Commutators in finite p-groups with 3-generator derived subgroup

TL;DR

This paper extends previous results by showing that in finite p-groups with a 3-generator derived subgroup, all elements are commutators, removing the need for the derived subgroup to be abelian.

Contribution

It generalizes Guralnick's theorem by removing the abelian condition and explores similar properties in pro-p groups, completing the understanding of commutators in these groups.

Findings

All elements of the derived subgroup are commutators without the abelian condition.

The result holds when the action on G' is uniserial modulo (G')^p and |G':(G')^p| ≤ p^{p-1}.

Analogous results are valid for pro-p groups.

Abstract

It is well known that, in general, the set of commutators of a group $G$ may not be a subgroup. Guralnick showed that if $G$ is a finite $p$-group with $p\ge 5$ such that $G'$ is abelian and $3$-generator, then all the elements of the derived subgroup are commutators. In this paper, we extend Guralnick's result by showing that the condition of $G'$ to be abelian is not needed. In this way, we complete the study of this property in finite $p$-groups in terms of the number of generators of the derived subgroup. We will also see that the same result is true when the action of $G$ on $G'$ is uniserial modulo $(G')^p$ and $|G':(G')^p|$ does not exceed $p^{p-1}$. Finally, we will prove that analogous results are satisfied when working with pro-$p$ groups.

Equations

$$|L^{p^i}:N^{p^i}|\le |L:N|$$

$$L^{p^i}/N^{p^i}=⟨x_1^{p^i},\dots ,x_n^{p^i}⟩N^{p^i}.$$

$$[x,g{]}^{p^i}\equiv [x,g^{p^i}](\mathrm{mod}{N}^{p^i}),$$

$$[x,g{]}^{p^{i-1}}\equiv [x,g^{p^{i-1}}](\mathrm{mod}{N}^{p^{i-1}})$$

$$L^{p^i}/N^{p^i}=⟨[x,g^{p^{i-1}}{]}^p{N}^{p^i}⟩.$$

$$[x,g^{p^{i-1}}{]}^p\equiv [x,g^{p^i}](\mathrm{mod}{N}^{p^i}).$$

$$[x,g^{p^{i-1}}{]}^p=[x,g^{p^i}]c_2^{\left(\genfrac{}{}{0ex}{}{p}{2}\right)}{c}_3^{\left(\genfrac{}{}{0ex}{}{p}{3}\right)}\dots c_p,$$

$$c_p\in [L^{p^{i-1}},G,\stackrel{p-1}{\dots },G]\le [L^{p^{i-1}},G,\stackrel{d+1}{\dots },G]\le [N^{p^{i-1}},G,\stackrel{d}{\dots },G]\le N^{p^i}.$$

$$[H,G^{p^k}]\le [H,G{]}^p[H,G,\stackrel{p^k}{\dots },G],$$

$$\begin{array}{rl}{c}_p\in [L^{p^{i-1}},G^{p^k},\stackrel{p-1}{\dots },G^{p^k}]& \le [L^{p^{i-1}},G{]}^p[L^{p^{i-1}},G,\stackrel{(p-1)p^k}{\dots },G]\\ & \le N^{p^i}[N^{p^{i-1}},G,\stackrel{(p-1)p^k-1}{\dots },G]\le N^{p^i},\end{array}$$

$$((G^{\prime }{)}^{p^k}{)}^p=N_d<N_{d-1}<\cdots <N_0=(G^{\prime }{)}^{p^k}$$

$$N_j^{p^i}/N_{j+1}^{p^i}\subseteq K_{x{N}_{j+1}^{p^i}}(G/N_{j+1}^{p^i}).$$

$$D(T)/T=Z(G/T),$$

$$|[N,G,\stackrel{i}{\dots },G]:[N,G,\stackrel{i+1}{\dots },G]|\le p$$

$$C_i=C_G({\gamma }_i(G)(G^{\prime }{)}^p/{\gamma }_{i+2}(G)(G^{\prime }{)}^p)$$

$$(G^{\prime }{)}^p=G_{d+2}\le G_{d+1}\le \dots \le G_3\le G_2=G^{\prime }$$

$$(xy{)}^p=x^p{y}^p{c}_2^{\left(\genfrac{}{}{0ex}{}{p}{2}\right)}{c}_3^{\left(\genfrac{}{}{0ex}{}{p}{3}\right)}\dots c_p$$

$$c_p\in [G_i,\stackrel{p}{\dots },G_i]\le [G_i,G,\stackrel{2(p-1)}{\dots },G]\le G_{i+1}^p,$$

$$|G_i:G_{i+1}^p|=|G_i:G_i^p||G_i^p:G_{i+1}^p|\le p^{d+1}$$

$$((G^{\prime }{)}^p{)}^p=G_{d+2}^p\le G_{d+1}^p\le \dots \le G_3^p\le G_2^p=(G^{\prime }{)}^p$$

$$[x,g_i{]}^p\equiv [x,g_i^p](\mathrm{mod}{G}_{i+2}^p)$$

$$[x,g_1{]}^p=[x,g_1^p]c_2^{\left(\genfrac{}{}{0ex}{}{p}{2}\right)}{c}_3^{\left(\genfrac{}{}{0ex}{}{p}{3}\right)}\dots c_p$$

$$c_p\in [G,C_2,\stackrel{p}{\dots },C_2]\le [G_4,C_2,\stackrel{p-2}{\dots },C_2].$$

$$\begin{array}[]{ccc}G&\longrightarrow&[x,G](G^{\prime})^{p}/(G^{\prime})^{p}\\ g&\longmapsto&[x,g](G^{\prime})^{p}\end{array}$$

$$G^{\prime }=⟨[u,v],[x,u{v}^i],[x,v]⟩.$$

$$(G^{\prime }{)}^p=⟨[u,v{]}^p,[x,u{v}^i{]}^p,[x,v{]}^p⟩.$$

$$[u,v{]}^p\equiv [u{v}^i,v^p](\mathrm{mod}([x,G](G^{\prime }{)}^p{)}^p)$$

$$[x,v{]}^p\equiv [x,v^p](\mathrm{mod}(G^{\prime }{)}^{p^2}).$$

$$(G^{\prime }{)}^p=⟨[u{v}^i,v^p],[x,v^p],[x,u{v}^i{]}^p⟩\le H_i^{\prime },$$

$${\Gamma }^{p^{i-1}}/(G^{\prime }{)}^{p^i}=\{[x,y,h^{p^{i-1}}](G^{\prime }{)}^{p^i}\mid h\in G\}.$$

Full text

The author is supported by the Spanish Government grant MTM2017-86802-P and by the Basque Government grant IT974-16. He is also supported by a predoctoral grant of the University of the Basque Country

1. Introduction

Let $K(G)$ denote the set of commutators of a group $G$. It is a well-known problem deciding whether the derived subgroup $G^{\prime}=\langle K(G)\rangle$ equals $K(G)$. Indeed, in general this equality does not hold since the product of two commutators need not be a commutator. As an easy example, we can consider the group $G=F_{d}/\gamma_{3}(F_{d})F_{d}^{p}$, where $F_{d}$ is the free group on $d\geq 6$ generators and $p>2$ is a prime. It is immediate by [11, Theorem 3.1] that this group does not satisfy the property, even if the nilpotency class of $G$ is 2 and the exponent of $G$ is $p$.

We need, then, to restrict our choice of the group $G$ to some particular family of groups if we want it to satisfy the desired property. For instance, Liebeck, O’Brien, Shalev and Tiep proved in [13] that if $G$ is a finite simple group, then $G^{\prime}=K(G)$, thereby proving the so-called Ore Conjecture. On the other hand, Guralnick ([6, Theorem 1]) and Kappe and Morse ([11, Theorem 3.4 and Theorem 4.2]) found some upper bounds for the order of $G$, for the order of $G^{\prime}$ and for $n$ if $G$ is a $p$-group of order $p^{n}$, in such a way that a group $G$ will satisfy the equality whenever it satisfies one of these bounds.

However, we will focus on restrictions on the number of generators of the derived subgroup $G^{\prime}$. In this direction, Macdonald proved in [14] that even if $G^{\prime}$ is cyclic the property may fail. Actually, he showed that for every $n\in\mathbb{N}$ we can find a group $G$ such that $G^{\prime}$ is cyclic but cannot be generated by less than $n$ commutators. This shows how delicate the equality $G^{\prime}=K(G)$ can be. The situation, fortunately, is much better when working with nilpotent groups. In that case, Rodney proved in [17] that if $G$ is nilpotent with cyclic derived subgroup, then $G^{\prime}=K(G)$.

The study of this property for finite nilpotent groups is clearly reduced to finite $p$-groups. If $G$ is a finite $p$-group with 2-generator abelian derived subgroup, Guralnick proved in [5, Theorem A] that $G^{\prime}$ consists only of commutators. In [3] Fernández-Alcober and the author extended this result, showing that the condition that $G^{\prime}$ is abelian is not necessary.

Theorem 1.1** ([3], Theorem A).**

Let $G$ be a finite $p$-group. If $G^{\prime}$ can be generated by $2$ elements, then $G^{\prime}=\{[x,g]\mid g\in G\}$ for a suitable $x\in G$.

Methods and strategies developed in [3] for the proof of Theorem 1.1 will have a great importance when proving Theorem A below. This theorem concerns finite $p$-groups with 3-generator derived subgroup. In this context Rodney addressed the simplest cases, namely, the one when the nilpotency class of $G$ is 2 ([16, Theorem A]) and the one when $G^{\prime}$ is elementary abelian of rank 3 ([16, Theorem B]), showing that in both cases we have $G^{\prime}=K(G)$. Notice, however, that Rodney’s results involve only groups for which $G^{\prime}$ is abelian. Thus, Guralnick generalized these results for $p\geq 5$, proving that if $G$ is a finite $p$-group with $G^{\prime}$ abelian and 3-generator, then $G^{\prime}=K(G)$ ([5, Theorem B]). Moreover, he found counterexamples showing that the result is false for $p=2$ or $p=3$, even if $G^{\prime}$ is abelian ([5], Example 3.5 and Example 3.6). In Theorem A we generalize Guralnick’s result to groups in which $G^{\prime}$ need not be abelian.

Theorem A**.**

Let $G$ be a finite $p$-group with $p\geq 5$. If $G^{\prime}$ can be generated by 3 elements, then $G^{\prime}$ consists only of commutators.

In this case, as shown in Remark 4.4, it is not true, in general, that there exists a fixed element $x\in G$ such that $G^{\prime}=\{[x,g]\mid g\in G\}$, as we have in Theorem 1.1 or Theorem B below.

Macdonald ([15, Exercise 5, page 78]) and Kappe and Morse ([11, Example 5.4]) showed that for every prime $p$ there exist finite $p$-groups with 4-generator abelian derived subgroup such that $G^{\prime}\neq K(G)$. These examples show that the property may fail if the derived subgroup has more than 3 generators. Therefore, with Theorem A and Theorem 1.1, we close the gap between the case when $G^{\prime}$ is abelian and can be generated by 3 elements and the case when $G^{\prime}$ is generated by more than 3 elements. Thus, the study of the condition $G^{\prime}=K(G)$ in terms of the number of generators of the derived subgroup is complete for finite $p$-groups. In Theorem B we show that with some additional restriction, groups with $d(G^{\prime})\geq 4$ satisfy the desired equality.

Theorem B**.**

Let $G$ be a finite $p$-group and write $d=\log_{p}|G^{\prime}:(G^{\prime})^{p}|$. If $d\leq p-1$ and the action of $G$ on $G^{\prime}$ is uniserial modulo $(G^{\prime})^{p}$, then there exists $x\in G$ such that $G^{\prime}=\{[x,g]\mid g\in G\}$.

More information about the condition $G^{\prime}=K(G)$ can be found in the papers [10] and [11].

Finally, we show that analogous results to Theorem A and Theorem B are satisfied when working with pro-$p$ groups. Recall that if a pro-$p$ group $G$ is topologically finitely generated, then the index of $G^{p}$ in $G$ is a $p$-power (and in particular finite).

Theorem A*′*.

Let $G$ be a pro-$p$ group with $p\geq 5$. If $G^{\prime}$ can be topologically generated by 3 elements, then $G^{\prime}$ consists only of commutators.

Theorem B*′*.

Let $G$ be pro-$p$ group with topologically finitely generated derived subgroup. Write $d=\log_{p}|G^{\prime}:(G^{\prime})^{p}|$. If $d\leq p-1$ and the action of $G$ on $G^{\prime}$ is uniserial modulo $(G^{\prime})^{p}$, then there exists $x\in G$ such that $G^{\prime}=\{[x,g]\mid g\in G\}$.

Notation and organization. Let $G$ be a group, and let $H\leq G$. We write $H\max G$ to denote that $H$ is maximal in $G$. If $x\in G$, then we set $K_{x}(H)=\{[x,h]\mid h\in H\}$ and $[x,H]=\langle K_{x}(H)\rangle$. We denote the Frattini subgroup of $G$ by $\Phi(G)$. If $G$ is finitely generated, $d(G)$ stands for the minimum number of generators of $G$. Finally, if $G$ is a topological group and $H\leq G$, we write $\operatorname{Cl}_{G}(H)$ to refer to the topological closure of $H$ in $G$ and we write $H\trianglelefteq_{\mathrm{o}}G$ to denote that $H$ is an open normal subgroup of $G$.

We start with some preliminary results in Section 2. Theorem B will be used in the proof of Theorem A, so it will be proved before Theorem A in Section 3. We then split the proof of Theorem A into two sections, dealing separately with the following two cases: in Section 4 we prove the result when $G^{\prime}$ is powerful and in Section 5 we prove it in the general case. Finally, in Section 6 we prove Theorem A*′* and Theorem B*′*.

2. Preliminary Results

In the proof of Theorem 1.1 ([3, Theorem A]), the authors rely on a result by Blackburn, according to which $G^{\prime}$ is powerful whenever $d(G^{\prime})\leq 2$ ([1, Theorem 1]). In this way, they reduce the proof to the case in which $G^{\prime}$ is powerful. Unfortunately, this is not true when $d(G^{\prime})=3$, as Example 5.1 below shows. However, we will see in Section 5 that the groups in which $d(G^{\prime})=3$ but $G^{\prime}$ is non-powerful are very specific. Powerful groups, then, will be essential in this paper. Background on such groups can be found in [4, Chapter 2] or [12, Chapter 11]. These groups are usually seen as a generalization of abelian groups since they satisfy, among others, the following properties:

1. (i)

$\Phi(G)=G^{p}$. In particular $|G:G^{p}|=p^{d(G)}$. 2. (ii)

$d(H)\leq d(G)$ for every $H\leq G$. 3. (iii)

$G^{p}=\{g^{p}\mid g\in G\}$. 4. (iv)

If $G=\langle x_{1},\ldots,x_{n}\rangle$, then $G^{p}=\langle x_{1}^{p},\ldots,x_{n}^{p}\rangle$. 5. (v)

The power map from $G^{p^{i-1}}/G^{p^{i}}$ to $G^{p^{i}}/G^{p^{i+1}}$ that sends $gG^{p^{i}}$ to $g^{p}G^{p^{i+1}}$ is an epimorphism for every $i\geq 0$.

Remark 2.1*.*

Property (v) implies that if $G^{p}\leq N\leq L\leq G$, then

[TABLE]

(and hence $|N:N^{p^{i}}|\leq|L:L^{p^{i}}|$), and if $L/N=\langle x_{1},\ldots,x_{n}\rangle N$, then

[TABLE]

We can generalize this concept even more with the notion of potent $p$-groups, which will also have an important role in the paper. For instance, as we will see in the proof of Theorem B, if a group satisfies the conditions of the theorem, then its derived subgroup is potent. A finite $p$-group $G$ is said to be potent if $\gamma_{p-1}(G)\leq G^{p}$ for odd $p$ or if $G^{\prime}\leq G^{4}$ for $p=2$. In this context, the following lemma, which is a reduced version of a theorem by González-Sánchez and Jaikin-Zapirain, will be particularly helpful. First, recall that a group $G$ is said to be power abelian if it satisfies the following three properties for all $i\geq 0$:

1. (i)

$G^{p^{i}}=\{g^{p^{i}}\mid g\in G\}$. 2. (ii)

$\Omega_{i}(G)=\{g\in G\mid o(g)\leq p^{i}\}$. 3. (iii)

$|G:G^{p^{i}}|=|\Omega_{i}(G)|$.

Lemma 2.2** ([9], Theorem 1.1).**

Let $G$ be a potent $p$-group with $p>2$. Then:

1. (i)

If $N\trianglelefteq G$ then $N$ is power abelian. 2. (ii)

If $N\leq G^{p}$ and $N\trianglelefteq G$, then $N$ is powerful.

Following the strategy developed in [3], the next two lemmas will be crucial. Lemma 2.3 says that if we want to show that a subgroup contains only commutators with a fixed element in the first position, we only have to care about the factors of a normal series. Lemma 2.4 shows that, actually, it suffices to find some suitable generators for such factors.

Lemma 2.3** ([3], Lemma 2.3).**

Let $G$ be a group and let $N\leq L\leq G$, with $N$ normal in $G$. Suppose that for some $x\in G$ the following two conditions hold:

1. (i)

$L/N\subseteq K_{xN}(G/N)$. 2. (ii)

$N\subseteq K_{x}(G)$.

Then $L\subseteq K_{x}(G)$.

Lemma 2.4** ([3], Lemma 2.4).**

Let $G$ be a group and let $N\leq L\leq G$, with $N$ normal in $G$. If $L/N=\langle[x,s]N\mid s\in S\rangle$ for some $x\in G$ and some $S\subseteq G$ with $[L,S]\subseteq N$, then $L/N\subseteq K_{xN}(\langle S\rangle N/N)\subseteq K_{xN}(G/N)$.

In order to apply these lemmas we will use the following result.

Lemma 2.5**.**

Let $G$ be a finite $p$-group with $p\geq 3$ and $(G^{\prime})^{p^{k}}$ powerful for some $k\geq 0$, and let $L,N$ be two normal subgroups of $G$ such that $((G^{\prime})^{p^{k}})^{p}\leq N\leq L\leq(G^{\prime})^{p^{k}}$. Write $d=d((G^{\prime})^{p^{k}})$ and suppose $d\leq p^{k+1}-p^{k}-1$. If $L/N=\langle[x,g]N\rangle$ where $x\in G$ and $g\in G^{p^{k}}$, then

[TABLE]

and $L^{p^{i}}/N^{p^{i}}=\langle[x,g^{p^{i}}]N^{p^{i}}\rangle$ for every $i\geq 0$.

Proof.

We will argue by induction on $i$. If $i=0$ there is nothing to prove, so assume $i\geq 1$ and suppose

[TABLE]

and $L^{p^{i-1}}/N^{p^{i-1}}=\langle[x,g^{p^{i-1}}]N^{p^{i-1}}\rangle$. By Lemma 2.2, $L$ and $N$ are power abelian, so $(L^{p^{i-1}})^{p}=L^{p^{i}}$ and $(N^{p^{i-1}})^{p}=N^{p^{i}}$. Since $((G^{\prime})^{p^{k}})^{p^{i-1}}$ is powerful, Remark 2.1 yields

[TABLE]

Thus, we only have to prove that

[TABLE]

By the Hall-Petresco Identity,

[TABLE]

where $c_{j}\in\gamma_{j}(\langle[x,g^{p^{i-1}}],g^{p^{i-1}}\rangle)\leq[L^{p^{i-1}},G^{p^{k}},\overset{j-1}{\ldots},G^{p^{k}}]$ for every $2\leq j\leq p$. Note that $L/N$ is cyclic of exponent $p$, so $|L:N|\leq p$ and by Remark 2.1 we have $|L^{p^{i-1}}:N^{p^{i-1}}|\leq p$, so that $[L^{p^{i-1}},G]\leq N^{p^{i-1}}$. Hence, since $N$ is power abelian, if $2\leq j\leq p-1$ we have $c_{j}^{\binom{p}{j}}\in N^{p^{i}}$.

If $j=p$, then $c_{p}\in[L^{p^{i-1}},G^{p^{k}},\overset{p-1}{\ldots},G^{p^{k}}]$. Recall that $(G^{\prime})^{p^{k}}$ is powerful, so we have $|((G^{\prime})^{p^{k}})^{p^{i-1}}:((G^{\prime})^{p^{k}})^{p^{i}}|\leq p^{d}$, and hence $|N^{p^{i-1}}:N^{p^{i}}|\leq p^{d}$ by Remark 2.1. If $k=0$, since $d\leq p-2$, we get

[TABLE]

If $k\geq 1$, then it can be proved using again the Hall-Petresco Identity that for every normal subgroup $H$ of $G$ we have

[TABLE]

so

[TABLE]

where the last equality holds since $d\leq p^{k+1}-p^{k}-1$. The result follows. ∎

Thus, combining Lemma 2.3, Lemma 2.4 and Lemma 2.5 we get the following useful result.

Lemma 2.6**.**

Let $G$ be a finite $p$-group with $p\geq 3$ and $(G^{\prime})^{p^{k}}$ powerful for some $k\geq 0$. Write $d((G^{\prime})^{p^{k}})=d$ and suppose $d\leq p^{k+1}-p^{k}-1$. If there exist $x\in G$, $g_{0},\ldots,g_{d-1}\in G^{p^{k}}$ and a series from $(G^{\prime})^{p^{k}}$ to $((G^{\prime})^{p^{k}})^{p}$

[TABLE]

in which each factor $N_{j}/N_{j+1}$ is a chief factor of $G$ generated by the commutator $[x,g_{j}]N_{j+1}$, then $(G^{\prime})^{p^{k}}=K_{x}(G)$.

Proof.

Since $(G^{\prime})^{p^{k}}$ is powerful we have $|(G^{\prime})^{p^{k}}:((G^{\prime})^{p^{k}})^{p}|=p^{d}$. By Remark 2.1, we have $|N_{j}^{p^{i}}/N_{j+1}^{p^{i}}|\leq p$ for every $i\geq 0$, and furthermore, by Lemma 2.5, this quotient is generated by $[x,g_{j}^{p^{i}}]N_{j+1}^{p^{i}}$ for every $i$ and $j$. Therefore, by Lemma 2.4, it follows that

[TABLE]

Thus, we have a series from $(G^{\prime})^{p^{k}}$ to $1$ in which all factors are chief factors of $G$ and all elements of each chief factor are images of commutators of the form $[x,g]$ with $g\in G$. The result follows by applying Lemma 2.3 again and again. ∎

Remark 2.7*.*

Lemma 2.6 (and hence also Lemma 2.5) will be used with $k\neq 0$ only when proving Theorem B, where we use it with $k=1$. The general result has been proved for completeness.

As in [3] the subgroups below will have an essential role in the paper.

Definition 2.8**.**

Let $G$ be a non-abelian finite $p$-group. For every $T\max G^{\prime}$ with $T\trianglelefteq G$ we define the subgroup $D(T)$ by the condition

[TABLE]

that is, $D(T)$ is the largest subgroup of $G$ satisfying $[D(T),G]\leq T$. We set $D=\cup\{D(T)\mid T\max G^{\prime}\text{ with }T\trianglelefteq G\}$.

Definition 2.9**.**

Let $G$ be a finite $p$-group with $G^{\prime}$ powerful. We define $C=C_{G}(G^{\prime}/(G^{\prime})^{p})$.

Recall that the action of $G$ on a normal subgroup $N$ of $G$ is uniserial if

[TABLE]

for every $i\geq 0$. We also define the following subgroups, which are just the so-called two-step centralizers modulo $(G^{\prime})^{p}$.

Definition 2.10**.**

Let $G$ be a finite $p$-group such that the action of $G$ on $G^{\prime}$ is uniserial modulo $(G^{\prime})^{p}$. Then, we define

[TABLE]

for every $i\geq 2$ such that $\gamma_{i+1}(G)\not\leq(G^{\prime})^{p}$.

Remark 2.11*.*

In the situation above, the subgroups $C_{i}$ are all maximal in $G$ since $|\gamma_{i}(G)(G^{\prime})^{p}:\gamma_{i+2}(G)(G^{\prime})^{p}|=p^{2}$ and $[\gamma_{i}(G)(G^{\prime})^{p},G]\not\leq\gamma_{i+2}(G)(G^{\prime})^{p}$.

We prove the following result exactly in the same way as in [3, Lemma 2.9], even if $d(G^{\prime})\neq 2$.

Lemma 2.12**.**

If $G$ is a non-abelian finite $p$-group then $[x,G]=G^{\prime}$ if and only if $x\not\in D$. Furthermore, for every $T\max G^{\prime}$ with $T\trianglelefteq G$, we have $\Phi(G)\leq D(T)$ and $\log_{p}|G:D(T)|$ is even.

Proof.

Since $[x,G]$ is a normal subgroup of $G$, we have $[x,G]<G^{\prime}$ if and only if $x\in D(T)$ for some $T\max G^{\prime}$ with $T\trianglelefteq G$, and the first assertion follows.

On the other hand, let $T\max G^{\prime}$ with $T\trianglelefteq G$. We have $[\Phi(G),G]=[G^{p},G]\gamma_{3}(G)=(G^{\prime})^{p}\gamma_{3}(G)\leq T$, and so $\Phi(G)\leq D(T)$. Thus $G/D(T)$ can be seen as an $\mathbb{F}_{p}$-vector space. In addition, the commutator map in $G/T$ induces a a non-degenerate alternating form on $G/D(T)$, and thus $\dim_{\mathbb{F}_{p}}G/D(T)$ is even. ∎

3. Proof of Theorem B

Before proving Theorem B we need the following simple lemma, according to which the first part of Remark 2.1 can be stated in a more general way, even if $G$ is potent.

Lemma 3.1**.**

Let $G$ be a potent $p$-group with $p\geq 3$. If $N\leq L$ are two normal subgroups of $G$, then $|N:N^{p^{i}}|\leq|L:L^{p^{i}}|$ for all $i\geq 0$. In particular $|L^{p^{i}}:N^{p^{i}}|\leq|L:N|$.

Proof.

By Lemma 2.2, the subgroups $N$ and $L$ are power abelian, so in particular $|N:N^{p^{i}}|=|\Omega_{i}(N)|$ and $|L:L^{p^{i}}|=|\Omega_{i}(L)|$. Since obviously $|\Omega_{i}(N)|\leq|\Omega_{i}(L)|$, the result follows. ∎

Proof of Theorem B.

If $d=1$, then $G^{\prime}$ is cyclic and the result follows from Theorem 1.1, so assume $d\geq 2$ (and in particular $p\geq 3$). For the sake of simplicity we will write $G_{i}=\gamma_{i}(G)(G^{\prime})^{p}$, so that

[TABLE]

is a series from $G^{\prime}$ to $(G^{\prime})^{p}$ such that $|G_{i}:G_{i+1}|=p$ for all $2\leq i\leq d+1$. Note that if $N\max G^{\prime}$ with $N\trianglelefteq G$, then $G_{3}\leq N$. Therefore, $N=G_{3}$ and $G_{3}$ is the unique subgroup satisfying those conditions. Hence, $D$ is a subgroup of $G$ whose index is greater than $p$ by Lemma 2.12. Note also that there are only $d-1\leq p-2$ two-step centralizers, which are maximal by Remark 2.11. Thus, we can take $x\in G\setminus(D\cup C_{2}\cup\ldots\cup C_{d})$. By Lemma 2.12 we have $G^{\prime}=[x,G]$ and since $C_{2}$ is maximal in $G$ we have $G^{\prime}=[x,G]=[x,\langle x\rangle C_{2}]=[x,C_{2}]$. In particular $G^{\prime}/G_{3}=\langle[x,g_{1}]G_{3}\rangle$ for some $g_{1}\in C_{2}$. Furthermore, since $x\not\in C_{i}$ for $2\leq i\leq d$, we also have $G_{i+1}/G_{i+2}=\langle[x,g_{i}]G_{i+2}\rangle$ for some suitable $g_{i}\in G_{i}$. It follows from Lemma 2.4 and Lemma 2.3 that $G^{\prime}/(G^{\prime})^{p}\subseteq K_{x(G^{\prime})^{p}}(G/(G^{\prime})^{p})$.

Recall that $\gamma_{d+2}(G)\leq(G^{\prime})^{p}$, and since $d\leq p-1$, it follows that $\gamma_{p-1}(G^{\prime})\leq\gamma_{2(p-1)}(G)\leq\gamma_{2d}(G)$. Thus, since $2d\geq d+2$ we have $\gamma_{p-1}(G^{\prime})\leq(G^{\prime})^{p}$, so that $G^{\prime}$ is potent. In this case the power map from $G^{\prime}/(G^{\prime})^{p}$ to $(G^{\prime})^{p}/(G^{\prime})^{p^{2}}$ defined above Remark 2.1 need not be a homomorphism. However, we can restrict its domain and codomain in order for it to be so. We claim that the map from $G_{i}/G_{i+1}$ to $G_{i}^{p}/G_{i+1}^{p}$ sending $gG_{i+1}$ to $g^{p}G_{i+1}^{p}$ is an epimorphism for every $2\leq i\leq d+1$.

Take $x,y\in G_{i}$. By the Hall-Petresco Identity we have

[TABLE]

with $c_{j}\in\gamma_{j}(G_{i})$. Obviously if $2\leq j\leq p-1$ then $c_{j}^{\binom{p}{j}}\in G_{i+1}^{p}$. Besides, if $j=p$, since $G_{i}\leq G^{\prime}$, we have

[TABLE]

where the last inequality holds since by Lemma 3.1 we have

[TABLE]

and $d+1\leq 2(p-1)$. Moreover, since $G^{\prime}$ is potent it follows that $G_{i}$ is power abelian, so the map must be an epimorphism. The claim is proved.

Thus, by Lemma 3.1 it follows that we have a series

[TABLE]

in which each factor $G_{i+1}^{p}/G_{i+2}^{p}$ has order less than or equal to $p$ and is generated by the image of $[x,g_{i}]^{p}$ for every $1\leq i\leq d$. In order to apply Lemma 2.6 let us prove that

[TABLE]

for every $i$. Assume first $i=1$. We will use again the Hall-Petresco Identity so that

[TABLE]

with $c_{j}\in\gamma_{j}(\langle[x,g_{1}],g_{1}\rangle)\leq[G,C_{2},\overset{j}{\ldots},C_{2}]$. If $2\leq j\leq p-1$ then $c_{j}^{\binom{p}{j}}\in G_{3}^{p}$. If $j=p$, we have

[TABLE]

Lemma 3.1 yields $|G_{4}:G_{3}^{p}|\leq p^{d-1}$, and since $d-1\leq p-2$, we conclude $c_{p}\in G_{3}^{p}$. For $i\geq 2$ we have $g_{i}\in G^{\prime}$, so the claim follows more easily applying the same method.

Now, $d\leq p-1\leq p^{2}-p-1$, so we apply Lemma 2.6 with $j=1$ and we get $(G^{\prime})^{p}\subseteq K_{x}(G)$. Since $G^{\prime}/(G^{\prime})^{p}\subseteq K_{x(G^{\prime})^{p}}(G/(G^{\prime})^{p})$, we conclude by Lemma 2.3. ∎

Remark 3.2*.*

If the exponent of $G^{\prime}$ is $p$, that is, if $(G^{\prime})^{p}=1$, then, following the same method, Theorem B can be stated for $d\leq p+1$. Indeed, if $G$ is the union of $p+1$ proper subgroups, then all of them must be maximal.

4. Proof of Theorem A when $G^{\prime}$ is Powerful

In order to prove Theorem A we need the following technical lemma, which will be very helpful when using induction on the order of the group.

Lemma 4.1**.**

Let $G$ be a finite $p$-group with $p\geq 5$, $G^{\prime}$ powerful and $d(G^{\prime})=3$. Assume there exist $x,u,v\in G$ such that $G^{\prime}=\langle[u,v],[x,u],[x,v]\rangle$, $G^{\prime}\neq[x,G]$ and $[x,G,G]\leq(G^{\prime})^{p}$. Then, there exists a family of proper subgroups of $G$ such that $[x,G](G^{\prime})^{p}$ equals the union of their derived subgroups. Moreover, each of these derived subgroups is powerful.

Proof.

Consider the subgroups $H_{i}=\langle x,uv^{i},v^{p}\rangle$ for $0\leq i\leq p-1$ and $H_{p}=\langle x,v,u^{p}\rangle$. Let us prove that $H_{i}^{\prime}=\langle[x,uv^{i}]\rangle(G^{\prime})^{p}$ for $0\leq i\leq p-1$ and that $H_{p}^{\prime}=\langle[x,v]\rangle(G^{\prime})^{p}$.

Suppose first $i\neq p$. Since $G^{\prime}=\langle[u,v],[x,u],[x,v]\rangle$ and $G^{\prime}\neq[x,G]$, we have $|G^{\prime}:[x,G](G^{\prime})^{p}|=p$, and since $[x,G,G]\leq(G^{\prime})^{p}$, the map

[TABLE]

is a homomorphism. Therefore, we can write

[TABLE]

Thus, since $G^{\prime}$ is powerful, we have

[TABLE]

The subgroups $[x,G](G^{\prime})^{p}$ and $\langle[x,v]\rangle(G^{\prime})^{p}$ are normal in $G$ since $[x,G,G]\leq(G^{\prime})^{p}$, so taking $k=0$ in Lemma 2.5, it follows that

[TABLE]

and

[TABLE]

Hence,

[TABLE]

so that $\langle[x,uv^{i}]\rangle(G^{\prime})^{p}=H_{i}^{\prime}$, as asserted. Similar arguments imply that $H_{p}^{\prime}=\langle[x,v]\rangle(G^{\prime})^{p}$.

It is easy to see now that $[x,G](G^{\prime})^{p}=\bigcup_{i=0}^{p}H_{i}^{\prime}$ (just observe that the $H_{i}^{\prime}$ are precisely the subgroups between $[x,G](G^{\prime})^{p}$ and $(G^{\prime})^{p}$). Finally, notice that $|H_{i}^{\prime}:(G^{\prime})^{p}|=p$ for every $i$, so since $(G^{\prime})^{p}$ is powerfully embedded in $G^{\prime}$, it follows by [12, Lemma 11.7] that $H_{i}^{\prime}$ is powerful. Thus, the proof is complete. ∎

We are now in a position to prove Theorem A in the case that $G^{\prime}$ is powerful.

Theorem 4.2**.**

Let $G$ be a finite $p$-group with $G^{\prime}$ powerful, $d(G^{\prime})\leq 3$ and $p\geq 5$. Then, $G^{\prime}=K(G)$.

Proof.

We proceed by induction on the order of $G$. For $d(G^{\prime})\leq 2$ the result follows from Theorem 1.1. Now assume that $d(G^{\prime})=3$ and note that we have $|G^{\prime}:(G^{\prime})^{p}|=p^{3}$. We will consider three different cases depending on the position of the subgroup $\Gamma=(G^{\prime})^{p}\gamma_{3}(G)$.

Case 1. $|G^{\prime}:\Gamma|=p$.

If $|\Gamma:\gamma_{4}(G)(G^{\prime})^{p}|=p$, then the action of $G$ on $G^{\prime}$ is uniserial modulo $(G^{\prime})^{p}$ and the result follows from Theorem B.

Assume then $\gamma_{4}(G)\leq(G^{\prime})^{p}$. If $G^{\prime}=K_{x}(G)$ for some $x\in G$, then, of course, we are done, so assume $G^{\prime}\neq K_{x}(G)$ for every $x\in G$. We claim that there exist $u,v\in G$ such that $G^{\prime}=\langle[u,v],[u,v,u],[u,v,v]\rangle$. For that purpose we can suppose that $(G^{\prime})^{p}=1$. As seen in the proof of Theorem B we have $D=D(\Gamma)$, and since $C$ is also a proper subgroup of $G$ (otherwise $\gamma_{3}(G)\leq(G^{\prime})^{p}$), we can take $u\not\in C\cup D$. Then, $G^{\prime}=[u,G]$ by Lemma 2.12 and $C_{G/\Gamma}(u\Gamma)\neq G/\Gamma$. Let us write $C^{*}/\Gamma=C_{G/\Gamma}(u\Gamma)$.

Since $u\not\in C$ we have $[u,G^{\prime}]\neq 1$. If $[u,G^{\prime}]=\Gamma$, then, we can find a series of normal subgroups of $G$ from $G^{\prime}$ to $(G^{\prime})^{p}$ such that all factors have order $p$ and are generated by images of elements of the form $[u,g]$ for some suitable $g\in G$. Thus, Lemma 2.6 implies $G^{\prime}=K_{u}(G)$, which is a contradiction. Therefore, we have $|[u,G^{\prime}]|=p$ and hence $C_{G}(G^{\prime}/[u,G^{\prime}])\neq G$. Take thus $v\in G\setminus(C_{G}(G^{\prime}/[u,G^{\prime}])\cup C^{*})$. Then, $G^{\prime}/\Gamma=\langle[u,v]\Gamma\rangle$ (because $v\not\in C^{*})$, and again, as we have seen for $u$, we also have $|[v,G^{\prime}]|=p$. It follows that $[u,v,u],[u,v,v]\neq 1$. Furthermore, since $v\not\in C_{G}(G^{\prime}/[u,G^{\prime}])$, we have $[u,G^{\prime}]\neq[v,G^{\prime}]$, and we conclude that $G^{\prime}=\langle[u,v],[u,v,u],[u,v,v]\rangle$. This proves the claim.

Remove now the assumption of $(G^{\prime})^{p}=1$ and observe that $[[u,v],G,G]\leq(G^{\prime})^{p}$, so we are in the situation of Lemma 4.1. It follows then that $\Gamma$ is the union of the derived subgroups of some proper subgroups of $G$. These derived subgroups are all powerful, and since $d(G^{\prime})=3$, they all can be generated by $3$ elements. So, by induction, $\Gamma\subseteq K(G)$.

Take now $g\in G^{\prime}\setminus\Gamma$ arbitrary. We claim that $g$ is a commutator modulo $\Gamma^{p^{i}}$ for every $i\geq 0$ (and hence that $g$ is a commutator). We proceed by induction on $i$. Clearly, we have $g=[x,y]z$ for some $x,y\in G$, $z\in\Gamma$, so the case $i=0$ is satisfied. Assume then that $i\geq 1$ and $g=[x,y]z_{1}$ where $x,y\in G$ and $z_{1}\in\Gamma^{p^{i-1}}$.

Note that $G^{\prime}/\Gamma=\langle[x,y]\Gamma\rangle$, so since $[\Gamma,G]\leq(G^{\prime})^{p}$, we have $\Gamma/(G^{\prime})^{p}=\{[x,y,h](G^{\prime})^{p}\mid h\in G\}$. Besides, since $G^{\prime}$ is powerful, the power map from $G^{\prime}/(G^{\prime})^{p}$ to $(G^{\prime})^{p^{i-1}}/(G^{\prime})^{p^{i}}$ is an epimorphism, so that $\Gamma^{p^{i-1}}/(G^{\prime})^{p^{i}}=\{[x,y,h]^{p^{i-1}}(G^{\prime})^{p^{i}}\mid h\in G\}$. By Lemma 2.5 we have

[TABLE]

Thus,

[TABLE]

for some $h^{*}\in G$ and $z_{2}\in(G^{\prime})^{p^{i}}$. We rewrite, in order to simplify the notation, $x$ instead of $x^{h^{*}}$ and $y$ instead of $y^{h^{*}}$, so that $g=[x,y]z_{2}$.

Note again that $G^{\prime}/\Gamma=\langle[x,y]\Gamma\rangle$, so it follows that

[TABLE]

Therefore,

[TABLE]

with $j\geq 0$ and $z_{3}\in\Gamma^{p^{i}}$. Now, by the last theorem in [7], there exist $x^{\prime},y^{\prime}\in G$ such that $[x,y]^{1+jp^{i}}=[x^{\prime},y^{\prime}]$, so $g=[x^{\prime},y^{\prime}]z_{3}$ with $z_{3}\in\Gamma^{p^{i}}$, as claimed.

Case 2. $|G^{\prime}:\Gamma|=p^{2}$.

Let us prove that $C\cup D\neq G$. On the one hand, as seen before, $\Gamma\leq N$ for all $N\max G^{\prime}$ which are normal in $G$, and since $|G^{\prime}:\Gamma|=p^{2}$, there are exactly $p+1$ subgroups between $G^{\prime}$ and $\Gamma$. Furthermore, since they are central over $\Gamma$, they are all normal in $G$. Thus, $D=D(U_{1})\cup\ldots\cup D(U_{p+1})$, where $U_{1},\ldots,U_{p+1}\max G^{\prime}$ are these normal subgroups. In addition, it follows from Lemma 2.12 that $|G:D(U_{i})|\geq p^{2}$ for every $i$.

On the other hand, observe again that $C\neq G$. Hence, if we write $|G|=p^{n}$, we have

[TABLE]

as we wanted. Take now $x\not\in C\cup D$. Since $x\not\in D$ we have $G^{\prime}/\Gamma=[x,G]\Gamma/\Gamma$ by Lemma 2.12, and since $x\not\in C$ we have $\Gamma/(G^{\prime})^{p}=[x,G^{\prime}](G^{\prime})^{p}/(G^{\prime})^{p}$. Thus, since all subgroups between $G^{\prime}$ and $\Gamma$ are central and hence normal in $G$, we can construct a series from $G^{\prime}$ to $(G^{\prime})^{p}$ where all factors have order $p$ and are generated by images of commutators of the form $[x,g]$ with $g\in G$. Again, the result follows from Lemma 2.6.

Case 3. $\gamma_{3}(G)\leq(G^{\prime})^{p}$.

If $G^{\prime}=[x,G](G^{\prime})^{p}$ for some $x$, again, all the subgroups between $G^{\prime}$ and $(G^{\prime})^{p}$ are normal in $G$, so we could construct a series from $G^{\prime}$ to $(G^{\prime})^{p}$ in such a way that we would be done by Lemma 2.6. Therefore, assume $[x,G](G^{\prime})^{p}<G^{\prime}$ for every $x\in G$. By [16, Theorem B] the result is satisfied for $G/(G^{\prime})^{p}$, so we have $G^{\prime}=\bigcup_{x\in G}[x,G](G^{\prime})^{p}$. Thus, it suffices to prove that $[x,G](G^{\prime})^{p}\subseteq K(G)$ for every $x\in G$.

Suppose first $|[x,G](G^{\prime})^{p}:(G^{\prime})^{p}|=p$. We claim that there always exists $y\in G$ such that $[x,G](G^{\prime})^{p}\leq[y,G](G^{\prime})^{p}\max G^{\prime}$. For that purpose, we assume $(G^{\prime})^{p}=1$. Note that $|G^{\prime}/[x,G]|=p^{2}$, so by Theorem 1.1, there exists $u\in G$ such that $G^{\prime}/[x,G]=[u,G][x,G]/[x,G]$. Hence $G^{\prime}=[u,G][x,G]$ with $|[u,G]|=p^{2}$. Observe that $C_{G}(u),C_{G}(x)\neq G$, so take $y\not\in C_{G}(u)\cup C_{G}(x)$. Thus, $[x,G]=\langle[x,y]\rangle$, and $[u,y]\neq 1$. If $[u,y]\in\langle[x,y]\rangle$, then $[x,y]\in[u,G]$, a contradiction. Observe, however, that $[x,y],[u,y]\in[y,G]$, so $|[y,G]|=p^{2}$. Since $[x,G]\leq[y,G]$, the claim is proved.

Hence, we only have to consider the case $|[x,G](G^{\prime})^{p}:(G^{\prime})^{p}|=p^{2}$. We claim now that there exist $u,v\in G$ such that $G^{\prime}=\langle[u,v],[x,v],[x,u]\rangle$. Assume again that $(G^{\prime})^{p}=1$. Since $|[x,G]|=|\{[x,g]\mid g\in G\}|=p^{2}$, we have $|G:C_{G}(x)|=p^{2}$, and we can consider a maximal subgroup $M$ such that $C_{G}(x)<M<G$. Observe that $G^{\prime}=[G,G]=[G,M]$, $G=\langle G\setminus M\rangle$ and $M=\langle M\setminus C_{G}(x)\rangle$. Hence, there exist $u\in G\setminus M$ and $v\in M\setminus C_{G}(x)$ such that $[u,v]\not\in[x,G]$. Furthermore, $[x,G]=\langle[x,u],[x,v]\rangle$, so $G^{\prime}=\langle[u,v],[x,u],[x,v]\rangle$, as claimed.

Remove now the assumption of $(G^{\prime})^{p}=1$ and note that we are in the situation of Lemma 4.1 since $[x,G,G]\leq\gamma_{3}(G)\leq(G^{\prime})^{p}$. Hence we have $[x,G](G^{\prime})^{p}\subseteq K(G)$, as we wanted. ∎

Remark 4.3*.*

Case 2 can be generalized for $p\geq 3$ using a slightly different version of Lemma 2.6, but one must be more selective in the choice of $x$.

Remark 4.4*.*

It is not true that, in general, if $d(G^{\prime})=3$ we have $G^{\prime}=K_{x}(G)$ for some $x\in G$, as we had in Theorem 1.1 or Theorem B. Indeed, let $G=F_{3}/\gamma_{3}(F_{3})F_{3}^{p}$, where $F_{3}$ is the free group on 3 generators and $p\geq 3$ is a prime. Note that $G^{\prime}$ is 3-generator and $|G:Z(G)|=p^{3}$. Now, if $x\in Z(G)$, then $K_{x}(G)=1$, and if $x\not\in Z(G)$, then $|K_{x}(G)|=|G:C_{G}(x)|\leq p^{2}$ since $\langle Z(G),x\rangle\leq C_{G}(x)$.

5. Proof of Theorem A when $G^{\prime}$ is Non-Powerful

The following example, taken directly from [8, Example 14.24, page 376], shows that unlike the case when $d(G^{\prime})\leq 2$, it may happen that $G^{\prime}$ is non-powerful when $d(G^{\prime})=3$.

Example 5.1**.**

Let $p\geq 5$ and consider the groups $A=\langle a_{1}\rangle\times\langle a_{2}\rangle\times\langle a_{3}\rangle\cong C_{p}\times C_{p}\times C_{p}$ and $B=\langle b_{1}\rangle\times\langle b_{2}\rangle\cong C_{p}\times C_{p}.$ Define $Y=A\rtimes B$ via the automorphisms

[TABLE]

Now, consider $X=\langle x\rangle\cong C_{p}$ and define $G=Y\rtimes X$ via the automorphism

[TABLE]

The group $G$ is a $p$-group of maximal class of order $p^{6}$ and exponent $p$ such that $d(G^{\prime})=3$ and $G^{\prime\prime}=\gamma_{5}(G)\neq 1$.

We will start, hence, analyzing which kind of groups may arise when $G^{\prime}$ is non-powerful. Actually, we will see that in such a case, $G/(G^{\prime})^{p}$ must be a very special kind of $p$-group, namely, a CF$(m,p)$-group. These groups were introduced by Blackburn in [2] and are defined as follows.

Definition 5.2**.**

Let $m\in\mathbb{N}$ such that $m\geq 3$. A $p$-group $G$ is said to be a CF$(m,p)$-group if the nilpotency class of $G$ is $m-1$ and the action of $G$ on $G^{\prime}$ is uniserial.

We next define the degree of commutativity on CF($m,p$)-groups exactly in the same way as for groups of maximal class.

Definition 5.3**.**

Let $G$ be a CF$(m,p)$-group. The degree of commutativity of $G$ is defined as

[TABLE]

where $G_{1}=C_{2}$ and $G_{i}=\gamma_{i}(G)$ for all $i\geq 2$.

Lemma 5.5 below shows that if $G^{\prime}$ is non-powerful, then we can reduce our proof to a very particular group which is a CF($6,p$)-group modulo $(G^{\prime})^{p}$. The key part of the proof is the following lemma due to Blackburn.

Lemma 5.4** ([2], Theorem 2.11).**

Let $G$ be a CF$(m,p)$-group with $m$ odd and $5\leq m\leq 2p+1$. Then $G$ has degree of commutativity greater than 0.

Lemma 5.5**.**

Let $G$ be a finite $p$-group with $p\geq 3$, $d(G^{\prime})=3$ and $G^{\prime}$ non-powerful. Then $G/(G^{\prime})^{p}$ is a CF$(6,p)$-group.

Proof.

Clearly we can assume $(G^{\prime})^{p}=1$ and $G^{\prime\prime}\neq 1$. Thus, the Frattini subgroup of $G^{\prime}$ is $G^{\prime\prime}$, and since $d(G^{\prime})=3$, then $|G^{\prime}:G^{\prime\prime}|=p^{3}$. Note that $G^{\prime\prime}\leq\gamma_{4}(G)$, so the only possibilities for $\gamma_{3}(G)$ are $|G^{\prime}:\gamma_{3}(G)|=p^{2}$ or $|G^{\prime}:\gamma_{3}(G)|=p$.

Assume first $|G^{\prime}:\gamma_{3}(G)|=p^{2}$. Then, since $G^{\prime\prime}\leq\gamma_{4}(G)$ and $|G^{\prime}:G^{\prime\prime}|=p^{3}$ we have $G^{\prime\prime}=\gamma_{4}(G)$. In addition, $G^{\prime}$ has two generators modulo $\gamma_{3}(G)$, which implies that $|G^{\prime\prime}:\gamma_{5}(G)|=p$ (recall that $(G^{\prime})^{p}=1$). Consider the subgroup $C_{3}$ and recall it is maximal by Remark 2.11. In the same way as in Case 2 of Theorem 4.2, it can be seen that there are only $p+1$ maximal subgroups of $G^{\prime}$ that are normal in $G$. Hence, making the same computations, it follows that $D\cup C_{3}\neq G$.

Thus, we can pick $x\in G\setminus(D\cup C_{3})$, and we have $G^{\prime}=[x,G]=[x,\langle x\rangle C_{3}]=[x,C_{3}]$. We can then find $y,z\in C_{3}$ such that $G^{\prime}=\langle[x,y],[x,z],\gamma_{3}(G)\rangle$. We write $a=[x,y]$ and $b=[x,z]$ for simplicity. Thus, $\gamma_{4}(G)=\langle[a,b],\gamma_{5}(G)\rangle$, and we write, again for simplicity, $d=[a,b]$.

On the one hand,

[TABLE]

so that $[b,y,x]\equiv d\pmod{\gamma_{5}(G)}$. Similarly we get

[TABLE]

and so $[z,a,x]\equiv d\pmod{\gamma_{5}(G)}$. In particular $[b,y],[z,a]\not\in\gamma_{4}(G)$, and since $\gamma_{3}(G)/\gamma_{4}(G)$ is of order $p$, we have $[z,a]\equiv[b,y]^{i}\pmod{\gamma_{4}(G)}$ for some $1\leq i\leq p-1$. Note, however, that

[TABLE]

so we get $i=1$ and thus

[TABLE]

On the other hand, we have $[G^{\prime},C_{3}^{\prime}]\leq[G^{\prime},C_{3},C_{3}]\leq\gamma_{5}(G)$. Denote $Z/\gamma_{5}(G)=Z(G^{\prime}/\gamma_{5}(G))$. We have $|G^{\prime}:Z|\geq p^{2}$, and since $[G^{\prime},\gamma_{3}(G)]\leq\gamma_{5}(G)$, we get $Z=\gamma_{3}(G)$. In particular, we get $C_{3}^{\prime}\leq\gamma_{3}(G)$, and the nilpotency class of $C_{3}$ is less than or equal to 2. Now,

[TABLE]

so that $[y,z,x]\equiv[b,y][z,a]d\pmod{\gamma_{5}(G)}$. This is a contradiction since $[b,y][z,a]\in\gamma_{3}(G)\setminus\gamma_{4}(G)$ but $[y,z,x],d\in\gamma_{4}(G)$.

Therefore we must have $|G^{\prime}:\gamma_{3}(G)|=p$. Thus

[TABLE]

and since $|G^{\prime}:G^{\prime\prime}|=p^{3}$, we have $|\gamma_{3}(G):\gamma_{4}(G)|=|\gamma_{4}(G):\gamma_{5}(G)|=p$ and $G^{\prime\prime}=\gamma_{5}(G)$. Let us write $\overline{G}=G/\gamma_{7}(G)$. Note that

[TABLE]

so $\overline{\gamma_{3}(G^{\prime})}=\overline{1}$ and since $d(\overline{G}^{\prime})=3$, then $d(\overline{G}^{\prime\prime})\leq 2$. Indeed, we can write $G^{\prime}=\langle a,b,c\rangle$ with $a\in G^{\prime}\setminus\gamma_{3}(G)$, $b\in\gamma_{3}(G)\setminus\gamma_{4}(G)$ and $c\in\gamma_{4}(G)\setminus\gamma_{5}(G)$, and so the generators of $\overline{G}^{\prime\prime}$ are $\overline{[a,b]}\in\gamma_{5}(\overline{G})$ and $\overline{[a,c]}\in\gamma_{6}(\overline{G})$ (note that $\overline{[b,c]}\in\gamma_{7}(\overline{G})=\overline{1}$). Hence $|\gamma_{5}(\overline{G}):\gamma_{6}(\overline{G})|=p$ and $|\gamma_{6}(\overline{G})|\leq p$.

If $|\gamma_{6}(\overline{G})|=\overline{1}$ then $\gamma_{6}(G)=\gamma_{7}(G)=1$ and we are done, so assume $|\gamma_{6}(\overline{G})|=p$. Thus, $\overline{G}$ is a CF($7,p$)-group, and since $p\geq 3$, by Lemma 5.4 it follows that the degree of commutativity of $\overline{G}$ is greater than 0. In particular we have $\overline{G}^{\prime\prime}=[\gamma_{2}(\overline{G}),\gamma_{3}(\overline{G})]\leq\gamma_{6}(\overline{G})$, which is a contradiction. The lemma follows. ∎

With all this, the second part of the proof of Theorem A follows easily.

Theorem 5.6**.**

Let $G$ be a finite $p$-group with $G^{\prime}$ non-powerful, $d(G^{\prime})\leq 3$ and $p\geq 5$. Then, $G^{\prime}=K(G)$.

Proof.

We assume $d(G^{\prime})=3$ by Theorem 1.1. By Lemma 5.5 the action of $G$ on $G^{\prime}$ is uniserial modulo $(G^{\prime})^{p}$ and, in addition, $|G^{\prime}:(G^{\prime})^{p}|=p^{4}\leq p^{p-1}$ since $p\geq 5$. The result follows directly from Theorem B. ∎

Thus, combining Theorem 4.2 and Theorem 5.6 we establish Theorem A.

6. Proof of Theorems A*′* and B*′*

The analogous to Theorem A and Theorem B for pro-$p$ groups can be easily proved. Let $\mathcal{A}$ and $\mathcal{B}$ be the family of groups satisfying the conditions of Theorem A and Theorem B respectively. Using this notation, we prove now Theorem A*′* and Theorem B*′* together.

Proof of Theorem A′ and Theorem B′.

Let $G$ be a pro-$p$ group satisfying the conditions of Theorem A*′* or Theorem B*′*. Then, for every open normal subgroup $N$ of $G$ we have $G/N\in\mathcal{A}$ or $G/N\in\mathcal{B}$, and therefore, by Theorem A or Theorem B respectively, we have $G^{\prime}N/N=K(G)N/N$ or $G^{\prime}N/N=K_{x_{N}}(G)N/N$ for some $x_{N}\in G$ depending on the normal subgroup $N$.

In the case of Theorem B*′, let $X_{N}=\{x\in G\mid(G/N)^{\prime}=K_{xN}(G/N)\}$, which is closed in $G$, being a union of cosets of $N$. Clearly, the family $\{X_{N}\}_{N\trianglelefteq_{o}G}$ has the finite intersection property and, since $G$ is compact, $\cap_{N\trianglelefteq_{o}G}\,X_{N}\neq\varnothing$. If $x$ belongs to this intersection, then $(G/N)^{\prime}=K_{xN}(G/N)$ for all $N\trianglelefteq_{o}G$. Thus, let us write $\mathcal{K}(G)$ to refer to the subset $K(G)$ if we are in the situation of Theorem A′, or to the subset $K_{x}(G)$ if we are in the situation of Theorem B′*. Note that in both cases $\mathcal{K}(G)$ is closed in $G$, being the image of a continuous function.

On the other hand, since $G^{\prime}$ is topologically finitely generated we have

[TABLE]

for suitable $n\geq 1$ and $x_{i}\in G$ with $1\leq i\leq 2n$. Define $H=\operatorname{Cl}_{G}(\langle x_{1},\ldots,x_{2n}\rangle)$. Since $H$ is topologically finitely generated, it follows that $H^{\prime}$ is closed in $H$, and hence in $G$. Thus, $H^{\prime}$ is also closed in $G^{\prime}$, so $G^{\prime}=H^{\prime}$. Therefore, $G^{\prime}$ is closed in $G$. Now,

[TABLE]

and the proof is complete. ∎