# Powerfully nilpotent groups of maximal powerful class

- 48 Downloads

## Abstract

In this paper we continue the study of powerfully nilpotent groups started in Traustason and Williams (J Algebra 522:80–100, 2019). These are powerful *p*-groups possessing a central series of a special kind. To each such group one can attach a powerful class that leads naturally to the notion of a powerful coclass and classification in terms of an ancestry tree. The focus here is on powerfully nilpotent groups of maximal powerful class but these can be seen as the analogs of groups of maximal class in the class of all finite *p*-groups. We show that for any given positive integer *r* and prime \(p>r\), there exists a powerfully nilpotent group of maximal powerful class and we analyse the structure of these groups. The construction uses the Lazard correspondence and thus we construct first a powerfully nilpotent Lie ring of maximal powerful class and then lift this to a corresponding group of maximal powerful class. We also develop the theory of powerfully nilpotent Lie rings that is analogous to the theory of powerfully nilpotent groups.

## Keywords

Powerful*p*-group Nilpotent p-group Coclass Maximal class

## Mathematics Subject Classification

20D15 20F40## 1 Introduction

In this paper we continue the study of powerfully nilpotent *p*-groups started in [6] and continued in [7]. Powerful *p*-groups were introduced by Lubotzky and Mann in [5]. The class of powerfully nilpotent groups is a special subclass of these, containing groups that possess a central series of a special kind. We start by recalling the basic terms. Let *G* be a finite *p*-group where *p* is a prime.

### Definition

*powerfully central in G*if \([H_{i},G]\le H_{i-1}^{p}\) for \(i=1,\ldots ,n\). Here

*n*is called the

*length*of the chain.

### Definition

*p*-group

*G*is

*powerfully nilpotent*if it has an ascending chain of subgroups of the form

*G*.

### Definition

If *G* is powerfully nilpotent then the *powerful nilpotence class* of *G* is the shortest length that a powerfully central chain of *G* can have.

### Definition

We say that a finite *p*-group *G* is *strongly powerful* if \([G,G]\le G^{p^{2}}\).

In [6] we showed that a strongly powerful *p*-group is powerfully nilpotent of powerful class at most \(e-1\) where \(p^{e}\) is the exponent of the group.

**The upper powerfully central series**. This is defined recursively as follows: \(\hat{Z}_{0}(G)=\{1\}\) and for \(n\ge 1\),

### Definition

Let *G* be a powerfully nilpotent *p*-group of powerful class *c* and order \(p^{n}\). We define the *powerful coclass* of *G* to be the number \(n-c\).

A natural approach is to develop something that corresponds to a coclass theory for finite *p*-groups where coclass is replaced by powerful coclass and in [6] we proved that there are indeed, for any fixed prime *p*, finitely many powerfully nilpotent *p*-groups of any given powerful coclass. More precisely, if *G* is a powerfully nilpotent *p*-group of rank *r* and exponent \(p^{e}\) then we showed that \(r\le n-c+1\) and \(e\le n-c+1\). This together with \(n\le re\) gives the result.

**The ancestry tree**. Let *p* be a fixed prime. The *vertices* of the ancestry tree are all the powerfully nilpotent *p*-groups (one for each isomorphism class). Two vertices *G* and *H* are joined by a directed edge from *H* to *G* if and only if \(H\cong G/Z(G)^{p}\) and *G* is not abelian. Notice that this implies that \(Z(G)^{p}\not =\{1\}\) and thus the powerful class of *G* is one more than that of *H*. We then also say that *G* is a *direct ancestor* of *H* or that *H* is a *direct descendant* of *G*, and we write \(H\rightarrow G\).

Let *c*(*H*), *c*(*G*) be the powerfully nilpotent classes of *H* and *G* and let *d*(*H*), *d*(*G*) be the powerful coclasses, that is \(d(H)=n(H)-c(H)\) and \(d(G)=n(G)-c(G)\) where the orders of *H* and *G* are \(p^{n(H)}\) and \(p^{n(G)}\). Notice that \(c(H)=c(G)+1\) and that \(d(H)\ge d(G)\) with equality if and only if \(|Z(G)^{p}|=p\).

We will now recall some more terms from [6].

### Definition

*G*be a powerfully nilpotent

*p*-group and let

*k*be the largest non-negative integer such that

*tail*of

*G*and

*k*as the

*length of the tail*.

### Remark

If *G* has a tail of length *k* then \(G, G/\hat{Z}_{1}(G)^{p}, G/\hat{Z}_{2}(G)^{p},\ldots ,G/\hat{Z}_{k}(G)^{p}\) all have the same powerful coclass.

### Definition

Let *G* be a powerfully nilpotent *p*-group. We say that *G* has *maximal tail* if the tail of *G* is \(G^{p}\).

The following summarises some of the properties of groups with a maximal tail.

### Theorem

*G*be a powerfully nilpotent group of rank \(r\ge 2\) that has a maximal tail. Suppose

*G*has order \(p^{n}\), powerful class

*c*and exponent \(p^{e}\). Let

*t*be the length of the tail.

- (a)
We have \(c-1\le t\le c\). It follows also that \(n-c\le r\le n-c+1\).

- (b)
We have \(t\le 1+\frac{r(r-1)}{2}\).

- (c)
We have \(\text{ rank }(G)>\text{ rank }(G^{p})>\cdots >\text{ rank }(G^{p^{e-2}})\).

We conjectured in [6] that the bound given in (b) is attained and we will show in this paper that this is the case when \(p>r\). A powerfully nilpotent group, for which this bound is attained, will be called a group of maximal powerful class.

The paper is organised as follows. In Sect. 2 we will develop the theory of powerfully nilpotent Lie rings that is analogous to the theory of powerfully nilpotent groups with similar results. In Sect. 3 we show the existence of a powerfully nilpotent Lie ring of maximal powerful class for any prime *p* and any rank *r* and we show that the structure of Lie rings of maximal powerful class is constrained. In the final section we then use the Lazard correspondence to obtain the analogous results for powerfully nilpotent *p*-groups of maximal powerful class when \(p>r\). The main structure result, Theorem 4.2, shows some parallels with the structure of groups with maximal class (see for example [1] or Chapter 3.14 in [2]) in the class of all finite *p*-groups. We refer to [4] for an account of coclass theory for finite *p*-groups.

## 2 Powerfully nilpotent Lie rings

### Definition

*L*be a finite Lie ring of

*p*-power order and let

*M*,

*N*be subrings where \(M\le N\le L\). We say that an ascending chain of subrings

*powerfully central*in

*L*if \([J_{i},L]\le pJ_{i-1}\) for \(i=1,\ldots ,n\). Here

*n*is called the

*length*of the chain.

### Definition

*L*be a finite Lie ring of

*p*-power order. We say that

*L*is

*powerfully nilpotent*if it has an ascending chain of ideals of the form

*L*.

### Definition

If *L* is powerfully nilpotent then the *powerful nilpotence class* of *L* is the shortest length that a powerfully central chain of *L* can have.

### Definition

Let *L* be a finite Lie ring of *p*-power order. We say that *L* is *powerful* if \([L,L]\le pL\) and *strongly powerful* if \([L,L]\le p^{2}L\).

### Remark

Clearly a powerfully nilpotent Lie ring is powerful. The next lemma shows that a strongly powerful Lie ring is powerfully nilpotent.

### Lemma 2.1

Let *L* be a strongly powerful Lie ring of characteristic \(p^{e}\). Then *L* is powerfully nilpotent. Furthermore the powerful class is at most \(e-1\) if \(e\ge 2\) and the powerful class is equal to *e* if \(e\le 1\).

### Proof

**The upper powerfully central series**. This is defined recursively as follows: \(\hat{Z}_{0}(L)=\{0\}\) and for \(n\ge 1\),

### Remark

The upper powerfully central series is clearly the fastest ascending series that is powerfully central. It is easy to see that *L* is powerfully nilpotent if and only if \(L=\hat{Z}_{n}(L)\) for some *n* and the smallest such *n* is then the powerful class.

The next result is analogous to a corresponding result from [6].

### Proposition 2.2

Let *L* be a finite Lie ring of characteristic \(p^{e}\) where \(e\ge 2\). If \(L/p^{2}L\) is powerfully nilpotent, then *L* is powerfully nilpotent. Furthermore if \(L/p^{2}L\) has powerful class *m*, then the powerful class of *L* is at most \((e-1)m\).

### Proof

*L*is powerful. Thus in particular \([pL,L]=p[L,L]\le p^{2}L\) and \(pL\le J_{m-1}\). Hence

*L*. Consider the descending chain

*L*. Thus \([p^{k}J_{i},L]=p^{k}[J_{i},L]\le p^{k+1}J_{i+1}\) for \(0\le i\le m-1\) and \(0\le k\le e-2\). It follows that we have a powerfully central chain in

*L*. Notice also that \([p^{e-2}J_{m-1},L]=p^{e-2}(p^{2}L)=\{0\}\) and thus \(p^{e-2}J_{m-1}\le Z(L)\). It follows that

*L*is powerful with powerful class \((e-1)m\). \(\square \)

Like for powerfully nilpotent groups there is a natural classification of powerfully nilpotent Lie rings in terms of a powerful coclass and an ancestry tree.

### Definition

Let *L* be a powerfully nilpotent Lie ring of powerful class *c* and order \(p^{n}\). We define the *powerful coclass* of *L* to be the number \(n-c\).

**The ancestry tree**. Let *p* be a fixed prime. The *vertices* of the ancestry tree are all the powerfully nilpotent Lie rings of prime power order (one for each isomorphism class). Two vertices *L* and *M* are joined by a directed edge from *M* to *L* if and only if \(M\cong L/pZ(L)\) and *L* is not abelian. Notice that this implies that \(pZ(L)\not =\{0\}\) and thus the powerful class of *L* is one more than that of *M*. We then say that *L* is a direct ancestor of *M* or that *M* is a direct descendant of *L* and write \(M\rightarrow L\).

### Remark

Suppose *M* is a powerfully nilpotent Lie ring of order \(p^{n(M)}\) and powerful class *c*(*M*) and suppose that *L* is a direct ancestor of *M*. Then *L* has powerful class \(c(M)+1\) and order \(p^{n(L)}=|L/pZ(L)|\cdot |pZ(L)|=p^{n(H)+k}\) where \(|pZ(L)|=p^{k}\). Thus the powerful coclass of *L* is \(d(L)=n(M)+k-(c(M)+1)= (n(M)-c(M))+(k-1)\) and thus \(d(L)\ge d(M)\) where *d*(*M*) is the powerful class of *M*. Notice that we have equality if and only if \(|pZ(L)|=p\).

### Lemma 2.3

*L*be a powerfully nilpotent Lie ring of powerful class \(c\ge 2\), then

### Proof

Suppose \(2\le j\le c\). If \([\hat{Z}_{j}(L),L]=[\hat{Z}_{j-1}(L),L]\), then \([\hat{Z}_{j}(L),L]\le p\hat{Z}_{j-2}(L)\) and thus we get the contradiction that \(\hat{Z}_{j}(L)\le \hat{Z}_{j-1}(L)\). The proof of the latter strict inequalities is similar. Let \(1\le j\le c-1\). If \(p\hat{Z}_{j}(L)= p\hat{Z}_{j-1}(L)\) then \([\hat{Z}_{j+1}(L), L]\le p\hat{Z}_{j-1}(L)\) and thus \(\hat{Z}_{j+1}(L)=\hat{Z}_{j}(L)\) that gives the contradiction that the powerful class is at most \(j\le c-1\). \(\square \)

We are going to see that, as for powerfully nilpotent groups, there are for a fixed prime *p* only finitely many powerfully nilpotent Lie rings of *p*-power order of any given coclass. This will follow from specific bounds for the rank and exponent in terms of the coclass. We start with the rank.

### Proposition 2.4

Let *L* be a powerfully nilpotent Lie ring of rank *r*, order \(p^{n}\) and powerful class *c*. Then \(r\le n-c+1\).

### Proof

It follows from Lemma 2.3 that \(p^{c-1}\le |pL|=p^{n-r}\). The result follows. \(\square \)

In order to get the bound for the exponent in terms of the coclass we need first some more structure results for powerfully nilpotent Lie rings.

*L*be any powerfully nilpotent Lie ring of characteristic \(p^{e}\), order \(p^{n}\) and rank

*r*. Suppose

*r*such that the factors are of order

*p*. Without loss of generality we can thus assume that we have a powerfully central chain

*L*such that

*L*. As we saw in the proof of Proposition 2.2, we then get a powerfully central series

*p*in front of the others (keeping the previous order unchanged otherwise) and still have that (1) gives us a powerfully central chain. We can thus assume that for some \(0\le s\le r\) we have \(pa_{1}=\cdots =pa_{s}=0\) and that \(pL=pJ_{s}>pJ_{s+1}> \cdots >pJ_{r}=p^{2}L\). Notice that the rank of

*pL*is the number of generators of order at least \(p^{2}\). We have \(pL=0\) if \(s=r\) and otherwise \(0\le s<r\) and \(\{0\}<pL=pJ_{s}\). Suppose we are in the latter situation. In this case we have for \(s\le j\le r-1\) that \(p^{2}J_{i}=p^{2}J_{i+1}\) if and only if there exists \(x\in J_{i}{\setminus } J_{i+1}\) such that \(p^{2}x=0\). We can thus choose our generators such that furthermore \(p^{2}J_{i}=p^{2}J_{i+1}\) if and only if \(a_{i+1}\) has order \(p^{2}\). Notice again that the rank of \(p^{2}L\) is then the number of generators of order at least \(p^{3}\). Continuing in this manner, considering next \(p^{3}J_{0}\ge p^{3}J_{1}\ge \cdots \ge p^{3}J_{r}=p^{4}L\) and then the \(p^{4}\)th powers and so on, we eventually arrive at a set of generators \(a_{1},\ldots ,a_{r}\) with some specific properties. If for \(0\le i\le r\) we let

*s*(

*i*) be the number of generators of order \(p^{i}\) then \(|p^{i-1}L/p^{i}L|=p^{s(i)+s(i+1)+\cdots + s(e)}\). Then

### Proposition 2.5

*L*be a powerfully nilpotent Lie ring of rank

*r*, characteristic \(p^{e}\) and order \(p^{n}\). Then we can choose generators for

*L*as a Lie ring such that we get a direct sum of cyclic groups

### Remark

If we omit repetitions in the chain above, then all the factors have order *p*.

### Lemma 2.6

Let *L* be a powerful Lie ring of *p*-power order. If \(p^{k}L\) is a cyclic group, then \(p^{k}L\le Z(L)\).

### Proof

Suppose \(p^{k}L={\mathbb {Z}}p^{k}c\). Then \([p^{k}L,L]=[p^{k}c,L]=[c,p^{k}L]={\mathbb {Z}}[c,p^{k}c]=\{0\}\). \(\square \)

### Lemma 2.7

*L*be a powerful Lie ring of

*p*-power order and suppose that \(p^{k}L\) has rank \(s\ge 2\). Let \(a_{1},\ldots ,a_{r}\) be as in Proposition 2.5 and that \(p^{k}L={\mathbb {Z}}p^{k}a_{i_{1}}+\cdots +{\mathbb {Z}}p^{k}a_{i_{s}}\) with \(i_{1}<i_{2}< \cdots <i_{s}\). Then the chain

*L*.

### Proof

*L*. Then, using the fact that \([a_{i_{1}},a_{i_{1}}]=0\),

### Remark

### Theorem 2.8

Let *L* be a powerfully nilpotent Lie ring of order \(p^{n}\), powerful class *c* and characteristic \(p^{e}\). Then \(e\le n-c+1\).

### Proof

*L*is of rank 1 so we can assume that the rank of

*L*is at least 2. Let

*k*be the largest non-negative integer such that the rank of \(p^{k}L\) is greater than or equal to 2. Let \(r_{i}\) be the rank of \(p^{i}L\) for \(i=0,1,\ldots ,k\) and let \(p^{n_{0}}= |p^{k+1}L|\). Notice then that

*L*. Adding up for \(j=0,1,\ldots ,k\) and using the fact from Lemma 2.6 that \(p^{k+1}L\le Z(L)\), we get a central chain for

*L*that is of length \((r_{0}-1)+(r_{1}-1)+\cdots + (r_{k}-1)+1\). Hence \(c\le (r_{0}-1)+(r_{1}-1)+\cdots +(r_{k}-1)+1\) and we conclude that

As a corollary we get a theorem that is analogous to one of the main results from [6]

### Theorem 2.9

For each prime *p* and non-negative integer *d*, there are only finitely many powerfully nilpotent Lie rings of *p*-power order that have powerful coclass *d*.

### Proof

Let *L* be a powerfully nilpotent Lie ring of order \(p^{n}\), rank *r* and exponent \(p^{e}\). Then \(n\le re\le (d+1)(d+1)\) and thus the order of *L* is bounded by the coclass. \(\square \)

## 3 Powerfully nilpotent Lie rings of maximal powerful class

### Definition

*L*be a powerfully nilpotent Lie ring of

*p*-power order and let

*k*be the largest non-negative integer such that

*tail*of

*L*and

*k*as the

*length of the tail*.

### Remark

If *L* has a tail of length *k*, then \(L,L/p\hat{Z}_{1}(L),L/p\hat{Z}_{2}(L),\ldots ,L/p\hat{Z}_{k}(L)\) all have the same powerful coclass.

### Lemma 3.1

*pL*is

*L*is \(p\hat{Z}_{k}(L)\), then \(M_{i}=p\hat{Z}_{i}(L)\) for \(i=0,\ldots ,k\).

### Proof

*j*. Let

*q*be the largest and then, for that

*q*,

*i*be the largest such that \(M_{j}=p^{q}J_{i}\). Then \(0\le i\le r-1\) and \(p^{q}J_{i+1}=M_{j-1}\). Thus

### Definition

We say that a powerfully nilpotent Lie ring *L* has *maximal tail* if the tail of *L* is *pL*.

### Remark

*L*is abelian then

*L*has maximal tail if and only if \(|pL|=p\) that happens if and only if

*p*-power order that has rank 1 and maximal tail.

### Theorem 3.2

*L*be a powerfully nilpotent Lie ring of rank \(r\ge 2\) that has maximal tail. Suppose that

*L*has order \(p^{n}\), powerful class

*c*and characteristic \(p^{e}\). Let

*t*be the length of the tail.

- (a)
We have \(\text{ rank }(L)>\text{ rank }(pL)>\cdots >\text{ rank }(p^{e-2}L)\).

- (b)
We have that \(c-1\le t\le c\).

- (c)
We have \(t\le 1+\frac{r(r-1)}{2}\).

### Proof

- (a)For \(e\le 2\) there is nothing to prove and we can thus assume that \(e\ge 3\). Notice first that the rank of \(p^{e-3}L\) must be at least 2. Otherwise, by Lemma 2.6, we would have that \(p^{e-3}L\le Z(L)\) and thus \(p^{e-2}L\le pZ(L)\). As \(|pZ(L)|=p\) we would then get the contradiction that \(p^{e-1}L=\{0\}\). Thus \(L,pL,\ldots ,p^{e-3}L\) all have rank at least 2. Let \(0\le k\le e-3\) and consider the chain from Proposition 2.5. The subchainis powerfully central in$$\begin{aligned} p^{k}L=p^{k}J_{0}\ge p^{k}J_{1}\ge \cdots \ge p^{k}J_{r}=p^{k+1}L \end{aligned}$$
*L*. Suppose that the rank of \(p^{k}L\) is \(s\ge 2\). Omitting repetitions we get a chainwith \(1\le i_{1}<i_{2}<\cdots <i_{s}\le r\). We know from Lemma 2.7 that we can omit the 2nd term and still have a powerfully central chain. Let \(E=p^{k}L\) and \(F={\mathbb {Z}}p^{k}a_{i_{3}}+\cdots +{\mathbb {Z}}p^{k}a_{i_{s}}+p^{k+1}L\). We know from Lemma 3.1 that \(pF=p\hat{Z}_{l}(L)\) for some \(0\le l\le t\). As \([E,L]\le pF\), it follows that \(E\le \hat{Z}_{l+1}(L)\) and thus \(pE\le p\hat{Z}_{l+1}(L)\). We show that \([pE:pF]\le p\). This is of course clear if \(pE=pF\). Otherwise \(p\hat{Z}_{l+1}(L)\ge pE>pF=p\hat{Z}_{l}(L)\) and thus we know from Lemma 3.1 that \([pE:pF]=[p\hat{Z}_{l+1}(L):p\hat{Z}_{l}(L)]=p\). Thus [$$\begin{aligned} p^{k}L= & {} {\mathbb {Z}}p^{k}a_{i_{1}}+\cdots +{\mathbb {Z}}p^{k}a_{i_{s}}+p^{k+1}L> {\mathbb {Z}}p^{k}a_{i_{2}}+\cdots +{\mathbb {Z}}p^{k}a_{i_{s}}+p^{k+1}L \\&>\cdots> {\mathbb {Z}}p^{k}a_{i_{s}}+p^{k+1}L>p^{k+1}L \end{aligned}$$*pE*:*pF*] is either 1 or*p*and thus the rank of \(p^{k+1}L\) is at most \(s-1\). - (b)
By Lemma 2.3 we know that \(p\hat{Z}_{c-2}(L)<p\hat{Z}_{c-1}(L)\le p\hat{Z}_{c}(L)=pL\). Thus \(c-1\le t\le c\).

- (c)By part (a) the largest potential tail occurs when \(\text{ rank }(L)=r, \text{ rank }(pL)=r-1,\ldots ,\text{ rank }(p^{r-1}L)=1\). By Lemma 2.6 we then have \(p^{r-1}L\le Z(G)\) and thus \(p^{r}L\le pZ(G)\). For there to be a tail of length greater than 0 we need \(|pZ(L)|=p\) and then \(|p^{r}L|\le p\). The tail can’t thus be larger thanThis finishes the proof. \(\square \)$$\begin{aligned} |pL|=|pL/p^{2}L|\cdot |p^{2}L/p^{3}L|\cdots |p^{r-2}L/p^{r-1}L|\cdot |p^{r-1}L|=p^{(r-1)+\cdots + 2+2}. \end{aligned}$$

### Remark

*L*has rank

*r*and maximal tail, then the length of the tail is at most \(t=1+r(r-1)/2\). We have also seen that in order for this upperbound to be attained we would need \(\text{ rank }(L)=r, \text{ rank }(pL)=r-1, \ldots ,\text{ rank }(p^{r-2}L)=2\) and that \(p^{r-1}L\) is of rank 1 and order \(p^{2}\). In particular, we must have \(e=r+1\). Clearly \(t=n-r\) and as, by Theorem 3.2, \(c-1\le t\le c\), it follows that \(n-c\le r\le n-c+1\). As \(e=r+1\) and \(e\le n-c+1\) we then must have

*r*that has maximal tail.

### Lemma 3.3

Suppose *L* is a powerfully nilpotent Lie ring of *p*-power order that has maximal tail, is of rank *r* and has powerful class \(c=1+r(r-1)/2\). Then *L* has tail of length \(t=c\).

### Proof

We argue by contradiction and suppose \(t\not =c\). By Theorem 3.2 (b), we then have \(t=c-1=r(r-1)/2\). Then \(e\le n-c+1= t+r-(c-1)=r\). By Theorem 3.2 (a), the only way of getting tail of length \(t=1+2+\cdots +(r-1)\) is if \(\text{ rank }(L)=r, \text{ rank }(pL)=r-1,\ldots ,\text{ rank }(p^{r-1}L)=1\).

*L*is a most the length of chain namely \((r-1)+(r-2)+\cdots +1=r(r-1)/2\) contradicting the assumption that the powerful class is \(1+r(r-1)/2\). \(\square \)

### Definition

Let *L* be a powerfully nilpotent Lie ring with maximal tail. We say that *L* has *maximal powerful class* if the powerful class is \(c=1+r(r-1)/2\).

### Remark

*L*is a powerfully nilpotent Lie ring of rank

*r*which has maximal powerful class, then \(t=c=1+r(r-1)/2\) and from the analysis above [see the proof of Theorem 3.2 (c)] we know that as an abelian group we can write

*Z*(

*L*). We also know from Lemma 2.7 that if we omit repetitions then the rows give us subchains of lengths \(r,r-1,\ldots ,2\) and that we still get a powerfully central chain if we omit the 2nd term in each line. That would give us a powerfully central chain of length \((r-1)+(r-2)+\cdots +1+1=1+r(r-1)/2=c\) (adding \(p^{r-1}L>\{0\}\)). As this is the powerful class there can’t be any strict subchain that is powerfully central. In particular

*p*, as if \(o(a_{j})=p\) then \([J_{j-1},L]\le pJ_{j}=pJ_{j+1}\) giving us a strict subchain that is powerfully central. It also follows that none of \(a_{2},\ldots ,a_{r-1}\) can have order \(p^{r+1}\). To see this we observe that if \(a_{j}\) is of order \(p^{r+1}\) for some \(j\ge 2\) then by (2) there are some \(0\le k<l\le r-1\) such that

*p*. As \( p[a_{0},a_{1}]=0\) we then can’t have that \(pa_{2}\) has order greater than

*p*and thus \(a_{2}\) must have order exactly \(p^{2}\). Next we use

### Lemma 3.4

Let *L* be a powerfully nilpotent Lie ring of rank *r* and maximal powerful class. We can pick our generators \(a_{0},a_{1},\ldots , a_{r-1}\) such that \(o(a_{1})=p, \ldots ,o(a_{r-1})=p^{r-1}\) and \(o(a_{0})=p^{r+1}\).

### Remark

In fact our proof also shows that we must have \(o(a_{j})=p^{j}\) for \(2\le j\le r-1\). For the following we will denote \(a_{0}\) by *x*.

### Lemma 3.5

We have \([a_{i},x]\in pJ_{i+1}{\setminus } pJ_{i+2}\) for \(i=1,\ldots ,r-2\) and \([a_{r-1},x]\in p^{2}J_{0}{\setminus } p^{2}J_{1}\).

### Proof

### Theorem 3.6

*L*be a powerfully nilpotent Lie ring of

*p*-power order and rank

*r*that has maximal powerful class \(1+r(r-1)/2\). There exist generators \(a_{1},\ldots ,a_{r-1},x\) such that we get a direct sum of \({\mathbb {Z}}\)-modules

- (a)
\([a_{1},x]=pa_{2}, [a_{2},x]=pa_{3},\ldots ,[a_{r-2},x]=pa_{r-1}, [a_{r-1},x]=p^{2}x.\)

- (b)
\(H=\Omega _{r}(L)={\mathbb {Z}}px+{\mathbb {Z}}a_{1}+\cdots + {\mathbb {Z}}a_{r-1}\) is powerfully embedded in

*L*and strongly powerful. - (c)
\(p^{r-1}L\le Z(L)\).

### Proof

*p*divides \(\alpha \) and as \(p^{j-1}[a_{i},a_{j}]=0\) we must have that

*p*divides \(\alpha _{j+1},\ldots ,a_{r-1}\). Thus \([a_{i},a_{j}]\le p^{2}H\). Furthermore \([a_{i},px]=p^{2}a_{i+1}\) when \(1\le i\le r-2\) whereas \([a_{r-1},px]=p^{3}x\). Hence \([H,H]\le p^{2}H\) and

*H*is strongly powerful. Also \([a_{i},x]=pa_{i+1}\) for \(1\le i\le r-2\) and \([a_{r-1},x]=p^{2}x\) and thus \([H,L]\le pH\) that shows that

*H*is powerfully embedded in

*L*. \(\square \)

### Remark

*x*from the right induces a surjective map

*p*-groups of maximal class. Notice also that Lemma 2.1 tells us that the class of

*H*is at most

*r*that is small compared to the class of

*L*.

Having analysed the structure of powerfully nilpotent Lie rings of maximal powerful class, we show that for each *r* there exists such a Lie ring.

*r*be a positive integer and let

*p*be a prime. Consider a \({\mathbb {Z}}\)-module

*x*where this is the linear map induced by

*x*. Notice that the image of

*x*is \(pA+{\mathbb {Z}}p^{2}x\) that has the same order as

*A*, namely \(pp^{2}\cdots p^{r-1}\), and thus \(x:A\rightarrow L\) is injective.

Now consider an alternating product on *A* such that \([A, A]\subseteq B\) where \(B={\mathbb {Z}}a_{1}+\cdots +{\mathbb {Z}}a_{r-2}\). Our aim is to find such an alternating product that together with the right multiplication of *x* will give us a Lie ring.

### Proposition 3.7

*A*with the following properties

*A*together with the right multiplication by

*x*gives

*L*the structure of a Lie ring. Also \([A, A]\subseteq p^{2}B\).

### Proof

*x*induces a bijection from

*B*to

*pA*. Thus for any \(u\in pA\) there exists a unique \(v\in B\) such that \([v,x]=u\). We will be making use of this property. We will first establish (1) and (2) as well as \([A, A]\subseteq p^{2}B\). Let \(1\le i<j\le r-1\). We define \([a_{i},a_{j}]\) by reverse induction on \(3\le i+j\le 2r-3\). We will show that there is a unique \(u\in B\) such that \(p^{i}u=0\) and

*B*such that \([u,x]=p^{3}a_{r-1}\). Notice that \(u\in p^{2}B\) and that \(p^{r-2}u=0\). We define \([a_{r-2},a_{r-1}]=p^{2}a_{r-2}\).

*x*we know that we also have \(p^{i}u=0\).

*x*. This finishes the inductive proof.

*A*such that (1) and (2) hold and that furthermore \([A, A]\subseteq p^{2}B\) we extend this into an alternating product on

*L*using the right multiplication on

*A*by

*x*. Notice that by (2) we know that

*x*acts as derivation on

*A*. In order to show that the Jacobi identity \(J(x,y,z)=0\) holds where

*L*is at most

*r*. In fact \([a_{1},_{r-1}x]=p[a_{2},_{r-2}x]=\cdots =p^{r-2}[a_{r-1},x]=p^{r}x\not =0\) and the class is thus exactly

*r*.

### Theorem 3.8

The Lie ring *L* is a powerfully nilpotent Lie ring with maximal tail.

### Proof

*L*is powerfully nilpotent. In order to determine the upper powerfully central series we make again use of \([A,A]\le p^{2}B\) as well as the fact that \(x:A\rightarrow pA+{\mathbb {Z}}p^{2}x\) is injective. From this one sees readily that the following ascending sequences are \((\hat{Z}_{i}(L))\) and \((p\hat{Z}_{i}(L)))\):

*p*and thus

*L*has maximal tail of length \(1+(1+2+\cdots + r-1)=1+r(r-1)/2\). \(\square \)

## 4 Powerfully nilpotent groups with maximal powerful class

In this section we want to prove results on powerfully nilpotent groups with maximal powerful class that are analogous to those we have obtained for the powerfully nilpotent Lie rings.

Let *p* be a prime. We will make use of the Lazard 1–1 correspondence between all finite *p*-groups of nilpotency class at most \(p-1\) and all finite Lie rings of class at most \(p-1\) whose order is a power of *p*.

*L*be a Lie ring of order \(p^{n}\) and nilpotency class \(m<p\). The corresponding group \(G=\{e^{a}:a\in L\}\) is then a finite group of the same order and nilpotency class consisting of formal expressions

*E*is the associative envelping algebra of

*L*.) The structure of

*G*is derived from the structure of

*L*, via the Baker–Hausdorff formula. Thus

*l*(

*x*,

*y*) is a word in the free Lie algebra over \({\mathbb {Q}}\) of class

*m*provided by the Lazard correpondence and

*l*(

*a*,

*b*) is the value of

*l*(

*x*,

*y*) in

*L*. In the following we will use the bracket notion both for group commutators and for the Lie product as there will be no ambiguity regarding this. Under the Lazard correspondence, one has in particular

Under the Lazard correspondence, sub Lie rings of *L* correspond to subgroups of *G* and ideals of *L* correspond to normal subgroups of *G*. Notice also that if *M* is a subring of *L* and \(H=e^{M}\) is the corresponding subgroup of *G*, then the subgroup corresponding to the subring *pM* is \(e^{pM}=\{h^{p}:\,h\in H\}=H^{p}\).

*M*,

*N*be subrings of

*L*. Notice that

*L*is powerfully nilpotent with a powerfully central series

*G*is powerfully nilpotent with a powerfully central series

*L*has maximal tail if and only if

*G*has maximal tail and that the two tails have then the same length.

In the previous section we constructed a powerfully nilpotent Lie ring of *p*-power order and rank *r* that is of maximal powerful class and saw that the real class of *L* was *r*. In the case when \(p>r\), we can thus use the Lazard Correspondence to get a corresponding example \(G=e^{L}\) that is a powerfully nilpotent group of maximal powerful class.

### Theorem 4.1

Let *r* be a positive integer and \(p>r\) a prime. There exists a powerfully nilpotent *p*-group of rank *r* that is of maximal powerful class.

*p*-power order that is of rank

*r*and maximal powerful class. We let \(G=e^{L}\) be the corresponding powerfully nilpotent

*p*-group that is of maximal powerful class. Recall that

*x*has the property that Lie multiplication by

*x*on the right induces surjective maps

### Theorem 4.2

*G*be a powerfully nilpotent

*p*-group of rank

*r*where \(p>r\) that has maximal powerful class \(1+r(r-1)/2\). There exist generators \(b_{1},\ldots ,b_{r-1},y\) such that

*G*is a product of the corresponding cyclic groups

- (a)
\([b_{1},y]=b_{2}^{p}, [b_{2},y]=b_{3}^{p},\ldots [b_{r-2},y]=b_{r-1}^{p}, [b_{r-1},y]=y^{p^{2}}.\)

- (b)
\(H=\Omega _{r}(G)=\langle y^{p}\rangle \cdot \langle b_{1}\rangle \cdots \langle b_{r-1}\rangle \) is powerfully embedded in

*G*and strongly powerful. - (c)
\(G^{p^{r-1}}\le Z(G)\).

### Remark

The fact that *H* is strongly powerful of exponent \(p^{r}\) implies that its powerful class is at most *r* and thus small compared to the powerful class of *G*. The situation is thus analogous to the structure of *p*-groups of maximal class as we have a subgroup of index *p* with relatively small powerful class and a ‘uniform’ element *y*.

## Notes

### Acknowledgements

This paper originated from the first author’s visit to the University of the Basque Country. He thanks the department of mathematics for their great hospitality during the visit and in particular G. A. Fernanández-Alcober and M. Noce for fruitful discussions.

## References

- 1.Fernández-Alcober, G.A.: An introduction to finite \(p\)-groups: regular \(p\)-groups and groups of maximal class. In: Mathemática Contemporânea, vol. 20. XVI Escola de Álgebra (Part 1), pp. 155–225 (2001)Google Scholar
- 2.Huppert, B.: Endliche Gruppen I, Die Grundlehren der Mathematishcen Wissenshaften, vol. 134. Springer, Berlin (1967)CrossRefGoogle Scholar
- 3.Khukhro, E.I.: \(p\)-Automorphisms of Finite \(p\)-Groups, LMS Lecture Note Series 246. Cambridge University Press, Cambridge (1998)CrossRefGoogle Scholar
- 4.Leedham-Green, C.R., McKay, S.: The Structure of Groups of Prime Power Order, LMS Monographs (New Series) 27. Oxford University Press, Oxford (2002)zbMATHGoogle Scholar
- 5.Lubotzky, A., Mann, A.: Powerful \(p\)-groups. J. Algebra
**105**, 485–505 (1987)MathSciNetzbMATHGoogle Scholar - 6.Traustason, G., Williams, J.: Powerfully nilpotent groups. J. Algebra
**522**, 80–100 (2019) MathSciNetCrossRefzbMATHGoogle Scholar - 7.Williams, J.: Omegas of agemos in powerful groups. Int. J. Group Theory
**(to appear)**Google Scholar

## Copyright information

**Open Access**This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.