Proof of a Conjecture of Wiegold for nilpotent Lie algebras

TL;DR

This paper confirms an analog of Wiegold's conjecture specifically for finite-dimensional nilpotent Lie algebras, contributing to the understanding of their structural properties.

Contribution

It provides the first proof of Wiegold's conjecture analog in the context of finite-dimensional nilpotent Lie algebras.

Findings

Confirmed the conjecture for nilpotent Lie algebras

Extended understanding of algebraic structure

Established new theoretical results

Abstract

In this short note we confirm an analog of a conjecture of James Wiegold for finite dimensional nilpotent Lie algebras.

Full text

1. Introduction

In this short note we confirm an analog of a conjecture of James Wiegold [1, 4.69] for finite dimensional nilpotent Lie algebras.

Conjecture 1.1** (Wiegold’s conjecture).**

Let $G$ be a finite $p$-group and let $|G^{\prime}|>p^{n(n-1)/2}$ for some non-negative integer $n$. Then the group $G$ can be generated by the elements of breadth at least $n$. The breadth $b(x)$ of an element $x$ of a finite $p$-group $G$ is defined by the equation $|G:C_{G}(x)|=p^{b(x)}$, where $C_{G}(x)$ is the centralizer of $x$ in $G$.

An overview of this problem can be found in [3]. In [2] M.R.Vaughan-Lee proved that for a finite $p$-group $G$ of breadth $b=\max_{g\in G}b(g)$, we have $|G^{\prime}|\leq p^{b(b-1)/2}$. He also showed that for a finite dimensional nilpotent Lie algebra $\mathfrak{g}$ of breadth $b=\max_{g\in\mathfrak{g}}b(g)$, $\dim[\mathfrak{g},\mathfrak{g}]\leq b(b+1)/2$.

Conjecture 1.1 was proved by the Author in [5]. The goal of this paper is to prove the Lie algebra analog of this conjecture.

The breadth $b(x)$ of an element $x$ of a finite dimensional Lie algebra $\mathfrak{g}$ over a field $\mathbb{F}$ is defined by the equation $\dim\mathfrak{g}-\dim C_{\mathfrak{g}}(x)=b(x)$, where $C_{\mathfrak{g}}(x)$ is the Lie centralizer of $x$ in $\mathfrak{g}$. We prove the following

Conjecture 1.2** (Wiegold’s conjecture for nilpotent Lie algebras).**

Let $\mathfrak{g}$ be a nilpotent Lie algebra over a field $\mathbb{F}$ and let $\dim\mathfrak{g}^{\prime}>n(n-1)/2$ for some non-negative integer $n$. Then the Lie algebra $\mathfrak{g}$ can be generated by the elements of breadth at least $n$.

In this article we confirm the Conjecture 1.2 and also show that the more general results are true:

Theorem 1.1**.**

Let $\mathfrak{g}$ be a nilpotent Lie algebra over an infinite field $\mathbb{F}$ and let $\dim\mathfrak{g}^{\prime}>n(n-1)/2$ for a non-negative $n$. Then the set of elements of breadth at least $n$ cannot be covered by a finite number of proper Lie subalgebras in $\mathfrak{g}$.

Theorem 1.2**.**

Let $\mathfrak{g}$ be a nilpotent Lie algebra over a finite field $\mathbb{F}\not=\mathbb{F}_{2}$ and let $\dim\mathfrak{g}^{\prime}>n(n-1)/2$ for some non-negative integer $n$. Then the set of elements of breadth at least $n$ cannot be covered by $|\mathbb{F}|-1$ proper Lie subalgebras in $\mathfrak{g}$.

In the particular case $\mathbb{F}=\mathbb{F}_{2}$ we also get a more general result:

Theorem 1.3**.**

Let $\mathfrak{g}$ be a nilpotent Lie algebra over a field $\mathbb{F}_{2}$ and let $\dim\mathfrak{g}^{\prime}>n(n-1)/2$ for some non-negative integer $n$. Then the set of elements of breadth at least $n$ cannot be covered by two proper Lie subalgebras in $\mathfrak{g}$, one of which has codimension at least 2 in $\mathfrak{g}$.

2. Formulations and proofs of the main Lemmas

The breadth $b_{\mathfrak{h}}(x)$ of an element $x$ of a finite dimensional Lie algebra $\mathfrak{g}$ with respect to a subalgebra $\mathfrak{h}\subseteq\mathfrak{g}$ is defined by the equation $\dim\mathfrak{h}-\dim C_{\mathfrak{h}}(x)=b_{\mathfrak{h}}(x)$, where $C_{\mathfrak{h}}(x)=\{h\in\mathfrak{h}|[x,h]=0\}$ is the centralizer of $x$ in $\mathfrak{h}$. By definition, $b(x)=b_{\mathfrak{g}}(x)$.

The following two lemmas are the well-known facts in theory of Lie algebras, so we state them without proof.

Lemma 2.1**.**

Let $\mathfrak{g}$ be a Lie algebra over a finite field $\mathbb{F}$. Then $\mathfrak{g}$ cannot be covered by $|\mathbb{F}|$ proper subalgebras. Additionally, if $\mathfrak{g}$ is covered by $|\mathbb{F}|+1$ proper subalgebras $\mathfrak{h}_{1}$, $\mathfrak{h}_{2}$, $\ldots$, $\mathfrak{h}_{|\mathbb{F}|+1}$, then each of these $\mathfrak{h}_{i}$ must have codimension $1$ in $\mathfrak{g}$.

Lemma 2.2**.**

Let $\mathfrak{g}$ be a nilpotent Lie algebra over a field $\mathbb{F}$, such that there exists a central subalgebra $\mathfrak{f}$ of codimension $2$ in $\mathfrak{g}$. Then $\dim\mathfrak{g}^{\prime}\leq 1$.

Lemma 2.3**.**

Consider an ideal $\mathfrak{h}$ of a finite dimensional nilpotent Lie algebra $\mathfrak{g}$. Let $\mathfrak{f}$ be the ideal of $\mathfrak{g}$ generated by elements $g$ from $\mathfrak{h}$, such that $b_{\mathfrak{h}}(g)=b(g)$. Then

1. (1)

If $\mathfrak{f}=\mathfrak{h}$, then $\mathfrak{g}^{\prime}=\mathfrak{h}^{\prime}$; 2. (2)

If $\mathfrak{f}$ has a codimension at most $1$ in $\mathfrak{h}$, then $\mathfrak{h}^{\prime}$ has a codimension at most $1$ in $\mathfrak{g}^{\prime}$.

Proof.

Consider the factorization homomorphism $\pi:\mathfrak{g}\to\mathfrak{g}/\mathfrak{h}^{\prime}$. For each element $g$ in $\mathfrak{h}$, such that $b_{\mathfrak{h}}(g)=b(g)$ we have $\{[g,g^{\prime}]|g^{\prime}\in\mathfrak{g}\}=\{[g,h]|h\in\mathfrak{h}\}\subseteq\mathfrak{h}^{\prime}$, so $\pi(g)$ lies in the center of $\pi(\mathfrak{g})$ and $\pi(\mathfrak{f})$ is the central Lie subalgebra of $\pi(\mathfrak{g})$. If $\mathfrak{f}=\mathfrak{h}$, then $\pi(\mathfrak{f})=\pi(\mathfrak{h})$ is the central Lie subalgebra of $\pi(\mathfrak{g})$ of codimension $1$, so $\pi(\mathfrak{g})$ is abelian and $\mathfrak{g}^{\prime}=\mathfrak{h}^{\prime}$. If $\mathfrak{f}$ has a codimension at most $1$ in $\mathfrak{h}$, then $\pi(\mathfrak{f})$ is the central Lie subalgebra of $\pi(\mathfrak{g})$ of codimension at most $2$, so from Lemma 2.2 we get $\dim\mathfrak{g}^{\prime}\leq 1$ and $\mathfrak{h}^{\prime}$ has a codimension at most $1$ in $\mathfrak{g}^{\prime}$.

∎

Lemma 2.4**.**

Let $\mathfrak{g}$ be a finite dimensional Lie algebra and let $\mathfrak{h}$ be its ideal of codimension $1$ in $\mathfrak{g}$. Then for any element $x$ from the set $\mathfrak{g}\setminus\mathfrak{h}$ we have $\dim\mathfrak{g}^{\prime}\leq b(x)+\dim\mathfrak{h}^{\prime}$.

Proof.

The set $V=\{[x,h]|h\in\mathfrak{h}\}$ form a vector space with the dimension not bigger than $b(x)$. So it is enough to prove that $\mathfrak{g}^{\prime}=V+\mathfrak{h}^{\prime}$. This follows from the following properties of the set $V+\mathfrak{h}^{\prime}$ :

1. (1)

$V+\mathfrak{h}^{\prime}$ is a vector space; 2. (2)

$V+\mathfrak{h}^{\prime}$ is an ideal in $\mathfrak{g}$ : $[[x,h],g]+\mathfrak{h}^{\prime}=\overbrace{[x,\underbrace{[h,g]}_{\in\mathfrak{h}}]}^{\in V}+\overbrace{[h,\underbrace{[x,g]}_{\in\mathfrak{h}}]}^{\in\mathfrak{h}^{\prime}}+\mathfrak{h}^{\prime}$; 3. (3)

Lie algebra $\mathfrak{g}/(V+\mathfrak{h}^{\prime})$ is abelian.

These facts imply the Lemma 2.4.

∎

3. Proof of Theorem 1.1

Assume the converse. Let the proper subalgebras $\mathfrak{h}_{1}$, $\mathfrak{h}_{2}$, $\ldots$, $\mathfrak{h}_{k}$ cover all the elements of breadth at least $n$. We will prove that $\dim\mathfrak{g}^{\prime}\leq n(n-1)/2$. Considering a sufficiently large finite-dimensional subalgebra in $\mathfrak{g}$, we can assume that $\mathfrak{g}$ is finite-dimensional. The proof is by induction on $\dim\mathfrak{g}$. We can assume that $\mathfrak{h}_{i}$ are the maiximal ideals of codimension $1$ in $\mathfrak{g}$ (because every proper subalgebra in the nilpotent and finite dimensional Lie algebra is contained in the ideal of codimension $1$). Consider any ideal $I\not=\mathfrak{h}_{i}$ of codimension $1$ in $\mathfrak{g}$. Denote by $J$ the ideal of $\mathfrak{g}$ generated by elements $g$ from $I$ such that $b_{I}(g)=b(g)$. Consider the case when $J=I$. Applying the Lemma 2.3 we conclude that $I^{\prime}=\mathfrak{g}^{\prime}$. The rest follows from the induction hypothesis : the Lie algebra $I$ has smaller dimension and all its elements of breadth at least $n$ are contained in the union of proper subalgebras $I\cap\mathfrak{h}_{i}$ (because $b_{I}(x)\leq b(x)$). Now consider the case $I\not=J$. Notice that the set $I\setminus\left(J\cup_{i}\mathfrak{h}_{i}\right)$ contain only the elements of breadth at most $n-2$ (because $J$ is generated by elements $\{g\in I|b_{I}(g)=b(g)\}$). Thus, from the induction hypothesis applied to $I$ and its $k+1$ proper subalgebras $J$, $I\cap\mathfrak{h}_{1}$, $I\cap\mathfrak{h}_{2}$, $\ldots$, $I\cap\mathfrak{h}_{k}$ we conclude that $\dim I^{\prime}\leq(n-1)(n-2)/2$. Finally, consider any element $a$ not lying in $I\cup_{i}\mathfrak{h}_{i}$, its breadth is less than $n$, so from the Lemma 2.4 we get $\dim\mathfrak{g}^{\prime}\leq b(a)+\dim I^{\prime}\leq\frac{n(n-1)}{2}$.

$\Box$

4. Proof of Theorem 1.2

Assume the converse. Let the proper subalgebras $\mathfrak{h}_{1}$, $\mathfrak{h}_{2}$, $\ldots$, $\mathfrak{h}_{|\mathbb{F}|-1}$ cover all the elements of breadth at least $n$. We will prove that $\dim\mathfrak{g}^{\prime}\leq n(n-1)/2$. Considering a sufficiently large finite-dimensional subalgebra in $\mathfrak{g}$, we can assume that $\mathfrak{g}$ is finite-dimensional. The proof is by induction on $\dim\mathfrak{g}$. We can assume that $\mathfrak{h}_{i}$ are the maiximal ideals of codimension $1$ in $\mathfrak{g}$ (because every proper subalgebra in the nilpotent and finite dimensional Lie algebra is contained in the ideal of codimension $1$) and $\mathfrak{h}_{1}\not=\mathfrak{h}_{2}$ (because in every non-abelian finite dimensional and nilpotent Lie algebra $\mathfrak{g}$, there are at least two different maximal proper subalgebras). Consider any ideal $I\not=\mathfrak{h}_{i}$ of codimension $1$ in $\mathfrak{g}$ such that $I\cap\mathfrak{h}_{1}=I\cap\mathfrak{h}_{2}=\mathfrak{h}_{1}\cap\mathfrak{h}_{2}$. Denote by $J$ the ideal of $\mathfrak{g}$ generated by elements $g$ from $I$ such that $b_{I}(g)=b(g)$. Consider the case when $J=I$. Applying the Lemma 2.3 we conclude that $I^{\prime}=\mathfrak{g}^{\prime}$. The rest follows from the induction hypothesis : the Lie algebra $I$ has smaller dimension and all its elements of breadth at least $n$ are contained in the union of proper subalgebras $I\cap\mathfrak{h}_{i}$ (because $b_{I}(x)\leq b(x)$). Now consider the case $I\not=J$. Notice that the set $I\setminus\left(J\cup_{i}\mathfrak{h}_{i}\right)$ contain only the elements of breadth at most $n-2$ (because $J$ is generated by elements $\{g\in I|b_{I}(g)=b(g)\}$). Thus, from the induction hypothesis applied to $I$ and its $|\mathbb{F}|-1$ proper subalgebras $J$, $I\cap\mathfrak{h}_{1}=I\cap\mathfrak{h}_{2}$, $I\cap\mathfrak{h}_{3}$ $\ldots$, $I\cap\mathfrak{h}_{|\mathbb{F}|-1}$ we conclude that $\dim I^{\prime}\leq(n-1)(n-2)/2$. Finally, consider any element $a$ not lying in $I\cup_{i}\mathfrak{h}_{i}$ (such element exists because of the Lemma 2.1), its breadth is less than $n$, so from the Lemma 2.4 we get $\dim\mathfrak{g}^{\prime}\leq b(a)+\dim I^{\prime}\leq\frac{n(n-1)}{2}$. $\Box$

5. Proof of Theorem 1.3

Theorem 5.1**.**

Let $\mathfrak{g}$ be a nilpotent Lie algebra over a finite field $\mathbb{F}$. Let the next two conditions hold for some integers $n\leq k+1$

1. (1)

The set of all elements of the breadth at least $n$ can be covered by $|\mathbb{F}|$ subalgebras of $\mathfrak{g}$; 2. (2)

The set of elements of the breadth at most $k$ generates $\mathfrak{g}$.

Then $\dim\mathfrak{g}^{\prime}\leq\frac{(n-1)(n-2)}{2}+k$.

Proof.

Considering a sufficiently large finite-dimensional subalgebra in $\mathfrak{g}$, we can assume that $\mathfrak{g}$ is finite-dimensional. The proof is by induction on $\dim\mathfrak{g}$. Let the proper subalgebras $\mathfrak{h}_{1}$, $\mathfrak{h}_{2}$, $\ldots$, $\mathfrak{h}_{|\mathbb{F}|}$ cover all the elements of breadth at least $n$. We can assume that $\mathfrak{h}_{i}$ are the maiximal ideals of codimension $1$ in $\mathfrak{g}$ (because every proper subalgebra in the finite dimensional and nilpotent Lie algebra is contained in the ideal of codimension $1$) and $\mathfrak{h}_{1}\not=\mathfrak{h}_{2}$ (because in every non-abelian finite dimensional and nilpotent Lie algebra $\mathfrak{g}$, there are at least two different maximal proper subalgebras). Consider any ideal $I\not=\mathfrak{h}_{i}$ of codimension $1$ in $\mathfrak{g}$ such that $I\cap\mathfrak{h}_{1}=I\cap\mathfrak{h}_{2}=\mathfrak{h}_{1}\cap\mathfrak{h}_{2}$. Denote by $J$ the ideal of $\mathfrak{g}$ generated by elements $g$ from $I$, such that $b_{I}(g)=b(g)$. Consider the case when $J=I$. Applying the Lemma 2.3 we conclude that $I^{\prime}=\mathfrak{g}^{\prime}$. The rest follows from the induction hypothesis : the Lie algebra $I$ has smaller dimension and all its elements of breadth at least $n$ are contained in the union of proper subalgebras $I\cap\mathfrak{h}_{i}$ (because $b_{I}(x)\leq b(x)$), also the set $I\setminus\cup_{i\geq 2}\mathfrak{h}_{i}$ generates $I$ (because of the Lemma 2.1) and consists only of elements of breadth at most $n-1\leq k$.

So we can assume that $J$ is a proper ideal of $I$. Consider the case when the codimension of $J$ in $I$ is $1$. Apply the Lemma 2.3 to $I$, we conclude that $\dim\mathfrak{g}^{\prime}\leq\dim I^{\prime}+1$. Apply the induction hypothesis to the Lie algebra $I$ and to its proper subalgebras $J$, $\mathfrak{h}_{1}\cap\mathfrak{h}_{2}$, $\mathfrak{h}_{3}$, $\ldots$, $\mathfrak{h}_{|\mathbb{F}|}$ so we get $\dim I^{\prime}\leq\frac{(n-3)(n-2)}{2}+n-1$ (every element from the set $I\setminus\left((\mathfrak{h}_{1}\cap\mathfrak{h}_{2})\cup\mathfrak{h}_{3}\cup\ldots\cup\mathfrak{h}_{|\mathbb{F}|}\right)$ is of the breadth at most $n-1$ in $I$ and this set generates $I$ (because of the Lemma 2.1), also for any element $g$ from the set $I\setminus\left(J\cup(\mathfrak{h}_{1}\cap\mathfrak{h}_{2})\cup\mathfrak{h}_{3}\cup\ldots\cup\mathfrak{h}_{|\mathbb{F}|}\right)$, we have $b_{I}(g)\leq n-2$). So if $k\geq 2$ we get $\dim\mathfrak{g}^{\prime}\leq\dim I^{\prime}+1\leq\frac{(n-3)(n-2)}{2}+n=\frac{(n-2)(n-1)}{2}+2\leq\frac{(n-2)(n-1)}{2}+k$ and the induction step is clear. In the case $k\leq 1$, we get $n=2$ (the case $n\leq 1$ is trivial) and the set $I\setminus\left(J\cup(\mathfrak{h}_{1}\cap\mathfrak{h}_{2})\cup\mathfrak{h}_{3}\cup\ldots\cup\mathfrak{h}_{|\mathbb{F}|}\right)$ is contained in the center of $I$. It is clear that the set $I\setminus\left(J\cup(\mathfrak{h}_{1}\cap\mathfrak{h}_{2})\cup\mathfrak{h}_{3}\cup\ldots\cup\mathfrak{h}_{|\mathbb{F}|}\right)$ generates the central subalgebra of the codimension at most $1$ in $I$ (Lemma 2.1), so $I$ is abelian. From the conditions of Theorem 5.1 there exists an element $g\notin I$ such that $b(g)\leq k\leq 1$. So the Lie algebra $C_{\mathfrak{g}}(g)$ has codimension at most $1$ in $\mathfrak{g}$. Also the subalgebra $C_{\mathfrak{g}}(g)\cap I$ has codimension at most $2$ in $\mathfrak{g}$ and is central in $\mathfrak{g}$. Apply Lemma 2.2 to the Lie algebra $\mathfrak{g}$, so we conclude that $\dim\mathfrak{g}^{\prime}\leq 1\leq\frac{(n-1)(n-2)}{2}+k$.

Eventually, we can assume that the codimension of $J$ in $I$ is greater or equal to $2$. So the set $I\setminus\left(J\cup(\mathfrak{h}_{1}\cap\mathfrak{h}_{2})\cup\mathfrak{h}_{3}\cup\ldots\cup\mathfrak{h}_{|\mathbb{F}|}\right)$ generates $I$ (Lemma 2.1), thus, $I$ is generated by the elements of the breadth at most $n-2$ in $I$. And we can apply the induction hypothesis to $I$ and its subalgebras $J$, $\mathfrak{h}_{1}\cap\mathfrak{h}_{2}$, $\mathfrak{h}_{3}$, $\ldots$, $\mathfrak{h}_{|\mathbb{F}|}$ and conclude that $\dim I^{\prime}\leq\frac{(n-3)(n-2)}{2}+n-2=\frac{(n-2)(n-1)}{2}$. Also from the conditions of the Theorem 5.1 there exists an element $a\notin I$ such that $b(a)\leq k$. Applying the Lemma 2.4 to $I$ and $a$, we conclude that $\dim\mathfrak{g}^{\prime}\leq b(a)+\dim I^{\prime}\leq\frac{(n-2)(n-1)}{2}+k$.

∎

5.1. Proof of Theorem 1.3

Assume the converse. Let the proper subalgebras $\mathfrak{h}_{1}$ and $\mathfrak{h}_{2}$ cover all the elements of breadth at least $n$ and $\mathfrak{h}_{2}$ has codimension bigger or equal to $2$ in $\mathfrak{g}$. We will prove that $\dim\mathfrak{g}^{\prime}\leq n(n-1)/2$. From Lemma 2.1 the set $\mathfrak{g}\setminus(\mathfrak{h}_{1}\cup\mathfrak{h}_{2})$ generates $\mathfrak{g}$. All the elements from $\mathfrak{g}\setminus(\mathfrak{h}_{1}\cup\mathfrak{h}_{2})$ have breadth at most $n-1$ in $\mathfrak{g}$, thus, we can apply Theorem 5.1 to $\mathfrak{g}$, $\mathfrak{h}_{1}$, $\mathfrak{h}_{2}$, $k=n-1$ and get $\dim\mathfrak{g}^{\prime}\leq\frac{(n-1)(n-2)}{2}+n-1=\frac{n(n-1)}{2}$. $\Box$

Remark 5.1**.**

In fact, Theorem 1.2 is also a consequence of Theorem 5.1.

6. Acknowledgements

I am grateful to Anton A. Klyachko for stating the problem.