Finite 𝑝-Group with Small Abelian Subgroups

Abstract

A finite 𝑝-group 𝐺 is said to have the property 𝑃, if, for any abelian subgroup 𝑀 of 𝐺, there is |𝑀𝑍(𝐺)/𝑍(𝐺)|≤𝑝. We show that if 𝐺 satisfies 𝑃, then 𝐺 has the following two types: (1) 𝐺 is isoclinic to some stem groups of order 𝑝⁵, which form an isoclinic family. (2) 𝐺 is isoclinic to a special 𝑝-group of exponent 𝑝. Elementary structures of groups with 𝑃 are determined.

1. Preliminaries

Let 𝑝 be a prime and 𝐺 be a finite 𝑝-group. A group 𝐺 is said to have the property 𝑃, if any abelian subgroup 𝑀 of 𝐺 satisfies |𝑀𝑍(𝐺)/𝑍(𝐺)|≤𝑝. It is well known that if 𝑀𝑍(𝐺)/𝑍(𝐺) is cyclic, then 𝑀 is abelian. It rises to consider the groups with property 𝑃.

Mann [1] obtained the structure of 2-groups with property 𝑃; leting any abelian subgroup 𝑀 of 2-group 𝐺 satisfy |𝑀𝑍(𝐺)/𝑍(𝐺)|≤𝑝, then 𝐺 is isoclinic to a dihedral group. And for any odd prime 𝑝 he showed that if any abelian subgroup 𝑀 of 𝐺 satisfies |𝑀𝑍(𝐺)/𝑍(𝐺)|≤𝑝, then 𝐺/𝑍(𝐺) is an elementary abelian 𝑝-group or a nonabelian group of order 𝑝³ and exponent 𝑝.

Lemma 1 (see [1]). Let 𝐺 be a finite 𝑝-group which satisfies the property 𝑃. Then 𝐺/𝑍(𝐺) is an elementary abelian 𝑝-group or a nonabelian group of order 𝑝³ and exponent 𝑝.

This paper further discusses groups with property 𝑃. In the following all considered groups are finite 𝑝-groups among which 𝑝 is an odd prime.

First we state some notions and lemmas.

Definition 2. Groups 𝐺 and 𝐻 are said to be isoclinic, if there exist isomorphisms 𝜎∶𝐺/𝑍(𝐺)→𝐻/𝑍(𝐻) and 𝜃∶𝐺^′→𝐻^′ which are compatible, in the sense that (𝑎𝑍(𝐺))^𝜎=𝑐𝑍(𝐻) and (𝑏𝑍(𝐺))^𝜎=𝑑𝑍(𝐻) (𝑐,𝑑∈𝐻−𝑍(𝐻)) imply that [𝑎,𝑏]^𝜃=[𝑐,𝑑].

Definition 3. A finite 𝑝-group 𝐺 is called a stem group, if 𝑍(𝐺)≤𝐺^′.

Definition 4. Group 𝐺 is called a special 𝑝-group, if 𝐺 is an elementary abelian 𝑝-group or 𝐺^′=𝑍(𝐺)=Φ(𝐺) is an elementary abelian 𝑝-group.

Lemma 5 (see [2]). Finite 𝑝-groups are isoclinic to stem groups.

Lemma 6 (see [3]). Let 𝐺 be a finite regular 𝑝-group, 𝑠≥0; then for any 𝑎,𝑏∈𝐺, 𝑎^𝑝𝑠=𝑏^𝑝𝑠 if and only if (𝑎⁻¹𝑏)^𝑝𝑠=1.

Lemma 7 (see [4]). Let 𝐺 be a regular 𝑝-group; then for any 𝑖, Ω_𝑖(𝐺) is the set of elements of order 𝑝^𝑖.

2. Main Results

We need the following result.

Lemma 8. Suppose that 𝐺 and 𝐻 are isoclinic. If 𝐺 satisfies 𝑃, then so does 𝐻.

Proof. Assume that isomorphisms 𝜎∶𝐺/𝑍(𝐺)→𝐻/𝑍(𝐻) and 𝜃∶𝐺^′→𝐻^′ are compatible. And assume that 𝑀 is an abelian subgroup of 𝐻. Then there exists a subgroup 𝑁 of 𝐺 such that (𝑁/𝑍(𝐺))^𝜎=𝑀𝑍(𝐻)/𝑍(𝐻). Suppose that 𝑐,𝑑∈𝑁⧵𝑍(𝐺) and (𝑐𝑍(𝐺))^𝜎=𝑎𝑍(𝐻),(𝑑𝑍(𝐺))^𝜎=𝑏𝑍(𝐻). Then 𝑎,𝑏∈𝑀⧵𝑍(𝐻). Then [𝑐,𝑑]^𝜃=[𝑎,𝑏]=1 since 𝜎 and 𝜃 are compatible. It follows that |𝑁/𝑍(𝐺)|⩽𝑝 since 𝐺 satisfies 𝑃. Hence |𝑀𝑍(𝐻)/𝑍(𝐻)|⩽𝑝. This completes the proof.

Now we state our main result.

Theorem 9. Let 𝐺 be a finite 𝑝-group satisfying property 𝑃, then
(1)𝐺is isoclinic to some stem groups of order 𝑝⁵, which form an isoclinic family, or(2)𝐺 is isoclinic to a special 𝑝-group of exponent 𝑝.

Proof. By Lemma 1, 𝐺/𝑍(𝐺) is an elementary abelian 𝑝-group or a nonabelian group of order 𝑝³ and exponent 𝑝.
(1) Suppose that 𝐺/𝑍(𝐺) is not abelian.
Then 𝐺/𝑍(𝐺) is of order 𝑝³ and exponent 𝑝. By Lemma 5 we may suppose that 𝑍(𝐺)⩽𝐺^′. Noting that the nilpotent class of 𝐺 is 3, thus 𝑍(𝐺)≠𝐺^′. It follows that 𝐺^′ is abelian since |𝐺^′/𝑍(𝐺)|=𝑝. Assuming that 𝐺/𝑍(𝐺)=⟨𝑎𝑍(𝐺),𝑏𝑍(𝐺)⟩, then 𝐺=⟨𝑎,𝑏⟩ since 𝑍(𝐺)<𝐺^′. Let 𝐻=⟨𝑎,𝑎^𝑏⟩=⟨𝑎,[𝑎,𝑏]⟩. Then 𝐻 is nilpotent of class 2 and thus 𝐻 is a regular 𝑝-group. Note that 𝑎^𝑝∈𝑍(𝐺) since 𝐺/𝑍(𝐺) is of exponent 𝑝. So (𝑏⁻¹𝑎𝑏)^𝑝=𝑏⁻¹𝑎^𝑝𝑏=𝑎^𝑝, following that [𝑎,𝑏]^𝑝=(𝑎⁻¹𝑏⁻¹𝑎𝑏)^𝑝=1 by Lemma 6. Hence 𝐺^′=⟨[𝑎,𝑏]⟩^𝐺=⟨[𝑎,𝑏],[𝑎,𝑏,𝑎],[𝑎,𝑏,𝑏]⟩ is abelian and |𝐺^′|⩽𝑝³. If |𝐺^′|=𝑝, then 𝐺 is inner-abelian and 𝐺/𝑍(𝐺)≅𝑍²_𝑝, a contradiction. If |𝐺^′|=𝑝², there are two cases.
Case  1  ([𝑎,𝑏,𝑎]=1 or [𝑎,𝑏,𝑏]=1). If [𝑎,𝑏,𝑎]=1, then ⟨𝐺^′,𝑎⟩ is abelian and ⟨𝐺^′,𝑎⟩/𝑍(𝐺)≅𝑍²_𝑝, a contradiction. For [𝑎,𝑏,𝑏]=1, then ⟨𝐺^′,𝑏⟩ is abelian and ⟨𝐺^′,𝑏⟩/𝑍(𝐺)≅𝑍²_𝑝, a contradiction.
Case  2  ([𝑎,𝑏,𝑎]=[𝑎,𝑏,𝑏]). Then we have [𝑎,𝑏,𝑎][𝑎,𝑏,𝑏⁻¹]=[𝑎,𝑏,𝑎𝑏⁻¹]=1. Thus 𝐻=⟨[𝑎,𝑏],𝑎𝑏⁻¹,𝑍(𝐺)⟩ is abelian and 𝐻/𝑍(𝐺)≅𝑍²_𝑝, a contradiction. Hence |𝐺^′|=𝑝³ and then |𝐺|=𝑝⁵.
Now we show that all groups of order 𝑝⁵ with 𝑃 are isoclinic.
Suppose that 𝐺=⟨𝑎₁,𝑏₁⟩, 𝐻=⟨𝑎₂,𝑏₂⟩ are groups of order 𝑝⁵ with 𝑃. Then 𝐺/𝑍(𝐺)=⟨𝑎₁𝑍(𝐺),𝑏₁𝑍(𝐺)⟩≅𝐸_𝑝3 and 𝐻/𝑍(𝐻)=⟨𝑎₂𝑍(𝐻),𝑏₂𝑍(𝐻)⟩≅𝐸_𝑝3.
Set 𝑐₁=[𝑎₁,𝑏₁], 𝑐₂=[𝑎₂,𝑏₂] and define 𝜎∶𝐺/𝑍(𝐺)→𝐻/𝑍(𝐻) such that (𝑎₁𝑍(𝐺))^𝜎=𝑎₂𝑍(𝐻), (𝑏₁𝑍(𝐺))^𝜎=𝑏₂𝑍(𝐻).
Then 𝜎 spans an isomorphism and (𝑐₁𝑍(𝐺))^𝜎=[𝑎₁𝑍(𝐺),𝑏₁𝑍(𝐺)]^𝜎=[𝑎₂𝑍(𝐻),𝑏₂𝑍(𝐻)]=𝑐₂𝑍(𝐻). Note that 𝑍(𝐺)=⟨[𝑐₁,𝑎₁]⟩×⟨[𝑐₁,𝑏₁]⟩,𝑍(𝐻)=⟨[𝑐₂,𝑎₂]⟩×⟨[𝑐₂,𝑏₂]⟩ and 𝐺^′=⟨𝑐₁⟩×𝑍(𝐺), 𝐻^′=⟨𝑐₂⟩×𝑍(𝐻).
Setting 𝑐^𝜃₁=𝑐₂, [𝑐₁,𝑎₁]^𝜃=[𝑐₂,𝑎₂], and [𝑐₁,𝑏₁]^𝜃=[𝑐₂,𝑏₂], then 𝜃 deduces an isomorphism from 𝐺^′ onto 𝐻^′. Now we show that 𝜎 and 𝜃 are compatible.
For any 𝑥,𝑦∈𝐺, write 𝑥=𝑎^𝑖1₁𝑏^𝑖2₁𝑐^𝑖3₁𝑧₁, 𝑦=𝑎^𝑗1₁𝑏^𝑗2₁𝑐^𝑗3₁𝑧₂, where 𝑧₁,𝑧₂∈𝑍(𝐺). Then (𝑥𝑍(𝐺))^𝜎=𝑎^𝑖1₁𝑏^𝑖2₁𝑐^𝑖3₁𝑍(𝐺)^𝜎=𝑎^𝑖1₁𝑍(𝐺)^𝜎𝑏^𝑖2₁𝑍(𝐺)^𝜎𝑐^𝑖3₁𝑍(𝐺)^𝜎=𝑎^𝑖1₂𝑍(𝐻)𝑏^𝑖2₂𝑍(𝐻)𝑐^𝑖3₂𝑍(𝐻)=𝑎^𝑖1₂𝑏^𝑖2₂𝑐^𝑖3₂𝑍(𝐻). (1)
Similarly, (𝑦𝑍(𝐺))^𝜎=𝑎^𝑗1₂𝑏^𝑗2₂𝑐^𝑗3₂𝑍(𝐻). Note that 𝑥,𝑦=𝑎^𝑖1₁𝑏^𝑖2₁𝑐^𝑖3₁𝑧₁,𝑎^𝑗1₁𝑏^𝑗2₁𝑐^𝑗3₁𝑧₂=𝑎^𝑖1₁𝑏^𝑖2₁𝑐^𝑖3₁,𝑎^𝑗1₁𝑏^𝑗2₁𝑐^𝑗3₁=𝑎^𝑖1₁𝑏^𝑖2₁,𝑎^𝑗1₁𝑏^𝑗2₁𝑐^𝑗3₁𝑐^𝑖3₁,𝑎^𝑗1₁𝑏^𝑗2₁𝑐^𝑗3₁×𝑎^𝑖1₁𝑏^𝑖2₁,𝑎^𝑗1₁𝑏^𝑗2₁𝑐^𝑗3₁,𝑐^𝑖3₁=𝑎^𝑖1₁𝑏^𝑖2₁,𝑎^𝑗1₁𝑏^𝑗2₁𝑎^𝑖1₁,𝑐^𝑗3₁𝑏^𝑖2₁,𝑐^𝑗3₁𝑐^𝑖3₁,𝑎^𝑗1₁×𝑐^𝑖3₁,𝑏^𝑗2₁=𝑐₁,𝑎₁^{𝑖3𝑗1−𝑖1𝑗3}𝑐₁,𝑏₁^{𝑖3𝑗2−𝑖2𝑗3}𝑎^𝑖1₁𝑏^𝑖2₁,𝑎^𝑗1₁𝑏^𝑗2₁=𝑐₁,𝑎₁^{𝑖3𝑗1−𝑖1𝑗3}𝑐₁,𝑏₁^{𝑖3𝑗2−𝑖2𝑗3}×𝑎^𝑖1₁𝑏^𝑖2₁,𝑎^𝑗1₁𝑎^𝑖1₁𝑏^𝑖2₁,𝑏^𝑗2₁𝑎^𝑖1₁𝑏^𝑖2₁,𝑎^𝑗1₁,𝑏^𝑗2₁=𝑐₁,𝑎₁^{𝑖3𝑗1−𝑖1𝑗3}𝑐₁,𝑏₁^{𝑖3𝑗2−𝑖2𝑗3}𝑐₁,𝑎₁^𝑖12𝑖21×𝑐₁,𝑏₁^𝑖11𝑖22×𝑐₁,𝑎₁^𝑖21𝑗12𝑐₁,𝑏₁^𝑗11𝑖22×𝑐₁,𝑏₁^{𝑖1𝑖2𝑗2−𝑖2𝑗1𝑗2}𝑐^{𝑖1𝑗2−𝑖2𝑗1}₁=𝑐₁,𝑎₁^{𝑖3𝑗1−𝑖1𝑗3+𝑖21𝑗12+𝑖12𝑗21}×𝑐₁,𝑏₁^{𝑖3𝑗2−𝑖2𝑗3+𝑖1𝑖2𝑗2−𝑖2𝑗1𝑗2+𝑗11𝑖22+𝑖11𝑗22}𝑐^{𝑖1𝑗2−𝑖2𝑗1}₁. (2)
Set 𝑓₁=𝑖₃𝑗₁−𝑖₁𝑗₃+𝑖₂1𝑗₁2+𝑖₁2𝑗₂1,𝑓₂=𝑖₃𝑗₂−𝑖₂𝑗₃+𝑖₁𝑖₂𝑗₂−𝑖₂𝑗₁𝑗₂+𝑗₁1𝑖₂2+𝑖₁1𝑗₂2,𝑓₃=𝑖₁𝑗₂−𝑖₂𝑗₁. (3)
Then [𝑥,𝑦]=𝑐^𝑓3₁[𝑐₁,𝑎₁]^𝑓1[𝑐₁,𝑏₁]^𝑓2. Thus [𝑥,𝑦]^𝜃=𝑐^𝑓3₂[𝑐₂,𝑎₂]^𝑓1[𝑐₂,𝑏₂]^𝑓2=[𝑎^𝑖1₂𝑏^𝑖2₂𝑐^𝑖3₂,𝑎^𝑗1₂𝑏^𝑗2₂𝑐^𝑗3₂]. Hence 𝜎 and 𝜃 are compatible.
(2)   Suppose that 𝐺/𝑍(𝐺) is an elementary abelian 𝑝-group.
Note that 𝐺^′≤𝑍(𝐺) since 𝐺/𝑍(𝐺) is abelian. By Lemma 5 we may suppose that 𝑍(𝐺)=𝐺^′. If exp𝐺=𝑝, then 𝐺 is a special 𝑝-group of exponent 𝑝. So suppose that exp𝐺>𝑝.
Let 𝐺/𝑍(𝐺)=⟨𝑎₁⟩×⟨𝑎₂⟩×⋯×⟨𝑎_𝑛⟩, where 𝑎_𝑖=𝑎_𝑖𝑍(𝐺). Then 𝐺=⟨𝑎₁,𝑎₂,…,𝑎_𝑛⟩ and 𝑎^𝑝_𝑖∈𝑍(𝐺). Note that 𝐺^′=⟨[𝑎_𝑘,𝑎_𝑙]⟩^𝐺 and [𝑎_𝑘,𝑎_𝑙]^𝑝=[𝑎^𝑝_𝑘,𝑎_𝑙]=1. Hence 𝐺^′ is an elementary abelian 𝑝-group of exponent 𝑝. Suppose that 𝐺^′=⟨𝑐_𝑘𝑙∣𝑓_𝑚(𝑐_𝑘𝑙)=1⟩, where 𝑐_𝑘𝑙=[𝑎_𝑘,𝑎_𝑙] and 𝑓_𝑚(𝑐_𝑘𝑙)=1 indicate the laws of 𝐺^′. Suppose that 𝐾=⟨𝑑_𝑘𝑙∣𝑓_𝑚(𝑑_𝑘𝑙)=1⟩ and 𝐾≅𝐺^′. By Schreier group expansion theories, we can add elements 𝑏₁,𝑏₂,…,𝑏_𝑛 of order 𝑝 such that [𝑏_𝑘,𝑏_𝑙]=𝑑_𝑘𝑙 into group 𝐾, note that 𝐻={𝑏₁,𝑏₂,…,𝑏_𝑛∣𝑏^𝑝_𝑖=1,[𝑏_𝑘,𝑏_𝑙]=𝑑_𝑘𝑙}. Then 𝐻 is a group by Schreier group expansion theories.
Define a map 𝜎∶𝑐_𝑘𝑙↦𝑑_𝑘𝑙 for 𝑘,𝑙=1,…,𝑛. Then 𝜎 deduces an isomorphism from 𝐺^′ onto 𝐻^′. Note that 𝐻/𝐻^′=⟨𝑏₁⟩×⟨𝑏₂⟩×⋯×⟨𝑏_𝑛⟩ and 𝐺/𝐺^′=⟨𝑎₁⟩×⟨𝑎₂⟩×⋯×⟨𝑎_𝑛⟩.
Setting 𝜎₁∶𝑎_𝑖𝐺^′↦𝑏_𝑖𝐻^′, then 𝜎₁ spans an isomorphism from 𝐺/𝐺^′ onto 𝐻/𝐻^′.
For any 𝑥𝐺^′,𝑦𝐺^′∈𝐺/𝐺^′, assume that 𝑥=𝑎^𝑖1₁𝑎^𝑖2₂⋯𝑎^𝑖𝑛_𝑛𝑐₁,𝑦=𝑎^𝑗1₁𝑎^𝑗2₂⋯𝑎^𝑗𝑛_𝑛𝑐₂, where 𝑐₁,𝑐₂∈𝐺^′.
Then (𝑥𝐺^′)^𝜎1=(𝑎^𝑖1₁𝑎^𝑖2₂⋯𝑎^𝑖𝑛_𝑛𝐺^′)^𝜎1=𝑏^𝑖1₁𝑏^𝑖2₂⋯𝑏^𝑖𝑛_𝑛𝐻^′(𝑦𝐺^′)^𝜎1=(𝑎^𝑗1₁𝑎^𝑗2₂⋯𝑎^𝑗𝑛_𝑛𝐺^′)^𝜎1=𝑏^𝑗1₁𝑏^𝑗2₂⋯𝑏^𝑗𝑛_𝑛𝐻^′. So [𝑥,𝑦]^𝜎2=[𝑎^𝑖1₁𝑎^𝑖2₂⋯𝑎^𝑖𝑛_𝑛𝑐₁,𝑎^𝑗1₁𝑎^𝑗2₂⋯𝑎^𝑗𝑛_𝑛𝑐₂]^𝜎2=(∏_{1⩽𝑘⩽𝑙⩽𝑛}[𝑎_𝑘,𝑎_𝑙]^{𝑖𝑘𝑗𝑙−𝑖𝑙𝑗𝑘})^𝜎2=∏_{1⩽𝑘⩽𝑙⩽𝑛}[𝑏_𝑘,𝑏_𝑙]^{𝑖𝑘𝑗𝑙−𝑖𝑙𝑗𝑘}=[𝑏^𝑖1₁𝑏^𝑖2₂⋯𝑏^𝑖𝑛_𝑛,𝑏^𝑗1₁𝑏^𝑗2₂⋯𝑏^𝑗𝑛_𝑛].
Thus 𝜎₁ and 𝜎₂ are compatible.
For any 𝑥∈𝑍(𝐻), write 𝑥=𝑏^𝑖1₁𝑏^𝑖2₂⋯𝑏^𝑖𝑛_𝑛𝑐, where 𝑐∈𝐻^′. Then [𝑏^𝑖1₁𝑏^𝑖2₂⋯𝑏^𝑖𝑛_𝑛𝑐,𝑏_𝑖]=1.
Note that (𝑏^𝑖1₁𝑏^𝑖2₂⋯𝑏^𝑖𝑛_𝑛𝐻^′)^𝜎1−1=(𝑏^𝑖1₁𝑏^𝑖2₂⋯𝑏^𝑖𝑛_𝑛)^𝜎1−1=𝑎^𝑖1₁𝑎^𝑖2₂⋯𝑎^𝑖𝑛_𝑛=𝑎^𝑖1₁𝑎^𝑖2₂⋯𝑎^𝑖𝑛_𝑛𝐺^′, and similarly (𝑏_𝑖𝐻^′)^𝜎1−1=𝑎_𝑖𝐺^′. Then [𝑏^𝑖1₁𝑏^𝑖2₂⋯𝑏^𝑖𝑛_𝑛,𝑏_𝑖]^𝜎1−1=[𝑎^𝑖1₁𝑎^𝑖2₂⋯𝑎^𝑖𝑛_𝑛,𝑎_𝑖]=1. Hence 𝑎^𝑖1₁𝑎^𝑖2₂⋯𝑎^𝑖𝑛_𝑛∈𝑍(𝐺)=𝐺^′.
It follows that 𝑏^𝑖1₁𝑏^𝑖2₂⋯𝑏^𝑖𝑛_𝑛𝐻^′=(𝐺^′)^𝜎1=𝐻^′ and thus 𝑏^𝑖1₁𝑏^𝑖2₂⋯𝑏^𝑖𝑛_𝑛∈𝐻^′. So 𝑍(𝐻)⩽𝐻^′. As a result 𝐻^′=𝑍(𝐻). Since 𝐻={𝑏₁,𝑏₂,…,𝑏_𝑛∣𝑏^𝑝_𝑖=1,[𝑏_𝑘,𝑏_𝑙]=𝑑_𝑘𝑙}. We know that ℧₁(𝐻)=⟨𝑏^𝑝₁,𝑏^𝑝₂,…,𝑏^𝑝_𝑛⟩=1; then exp𝐻=𝑝. Hence 𝐺 is isoclinic to a special 𝑝-group of exponent 𝑝.

Note. In the sense of isoclinism, to investigate groups with 𝑃 is to consider the special 𝑝-groups of exponent 𝑝.

We need the following result.

Lemma 10. Let 𝐺 be a special 𝑝-group of exponent 𝑝. Then(1)if 𝑑(𝐺)=𝑛, then |𝐺^′|≤𝑝^{𝑛(𝑛−1)/2};(2)for any 𝑥,𝑦∈𝐺, if ⟨𝑥𝑍(𝐺)⟩=⟨𝑦𝑍(𝐺)⟩, then 𝐺 satisfies 𝑃.

Then we deduce the following.

Theorem 11. Let 𝐺 be a special 𝑝-group of exponent 𝑝 and 𝑑(𝐺)=𝑛, then(1)if |𝐺^′|<𝑝^𝑛−1, then 𝐺 does not satisfy 𝑃.(2)if |𝐺^′|=𝑝^{𝑛(𝑛−1)/2}, then 𝐺 satisfies 𝑃.

Proof. Suppose that 𝐺=⟨𝑎₁,𝑎₂,…,𝑎_𝑛⟩. Since 𝐺^′ is an elementary abelian 𝑝-group, we may see 𝐺^′ as an additive group of the vector space on 𝐺𝐹(𝑝).
(1) Assume that |𝐺^′|<𝑝^𝑛−1. Then [𝑎₁,𝑎₂],[𝑎₁,𝑎₃],…,[𝑎₁,𝑎_𝑛] are linearly dependent. So there exist some integers which are not all 0 mod 𝑝 such that [𝑎₁,𝑎₂]^𝑘1[𝑎₁,𝑎₃]^𝑘2⋯[𝑎₁,𝑎_𝑛]^𝑘𝑛−1=1. Thus ⟨𝑎₁,𝑎^𝑘1₂𝑎^𝑘2₃⋯𝑎^𝑘𝑛−1_𝑛⟩ is abelian. However, note that |⟨𝑎₁,𝑎^𝑘1₂𝑎^𝑘2₃⋯𝑎^𝑘𝑛−1_𝑛⟩𝑍(𝐺)/𝑍(𝐺)|≠𝑝 since 𝑎₁,𝑎₂,…,𝑎_𝑛 are generators of 𝐺. So 𝐺 does not have 𝑃.
(2) Assume that |𝐺^′|=𝑝^{𝑛(𝑛−1)/2}. Then 𝐺^′=⟨[𝑎₁,𝑎₂],…,[𝑎₁,𝑎_𝑛],…,[𝑎_𝑛−1,𝑎_𝑛]⟩. For any 𝑥,𝑦∈𝐺, assume that 𝑥=𝑎^𝑖1₁𝑎^𝑖2₂⋯𝑎^𝑖𝑛_𝑛𝑐₁ and 𝑦=𝑎^𝑗1₁𝑎^𝑗2₂⋯𝑎^𝑗𝑛_𝑛𝑐₂, where 𝑐₁,𝑐₂∈𝐺^′. Then [𝑥,𝑦]=[𝑎^𝑖1₁𝑎^𝑖2₂⋯𝑎^𝑖𝑛_𝑛𝑐₁,𝑎^𝑗1₁𝑎^𝑗2₂⋯𝑎^𝑗𝑛_𝑛𝑐₂]=[𝑎^𝑖1₁𝑎^𝑖2₂⋯𝑎^𝑖𝑛_𝑛,𝑎^𝑗1₁𝑎^𝑗2₂⋯𝑎^𝑗𝑛_𝑛]=∏_{1⩽𝑘⩽𝑙⩽𝑛}[𝑎_𝑘,𝑎_𝑙]^{𝑖𝑘𝑗𝑙−𝑖𝑙𝑗𝑘}.
Looking at 𝐺^′ as an additive group of the vector space on 𝐺𝐹(𝑝), then [𝑎₁,𝑎₂],…,[𝑎₁,𝑎_𝑛],…,[𝑎_𝑛−1,𝑎_𝑛] are linearly independent. If [𝑥,𝑦]=1, then 𝑖_𝑘𝑗_𝑙−𝑖_𝑙𝑗_𝑘=0mod 𝑝 for 1⩽𝑘⩽𝑙⩽𝑛. Thus 𝑖_𝑙/𝑗_𝑙=𝑖₂/𝑗₂⋯=𝑖_𝑛/𝑗_𝑛. Setting 𝑖_𝑙/𝑗_𝑙=𝑚, then 𝑥=𝑎^𝑗1𝑚₁𝑎^𝑗2𝑚₂⋯𝑎^𝑗𝑛𝑚_𝑛𝑐₁=(𝑎^𝑗1₁𝑎^𝑗2₂)^𝑚𝑐^𝑚2₂𝑎^𝑗3𝑚₃⋯𝑎^𝑗𝑛𝑚_𝑛𝑐₁=(𝑎^𝑗1₁𝑎^𝑗2₂,…,𝑎^𝑗𝑛_𝑛𝑐₂)^𝑚𝑐=𝑦^𝑚𝑐, where 𝑐∈𝐺^′. Hence ⟨𝑥𝑍(𝐺)⟩=⟨𝑦^𝑚𝑍(𝐺)⟩⩽⟨𝑦𝑍(𝐺)⟩. By Lemma 10  𝐺 satisfies the property 𝑃.

Corollary 12. Let 𝐺 be a special 𝑝-group of exponent 𝑝. Then(1)if 𝑑(𝐺)=2, then 𝐺 satisfies the property 𝑃.(2)if 𝑑(𝐺)=3, then 𝐺 satisfies the property 𝑃 if and only if |𝐺^′|=𝑝³.

Proof. (1) If 𝑑(𝐺)=2, then 𝐺 is an inner-abelian 𝑝-group. Obviously, 𝐺 has the property 𝑃.
(2) Assume that 𝑑(𝐺)=3 and 𝐺=⟨𝑎₁,𝑎₂,𝑎₃⟩. By Theorem 11   𝐺 satisfies 𝑃 if |𝐺^′|=𝑝³, but it does not if |𝐺^′|=𝑝. Assuming that |𝐺^′|=𝑝², then [𝑎₁,𝑎₂], [𝑎₁,𝑎₃], [𝑎₂,𝑎₃] are linearly dependent. So there exist integers 𝑘₁,𝑘₂,𝑘₃ which are not all 0 mod 𝑝 such that [𝑎₁,𝑎₂]^𝑘1[𝑎₁,𝑎₃]^𝑘2[𝑎₂,𝑎₃]^𝑘3=1. Hence ⟨𝑎₁,𝑎^𝑘1₂𝑎^𝑘2₃𝑎^𝑖𝑘3₂𝑎^𝑗𝑘3₃⟩ is abelian.
Since 𝑎₁, 𝑎₂, and 𝑎₃ are generators of 𝐺, |⟨𝑎₁,𝑎^𝑘1₂𝑎^𝑘2₃𝑎^𝑖𝑘3₂𝑎^𝑗𝑘3₃⟩𝑍(𝐺)/𝑍(𝐺)|≠𝑝, then 𝐺 does not satisfy 𝑃;

Theorem 13. Assume that 𝐺=⟨𝑎₁,…,𝑎_𝑛⟩ (𝑛≥4) is a special 𝑝-group of exponent 𝑝 and |𝐺^′|=𝑝^𝑚, where 𝑛−1≤𝑚<𝑛(𝑛−1)/2. Suppose that 𝐺^′=⟨[𝑎_𝑘1,𝑎_𝑙1],…,[𝑎_𝑘𝑛−1,𝑎_𝑙𝑛−1],…,[𝑎_𝑘𝑚,𝑎_𝑙𝑚]⟩ for 𝑘_𝑖,𝑙_𝑖∈{1,2,…,𝑛}. If 𝐺 satisfies the following properties, then 𝐺 has 𝑃.(1)[𝑎_𝑘,𝑎_𝑙]≠1 for any 𝑘≠𝑙.(2)For any distinct 𝑙≠𝑚∈{1,…,𝑛} and for not all 0 mod  𝑝 integers 𝑖,𝑗,[𝑎_𝑘,𝑎_𝑙]^𝑖≠[𝑎_𝑘,𝑎_𝑚]^𝑗.(3)For [𝑎_𝑘1,𝑎_𝑙1],[𝑎_𝑘2,𝑎_𝑙2],…,[𝑎_𝑘𝑛−1,𝑎_𝑙𝑛−1], which satisfy {𝑘₁,𝑙₁,𝑘₂,𝑙₂,…,𝑘_𝑛−1,𝑙_𝑛−1}={1,2,…,𝑛} and 𝑘_𝑖+1∈{𝑘_𝑖,𝑙_𝑖} or 𝑙_𝑖+1∈{𝑘_𝑖,𝑙_𝑖}, where 𝑖=1,2,…,𝑛−2.(4)Any [𝑎_𝑘𝑖,𝑎_𝑙𝑖] of [𝑎_𝑘𝑚+1,𝑎_𝑙𝑚+1],…,[𝑎_{𝑘𝑛(𝑛−1)/2},𝑎_{𝑙𝑛(𝑛−1)/2}] can be expressed as [𝑎_𝑘𝑖,𝑎_𝑙𝑖]=[𝑎_𝑘𝑛,𝑎_𝑙𝑛]^𝑡1𝑖[𝑎_𝑘𝑛+1,𝑎_𝑙𝑛+1]^𝑡2𝑖⋯[𝑎_𝑘𝑚,𝑎_𝑙𝑚]^{𝑡(𝑚−𝑛+1)𝑖}, where 𝑡_1𝑖,𝑡_2𝑖,…,𝑡_{(𝑚−𝑛+1)𝑖} are integers.

Proof. For any 𝑥,𝑦∈𝐺, write 𝑥=𝑎^𝑖1₁𝑎^𝑖2₂⋯𝑎^𝑖𝑛_𝑛𝑐₁ and 𝑦=𝑎^𝑗1₁𝑎^𝑗2₂⋯𝑎^𝑗𝑛_𝑛𝑐₂, where 𝑐₁,𝑐₂∈𝐺^′. Then 𝑥,𝑦=𝑎^𝑖1₁𝑎^𝑖2₂⋯𝑎^𝑖𝑛_𝑛𝑐₁,𝑎^𝑗1₁𝑎^𝑗2₂⋯𝑎^𝑗𝑛_𝑛𝑐₂=𝑎_𝑘1,𝑎_𝑙1^{𝑖𝑘1𝑗𝑙1−𝑖𝑙1𝑗𝑘1}𝑎_𝑘2,𝑎_𝑙2^{𝑖𝑘2𝑗𝑙2−𝑖𝑙2𝑗𝑘2}⋯𝑎_𝑘𝑛−1,𝑎_𝑙𝑛−1^𝑓1×𝑎_𝑘𝑛,𝑎_𝑙𝑛^𝑓2⋯𝑎_𝑘𝑚,𝑎_𝑙𝑚^𝑓3, (4)
where 𝑓₁=𝑖_𝑘𝑛−1𝑗_𝑙𝑛−1−𝑖_𝑙𝑛−1𝑗_𝑘𝑛−1,𝑓₂=𝑖_𝑘𝑛𝑗_𝑙𝑛−𝑖_𝑙𝑛𝑗_𝑘𝑛+^{𝑛(𝑛−1)/2}_𝑠=𝑚+1𝑡_1𝑠𝑖_𝑘𝑠𝑗_𝑙𝑠−𝑖_𝑙𝑠𝑗_𝑘𝑠,𝑓₃=𝑖_𝑘𝑚𝑗_𝑙𝑚−𝑖_𝑙𝑚𝑗_𝑘𝑚+^{𝑛(𝑛−1)/2}_𝑠=𝑚+1𝑡_{(𝑛−𝑚+1)𝑠}𝑖_𝑘𝑠𝑗_𝑙𝑠−𝑖_𝑙𝑠𝑗_𝑘𝑠. (5)
Note that [𝑎_𝑘1,𝑎_𝑙1],[𝑎_𝑘2,𝑎_𝑙2],…,[𝑎_𝑘𝑛−1,𝑎_𝑙𝑛−1],[𝑎_𝑘𝑛,𝑎_𝑙𝑛],…,[𝑎_𝑘𝑚,𝑎_𝑙𝑚] are linearly independent.
Hence if [𝑥,𝑦]=1 then 𝑖_𝑘𝑟𝑗_𝑙𝑟−𝑖_𝑙𝑟𝑗_𝑘𝑟=𝑖_𝑘𝑟𝑗_𝑙𝑟−𝑖_𝑙𝑟𝑗_𝑘𝑟+∑^{𝑛(𝑛−1)/2}_𝑠=𝑚+1𝑡_𝑟𝑠(𝑖_𝑘𝑠𝑗_𝑙𝑠−𝑖_𝑙𝑠𝑗_𝑘𝑠)=0 mod 𝑝, where 𝑟=1,2,…,𝑛,…,𝑚. It follows that 𝑖_𝑙/𝑗_𝑙=𝑖₂/𝑗₂⋯=𝑖_𝑛/𝑗_𝑛 by (1) and (2). Supposing that 𝑖_𝑙/𝑗_𝑙=𝑚, then 𝑥=𝑦^𝑚𝑐 and thus ⟨𝑥𝑍(𝐺)⟩=⟨𝑦^𝑚𝑍(𝐺)⟩⩽⟨𝑦𝑍(𝐺)⟩. Hence |⟨𝑥,𝑦,𝑍(𝐺)⟩/𝑍(𝐺)|=|⟨𝑦,𝑍(𝐺)⟩/𝑍(𝐺)|=𝑝, following that 𝐺 has 𝑃.

Corollary 14. Let 𝐺 be a special 𝑝-group of exponent 𝑝 and 𝑑(𝐺)=𝑛 (𝑛≥4). If |𝐺^′|=𝑝^𝑛−1 or 𝑝^𝑛, then 𝐺 does not satisfy the property 𝑃.

Proof. By Theorem 13 if |𝐺^′|=𝑝^𝑛−1 or 𝑝^𝑛, then 𝐺 does not satisfy the conditions of Theorem 9. So 𝐺 does not have 𝑃.

Acknowledgments

The authors thank Professor Mingyao Xu for his valuable suggestions. This work was supported by the Natural Science Foundation of Shanxi (Grant no. 10771132), the Natural Science Foundation of the Ministry of Education of China for the Returned Overseas Scholars (Grant no. 2008101), and the Natural Science Foundation of Shanxi for the Returned Overseas Scholars (Grant no. 201199).