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\markboth{T.~H.~Dung and N.~T.~T.~Ha}{Covering numbers with involutions}
\title[Covering numbers with involutions]{Covering numbers with involutions   in decomposing infinite matrices}

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%\author[A. Author]{Author Author\corrauth}
%\address{School of Applied Mathematics and Computer Science, University of Osijek, Trg Ljudevita Gaja 6, HR-31 000 Osijek, Croatia}
%\email{{\tt author@mathos.hr} (A. Author)}


% multiple authors:
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\author[T.~H.~Dung,N.~T.~T.~Ha]{Truong Huu Dung\affil{1} and Nguyen Thi Thai  Ha\affil{2,3,4}\comma\corrauth}
\address{\affilnum{1}\  Department of Mathematics, Dong Nai University,
%9 Le Quy Don Str., Tan Hiep Ward,
810\,000 Dong Nai Province, %Dong Nai Province,
  Vietnam\\
\affilnum{2}\ Faculty of Mathematics and Computer Science, University of Science, 700\,000 Ho Chi Minh City, Vietnam \\
\affilnum{3}\ Vietnam National University,
			700\,000 Ho Chi Minh City, Vietnam \\
\affilnum{4}\ Campus in Ho Chi Minh City, University of Transport and Communications,  700\,000 Ho Chi Minh City, Vietnam}

\emails{{\tt thdung@dnpu.edu.vn or dungth0406@gmail.com} (T.~H.~Dung), {\tt hantt\_ph@utc.edu.vn} (N.~T.~T.~Ha)}



%%%%% Begin Abstract %%%%%%%%%%%
\begin{abstract}
	Let $D$ be a division ring. The  aim of this paper is to explore the problem of decomposing an infinite matrix over $D$ into a product of involutions and a product of commutators of involutions  within the context of covering numbers.  Specifically, we focus on decomposing  matrices in the commutator subgroup $\mathrm{SL}_{VK,\infty}(D)$ of the Vershik–Kerov group and in the subgroup $\mathrm{SL}_{\infty}(D)$ of the stable general linear group $\mathrm{GL}_{\infty}(D)$.
\end{abstract}
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%%%%% Keywords %%%%%%%%%%%
\keywords{division ring, matrix decomposition, commutator, involution, infinite matrix }

%%%% AMS subject classifications %%%%
\ams{12E15, 15A23, 15B99}

%%%% maketitle %%%%%
\maketitle

%%%% Start %%%%%%
\section{Introduction}
	Let $G$ be a group with the identity element $1$ and let $X$ be a set of generators of $G$ such that $x^{-1}\in X$ for every $x\in X$. Then, for every $g \in G$, there exist $x_1,\ldots, x_k \in X$ such that $g=x_{1}x_{2}\cdots x_{k}$, where k is a positive integer.	 We denote $X^k =\{x_{1}x_{2}\cdots x_{k} \mid x_i\in X, i=1,\ldots,k \}$. The {\it  covering number of $G$ by $X$}, denoted by $\mathrm{cn}_X(G)$, is defined to be the smallest integer $k$  such that $X^k =G$ or $\infty$ if no such $k$ exists. For instance, if  $\mathcal{C}$ is the set of all commutators $aba^{-1}b^{-1}$, where $a, b$ range over $G$ and $G'=[G,G]$ is the commutator subgroup of $G$, then the covering number $\mathrm{cn}_{\mathcal{C}}(G')$    is called the {\it commutator width}  of $G$.		


An element $x$ in the group $G$ is called an \textit{involution} if $x^2 = 1$. If $a$ and $b$ are involutions in $G$, then $[a,b] = aba^{-1}b^{-1}$ is called the \textit{commutator of involutions}. We denote $\mathcal{I}$ and $\mathcal{CI}$ as the sets of involutions and the commutators of involutions in the group $G$, respectively. Clearly,  $\mathcal{I}$ and $\mathcal{CI}$ are closed under taking the inverse. In this paper, we will evaluate the covering numbers of the subgroup $\mathrm{SL}_{\infty}(D)$  of the stable general linear group $\mathrm{GL}_{\infty}(D)$ and the commutator subgroup $\mathrm{SL}_{VK,\infty}(D)$ of the Vershik–Kerov group by these sets. To evaluate such covering numbers, we will decompose infinite matrices within these groups into  a product of involutions and a product of commutators of involutions.

The decomposition of elements in a group, especially linear groups, into products of involutions has received significant attention from the mathematical community, e.g. see \cite{Pa_BiDuHa, Pa_BiDuHaSo_2023 , Pa_BiDuHaSo_2022, Gu_Ha_76, Pa_Gvo_20, Pa_SoDuHaBi_2021, Pa_Zh_02}.		
Assume that $R$ is a unitary associative ring. The notation $\mathrm{GL}_n(R)$ denotes the group of invertible $n\times n$ matrices over $R$, and $\mathrm{SL}_n(R)$ is the commutator subgroup of $\mathrm{GL}_n(R)$.
A matrix $A \in \mathrm{GL}_n(R)$ is called an \textit{involution} if $A^2 = \mathbf{1}_{n}$, where $\mathbf{1}_{n}$ is the identity matrix.
Over an arbitrary field, every matrix with determinant $\pm 1$ can be expressed as a product of at most four involutions \cite[Theorem]{Gu_Ha_76}. Note that if $A$ is an involution, then both $A^{-1}$ and $B^{-1}AB$ are also involutions. Therefore, a commutator of involutions is essentially the product of two involutions. In connection with this topic, X. Hou in \cite{Pa_Ho_20} and T. N. Son et al. in \cite{Pa_SoDuHaBi_2021} proved that a matrix over a field is a product of at most two commutators of involutions. These results have been extended to division rings in \cite[Theorem 4.5 and Theorem 6.3]{Pa_BiDuHaSo_2023 }.

Let $D$ be a division ring. We define the notation  $\mathrm{GL}_{c,\infty}(D)$ to denote the group consisting of all countable-dimensional column-finite invertible matrices. Moreover, we introduce $\mathbf 1_\infty \in \mathrm{GL}_{c,\infty}(D)$   to denote  the diagonal matrix with $1$ entries along its diagonal. 	
If we consider a matrix $A\in \mathrm{GL}_n(D)$ as the  matrix $\begin{pmatrix}A& 0\\ 0& \mathbf 1_\infty\end{pmatrix}\in \mathrm{GL}_{c,\infty}(D)$, then $\mathrm{GL}_n(D)$ becomes a subgroup of $\mathrm{GL}_{c,\infty}(D)$.
A   matrix in \(\mathrm{GL}_{c,\infty}(D)\) is considered  \textit{unitriangular} if  it is upper triangular and has diagonal entries equal to $1$. The subgroup  $\mathrm{T}_{\infty}(D)$ of \(\mathrm{GL}_{c,\infty}(D)\) consists of all upper triangular matrices, while $ \mathbb{UT}_{\infty}(D)$ denotes  the subgroup  of upper unitriangular matrices.
Recall that  the \textit{Vershik–Kerov group}, denoted as $\mathrm{GL}_{VK,\infty}(D),$   is a subgroup of $\mathrm{GL}_{c,\infty}(D)$. This subgroup consists of matrices in the form $A=\begin{pmatrix}A_1& A_2\\ 0& A_3\end{pmatrix}$, where $A_1\in \mathrm{GL}_n(D), A_3\in \mathrm{T}_{\infty}(D)$ and $A_2$ has the size $n\times \mathbb{N}$. We denote $\mathrm{SL}_{\infty, n}(D)$ as a  subset of $\mathrm{GL}_{VK,\infty}(D)$, where $A_1\in \mathrm{SL}_n(D)$ and $A_3\in \mathrm{T}_{\infty}(D)$ with the main diagonal entries represented by elements $s_i \in D'$.   Let $\mathrm{SL}_{VK,\infty}(D)=\bigcup \limits_{n\ge 1}\mathrm{SL}_{\infty, n}(D).$ According to
\cite[Corollary 1.3]{Pa_BiDuHa},   if $D$ is a   centrally finite division ring with more than three elements, then the commutator subgroup of $\mathrm{GL}_{VK,\infty}(D)$ is equal to $ \mathrm{SL}_{VK,\infty}(D)$.

In Section~\ref{vk}, we prove that every matrix in $\mathrm{SL}_{VK, \infty} (D)$ can be expressed as a product of at most $8s+4$ involutions in $\mathrm{GL}_{VK, \infty} (D)$ provided that $\mathrm{cn}_{\mathcal{C}}(D')=s$, where $D$ is a centrally finite division ring with more than three elements. We also prove that $\mathrm{cn}_{\mathcal{CI}}(\mathrm{SL}_{VK, \infty}(D)) \le 9s+2$ if $D$ is a noncommutative centrally finite division ring of characteristic different from $2$ and $\mathrm{cn}_{\mathcal{C}}(D')=s$.

Recall that if $D$ is a division ring, then the direct limit $\mathrm{GL}_{\infty}(D) = \underrightarrow{\lim}\ \mathrm{GL}_{n}(D)$ with respect to the transition homomorphisms $\mathrm{GL}_{n}(D) \longrightarrow \mathrm{GL}_{n+1}(D)$ given by $A \mapsto \begin{pmatrix} A & 0 \\ 0 & 1 \end{pmatrix}$ is called the \textit{stable general linear group} over $D$.	
%The purpose of this section is to express matrices in
The subgroup $\mathrm{SL}_{\infty}(D)$ of $\mathrm{GL}_{\infty}(D)$ is defined as $\mathrm{SL}_{\infty}(D) = \underrightarrow{\lim}\ \mathrm{SL}_{n}(D)$.

In Section~\ref{sl}, we prove that $\mathrm{cn}_{\mathcal{I}}(\mathrm{SL}_{\infty} (D)) \le 4$ and $\mathrm{cn}_{\mathcal{CI}}(\mathrm{SL}_{\infty}(D)) \le 5$ when $D$ is a noncommutative centrally finite division ring such that $\mathrm{cn}_{\mathcal{C}}(D') < \infty$.











We present some remarks frequently utilized in this paper. The proofs of these claims are simple and for convenience we provide them here.

\begin{remark}\label{chung}
	Assume that $D$ is a division ring. Then,
	\begin{enumerate}[\rm (i)]
		\item The matrix $\begin{pmatrix}a&0\\0&a^{-1}\end{pmatrix}$ can be decomposed into a product of at most two involutions, where $a \in D^*$.
		\item If $s$ is a commutator, then $\mathrm{diag}(1,s)$ can be decomposed into a product of at most four involutions.
		\item If $B_i$ is an involution, then $\oplus_{i\in \mathbb{N}} B_i$ is also an involution for every $i \in \mathbb{N}$.
		\item If $A \in \mathrm{GL}_{n}(D)$ and $B \in \mathrm{GL}_{m}(D)$ are products of $k$ and $\ell$ involution matrices, respectively, then $A\oplus B$ is the product of at most $\max\{k,\ell\}$ involution matrices in $\mathrm{GL}_{n+m}(D)$.
	\end{enumerate}
\end{remark}
\begin{proof}
	
	(i) It is demonstrated in \cite[Lemma 5.2]{Pa_BiDuHaSo_2023 }.
	
	(ii) It follows from \cite[Lemma 4.1]{Pa_BiDuHaSo_2023 }.
	
	(iii) We have $B_i ^2 = \mathbf{1}_n$ for every $i\in \mathbb{N}$, leading to $\left(\oplus_{\lambda\in\mathbb{N}}B_i \right) ^2 = \oplus_{\lambda\in\mathbb{N}}B_i^2 = \mathbf{1}_{\infty}$, which is also an involution.
	
	(iv)	We can assume that $ k \leq \ell$. For every $i=1,\ldots,k; j=1,\ldots,\ell$ assume $A_i, B_j $ are  involution matrices such that $A = A_1 \ldots A_k$ and $B = B_1 \ldots B_{\ell}$. Then, $A\oplus B = C_1\oplus \ldots\oplus C_{\ell}$ is a product of $\ell$ involutions, in which  $C _i = A_i\oplus B_i$ for $i = 1, \ldots,k$ and  $C_i = \mathbf 1_n\oplus B_i$ for $i = k+1, \ldots, \ell.$  	
\end{proof}
\begin{remark}\label{baiso2}
	Assume that $D$ is a noncommutative division ring.
	\begin{enumerate}[\rm (i)]
		\item Suppose  $D$ is a centrally finite division ring such that $\mathrm{cn}_{\mathcal{C}}(D')<\infty$. In this case, $\mathrm{diag}(1,\dots,1,s)\in \mathrm{GL}_n(D)$ is a product of at most $3\mathrm{cn}_{\mathcal{C}}(D')$ commutators of involutions  if $n\geq 3$ or $\mathrm{char}D \neq 2$. Particularly, if $s$ is a commutator, then the matrix $\mathrm{diag}(1,\ldots,1,s)$ is a product of at most three commutators of involutions.
		\item If $A_i$ is a commutator of involutions for each $i \in \mathbb{N}$, then $\oplus_{i \in \mathbb{N}}A_i$ is also a commutator of involutions.
		\item If $A \in \mathrm{GL}_{n}(D)$ and $B\in \mathrm{GL}_{m}(D)$ are each expressed as products  of $k$ and $\ell$ commutators of involutions, respectively, then $A \oplus B$ can be decomposed into a product of at most $\max\{k,\ell\}$ commutators of involutions in $\mathrm{GL}_{n+m}(D)$.
	\end{enumerate}
\end{remark}
\begin{proof}
	The first statement is established in \cite[Lemma 4.3 and Lemma 4.4]{Pa_BiDuHaSo_2022}. The last two statements can be proven similarly to (iii) and (iv) of Remark \ref{chung}.
\end{proof}

In this paper, we define a centrally finite division ring as one that has finite dimensionality over its center. We use the following notations: Let $D$ be a division ring, and denote $D' = [D^*, D^*]$, where $D^* =D\setminus\{0\}$.
We represent the diagonal matrix with elements $a_1,\ldots,a_n \in D$ on the main diagonal as  $\mathrm{diag}(a_1,\ldots,a_n).$


	\section{Decompositions of matrices in $\mathrm{SL}_{VK,\infty}(D)$}\label{SL_VK} \label{vk}

In this section, we decompose matrices in  the  subgroup $\mathrm{SL}_{VK,\infty}(D)$ of the Vershik–Kerov group, where $D$ is a centrally finite division ring containing    more than three elements, and evaluate the covering numbers of the subgroup $\mathrm{SL}_{VK,\infty}(D)$ by the set of commutators of involutions $\mathcal{CI}$.





Assume that $(N_{\lambda})_{\lambda\in\Lambda}$, where $\Lambda \subseteq \mathbb{N}$ is a partition of $\mathbb{N}$. Then, a \textit{finite or infinite Jordan block} is denoted as
\[
J_{\mid N_{\lambda}\mid}(1,1) =
\begin{pmatrix}
1 & 1 \\
& 1 & 1 \\
&& \ddots & \ddots & \\
&&& 1 & 1 \\
&&&& 1
\end{pmatrix}.
\]


\begin{lemma}\label{truong}
	Assume that $D$ is a division ring, and $A \in \mathbb{UT}_{\infty}(D)$. Then, $A$ is similar to an infinite Jordan block $\oplus_{\lambda\in\Lambda} J_{\mid N_{\lambda} \mid}(1,1)$, where $(N_{\lambda})_{\lambda \in \Lambda}$ is a partition of $\mathbb{N}$, and $\Lambda \subseteq \mathbb{N}$ is a subset of the natural numbers.
\end{lemma}
\begin{proof}
	The lemma is established in \cite[Corollary 3.4]{Pa_BiSoThuyTru_2023}.
\end{proof}



Suppose that $R$ is a unitary associative ring. According to \cite[Theorem 1.1]{Hou2019}, if $2$ is invertible in $R$, then every matrix in the groups $\mathbb{UT}_{n}(R)$ and $\mathbb{UT}_\infty(R)$ can be written as a product of at most two commutators of involutions in $\mathrm{T}_\infty(R)$.
Recently,  we have shown that if $R$ is a division ring, then every matrix in the group $\mathbb{UT}_{n}(R)$ can be expressed as a product of two involutions, which is a special case of \cite[Lemma 4.3]{Pa_BiDuHaSo_2023 }.

In the following lemma, we continue considering  the group $\mathbb{UT}_\infty(R)$,   where $R$ is a division ring. Our goal is to reduce  the number of involutions in the decomposition to $2$, and the number of commutators of involutions to $1$.
\begin{lemma}\label{cuoi}
	Assume that $D$ is a division ring and  $A \in \mathbb{UT}_{\infty}(D) $. Then,
	
	\begin{enumerate}[\rm (i)]
		\item Every matrix in $\mathbb{UT}_{\infty}(D)$ can be expressed as a product of at most two involutions.
			\item Every matrix in $\mathbb{UT}_{\infty}(D)$ can be written as a commutator of involutions,			provided that $\mathrm{char}D \neq 2$.
	\end{enumerate}
\end{lemma}

\begin{proof} 	
	Assume that $A \in \mathbb{UT}_{\infty}(D)$. According to Lemma \ref{truong}, the matrix $A$ is similar to $\oplus_{\lambda\in\Lambda}J_{\mid N_{\lambda} \mid}(1,1)$.
	
	(i) According to \cite[Theorem 2.3]{Pa_Ho_Li_Zhe_17}, the matrix $J_{\mid N_{\lambda} \mid}(1,1)$ is a product of two involutions in $\mathbb{UT}_{\infty}(D)$. Therefore, $A$ can be decomposed into a product of two involutions
	according to Remark \ref{chung}.
	
	(ii) According to \cite[Lemma 7]{Pa_Zh_02} and \cite[Corollary 2.7]{Hou2019}, the matrix $J_{\mid N_{\lambda} \mid}(1,1)$ is a commutator of involutions if $\mathrm{char}D \neq 2$, so $A$ is also a commutator of involutions.
\end{proof}	


It is known that  if $D$ is a field with characteristic different from $2$, then every matrix in $\mathrm{SL}_{VK,\infty}(D)$ can be expressed as a product of at most two commutators of involutions according to \cite[Theorem 1.3]{Hou2019}. Since matrices similar to involutions are also involutions, every matrix in $\mathrm{SL}_{VK,\infty}(D)$ can be expressed as a product of at most four involutions.  The results presented below address this problem for division rings.



\begin{theorem}\label{lemma5.6}
	Let $D$ be  a centrally finite division ring  with more than three elements. If $\mathrm{cn}_{\mathcal{C}}(D') = s$, then every element in $\mathrm{SL}_{VK,\infty}(D)$ can be decomposed into a product of at most $8s+4$ involutions in $\mathrm{GL}_{VK,\infty}(D)$.
\end{theorem}

\begin{proof}
	Let $A \in \mathrm{SL}_{VK,\infty}(D)$ and $F=Z(D)$. If $A$ is central in $\mathrm{SL}_{VK,\infty}(D)$, then according to \cite[Lemma 2.6]{Pa_BiDuHa}, $A=\lambda \mathbf 1_{\infty}$ for $\lambda\in F\cap D'$. Furthermore,
	\[	A = \mathrm{diag}(\lambda,1,\lambda,1,\dots)\mathrm{diag}(1,\lambda,1,\lambda,\dots).\]
	Since $\lambda \in D'$, there exist commutators $\lambda_1, \dots, \lambda_s$ such that $\lambda = \lambda_1 \ldots \lambda_s$. Then,
	\[\mathrm{diag}(\lambda,1) = \mathrm{diag}(\lambda_1,1)\mathrm{diag}(\lambda_2,1)\ldots\mathrm{diag}(\lambda_s,1),\]
	where each $\lambda_i$ is a commutator for $i=1,\dots,s$. By Remark \ref{chung}, $\mathrm{diag}(\lambda,1)$ can be expressed as a product of at most $4s$ involutions, and the same holds for \[\mathrm{diag}(\lambda,1,\lambda,1\dots).\] Therefore, $A$ is a product of at most $8s$ involutions.
	
	Now, assume $A$ is noncentral in $\mathrm{SL}_{VK,\infty}(D)$. In this case, $A=\begin{pmatrix}A_1&A_2\\0&A_3\end{pmatrix}$, where $A_1$ is a matrix in $\mathrm{SL}_n(D)$, $A_2$ is an $n \times \mathbb{N}$ matrix, and $A_3\in\mathrm{T}(\infty,D)$ with diagonal entries $\alpha_i \in D'$. We can express $A = MN$, where
	\[ M = \begin{pmatrix}\mathbf 1_n&A_2A_3^{-1}\\0&\mathbf 1_\infty\end{pmatrix} \text{ and } N = \begin{pmatrix}A_1&0\\0&A_3\end{pmatrix}.\]  We have $A_3=U D$, where $D = \mathrm{diag}(\alpha_1,\alpha_2,\dots)$ and $U \in \mathbb{UT}_\infty(D)$. According to Lemma~\ref{cuoi}, the matrix $U$ can be decomposed into a product of at most two involutions. Moreover,
	\[D =\mathrm{diag}(\alpha_1,1,\alpha_3,1,\dots) \mathrm{diag}(1,\alpha_2,1,\alpha_4,\dots).\]
	Since $\alpha_i \in D'$, there exist commutators $a_1^i,\dots, a_s^i$ such that $\alpha_i = a_1^i \ldots a_s^i$. Thus,
	\[\mathrm{diag}(1,\alpha_i) =\mathrm{diag}(1, a_1^i)\ldots\mathrm{diag}(1,a_s^i).\] For each $k=1,\dots,s$, the matrix  $\mathrm{diag}(1, a_k^i)$ is a product of at most four involutions according to Remark \ref{chung}. Therefore, $\mathrm{diag}(1,\alpha_i)$ is a product of $4s$ involutions. Similarly, according to Remark \ref{chung}, the matrix $\mathrm{diag}(\alpha_1,1,\alpha_3,1,\ldots)$ is a product of at most $4s$ involutions. Hence, the matrix $D$ is a product of at most $8s$ involutions. It follows that $A_3$ can be decomposed into a product  of at most $8s+2$ involutions.	
	Since $A_1 \in \mathrm{SL}_n(D)$, by \cite[Theorem 4.5]{Pa_BiDuHaSo_2023 }, the matrix $A_1$ is a product of at most $4s+4$ involutions. Thus, $N$ is a product of at most $8s+2$ involutions. Therefore, $A$ can be decomposed into   a product of at most $8s+4$ involutions, as $M$ can be expressed as a product of at most two involutions according to Lemma~\ref{cuoi}.	
\end{proof}

Below is an alternative version of \cite[Theorem 1.3]{Hou2019} for a noncommutative division ring of characteristic different from $2$.


\begin{theorem}\label{lemma5.4}
	Let $D$ be a noncommutative  centrally finite division ring  of characteristic different from $2$ and  $\mathrm{cn}_{\mathcal{C}}(D') = s$. Then, $\mathrm{cn}_{\mathcal{CI}}(\mathrm{SL}_{VK,\infty}(D)) \leq 9s+2.$
\end{theorem}
\begin{proof}
	Assume $A \in \mathrm{SL}_{VK,\infty}(D)$ and $F=Z(D)$. If $A$ is central in $\mathrm{SL}_{VK,\infty}(D)$, then $A=\lambda\mathbf{1}_{\infty}$ for $\lambda\in F\cap D'$. We have
	{\fontsize{11pt}{11pt}	
	\[
		A = \mathrm{diag}(\lambda,1,1,\lambda,1,1,\dots)\mathrm{diag}(1,\lambda,1,1,\lambda,1,\dots) \mathrm{diag}(1,1,\lambda,1,1,\lambda,\dots).
		\]}
	According to Remark \ref{baiso2}, we observe that $\mathrm{diag}(\lambda,1,1)$ can be decomposed into a product of at most $3s$ commutators of involutions, and the same holds for \[\mathrm{diag}(\lambda,1,1,\lambda,1,1,\dots).\] Similarly, $\mathrm{diag}(1,\lambda,1,1,\lambda,1,\dots)$ and $\mathrm{diag}(1,1,\lambda,1,1,\lambda,\dots)$ are also a product of $3s$ commutators of involutions. Therefore, $A$ is a product of at most $9s$ commutators of involutions.
	
	Now consider $A$ to be  noncentral in $\mathrm{SL}_{VK,\infty}(D)$. Similarly, we have $A=\begin{pmatrix}
	A_1&A_2\\0&A_3
	\end{pmatrix}$, where $A_1$ is a matrix in $\mathrm{SL}_n(D)$, and $A_3\in\mathrm{T}_{\infty}(D)$ with entries on the main diagonal being $s_i \in D'$. Again, we have $A = MN$ with
	\[
	M = \begin{pmatrix}
	\mathbf 1_n&A_2A_3^{-1}\\0&\mathbf 1_\infty
	\end{pmatrix};\quad N = \begin{pmatrix}
	A_1&0\\0&A_3
	\end{pmatrix}.
	\]
	Since $M \in \mathbb{UT}_{\infty}(D)$, according to Lemma~\ref{cuoi}, $M$ can be written as a commutator of involutions. Consider the matrix $A_3=U D$ with $D = \mathrm{diag}(s_1,s_2,\ldots)$ and $U \in \mathbb{UT}_\infty(D)$. Again, according to Lemma~\ref{cuoi}, the matrix $U$ is a commutator of involutions. Moreover,
	{
		\[
		D =\mathrm{diag}(s_1,1,1,s_4,1,1,\ldots) \mathrm{diag}(1,s_2,1,1,s_5,1,\ldots)\mathrm{diag}(1,1,s_3,1,1,s_6,\ldots).
		\]}
	Using an argument similar to the above, $D$ can be decomposed into a product of at most $9s$ commutators of involutions. Therefore, $A_3$ can be represented as a product of at most $9s+1$ commutators of involutions. Furthermore, according to \cite[Theorem 6.3]{Pa_BiDuHaSo_2023 }, the matrix $A_1$ can be expressed  as a product of at most $3s+2$ commutators of involutions. Thus, $N$ can be decomposed into a product of at most $9s+1$ commutators of involutions. Therefore, $A$ is a product of at most $9s+2$ commutators of involutions. Furthermore, according to  \cite[Corollary 1.3]{Pa_BiDuHa} a product of commutators of involutions belongs to  $\mathrm{SL}_{VK, \infty}(D).$ Therefore, $\mathrm{cn}_{\mathcal{CI}}(\mathrm{SL}_{VK,\infty}(D)) \leq 9s+2.$
\end{proof}

\section{Decompositions of matrices in $\mathrm{SL}_{\infty}(D)$} \label{sl}




In this section, we will evaluate the covering numbers of the subgroup $\mathrm{SL}_{\infty}(D)$  of the stable general linear group $\mathrm{GL}_{\infty}(D)$ by the set of involutions $\mathcal{I}$ and the set of commutators of involutions $\mathcal{CI}$. We need several lemmas for this purpose. To do this, we introduce the notation $\mathbb{LT}_n(D)$ (respectively, $\mathbb{UT}_n(D)$) to denote the group of lower (respectively, upper) unitriangular matrices in $\mathrm{GL}_n(D)$, where each matrix has elements on the main diagonal equal to $1$ and $(x)$  is a  square matrix of size $1$.



\begin{lemma}[see \cite{Pa_Gvo_20}, Lemma 10]\label{lemma5.1}
	Let $c=d_1+\cdots+d_n$ be a partition of the number $c$, where $d_n\in\mathbb{N}$. If $g\in \mathrm{SL}_{n}(D)\setminus Z(\mathrm{SL}_{n}(D))$, then there is $\gamma\in \mathrm{SL}_{n}(D)$ such that $\gamma g\gamma^{-1}=vhu$, where $v\in \mathbb{LT}_n(D) ,u\in \mathbb{UT}_n(D)$ and $h=\mathrm{diag}(\epsilon_1,\cdots,\epsilon_n)$, where $\epsilon_i\in D'$ and each $\epsilon_i$ is a product of at most $d_i$ commutators for all $1\leq i\leq n$.
\end{lemma}

\begin{lemma}\label{lemma_cheohoa_comment}  Let $D$ be a division ring and $n \geq 1$. If  $A$ is a noncentral matrix in $\mathrm{SL}_n(D)$, then there exists $P\in \mathrm{GL}_n(D)$ and $s\in D'$ such that    \[P^{-1}AP=XHY,\] where $X\in \mathbb{LT}_n(D), Y\in \mathbb{UT}_n(D)$ and $H=\mathrm{diag}(1,1,\dots,s)$.
	In specific cases, when $D$  is finite
	dimensional over its center and $A$ represents a  lower or upper triangular matrix with pairwise nonconjugate diagonal entries $a_{11},\ldots,a_{nn}\in D$, it follows that $A$ is similar to the diagonal matrix $\mathrm{diag}(a_{11},\ldots,a_{nn})$.
\end{lemma}
\begin{proof}
	The first part of this lemma follows from \cite[Theorem 2.1]{Pa_EgGo_19}. The remaining part is derived from \cite[Lemma 3.2]{Pa_BiDuHaSo_2022}.
\end{proof}

We know that if $D$ is a centrally finite division ring, then every matrix in the group $\mathrm{SL}_{\infty}(D)$ is a commutator in $\mathrm{SL}_{\infty}(D)$, as stated in \cite[Corollary 5]{Pa_Gvo_20}. Assume that $D$ is a field; it has been shown in \cite[Theorem 1.1]{Paz_19} that every $A \in \mathrm{GL}_{\infty}(D)$ can be expressed as a product of three involutions if and only if $\det(A) = \pm 1$. The following theorem is an example showing that the result of \cite[Theorem 1.1]{Paz_19} does not hold when $D$ is a noncommutative division ring.

 For each element $a$ in the division ring $D$, we denote $N(a)$ as the norm of the element $a$. For further details on the norm of elements in finite dimensional division rings, please refer to  \cite[p. 143]{Bo_Dra_80}. Before presenting the main result, we state the following lemma.
\begin{lemma}\label{3.3}
	Let $D$  be a noncommutative centrally finite  division ring and $ n\geq 1$ an integer. Assume $ g_1, g_2, \dots, g_n\in D \setminus F^*$ are pairwise non-conjugate elements. Then  there exist $\alpha \in F^*$ such that $\alpha g_1, \alpha g_2,   \dots, \alpha g_n, (\alpha g_1)^{-1}, \ldots, (\alpha g_n)^{-1}$ are pairwise non-conjugate.
\end{lemma}
\begin{proof} Let $F$ be the center of  $D$ and  $S=\{t \in F : t^{2m} \in T \}$, in which   $m^2 = \mathrm{dim}_FD$ and $T = \left\{  N(g_i^{-1}) N(g_j^{-1}) : 1 \leq i, j \leq n \right\}$.  For each pair $1 \leq i, j \leq n, $ the equation  $t^{2m} =  N(g_i^{-1}) N(g_j^{-1})$ represents a polynomial of degree $2m$ over the field $F$. It is well known that this equation can have at most $2m$ roots in $ F$, thus  the set $S$ is finite.   Because $D$ is a noncommutative   division ring and $\mathrm{dim}_FD$ is finite,  $F$ is infinite (see \cite[Theorems 13.11 and 15.13]{Bo_La_99} ). Let  $\alpha \in F^*\setminus S$. We shall show that
	$\alpha$ satisfies the required condition. Indeed, for every $i \ne j,$ since $ g_i$ and $g_j$ are non-conjugate,   $\alpha g_i$ and $\alpha g_j$ are also non-conjugate, and similarly,  $(\alpha g_i)^{-1}$ and $(\alpha g_j)^{-1}$ are non-conjugate as well. Next, we will prove that   $\alpha g_i$ and  $(\alpha g_j)^{-1}$ are  non-conjugate by using the method of contradiction. Assume  $\alpha g_i$ and $(\alpha g_j)^{-1}$  are conjugate for every  $1 \leq i; j \leq n.$ Then, there exists $h \in D^*$ such that $\alpha g_i = h^{-1}(\alpha g_j)^{-1}h$, implying $\alpha^2 = h^{-1}g_j^{-1}hg_i^{-1}$.  Consequently, $N(\alpha^2) = N(h^{-1})N(g_j^{-1})N(h)N(g_i^{-1})=N(g_j^{-1})N(g_i^{-1})$. By \cite[p. 143]{Bo_Dra_80}, we have $\alpha^{2m} = N(g_j^{-1})N(g_i^{-1})$. This implies that $\alpha \in S$, which contradicts our initial choice of $\alpha$.  Therefore,  $\alpha g_i$ and $(\alpha g_j)^{-1}$ are non-conjugate for every $1 \leq i; j \leq n.$  Hence, $\alpha g_1, \alpha g_2,   \dots, \alpha g_n, (\alpha g_1)^{-1}, \ldots, (\alpha g_n)^{-1}$ are pairwise non-conjugate.
\end{proof}
By applying the above lemma, we obtain the following result.





\begin{theorem}\label{EnD}
	Let $D$ be a noncommutative centrally finite  division ring such that $\mathrm{cn}_{\mathcal{C}}(D') < \infty$. Then,
	
	\begin{enumerate}[\rm (i)]
			\item $\mathrm{cn}_{\mathcal{I}}(\mathrm{SL}_{\infty}(D)) \leq 4$.
			\item $\mathrm{cn}_{\mathcal{CI}}(\mathrm{SL}_{\infty}(D)) \leq 5$.
	\end{enumerate}
\end{theorem}

\begin{proof}
(i)
Assume $A \in \mathrm{SL}_{\infty}(D)$. If $A = \mathbf{1}_{\infty}$, then $A$ is always an involution. If $A \neq \mathbf{1}_{\infty}$, then there exists $n \geq 2$ such that $A \in \mathrm{SL}_n(D)$ and $A$ is a noncentral element in $\mathrm{SL}_n(D)$. By Lemma \ref{lemma_cheohoa_comment}, there exists $P \in \mathrm{GL}_n(D)$ such that $P^{-1}AP=UHV,$ where $U \in \mathbb{LT}_n(D)$, $V \in \mathbb{UT}_n(D)$, and $H=\mathrm{diag}(1,1,\dots,1,x)$ for some $x \in D'$.

Since $\mathrm{cn}_{\mathcal{C}}(D') = s < \infty$, there exist elements $a_1, \dots, a_s$ and $b_1, \dots, b_s$ in $D^*$ such that $x = [a_1,b_1] \dots [a_s, b_s]$.  Now, we shall show $\mathrm{SL}_{\infty}(D)\subseteq\mathcal{I}^4$, where $\mathcal{I}$ is the set of involutions of $\mathrm{SL}_{\infty}(D).$

Assume $n\leq 2s$. By \cite[Lemma 2.3]{Pa_BiDuHa}, we can choose elements $\lambda_1, \lambda_2, \ldots, \lambda_s$ in $F$ such that $\lambda_1 b_1, \lambda_2 b_2, \ldots, \lambda_s b_s$ are pairwise non-conjugate. Since $[a_i, b_i] = [a_i, \lambda_i b_i]$, without loss of generality, we assume that $b_1, \dots, b_s$ are pairwise non-conjugate. By Lemma \ref{3.3}, there exists  $\alpha \in F^*$ such that $\alpha b_1, \alpha b_2,   \dots, \alpha b_s, (\alpha b_1)^{-1},$ $(\alpha b_2)^{-1}, \ldots, (\alpha b_s)^{-1}$ are pairwise non-conjugate. By appropriately adjusting $n$, specifically by inserting $2s-n$ elements 1 into the main diagonal of $A$, we then consider $A$ as a matrix of size $n=2s$.	 By  Lemma \ref{lemma5.1}, the matrix $A$ is similar to $U_1H_1V_1$, where $U_1 \in \mathbb{LT}_n(D)$, $V_1 \in \mathbb{UT}_n(D)$, and $H_1=\mathrm{diag}(1,[a_1,b_1],1,[a_2,b_2],\dots,1,[a_s, b_s])$.

Put $S= \mathrm{diag}(1,a_1,1,a_2,\dots,1, a_s)$. Then \[U_1H_1V_1 = SXS^{-1}Y,\] where  	
\[ X= \left( \begin{smallmatrix}(\alpha b_1)^{-1} \\  &\alpha b_1\\ & &  \ddots  \\  &\text{\LARGE *} & & (\alpha b_s)^{-1} \\  & & &&\alpha b_s\end{smallmatrix}\right);
Y =  \left(\begin{smallmatrix}\alpha b_1 \\ &(\alpha b_1)^{-1} \\&&\ddots&&\text{\LARGE *} \\ &&&\alpha b_s  \\ &&&&(\alpha b_s)^{-1}\end{smallmatrix}\right).
\]

Furthermore, by Lemma \ref{lemma_cheohoa_comment}, $X$ is similar to $ \mathrm{diag}((\alpha b_1)^{-1},\alpha b_1, \dots, (\alpha b_s)^{-1},\alpha b_s)$. Then, according to Remark \ref{chung}, $X$ is a product of two involutions. Similarly, $Y$ is also a product of two involutions. Therefore, $A$ can be expressed as a product of at most four involutions. This leads to $\mathrm{SL}_{\infty}(D)\subseteq\mathcal{I}^4$.

Assume $n >2s$ and $n$ is even.  By similar reasoning, we have  $x = [a_1,b_1] \dots [a_s, b_s]$, in which $b_1, \ldots, b_s$ in $D^*\setminus F$. We choose $b_{s+1}, b_{s+2}, \ldots, b_{\frac{n}{2}}$  in $D^*\setminus F.$  By Lemma \ref{3.3},  there exists $\lambda \in F^*$ such that  $\lambda  b_1, \lambda b_2,   \dots, \lambda b_{\frac{n}{2}}, (\lambda b_1)^{-1},$ $ \lambda b_1 \ldots, (\lambda b_{\frac{n}{2}})^{-1}$ are non-conjugate.
By Lemma \ref{lemma5.1}, the matrix $A$ is similar to $U_1H_1V_1$, where   $U_1 \in \mathbb{LT}_n(D)$,  $V_1 \in \mathbb{UT}_n(D)$ and $H_1=\mathrm{diag}(1,\ldots, 1,[a_1,b_1],1,[a_2,b_2],\dots,1,[a_s, b_s])$.
	Similarly, put
	$S= \mathrm{diag}(1, \ldots,1,a_1,1,a_2,\dots,1, a_s)$. Then \[U_1H_1V_1 = SXS^{-1}Y,\] where  	
	\[ X=\left( \begin{smallmatrix}(\lambda b_{s+1})^{-1}&&& &&&&& \\  &\lambda b_{s+1}&&&&& \\& &\ddots&& \\&&&(\lambda b_{\frac{n}{2}})^{-1}& &&&&&&& \\ &&&&\lambda b_{\frac{n}{2}}&&&&&&& \\  &   &&&&(\lambda b_1)^{-1}&&&& \\ & & &&&&\lambda b_1&&&\\ && \text{\LARGE *}&& &&&&\ddots&& \\  & & & & &&&&&(\lambda b_s)^{-1}& \\  & & & & &&&&&&\lambda b_s\end{smallmatrix}\right); \]
	and \[Y=\left( \begin{smallmatrix}\lambda b_{s+1}&&& &&&&& \\  &(\lambda b_{s+1})^{-1}&&&&& \\& &\ddots&& \\&&&\lambda b_{\frac{n}{2}} & &&&&&&& \\ &&&&(\lambda b_{\frac{n}{2}})^{-1}&&&&\text{\LARGE *}&&& \\  &   &&&& \lambda b_1 &&&& \\ & & &&&&(\lambda b_1)^{-1}&&&\\ && && &&&&\ddots&& \\  & & & & &&&&&\lambda b_s& \\  & & & & &&&&&&(\lambda b_s)^{-1}\end{smallmatrix}\right); \]
	By Lemma \ref{lemma_cheohoa_comment}, $X$ is similar to \[ \mathrm{diag}\big((\lambda b_{s+1})^{-1},\lambda b_{s+1}, \dots, (\lambda b_{\frac{n}{2}})^{-1},\lambda b_{\frac{n}{2}}, (\lambda b_1)^{-1},\ldots,\lambda b_s \big).\]  Therefore, $X$ is a product of two involutions and $Y$ as well.  When  $n$ is odd, by a similar argument, we obtain that $X$ is similar to \[ \mathrm{diag}\big(1, (\lambda b_{s+1})^{-1},\lambda b_{s+1}, \dots, (\lambda b_{\frac{n}{2}})^{-1},\lambda b_{\frac{n}{2}}, (\lambda b_1)^{-1},\ldots,\lambda b_s \big).\] This leads to $\mathrm{SL}_{\infty}(D)\subseteq\mathcal{I}^4$.


 Next, we shall show that $\mathcal{I}^4\subseteq\mathrm{SL}_{\infty}(D)$. Let  $A \in \mathrm{SL}_{\infty}(D),$ and for each $i = 1,\ldots, 4$ suppose $X_i$ is an involution matrix such that $A = X_1X_2X_3X_4$. We shall show  $X_i \in \mathrm{SL}_{\infty}(D)$. Indeed, we see that $A \in \mathrm{SL}_n(D) \setminus Z(\mathrm{SL}_n(D))$, where  $n\geq 3$. According to \cite[Corollary 2.4]{Pa_BiDuHaSo_2023 }, a matrix is a product of involutions if and only if its Dieudonné determinant equals $\overline{\pm1}$. Therefore, we have the following cases:
\bigskip

\noindent\textit{Case 1. $\mathrm{det}(X_i)=\overline{1}$ for all $i=1,\ldots, 4$.}
Then, it is evident that $X_i \in \mathrm{SL}_{\infty}(D).$
\bigskip

\noindent\textit{Case 2. There exist two matrices with determinants equal to $\overline{-1}$.} Without loss of generality, we can assume that $X_1, X_2 \in \mathrm{GL}_n(D)$ and $\mathrm{det}(X_1)=\mathrm{det}(X_2)=\overline{-1}$. Increasing the number $n$, and we rewrite $A$ as \[A = [X_1\oplus (-1)][(X_2\oplus (-1)][X_3\oplus (1)][X_4\oplus (1)].\] Because  $\mathrm{det}[X_1\oplus (-1)]=\overline{1}$, we have $ X_1\oplus (-1) \in \mathrm{SL}_{n+1}(D)$. Similarly,  $  X_2\oplus (-1), X_3\oplus (1), X_4\oplus (1)$ also belong to $ \mathrm{SL}_{n+1}(D)$. Therefore,    $X_i \in \mathrm{SL}_{\infty}(D)$ for each $i= 1,\ldots, 4$.
\bigskip

\noindent\textit{Case 3. $\mathrm{det}(X_i)=\overline{-1}$ for all $i =1,\ldots, 4$.}  Using the same argument  as in the proof of  Case 2 and expressing $A$ in the form \[A = [X_1\oplus (-1)][(X_2\oplus (-1)][X_3\oplus (-1)][X_4\oplus (-1)],\] we conclude that $X_i \in \mathrm{SL}_{\infty}(D).$

Therefore, $\mathrm{SL}_{\infty}(D)=\mathcal{I}^4.$
	
	
	
	(ii) According to the above, there exists $Q \in \mathrm{GL}_n(D)$ such that $ Q^{-1}AQ = U_1H_1V_1.$ Put $X=U_1, Y=H_1V_1H_1^{-1}$, and $Z=H_1$. Then \[ Q^{-1}AQ = XYZ.\] Furthermore, according to Remark \ref{baiso2}, $Z$ is a product of at most three commutators of involutions. Since $X\in \mathbb{UT}_n(D),$ and $Y\in \mathbb{LT}_n(D)$, according to \cite[Theorem 3.4]{Pa_BiDuHaSo_2022}, $XY$ is a product of at most two commutators of involutions. Therefore, $A$ can be expressed as a product of at most five commutators of involutions. 	Therefore, $\mathrm{SL}_{\infty}(D) \subseteq \mathcal{CI}^5.$
	Furthermore, every commutator of involutions belongs to
	$\mathrm{SL}_{\infty}(D)$, so $\mathrm{cn}_{\mathcal{CI}}(\mathrm{SL}_{\infty}(D)) = 5.$
\end{proof}	


	\section*{Acknowledgements}
The authors sincerely thank the referee for his/her valuable comments and suggestions, which significantly improved this paper.
This research is funded by Vietnam National Foundation for Science and Technology Development (NAFOSTED) under Grant No. 101.04-2023.18.



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