On an Analogue of a Property of Singular $M$-matrices, for the Lyapunov and
the Stein Operators

A.M. Encinas 1, Samir Mondal 2 and K.C. Sivakumar 2

11footnotetext: Department of Mathematics, Polytechnic University of
Catalunya, Barcelona, Spain (andres.marcos.encinas@upc.edu).22footnotetext:
Department of Mathematics, Indian Institute of Technology Madras, Chennai
600036, India<EMAIL_ADDRESS>kcskumar@iitm.ac.in).

###### Abstract

In the setting of real square matrices, it is known that, if $A$ is a singular
irreducible $M$-matrix, then the only nonnegative vector that belongs to the
range space of $A$ is the zero vector. In this paper, we prove an analogue of
this result for the Lyapunov and the Stein operators.

Keywords: $M$-matrix, Singular irreducible $M$-matrix, Almost monotonicity,
Lyapunov operator, Stein operator.

AMS Subject Classifications: 15A48, 15A23, 15A09, 15A18

##  1 Introduction

The set of all real matrices of order $n\times n$ will be denoted by
$\mathbf{M}_{n}\,(\mathbb{R})$. For a matrix
$X\in\mathbf{M}_{n}\,(\mathbb{R})$, we use $X\geq 0$ to denote the fact that
all the entries of $X$ are nonnegative. If all the entries of $X$ are
positive, we denote that by $X>0$. We use the same notation for vectors.

As usual, for any $A\in\mathbf{M}_{n}\,(\mathbb{R})$, $N(A)$ and $R(A)$ denote
the null and the range spaces of $A\in\mathbf{M}_{n}\,(\mathbb{R})$,
respectively. The rank of $A$ is ${\rm rk}(A)$, the dimension of $R(A)$. In
addition, $\rho(A)$ denotes the spectral radius of $A$; that is, the maximum
of the absolute values of its eigenvalues.

A matrix $A\in\mathbf{M}_{n}\,(\mathbb{R})$ is said to be reducible if there
is a permutation matrix $P\in\mathbf{M}_{n}\,(\mathbb{R})$ such that $P^{T}AP$
has the form

$\begin{bmatrix}S_{11}&S_{12}\\\ 0&S_{22}\end{bmatrix}$

for some square matrices $S_{11}$ and $S_{22}$ of order at least one. A matrix
is irreducible if it is not reducible.

The group inverse of a matrix $A\in\mathbf{M}_{n}\,(\mathbb{R})$ is the unique
matrix $X\in\mathbf{M}_{n}\,(\mathbb{R})$, if it exists, that satisfies the
equations $AXA=A,XAX=X$ and $AX=XA$. If it exists, then the group inverse is
denoted by $A^{\\#}$. Of course, when $A$ is nonsingular, then $A^{\\#}$
exists and moreover $A^{\\#}=A^{-1}$.

A necessary and sufficient condition for the group inverse of a matrix $A$ to
exist is that $A$ has index $1$; that is, $R(A^{2})=R(A)$, which is equivalent
to the condition $N(A^{2})=N(A)$, see [1, Theorem 2, Section 4.4]. Another
characterization is that $A^{\\#}$ exists iff $R(A)$ and $N(A)$ are
complementary subspaces of $\mathbb{R}^{n}$. It is easy to show that any
symmetric matrix has group inverse, that a nilpotent matrix $A$ (viz.,
$A^{n}=0$), does not have group inverse, whereas the group inverse of an
idempotent matrix (viz., $A^{2}=A$), (exists and) is itself. We refer the
reader to [1, Chapter 4] for more details.

If ${\cal S}^{n}(\mathbb{R})$ is the subspace of symmetric matrices in
$\mathbf{M}_{n}\,(\mathbb{R})$, for $X\in{\cal S}^{n}(\mathbb{R}),$ let us
signify $X\succeq 0$ to denote the fact that $X$ is a positive semidefinite
matrix. We use $X\succ 0,$ when $X$ is positive definite.

######  Definition 1.1.

A matrix $A\in\mathbf{M}_{n}\,(\mathbb{R})$ is called a $Z$-matrix, if all the
off-diagonal entries of $A$ are nonpositive. Any $Z$-matrix $A$ has the
representation $A=sI-B,$ where $s\geq 0$ and $B\geq 0$. If $s\geq\rho(B)$,
then $A$ is called an $M$-matrix.

The notion of $M$-matrix introduced by A. Ostrowski in 1935 in honor of H.
Minkowski who worked with this class of matrices around 1900. $M$-matrices
possess many interesting nonnegativity properties. For instance, in the
representation as above, let $s>\rho(B).$ Then,
$A=s\big{(}I-\frac{B}{s}\big{)}$ and considering the Neumann series for
$\frac{B}{s}$; that is,

$\Big{(}I-\frac{B}{s}\Big{)}^{-1}=\sum_{m=0}^{\infty}\Big{(}\frac{B}{s}\Big{)}^{m}\geq
0.$

we conclude that $A$ is invertible and moreover
$A^{-1}=\frac{1}{s}\Big{(}I-\frac{B}{s}\Big{)}^{-1}\geq 0$. In such a case, we
shall refer to $A$ as an invertible $M$-matrix. Otherwise, we call $A$, a
singular $M$-matrix.

Here is a characterization for invertible $M$-matrices. For a proof, we refer
the reader to the book [2, Chapter 6], where fifty different characterizations
are given.

######  Theorem 1.2.

[2, Theorem 2.3, Chapter 6]
Let $A$ be a $Z$-matrix. Then the following statements are equivalent:
(a) $A$ is an invertible $M$-matrix.
(b) There exists $x>0$ such that $Ax>0$.
(c) For every $q>0,$ there exiss $x>0$ such that $Ax=q.$
(d) $A$ is monotone, i.e. $Ax\geq 0\Longrightarrow x\geq 0$.
$A$ is inverse-nonnegative, i.e. $A$ is invertible and $A^{-1}\geq 0$.
(f) $A$ is a $P$-matrix, i.e. all principal minors of $A$ are positive.
(g) $A$ is positive stable i.e. if the real part of each of its eigenvalues is
positive.
Suppose that $A$ has the representation $A=sI-B$, with $B\geq 0$ and
$s\geq\rho(B)$. Then each of the above statements is equivalent to:
(h) $s<\rho(B)$.
When in addition, $A$ is irreducible the above statements are equivalent to:
(e’) $A$ is inverse-positive, i.e. $A$ is invertible and $A^{-1}>0$.

We will have the occasion to consider versions of Theorem 1.2 for two special
classes of operators on $S^{n}(\mathbb{R})$ (Theorem 2.1 and Theorem 2.2), in
the next section.

Next, let us turn our attention to the case singular $M$-matrices. A
distinguished subclass of such matrices, due to their relevance in many
applications, is the set of singular irreducible $M$-matrices. For such
matrices, we us recall a well known result, which will serve to motivate the
contents of this article.

######  Theorem 1.3.

[2, Theorem 4.16, Chapter 6],[8, Theorem 3]
Let $A\in\mathbf{M}_{n}\,(\mathbb{R})$ be a singular irreducible $M$-matrix.
Then the following hold:
(a) $\rm rk(A)=n-1.$
(b) There exists a vector $x>0$ such that $Ax=0.$
(c) $A^{\\#}$ exists and is nonnegative on $R(A)$, i.e. $x\geq 0,x\in
R(A)\Longrightarrow A^{\\#}x\geq 0$.
(d) All the principal submatrices of $A$, except $A$ itself is an invertible
$M$-matrix.
(e) $A$ is almost monotone, i.e. $Ax\geq 0\Longrightarrow Ax=0.$

We remark that in [2, Theorem 4.16, Chapter 6] statement (c) was established
as $A$ has ”property c”, and that the equivalence with (c) in Theorem 1.3 was
proved in [8, Theorem 3]. Recall that a singular $M$-matrix $A$ is said to
have ”property c”, if $A=sI-B$, where $B\geq 0,s>0$ and the matrix
$\frac{1}{s}B$ is semi-convergent. Semi-convergence of a matrix $X$ means that
the matrix sequence $\\{X^{k}\\}$ converges. Note that for an invertible
$M$-matrix, such a sequence converges to zero. One of the most prominent
situations where such matrices arise concerns irreducible Markov processes,
where one deals with matrices of the form $I-T$, where $T$ is an irreducible
column stochastic matrix. We refer the reader to [6] for additional results
that characterize singular irreducible $M$-matrices.

######  Definition 1.4.

A matrix $A$ is referred to as range monotone (see, for instance [7]), if

$Ax\geq 0,x\in R(A)\Longrightarrow x\geq 0.$

If $A$ is range monotone, then we say that $A$ has the range monotonicity
property.

Observe that range monotonicity of $A$ is equivalent to: $A^{2}x\geq
0\Longrightarrow Ax\geq 0.$ Moreover, [8, Theorem 3 (e)] shows that statement
$(c)$ of Theorem 1.3 is equivalent to the range monotonicity of $A$.

Let us recall a notion that is stronger than range monotonicity. This will
play a central role in this article.

######  Definition 1.5.

Let $A\in\mathbf{M}_{n}\,(\mathbb{R})$. Then $A$ is called trivially range
monotone if

$Ax\geq 0,x\in R(A)\Longrightarrow x=0.$

Trivial range monotonicity of $A$ is the same as: $A^{2}x\geq 0\Longrightarrow
Ax=0.$ It is easy to verify that the matrix
$A=\begin{pmatrix}\leavevmode\nobreak\ \leavevmode\nobreak\ 1&-1\\\
-1&\leavevmode\nobreak\ \leavevmode\nobreak\ 1\end{pmatrix}$ is trivially
range monotone.

######  Theorem 1.6.

Let $A\in\mathbf{M}_{n}\,(\mathbb{R})$ be a singular irreducible $M$-matrix.
Then $A$ is trivially range monotone.

Proof. Suppose that $Ax\geq 0,x\in R(A).$ Then by the almost monotonicity of
$A$ (item (e), Theorem 1.3), it follows that $Ax=0$. So, $x\in N(A)\cap R(A)$
and since the group inverse $A^{\\#}$ exists, we have $x=0.$ $\Box$

In this work, we shall be interested in the problem of determining trivial
range monotonicity for the Lyapunov and the Stein operators on the space of
real symmetric matrices. We consider four classes of matrices that give rise
to specific instances of these operators, for which we give
affirmative/negative answers. The main results are presented in Theorem 3.10,
Theorem 3.15 and Theorem 3.17. Pertinent counterexamples are presented in
Section 4, followed by a summary table consolidating the findings.

##  2 Preliminaries

A nonempty subset $C$ of a finite dimensional Hilbert space $H$ is said to be
a proper cone if $C+C=C,\alpha C\subseteq C$, for all $\alpha\geq 0$,
$C\cap-C=\\{0\\}$ and $C$ has a nonempty interior. Given a cone $C$, we define
its dual cone $C^{*}$ by

$C^{*}:=\\{y\in\mathbb{R}^{n}:\langle x,y\rangle\geq 0,\leavevmode\nobreak\
\forall x\in C\\}.$

It follows that $C^{*}$ is indeed a proper cone. A well known example of a
proper cone is the nonnegative orthant $\mathbb{R}^{n}_{+}$, viz., the cone of
nonnegative vectors. Let the space of all real symmetric matrices of order
$n,$ be denoted by ${\cal S}^{n}$. Then ${\cal S}^{n}$ is a real Hilbert space
with the trace inner product, i.e., $\langle
A,B\rangle=tr(AB),\leavevmode\nobreak\ A,B\in{\cal S}^{n}.$ In ${\cal S}^{n},$
the set of all positive semidefinite matrices denoted by ${\cal S}^{n}_{+}$,
is a cone. Both these cones satisfy the condition of self-duality, viz.,
$C^{*}=C,$ when $C=\mathbb{R}^{n}_{+}$ or ${\cal S}^{n}_{+}$.

Next, let us observe that if $A$ is a $Z$-matrix, then $\langle
Ae_{j},e_{i}\rangle=a_{ij}\leq 0,\;i\neq j$, where $e_{i}$ denotes the $i$th
standard basis vector of $\mathbb{R}^{n}$. These basis vectors are mutually
orthogonal and positive, as well. It then follows that an equivalent
formulation for a matrix $A$ to be a $Z$-matrix is:

$x\geq 0,\,y\geq 0$ and $\langle x,y\rangle=0\;\Longrightarrow\;\langle
Ax,y\rangle\leq 0$.

This in turn is the same as saying (with $C=\mathbb{R}^{n}_{+}$):

$x\in C,\,y\in C^{*}$ and $\langle x,y\rangle=0\;\Longrightarrow\;\langle
Ax,y\rangle\leq 0$.

Motivated by the reformulation above, a linear operator $T:{\cal
S}^{n}\longrightarrow{\cal S}^{n}$ is called a $Z$-operator if it satisfies:

$X\succeq 0,Y\succeq 0$ and $\langle X,Y\rangle=0\;\Longrightarrow\;\langle
T(X),Y\rangle\leq 0$,

where $U\succeq 0$ stands for the fact that the symmetric matrix $U$ is
positive semidefinite. When $-U\succeq 0,$ we will use the notation $U\preceq
0.$ In this article, we shall be concerned with two important classes of
operators on ${\cal S}^{n}$. These are the Lyapunov operator and the Stein
operator. Let us first look at their definitions. Given
$A\in\mathbf{M}_{n}\,(\mathbb{R})$, the Lyapunov operator $L_{A}$ on ${\cal
S}^{n}$ is the operator

$L_{A}(X):=AX+XA^{T},\;X\in{\cal S}^{n}$

and the Stein operator $S_{A}$ is defined by

$S_{A}(X):=X-AXA^{T},\;X\in{\cal S}^{n}.$

Let $L$ denote either the Lyapunov or the Stein operator for a given matrix
$A$. Then $L$ satisfies the following (independent of $A$):

$X\succeq 0,\,Y\succeq 0$ and $\langle X,Y\rangle=0\;\Longrightarrow\;\langle
L(X),Y\rangle\leq 0.$

Thus, both these operators could be thought of as analogues of $Z$-matrices,
for linear maps on ${\cal S}^{n}$.

An operator $T$ on ${\cal S}^{n}$ will be called a positive stable
$Z$-operator if $T$ is an invertible $Z$-operator and $T^{-1}({\cal
S}^{n}_{+})\subseteq{\cal S}^{n}_{+}$. Note that this generalizes what we know
for an invertible $M$-matrix, namely that it is invertible and that its
inverse is nonnegative, i.e. leaves the cone $\mathbb{R}^{n}_{+}$ invariant.

The following result is a version of (items (b) and (c) of) Theorem 1.2 for
the Lyapunov operator [5]. The notation $U\succ 0$ denotes the fact that the
symmetric matrix $U$ is positive definite.

######  Theorem 2.1.

([5, Theorem 5])
For $A\in\mathbf{M}_{n}\,(\mathbb{R})$, the following statements are
equivalent:
(a) There exists $E\succ 0$ such that $L_{A}(E)\succ 0$.
(b) For every $Q\succ 0$ there exists $X\succ 0$ such that $L_{A}(X)=Q$.
(c) $A$ is positive stable.

A version for the Stein operator is stated next. The original result holds for
complex matrices [4]. We present only the real case. Recall that a matrix $A$
is called Schur stable if all its eigenvalues lie in the open unit disc of the
complex plane.

######  Theorem 2.2.

([4, Theorem 11])
For $A\in\mathbf{M}_{n}\,(\mathbb{R})$, the following statements are
equivalent:
(a) There exists $E\succ 0$ such that $S_{A}(E)\succ 0$.
(b) For every $Q\succ 0$ there exists $X\succ 0$ such that $S_{A}(X)=Q$.
(c) $A$ is Schur stable.

These results may be considered as analogues of Theorem 1.2. Let $L$ stand for
either the Lyapunov operator or the Stein operator. Then, $(b)$ of Theorem 2.1
or Theorem 2.2 states that the range of the operator $L$ contains the open set
of all (symmetric) positive definite matrices. Hence $L$ is surjective and so,
it is injective, too. Thus, $L$ is invertible. By the same statement, it also
follows that $L^{-1}({\cal S}^{n}_{+})\subseteq{\cal S}^{n}_{+}$. Thus, these
two operators are examples of positive stable $Z$-operators.

##  3 Main Results

We are interested in identifying some matrix classes for which the Lyapunov
operator and/or the Stein operator are/is trivially range monotone. Let us
define this notion, first. Motivated by the definitions for matrices, we call
a linear operator $T:{\cal S}^{n}\rightarrow{\cal S}^{n}$ range monotone if

$T(X)\succeq 0,X\in R(T)\Longrightarrow X\succeq 0$

and refer to $T$ as trivially range monotone if

$T(X)\succeq 0,X\in R(T)\Longrightarrow X=0.$

As in the case of matrices, one may observe that range monotonicity of a
linear operator $T$ is equivalent to the condition: $T^{2}(X)\succeq
0\Longrightarrow T(X)\succeq 0$ and that trivial range monotonicity is the
same as: $T^{2}(X)\succeq 0\Longrightarrow T(X)=0.$

Next, we give an example of a range monotone operator and also one which is
not range monotone.

######  Example 3.1.

Let $A=\begin{pmatrix}1&1\\\ 0&0\end{pmatrix}.$ The corresponding Lyapunov
operator is given by

$L_{A}(X)=\begin{pmatrix}2(x_{11}+x_{12})&x_{12}+x_{22}\\\
x_{12}+x_{22}&0\end{pmatrix},$

where $X=\begin{pmatrix}x_{11}&x_{12}\\\ x_{12}&x_{22}\end{pmatrix}.$ The
choice $x_{11}=x_{22}=1,x_{12}=-1$ shows that $L_{A}$ is not invertible. It is
easily seen that if $X\in R(L_{A})$, then $x_{22}=0.$ Next, for such an $X$,
if $L_{A}(X)$ is a positive semidefinite matrix, then
$X=\begin{pmatrix}x_{11}&0\\\ 0&0\end{pmatrix}$ is also positive semidefinite.
Thus, $L_{A}$ is a range monotone operator. It is not trivially range
monotone, as $X=\begin{pmatrix}1&0\\\ 0&0\end{pmatrix}\in R(L_{A})\cap{\cal
S}^{n}_{+}.$

######  Example 3.2.

Let $A=\begin{pmatrix}1&1\\\ 0&1\end{pmatrix}.$ Then the associated Stein
operator is given by

$S_{A}(X)=-\begin{pmatrix}2b+c&c\\\ c&0\end{pmatrix},$

given $X=\begin{pmatrix}a&b\\\ b&c\end{pmatrix}.$ Set
$X=\begin{pmatrix}-1&0\\\ \leavevmode\nobreak\ \leavevmode\nobreak\
0&0\end{pmatrix}$ and $Y=\begin{pmatrix}0&\frac{1}{2}\\\
\frac{1}{2}&0\end{pmatrix}.$ Then $S_{A}(Y)=X,$ so that $X\in R(S_{A})$.
Further, $S_{A}(X)=0$. This shows that $S_{A}$ is not range monotone.

It is easy to prove that if an operator is idempotent, then it is range
monotone. The only fact that is used, is that an idempotent operator acts like
the identity operator on its range space. Let us record this statement.

######  Lemma 3.3.

Let $T:V\rightarrow V$ be idempotent. Then $T$ is range monotone.

As a first step, we address the question of when the Lyapunov and the Stein
operators are idempotent. The relevant results are Theorem 3.6 and Theorem
3.9. It appears that this question has not been addressed, to the best of our
knowledge.

Let us fix some more notation. The all ones vector in $\mathbb{R}^{n}$ is
denoted by $e$. For any $j=1,\ldots,n$, we define the symmetric matrix
$E_{jj}=[0,\ldots,\stackrel{{\scriptstyle
j\atop\downarrow}}{{e_{j}}},\ldots,0]$ and for any $1\leq i<j\leq n$ we also
define the symmetric matrix $E_{ij}=[0,\ldots,\stackrel{{\scriptstyle
i\atop\downarrow}}{{e_{j}}},\ldots,\stackrel{{\scriptstyle
j\atop\downarrow}}{{e_{i}}},\ldots,0]$. It is clear that $\\{E_{ij}\\}_{1\leq
i\leq j\leq n}$ is a basis of $\mathcal{S}^{n}$. Thus, if
$A=[c_{1},\ldots,c_{n}]\in\mathbf{M}_{n}\,(\mathbb{R})$, we have,
$AE_{jj}=[0,\ldots,\stackrel{{\scriptstyle
j\atop\downarrow}}{{c_{j}}},\ldots,0]$, for any $j$ whereas,
$AE_{ij}=[0,\ldots,\stackrel{{\scriptstyle
i\atop\downarrow}}{{c_{j}}},\ldots,\stackrel{{\scriptstyle
j\atop\downarrow}}{{c_{i}}},\ldots,0]$, for any $1\leq i<j\leq n$.

In what follows, we collect some properties of the Lyapunov operator. We only
prove the second item, as the others are straightforward.

######  Lemma 3.4.

(a) For any $A,B\in\mathbf{M}_{n}\,(\mathbb{R})$ and any $s,t\in\mathbb{R}$,
$L_{tA+sB}=tL_{A}+sL_{B}$.
(b) $L_{A}$ is the zero operator iff $A=0$ and $L_{A}$ is the identity
operator iff $A=\frac{1}{2}I$.
(c) For any $A\in\mathbf{M}_{n}\,(\mathbb{R})$, $L_{A}(I)=A+A^{T}$ and hence
$L_{A}(A+A^{T})=L_{A}^{2}(I).$
(d) For any orthogonal matrix $P$, we have:
$L_{PAP^{T}}(PXP^{T})=PL_{A}(X)P^{T}.$

Proof. (b) Let $c_{j}=(c_{1j},\ldots,c_{nj})^{T}$ be the $j$-th column of $A$,
for $1\leq j\leq n.$ Then

$L_{A}(E_{jj})=[0,\ldots,c_{j},\ldots,0]+[0,\ldots,c_{j},\ldots,0]^{T}.$

Therefore, when $L_{A}=0,$ we infer that $c_{ij}=0$, $j\not=i$ and
$2c_{jj}=0$. When $L_{A}=I$, then $c_{ij}=0$, $j\not=i$ and $2c_{jj}=1$.
$\Box$

######  Definition 3.5.

A square matrix $A\in\mathbf{M}_{n}\,(\mathbb{R})$ is called Lyapunov
idempotent or $L$-idempotent, for short, iff $L_{A}$ is idempotent; that is,
$L_{A}^{2}=L_{A}$.

It is clear that $A$ is $L$-idempotent when $A=0$ and also when
$A=\frac{1}{2}I$. Moreover, when $L_{A}$ is idempotent, then $L_{tA}$ is not,
except when $A=0$ or $t\neq 0,1$: Since $L_{tA}=tL_{A}$ for any
$t\in\mathbb{R}$, we have that $L_{tA}^{2}=t^{2}L_{A}^{2}=t^{2}L_{A}$.
Therefore, if $tA$ is $L$-idempotent, then
$tL_{A}=L_{tA}=L_{tA}^{2}=t^{2}L_{A}$, which implies that either $t=0$ or
$L_{A}=tL_{A}$. When $t\not=0$, the above equality implies that either $t=1$
or $L_{A}=0$, which is equivalent to: $A=0$.

######  Theorem 3.6.

A matrix $A\in\mathbf{M}_{n}\,(\mathbb{R})$ is $L$-idempotent iff it is
diagonal and moreover $\sigma(A)\subseteq\big{\\{}0,\frac{1}{2}\big{\\}}$. In
particular, the only nonsingular $L$-idempotent matrix is $\frac{1}{2}I$.

Proof. First, $A$ is $L$-idempotent iff for any $X\in\mathcal{S}^{n}$ we have
that

$A^{2}X+2AXA^{T}+X(A^{T})^{2}=AX+XA^{T}.$

In particular, if $A^{2}=[\hat{c}_{1},\ldots,\hat{c}_{n}]$, choosing
$X=E_{jj}$, $j=1,\ldots,n$, we obtain, for the left hand side, the expression

$[0,\ldots,\hat{c}_{j},\ldots,0]+[0,\ldots,\hat{c}_{j},\ldots,0]^{T}+2[c_{1j}c_{j},\ldots,c_{jj}c_{j},\ldots,c_{nj}c_{j}]^{T},$

whereas, the right hand side reduces to

$[0,\ldots,c_{j},\ldots,0]+[0,\ldots,c_{j},\ldots,0]^{T}.$

This implies that $2c_{ij}c_{kj}=0$ for $i,k\not=j$. In particular,
$c_{ij}^{2}=0$ for $i\not=j$ and so $c_{j}=c_{jj}e_{j}$, which in turn,
implies that $A$ must be diagonal and hence symmetric. Now, let
$d_{1},\ldots,d_{n}$ be the diagonal entries of $A$. Finally, by setting $X=I$
in the identity, we conclude that $4A^{2}=2A$, that is $d_{j}(2d_{j}-1)=0$ for
any $j=1,\ldots,n$, completing the proof. $\Box$

######  Lemma 3.7.

The Stein operator satisfies the following basic properties:
(a) For any $A\in\mathbb{R}^{n\times n}$, $S_{A}$ is linear.
(b) If $P$ is an orthogonal matrix, then,
$S_{PAP^{T}}(PXP^{T})=PS_{A}(X)P^{T}.$
(c) $S_{A}$ is the identity operator iff $A=0$ and $S_{A}$ is the zero
operator iff $A=\pm I$.

Proof. We only prove the property (c), since the others are straightforward.
For the first part, clearly, $S_{A}$ is the identity iff $AXA^{T}=0$ for any
symmetric matrix $X$. In particular, taking $X=I$ we get that $AA^{T}=0$ and
so $A=0$.
Let us prove the second part. Observe that, if for any $j=1,\ldots,n$,
$c_{j}=(c_{1j},\ldots,c_{nj})^{T}$ is the $j$-th column of $A$, then

$S_{A}(E_{jj})=[0,\ldots,e_{j},\ldots,0]-[c_{1j}c_{j},\ldots,c_{jj}c_{j},\ldots,c_{nj}c_{j}]^{T}.$

Therefore, when $S_{A}=0$, this implies that $c_{jj}c_{j}=e_{j}$. Thus,
$c_{jj}^{2}=1$ and $c_{ij}=0$, $i\not=j$.

Finally, when $i<j$, taking into account that
$A=[c_{11}e_{1},\ldots,c_{nn}e_{n}]$, we obtain that

$0=S_{A}(E_{ij})=[0,\ldots,e_{j},\ldots,e_{i},\ldots,0]-[0,\ldots,c_{ii}c_{jj}e_{j},\ldots,c_{jj}c_{ii}e_{i}\ldots,0],$

which implies that $c_{ii}c_{ij}=1$, thereby showing that $A=\pm I$. $\Box$

######  Definition 3.8.

A square matrix $A\in\mathbf{M}_{n}\,(\mathbb{R})$ is called $S$-idempotent
iff $S_{A}$ is idempotent; that is, $S_{A}^{2}=S_{A}$.

It is clear that $A$ is $S$-idempotent when $A=0$ and also when $A=\pm I$.

######  Theorem 3.9.

A matrix $A\in\mathbf{M}_{n}\,(\mathbb{R})$ is $S$-idempotent iff $A^{2}=\pm
A$. In particular, the only nonsingular $S$-idempotent matrices are $\pm I$.

Proof. First, for any $X\in\mathcal{S}^{n}(\mathbb{R})$ we have that

$\displaystyle S^{2}_{A}(X)=S_{A}(S_{A}(X))$ $\displaystyle=$ $\displaystyle
S_{A}(X)-AS_{A}(X)A^{T}$ $\displaystyle=$ $\displaystyle
X-AXA^{T}-A(X-AXA^{T})A^{T}$ $\displaystyle=$ $\displaystyle
X-2AXA^{T}+A^{2}X(A^{T})^{2}$

and hence $A$ is $S$-idempotent iff

$AXA^{T}=A^{2}X(A^{T})^{2}$

for any $X\in\mathcal{S}^{n}(\mathbb{R})$, which in particular implies that

$AE_{jj}A^{T}=A^{2}E_{jj}(A^{T})^{2}$

for any $j=1,\ldots,n$. Therefore, if for any $j=1,\ldots,n$ we consider
$c_{j}=(c_{1j},\ldots,c_{nj})^{T}$, the $j$-th column of $A$ and
$\hat{c}_{j}=(\hat{c}_{1j},\ldots,\hat{c}_{nj})^{T}$, the $j$-th column of
$A^{2}$, then

$[c_{1j}c_{j},\ldots,c_{nj}c_{j}]=[\hat{c}_{1j}\hat{c}_{j},\ldots,\hat{c}_{nj}\hat{c}_{j}],\hskip
7.11317ptj=1,\ldots,n.$

For a fixed $j=1,\ldots,n$, the above equality implies that $\hat{c}_{j}=0$
iff $c_{j}=0$ and moreover $\hat{c}_{j}=d_{j}c_{j}$, where
$d_{j}\in\mathbb{R}$ and $d_{j}\not=0$. Moreover, for any $k=1,\ldots,n$ we
also have that $c_{kj}c_{j}=\hat{c}_{kj}d_{j}c_{j}$, so that when
$c_{j}\not=0$ this implies that $c_{kj}=d_{j}\hat{c}_{kj}$. Hence
$c_{j}=d_{j}\hat{c}_{j}$. In conclusion, if $c_{j}\not=0$ we have
$c_{j}=d_{j}\hat{c}_{j}=d_{j}^{2}c_{j}$ and hence $d_{j}^{2}=1$. Since this
statement is also true when $c_{j}=0$ because then $\hat{c}_{j}=0$, we
conclude that $A^{2}=AD$, where $D^{2}=I$.

Finally, suppose that $d_{i}d_{j}=-1$ when $1\leq i<j\leq n$. Since

$AE_{ij}A^{T}=A^{2}E_{ij}(A^{T})^{2},\hskip 4.26773pt1\leq i\leq j\leq n.$

we also have

$[c_{1j}c_{i}+c_{1i}c_{j},\ldots,c_{nj}c_{i}+c_{ni}c_{j}]=[\hat{c}_{1j}\hat{c}_{i}+\hat{c}_{1i}\hat{c}_{j},\ldots,\hat{c}_{nj}\hat{c}_{i}+\hat{c}_{ni}\hat{c}_{j}],$

and hence, for any $k=1,\ldots,n$

$c_{kj}c_{i}+c_{ki}c_{j}=d_{i}d_{j}(c_{kj}c_{i}+c_{ki}\
c_{j})=-c_{kj}c_{i}-c_{ki}c_{j}$

which implies that $c_{kj}c_{i}+c_{ki}c_{j}=0$, which in turn, leads us to
conclude that $c_{kj}c_{ki}=0$. If $c_{i}\not=0$, then $c_{ki}\not=0$ for some
$k=1,\ldots,n$ and hence the above identities imply that $c_{kj}=0$ and hence
$c_{j}=0$. So, when $d_{i}d_{j}=-1$ then either $c_{i}$ or $c_{j}$ must be the
zero vector.

Let $i=\min\\{k=1,\ldots,n:c_{k}\not=0\\}$. Then,

$AD=[d_{1}c_{1},\ldots,d_{i}c_{i},\ldots,d_{n}c_{n}]=d_{i}[c_{1},\ldots,c_{i},\ldots,c_{n}]=d_{i}A,$

since when $1\leq j<i$, $0=c_{j}=d_{j}c_{j}=d_{i}c_{j}$, whereas when $j>i$
and $d_{j}\not=d_{i}$, then $c_{j}=0$. Hence $0=c_{j}=d_{j}c_{j}=d_{i}c_{j}$.
$\Box$

In view of Lemma 3.3, Theorem 3.6 and Theorem 3.9, we obtain the following
result. We skip the proof.

######  Theorem 3.10.

Let $A\in\mathbf{M}_{n}\,(\mathbb{R}).$ We then have the following:
(a) If $A$ is a diagonal matrix, whose diagonal entries are either $0$ or
$\frac{1}{2}$, then the Lyapunov operator $L_{A}$ is range monotone.
(c) If $A$ satisfies $A^{2}=\pm A$, then the Stein operator $S_{A}$ is range
monotone.

Let us briefly discuss the notion of the group inverse of a linear operator on
a finite dimensional vector space $V$ [9]. Let $T:V\rightarrow V$ be linear.
$T$ is said to have a group inverse $L:V\rightarrow V$ if $TLT=T,LTL=L$ and
$TL=LT$. The group inverse of $T$ need not exist, but when it does exist, it
is unique and will be denoted by $T^{\\#}.$ We make use of the following in
our discussion.

######  Theorem 3.11.

[9, Theorem 5] Let $T:V\rightarrow V$ be linear. Then the following are
equivalent:
(a) $T^{\\#}$ exists.
(b) The subspaces $R(T)$ and $N(T)$, of $V$, are complementary.
(c) $R(T^{2})=R(T).$
(d) $N(T^{2})=N(T).$

Let us recall that an operator $T$, on an inner product space, is called
normal if $TT^{\ast}=T^{\ast}T$, where $T^{\ast}$ denotes the adjoint of $T$.
It is easy to see that if $T$ is normal, then $R(T^{\ast})=R(T)$. So, if $T$
is normal, then the subspaces $R(T)$ and $N(T)$ are complementary. Thus, we
have the following consequence of Theorem 3.11.

######  Corollary 3.12.

Let $T:V\rightarrow V$ be linear, where $V$ is a finite dimensional inner
product space. If $T$ is a normal operator, then $T^{\\#}$ exists.

######  Definition 3.13.

An operator $T:{\cal S}^{n}\longrightarrow{\cal S}^{n}$, is called generalized
$k$-potent ($k\geq 2$) if there exists a constant $\alpha$ such that
$T^{k}=\alpha T.$

######  Proposition 3.14.

Let $T:{\cal S}^{n}\rightarrow{\cal S}^{n}$ be a generalized $k$-potent. Then
$T^{\\#}$ exists. In particular, any idempotent operator is group invertible.

Proof. We have $T^{k}=\alpha T$, for some $k\geq 2.$ Let $T^{2}(x)=0.$ Then
$0=T^{k}(x)=0=\alpha T(x)=0,$ proving that $T(x)=0.$ Thus $N(T^{2})=N(T).$ So,
the group inverse of $T$ exists. $\Box$

In the next result, we identify four classes of matrices for which the
Lyapunov as well as the Stein operators are group invertible. We shall make
use of the following observation. Let $A\in\mathbb{R}^{n\times n}$. Then,
$L_{A}^{\ast}=L_{A^{T}}$ and $S_{A}^{\ast}=S_{A^{T}}.$

######  Theorem 3.15.

The group inverses $L_{A}^{\\#}$ and $S_{A}^{\\#}$ exist, if $A$ satisfies any
of the following conditions:
(a) $A^{2}=-I$.
(b) $A^{2}=I.$
(c) $A^{T}=-A$.
(d) $A^{T}=A.$

Proof. (a) Let $A^{2}=-I$. First, we consider the Lyapunov operator $L_{A}$.
We have

$L_{A}(X)=AX+XA^{T},X\in{\cal S}^{n}.$

Thus,

$\displaystyle L_{A}^{2}(X)$ $\displaystyle=$ $\displaystyle
A^{2}X+AXA^{T}+AXA^{T}+X(A^{T})^{2}$ $\displaystyle=$
$\displaystyle-2(X-AXA^{T}).$

So,

$\displaystyle L_{A}^{3}(X)$ $\displaystyle=$
$\displaystyle-2(L_{A}(X)-L_{A}(AXA^{T})$ $\displaystyle=$
$\displaystyle-2(AX+XA^{T}-A^{2}XA^{T}-AX(A^{T})^{2})$ $\displaystyle=$
$\displaystyle-4(AX+XA^{T})$ $\displaystyle=$ $\displaystyle-4L_{A}(X).$

Thus, $L_{A}^{3}=-4L_{A}$ and so by Proposition 3.14, $L_{A}^{\\#}$ exists.
Next, we take the case of the Stein operator. We have

$S_{A}(X)=X-AXA^{T},X\in{\cal S}^{n}.$

So,

$\displaystyle S_{A}^{2}(X)$ $\displaystyle=$ $\displaystyle S_{A}(X-AXA^{T})$
$\displaystyle=$ $\displaystyle X-AXA^{T}-AXA^{T}+A^{2}X(A^{T})^{2}$
$\displaystyle=$ $\displaystyle 2(X-AXA^{T})$ $\displaystyle=$ $\displaystyle
2S_{A}(X).$

So, $S_{A}^{2}=S_{A}$ and again, by Proposition 3.14 it follows that
$S_{A}^{\\#}$ exists. In fact, in this case, $S_{A}^{\\#}=2S_{A}$. This proves
(a).
(b) Let $A^{2}=I.$ Then, a calculation done as earlier, leads to the formula

$L_{A}^{2}(X)=2(X+AXA^{T}).$

This, in turn, implies that

$L_{A}^{3}=4L_{A}.$

The conclusion on the existence of the group inverse of $L_{A}$, now follows.
For the Stein operator, just as in the case $A^{2}=-I,$ it follows that

$S_{A}^{2}=2S_{A}$

and so, $S_{A}^{\\#}$ exists.
(c) Since $A^{T}=-A,$ we have $L_{A}(X)=AX-XA.$ Now, for $X,Y\in{\cal S}^{n},$

$\displaystyle\langle X,L_{A}(Y)\rangle$ $\displaystyle=$ $\displaystyle\rm
tr(X(AY-YA))$ $\displaystyle=$ $\displaystyle\rm tr(XAY)-\rm tr(XYA)$
$\displaystyle=$ $\displaystyle-(\rm tr(XYA)-\rm tr(XAY))$ $\displaystyle=$
$\displaystyle-(\rm tr(AXY)-\rm tr(XAY))$ $\displaystyle=$ $\displaystyle-(\rm
tr((AX-XA)Y))$ $\displaystyle=$ $\displaystyle-\langle L_{A}(X),Y\rangle.$

Hence $L_{A}^{*}=-L_{A},$ which means that $R(L_{A})=R(L_{A}^{\ast})$. Thus
$R(L_{A})$ and $N(L_{A})$ are complementary subspaces. Thus, $L_{A}^{\\#}$
exists.

Next, let us consider the Stein operator. Since $A^{T}=-A,$ we have
$S_{A}(X)=X+AXA.$ Now,

$\displaystyle\langle X,S_{A}(Y)\rangle$ $\displaystyle=$ $\displaystyle\rm
tr(X(Y+AYA))$ $\displaystyle=$ $\displaystyle\rm tr(XY+XAYA)$ $\displaystyle=$
$\displaystyle\rm tr(XY)+\rm tr(AXAY)$ $\displaystyle=$ $\displaystyle\rm
tr((X+AXA)Y)$ $\displaystyle=$ $\displaystyle\langle S_{A}(X),Y\rangle.$

Hence $S_{A}^{*}=S_{A}.$ Once again, the subspaces $R(S_{A})$ and $N(S_{A})$
are complementary and so $S_{A}^{\\#}$ exists.
(d) When $A=A^{T},$ as was remarked earlier, we have
$L_{A}^{*}=L_{A^{T}}=L_{A}$ and $S_{A}^{*}=S_{A^{T}}=S_{A}.$ Thus,
$L_{A}^{\\#}$ and $S_{A}^{\\#}$ exist. $\Box$

Let $A\in\mathbf{M}_{n}\,(\mathbb{R})$. We say that $A$ is nonnegative stable,
if all the eigenvalues of $A$ have a nonnegative real part. $A$ will be
referred to as Shur semi-stable, if all its eigenvalues lie in the closed unit
disc of the complex plane.

######  Remark 3.16.

Some remarks are in order. Let matrix $A$ be nilpotent (so that $A$ is
nonnegative stable). Then it is easy to show that the operator $L_{A}$ is also
nilpotent. As noted earlier, it follows that the group inverse of $L_{A}$ does
not exist. So, $L_{A}$ is not trivially range monotone. Nevertheless, we
identify a class of matrices $A$, which are nonnegative stable, for which the
associated Lyapunov operator is trivially range monotone. Example 3.2 shows
that not all Schur semi-stable matrices are trivially range monotone. However,
we identify a class of Schur semi-stable matrices $A$, for which the Stein
operator is trivially range monotone. Both these are presented in the next
result.

In what follows, we consider the question of trivial range monotonicity of the
Lyapunov and the Stein operators corresponding to matrices that satisfy one of
the first three conditions of Theorem 3.15.

######  Theorem 3.17.

For the Lyapunov operator $L_{A}$ and Stein operator $S_{A}$ the following are
true:
$(a)$ Let $A^{2}=-I.$ Then $L_{A}$ and $S_{A}$ are trivially range monotone.
$(b)$ Let $A^{2}=I.$ Then $S_{A}$ is trivially range monotone.
$(c)$ Let $A^{T}=-A.$ then $L_{A}$ is trivially range monotone.

Proof. First, we observe that both the Lyapunov and the Stein operators are
group invertible, under any of the three conditions on the matrix $A$, as
above (in view of Theorem 3.15). Thus in all these cases, the range space and
the null space (of either the Lyapunov operator or the Stein operator) are
complementary subspaces of ${\cal S}^{n}.$ We shall make repeated use of this
fact, in our proofs.

$(a)$ Let $L_{A}(X)\succeq 0$ with $X\in R(L_{A}).$ Then $0\preceq
L_{A}^{3}(X)=-4L_{A}(X).$ This means that $L_{A}(X)=0.$ Thus, $X\in
R(L_{A})\cap N(L_{A})=\\{0\\}$, proving that $L_{A}$ is trivially range
monotone.

Next, we show the trivial range monotonicity of the Stein operator. Let $X\in
R(S_{A})$ so that there exists $Y\in\mathcal{S}^{n}$ with $S_{A}(Y)=X$. Also,
let $S_{A}(X)\succeq 0.$ Then, from the calculation as earlier, we have

$\displaystyle 0\preceq S_{A}(X)=S_{A}^{2}(Y)=2S_{A}(Y)=2(Y-AYA^{T}).$

Set $Z:=Y-AYA^{T}.$ Then, $Z\succeq 0$. This means that

$0\preceq AZA^{T}=AYA^{T}-A^{2}Y(A^{T})^{2}=AYA^{T}-Y=-Z.$

Thus, $Z=0$, i.e. $S_{A}(X)=0,$ which in turn implies that that $X=0,$ as
$X\in R(S_{A})\cap N(S_{A}).$

$(b)$ The proof of the trivial range monotonicity of the Stein operator, is
entirely similar to the second part of the above result and is skipped.

$(c)$ Let $Z:=L_{A}(X)=AX-XA\succeq 0,$ with $X\in R(L_{A}).$ Then all the
eigenvalues of $Z$ are non-negative. Further, $\rm tr(Z)=\rm tr(AX-XA)=0$ and
so, all the eigenvalues of $Z$ are zero. Since $Z$ is diagonalizable,
$0=Z=L_{A}(X).$ Therefore $X=0,$ as $X\in R(L_{A})\cap N(L_{A}).$ This shows
the trivial range monotonicity property of $L_{A}$.
$\Box$

Let matrix $A$ be such that either $A^{2}=I$ or $A^{2}=-I.$ Then $A^{-1}$
exists and shares the same property as that of $A$. If $A$ is skew-symmetric,
then $A^{\\#}$ exists (since $R(A)$ and $N(A)$ are complementary) and
$A^{\\#}$ is also skew-symmetric. Thus we have the following immediate
consequence of Theorem 3.17.

######  Corollary 3.18.

Let $A^{2}=-I.$ Then $L_{A^{-1}}$ and $S_{A^{-1}}$ are trivially range
monotone. If $A^{2}=I,$ then $S_{A^{-1}}$ is trivially range monotone. Let
$A^{T}=-A.$ Then $L_{A^{\\#}}$ is trivially range monotone.

The first example below illustrates (a) of Theorem 3.17.

######  Example 3.19.

Let $A=\begin{pmatrix}\leavevmode\nobreak\ \leavevmode\nobreak\ 0&1\\\
-1&0\end{pmatrix},$ so that $A^{2}=-I.$ For any $X=\begin{pmatrix}a&b\\\
b&c\end{pmatrix}\in{\cal S}^{2},$ we have

$S_{A}(X)=\begin{pmatrix}a-c&2b\\\ 2b&c-a\end{pmatrix}.$

If $S_{A}(X)\succeq 0,$ then $a=c$ and so $b=0$. Thus,
$X=\begin{pmatrix}a&0\\\ 0&a\end{pmatrix}.$ If we impose the condition that
$X\in R(S_{A}),$ then $a=-a$, proving that $X=0.$ Thus, $S_{A}$ is trivially
range monotone.

The next example illustrates item (b) of Theorem 3.17.

######  Example 3.20.

Let $A=\begin{pmatrix}0&1\\\ 1&0\end{pmatrix},$ so that $A^{2}=I.$ For any
$X=\begin{pmatrix}a&b\\\ b&c\end{pmatrix}\in{\cal S}^{2},$ we have

$S_{A}(X)=\begin{pmatrix}a-c&0\\\ 0&c-a\end{pmatrix}.$

If $S_{A}(X)\succeq 0,$ then $a=c$ and so $X=\begin{pmatrix}a&b\\\
b&a\end{pmatrix}.$ Further, if $X\in R(S_{A}),$ then $a=-a$ and $b=0$, proving
that $X=0.$ Thus, $S_{A}$ is trivially range monotone.

Items (a) (for the Lyapunov operator) and (c) of Theorem 3.17 are illustrated,
next.

######  Example 3.21.

Let $A=\begin{pmatrix}\leavevmode\nobreak\ \leavevmode\nobreak\ 0&1\\\
-1&0\end{pmatrix}.$ Then $A$ is skew-symmetric and $A^{2}=-I.$ For any
$X=\begin{pmatrix}a&b\\\ b&c\end{pmatrix}\in{\cal S}^{2},$ we have

$L_{A}(X)=\begin{pmatrix}2b&c-a\\\ c-a&-2b\end{pmatrix}.$

The requirement that $L_{A}(X)\succeq 0$ yields $b=0$ and so $a=c$. Thus
$X=aI$. The condition that $X\in R(L_{A})$ then implies that $a=0$, so that
$X=0,$ proving the trivial range monotonicity of $L_{A}.$

##  4 Counterexamples

The Lyapunov operator, is not trivially range monotone, in general, when $A$
is an involutory matrix.

######  Example 4.1.

Let $A=\begin{pmatrix}0&1\\\ 1&0\end{pmatrix},$ so that $A^{2}=I$ and for any
$X=\begin{pmatrix}a&b\\\ b&c\end{pmatrix}\in{\cal S}^{2},$ we have

$L_{A}(X)=\begin{pmatrix}2b&a+c\\\ a+c&2b\end{pmatrix}.$

Note that, if $X=\begin{pmatrix}1&0\\\ 0&-1\end{pmatrix},$ then $L_{A}(X)=0,$
proving that $L_{A}$ is singular. Next, take $Y=A$ and
$U=\begin{pmatrix}1&0\\\ 0&0\end{pmatrix}.$ Then $L_{A}(U)=Y$ so that $Y\in
R(L_{A}).$ We have $L_{A}(Y)=2I\succeq 0$, but $Y\nsucceq 0.$ Thus, the
Lyapunov operator is not even range monotone, if $A$ is involutory.

The conclusion above holds for matrices of higher order, too. For instance, if
$A$ is involutory and if one sets $\tilde{A}=\begin{pmatrix}A&0\\\ 0&\pm
1\end{pmatrix},$ then $L_{\tilde{A}}$ not range monotone. We omit the details.

######  Example 4.2.

Consider the case, when $A^{T}=-A.$ For $n=1,S_{A}(X)=X.$ Here, $S_{A}$ is
even monotone. For $n=2,$ (after normalizing) $A$ must be of the following
form

$\begin{pmatrix}\leavevmode\nobreak\ 0&\pm 1\\\ \mp 1&\leavevmode\nobreak\
0\end{pmatrix}.$

Then, for any symmetric matrix $X=\begin{pmatrix}a&b\\\ b&d\end{pmatrix},$ we
have

$S_{A}(X)=\begin{pmatrix}a-d&2b\\\ 2b&d-a\end{pmatrix}.$

$S_{A}$ is singular since $S_{A}(I)=0.$ Next, if $S_{A}(X)\succeq 0,$, then
$a=d$ and so $b=0$. Further, if $X\in R(S_{A}),$ then we have $a=-d$, and so
$a=d=0,$ proving that $X=0$. This proves the trivial range monotonicity of
$S_{A},$ for $n=2.$

In the next two examples, we show that the Stein operator is not trivially
range monotone for $n=3,4$, for a skew-symmetric matrix $A$.

######  Example 4.3.

Let $A=\frac{1}{\sqrt{2}}\begin{pmatrix}\leavevmode\nobreak\
0&\leavevmode\nobreak\ 1&\leavevmode\nobreak\ 0\\\ -1&\leavevmode\nobreak\
0&\leavevmode\nobreak\ 1\\\ \leavevmode\nobreak\ 0&-1&\leavevmode\nobreak\
0\end{pmatrix},$ so that for any symmetric matrix $X=\begin{pmatrix}a&b&c\\\
b&d&e\\\ c&e&f\end{pmatrix},$ we have

$S_{A}(X)=\frac{1}{2}\begin{pmatrix}2a-d&\ 3b-e&2c+d\\\
3b-e&2d+2c-a-f&3e-2b\\\ 2c+d&3e-2b&2f-d\end{pmatrix}.$

$S_{A}$ is singular, since $S_{A}(X)=0,$ for
$X=\begin{pmatrix}-1&\leavevmode\nobreak\ 0&\leavevmode\nobreak\ 1\\\
\leavevmode\nobreak\ 0&-2&\leavevmode\nobreak\ 0\\\ \leavevmode\nobreak\
1&\leavevmode\nobreak\ 0&-1\end{pmatrix}.$ Next, if $U=\begin{pmatrix}2&0&1\\\
0&1&0\\\ 1&0&2\end{pmatrix}$ and $Y=\frac{1}{2}\begin{pmatrix}3&0&3\\\
0&0&0\\\ 3&0&3\end{pmatrix}$ then $S_{A}(U)=Y$ so that $Y\in\ R(S_{A}).$ We
have $S_{A}(Y)=\frac{1}{2}\begin{pmatrix}3&0&3\\\ 0&0&0\\\
3&0&3\end{pmatrix}\succeq 0$ and $Y\succeq 0.$ Since $Y\neq 0$, we conclude
that the Stein operator is not trivially range monotone.

######  Example 4.4.

Let $A=\begin{pmatrix}\leavevmode\nobreak\ 0&1&\leavevmode\nobreak\ 0&0\\\
-1&0&\leavevmode\nobreak\ 0&0\\\ \leavevmode\nobreak\ 0&0&\leavevmode\nobreak\
0&2\\\ \leavevmode\nobreak\ 0&0&-2&0\end{pmatrix},$ so that for any symmetric
matrix $X=\begin{pmatrix}a&b&c&d\\\ b&e&f&g\\\ c&f&h&i\\\
d&g&i&j\end{pmatrix},$ we have

$S_{A}(X)=\begin{pmatrix}a-e&2b&c-2g&d+2f\\\ 2b&e-a&f+2d&g-2c\\\
c-2g&f+2d&h-4j&5i\\\ d+2f&g-2c&5i&j-4h\end{pmatrix}.$

It is clear that, if $X=\begin{pmatrix}1&0&0&0\\\ 0&1&0&0\\\ 0&0&0&0\\\
0&0&0&0\end{pmatrix},$ then $S_{A}(X)=0,$ proving that $S_{A}$ is singular.
Next, if $Y=\begin{pmatrix}0&0&\leavevmode\nobreak\ 0&\leavevmode\nobreak\
0\\\ 0&0&\leavevmode\nobreak\ 0&\leavevmode\nobreak\ 0\\\
0&0&-3&\leavevmode\nobreak\ 0\\\ 0&0&\leavevmode\nobreak\ 0&-3\end{pmatrix}$
and $U=\begin{pmatrix}0&0&0&0\\\ 0&0&0&0\\\ 0&0&1&0\\\ 0&0&0&1\end{pmatrix},$
then $S_{A}(U)=Y$ so that $Y\in R(S_{A}).$ We have
$S_{A}(Y)=\begin{pmatrix}0&0&\leavevmode\nobreak\ 0&\leavevmode\nobreak\ 0\\\
0&0&\leavevmode\nobreak\ 0&\leavevmode\nobreak\ 0\\\ 0&0&\leavevmode\nobreak\
9&\leavevmode\nobreak\ 0\\\ 0&0&\leavevmode\nobreak\ 0&\leavevmode\nobreak\
9\end{pmatrix}\succeq 0.$ However, $Y\nsucceq 0,$ proving that the Stein
operator is not even range monotone.

######  Example 4.5.

For $n=5,$ consider the skew symmetric block diagonal matrix $B,$ whose
leading principal sub-block is $A,$ as in example 4.3 and whose trailing
principal sub-block matrix is $\begin{pmatrix}0&-1\\\ 1&\leavevmode\nobreak\
0\end{pmatrix}.$ Then, by the argument given as earlier, it may be shown that
$S_{B}$ is not trivially range monotone. This inductive argument allows us to
conclude that, in general, $S_{A}$ is not trivially range monotone when $A$ is
a skew symmetric matrix, of any order $n\geq 3$.

This finishes the discussion for skew-symmetric case.

Next, let $A$ be symmetric. Example 4.1, shows that $L_{A}$ is not trivially
range monotone. The following example shows that $S_{A}$ is not trivially
range monotone.

######  Example 4.6.

Consider the symmetric matrix $A=\begin{pmatrix}1&0\\\ 0&2\end{pmatrix},$ so
that for any $X=\begin{pmatrix}a&b\\\ b&c\end{pmatrix}\in\mathcal{S}^{2},$
$S_{A}$ is given by

$S_{A}(X)=\begin{pmatrix}\leavevmode\nobreak\ 0&-b\\\ -b&-3c\end{pmatrix}.$

$S_{A}$ is not invertible, as $S_{A}(X)=0,$ for $X=\begin{pmatrix}1&0\\\
0&0\end{pmatrix}.$ Note that, by item $(d)$ of Proposition 3.15, $S_{A}^{\\#}$
exists. Next, if $Y=\begin{pmatrix}0&\leavevmode\nobreak\ 0\\\
0&-1\end{pmatrix}$ then $S_{A}(U)=Y$, where $U=\begin{pmatrix}0&0\\\
0&\frac{1}{3}\end{pmatrix},$ showing that $Y\in R(S_{A}).$ Also, $Y\nsucceq
0,$ whereas, $S_{A}(Y)=\begin{pmatrix}0&0\\\ 0&3\end{pmatrix}\succeq 0$.

We summarize the findings in the following table. Note that, while ”Yes”
stands for an affirmative answer, ”No” means that counterexamples exist to
show that the answers are negative, in general.

Matrix classes | Trivial Range Monotonicity
---|---
| $L_{A}$ | $S_{A}$
$A^{2}=-I$ | Yes [Theorem 3.17(a)] | Yes [Theorem 3.17(a)]
$A^{2}=I$ | No [Example 4.1] | Yes [Theorem 3.17(b)]
$A^{T}=-A$ | Yes [Theorem 3.17(c)] | Yes ($n=2$) [Example 4.2]
|  | No ($n\geq 3$) [Examples 4.3, 4.4, 4.5]
$A^{T}=A$ | No [Example 4.1] | No [Example 4.6]

##  5 Concluding Remarks

We have studied four matrix classes in the context of the notion of trivial
range monotonicity of the associated Lyapunov and the Stein operators. The
motivation, as was mentioned in the introduction, is to obtain a
generalization of the trivial range monotonicity property of singular
irredudcible $M$-matrices. It would be interesting to bring the matrix classes
for which we could obtain affirmative results, under a more general framework.
Similarly, to unify the matrix classes for which negative results for operator
analogues, have been proved. It would be another interesting question to study
the notion of irreducibility of the Lyapunov operator and the Stein operator.
It is pertinent to point to the fact that there are different (possibly
nonequivalent) ways of defining irreducibility of a ”positive” operator. So
one could propose such a notion for the Stein operator (which is the
difference: $I$ minus a positive operator $X\rightarrow AXA^{T}$), simply by
introducing the assumption of irreducibility for the second term, which is
positive (meaning, that the matrix $AXA^{T}$ is symmetric and positive
semidefinite, whenever so is $X$). However, it is not clear how one could
propose irreducibility for the Lyapunov operator and the likes of it. While we
will pursue these questions in future, we note that, motivated by the
considerations of this article, the notion of irreducibility has been
investigated for $Z$-operators over Euclidean Jordan Algebras [3].

##  6 Acknowledgements

Samir Mondal acknowledges funding received from the Prime Minister’s Research
Fellowship (PMRF), Ministry of Education, Government of India, for carrying
out this work. The first and the third authors thank ASEM-DUO for financial
support enabling the former’s visit to India and the latter, to Spain. The
first author was also partially supported by the Spanish I+D+i program under
project PID2021-122501NB-I00. The authors thank Prof. M.S. Gowda for his
suggestions and comments that have helped in a clearer presentation of the
material. He suggested the nomenclature ”positive stable $Z$-operators”.

## References

  * [1] A. Ben-Israel and T. N. E. Greville, Generalized Inverses - Theory and Applications, CMS Books in Mathematics vol. 15. Springer-Verlag, New York, 2003.
  * [2] A. Berman and R.J. Plemmons, Nonnegative Matrices in the Mathematical Sciences, Classics in Applied Mathematics, 9 SIAM, Philadelphia, 1994.
  * [3] M.S. Gowda, Completely mixed linear games corresponding to $Z$-transformations over self-dual cones, April 2023, private communication.
  * [4] M.S. Gowda and T. Parthasarathy, Complementarity forms of theorems of Lyapunov and Stein, and related results, Linear Algebra Appl., 320 (2000), 131-144.
  * [5] M.S. Gowda and Y. Song, On semidefinite linear complementarity problems, Math. Program., 88 (2000), Ser. A, 575–587. Errata, Math. Program. 91 (2001), Ser. A, 199–200.
  * [6] Wen Li, Characterizations of singular irreducible M-matrices, Linear Mult. Algebra, 38 (1995), 241-247
  * [7] D. Mishra and K.C. Sivakumar, Generalizations of matrix monotonicity and their relationships with certain subclasses of proper splittings, Linear Algebra Appl., 436 (2012), 2604-2614.
  * [8] R.J. Plemmons, $M$-matrices leading to semiconvergent splittings, Linear Algebra Appl., 15 (1976), 243-252.
  * [9] P. Robert, On the group inverse of a linear transformation, J. Math. Anal. Appl., 22 (1968), 658-669.


# Central Products and the Chermak–Delgado Lattice

William Cocke  School of Computer and Cyber Sciences, Augusta University,
Augusta, GA 30901<EMAIL_ADDRESS>and  Ryan McCulloch  Division of
Mathematics & Natural Sciences, Elmira College, Elmira, NY 14901
<EMAIL_ADDRESS>
###### Abstract.

The Chermak–Delgado lattice of a finite group is a modular, self-dual
sublattice of the lattice of subgroups. We prove that the Chermak–Delgado
lattice of a central product contains the product of the Chermak–Delgado
lattices of the relevant central factors. Furthermore, we obtain information
about heights of elements in the Chermak–Delgado lattice relative to their
heights in the Chermak–Delgado lattices of central factors. We also explore
how the central product can be used as a tool in investigating Chermak–Delgado
lattices.

###### Key words and phrases:

finite group theory, group theory, Chermak-Delgado lattice, subgroup lattice,
central product

###### 2020 Mathematics Subject Classification:

Primary 20E15, 20D40

##  1\. Introduction.

The Chermak–Delgado lattice consists of subgroups of a finite group that have
maximal Chermak–Delgado measure. Due to the many unique properties of the
Chermak–Delgado lattice, it has attracted attention from researchers
interested in lattice theory, general finite group theory, and centralizers of
groups.

Originally defined by Chermak and Delgado [8], the Chermak–Delgado lattice for
a finite group $G$ is defined using the so-called Chermak–Delgado measure $m$
which takes subgroups of $G$ to positive integers via the formula

$m_{G}(H)=|H|\cdot|\textbf{C}_{G}(H)|.$

It is quite interesting, and perhaps counter-intuitive that the subgroups with
maximal value of $m$ in a group form a sublattice of the subgroup lattice of
$G$. Recall that the subgroup lattice of $G$ is the poset of subgroups of $G$
with the operations of meet and join defined by subgroup intersection and
subgroup generated by respectively, i.e., $H\wedge K=H\cap K$ and $H\vee
K=\langle H,K\rangle$. Surprisingly, when $H$ and $K$ are in the
Chermak–Delgado lattice, we have that $\langle H,K\rangle=HK$, i.e., the set
theoretic product is actually a subgroup. Furthermore if $H$ is in the
Chermak–Delgado lattice then $\textbf{C}_{G}(H)$ is in the Chermak–Delgado
lattice. In addition, we know that for all $H$ in the Chermak–Delgado lattice
of $G$, the subgroup $H$ must contain $\textbf{Z}(G)$, the center of $G$. The
many properties we presented in this paragraph can be found in Isaacs [12,
Section 1.G]. Because of these properties as well as current research about
the Chermak–Delgado lattice for various groups, questions about the
Chermak–Delgado lattice make for good research projects that can be accessible
to students [16, 7].

We write $m^{*}(G)$ for the maximum value that $m$ takes on a finite group
$G$, and $\mathcal{CD}(G)$ for the Chermak–Delgado lattice of $G$. Hence for a
subgroup $H$ of $G$ we have that $H\in\mathcal{CD}(G)$ if and only if
$m_{G}(H)=m^{*}(G)$.

Recent work on the Chermak–Delgado lattice can be broadly classified as coming
in two broad themes. While not exhaustive, we list some references below.

  1. (1)

Showing that certain types of structures occur as Chermak–Delgado lattices for
various group: chains [5], antichains [6], diamonds [3], all subgroups
containing $\textbf{Z}(G)$ in a group [15]; or how the Chermak–Delgado lattice
relates to certain families of groups [11], [14], [2], [13].

  2. (2)

Showing properties of the Chermak–Delgado lattice in general: the
Chermak–Delgado lattice of a direct product is the direct product of the
Chermak–Delgado lattices of the factors [4]. All subgroups of the
Chermak–Delgado lattice of a finite group are subnormal in that group with
subnormal depth bounded by their relevant position within the lattice [4, 9].
Most recently, the question was asked about how many groups are not contained
in the Chermak–Delgado lattice [10].

Related to the direct product is the central product. Recall that a group $G$
is a central product of two of its subgroups $A$ and $B$ if $G=AB$ and
$[A,B]=1,$ i.e., $ab=ba$ for all $a\in A$ and $b\in B$.

We prove that the Chermak–Delgado lattice of a central product contains the
product of the Chermak–Delgado lattices of the relevant central factors.

######  Theorem A.

Let $G$ be a finite group and $A,B\leq G$ such that $G$ is a central product
of $A$ and $B$. Then
$\mathcal{CD}(A)\cdot\mathcal{CD}(B)\subseteq\mathcal{CD}(G)$. Furthermore,
the top (resp. bottom) of $\mathcal{CD}(G)$ is equal to the product of the
tops (resp. bottoms) of $\mathcal{CD}(A)$ and $\mathcal{CD}(B)$.

Hence we have the following corollary.

######  Corollary 1.

If $G$ is a central product of $A$ and $B$ then $m^{*}(G)=\dfrac{m^{*}(A)\cdot
m^{*}(B)}{|A\cap B|^{2}}.$

In addition, we have the following structural information about the
Chermak–Delgado lattices of central products regarding the heights and depths
of elements in the product.

Recall that the _depth of_ $H\in\mathcal{CD}(G)$ is the length of a maximal
chain

$H=G_{0}<G_{1}<\dots<G_{n}=T_{G}$

of elements of $\mathcal{CD}(G)$ where $T_{G}$ is the top element; the _height
of_ $H\in\mathcal{CD}(G)$ is the length of a maximal chain

$B_{G}=G_{0}<G_{1}<\dots<G_{n}=H$

of elements of $\mathcal{CD}(G)$ where $B_{G}$ is the bottom element. The
_height of_ $\mathcal{CD}(G)$ is the height of the top element (or,
equivalently, the depth of the bottom element). These quantities are well-
defined since $\mathcal{CD}(G)$ is a modular lattice, and so all of the
maximal chains between two fixed elements are of the same length.

######  Theorem B.

If $G$ is a central product of $A$ and $B$, then the height of the
Chermak–Delgado lattice of $G$ is equal to the sum of the heights of the
Cheramk–Delgado lattices of $A$ and $B$ respectively. Moreover, an element
$HK\in\mathcal{CD}(G)$ with $A\cap B\leq H\leq A$ and $A\cap B\leq K\leq B$
has height (resp. depth) equal to the sum of the heights (resp. depths) of
$H\in\mathcal{CD}(A)$ and $K\in\mathcal{CD}(B)$.

In Section 3 we show how the central product can be used to prove results
about the Chermak–Delgado lattice.

##  2\. Central Products and the Chermak–Delgado Lattice.

A group $G$ is a central product of $A$ and $B$, which are subgroups of $G$,
if $G=AB$ and $[A,B]=1$. Note for every group $G$ we have the central product
$G=G\textbf{Z}(G)$. Also note that if $G=AB$ is a central product, then both
$A$ and $B$ are normal subgroups of $G$ and that $A\cap
B\subseteq\textbf{Z}(G)$. Also for any $X_{1}\leq A$ and for any $X_{2}\leq
B$, we have that
$\textbf{C}_{G}(X_{1}X_{2})=\textbf{C}_{A}(X_{1})\textbf{C}_{B}(X_{2})$. This
last observation is almost enough to prove Theorem A.

######  Proposition 1.

Suppose a finite group $G$ is a central product of $A$ and $B$. If
$HK\in\mathcal{C}\mathcal{D}(G)$ for some $H\leq A$ and some $K\leq B$, then
$m^{*}(G)=\dfrac{m^{*}(A)\cdot m^{*}(B)}{|A\cap B|^{2}}$,
$\mathcal{C}\mathcal{D}(A)\cdot\mathcal{C}\mathcal{D}(B)\subseteq\mathcal{C}\mathcal{D}(G)$,
and $H(A\cap B)\in\mathcal{C}\mathcal{D}(A)$ and $K(A\cap
B)\in\mathcal{C}\mathcal{D}(B)$.

###### Proof.

Suppose $HK\in\mathcal{C}\mathcal{D}(G)$. Then $A\cap B\leq\textbf{Z}(G)\leq
HK$. And so $HK=H(A\cap B)K(A\cap B)$. Also, $A\cap B\leq\textbf{C}_{A}(H)$
and $A\cap B\leq\textbf{C}_{B}(K)$.

Then

$m^{*}(G)=m_{G}(HK)=|HK|\cdot|\textbf{C}_{G}(HK)|=|H(A\cap B)K(A\cap
B)|\cdot|\textbf{C}_{A}(H)\textbf{C}_{B}(K)|=$

$\frac{|H(A\cap B)|\cdot|K(A\cap B)|}{|A\cap
B|}\cdot\frac{|\textbf{C}_{A}(H)|\cdot|\textbf{C}_{B}(K)|}{|A\cap
B|}=\frac{m_{A}(H(A\cap B))\cdot m_{B}(K(A\cap B))}{|A\cap B|^{2}}.$

Let $X\in\mathcal{C}\mathcal{D}(A)$ and $Y\in\mathcal{C}\mathcal{D}(B)$. Then
$A\cap B\leq\textbf{Z}(A)\leq X$, $A\cap B\leq\textbf{Z}(B)\leq Y$, $A\cap
B\leq\textbf{C}_{A}(X)$, and $A\cap B\leq\textbf{C}_{B}(Y)$. And so

$m_{G}(XY)=|XY|\cdot|\textbf{C}_{G}(XY)|=|XY|\cdot|\textbf{C}_{A}(X)\textbf{C}_{B}(Y)|=\frac{|X|\cdot|Y|}{|A\cap
B|}\cdot\frac{|\textbf{C}_{A}(X)|\cdot|\textbf{C}_{B}(Y)|}{|A\cap B|}=$

$\frac{m_{A}(X)\cdot m_{B}(Y)}{|A\cap B|^{2}}=\dfrac{m^{*}(A)\cdot
m^{*}(B)}{|A\cap B|^{2}}\geq\frac{m_{A}(H(A\cap B))\cdot m_{B}(K(A\cap
B))}{|A\cap B|^{2}}=m^{*}(G).$

Thus $m_{G}(XY)=\dfrac{m^{*}(A)m^{*}(B)}{|A\cap B|^{2}}=m^{*}(G)$, and we have
that
$\mathcal{C}\mathcal{D}(A)\cdot\mathcal{C}\mathcal{D}(B)\subseteq\mathcal{C}\mathcal{D}(G)$,
and also $H(A\cap B)\in\mathcal{C}\mathcal{D}(A)$ and $K(A\cap
B)\in\mathcal{C}\mathcal{D}(B)$. ∎

As it stands, Proposition 1 proves Theorem A modulo the existence of a
subgroup of the form $HK\in\mathcal{CD}(G)$. While Theorem A tells us what
such a subgroup could be, i.e., take the product of the tops of
$\mathcal{CD}(A)$ and $\mathcal{CD}(B)$, it is difficult to directly show that
this product is in $\mathcal{CD}(G)$. Instead, we utilize some lemmata
comparing information about how the Chermak–Delgado measure behaves across
various subgroups.

The following lemma is found in Issacs [12, Lemma 1.43].

######  Lemma 1.

Suppose that $G$ is a finite group. If $H,K\leq G$, then

$m_{G}(H)\cdot m_{G}(K)\leq m_{G}(\langle H,K\rangle)\cdot m_{G}(H\cap K).$

Moreover, equality holds if and only if $\langle H,K\rangle=HK$ and
$\textbf{C}_{G}(H\cap K)=\textbf{C}_{G}(H)\textbf{C}_{G}(K)$.

###### Proof.

$\displaystyle m_{G}(H)\cdot m_{G}(K)$
$\displaystyle=|H|\cdot|\textbf{C}_{G}(H)|\cdot|K|\cdot|\textbf{C}_{G}(K)|$
$\displaystyle=|H|\cdot|K|\cdot|\textbf{C}_{G}(H)|\cdot|\textbf{C}_{G}(K)|$
$\displaystyle=|HK|\cdot|H\cap
K|\cdot|\textbf{C}_{G}(H)\textbf{C}_{G}(K)|\cdot|\textbf{C}_{G}(H)\cap\textbf{C}_{G}(K)|$
$\displaystyle\leq|\langle
H,K\rangle|\cdot|\textbf{C}_{G}(H)\cap\textbf{C}_{G}(K)|\cdot|H\cap
K|\cdot|\textbf{C}_{G}(H)\textbf{C}_{G}(K)|$ $\displaystyle\leq|\langle
H,K\rangle|\cdot|\textbf{C}_{G}(H)\cap\textbf{C}_{G}(K)|\cdot|H\cap
K|\cdot|\textbf{C}_{G}(H\cap K)|$ $\displaystyle=m_{G}(\langle
H,K\rangle)\cdot m_{G}(H\cap K).$

∎

The following lemma generalizes a result of An [1, Lemma 3.2].

######  Lemma 2.

Suppose that $G$ is a finite group. If $K\leq X\leq H\leq G$, then

$\frac{m_{H}(K)}{m_{G}(K)}\leq\frac{m_{H}(X)}{m_{G}(X)}.$

Moreover, equality holds if and only if $\textbf{C}_{G}(K)\subseteq
H\textbf{C}_{G}(X)$.

###### Proof.

Note that

$\frac{m_{H}(K)}{m_{G}(K)}=\frac{|K|\cdot|\textbf{C}_{H}(K)|}{|K|\cdot|\textbf{C}_{G}(K)|}=\frac{|\textbf{C}_{H}(K)|}{|\textbf{C}_{G}(K)|}=\frac{|H\cap\textbf{C}_{G}(K)|}{|\textbf{C}_{G}(K)|}=\frac{|H|}{|H\textbf{C}_{G}(K)|}$

and similarly

$\frac{m_{H}(X)}{m_{G}(X)}=\frac{|X|\cdot|\textbf{C}_{H}(X)|}{|X|\cdot|\textbf{C}_{G}(X)|}=\frac{|\textbf{C}_{H}(X)|}{|\textbf{C}_{G}(X)|}=\frac{|H\cap\textbf{C}_{G}(X)|}{|\textbf{C}_{G}(X)|}=\frac{|H|}{|H\textbf{C}_{G}(X)|}.$

Since $K\leq X$, $\textbf{C}_{G}(X)\leq\textbf{C}_{G}(K)$. So
$H\textbf{C}_{G}(X)\subseteq H\textbf{C}_{G}(K)$ and we have
$|H\textbf{C}_{G}(X)|\leq|H\textbf{C}_{G}(K)|$, where equality holds if and
only if $\textbf{C}_{G}(K)\subseteq H\textbf{C}_{G}(X)$. Thus

$\frac{m_{H}(K)}{m_{G}(K)}=\frac{m_{H}(X)\cdot|H\textbf{C}_{G}(X)|}{m_{G}(X)\cdot|H\textbf{C}_{G}(K)|}\leq\frac{m_{H}(X)}{m_{G}(X)},$

where equality holds if and only if $\textbf{C}_{G}(K)\subseteq
H\textbf{C}_{G}(X)$. ∎

The last lemma we will use to prove Theorem A is one step below that of a
central product. Here the group $G$ will be equal to $H\textbf{C}_{G}(X)$
where $X\leq H\leq G$. Note that a group $G$ is a central product if
$G=H\textbf{C}_{G}(H)$ for some subgroup $H$ of $G$. This lemma generalizes a
result of An [1, Lemma 3.3].

######  Lemma 3.

Suppose that $G$ is a finite group and $X\leq H\leq G$ such that
$G=H\textbf{C}_{G}(X)$. If $X\in\mathcal{C}\mathcal{D}(H)$, then for every
$Y\in\mathcal{C}\mathcal{D}(G)$, we have $\langle
X,Y\rangle\in\mathcal{C}\mathcal{D}(G)$ and $X\cap
Y\in\mathcal{C}\mathcal{D}(H)$. Furthermore, $\langle X,Y\rangle=XY$ and
$\textbf{C}_{G}(X\cap Y)=\textbf{C}_{G}(X)\textbf{C}_{G}(Y)$.

###### Proof.

Let $Y\in\mathcal{C}\mathcal{D}(G)$. Since $\textbf{C}_{G}(X\cap Y)\leq
G=H\textbf{C}_{G}(X)$, by Lemma 2,

$\frac{m_{H}(X\cap Y)}{m_{G}(X\cap Y)}=\frac{m_{H}(X)}{m_{G}(X)}.$

Since $X\in\mathcal{C}\mathcal{D}(H)$, $m_{H}(X\cap Y)\leq m_{H}(X)$. It
follows that $m_{G}(X\cap Y)\leq m_{G}(X)$. By Lemma 1,

$m_{G}(X)\cdot m_{G}(Y)\leq m_{G}(\langle X,Y\rangle)\cdot m_{G}(X\cap Y).$

It follows that $m_{G}(\langle X,Y\rangle)\geq m_{G}(Y)=m^{*}(G)$. Hence
$m_{G}(\langle X,Y\rangle)=m_{G}(Y)$ and $\langle
X,Y\rangle\in\mathcal{C}\mathcal{D}(G)$. Hence $m_{G}(X)\leq m_{G}(X\cap Y)$,
and so $m_{G}(X)=m_{G}(X\cap Y)$.

And so

$m_{G}(X)\cdot m_{G}(Y)=m_{G}(\langle X,Y\rangle)\cdot m_{G}(X\cap Y),$

and by Lemma 1 we have that $\langle X,Y\rangle=XY$ and $\textbf{C}_{G}(X\cap
Y)=\textbf{C}_{G}(X)\textbf{C}_{G}(Y)$.

Finally, since

$\frac{m_{H}(X\cap Y)}{m_{G}(X\cap Y)}=\frac{m_{H}(X)}{m_{G}(X)},$

we conclude that $m_{H}(X\cap Y)=m_{H}(X)=m^{*}(H)$. Hence $X\cap
Y\in\mathcal{C}\mathcal{D}(H)$. ∎

We can now prove Theorem A which states that for a finite group $G=AB$ a
central product, we have
$\mathcal{CD}(A)\cdot\mathcal{CD}(G)\subseteq\mathcal{CD}(G)$, and furthermore
we have that the top (resp. bottom) of $\mathcal{CD}(G)$ is equal to the
product of the tops (resp. bottoms) of $\mathcal{CD}(A)$ and
$\mathcal{CD}(B)$.

###### Proof of Theorem A.

We write $T_{A}$, $T_{B}$, $T_{G}$ for the top elements of the Chermak–Delgado
lattices of $A$, $B$ and $G$ respectively. Similarly $B_{A},B_{B},B_{G}$ refer
to the bottom elements of these lattices.

For any $X_{1}\leq A$, $G=A\textbf{C}_{G}(X_{1})$, and for any $X_{2}\leq B$,
$G=B\textbf{C}_{G}(X_{2})$, and so Lemma 3 applies for any
$X_{1}\in\mathcal{C}\mathcal{D}(A)$ and any
$X_{2}\in\mathcal{C}\mathcal{D}(B)$, with any $Y\in\mathcal{C}\mathcal{D}(G)$.

By Lemma 3, $T_{A}T_{G}\in\mathcal{C}\mathcal{D}(G)$, and so $T_{A}\leq
T_{G}$. Similarly $T_{B}\leq T_{G}$. So $T_{A}T_{B}\leq T_{G}$.

By Lemma 3, $B_{A}\cap B_{G}\in\mathcal{C}\mathcal{D}(A)$, and so
$B_{A}=B_{A}\cap B_{G}$. By Lemma 3,
$\textbf{C}_{G}(B_{A})=\textbf{C}_{G}(B_{A}\cap
B_{G})=\textbf{C}_{G}(B_{A})\textbf{C}_{G}(B_{G})=\textbf{C}_{G}(B_{A})T_{G}$.
So $T_{G}\leq\textbf{C}_{G}(B_{A})=T_{A}B$. Similarly $T_{G}\leq AT_{B}$. We
see that $T_{G}\leq AT_{B}\cap T_{A}B=T_{A}T_{B}(A\cap B)$. And $A\cap
B\leq\textbf{Z}(A)\leq T_{A}$, and so $T_{G}\leq T_{A}T_{B}$.

Thus, $T_{A}T_{B}=T_{G}$, and so
$B_{G}=\textbf{C}_{G}(T_{G})=\textbf{C}_{A}(T_{A})\textbf{C}_{B}(T_{B})=B_{A}B_{B}$.
We apply Proposition 1 to complete the proof. ∎

We provide two examples of equality in Theorem A.

######  Proposition 2.

Suppose that a finite group $G=A\times B$ is a direct product. Then
$\mathcal{CD}(G)=\mathcal{CD}(A)\cdot\mathcal{CD}(B)$.

######  Proposition 3.

Suppose that a finite group $G=AB$ with $B\leq\textbf{Z}(G)$. Then
$\mathcal{CD}(G)=\mathcal{CD}(A)\cdot\\{B\\}$.

Proposition 2 appears in [4]. We shall prove Proposition 3 in a moment, but
first a few facts regarding central products and group products in general.

######  Lemma 4.

Suppose that a group $G=AB$. Suppose $A\cap B\leq H_{1}\leq A$, $A\cap B\leq
H_{2}\leq A$, $A\cap B\leq K_{1}\leq B$, and $A\cap B\leq K_{2}\leq B$. If
$H_{1}K_{1}\subseteq H_{2}K_{2}$, then $H_{1}\leq H_{2}$ and $K_{1}\leq
K_{2}$; and if equality holds then $H_{1}=H_{2}$ and $K_{1}=K_{2}$.

###### Proof.

Suppose $H_{1}K_{1}\subseteq H_{2}K_{2}$, and let $h_{1}\in H_{1}$ and
$k_{1}\in K_{1}$ be arbitrary. Then $h_{1}k_{1}=h_{2}k_{2}$ for some $h_{2}\in
H_{2}$ and some $k_{2}\in K_{2}$. And so $x=h_{2}^{-1}h_{1}=k_{2}k_{1}^{-1}\in
H_{1}\cap K_{1}\leq A\cap B$. And since $A\cap B\leq H_{2}$ and $A\cap B\leq
K_{2}$, we have that $h_{1}=h_{2}x\in H_{2}$ and $k_{1}=x^{-1}k_{2}\in K_{2}$.
Hence $H_{1}\leq H_{2}$ and $K_{1}\leq K_{2}$.

If $H_{1}K_{1}$ and $H_{2}K_{2}$ coincide, then $H_{1}K_{1}\subseteq
H_{2}K_{2}$ implies that $H_{1}\leq H_{2}$ and $K_{1}\leq K_{2}$, and
similarly $H_{2}K_{2}\subseteq H_{1}K_{1}$ implies that $H_{2}\leq H_{1}$ and
$K_{2}\leq K_{1}$. ∎

Note that Lemma 4 applies for any finite central product $G=AB$ with
$H_{1},H_{2}\in\mathcal{CD}(A)$ and $K_{1},K_{2}\in\mathcal{CD}(B)$, since
$A\cap B\leq Z(A)$ and $A\cap B\leq Z(B)$ and all subgroups in the
Chermak–Delgado lattice contain the center. It follows that for any finite
central product $G=AB$, we have that the sum of the heights of
$\mathcal{CD}(A)$ and $\mathcal{CD}(B)$ cannot exceed the height of
$\mathcal{CD}(G)$. For if $CD(A)$ has height $n$, and say
$B_{A}=G_{0}<G_{1}<\cdots<G_{n}=T_{A}$ is a chain in $\mathcal{CD}(A)$, and if
$CD(B)$ has height $m$, and say $B_{B}=H_{0}<H_{1}<\cdots<H_{m}=T_{B}$ is a
chain in $\mathcal{CD}(B)$, then we have by Lemma 4 that
$G_{0}H_{0}<G_{1}H_{0}<\cdots<G_{n}H_{0}<G_{n}H_{1}<\cdots<G_{n}H_{m}$, and by
Theorem A, this chain of length $n+m$ lives in $\mathcal{CD}(G)$. Of course,
our Theorem B gives the precise relationship between heights in
$\mathcal{CD}(A)$, $\mathcal{CD}(B)$, and $\mathcal{CD}(G)$.

If a group $G=AB$ is a central product, then for any $U\leq G$, we define
$\pi_{A}(U)=\\{a\,|\,g=ab\text{ with }g\in U\text{ and }a\in A\text{ and }b\in
B\\}$ and $\pi_{B}(U)=\\{b\,|\,g=ab\text{ with }g\in U\text{ and }a\in A\text{
and }b\in B\\}$.

######  Lemma 5.

Suppose a group $G=AB$ is a central product, and $U\leq G$. Then $\pi_{A}(U)$
is a subgroup of $A$ and, furthermore, $A\cap B\leq\pi_{A}(U)\leq A$. A
similar result is true for $\pi_{B}(U)$.

###### Proof.

We prove the assertion for $\pi_{A}(U)$, and the assertion for $\pi_{B}(U)$ is
similar.

Clearly $\pi_{A}(U)\subseteq A$. We show that $\pi_{A}(U)$ is a subgroup of
$A$. Note that $\pi_{A}(U)$ is nonempty as $1\in\pi_{A}(U)$. Let
$x,y\in\pi_{A}(U)$. We show that $xy^{-1}\in\pi_{A}(U)$.

So there exists $u\in U$ and $v\in U$ so that $u=xb_{1}$ and $v=yb_{2}$ with
$b_{1},b_{2}\in B$.

And so $v^{-1}=y^{-1}{b_{2}}^{-1}$, and we have
$uv^{-1}=xy^{-1}b_{1}{b_{2}}^{-1}$ with $xy^{-1}\in A$ and
$b_{1}{b_{2}}^{-1}\in B$. And since $uv^{-1}\in U$, it follows that
$xy^{-1}\in\pi_{A}(U)$.

Finally, we show that $A\cap B\leq\pi_{A}(U)$. Let $z\in A\cap B$. Then $z\in
A$ and $z=b$ for some $b\in B$. And so $z^{-1}=b^{-1}$. Let $h\in\pi_{A}(U)$.
Then there is $u\in U$ so that $u=hb^{\prime}$ for some $b^{\prime}\in B$. And
so $\displaystyle u=uzz^{-1}=hb^{\prime}zb^{-1}=hzb^{\prime}b^{-1}$ and we
have $hz\in A$ and $b^{\prime}b^{-1}\in B$. Hence $hz\in\pi_{A}(U)$. And since
$\pi_{A}(U)$ is a subgroup, $h^{-1}hz=z\in\pi_{A}(U)$. ∎

######  Lemma 6.

Suppose a group $G=AB$ is a central product. Suppose $K\leq B$ and suppose
$A\cap B\leq U\leq G$. If $K\leq U\leq AK$, then $U=\pi_{A}(U)K$.

###### Proof.

If $K\leq U\leq AK$, then, of course, $K\leq U\leq\pi_{A}(U)K$. And so to
prove that $U=\pi_{A}(U)K$, it suffices to prove that $\pi_{A}(U)\leq U$.

Let $ab\in U$ with $a\in A$ and $b\in B$. Since $U\leq AK$, $ab=a^{\prime}k$
for some $a^{\prime}\in A$ and some $k\in K$. So $a^{\prime}=az$ for some
$z\in A\cap B$. Since $K\leq U$, $a^{\prime}k{k}^{-1}=a^{\prime}=az\in U$.
Since $A\cap B\leq U$, $azz^{-1}=a\in U$. Hence $\pi_{A}(U)\leq U$. ∎

We now prove Proposition 3 which states that for a finite group $G=AB$ with
$B\leq\textbf{Z}(G)$, we have that
$\mathcal{CD}(G)=\mathcal{CD}(A)\cdot\\{B\\}$.

###### Proof of Proposition 3.

We have that $G=AB$ is a central product, and since $B$ is abelian,
$\mathcal{CD}(B)=\\{B\\}$. It follows from Theorem A that
$\mathcal{CD}(A)\cdot\\{B\\}\subseteq\mathcal{CD}(G)$.

Let $U\in\mathcal{CD}(G)$. Then $\textbf{Z}(G)\leq U$, and so $B\leq U$, and
so by Lemma 6, $U=\pi_{A}(U)B$. By Proposition 1, we have that
$\pi_{A}(U)(A\cap B)=\pi_{A}(U)\in\mathcal{CD}(A)$ and
$B\in\mathcal{CD}(B)=\\{B\\}$, and so
$U=\pi_{A}(U)B\in\mathcal{CD}(A)\cdot\\{B\\}$. ∎

######  Corollary 2.

Suppose a finite group $G=AB$ is a a central product. Then
$G\in\mathcal{CD}(G)$ if and only if $A\in\mathcal{CD}(A)$ and
$B\in\mathcal{CD}(B)$.

###### Proof.

If $A\in\mathcal{CD}(A)$ and $B\in\mathcal{CD}(B)$, then by Theorem A,
$AB=G\in\mathcal{CD}(G)$.

Conversely, suppose $G\in\mathcal{CD}(G)$. Then by Theorem A, we have
$T_{A}T_{B}=G\in\mathcal{CD}(G)$, where $T_{A}$ is the top element of
$\mathcal{CD}(A)$ and $T_{B}$ is the top element of $\mathcal{CD}(B)$. Now
$A\cap B\leq Z(A)\leq T_{A}$, and $A\cap B\leq Z(B)\leq T_{B}$. Since
$T_{A}T_{B}=AB$, it follows from Lemma 4 that $T_{A}=A$ and $T_{B}=B$. ∎

We say that a central product $G=AB$ is proper if $\textbf{Z}(G)<A<G$ and
$\textbf{Z}(G)<B<G$, and in such case we say that the group $G$ admits a
proper central product.

Note that if a group $G=AB$ is a central product with one of $A$ or $B$
abelian, say $B$, then $B\leq\textbf{Z}(G)$, and we are in the situation of
Proposition 3.

Given a finite group $G$ which admits a proper central product, one wonders if
$\mathcal{CD}(G)$ is equal to the subgroup collection of $G$ that is generated
by all $\mathcal{CD}(X)\cdot\mathcal{CD}(Y)$ where $G=XY$ a proper central
product. The answer is “no” in general.

######  Proposition 4.

There exists a finite group $G$ with $G\in\mathcal{CD}(G)$ so that
$\mathcal{CD}(G)$ has height $2$ and $\mathcal{CD}(G)$ possesses both abelian
and nonabelian subgroups of height $1$.

See [6, Proposition 10] for an explicit construction.

######  Proposition 5.

There exists a finite group $G$ that admits a proper central product, so that
for $\mathcal{C}$ the subgroup collection of $G$ generated by all
$\mathcal{CD}(X)\cdot\mathcal{CD}(Y)$ where $G=XY$ is a proper central
product, we have that $\mathcal{C}$ is not equal to $\mathcal{CD}(G)$.

###### Proof.

Let $G$ be a group as in Proposition 4. Then there exists
$H\in\mathcal{CD}(G)$ nonabelian of height $1$, and so
$\textbf{C}_{G}(H)\in\mathcal{CD}(G)$ is nonabelian of height $1$, and so the
central product $G=H\textbf{C}_{G}(H)$ is proper.

If $G=AB$ is an arbitrary proper central product, then by Corollary 2, we have
$A\in\mathcal{CD}(A)$ and $B\in\mathcal{CD}(B)$. By Theorem A we have that
$\mathcal{CD}(A)\cdot\mathcal{CD}(B)\subseteq\mathcal{CD}(G)$, and since
$\mathcal{CD}(G)$ has height $2$, it follows that
$\mathcal{CD}(A)=\\{A,\textbf{Z}(A)\\}$ and
$\mathcal{CD}(B)=\\{B,\textbf{Z}(B)\\}$ both have height $1$. Thus the only
abelian subgroup in $\mathcal{CD}(A)\cdot\mathcal{CD}(B)$ is
$\textbf{Z}(A)\textbf{Z}(B)=\textbf{Z}(G)$. And so if $\mathcal{C}$ is the
subgroup collection of $G$ generated by all
$\mathcal{CD}(X)\cdot\mathcal{CD}(Y)$ where $G=XY$ is a proper central
product, then the only abelian subgroup in $\mathcal{C}$ is $\textbf{Z}(G)$.
Thus $\mathcal{C}$ is not equal to $\mathcal{CD}(G)$. ∎

We continue with an application of Theorem A.

######  Proposition 6.

Suppose that $G$ is a finite group and suppose that
$\mathcal{CD}(G)=\\{\textbf{Z}(G),G\\}$. Then $G$ admits no proper central
product.

###### Proof.

Note that if $G=\textbf{Z}(G)$, then $G$ is abelian and the result is true.
Suppose $G=AB$ is an arbitrary central product. We will show that one of $A$
or $B$ is abelian, and thus $G$ admits no proper central product.

By Corollary 2, we have $A\in\mathcal{CD}(A)$ and $B\in\mathcal{CD}(B)$. By
Theorem A we have that
$\mathcal{CD}(A)\cdot\mathcal{CD}(B)\subseteq\mathcal{CD}(G)$, and since
$\mathcal{CD}(G)$ has height $1$, it follows that one of $\mathcal{CD}(A)$ or
$\mathcal{CD}(B)$ has height $0$, say $\mathcal{CD}(A)$. So
$\mathcal{CD}(A)=\\{A\\}$, and thus $A$ is abelian. ∎

We say that a subgroup $A\in\mathcal{CD}(G)$ is an atom if the height of
$A\in\mathcal{CD}(G)$ is $1$, and we say that a subgroup $B\in\mathcal{CD}(G)$
is a coatom if the depth of $B\in\mathcal{CD}(G)$ is $1$.

######  Lemma 7.

Suppose that $G$ is a finite group with $G\in\mathcal{CD}(G)$ and suppose that
$\mathcal{CD}(G)$ has height greater than $1$. If $A$ is a nonabelian atom in
$\mathcal{CD}(G)$, then $A\textbf{C}_{G}(A)$ is a proper central product.

###### Proof.

We note that $\textbf{C}_{G}(A)$ is a coatom of $\mathcal{CD}(G)$. Since $A$
is nonabelian it is not a subgroup of $\textbf{C}_{G}(A)$. Since
$\textbf{C}_{G}(A)$ is a coatom, the only element of $\mathcal{CD}(G)$
properly containing it is $G$. Thus $A\textbf{C}_{G}(A)$ is all of $G$.
Moreover, $\textbf{C}_{G}(A)$ is not abelian, else it would be contained in
$C_{G}(C_{G}(A))=A$, which would imply it is $A$ since $A$ is an atom, but
this implies that $A=\textbf{C}_{G}(A)$ which implies that $A$ is abelian. So
the central product $G=A\textbf{C}_{G}(A)$ is proper. ∎

######  Lemma 8.

Suppose that $G$ is a finite group with $G\in\mathcal{CD}(G)$ and suppose that
$\mathcal{CD}(G)$ has height greater than $1$. If $G$ admits no proper central
product, then every atom in $\mathcal{CD}(G)$ is abelian.

###### Proof.

If some atom $A\in\mathcal{CD}(G)$ is nonabelian, then $G=A\textbf{C}_{G}(A)$
is a proper central product. ∎

######  Lemma 9.

Suppose that $G$ is a finite group with $G\in\mathcal{CD}(G)$ and suppose that
$\mathcal{CD}(G)$ has height greater than $1$. If $C$ is a coatom in
$\mathcal{CD}(G)$, then either $\textbf{C}_{G}(C)=\textbf{Z}(C)$, or
$C\textbf{C}_{G}(C)$ is a proper central product.

We now prove Theorem B which states that for a finite group $G=AB$ a central
product, the height of the Chermak–Delgado lattice of $G$ is equal to the sum
of the heights of the Cheramk–Delgado lattices of $A$ and $B$ respectively.
Moreover, an element $HK\in\mathcal{CD}(G)$ with $A\cap B\leq H\leq A$ and
$A\cap B\leq K\leq B$ has height (resp. depth) equal to the sum of the heights
(resp. depths) of $H\in\mathcal{CD}(A)$ and $K\in\mathcal{CD}(B)$.

###### Proof of Theorem B.

We first prove the assertion in regards to height. It follows from Proposition
1 that given $HK\in\mathcal{CD}(G)$ with $A\cap B\leq H\leq A$ and $A\cap
B\leq K\leq B$, we have $H\in\mathcal{CD}(A)$ and $K\in\mathcal{CD}(B)$, and
it follows from Theorem A that such $HK\in\mathcal{CD}(G)$ exists, namely the
product of the top element of $\mathcal{CD}(A)$ and the top element of
$\mathcal{CD}(B)$.

Suppose $HK\in\mathcal{CD}(G)$ with $A\cap B\leq H\leq A$ and $A\cap B\leq
K\leq B$, and suppose $H\in\mathcal{CD}(A)$ has height $i$ and
$K\in\mathcal{CD}(B)$ has height $j$. We proceed by induction on $i+j$. If
$i+j=0$, the result is true by Theorem A, and we have $H=B_{A}$, $K=B_{B}$,
and $HK=B_{A}B_{B}=B_{G}$. Suppose $i+j>0$ and suppose without loss of
generality that $i>0$. Let $H_{0}\in\mathcal{CD}(A)$ of height $i-1$ with
$H_{0}<H$. Then by induction $H_{0}K\in\mathcal{CD}(G)$ is of height $i+j-1$.
Suppose $U\in\mathcal{CD}(G)$ with $H_{0}K<U\leq HK$ is arbitrary. By Lemma 6,
$U=\pi_{A}(U)K$. By Proposition 1, $\pi_{A}(U)(A\cap
B)=\pi_{A}(U)\in\mathcal{CD}(A)$. And since $H_{0},\pi_{A}(U),$ and $H$ all
contain $A\cap B$, by Lemma 4 we have that $H_{0}<\pi_{A}(U)\leq H$. Since
$H_{0}\in\mathcal{CD}(A)$ is of height $i-1$ and $H\in CD(A)$ is of height
$i$, we have that $\pi_{A}(U)=H$, and so $U=HK$. Since $U$ was arbitrary, it
follows that $HK\in\mathcal{CD}(G)$ is of height $i+j$.

We now establish the assertion in regards to depth. Let $n_{A}$, $n_{B}$, and
$n_{G}$ be the heights of $\mathcal{CD}(A)$, $\mathcal{CD}(B)$, and
$\mathcal{CD}(G)$, respectively. Suppose $H\in\mathcal{CD}(A)$ has height $i$
and $K\in\mathcal{CD}(B)$ has height $j$. Then we have just shown that
$HK\in\mathcal{CD}(G)$ has height $i+j$. And so $H\in\mathcal{CD}(A)$ has
depth $n_{A}-i$ and $K\in\mathcal{CD}(B)$ has depth $n_{B}-j$, and
$HK\in\mathcal{CD}(G)$ has depth $n_{G}-(i+j)$. Now we have established that
$n_{A}+n_{B}=n_{G}$, by applying the result to the heights of the top elements
of $\mathcal{CD}(A)$, $\mathcal{CD}(B)$, and $\mathcal{CD}(G)$. Observe that
$n_{A}-i+n_{B}-j=n_{A}+n_{B}-(i+j)=n_{G}-(i+j)$, and we have the result in
regards to depth. ∎

Our last application of this section relies on both Theorem A and Theorem B.

######  Proposition 7.

Suppose that $G$ is a finite group with $G\in\mathcal{CD}(G)$ and suppose that
$\mathcal{CD}(G)$ has height $2$ or $3$. Then $G$ admits no proper central
product if and only if every atom in $\mathcal{CD}(G)$ is abelian.

###### Proof.

One direction is true by Lemma 8.

Suppose that every atom in $\mathcal{CD}(G)$ is abelian, and let $G=AB$ be an
arbitrary central product. Since $G\in\mathcal{CD}(G)$, by Corollary 2 we have
that $A\in\mathcal{CD}(A)$ and $B\in\mathcal{CD}(B)$.

Now if one of $\mathcal{CD}(A)$ or $\mathcal{CD}(B)$ has height $0$, say
$\mathcal{CD}(A)$, then $\mathcal{CD}(A)=\\{A\\}$ and so $A$ is abelian, and
the result follows.

Otherwise, since $\mathcal{CD}(G)$ has height $2$ or $3$, we have that at
least one of $\mathcal{CD}(A)$ or $\mathcal{CD}(B)$ has height $1$, say
$\mathcal{CD}(A)$. So $\mathcal{CD}(A)=\\{A,\textbf{Z}(A)\\}$. By Theorem B we
have that $A\textbf{Z}(B)$ has height 1 in $\mathcal{CD}(G)$, and thus is a
nonabelian atom in $\mathcal{CD}(G)$, a contradiction. ∎

The fun stops after height $3$. Construct the central product $G=Q_{8}*Q_{8}$,
which is of course proper. Note that $\mathcal{CD}(G)$ has height $4$, and all
of the atoms have order $4$ so they are all abelian.

##  3\. Investigating the Chermak–Delgado Lattice via the Central Product.

In this section we show how using the central product and the results from
Section 2 we can obtain new proofs of some results about Chermak–Delgado
lattice.

Let $\mathcal{L}(G)$ denote the lattice of all subgroups of a group $G$. Given
a sublattice $\mathcal{C}$ of $\mathcal{L}(G)$, and given $H\leq K\leq G$, we
denote the interval in $\mathcal{C}$ between $H$ and $K$ by
$[H:K]_{\mathcal{C}}=\\{X\in\mathcal{C}\,\,|\,\,H\leq X\leq K\\}$. For
example, when $G=Q_{8}$ the quaternion group of order 8, then
$[\langle-1\rangle\ :G]_{\mathcal{L}(G)}$ contains 5 groups:
$\langle-1\rangle,\langle i\rangle,\langle j\rangle,\langle k\rangle,$ and
$G$.

The next lemma originates from [4, Proposition 1.5].

######  Lemma 10.

Let $G$ be a finite group and let $H\in\mathcal{CD}(G)$. Then
$\mathcal{CD}(H\textbf{C}_{G}(H))=[\textbf{Z}(H):H\textbf{C}_{G}(H)]_{\mathcal{CD}(G)}$.

###### Proof.

First note that $H\textbf{C}_{G}(H)$ is a central product and so
$\textbf{Z}(H\textbf{C}_{G}(H))=\textbf{Z}(H)\textbf{Z}(\textbf{C}_{G}(H))$.

Now since $H\in\mathcal{CD}(G)$, we have that
$\textbf{Z}(\textbf{C}_{G}(H))=\textbf{C}_{G}(\textbf{C}_{G}(H))\cap\textbf{C}_{G}(H)=H\cap\textbf{C}_{G}(H)=\textbf{Z}(H)$.
And so $\textbf{Z}(H\textbf{C}_{G}(H))=\textbf{Z}(H)$.

We establish that for any $X$ with $\textbf{Z}(H)\leq X\leq
H\textbf{C}_{G}(H)$, we have that
$\textbf{C}_{G}(X)=\textbf{C}_{H\textbf{C}_{G}(H)}(X)$.

Suppose that $H\cap\textbf{C}_{G}(H)\leq X\leq H\textbf{C}_{G}(H)$. Then
$\textbf{C}_{G}(H\textbf{C}_{G}(H))\leq\textbf{C}_{G}(X)\leq\textbf{C}_{G}(H\cap\textbf{C}_{G}(H))$,
and since $H\in\mathcal{CD}(G)$, we have that
$H\cap\textbf{C}_{G}(H)\leq\textbf{C}_{G}(X)\leq H\textbf{C}_{G}(H)$, and
hence $\textbf{C}_{G}(X)=\textbf{C}_{H\textbf{C}_{G}(H)}(X)$.

It follows that $X\in\mathcal{CD}(H\textbf{C}_{G}(H))$ if and only if
$X\in[\textbf{Z}(H):H\textbf{C}_{G}(H)]_{\mathcal{CD}(G)}$. ∎

The following corollary appears in An [1, Theorem 3.4] and [2, Theorem 4.4].
However, as stated at the start of the section, our goal is to show the power
of the central product when investigating the Chermak–Delgado lattice. Of
note, our proof utilizes Lemma 10 and Proposition 1 together with some
standard results about the Chermak–Delgado lattice.

######  Corollary 3.

Let $G$ be a finite group and let $H\in\mathcal{CD}(G)$. Then
$\mathcal{CD}(H)=[\textbf{Z}(H):H]_{\mathcal{CD}(G)}$.

###### Proof.

By Lemma 10,
$\mathcal{CD}(H\textbf{C}_{G}(H))=[\textbf{Z}(H):H\textbf{C}_{G}(H)]_{\mathcal{CD}(G)}$.

So for any $X\in[\textbf{Z}(H):H]_{\mathcal{CD}(G)}$,
$X\in\mathcal{CD}(H\textbf{C}_{G}(H))$. Applying Proposition 1 with the
central product $H\textbf{C}_{G}(H)$, we have that
$X(H\cap\textbf{C}_{G}(H))=X\textbf{Z}(H)=X\in\mathcal{CD}(H)$. Thus
$[\textbf{Z}(H):H]_{\mathcal{CD}(G)}\subseteq\mathcal{CD}(H)$.

To see the reverse containment, note that every subgroup of $\mathcal{CD}(H)$
contains $\textbf{Z}(H)$, and every subgroup of
$\mathcal{CD}(\textbf{C}_{G}(H))$ contains
$\textbf{Z}(\textbf{C}_{G}(H))=\textbf{C}_{G}(\textbf{C}_{G}(H))\cap\textbf{C}_{G}(H)=H\cap\textbf{C}_{G}(H)=\textbf{Z}(H)$.
By Proposition 1,
$\mathcal{CD}(H)\cdot\mathcal{CD}(\textbf{C}_{G}(H))\subseteq\mathcal{CD}(H\textbf{C}_{G}(H))$,
and since $\textbf{Z}(H)\in\mathcal{CD}(H\textbf{C}_{G}(H))$, it follows that
$\textbf{Z}(H)\in\mathcal{CD}(H)$ and
$\textbf{Z}(H)\in\mathcal{CD}(\textbf{C}_{G}(H))$. And so
$\mathcal{CD}(H)\subseteq\mathcal{CD}(H\textbf{C}_{G}(H))$. And since
$\mathcal{CD}(H\textbf{C}_{G}(H))=[\textbf{Z}(H):H\textbf{C}_{G}(H)]_{\mathcal{CD}(G)}$,
we have that $\mathcal{CD}(H)\subseteq[\textbf{Z}(H):H]_{\mathcal{CD}(G)}$. ∎

Of note, Corollary 3 can be used to argue that certain groups never appear in
a Chermak–Delgado lattice for any group $G$. For example, the alternating
group $A_{4}$ can never be in a Chermak–Delgado lattice since
$A_{4}\notin\mathcal{CD}(A_{4})$.

Taking the central product approach also allows us to prove a result of
Tărnăuceanu [15, Corollary 4]. Tărnăuceanu’s result occurs as a corollary to
their classification of groups $G$ satisfying
$\mathcal{CD}(G)=[\textbf{Z}(G):G]_{\mathcal{L}(G)}$. Once again, we are able
to present a central product based proof.

######  Corollary 4.

If $G$ is a finite group satisfying
$\mathcal{CD}(G)=[\textbf{Z}(G):G]_{\mathcal{L}(G)}$ and $H$ is a subgroup of
$G$, then $\mathcal{CD}(H)=[\textbf{Z}(H):H]_{\mathcal{L}(G)}$.

###### Proof.

By definition $\mathcal{CD}(H)\subseteq[\textbf{Z}(H):H]_{\mathcal{L}(G)}$.
For the reverse containment, suppose $X\in[\textbf{Z}(H):H]_{\mathcal{L}(G)}$.

Since $\textbf{Z}(G)\leq\textbf{C}_{G}(H)$, we have that
$H\textbf{C}_{G}(H)\in\mathcal{CD}(G)$. It follows from Corollary 3 that
$\mathcal{CD}(H\textbf{C}_{G}(H))=[\textbf{Z}(H\textbf{C}_{G}(H)):H\textbf{C}_{G}(H)]_{\mathcal{L}(G)}=[\textbf{Z}(H)\textbf{Z}(\textbf{C}_{G}(H)):H\textbf{C}_{G}(H)]_{\mathcal{L}(G)}$.

And so $X\textbf{C}_{G}(H)\in\mathcal{CD}(H\textbf{C}_{G}(H))$, and it follows
by Proposition 1 that
$X(H\cap\textbf{C}_{G}(H))=X\textbf{Z}(H)=X\in\mathcal{CD}(H)$. ∎

## References

  * An [2022a] Lijian An.  Groups whose Chermak-Delgado lattice is a subgroup lattice of an elementary abelian $p$-group.  _Communications in Algebra_ , 50(7):2846–2853, 2022a.
  * An [2022b] Lijian An.  Twisted centrally large subgroups of finite groups.  _Journal of Algebra_ , 604:87–106, 2022b.
  * An et al. [2015] Lijian An, Joseph Phillip Brennan, Haipeng Qu, and Elizabeth Wilcox.  Chermak–Delgado lattice extension theorems.  _Communications in Algebra_ , 43(5):2201–2213, 2015.
  * Brewster and Wilcox [2012] Ben Brewster and Elizabeth Wilcox.  Some groups with computable Chermak–Delgado lattices.  _Bulletin of the Australian Mathematical Society_ , 86(1):29–40, 2012.
  * Brewster et al. [2014a] Ben Brewster, Peter Hauck, and Elizabeth Wilcox.  Groups whose Chermak–Delgado lattice is a chain.  _Journal of Group Theory_ , 17(2):253–279, 2014a.
  * Brewster et al. [2014b] Ben Brewster, Peter Hauck, and Elizabeth Wilcox.  Quasi-antichain Chermak–Delgado lattices of finite groups.  _Archiv der Mathematik_ , 103(4):301–311, 2014b.
  * Brush et al. [2016] Erin Brush, Jill Dietz, Kendra Johnson-Tesch, and Brianne Power.  On the Chermak–Delgado lattices of split metacyclic $p$-groups.  _Involve_ , 9(5):765–782, 2016.
  * Chermak and Delgado [1989] Andrew Chermak and Alberto Delgado.  A measuring argument for finite groups.  _Proceedings of the American Mathematical Society_ , 107(4):907–914, 1989.
  * Cocke [2020] William Cocke.  Subnormality and the Chermak–Delgado lattice.  _Journal of Algebra and its Applications_ , 19(8):2050141, 2020.
  * Fasolă and Tărnăuceanu [2022] Georgiana Fasolă and Marius Tărnăuceanu.  Finite groups with large Chermak–Delgado lattices.  _Bulletin of the Australian Mathematical Society_ , pages 1–5, 2022\.  doi: 10.1017/S0004972722000806.  URL http://dx.doi.org/10.1017/S0004972722000806.
  * Glauberman [2006] George Glauberman.  Centrally large subgroups of finite $p$-groups.  _Journal of Algebra_ , 300:480–508, 2006.
  * Isaacs [2008] I. Martin Isaacs.  _Finite Group Theory_ , volume 92.  American Mathematical Soc., 2008.
  * McCulloch [2018] Ryan McCulloch.  Finite groups with a trivial Chermak–Delgado subgroup.  _Journal of Group Theory_ , 21(3):449–461, 2018\.
  * Morresi Zuccari et al. [2018] Alessandro Morresi Zuccari, Valentina Russo, and Carlo Maria Scoppola.  The Chermak–Delgado measure in finite $p$-groups.  _Journal of Algebra_ , 502:262–276, 2018.
  * Tărnăuceanu [2018] Marius Tărnăuceanu.  A note on the Chermak–Delgado lattice of a finite group.  _Communications in Algebra_ , 46(1):201–204, 2018\.
  * Wilcox [2016] Elizabeth Wilcox.  Exploring the Chermak–Delgado lattice.  _Mathematics Magazine_ , 89(1):38–44, 2016.