# Centralizers of the Riordan Group

Tian-Xiao He1 and Yuanziyi Zhang2
1Department of Mathematics
Illinois Wesleyan University, Bloomington, Illinois 61702, USA
2 Mathematics Department
University of Illinois at Urbana-Champaign, Urbana, IL 61801, USA

###### Abstract

In this paper, we discuss centralizers in the Riordan group. We will see that
Faà di Bruno’s formula is an application of the Fundamental Theorem of Riordan
arrays. Then the composition group of formal power series in ${\cal F}_{1}$ is
studied to construct the centralizers of Bell type and Lagrange type Riordan
arrays. Our tools are the $A$-sequences of Riordan arrays and Faà di Bruno’s
formula. Some combinatorial explanation and discussion about related algebraic
topics are also given.

##  1 Introduction

Riordan arrays are infinite, lower triangular matrices defined by the
generating function of their columns. They form a group, called the Riordan
group (see Shapiro, Getu, Woan and Woodson [21]).

More formally, let us consider the set of formal power series (f.p.s.)
${\mathcal{F}}={\mathbb{K}}[\\![$$z$$]\\!]$ (${{\mathbb{K}}}={{\mathbb{R}}}$
or ${{\mathbb{C}}}$); the _order_ of $f(z)\in{\mathcal{F}}$,
$f(z)=\sum_{k=0}^{\infty}f_{k}z^{k}$ ($f_{k}\in{{\mathbb{R}}}$), is the
minimal number $r\in{\mathbb{N}}$ such that $f_{r}\neq 0$; ${\mathcal{F}}_{r}$
is the set of formal power series of order $r$. Let $d(z)\in{\mathcal{F}}_{0}$
and $h(z)\in{\mathcal{F}}_{1}$; the pair $(d(z),\,h(z))$ defines the (proper)
Riordan array
$D=(d_{n,k})_{n,k\in\mbox{\scriptsize${\mathbb{N}}$}}=(d(z),h(z))$ having

$d_{n,k}=[z^{n}]d(z)h(z)^{k}$ (1)

or, in other words, having $d(z)h(z)^{k}$ as the generating function whose
coefficients make-up the entries of column $k$.

From the fundamental theorem of Riordan arrays (see [21]),

$(d(z),h(z))f(z)=d(z)f(h(z)),$

it is immediate to show that the usual row-by-column product of two Riordan
arrays is also a Riordan array:

$(d_{1}(z),\,h_{1}(z))(d_{2}(z),\,h_{2}(z))=(d_{1}(z)d_{2}(h_{1}(z)),\,h_{2}(h_{1}(z))).$
(2)

The Riordan array $I=(1,\,z)$ acts as an identity for this product, that is,
$(1,\,t)(d\left(z\right),\,h(z))=(d\left(z\right),\,h(z))(1\,,z)=(d\left(z\right),\,h(z))$.
Let $(d\left(z\right),\,h(z))$ be a Riordan array. when its inverse is

$(d\left(z\right),\,h(z))^{-1}=\left(\frac{1}{d(\overline{{h}}(z))},\overline{h}(z),\right)$
(3)

where $\overline{h}(z)$ is the compositional inverse of $h(z)$, i.e.,
$(h\circ\overline{h})(z)=(\overline{h}\circ h)(z)=z$. In this way, the set
$\mathcal{R}$ of proper Riordan arrays forms a group (see [21]).

Here is a list of six important subgroups of the Riordan group (see [21]),
where $d\in{\cal F}_{0}$ and $h\in{\cal F}_{1}$.

  * •

the Appell subgroup ${\cal A}=\\{(d(z),\,z)\\}$.

  * •

the Lagrange (associated) subgroup ${\cal L}=\\{(1,\,h(z))\\}$.

  * •

the Bell subgroup ${\cal B}=\\{(d(z),\,zd(z))\\}$.

  * •

the hitting-time subgroup ${\cal H}=\\{(zh^{\prime}(z)/h(z),\,h(z))\\}$.

  * •

the derivative subgroup ${\cal D}=\\{(h^{\prime}(z),\,h(z))\\}.$

  * •

the checkerboard subgroup ${\cal C}=\\{(d(z),\,h(z))\\},$ where $d$ is an even
function and $h$ is an odd function.

An infinite lower triangular array
$[d_{n,k}]_{n,k\in{{\mathbb{N}}}}=(d(z),h(z))$ is a Riordan array if and only
if a unique sequence $A=(a_{0}\not=0,a_{1},a_{2},\ldots)$ exists such that for
every $n,k\in{{\mathbb{N}}}$ (see Merlini, Rogers, Sprugnoli, and Verri [16])

$d_{n+1,k+1}=a_{0}d_{n,k}+a_{1}d_{n,k+1}+\cdots+a_{n}d_{n,n}.$ (4)

This is equivalent to

$h(z)=zA(h(z))\quad\text{or}\quad z=\bar{h}(z)A(z).$ (5)

Here, $A(z)$ is the generating function of the $A$-sequence. There exists a
unique sequence $Z=(z_{0},z_{1},z_{2},\ldots)$ such that every element in
column $0$ can be expressed as the linear combination

$d_{n+1,0}=z_{0}d_{n,0}+z_{1}d_{n,1}+\cdots+z_{n}d_{n,n},$ (6)

or equivalently,

$d(z)=\frac{1}{1-zZ(h(z))}.$ (7)

######  Definition 1.1.

The centralizer of a subset $S$ of group $G$ is defined to be

$C_{G}(S)=\\{g\in G:gs=sg\mbox{ for all}\,\,s\in S\\}$

It is worth noting that $S$ may be a singleton set. Thus, the centralizer of a
subset $S$ of the group $G$ is the set of all elements in $G$ that fix each
element of $S$ under conjugation, i.e. commute with each element in $S.$

Example 1.1 Let $S=\\{(1+z^{2},z)\\}$. For any $(d,h)\in{{\cal R}}$ in
$C_{{\cal R}}(S)$, we have

$(d,h)(1+z^{2},z)=(1+z^{2},z)(d,h),$

i.e.,

$(d(1+h^{2}),h)=((1+z^{2})d,h),$

which implies $h^{2}=z^{2}$, or equivalently, $h=\pm z$. Thus, $C_{{\cal
R}}(S)=\\{(d,h)\in{{\cal R}}:h=\pm z\\},$ the double Appell subgroup
$A^{\pm}=(d(z),\pm z)$ of the Riordan group ${{\cal R}}$. To guarantee
$C_{{\cal R}}(S)$ contains only elements of ${{\cal A}}$, the Appell subgroup
of ${{\cal R}}$, we need an additional condition, which will be presented in
Theorem 2.7.

A _generalized Riordan array_ with respect to the sequence
$(c_{n})_{n\in\mathbb{N}}$ ($c_{n}\not=0$) is a pair $(g(z),f(z))$ of formal
power series, where $g(z)=\sum_{k=0}^{\infty}g_{k}z^{k}/c_{k}$ and
$f(z)=\sum_{k=1}^{\infty}f_{k}z^{k}/c_{k}$ with $f_{1}\neq 0$. The Riordan
array $(g(z),f(z))$ defines an infinite, lower triangular array
$(d_{n,k})_{0\leq k\leq n<\infty}$ according to the rule:

$d_{n,k}=\left[{\frac{z^{n}}{c_{n}}}\right]g(z)\frac{(f(z))^{k}}{c_{k}},$ (8)

where the functions $g(z)(f(z))^{k}/c_{k}$ are column generating functions of
the Riordan array.

Let $D=(g(z),f(z))=(d_{n,k})_{n,k\in\mathbb{N}}$ be a Riordan array with
respect to $(c_{n})_{n\in\mathbb{N}}$ and let
$h(z)=\sum_{k=0}^{\infty}h_{k}z^{k}/c_{k}$ be the generating function of the
sequence $(h_{n})_{n\in\mathbb{N}}$. Then we have the fundamental theorem of
Riordan arrays (FTRA)

$\sum_{k=0}^{n}d_{n,k}h_{k}=\left[{\frac{z^{n}}{c_{n}}}\right]g(z)h(f(z))\,,$
(9)

or equivalently, $(g(z),f(z))*h(z)=g(z)h(f(z))$ .

With every formal power series $f(z)=\sum_{k=1}^{\infty}f_{k}z^{k}/c_{k}$, we
associate the following infinite lower _iteration matrix_ with respect to
$(c_{n})_{n\in\mathbb{N}}$ (see p. 145 of Comtet [3]):

$B(f(z)):=\left(\begin{array}[]{ccccccc}1&0&0&0&0&0&\cdots\\\
0&B_{1,1}&0&0&0&0&\cdots\\\ 0&B_{2,1}&B_{2,2}&0&0&0&\cdots\\\
0&B_{3,1}&B_{3,2}&B_{3,3}&0&0&\cdots\\\
0&B_{4,1}&B_{4,2}&B_{4,3}&B_{4,4}&0&\cdots\\\
0&B_{5,1}&B_{5,2}&B_{5,3}&B_{5,4}&B_{5,5}&\cdots\\\
\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\ddots\end{array}\right)\,,$

where $B_{n,k}=B_{n,k}(f_{1},f_{2},\ldots)$ is the Bell polynomial with
respect to $(c_{n})_{n\in\mathbb{N}}$, defined as follows:

$\frac{1}{c_{k}}(f(z))^{k}=\sum_{n=k}^{\infty}B_{n,k}\frac{z^{n}}{c_{n}}\,.$
(10)

The most obvious choices for $(c_{n})$ are $(1)$, $(n!)$, and $(n!n!)$ or
$(n!!)$.

Therefore, $B_{n,k}=[z^{n}/c_{n}](f(z))^{k}/c_{k}$, which implies that the
iteration matrix $B(f(z))$ is the Riordan array $(1,f(z))$. Now, the following
important property of the iteration matrix (see Theorem A on p. 145 of Comtet
[3], Roman [18], and Roman and Rota [19] )

$B(f(g(z)))=B(g(z))*B(f(z))$

is trivial in the context of the theory of Riordan arrays, i.e.,

$(1,f(g(z)))=(1,g(z))*(1,f(z))\,.$

In the case where $c_{n}=n!$ we obtain that the following well-known Faà di
Bruno’s formula is an application of the FTRA:

$\sum_{k=1}^{n}B_{n,k}(g_{1},g_{2},\ldots,g_{n-k+1})f_{k}=\left[{\frac{t^{n}}{n!}}\right]f(g(t))\,.$

More precisely, Let $f$ and $g$ be two formal (Taylor) series:

$f=\sum_{k\geq 0}f_{k}\frac{z^{k}}{k!},\quad g=\sum_{k\geq
0}g_{k}\frac{z^{k}}{k!},$

where $g_{0}=0$, and let $h=f\circ g$ with formal Taylor expansion

$h=f\circ g=\sum_{n\geq 0}h_{n}\frac{z^{n}}{n!}.$

Then the coefficients $h_{n}$ are given by the following expression (see
Theorem A on p. 137 of Comtet [3]):

$h_{0}=f_{0},\quad
h_{n}=\sum^{n}_{k=1}f_{k}B_{n,k}(g_{1},g_{2},\ldots,g_{n-k+1}),$

where the $B_{n,k}$ are the exponential Bell polynomials. $B_{n,k}$ can be
presented as

$B_{n,k}(g_{1},g_{2},\ldots,g_{n-k+1})=\sum\frac{n!}{c_{1}!c_{2}!\cdots(1!)^{c_{1}}(2!)^{c_{2}}\cdots}x_{1}^{c_{1}}x_{2}^{c_{2}}\cdots,$

where the summation takes place over all integers $c_{1}$, $c_{2}$,
$c_{3},\cdots\geq 0$ such that there hold (the partition of $n$ with $k$
parts)

$c_{1}+2c_{2}+3c_{3}+\cdots=n,\quad c_{1}+c_{2}+c_{3}+\cdots=k.$

Example 1.2 Consider the problem of finding the $n$-th derivative of
$F(x)=x^{\alpha x}$, where $x>0$ and $\alpha\not=0$ is a real number (cf.
Comtet [3]). By using the Stirling numbers of the first kind, we expand
$f(t)=F(x+t)$ as

$\displaystyle
f(t)=(x+t)^{\alpha(x+t)}=x^{\alpha(x+t)}\left(1+\frac{t}{x}\right)^{\alpha(x+t)}=x^{\alpha
x}x^{\alpha t}\left(1+\frac{t}{x}\right)^{\alpha x\left(1+\frac{t}{x}\right)}$
$\displaystyle=$ $\displaystyle F(x)\exp(\alpha t\log
x)\exp\left(ax\left(1+\frac{t}{x}\right)\log\left(1+\frac{t}{x}\right)\right).$

Notice that there is a well-known expansion on P. 139 of Comtet [3] shown as

$\frac{1}{k!}((1+z)\log(1+z))^{k}=\sum_{n\geq k}b(n,k)\frac{z^{n}}{n!},$

where $b(0,0)=1$ and

$b(n,k)=\sum_{k\leq\ell\leq n}\binom{\ell}{k}k^{\ell-k}s(n,\ell)$

with the Stirling numbers $s(n,\ell)$ of the first kind. The numbers $b(n,k)$,
called Lehmer-Comtet numbers of first kind, follow the recurrence relation

$b(n+1,k)=(k-n)b(n,k)+b(n,k-1)+nb(n-1,k-1),$

and the first few values are listed below

$\left(\begin{array}[]{cccccccc}1&0&0&0&0&0&0&\cdots\\\
0&1&0&0&0&0&0&\cdots\\\ 0&1&1&0&0&0&0&\cdots\\\ 0&-1&3&1&0&0&0&\cdots\\\
0&2&-1&6&1&0&0&\cdots\\\ 0&-6&0&5&10&1&0&\cdots\\\
0&24&4&-15&25&15&1&\cdots\\\
\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\ddots\end{array}\right)\,.$

More properties of Lehmer-Comtet numbers of first kind can be found in OEIS
A008296, P. 139 of [3], and Gould [7]. However, the combinatorial explanation
seems not to be known. By using Lehmer-Comtet numbers of first kind, we obtain
consequently

$f(t)=F(x)\sum_{j\geq 0,k\leq m}b(m,k)\frac{(\alpha t\log
x)^{j}(t/x)^{m}}{j!m!}(\alpha x)^{k},$

which implies

$f_{n}=\frac{d^{n}(x^{\alpha x})}{dx^{n}}=\alpha^{n}x^{\alpha
x}\sum^{n}_{j=0}\left[\binom{n}{j}u^{j}\sum^{n-j}_{k=0}b(n-j,n-k-j)v^{k}\right],$

where $u=\log x$ and $v=(\alpha x)^{-1}$. For instance, by using the above
table of $b(n.k)$ we get

$\displaystyle f_{4}=\frac{d^{4}(x^{\alpha x})}{dx^{4}}$ $\displaystyle=$
$\displaystyle\alpha^{4}x^{\alpha
x}\sum^{4}_{j=0}\left[\binom{4}{j}u^{j}\sum^{4-j}_{k=0}b(4-j,4-k-j)v^{k}\right]$
$\displaystyle=$ $\displaystyle\alpha^{4}x^{\alpha
x}\left[\sum^{4}_{k=0}b(4,4-k)v^{k}+4u\sum^{3}_{k=0}b(3,3-k)v^{k}\right.$
$\displaystyle\quad\left.+6u^{2}\sum^{2}_{k=0}b(2,2-k)v^{k}+3u^{3}\sum^{1}_{k=0}b(1,1-k)v^{k}+u^{4}\right]$
$\displaystyle=$ $\displaystyle\alpha^{4}x^{\alpha
x}\left[1+6v-v^{2}+2v^{3}+4u(1+3v-v^{2})+6u^{2}(1+v)+4u^{3}+u^{4}\right].$

Additionally, for any $f\in{\cal F}_{1}$, the inverse array
$(1,\bar{f})=B(\bar{f})$ is also an iteration matrix. Thus, the set of
iteration matrices with respect to $(c_{n})_{n\in\mathbb{N}}$, denoted by
$\mathcal{B}$, is a nonempty subset of the Riordan group $\mathcal{R}$ with
respect to $(c_{n})_{n\in\mathbb{N}}$, closed under multiplication and taking
inverses in $\mathcal{R}$. These indicate that $\mathcal{B}$ is a subgroup of
$\mathcal{R}$ and we call it the _associated subgroup_.

If ${\displaystyle f(x)=\sum_{n=0}^{\infty}{\frac{b_{n}}{n!}}x^{n}}$ and
${\displaystyle g(x)=\sum_{n=1}^{\infty}{\frac{a_{n}}{n!}}x^{n}}$ and
${\displaystyle f(g(x))=\sum_{n=0}^{\infty}{\frac{d_{n}}{n!}}x^{n},}$ then the
coefficient $d_{n}$ (which would be the nth derivative of $f\circ g$ evaluated
at $0$ if we were dealing with convergent series rather than formal power
series) is given by

${\displaystyle
d_{n}=\sum_{\pi=\left\\{B_{1},\ldots,B_{k}\right\\}}a_{\left|B_{1}\right|}\cdots
a_{\left|B_{k}\right|}b_{k}}$

where $\pi$ runs through the set of all partitions of the set $\\{1,...,n\\}$
and $B_{1}$, $\ldots$, $B_{k}$ are the blocks of the partition $\pi$, and
$|B_{j}|$ is the number of members of the $j$th block, for $j=1,\ldots,k.$
This version of the formula is equivalent to Faá di Bruno’s formula for
exponential formal power series shown before, i.e.,

${\displaystyle
f(g(x))=b_{0}+\sum_{n=1}^{\infty}{\frac{\sum_{k=1}^{n}b_{k}B_{n,k}(a_{1},\ldots,a_{n-k+1})}{n!}}x^{n},}$

where $B_{n,k}(a_{1},\ldots,a_{n-k+1})$ are Bell polynomials. Faá di Bruno’s
formula is useful in finding centralizers, commutative compositions, Riordan
involutions and pseudo-involutions. The centralizers related to the Riordan
pseudo-involution multiplication and palindromes are shown in a recent paper
by Shapiro and one of the authors [10].

In the next section, we will use Faá di Bruno’s formula and sequence
characterization of Riordan arrays to construct centralizers of Bell type
Riordan arrays and Riordan arrays of the form $(1/(1-kz),f)$,
$f\in{{\mathcal{F}}_{1}}$. Some combinatorial explanation are also given. In
Section $3$, we will study the method of the construction of centralizers of
the Riordan arrays related to the composition group of formal power series in
${\mathcal{F}}_{1}$ and its application for Lagrange type Riordan arrays and
Bell type Riordan arrays. Section $4$ diverts to some related topics such as
reversers, conjugates, commutators, twisted subgroup, and pseudo-Riordan
involutions.

##  2 Centralizers of the Riordan group

Suppose $f(x)=\sum_{n=0}^{\infty}{a_{n}}x^{n}$ and
$g(x)=\sum_{n=0}^{\infty}{b_{n}}x^{n}$ are formal power series and $b_{0}=0.$
Then the composition $f\circ g$ is again a formal power series,

$f(g(x))=\sum_{n=0}^{\infty}{c_{n}}x^{n},$ (11)

where $c_{0}=a_{0}$ and the other coefficients $c_{n}$ for $n\geq 1$ can be
expressed as a sum over compositions of $n$ or as an equivalent sum over
partitions of $n$:

$c_{n}=\sum_{\mathbf{i}\in{\mathcal{C}}_{n}}a_{k}b_{i_{1}}b_{i_{2}}\cdots
b_{i_{k}},$

where

${\mathcal{C}}_{n}=\\{(i_{1},i_{2},\dots,i_{k})\,:\ 1\leq k\leq n,\
i_{1}+i_{2}+\cdots+i_{k}=n\\}$

is the set of compositions of n with k denoting the number of parts, or

$c_{n}=\sum_{k=1}^{n}a_{k}\sum_{\mathbf{\pi}\in{\mathcal{P}}_{n,k}}{\binom{k}{\pi_{1},\pi_{2},...,\pi_{n}}}b_{1}^{\pi_{1}}b_{2}^{\pi_{2}}\cdots
b_{n}^{\pi_{n}},$ (12)

where

${\mathcal{P}}_{n,k}=\\{(\pi_{1},\pi_{2},\dots,\pi_{n})\,:\
\pi_{1}+\pi_{2}+\cdots+\pi_{n}=k,\ \pi_{1}\cdot 1+\pi_{2}\cdot
2+\cdots+\pi_{n}\cdot n=n\\}$

is the set of partitions of n into k parts, in frequency-of-parts form.

The first form is obtained by picking out the coefficient of $x^{n}$ in
$(b_{1}x+b_{2}x^{2}+\cdots)^{k}$ ”by inspection”, and the second form is then
obtained by collecting like terms, or alternatively, by applying the
multinomial theorem.

The product of two Riordan arrays is a Riordan array. Sprugnoli and one of the
authors [11] study the problem of how do the $A$-sequence and $Z$-sequence of
the product depend on the analogous sequences of the two factors. More
precisely, consider two proper Riordan arrays $D_{1}=(d_{1},h_{1})$ and
$D_{2}=(d_{2},h_{2})$ and their product,

$D_{3}=D_{1}D_{2}=(d_{1}d_{2}(h_{1}),h_{2}(h_{1})).$

######  Theorem 2.1.

[11] Let $A_{i}(z)$, $i=1,2,$ and $3$, be the generating functions of
$A$-sequences of $D_{i}$, $i=1,2,$ and $3$, respectively. Then

$A_{3}(z)=A_{2}(z)A_{1}\left(\frac{z}{A_{2}(z)}\right).$

Let $h\in{\cal F}_{1}$. We define

$A_{h}(z)=\frac{z}{\bar{h}(z)}$ (13)

where $\bar{h}$ is the compositional inverse of $h$. Thus, from Theorem 2.1,
we have the following result.

######  Theorem 2.2.

Let $A_{h}$ be defined by (13), and let $h,f\in{\cal F}_{1}$. Then

$A_{h\circ f}(z)=A_{f}(z)A_{h}\left(\frac{z}{A_{f}(z)}\right).$

######  Corollary 2.3.

Let $A_{h}$ be defined by (13), and let $h,f\in{\cal F}_{1}$. Then $h\circ
f=f\circ h$ if and only if

$A_{f}(z)A_{h}\left(\frac{z}{A_{f}(z)}\right)=A_{h}(z)A_{f}\left(\frac{z}{A_{h}(z)}\right).$
(14)

Particularly, $h\circ f=f\circ h=z$ if and only if

$A_{f}(z)A_{h}\left(\frac{z}{A_{f}(z)}\right)=A_{h}(z)A_{f}\left(\frac{z}{A_{h}(z)}\right)=1.$
(15)

######  Proposition 2.4.

Let $B$ be the Bell subgroup of the Riordan group, and let $(h/z,h)\in B$.
Then the centralizer of $(h/z,h)$ in the subgroup $B$ is

$C_{B}((h/z,h))=\\{(f/z,f)\in B:h\circ f=f\circ h\\}.$

###### Proof.

Let $(f/z,f)\in B$ with $h\circ f=f\circ h$. Then

$(h/z,h)(f/z,f)=((f\circ h)/z,f\circ h)=((h\circ f)/z,h\circ
f)=(f/z,f)(h/z,h),$

completing the proof. ∎

######  Theorem 2.5.

Let $S=\\{(1/(1-z),z/(1-z))\\}$, and let $B$ be the Bell subgroup of the
Riordan group. Then the centralizer

$C_{B}(S)=\\{(1/(1-rz),z/(1-rz)):r\in{{\mathbb{R}}}\\}.$ (16)

###### Proof.

From Proposition 2.4, to obtain (14), it is sufficient to prove the set of all
elements in ${\cal F}_{1}$ that commute with $h=z/(1-z)$ is

$E=\\{z/(1-rz):r\in{{\mathbb{R}}}\\}.$ (17)

Let $f\in{\cal F}_{1}$, and let $A_{f}$ and $A_{h}$ be defined by (13). Since
$\bar{h}=z/(1+z)$, $A_{h}(z)=(1+z)$. From Corollary 2.3, $h\circ f=f\circ h$
if and only if (14) holds, i.e.,

$A_{f}(z)\left(1+\frac{z}{A_{f}(z)}\right)=(1+z)A_{f}\left(\frac{z}{1+z}\right),$

or equivalently,

$A_{f}(z)+z=(1+z)A_{f}\left(\frac{z}{1+z}\right).$

Substituting $z=z/(1-z)$ into above equation yields

$(1-z)A_{f}\left(\frac{z}{1-z}\right)+z=A_{f}(z).$ (18)

Denote $A_{f}(z)=\sum_{k\geq 0}a_{k}z^{k}$ and substitute it into (18):

$(1-z)(a_{0}+a_{1}(z+z^{2}+\cdots)+a_{2}(z+z^{2}+\cdots)^{2}+\cdots)=a_{0}+a_{1}z+a_{2}z^{2}+\cdots.$

By comparing the coefficients of powers $z^{0}$, $z^{1}$, and $z^{2}$ on both
sides of the resulting equation, we obtain $a_{0}=a_{0}$,

$-a_{0}+a_{1}+1=a_{1}\quad\mbox{and}\quad a_{1}-a_{1}+a_{2}=a_{2},$

which implies $a_{0}=1$ and $a_{1}$ is free. To determine $a_{n}$ for $n\geq
2$, we use the Lagrange Inversion Formula

$[z^{n}]h(g(z))=\frac{1}{n}[z^{-1}]\frac{h^{\prime}(z)}{\bar{g}^{n}}$

for $n\geq 3$ to $A_{f}(z/(1-z))$ in (18), where $g\in{\cal F}_{1}$ and
$\bar{g}$ is the compositional inversion of $g$. Then we can write the
coefficients of $z^{n}$ on the lefthand side of (18) as

$\displaystyle[z^{n}]A_{f}\left(\frac{z}{1-z}\right)-[z^{n-1}]A_{f}\left(\frac{z}{1-z}\right)$
(19) $\displaystyle=$
$\displaystyle\frac{1}{n}[z^{-1}]\frac{A_{f}^{\prime}(Z)}{\left(\frac{z}{1+z}\right)^{n}}-\frac{1}{n-1}[z^{-1}]\frac{A_{f}^{\prime}(Z)}{\left(\frac{z}{1+z}\right)^{n-1}}$
$\displaystyle=$
$\displaystyle\frac{1}{n}[z^{n-1}](1+z)^{n}A_{f}^{\prime}(z)-\frac{1}{n-1}[z^{n-2}](1+z)^{n-1}A_{f}^{\prime}(Z)$
$\displaystyle=$
$\displaystyle\frac{1}{n}\sum^{n-1}_{k=0}\binom{n}{k}(n-k)a_{n-k}-\frac{1}{n-1}\sum^{n-2}_{k=0}\binom{n-1}{k}(n-k-1)a_{n-k-1}$
$\displaystyle=$
$\displaystyle\sum^{n-1}_{k=0}\binom{n-1}{k}a_{n-k}-\sum^{n-2}_{k=0}\binom{n-2}{k}a_{n-k-1}$
$\displaystyle=$ $\displaystyle
a_{n}+\sum^{n-1}_{k=1}\left(\binom{n-1}{k}-\binom{n-2}{k-1}\right)a_{n-k}$
$\displaystyle=$ $\displaystyle a_{n}+\sum^{n-2}_{k=1}\binom{n-2}{k}a_{n-k}.$

Comparing the coefficients of $z^{n}$ on both sides of (18), we have

$a_{n}+\sum^{n-2}_{k=1}\binom{n-2}{k}a_{n-k}=a_{n},$

or equivalently,

$\sum^{n-2}_{k=1}\binom{n-2}{k}a_{n-k}=0.$

Using the above equation and noting $a_{2}=0$ when $n=3$, we may prove all
$a_{n}=0$ for $n\geq 2$ inductively. Thus $A_{f}(t)=1+a_{1}t$ and
$f(t)=t/(1-rt)$ for $r=a_{1}\in{{\mathbb{R}}}$. ∎

We now give a combinatorial proof of Theorem 2.5 based on a combinatorial
interpretation of expression of (12).

######  Theorem 2.6.

Let $f(x)=\sum_{n=0}^{\infty}{a_{n}}x^{n}$ and
$g(x)=\sum_{n=0}^{\infty}{b_{n}}x^{n}$ be formal power series and $b_{0}=0$,
and let $f\circ g$ be the composition of $f$ and $g$ shown in (11),
$f(g(x))=\sum_{n=0}^{\infty}{c_{n}}x^{n}$, where $c_{0}=a_{0}$ and the other
coefficients $c_{n}$ for $n\geq 1$ are expressed as a sum over compositions of
$n$ or as an equivalent sum over partitions of $n$ shown in (12). Then

$c_{n}=\sum_{k=1}^{n}B_{n,k}a_{k},$ (20)

where $B_{n,k}=\sum b_{j_{1}}b_{j_{2}}...b_{j_{k}}$, where
$|j|=j_{1}+j_{2}+...+j_{k}=n$ and $B_{n,k}$ could be interpreted as the
combinations of ways to put $n$ non-distinguished objects into $k$ nonempty
distinguished (labeled) boxes.

###### Proof.

As an example of $B_{n,k}$, $B_{2,2}={b_{1}}^{2}$ can be interpreted as
combinations of ways to put $2$ objects into $2$ boxes. Similarly,
$B_{4,2}=2b_{1}b_{3}+{b_{2}}^{2}$. We now give a combinatorial proof of
Theorem 2.5. Let $h(t)=\frac{1}{1-t}=\sum_{k=1}^{\infty}a_{k}t^{k}$ where
$a_{k}=1$ for all $k\in\mathbb{N}$. We want to find elements $f$ of
$C_{B}(h)$, $f(t)=\sum_{k=1}^{\infty}b_{k}t^{k}$ where $b_{1}=1$ such that
$h(f)=f(h)$. Let $h(f)=\sum_{k=1}^{\infty}c_{k}t^{k}$ and
$f(h)=\sum_{k=1}^{\infty}\hat{c}_{k}t^{k}$. We first compute the first few
terms.

For $k=1$, we have

$c_{1}=B_{1,1}a_{1}=b_{1}a_{1}.$

Switch the terms accordingly, we obtain

$\hat{c}_{1}=a_{1}b_{1}.$

Then $c_{1}=\hat{c}_{1}$ naturally.

For $k=2$, we have

$c_{2}=B_{2,1}a_{1}+B_{2,2}a_{2}=b_{2}a_{1}+{b_{1}}^{2}a_{2},$
$\hat{c}_{2}=a_{2}b_{1}+{a_{1}}^{2}b_{2}.$

Plugging in $a_{k}=1$ and $b_{1}=1$, and then equate $c_{2}$ and $\hat{c}_{2}$
we have

$b_{2}+1=1+b_{2}.$

Therefore, we conclude that $b_{2}$ is free.

For $k=3$, we have

$\begin{split}c_{3}&=B_{3,1}a_{1}+B_{3,2}a_{2}+B_{3,3}a_{3}\\\
&=a_{1}b_{3}+a_{2}(b_{1}b_{2}+b_{2}b_{1})+a_{3}{b_{1}}^{3}\\\
&=a_{1}b_{3}+2a_{2}b_{1}b_{2}+a_{3}{b_{1}}^{3},\end{split}$
$\hat{c}_{3}=b_{1}a_{3}+2b_{2}a_{1}a_{2}+b_{3}{a_{1}}^{3}.$

Equate $c_{3}$ and $\hat{c}_{3}$, the equation holds naturally.

For $k=4$, we have

$\begin{split}c_{4}&=B_{4,1}a_{1}+B_{4,2}a_{2}+B_{4,3}a_{3}+B_{4,4}a_{4}\\\
&=a_{1}b_{4}+a_{2}(2b_{1}b_{3}+{b_{2}}^{2})+a_{3}(3b_{1}b_{1}b_{2})+a_{4}{b_{1}}^{4}\\\
&=a_{1}b_{4}+2a_{2}b_{1}b_{3}+a_{2}{b_{2}}^{2}+3a_{3}b_{1}b_{1}b_{2}+a_{4}{b_{1}}^{4},\end{split}$
$\hat{c}_{4}=b_{1}a_{4}+2b_{2}a_{1}a_{3}+b_{2}{a_{2}}^{2}+3b_{3}a_{1}a_{1}a_{2}+b_{4}{a_{1}}^{4}.$

Plugging in $a_{k}=1$ and $b_{1}=1$, and then equate $c_{4}$ and $\hat{c}_{4}$
we have

$b_{4}+2b_{3}+{b_{2}}^{2}+3b_{2}+1=1+2b_{2}+b_{2}+3b_{3}+b_{4},$
$2b_{3}+{b_{2}}^{2}=3b_{3}.$

Therefore, we conclude that $b_{3}={b_{2}}^{2}$.

For $k=5$, we have

$\begin{split}c_{5}&=B_{5,1}a_{1}+B_{5,2}a_{2}+B_{5,3}a_{3}+B_{5,4}a_{4}+B_{5,5}a_{5}\\\
&=a_{1}b_{5}+a_{2}(2b_{1}b_{4}+2b_{2}b_{3})+a_{3}({3b_{1}}^{2}b_{3}+3b_{1}{b_{2}}^{2})+a_{4}(4{b_{1}}^{3}b_{2})+a_{5}{b_{1}}^{5},\end{split}$
$\hat{c}_{5}=b_{1}a_{5}+b_{2}(2a_{1}a_{4}+2a_{2}a_{3})+b_{3}(3{a_{1}}^{2}a_{3}+3a_{1}{a_{2}}^{2})+b_{4}(4{a_{1}}^{3}a_{2})+b_{5}{a_{1}}^{5}.$

Equate $c_{5}$ and $\hat{c}_{5}$ we have

$2b_{4}+2b_{2}b_{3}+3b_{3}+3{b_{2}}^{2}=6b_{3}+4b_{4}.$

Therefore, we conclude that $b_{4}={b_{2}}^{3}$.

Base on the computational results above, we make the conjecture that
$b_{p}={b_{2}}^{p-1}$ for all $p\in\mathbb{N}$ and $p\geq 2$. We prove this
result by induction.

For $p=1$, the result holds naturally as demonstrated above. Suppose now the
result holds up to $p=q$. Since $a_{k}=1$ for all $k\in\mathbb{N}$, as a
matter of simplicity, we put the coefficients $B_{n,k}$ as the entries of a
lower triangular matrix below,

$\begin{pmatrix}1\\\ b_{2}&1\\\ {b_{2}}^{2}&2b_{2}&1\\\
{b_{2}}^{3}&3{b_{2}}^{2}&3b_{2}&1\\\ \vdots&&&&\ddots\\\
{b_{2}}^{q}&q{b_{2}}^{q-1}&\cdots&&\cdots&1\\\ \end{pmatrix}.$

Observe that the matrix above is a Riordan array with A-sequence
$a_{0}=1,a_{1}=b_{2}$.

Now we want to show that the result holds for $p=q+1$. It is sufficient to
show that adding another row and column to the matrix above, we still obtain a
Riordan array preserving the same A-sequence. First, we consider the
combinatorial interpretation of $B_{q+1,k}$ for $k\geq 3$. If we have $k$
boxes and $q+1$ objects, each of the boxes contains at most $q-k+2$ objects,
i.e., the subscript of the terms that appear in $B_{q+1,k}$ is at most $q-k$.
Since $q-k+2<q$, by induction, we conclude that the $k$th entries of the
$q+1$th row preserve the same A-sequence for $k\geq 3$.

It can be seen that the matrix of the size $(q+1)\times(q+1)$ is uniquely
determined. Now we need to check if the result holds for the two entries
$B_{q+1,1}$ and $B_{q+1,2}$. Again, by the combinatorial interpretation, if we
contract the objects of the entry $B_{q,1}$ into one, the entry $B_{q+1,1}$
can be interpreted as adding an object to the same box. Then it follows that
$B_{q+1,1}=B_{q,1}\times b_{2}$. Similarly, the entry $B_{q+1,2}$ can be
obtained by adding one object and one box to $B_{q,1}$ and adding one object
into one existing box of $B_{q,2}$. It follows that
$B_{q+1,2}=B_{q,1}+B_{q,2}\times b_{2}$.

Therefore, we conclude that the result indeed holds. Furthermore, if
$b_{p}={b_{2}}^{p-1}$, the corresponding power series is indeed
$\frac{t}{1-b_{2}t}$. ∎

We now return to the problem raised in Example $1.1$.

######  Theorem 2.7.

Let $S$ be a subset of Appell subgroup ${{\cal A}}$ of the Riordan group
${{\cal R}}$ defined by $S=\\{(g,f):(g,f)\in{{\cal
A}},g^{\prime}(0)\not=0\\}$. Then ${C}_{{\cal R}}(S)={{\cal A}}$.
Particularly, if $(g,f)$ does not satisfy the condition $g^{\prime}(0)\not=0$,
then $C_{{\cal R}}$ contains elements that do not belong to ${{\cal A}}$. In
general, $C_{{\cal R}}((g,t),g\in{\cal F}_{0})=A^{\pm}$, the double Appell
subgroup of ${{\cal R}}$.

###### Proof.

Let $(g,f)\in S\subset{{\cal A}}$ with $g^{\prime}(0)\not=0$. Then
$(g,f)=(g,t)$ and $g=\sum_{k\geq 0}g_{k}t^{k}$ with $g_{0},g_{1}\not=0$. A
Riordan array $(d,h)\in{{\cal C}}_{{\cal R}}(S)$ with $h=\sum_{k\geq
0}h_{k}t^{k}$ if and only if

$(d,h)(g,t)=(g,t)(d,h);\,\,i.e.,\,\,(d(g\circ h),h)=(gd,h).$

Hence, $g(h)=g$. From (11), we have

$g(h(t)=\sum_{n\geq 0}c_{n}t^{n},$ (21)

where, by denoting ${\mathcal{C}}_{n}=\\{(i_{1},i_{2},\dots,i_{k})\,:\ 1\leq
k\leq n,\ i_{1}+i_{2}+\cdots+i_{k}=n\\}$,

$\displaystyle
c_{n}=\sum_{\mathbf{i}\in{\mathcal{C}}_{n}}g_{k}h_{i_{1}}h_{i_{2}}\cdots
h_{i_{k}}$ $\displaystyle=$ $\displaystyle
g_{1}h_{n}+g_{n}h_{1}^{n}+\sum_{\mathbf{i}\in{\mathcal{C}}^{\prime}_{n}}g_{k}h_{i_{1}}h_{i_{2}}\cdots
h_{i_{k}},$

in which ${\mathcal{C}}^{\prime}_{n}=\\{(i_{1},i_{2},\dots,i_{k})\,:\ 2\leq
k\leq n-1,\ i_{1}+i_{2}+\cdots+i_{k}=n\\}$. It is clear that the summation in
the last equation contains terms involving only $h_{k}$ with $1\leq k\leq n-1$
because

$i_{1}+i_{2}+\cdots i_{k}=n$

and $k\geq 2$ imply $1\leq i_{j}\leq n-1$ for all $j=1,2,\ldots k$. In
addition, the summation contains terms involving $g_{k}$ with $2\leq k\leq
n-1$, but not $g_{1}$ and $g_{n}$. For instance,

$\displaystyle c_{1}=g_{1}h_{1},$ $\displaystyle
c_{2}=g_{1}h_{2}+g_{2}h_{1}^{2}$ $\displaystyle
c_{3}=g_{1}h_{3}+g_{3}h_{1}^{3}+2g_{2}h_{1}h_{2},\,\ldots$

Comparing the coefficients of the same powers of $g(h)=g$ yields

$g_{n}=c_{n}=g_{1}h_{n}+g_{n}h_{1}^{n}+\sum_{\mathbf{i}\in{\mathcal{C}}^{\prime}_{n}}g_{k}h_{i_{1}}h_{i_{2}}\cdots
h_{i_{k}}.$ (22)

Therefore, for $n=1$, we have

$g_{1}=g_{1}h_{1}.$

Since $g_{1}\not=0$, we obtain $h_{1}=1$. Considering $n=2$, we have

$g_{2}=c_{2}=g_{1}h_{2}+g_{2}h^{2}_{1}=g_{1}h_{2}+g_{2},$

which implies $h_{2}=0$. Similarly, we have $h_{3}=0$. Assume that $h_{k}=0$
for $2\leq k\leq n-1$. Then from

$\displaystyle
g_{n}=c_{n}=g_{1}h_{n}+g_{n}h_{1}^{n}+\sum_{\mathbf{i}\in{\mathcal{C}}^{\prime}_{n}}g_{k}h_{i_{1}}h_{i_{2}}\cdots
h_{i_{k}}$ $\displaystyle=$ $\displaystyle g_{1}h_{n}+g_{n}$

we obtain $h_{n}=0$. Thus, $h=t$, which means ${C}_{{\cal R}}(S)={{\cal A}}$.
Combining the above and Example 1.1, we complete the proof. ∎

Remark 2.1 In Theorem 2.7, the set $S$ may be a singleton set. From Example
$1.1$, we may know that the condition $g^{\prime}(0)\not=0$ is necessary to
insure that ${C}_{{\cal R}}(S)={{\cal A}}$. In addition, this condition can
not be replaced to be $g(z)\not=1$ since $C_{{\cal R}}(1+z^{2},z)={{\cal
A}}^{\pm}$, which is shown in Example $1.1$. For instance $(g,-z)\in{C}_{{\cal
R}}(1+z^{2},z)$. As examples of nontrivial $g$ with $g^{\prime}(0)=0$, we may
see $g(z)=(1-z+z^{2})/(1-z)$, $(1-2z-\sqrt{1-4z})/(2z)$, etc.

##  3 The subgroup of ${\cal F}_{1}$ and centralizers of Riordan group

Let $h(z)=z(1+\sum_{i\geq 1}b_{i}z^{i})$ and $g(z)=z(1+\sum_{i\geq
1}a_{i}z^{i})$. Then $(g\circ h)(z)=z(1+\sum_{i\geq 1}d_{i}z^{i})$.
Successively carrying out multiplications and combining similar terms, we
obtain the following expressions for the canonical coordinates $d_{i}$:

$\displaystyle d_{1}=a_{1}+b_{1},\quad(4_{1})$ $\displaystyle
d_{2}=a_{2}+b_{2}+2a_{1}b_{1},\quad(4_{2})$ $\displaystyle
d_{3}=a_{3}+b_{3}+2a_{1}b_{2}+3a_{2}b_{1}+a_{1}b_{1}^{2},\quad(4_{3})$
$\displaystyle\cdots\cdots$ $\displaystyle
d_{n}=a_{n}+b_{n}+\sum^{n-1}_{i=1}\alpha_{i}\phi_{n-i}(b_{1},b_{2},\ldots,b_{n-i}).\quad(4_{n})$

The expressions $\phi_{r}(b_{1},b_{2},\ldots,b_{r})$ are universal homogeneous
polynomials of multidegree $r$ with integer coefficients. It can be seen that
the formulae $(4_{k})$, $k=1,2,\ldots,n$, define a group structure on the
direct product ${{\mathbb{K}}}^{n}$. This group will be denoted by
${\mathcal{F}}^{(n)}_{1}({{\mathbb{K}}})$. Clearly,
${\mathcal{F}}^{(1)}_{1}({{\mathbb{K}}})$ is an Abelian group isomorphic to
the additive group of the ring ${{\mathbb{K}}}$. The group
${\mathcal{F}}^{(2)}_{1}({\mathbb{K}})$ is isomorphic to the additive group of
${{\mathbb{K}}}\oplus{{\mathbb{K}}}$, and the corresponding isomorphism
depends non-linearly on the canonical coordinates $(b_{1},b_{2})$, etc. (see
Babenko [2]).

######  Proposition 3.1.

$({\cal F}_{1},\circ)$ is a group with the identity $z$. For any $f\in{\cal
F}_{1}$, the inverse of $f$ is its compositional inverse.

The properties of the substitution group of power series are discussed by
Jennings [13], Johnson [14], and Babenko [2].

We denote the centralizers of a subset $S\in{\cal F}_{1}$ as $C_{{\cal
F}_{1}}(S)=\\{h\in{\cal F}_{1}:h\circ f=f\circ h\,\,\mbox{for all}\,\,f\in
S\\}.$ We now use the centralizers of ${\cal F}_{1}$ to describe some
centralizers of ${\cal R}$.

######  Proposition 3.2.

The centralizer of a Lagrange type Riordan array $(1,h)$ with $h\in{\cal
F}_{1}$, $h^{\prime}(0)=1$, is

$C_{{\cal R}}(1,h)=\left\\{\begin{array}[]{ll}{\cal R}&if\,h=z\\\
\\{(1,f):f\in C_{{\cal F}_{1}}(h)\\}&if\,h\not=z.\end{array}\right.$ (23)

###### Proof.

It is obvious that if $h=z$, then $C_{\cal R}(1,z)={\cal R}$.

Suppose $h(z)=\sum_{n\geq 1}h_{n}z^{n}\not=z$, thus there exists a positive
integer $\ell$ such that $h_{\ell}\not=0$ and
$h_{2}=h_{3}=\cdots=h_{\ell-1}=0$. Without loss of generality, we may assume
$h_{1}=1$. From compositional formula (11) of $g\in{\cal F}_{0}$, where
$g(z)=\sum_{n\geq 0}g_{n}z^{n}$ with $g_{0}=1$, and the above $h\in{\cal
F}_{1}$, we have $(g\circ h)(z)=\sum_{n\geq 0}d_{n}z^{n}$, where
$d_{0}=g_{0}=1$ and the other coefficient $c_{n}$ for $n\geq 1$ can be
expressed as a sum over compositions of $n$ or as an equivalent sum over
partitions of $n$:

$d_{n}=\sum_{\mathbf{i}\in{\mathcal{D}}_{n}}g_{k}h_{i_{1}}h_{i_{2}}\cdots
h_{i_{k}},$

where

${\mathcal{D}}_{n}=\\{(i_{1},i_{2},\dots,i_{k})\,:\ 1\leq k\leq n,\
i_{1}+i_{2}+\cdots+i_{k}=n\\}$

is the set of compositions of $n$ with $k$ denoting the number of parts. If
$(g,f)\in C_{\cal R}(1,h)$, then $(g,f)(1,h)=(1,h)(g,f)$ implies

$h(f)=f(h)\quad\mbox{and}\quad g=g(h),$ (24)

where the first equation of (24) implies $f\in C_{{\cal F}_{1}}(h)$, while the
second equation derives $[z^{n}]g=[z^{n}]g(h)$, i.e., $g_{n}=d_{n}$, which
implies

$g_{n}=\sum_{\mathbf{i}\in{\mathcal{D}}_{n}}g_{k}h_{i_{1}}h_{i_{2}}\cdots
h_{i_{k}}.$ (25)

For the $h$ assumed above, i.e., $h(z)=\sum_{n\geq 1}h_{n}z^{n}\not=z$ with
$h_{\ell}\not=0$ and $h_{2}=h_{3}=\cdots=h_{\ell-1}=0$, we substitute $n=\ell$
and the conditions of $h_{i_{1}},h_{i+2},\ldots,$ and $h_{i_{k}}$ into the
(25) and get

$g_{\ell}=g_{1}h_{\ell}+2g_{2}h_{1}h_{2}+\cdots+g_{\ell}h_{1}^{\ell}.$

In the sum on the right-hand side of the above equation, every term except the
first one and the last one contains $h_{2},h_{3},\ldots,$ and $h_{\ell-1}$.
Thus the equation can be reduced to

$g_{\ell}=g_{1}h_{\ell}+g_{\ell}h_{1}^{\ell}=g_{1}h_{\ell}+g_{\ell},$

which implies $g_{1}=0$ due to $h_{\ell}\not=0$. For $n=\ell$, (25) gives

$g_{\ell+1}=g_{1}h_{\ell+1}+2g_{2}h_{1}h_{\ell}+\cdots+g_{\ell+1}h_{1}^{\ell+1}=2g_{2}h_{\ell}+g_{\ell+1}$

because $g_{1}=0$, $h_{1}=1$, and $h_{2}=h_{3}=\cdots=h_{\ell-1}=0$. Thus from
the assumption of $h_{\ell}\not=0$, we obtain $g_{2}=0$. Similarly, by using
the induction assumption of $g_{2}=g_{3}=\cdots=g_{m}=0$, we may have
$g_{m+1}=0$ from

$\displaystyle g_{\ell+m}=g_{1}h_{\ell+k}+2g_{2}h_{1}h_{\ell+m-1}+\cdots$
$\displaystyle\quad+g_{m+1}((m+1)h_{1}^{m}h_{\ell}+\mbox{terms contain
$h_{2},h_{3},\ldots,h_{\ell-1}$})+\cdots$
$\displaystyle\quad+g_{\ell+m}h_{1}^{\ell+m}.$

More precisely, from the induction assumption and the conditions of
$h_{2}=h_{3}=\cdots=h_{\ell-1}=0$, the above equation can be reduced to

$g_{\ell+m}=(m+1)g_{m+1}h_{\ell}+g_{\ell+m},$

which implies $g_{m+1}=0$. Hence, we have proved that for $h(z)\not=z$, every
element $(g,f)$ of the centralizers of $(1,h)$ in Riordan group must has
$g=1$, completing the proof of the proposition. ∎

Similarly, we have the following result.

######  Proposition 3.3.

The centralizer of a Bell type Riordan array $(h/z,h)$ is

$C_{{\cal R}}(h/z,h)=\left\\{\begin{array}[]{ll}{\cal R}&if\,h=z\\\
\\{(f/z,f):f\in C_{{\cal F}_{1}}(h)\\}&if\,h\not=z.\end{array}\right.$

######  Corollary 3.4.

Let $S$, and $T$ be subsets of the Appel and Bell subgroups of ${\cal R}$,
which contain more than one element. Then we have $C_{{\cal
R}}(S)=\\{(1,f):f\in C_{{\cal F}_{1}}(h)\\}$ and $C_{{\cal
R}}(T)=\\{(f/z,f):f\in C_{{\cal F}_{1}}(h)\\}$, respectively.

Remark 3.1 Propositions 3.2 and 3.3 are not trivial. Although the maps
$\gamma_{1}:{\cal L}\to{\cal F}_{1}$ and $\gamma_{2}:{\cal B}\to{\cal F}_{1}$
defined by $\gamma_{1}(1,f)=f$ and $\gamma_{2}(f/z,f)=f$, respectively, for
all $f\in{\cal F}_{1}$ are group isomorphisms, $\gamma_{k}$, $k=1,2$, do not
transfer the centralizers from ${\mathcal{F}}_{1}$ to ${\mathcal{R}}$. This is
because ${\mathcal{F}}_{1}$ contains only one type formal power series,
$f\in\mathcal{F}_{1}$ with $f(0)=0$ and $f^{\prime}(0)\not=0$, while Riordan
array $(g,f)$ contains two type formal power series including the above $f$
and $g\in{\mathcal{F}}_{0}$ with $g(0)\not=0$.

##  4 Related topics

In this section, first we recall the definition of a reverser of a group and
study the collection of all reversers of a group as well as their applications
to the centralizers of a group. The part on the reversers is a slight
modification of a survey shown on pages 25-27 of O’Farrell and Short [17].
Secondly, we consider the set of all elements of a group that can be reversed
by an element of a group, particularly, the set of all elements of the Riordan
group ${{\cal R}}$ that can be reversed by $M(1,-z)$, i.e., the set of pseudo-
involutions of ${{\cal R}}$. Shapiro and one of the authors [10] found that
this set is a twisted subgroup of ${{\cal R}}$.

######  Definition 4.1.

An element $g$ of a group $G$ is said to be reversible in $G$ if there is
another element $h$ of $G$ such that

$hgh^{-1}=g^{-1}.$

We say that $h$ reverses $g$ or that $h$ is a reverser for $g$. We denote by
$R_{g}(G)$, or just $R_{g}$, the set of reversers of $g$.

Denote by $R_{G}(h)$ the set of all elements in $G$ that can be reversed by
$h$, namely,

$R_{G}(h)=\\{g\in G:hgh^{-1}=g^{-1}\\}.$ (26)

We call $R_{G}(h)$ the reverse set in $G$ reversed by $h$.

######  Definition 4.2.

Two elements $f$ and $g$ of a group $G$ are said to be conjugate in $G$ if
there is a third element $h$ such that

$f=h^{-1}gh,$

which can be denoted by $g^{h}$. Then $f$ is called the conjugate of $g$ by
$h$.

######  Definition 4.3.

The commutator of two elements, g and h, of a group G, is the element

$[g,h]=g^{-1}h^{-1}gh.$

It is equal to the group’s identity if and only if g and h commute. The
subgroup of G generated by all commutators is called the derived group or the
commutator subgroup of G. The commutator subgroup of the Riordan group ${{\cal
R}}$ is discussed in Luzón, Morón, and Prieto-Martinez [15].

The relationship between the conjugate of $g$ by $h$ and the commutator of $g$
and $h$ is

$g^{h}=g[g,h].$

Particularly, if $g\in C_{G}(h)$, then $[g,h]=e$ and $g^{h}=g$.

######  Proposition 4.4.

[17] Let $g$ be reversible in $G$, and suppose that $h\in R_{g}$. Then

$R_{g}=C_{g}h=hC_{g}.$

Here is an explanation of the last proposition. Let $f\in C_{g}$. Since $h\in
R_{g}$, then $fg=gf$ and $hgh^{-1}=g^{-1}$. We have

$(hf)g(hf)^{-1}=hfgf^{-1}h^{-1}=hgff^{-1}h^{-1}=hgh^{-1}=g^{-1}.$

Thus $hC_{g}\subset R_{g}$. Conversely, let $k\in R_{g}$, then

$kgk^{-1}=g^{-1}.$ (27)

We have

$(h^{-1}k)g(h^{-1}k)^{-1}=h^{-1}kgk^{-1}h=h^{-1}g^{-1}h=g.$

Thus $h^{-1}k\in C_{g}$, which implies that $k\in hC_{g}$. Hence,
$R_{g}\subset hC_{g}$. The same method could be applied to show
$R_{g}=C_{g}h$.

The above proposition motivates the definition of the extended centralizer:

$E_{g}(G):=\\{h\in G:g^{h}=g\quad or\quad g^{h}=g^{-1}\\}=C_{g}\cup R_{g}.$

This is also a subgroup of $G$.

######  Proposition 4.5.

[17] Let $g\in G$. Then

(i) if $g$ is not reversible, then $E_{g}=C_{g}$ and $R_{g}=\phi$.

(ii) if $g$ is an involution, then $E_{g}=C_{g}=R_{g}$.

######  Proposition 4.6.

[17] Let $g$ and $h$ be elements of a group $G$. Then (i) $C_{g^{-1}}=C_{g}$;
(ii) $R_{g^{-1}}=R_{g}$; (iii) $C_{hgh^{-1}}=hC_{g}h^{-1}$; (iv)
$R_{hgh^{-1}}=hR_{g}h^{-1}$.

The extended centralizer group $E_{g}$ consists of $C_{g}$ and a single coset
$R_{g}$. If $C_{g}$ is abelian, then we have a generalized dihedral group.
Hence, there is a question arises: are there examples where $C_{g}$ is non-
abelian? The answer is ”yes”. Here are two examples. If $G$ is non-abelian
group with the identity $e$, then the centralizer $C_{G}(e)=G$ is not abelian.
Another example is the dihedral group of order $4n$. This has an element, the
rotation of $180^{o}$, in its center.

We now give the definition of a twisted subgroup.

######  Definition 4.7.

The set of elements reversed by $h\in{G}$ forms a twisted subgroup $T\equiv
T_{h}:=\\{g\in G:hgh^{-1}=g^{-1}\\}$ of $G$ defined by (see Foguel and Ungar
[4, 5]):

(i) $e\in T$ ($h$ reverses $e$)

(ii) $T$ is closed under taking inverse (if $h$ reverses $g$, then $h$
reverses $g^{-1}$).

(iii) if $x,y\in T$, then $xyx\in T$.

The set $R(G)$ of reversible elements in $G$ is a union of twisted subgroups.
$R(G)$ contains the identity and is closed under taking inverse, but it is not
necessarily closed under composition. However, if $hfh^{-1}=f^{-1}$ and
$hgh^{-1}=g^{-1}$, i.e., $f,g\in R(G)$ and both are reversed by the same $h$,
then

$(hg)(fg)(hg)^{-1}=(fg)^{-1},$

i.e., $hg\in R_{fg}$.

Considering the set of the pseudo-involutions of Riordan group ${\cal R}$,
which means the set of all $D\in{\cal R}$ such that $MD$ (and $DM$) is an
involution, where $M=(1,-t)$. Then $MDM^{-1}=D^{-1}$. Shapiro [20] proved that
the centralizer of $M$ in ${{\cal R}}$ is ${\mathcal{C}}$, the checkerboard
subgroup (cf. also Jean-Louis and Nkwanta [12]). Hence, from [10] and [20] we
have the the first half and the second half, respectively, of the following
result.

######  Proposition 4.8.

The reserves set in ${{\cal R}}$ reversed by $M$, $R_{{\cal R}}(M)$, is a
collection of all elements of pseudo-involutions form a twisted subgroup,
which is denoted by $T_{M}$. The centralizers of $M$ in ${{\cal R}}$,
$C_{{\cal R}}(M)$, is the checkerboard subgroup of ${{\cal R}}$.

###### Proof.

For the sake of the readers’ convenience, we give a brief proof. First,
$I=(1,t)\in T_{M}$ because $MIM^{-1}=I$. Secondly, $D=(d,h)$ and $E=(g,f)\in
R_{{\cal R}}(M)$ implies that $DE^{-1}D\in R_{{\cal R}}(M)$ due to

$\displaystyle MDE^{-1}DM^{-1}=(MDM^{-1})(ME^{-1}M^{-1})(MDM^{-1})$
$\displaystyle=$ $\displaystyle D^{-1}ED^{-1}=(DE^{-1}D)^{-1}.$

Hence, $R_{{\cal R}}(M)$ is the twisted subgroup $T_{M}$.

If $D=(d,h)\in C_{G}(M)$, then $MD=DM$ implies that $(1,-t)(d,h)=(d,h)(1,-t)$,
i.e.,

$(d(-t),h(-t))=(d,-h).$

Hence, $d(-t)=d(t)$ and $h(-t)=-h(t)$, i.e., $(d,h)$ is in the checkerboard
subgroup of ${{\cal R}}$. ∎

Let $G$ be a group and let $T\subset G$ be a twisted subgroup, from Lemma
$1.2$ of Aschbacher [1] we have $\langle x\rangle\subset T$ for each $x\in T$.
Furthermore, if $T$ is a $2$-divisible twisted subgroup, i.e., for each $x\in
T$, there exists a unique element $x^{1/2}\in T$ such that $(x^{1/2})^{2}=x$,
from Corollary 3 of Glauberman [6], we have Lagrange’s Theorem on twisted
subgroup, i.e., for every $x\in T$, the order of $x$ divides $|T|$. Since
$T_{M}$ is not finite, the Lagrange’s Theorem is invalid on $T_{M}$.

Shapiro and one of the authors [10] presents a palindromic property of pseudo-
involutions. More precisely, if $A$ and $B$ are both pseudo-involutions, then
so is the triple product $ABA$. With this it follows that if are pseudo-
involutions so is any palindromic word using these symbols.

##  5 acknowledgements

The authors wish to express their sincere gratitude and appreciation to
Professor Louis Shapiro, the referees, and the handling editor for their
helpful comments and remarks that led to an revised version of the original
manuscript.

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