies. One of the classical results in the classification of p-groups is the one given by Burnsidet [2], which provides a classification of p-groups with a cyclic subgroup of index p. Naturally, one may inquire: What is the structure of a p-group G with an abelian subgroup of index p? Tuan [8] provides an elegant characterization for such p-groups G, namely |G|= p|Z(G)||G^{\prime }|. Subsequently, efforts have been made to classify p-groups with certain restrictions imposed. For instance, Blarkburn [1] classifies maximal class p-groups with an abelian subgroup of index p. Mann [4] classifies p-groups that are centrally cyclic and have an abelian subgroup of index p. In this paper, we study the structure of p-groups by imposing restrictions on the number of generators of their maximal abelian subgroups. Our research can be seen as an extension of Burnside’s classification results on p-groups with a cyclic subgroup of index p.

Notice that the number of generators for cyclic groups is 1. We introduce the following concept: Let G be a finite non-abelian p-group. If G contains a maximal abelian subgroup with a minimum number of generators being n, then G is called an M_n-group. Clearly, an M₁-group is exactly the finite non-abelian p-groups with cyclic maximal subgroups as classified by Burnside; an M₂-group is precisely a finite non-abelian p-group with maximal abelian subgroups generated by two elements and without cyclic maximal subgroups. For p = 2, Qu et al. [5] have classified M₂-groups. For odd prime p, this paper provides an isomorphic classification of M₂-groups, thus completing the classification of M₂-groups.

To classify M₂-groups G, we consider two cases for the maximal abelian subgroups of G: those with cyclic maximal subgroups and those without cyclic maximal subgroups. If G has maximal abelian subgroups with cyclic maximal subgroups, such p-groups have already been classified by Hua and Tuan [3], and we only need to identify the M₂-groups among them. This paper primarily investigates the case of maximal abelian subgroups without cyclic maximal subgroups for M₂-groups. That is, we only need to study M₂-groups with a maximal subgroup A ≌ C_pm × C_pn, where m\ge n\ge 2. It can be seen that the order of such M₂-groups is not less than p⁵.

Throughout this paper, we always assume that p is an odd prime. Unless otherwise stated, the symbols and concepts used are taken from [10].

2 Preliminary results

First, we introduce some key concepts and results used in this paper.

Definition 2.1　A finite p-group G is called regular if for any a,b\in G^{\mathrm{\prime }} and for every positive integer s, there exist elements c_3,c_4,\dots ,c_i\in ⟨a,b⟩ (where ⟨a,b⟩ denotes the subgroup generated by a and b), dependent on the choice of a and b, such that {\left(ab\right)}^{p^a} = a^{p^s}{b}^{p^s}{c_3}^{p^s}\cdots {c_m}^{p^s}. Particularly, if (ab)^p = a^pb^p, then G is called a p-commutative group.

Definition 2.2　Let G be a finite p-group. If for p>2,G^{\prime }\le {\mho }_1(G), and for p=2,G^{\prime }\le {\mho }_2(G), then G is called a power-commutative p-group.

Definition 2.3　Let G be a finite p-group, and N⊴G. If for p>2, [N,G]\le {\mho }_1(N), and for p = 2, [N,G]\le {\mho }_2(N), then we say N is power-embedded in G.

Lemma 2.1 [10, Theorem 5.4.17]　 Let G be a finite regular p-group, α, b\in G, and S, t be a non-negative integer. Then

[a^{p^s},b^{p^t}]=1\iff a,b{]}^{p^{s+t}}=1.

Lemma 2.2 [10, Proposition 5.1.13]　 A regular p-group G is p-commutative if and only if {\mho }_l\left(G^{\mathrm{\prime }}\right)=1.

Lemma 2.3 [10, Theorems 5.2.2 and 5.2.17]　 Let G be a finite p-group. If G satisfies one of the following conditions, then G is regular.

(1) ;

(2) .

Lemma 2.4 [8]　 Let A be a commutative normal subgroup of the non-abelian group G, and let its quotient group G/A=⟨xA⟩ be cyclic. Then:

(1) The mapping a\mapsto [a,x], a\in A, is a surjective homomorphism from A to G^{\prime };

(2) G{}^{\prime }\cong A/(A\cap Z(G\left)\right).

In particular, if G has a maximal abelian subgroup, then \left|G\right|=p\left|G{}^{\prime }\right|\left|Z\right(G\left)\right|.

Lemma 2.5 [9]　 For a finite p-group G, the following propositions are equivalent:

(1) G is an internally central group;

(2) d(G)=2and\left|G^{\mathrm{\prime }}\right|=p;

(3) d(G)=2andZ(G)=\mathrm{\Phi }(G).

Lemma 2.6 [10, Theorem 2.5.10]　 Let the finite p-group G be a finite p-group with a maximal abelian subgroup A, and d(G)=2. Then G/Z(G) is a maximal class p-group.

Lemma 2.7 [7]　 Let G be a finite power-commutative p-group, with M and N power-embedded in G^{\prime }. Then \left[{\mho }_1\right(M),N]={\mho }_1\left(\right[M,N\left]\right).

Lemma 2.8 [10, Theorem 2.5.6]　 For p > 2, let G be a non-abelian p-group of order pⁿ. Assuming G has a maximal abelian subgroup, then G is a maximal class p-group if and only if \left|G:G^{\mathrm{\prime }}\right|=p^2.

By examining the classification results in [3, 11] and [10, Theorem 2.6.4], we can obtain the following theorem.

Theorem 2.1　 Let G be a finite non-abelian p-group with maximal abelian subgroups that are cyclic. Then G is an M₂-group if and only if G is isomorphic to one of the following non-isomorphic groups:

(1) ⟨a,b,c|a^{p^m}=b^p=c^p=1,[a,b]=c,[a,c]=[b,c]=1⟩,m\ge 1.

(2) ⟨a,b|a^{p^m}=b^{p^2}=1,[a,b]=a^{p^{m-1}}⟩,m\ge 2.

(3) ⟨a,b,c|a^{p^m}=b^p=c^p=1,[a,b]=a^{p^{m-1}},[a,c]=[b,c]=1⟩,m\ge 2.

(4) ⟨a,b,c|a^{p^m}=b^p=c^p=1,[b,c]=a^{p^{m-1}},[a,b]=[a,c]=1⟩,m\ge 2.

(5) ⟨a,b,c|a^{p^m}=b^p=c^p=1,[a,b]=c,[c,b]=a^{p^{m-1}},[a,c]=1⟩,m\ge 2.

(6) ⟨a,b,c|a^{p^m}=b^p=c^p=1,[a,b]=c,[c,b]=a^{k{p}^{m-1}},[a,c]=1⟩,wherem\ge 2 and k is a non-residue modulo p².

(7) ⟨a,b|a^{p^m}=b^{p^2}=1,[b,a]=b^p⟩,m\ge 3.

(8) ⟨a,b,c|a^{3^2}=b^3=1,c^3=a^3,[a,b]=1,[a,c]=b,[b,c]=a^6⟩.

3 Classification of non-regular {\mathit{M}}_{\mathbf{2}}-groups

Lemma 3.1　 Suppose G=AB, where A and B are two distinct cyclic subgroups, and A\cap B>1. Then there exists no subgroup H such that H/Z(H) ≌ G.

Proof　If there exists Η such that H/Z(H) ≌ G^{\prime }, then we can assume H/Z(H) = A/Z(H)·B/Z(H), and (A/Z(H)) ∩ (B/Z(H)) > 1. Thus, A\cap B> Z(H). Since A/Z(H) and B/Z(H) are cyclic, both A and B commute. From G = AB, we have A\cap B\le Z(H), leading to a contradiction.

Lemma 3.2　 Suppose G is a non-regular M₂-group. Then

(1) p = 3;

(2) d(G) = 2;

(3) either G has a cyclic second maximal subgroup or c(G)\ge 4.

Proof　(1) Let A be the maximal abelian subgroup of G. Since G is an M₂-group, d(A) = 2 implies

|G:{\mho }_1(G\left)\right|\le |G:{\mho }_1(A\left)\right|=|G:A||A:{\mho }_1(A\left)\right|=p\cdot p^2=p^3.

On the other hand, as G is non-regular, by Lemma 2.3 (2), we have \left|G:{\mho }_1(G)\right|\ge p^p. Combining the two, we can get p\le 3. Since p is an odd prime, p = 3.

(2) Assume the contrary. As G is an M₂-group, d(G) = 3. Let A be the maximal abelian subgroup of G generated by two elements. Then

p^3=|G:\mathrm{\Phi }(G\left)\right|\le |G:{\mho }_1(G\left)\right|\le |G:{\mho }_1(A\left)\right|=|G:A||A:{\mho }_1(A\left)\right|=p^3.

This forces \mathrm{\Phi }\left(G\right)={\mho }_1\left(G\right)={\mho }_1\left(A\right). Since [\mathrm{A},G]\le G^{\mathrm{\prime }}\le \mathrm{\Phi }(G)={\mho }_1(G)={\mho }_1(A), both A and G are power-embedded in G. By Lemma 2.7 (Shalev’s commutator lemma), we have

\left[\mathrm{\Phi }\right(G),G]=\left[{\mho }_1\right(A),G]={\mho }_1\left(\right[A,G\left]\right)=[A,{\mho }_1(G\left)\right]=1.

This indicates c(G)\le 2. Since p is an odd prime, G is regular, leading to a contradiction.

(3) Since G is an M₂-group, let’s assume without loss of generality that A is the maximal abelian subgroup of G. For convenience, let A\cong {\mathrm{C}}_{p^m}\times {\mathrm{C}}_{p^n}, where m\ge n. If G has no cyclic second maximal subgroup, then n ≥ 2. We aim to prove that in this case, c(G)\ge 4. Suppose otherwise, that is, if G is non-regular, by Lemma 2.3 (1), we must have c(G)=3. From Lemma 2.6, we know that G/Z(G) is a maximal class p-group. Note that c(G/Z(G))=2. This implies that G/Z(G) is a non-abelian group of order p³. Furthermore, using Lemma 3.1, we find exp(G/Z(G)) =p. Particularly, . By Lemma 2.4, we have

[a,x{]}^p=\left[a^p,x\right]=1,\mathrm{\forall }\mathrm{a}\in A,x\in G\setminus A.

In particular, exp\left(G^{\mathrm{\prime }}\right)=p. Therefore, G^{\mathrm{\prime }}\le {\mathrm{\Omega }}_1(A). Since n\ge 2, we have {\mathrm{\Omega }}_1(A)\le {\mho }_1(A). Hence, G^{\mathrm{\prime }}\le Z(G). This contradicts c(G)=3.

Theorem 3.1　 Suppose G has no cyclic second maximal subgroup. Then G is a non-regular M₂-group if and only if G is a maximal class 3-group with a maximal abelian subgroup of order greater than 3⁴.

Proof　(\Leftarrow) It is easy to see from the classification results in [10, Theorem 8.3.6] that G is a non-regular M₂-group.

(\Rightarrow) Let A be the maximal abelian subgroup of G. First, we prove that Z(G)\le G^{\mathrm{\prime }}. If Z\left(G\right)\nleqq G{}^{\prime }, then G^{\mathrm{\prime }}\ne Z(G)G^{\mathrm{\prime }}. Consequently, Z\left(G\right)\nleqq \mathrm{\Phi }\left(Z\right(G\left)G{}^{\prime }\right)G{}^{\prime }. Since d\left(\mathrm{Z}(G)G^{\mathrm{\prime }}\right)\le d(A)=2, \left(Z(G)G^{\mathrm{\prime }}\right)/Z(G) is cyclic. By Lemma 2.6 and Lemma 3.2 (2), we deduce that G/Z(G) is a maximal class p-group.

However, all normal subgroups of odd-order maximal class p-groups with order greater than p are non-cyclic. This forces the order of (G/Z(G{)}^{\mathrm{\prime }}=\left(Z(G)G^{\mathrm{\prime }}\right)/Z(G) to not exceed p. In particular, c(G/Z(G)\le 2. Hence, c(G)\le 3. This contradicts our initial assumption. Therefore, Z(G)\le G^{\mathrm{\prime }}. Thus, |G:G^{\prime }|=\left|G/Z(G):(G/Z(G)){}^{\mathrm{\prime }}\right|=p^2. By Lemma 2.8, G is a maximal class p-group. Moreover, by Lemma 3.2 (1), we know that p = 3. The maximal class 3-groups with maximal abelian subgroups have been classified by Blackburn [1]. The classification results can be found in [10, Theorem 8.3.6]. Note that since G has no cyclic second maximal subgroup, we have |G|>3^4.

4 Classification of regular {\mathit{M}}_{\mathbf{2}}-groups

Theorem 4.1　 Suppose G is a regular M₂-group, A is a maximal subgroup of G with A\cong C_{p^m}\times C_{p^n}, where m\ge n>2. Then G has the following properties:

(1) exp(G^{\mathrm{\prime }}) = p, particularly, .

(2) G is p-central.

(3) {\mho }_1(G)\le Z(G).

(4) c(G)=2.

(5) d(G)\le 3. If d(G)=2, then G is internally central; if d(G)=3, then there exists an element of order p in G\setminus A.

Proof　(1) Take any x\in G\setminus A. By Lemma 2.4, we have G^{\mathrm{\prime }}=\{[a,x]\mid a\in A\}. Further, by Lemma 2.1, [a, x]^p = 1 if and only if [a, x^p] = 1. Note that x^p\in A and A is abelian, implying exp\left(G^{\mathrm{\prime }}\right)\le p. Since G is non-abelian, exp\left(G^{\mathrm{\prime }}\right)=p. Clearly G^{\mathrm{\prime }}\le A, which implies .

(2) From (1), we have exp\left(G^{\mathrm{\prime }}\right)=p. Furthermore, by Lemma 2.2, the result is established.

(3) By Lemma 2.1, [x^p,y]=1\iff [x,y{]}^p=1. By (1), we conclude that {\mho }_1(G)\le Z(G).

(4) Since m\ge n\ge 2, we have {\mathrm{\Omega }}_1(A)\le {\mho }_1(A). By (1) and (3), we have G^{\mathrm{\prime }}\le Z(G), and as G is non-abelian, c(G)=2.

(5) Since A is a maximal subgroup of G and d(A)=2, it follows that d(G)\le 3. If d(G)=2, then by (4), G^{\mathrm{\prime }}=⟨[a,b]⟩. Further, by (1), we know \left|G^{\mathrm{\prime }}\right|=p. By Lemma 1.5, G is internally central. If d(G)=3, then \mathrm{\Phi }(G)=\mathrm{\Phi }(A)={\mho }_1(A). Take any x\in G\setminus A, there exists \mathrm{a}\in A such that x^p={\mathrm{a}}^p. Let y=x{a}^{-1}. Then y\in G\setminus A. By (2), we have y^p=x^p{\mathrm{a}}^{-p}=1. Thus, y is the desired element of order p.

Based on Theorem 4.1 (5), we discuss the cases of d(G)=2 and d(G)=3 separately.

Theorem 4.2　 Suppose G is a regular M₂-group satisfying d(G)=2 if and only if G is isomorphic to one of the following groups:

(1)\begin{array}{c}{M}_p(m,n)=⟨a,b|a^{p^m}=b^{p^n}=1,[a,b]=a^{p^{m-1}}⟩,m>2,n>2.\end{array}

(2) M_p(m,1,1)=⟨a,b,c|a^{p^m}=b^p=c^p=1,[a,b]=c,[a,c]=[b,c]=1⟩,m\ge 3.

Proof　(\Leftarrow) As the verification process is trivial, it is omitted.

(\Rightarrow) By Theorem 4.1 (5), G is an inner abelian group. Internally central p-groups have been classified in [6], and the classification results are also available in [10, Theorem 2.3.7]. By examining the groups in [10, Theorem 2.3.7], it’s straightforward to confirm that G belongs to one of the groups stated in Theorem 4.2. The verification process is trivial, and details are omitted.

Note　The abelian maximal subgroups of the type (2) groups in Theorem 4.2 have cyclic maximal subgroups.

Next, we classify the M₂-groups G with d(G)=3. Let A be the maximal abelian subgroup of G. Then A\cong {\mathrm{C}}_{p^m}\times {\mathrm{C}}_{p^n}, where m\ge n\ge 2. Since G^{\mathrm{\prime }}\le {\mathrm{\Omega }}_1(A)\le {\mathrm{C}}_p\times {\mathrm{C}}_p, we discuss two cases: when G^{\mathrm{\prime }}\cong {\mathrm{C}}_p and when G^{\mathrm{\prime }}\cong {\mathrm{C}}_p\times {\mathrm{C}}_p.

Theorem 4.3　 G is a regular M₂-group satisfying d(G)=3 and G^{\mathrm{\prime }}\cong {\mathrm{C}}_p if and only if G is isomorphic to one of the following groups:

(1) G=⟨a,b,c|a^{p^m}=b^{p^n}=c^p=1,[b,a]=1,[a,c]=1,[b,c]=b^{p^{n-1}}⟩,m>n.

(2) G=⟨a,b,c|a^{p^m}=b^{p^n}=c^p=1,[b,a]=1,[a,c]=1,[b,c]=a^{p^{m-1}}⟩,m>n.

(3) G=⟨a,b,c|a^{p^m}=b^{p^n}=c^p=1,[b,a]=1,[b,c]=1,[a,c]=b^{p^{m-1}}⟩,m>n.

(4) G=⟨a,b,c|a^{p^m}=b^{p^n}=c^p=1,[b,a]=1,[b,c]=1,[a,c]=a^{p^{m-1}}⟩,m>n.

(5) ⟨a,b,c|a^{p^m}=b^{p^m}=c^p=1,[b,a]=1,[b,c]=1,[a,c]=b^{p^{m-1}}⟩.

(6) ⟨a,b,c|a^{p^m}=b^{p^m}=c^p=1,[b,a]=1,[b,c]=1,[a,c]=a^{p^{m-1}}⟩.

Proof　(\Leftarrow) As the verification process is trivial, it is omitted.

(\Rightarrow) Let A=⟨\mathrm{a}⟩\times ⟨\mathrm{b}⟩, where a is of order p^m and b is of order pⁿ. By Theorem 4.1(5), there exists c\in G\setminus A such that c is of order p. Thus, G=(⟨\mathrm{a}⟩\times ⟨\mathrm{b}⟩)⋊⟨\mathrm{C}⟩, and G^{\mathrm{\prime }}=⟨[\mathrm{a},\mathrm{c}],[\mathrm{b},\mathrm{c}]⟩.

Case 1　 m>n\ge 2.

If [b, c] = 1, then [a, c] ≠ 1. By Theorem 4.1(1), G^{\mathrm{\prime }}\le {\mathrm{\Omega }}_1(A). Therefore, we can assume [a,c]=a^{k{p}^{m-1}}{b}^{l{p}^{n-1}}, where k,l\in {\mathbb{Z}}_p and are not simultaneously zero. If l = 0, replacing c with c^{k^{-1}}, we obtain group (4). If l ≠ 0, because m>n, replacing b with a^{k{p}^{m-n}}{b}^l, then

[a,c]=a^{k{p}^{m-1}}{b}^{l{p}^{n-1}}=(a^{k{p}^{m-n}}{b}^l{)}^{p^{n-1}}=b_1^{p^{n-1}}.

Let b_1=a^{k{p}^{m-n}}{b}^l. We obtain group (3).

If [b, c] ≠ 1, we can assume [b,c]=a^{k{p}^{m-1}}{b}^{l{p}^{n-1}}, where k,l\in {\mathbb{Z}}_p and are not simultaneously zero. By choosing an appropriate i, replacing a with abⁱ, let a₁ =abⁱ, then [a₁, c] = 1. If l = 0, replacing a with a^k, then [b,c]=(a^k{)}^{p^{m-1}}. Let a₁ =a^k. We obtain group (2). If l ≠ 0, because m>n, replacing b with a^{k{p}^{m-n}}{b}^l and c with c^{l^{-1}}, then

[a^{k{p}^{m-n}}{b}^l,c^{l^{-1}}]=[b^l,c^{l^{-1}}]=[b,c]=(a^{k{p}^{m-n}}{b}^l{)}^{p^{n-1}}.

Let a^{k{p}^{m-n}}{b}^l=b_1 and c^{l^{-1}}=c₁. We obtain group (1).

Case 2　 m=n\ge 2.

If [b, c] = 1, then [a, c] ≠ 1. Let [a,c]=a^{k{p}^{m-1}}{b}^{l{p}^{m-1}}=(a^k{b}^l{)}^{p^{m-1}}, where \begin{array}{c}k,l\in {\mathbb{Z}}_p\end{array} and are not simultaneously zero. If k = 0, replacing b with b^l, we have [a,c]=(b^l{)}^{p^{m-1}}. Let b₁ = b^l,we obtain group (5). If k ≠ 0, replacing a with a^kb^l and c with c^{k^{-1}}, then [a^k{b}^l,c^{k^{-1}}]=[a,c]=(a^k{b}^l{)}^{p^{m-1}}. Let a₁ =a^kb^l and c₁ =c^{k^{-1}}, we obtain group (6).

If [b, c] ≠ 1, assume

[b,c]=a^{k{p}^{m-1}}{b}^{l{p}^{m-1}}=(a^k{b}^l{)}^{p^{m-1}},

where \begin{array}{c}k,l\in {\mathbb{Z}}_p\end{array} and are not simultaneously zero. By choosing an appropriate i, replacing a with abⁱ, let a₁ =abⁱ, then [a₁, c] = 1. If l ≠ 0, replacing b with a^kb^l and then replacing c with c^{l^{-1}}, then

[a^k{b}^l,c^{l^{-1}}]=[b^l,c^{l^{-1}}]=[b,c]=(a^k{b}^l{)}^{p^{n-1}}.

Let b₁ = a^kb^l and c₁ = c^{k^{-1}}. Swapping a₁, and b₁ in the group yields group (6). If l = 0, replacing a with a^k, then [b,c]=(a^k{)}^{p^{m-1}}.

Let a₁ = a^k. Swapping a₁ and b₁ in the group yields group (5).

Next, we prove that the groups listed in the theorem are non-isomorphic.

Note that G/G^{\mathrm{\prime }} and Z(G) are both abelian groups. By examining their invariant types, it can be seen that when m>n, the groups (1)−(4) listed in the theorem are non-isomorphic. In fact, for types (1) and (2), we have Z(G)=⟨a,b^p⟩ . In type (1) group, G/G^{\mathrm{\prime }} is of type (p^m, pⁿ⁻¹, p), while in type (2) group, G/G^{\mathrm{\prime }} is of type (p^m−1, pⁿ, p). For types (3) and (4), we also have Z(G)=⟨a,b^p⟩. In type (3) group, G/G^{\mathrm{\prime }} is of type (p^m, pⁿ⁻¹, p), while in type (4) group, G/G^{\mathrm{\prime }} is of type (p^m−1, pⁿ, p).

When m = n, the derived subgroups of the groups (5) and (6) listed in the theorem are ⟨a^{p^{m-1}}⟩ and ⟨b^{p^{m-1}}⟩ respectively. Groups (5) and (6) are non-isomorphic.

Next, we study the case where G{}^{\prime }\cong {\mathrm{C}}_p\times {\mathrm{C}}_p. For the case m=n\ge 2, by Theorem 4.1(1), we can assume

\begin{array}{rl}G& =⟨a,b,c|a^{p^m}=b^{p^m}=c^p=1,[a,b]=1,\\ [a,c]& =a^{i{p}^{m-1}}{b}^{j{p}^{m-1}},[b,c]=a^{k{p}^{m-1}}{b}^{l{p}^{m-1}}⟩,\end{array}

where m\ge 2,i,j,k,l\in {\mathbb{Z}}_p, and il-jk\ne 0.

For convenience, we call the matrix \left(\begin{array}{cc}i& j\\ k& l\end{array}\right) the type matrix of G. Clearly, the structure of G is entirely determined by its type matrix.

Lemma 4.1　 Let G₁ and G₂ be groups with type matrices X₁ and X₂, respectively. Then G₁ ≌ G₂ if and only if there exists an invertible matrix T over {\mathbb{Z}}_p and a nonzero element λ, such that X₂ = λT X₁T⁻¹.

Proof　To distinguish between G₁ and G₂, let us denote

G_1=⟨a_1,b_1,c_1⟩,G_2=⟨a_2,b_2,c_2⟩,

X_1=\left(\begin{array}{cc}{i}_1& j_1\\ k_1& l_1\end{array}\right),X_2=\left(\begin{array}{cc}{i}_2& j_2\\ k_2& l_2\end{array}\right).

Let σ be an isomorphism from G₂ to G₁. Note that A_i=⟨a_i,b_i⟩ are the unique maximal abelian subgroups of G_i (i = 1, 2). Therefore, a_2^{\sigma },b_2^{\sigma }\in A_1=⟨a_1,b_1⟩. Furthermore, since c_2\in {\mathrm{\Omega }}_{\mathrm{i}}\left(G_2\right)\setminus {\mathrm{\Omega }}_1\left(A_2\right), we have

c_2^{\sigma }\in {\mathrm{\Omega }}_1\left(G_1\right)\setminus {\mathrm{\Omega }}_1\left(A_1\right).

Thus, we can assume

a_2^{\sigma }=a_1^u{b}_1^v\text{,}{b}_2^{\sigma }=a_1^s{b}_1^t\text{,}{c}_2^{\sigma }=x{c}_1^{\lambda },

where u, υ, s, t, λ∈ {\mathbb{Z}}_p, and x=a^{k{p}^{m-1}}{b}^{l{p}^{m-1}}\in {\mho }_1\left(G_1\right)\le Z\left(G_1\right). Since σ is surjective, we require λ ≠ 0 and T=\left(\begin{array}{cc}u& v\\ s& t\end{array}\right) to be an invertible matrix. As σ is an isomorphism, we have

[a_2,c_2{]}^{\sigma }=[a_2^{\sigma },c_2^{\sigma }],[b_2,c_2{]}^{\sigma }=[b_2^{\sigma },c_2^{\sigma }].

Note that A_i is an abelian group. For convenience, we denote the operations on A_i as an addition. Then we have \left(\begin{array}{c}{a}_2^{\sigma }\\ b_2^{\sigma }\end{array}\right)=T\left(\begin{array}{c}{a}_1\\ b_1\end{array}\right), and consequently \left(\begin{array}{c}{a}_1\\ b_1\end{array}\right)=T^{-1}\left(\begin{array}{c}{a}_2^{\sigma }\\ b_2^{\sigma }\end{array}\right). Note that x\in Z(G), we have

\begin{array}{rl}{X}_2\left(\begin{array}{c}{p}^{m-1}{a}_2^{\sigma }\\ p^{m-1}{b}_2^{\sigma }\end{array}\right)& =\left(\begin{array}{c}{[a}_2,c_2{]}^{\sigma }\\ {[b}_2,c_2{]}^{\sigma }\end{array}\right)=\left(\begin{array}{c}{a}_2^{\sigma },c_2^{\sigma }\\ b_2^{\sigma },c_2^{\sigma }\end{array}\right)=\lambda T\left(\begin{array}{c}{a}_1,c_1\\ b_1,c_1\end{array}\right)\\ & =\lambda T{X}_1\left(\begin{array}{c}{p}^{m-1}{a}_1\\ p^{m-1}{b}_1\end{array}\right)=\lambda T{X}_1{T}^{-1}\left(\begin{array}{c}{p}^{m-1}{a}_2^{\sigma }\\ p^{m-1}{b}_2^{\sigma }\end{array}\right).\end{array}

Thus, we obtain X₂ = \lambda T{X}_1{T}^{-1}. Conversely, if there exist λ and T satisfying the conditions, from the above derivation process, it can be seen that the corresponding isomorphism σ exists. Hence, the proof is complete.

Theorem 4.4　 Let G be a finite non-abelian p-group, A be the maximal abelian subgroup of G, and A\cong C_{p^m}\times C_{p^n}, where m\ge 2. Then, G is a regular M₂-group satisfying d(G)=3 and G{}^{\prime }\cong C_p\times C_p, if and only if G is one of the following non-isomorphic groups:

(1) G=⟨a,b,c|a^{p^m}=b^{p^m}=c^p=1,[a,b]=1,[a,c]=a^{p^{m-1}},[b,c]=b^{p^{m-1}}⟩.

(2) G=⟨a,b,c|a^{p^m}=b^{p^m}=c^p=1,[a,b]=1,[a,c]=b^{k{p}^{m-1}},[b,c]=a^{p^{m-1}}{b}^{t{p}^{m-1}}⟩, where k = 1 or a fixed non-residue modulo p², and t\in \left\{0,1,\dots ,\frac{p-1}{2}\right\}.

Proof　(\Leftarrow) As the verification process is trivial, it is omitted.

(\Rightarrow) Suppose G=⟨a,b,c|a^{p^m}=b^{p^m}=c^p=1,[a,b]=1,[a,c]=a^{i{p}^{m-1}}{b}^{j{p}^{m-1}},[b,c]=a^{k{p}^{m-1}}\cdot b^{l{p}^{m-1}}⟩, where m\ge 2,i,j,k,l\in {\mathbb{Z}}_p, and il-jk\ne 0. According to Lemma 3.1, the matrix X=\left(\begin{array}{cc}i& j\\ k& l\end{array}\right) completely determines the structure of G. Let p(x) be the minimal polynomial of X. We discuss the following cases.

Case 1　 p(x) is a linear polynomial.

Let p(x) = X − λ. In this case, X=\left(\begin{array}{cc}\lambda & 0\\ 0& \lambda \end{array}\right), according to Lemma 4.1, X is isomorphic to the group corresponding to the identity matrix E, which is Group (1).

Case 2　 p(x) is a quadratic polynomial.

Let p\left(x\right)=x^2-{\lambda }_{1}x-{\lambda }_0, where λ₁ ≠ 0 and λ₀ ≠ 0. In this case, X is similar to (\begin{array}{cc}0& {\lambda }_0\\ 1& {\lambda }_1\end{array}). Choosing x ≠ 0, let T=\left(\begin{array}{cc}x& 0\\ 0& 1\end{array}\right). Then

xT\left(\begin{array}{cc}0& {\lambda }_0\\ 1& {\lambda }_1\end{array}\right)T^{-1}=\left(\begin{array}{cc}0& {\lambda }_0{x}^2\\ 1& {\lambda }_{1}x\end{array}\right).

By selecting an appropriate x, we can always turn {\lambda }_0{x}^2 into 1 or a fixed quadratic non-residue k. Thus, X is isomorphic to the group corresponding to \left(\begin{array}{cc}0& k\\ 1& {\lambda }_1^{\prime }\end{array}\right), where k = 1 or a fixed residue modulo p², {\lambda }_1^{\prime }\in {\mathbb{Z}}_p.

Furthermore, we can choose x = -1 to change {\lambda }_1^{\prime } to {-\lambda }_1^{\prime }. Hence, X is isomorphic to the group corresponding to (\begin{array}{cc}0& k\\ 1& t\end{array}), where k = 1 or a fixed residue modulo p², and t\in \left\{0,1,\dots ,\frac{p-1}{2}\right\}, which gives Group (2).

Next, we prove that the different groups corresponding to different parameters in the theorem are mutually non-isomorphic. According to Lemma 4.1, the degree of the minimal polynomial corresponding to isomorphic groups’ type matrices is the same. Hence, (1) and (2) are non-isomorphic. Similarly, according to Lemma 4.1, the determinant values of the type matrices corresponding to isomorphic groups differ by a factor of λ², so the groups corresponding to k = 1 and k being a quadratic non-residue in (2) are non-isomorphic. We will now prove that for the same k value, different t values correspond to non-isomorphic groups. Let G_1=⟨a_1,b_1,c_1⟩ and G_2=⟨a_2,b_2,c_2⟩, X_1=\left(\begin{array}{cc}0& k\\ 1& t_1\end{array}\right), and X_2=\left(\begin{array}{cc}0& k\\ 1& t_2\end{array}\right). According to Lemma 4.1, if G₁ ≌ G₂, then there exists a non-zero element λ such that det(X₂) = λ²det(X₁), and X₂ is similar to λX₁, thus X₂ and λX₁ have the same characteristic polynomial. The characteristic polynomial of X₂ is x² − t₂x − k, and the characteristic polynomial of λX₁ is x² − λt₁x − λ²k. Thus λ = 1 and t₁ = t₂ or λ = -1 and t₁ = -t_2. Note that t takes values in the range between 0 and \frac{p-1}{2}, so t₁ = -t_2 does not occur. Hence, t₁ = t₂. This completes the proof.

Theorem 4.5　 Let G be a finite non-abelian p-group, A be the maximal abelian subgroup of G, and A\cong C_{p^m}\times C_{p^n}, m>n\ge 2. Then G is a regular M₂-group satisfying d(G)=3 and G{}^{\prime }\cong C_p\times C_p if and only if G is isomorphic to one of the following non-isomorphic groups:

(1) G=⟨a,b,c|a^{p^m}=b^{p^n}=c^p=1,[a,b]=1,[a,c]=a^{i{p}^{m-1}},[b,c]=b^{p^{n-1}}⟩,1\le i\le p-1.

(2) G=⟨a,b,c|a^{p^m}=b^{p^n}=c^p=1,[a,b]=1,[a,c]=b^{j{p}^{n-1}},[b,c]=a^{p^{m-1}}⟩, where j = 1 or a fixed quadratic non-residue modulo p.

Proof　(\Leftarrow) As the verification process is trivial, it is omitted.

(\Rightarrow) Without loss of generality, let’s assume

G=(⟨a⟩\times ⟨b⟩)⋊⟨c⟩,G{}^{\prime }=⟨[a,c]⟩\times ⟨[b,c]⟩,

where

o\left(a\right)=p^m,o\left(b\right)=p^n,o(c)=p.

In this case, G^{\mathrm{\prime }}={\mathrm{\Omega }}_1(A), so we can assume

\left[a,c\right]=a^{i{p}^{m-1}}{b}^{j{p}^{n-1}},[b,c]=a^{k{p}^{m-1}}{b}^{l{p}^{n-1}},

where i,j,k,l\in {\mathbb{Z}}_p and il-jk\ne 0.

If l ≠ 0, replacing b with b{a}^{k{l}^{-1}}{p}^{m-n}, we get

[b{a}^{k{l}^{-1}{p}^{m-n}},c]=[b,c]=(a^{k{l}^{-1}{p}^{m-n}}b{)}^{l{p}^{n-1}}.

Let b_1=b{a}^{k{l}^{-1}{p}^{m-n}}, then [b_1,c]={b_1}^{l{p}^{n-1}}. Replacing c with c^{l-1}, let c₁ =c^{l-1}, then [b_1,c_1]={b_1}^{p^{n-1}}. Therefore, we can assume k = 0 and l = 1. Replacing a with a{b}^{-j}, we get

[a{b}^{-j},c]=[a,c][b,c{]}^{-j}=a^{i{p}^{m-1}}=(a{b}^{-j}{)}^{i{p}^{m-1}}.

Let a₁= a{b}^{-j}, which turns j into 0. Thus, we obtain group (1).

Now let’s assume l = 0. In this case, by appropriately replacing c with a suitable power, we can set k = 1, resulting in [b,c]=a^{p^{m-1}}. Then replacing a with a{b}^{-i}, we can obtain [a{b}^{-i},c]=[a,c][b,c{]}^{-i}=b^{j{p}^{n-1}}. Let a₁= a{b}^{-j}, which sets i to 0. Finally, by replacing a with a^x and c with c^x, we can set j to j{x}^2. By choosing an appropriate x, we can always set j to 1 or a fixed quadratic non-residue. Thus, we obtain group (2).

Next, we prove that the groups listed in the theorem with different parameters are non-isomorphic.

Let L/{\mho }_{m-1}\left(A\right)=Z(G/{\mho }_{m-1}(A\left)\right). Then L is a characteristic subgroup of G. For the (1)-type group, L=⟨a,b^p⟩, and particularly, exp(L) = p^m. For the (2)-type group, L=⟨a^p,b⟩, and particularly, exp(L) = p^m−1. Hence, the (1)-type group and the (2)-type group are non-isomorphic.

Now we prove that for the (1)-type group, the groups corresponding to different values of i are non-isomorphic. Indeed, when i₁ and i₂ are different, suppose G_1=⟨a_1,b_1,c_1⟩ and G_2=⟨a_2,b_2,c_2⟩. If G_1\cong G_2, suppose

a_2^{\sigma }=a_1^{k_1}{b}_1^{l_1},b_2^{\sigma }=a_1^{k_2{p}^{m-n}}{b}_1^{l_2},c_2^{\sigma }=c_1^s,(s,p)=1.

From [b_2,c_2{]}^{\sigma }=\left[b_2^{\sigma },c_2^{\sigma }\right], we have

(a_1^{k_2{p}^{m-n}}{b}_1^{l_2}{)}^{p^{n-1}}=[b_1,c_1{]}^{l_{2}s}={b_1}^{l_{2}s{p}^{n-1}}.

This implies p\left|k_2,p\right|l_2(\mathrm{s}-1). However, p cannot simultaneously divide k_2 and l_2, so s = 1. Furthermore, from [a_2,c_2{]}^{\sigma }=[a_2^{\sigma },c_2^{\sigma }], we have

(a_2^{i_2{p}^{m-1}}{)}^{\sigma }=[a_1^{k_1}{b}_1^{l_1},c_1^s]=[a_1,c_1{]}^{k_{1}s}[b_1,c_1{]}^{l_{1}s},

(a_1^{k_1{i}_2{p}^{m-1}})=(a_1{)}^{i_1{k}_{1}s{p}^{m-1}}(b_1{)}^{l_{1}s{p}^{n-1}}.

This implies p\left|l_1,p\right|k_1\left(i_2-i_1\right). Since i₁ and i₂ are different, p|k_1 and p|l_1, which is a contradiction.

We then prove that for the (2)-type group, when j = 1 or j is a fixed quadratic non-residue modulo p, the corresponding groups are non-isomorphic. Suppose

G_1=⟨a_1,b_1,c_1⟩,G_2=⟨a_2,b_2,c_2⟩,

where G₁ has j = 1 and G₂ has j as a quadratic non-residue modulo p. If G_1\cong G_2, suppose

a_2^{\sigma }=a_1^{k_1}{b}_1^{l_1},b_2^{\sigma }=a_1^{k_2{p}^{m-n}}{b}_1^{l_2},c_2^{\sigma }=c_1^s,(s,p)=1.

From [b_2,c_2{]}^{\sigma }=\left[b_2^{\sigma },c_2^{\sigma }\right], we have

(a_2^{p^{m-1}}{)}^{\sigma }=[a_1^{k_2{p}^{m-n}}{b}_1^{l_2},c_1^s]=[b_1,c_1{]}^{l_{2}s}=({a_1}^{p^{m-1}}{)}^{l_{2}s}.

Thus, we have {a_1}^{k_1{p}^{m-1}}={a_1}^{l_{2}s{p}^{m-1}}. From this, it follows that k₁ ≡ l₂s. Furthermore, from [a_2,c_2{]}^{\sigma }=[a_2^{\sigma },c_2^{\sigma }], we have

(b_2^{j{p}^{n-1}}{)}^{\sigma }=[a_1^{k_1}{b}_1^{l_1},c_1^s]=[a_1,c_1{]}^{k_{1}s}[b_1,c_1{]}^{l_{1}s}=b_1^{k_{1}s{p}^{n-1}}{a}_1^{l_{1}s{p}^{m-1}},

implying a_1^{j{k}_2{p}^{m-1}}{b}_1^{j{l}_2{p}^{n-1}}={b_1}^{k_{1}s{p}^{n-1}}{a_1}^{l_{1}s{p}^{m-1}}. From this, we can deduce that k₁s = jl₂ and l₁s = jk₂. Since p does not divide k₂, l₂, or s, taking k₁= l₂s, we have l_2{s}^2\equiv j{l}_2, which means s²= j, a contradiction.

In summary, we have the following theorem.

Theorem 4.6　 Let G be a finite non-abelian p-group, where p is an odd prime. Then G is an M₂-group if and only if G is one of the groups listed in Theorems 2.1, 3.1 and 4.2‒4.5.

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