A criterion for metanilpotency of a finite group

TL;DR

This paper establishes a criterion for metanilpotency in finite groups by linking the nilpotency of the lower central series' terms to a specific order condition on certain commutators.

Contribution

It provides a new characterization of metanilpotency in finite groups based on the order behavior of $ ext{γ}_k$-commutators.

Findings

The $k$th term of the lower central series is nilpotent if and only if $|ab|=|a||b|$ for coprime $ ext{γ}_k$-commutators.

The criterion offers a practical test for metanilpotency in finite groups.

The result deepens understanding of the structure of finite groups through commutator properties.

Abstract

We prove that the $k$th term of the lower central series of a finite group $G$ is nilpotent if and only if $|ab|=|a||b|$ for any $\gamma_k$-commutators $a,b\in G$ of coprime orders.

Equations

$${\gamma }_1=x_1,\text{and}{\gamma }_k=[{\gamma }_{k-1},x_k]=[x_1,\dots ,x_k]\text{for k≥2.}$$

$$1\ne [H,a]=[H,\underset{(k-1)times}{\underbrace{a,\dots ,a}}]\le N_p,$$

$$P_2=[P_2,\underset{(k-1)times}{\underbrace{P_1,\dots ,P_1}}].$$

$$[y,\underset{ktimes}{\underbrace{x,\dots ,x}}]\in {\gamma }_k(R)\le F(G)=Z(G).$$

Full text

1. Introduction

All groups considered in this article are finite. The following sufficient condition for nilpotency of a group $G$ was discovered by B. Baumslag and J. Wiegold [2].

Let $G$ be a group in which $|ab|=|a||b|$ whenever the elements $a,b$ have coprime orders. Then $G$ is nilpotent.

Here the symbol $|x|$ stands for the order of the element $x$ in a group $G$. In [1] a similar sufficient condition for nilpotency of the commutator subgroup $G^{\prime}$ was established.

Let $G$ be a group in which $|ab|=|a||b|$ whenever the elements $a,b$ are commutators of coprime orders. Then $G^{\prime}$ is nilpotent.

Of course, the conditions in both above results are also necessary for the nilpotency of $G$ and $G^{\prime}$, respectively. In the present article we extend the above results as follows.

Given an integer $k\geq 1$, the word $\gamma_{k}=\gamma_{k}(x_{1},\dots,x_{k})$ is defined inductively by the formulae

[TABLE]

The subgroup of a group $G$ generated by all values of the word $\gamma_{k}$ is denoted by $\gamma_{k}(G)$. Of course, this is the familiar $k$th term of the lower central series of $G$. If $k=2$ we have $\gamma_{k}(G)=G^{\prime}$. In the sequel the values of the word $\gamma_{k}$ in $G$ will be called $\gamma_{k}$-commutators.

Theorem 1**.**

The $k$th term of the lower central series of a group $G$ is nilpotent if and only if $|ab|=|a||b|$ for any $\gamma_{k}$-commutators $a,b\in G$ of coprime orders.

Recall that a group $G$ is called metanilpotent if there is a normal nilpotent subgroup $N$ such that $G/N$ is nilpotent. The following corollary is immediate.

Corollary 2**.**

A group $G$ is metanilpotent if and only if there exists a positive integer $k$ such that $|ab|=|a||b|$ for any $\gamma_{k}$-commutators $a,b\in G$ of coprime orders.

We suspect that a similar criterion of nilpotency of the $k$th term of the derived series of $G$ can be established. On the other hand, Kassabov and Nikolov showed in [4] that for any $n\geq 7$ the alternating group $A_{n}$ admits a commutator word all of whose nontrivial values have order 3. Thus, the verbal subgroup $w(G)$ need not be nilpotent even if all $w$-values have order dividing 3.

2. Proofs

As usual, if $\pi$ is a set of primes, we denote by $\pi^{\prime}$ the set of all primes that do not belong to $\pi$. For a group $G$ we denote by $\pi(G)$ the set of primes dividing the order of $G$. The maximal normal $\pi$-subgroup of $G$ is denoted by $O_{\pi}(G)$. The Fitting subgroup of $G$ is denoted by $F(G)$. The Fitting height of $G$ is denoted by $h(G)$. Throughout the article we use without special references the well-known properties of coprime actions: if $\alpha$ is an automorphism of a finite group $G$ of coprime order, $(|\alpha|,|G|)=1$, then $C_{G/N}(\alpha)=C_{G}(\alpha)N/N$ for any $\alpha$-invariant normal subgroup $N$, the equality $[G,\alpha]=[[G,\alpha],\alpha]$ holds, and if $G$ is in addition abelian, then $G=[G,\alpha]\times C_{G}(\alpha)$. Here $[G,\alpha]$ is the subgroup of $G$ generated by the elements of the form $g^{-1}g^{\alpha}$, where $g\in G$.

For elements $x,y$ of a group $G$ write $[x,{}_{0}y]=x$ and $[x,{}_{i+1}y]=[[x,{}_{i}y],y]$ for $i\geq 0$. An element $y\in G$ is called Engel if for any $x\in G$ there is a positive integer $n=n(x)$ such that $[x,{}_{n}y]=1$.

The following lemma is well-known.

Lemma 3**.**

Let $p$ be a prime and $G$ a metanilpotent group. Suppose that $x$ is a $p$-element in $G$ such that $[O_{p^{\prime}}(F(G)),x]=1$. Then $x\in F(G)$.

Proof.

Since all Engel elements of a finite group lie in the Fitting subgroup [5, 12.3.7], it is sufficient to show that $x$ is an Engel element. Let $F=F(G)$ and $P$ be the Sylow $p$-subgroup of $F$. We have $F=P\times O_{p^{\prime}}(F)$. By hypothesis, $G/F$ is nilpotent of class $n$ for some positive integer $n$. We deduce that $[G,{}_{n}x]\leq F$ and so $[G,_{n+1}x]\leq P$. Therefore $\langle[G,_{n+1}x],x\rangle$ is contained in a Sylow $p$-subgroup of $G$. Hence, $x$ is an Engel element in $G$ and so the lemma follows. ∎

Lemma 4**.**

Let $k$ be a positive integer and $G$ a group such that $G=G^{\prime}$. Let $q\in\pi(G)$. Then $G$ is generated by $\gamma_{k}$-commutators of $p$-power order for primes $p\neq q$.

Proof.

For each prime $p\in\pi(G)\setminus\{q\}$ let $N_{p}$ denote the subgroup generated by all $\gamma_{k}$-commutators of $p$-power order. Let us show first that for each $p$ the Sylow $p$-subgroups of $G$ are contained in $N_{p}$. Suppose that this is false and choose $p$ such that a Sylow $p$-subgroup of $G$ is not contained in $N_{p}$. We can pass to the quotient $G/N_{p}$ and assume that $N_{p}=1$. Since $G=G^{\prime}$, it is clear that $G$ does not possess a normal $p$-complement. Therefore the Frobenius Theorem [3, Theorem 7.4.5] shows that $G$ has a $p$-subgroup $H$ and a $p^{\prime}$-element $a\in N_{G}(H)$ such that $[H,a]\neq 1$. We have

[TABLE]

a contradiction. Therefore indeed $N_{p}$ contains the Sylow $p$-subgroups of $G$. Let $T$ be the product of all $N_{p}$ for $p\neq q$. We see that $G/T$ is a $q$-group. Since $G=G^{\prime}$, we conclude that $G=T$. The proof is complete. ∎

Let us call a subgroup $H$ of $G$ a tower of height $h$ if $H$ can be written as a product $H=P_{1}\cdots P_{h}$, where

(1) $P_{i}$ is a $p_{i}$-group ($p_{i}$ a prime) for $i=1,\dots,h$.

(2) $P_{i}$ normalizes $P_{j}$ for $i<j$.

(3) $[P_{i},P_{i-1}]=P_{i}$ for $i=2,\dots,h$.

It follows from (3) that $p_{i}\neq p_{i+1}$ for $i=1,\dots,h-1$. A finite soluble group $G$ has Fitting height at least $h$ if and only if $G$ possesses a tower of height $h$ (see for example [6]).

Throughout the remaining part of the article $G$ denotes a finite group for which there exists $k\geq 1$ such that $|ab|=|a||b|$ for any $\gamma_{k}$-commutators $a,b\in G$ of coprime orders. We denote by $X$ the set of all $\gamma_{k}$-commutators in $G$.

Lemma 5**.**

Let $x\in X$ and $N$ be a subgroup normalized by $x$. If $(|N|,|x|)=1$, then $[N,x]=1$.

Proof.

Choose $y\in N$. The order of the $\gamma_{k}$-commutator $[x,y]$ is prime to that of $x$. Therefore we must have $|x[x,y]|=|x||[x,y]|$. However $x[x,y]=y^{-1}xy$. This is a conjugate of $x$ and so $|x[x,y]|=|x|$. Therefore $[x,y]=1$. ∎

Lemma 6**.**

If $G$ is soluble, then the subgroup $\gamma_{k}(G)$ is nilpotent.

Proof.

Set $h=h(G)$ and $F=F(G)$. If $G$ is nilpotent, the result is immediate so we assume that $h\geq 2$.

We first examine the case $h=2$. If $G/F$ has nilpotency class at most $k-1$, then $\gamma_{k}(G)\leq F$ is nilpotent. So we will assume that $G/F$ is nilpotent of class at least $k$. Hence, the image of some Sylow $p$-subgroup $P$ of $G$ in $G/F$ has nilpotency class at least $k$. Therefore, there exists a $\gamma_{k}$-commutator $x$ in elements of $P$ which does not belong to $F$. By Lemma 5 $[O_{p^{\prime}}(F),x]=1$, whence by Lemma 3, $x\in F$. This is a contradiction.

Now assume that $h\geq 3$. By [6, Lemma 1.9], there exists a tower $P_{1}P_{2}P_{3}\ldots P_{h}$ of height $h$ in $G$. Since $P_{2}=[P_{2},P_{1}]$, it follows that

[TABLE]

Combining Lemma 5 with the fact that $P_{2}$ is generated by $\gamma_{k}$-commutators of $G$ of $p_{2}$-orders, we deduce that $P_{3}$ commutes with $P_{2}$. On the other hand, $[P_{3},P_{2}]=P_{3}$, because $P_{1}P_{2}P_{3}\ldots P_{h}$ is a tower. This is a contradiction. The proof is complete. ∎

We are now in a position to complete the proof of Theorem 1.

Proof of Theorem 1.

It is clear that if $\gamma_{k}(G)$ is nilpotent, then $|ab|=|a||b|$ for any $\gamma_{k}$-commutators $a,b\in G$ of coprime orders. So we only need to prove the converse. Since the case where $k\leq 2$ was considered in [1] and [2], we will assume that $k\geq 3$.

Suppose that the theorem is false and let $G$ be a counterexample of minimal order. In view of Lemma 6 $G$ is not soluble while all proper subgroups of $G$ are. Therefore $G=G^{\prime}$. Let $R$ be the soluble radical of $G$. It follows that $G/R$ is nonabelian simple. By Lemma 6 $\gamma_{k}(R)$ is nilpotent. Suppose that $R\neq 1$.

Choose $q\in\pi(F(G))$. According to Lemma 4 $G$ is generated by the $\gamma_{k}$-commutators of $p$-power order for primes $p\neq q$. Let $Q$ be the Sylow $q$-subgroup of $F(G)$. By Lemma 5, $[Q,x]=1$, for every $\gamma_{k}$-commutator $x$ of $q^{\prime}$-order. Therefore $Q\leq Z(G)$. This happens for each choice of $q\in\pi(F(G))$ so we conclude that $F(G)\leq Z(G)$.

For each $x\in R$ and $y\in G$ we have

[TABLE]

Consequently, every element $x\in R$ is Engel. Therefore $R\leq F(G)$ ([5, 12.7.4]). Hence, $R=Z(G)$ and $G$ is quasisimple.

Since $G$ does not possess a normal 2-complement, it follows from the Frobenius Theorem [3, Theorem 7.4.5] that $G$ contains a 2-subgroup $H$ and an element of odd order $b\in N_{G}(H)$ such that $[H,b]\neq 1$. In view of Thompson’s Theorem [3, Theorem 5.3.11] we can assume that $H$ is of nilpotency class at most two and $H/Z(H)$ is elementary abelian. We claim that $G$ contains an element $a\in X$ such that $a$ is a 2-element and $a$ has order 2 modulo $Z(G)$.

Recall that $k\geq 3$. Let $Y$ be the set of all elements $[h,{}_{(k-2)}b]$, where $h\in H$. Since $H$ is nilpotent of class at most $2$, it follows that for every $y\in Y$ all elements in $[H,y]$ are $\gamma_{k}$-commutators. If for some $y\in Y$ the subgroup $[H,y]$ does not lie in $Z(G)$, every element of $[H,y]$ having order $2$ modulo $Z(G)$ enjoys the required properties. Therefore we assume that $[H,y]\leq Z(G)$ for every $y\in Y$. Let $K$ be the subgroup generated by $Y$. Obviously, $K=[K,b]$ and $K^{\prime}\leq Z(G)$. It follows that $C_{K}(b)\leq Z(G)$ and all elements in $K$ are $\gamma_{k}$-commutators modulo $Z(G)$. If $d\in K$ such that $d\not\in Z(G)$ and $d^{2}\in Z(G)$, then the commutator $[d,b]$ is as required.

Now we fix an element $a$ with the above properties. Since $G/Z(G)$ is nonabelian simple, it follows from the Baer-Suzuki Theorem [3, Theorem 3.8.2] that there exists an element $t\in G$ such that the order of $[a,t]$ is odd. On the one hand, it is clear that $a$ inverts $[a,t]$. On the other hand, by Lemma 5, $a$ centralizes $[a,t]$. This is a contradiction. ∎

3. Acknowledgment

The work of the first and the third authors was supported by CNPq-Brazil and FAPDF. The second author was partially supported by the National Group for Algebraic and Geometric Structures, and their Applications (GNSAGA – INdAM). This article was written during the second author’s visit to the University of Brasilia. He wishes to thank the Department of Mathematics for excellent hospitality.