Algebraic properties of Riordan subgroups

Abstract

We present properties of the group structure of Riordan arrays. We examine similar properties among known Riordan subgroups, and from this, we define H[r, s, p], a family of Riordan arrays. We generalize conditions for involutions, and pseudo-involutions of H[r, s, p], and we present stabilizers of this family. We find abelian subgroups as intersections of Riordan subgroups and show some alternative semi-direct products of the Riordan group.

1 Introduction

The group structure of the set of Riordan arrays has been the subject of a number of papers [3, 9, 12, 15, 19], where Riordan subgroups and their group theoretical properties are presented. In this paper, we exclusively focus on the algebraic elements of Riordan arrays, by providing new findings. The motivation of our research was “Some algebraic structure of the Riordan group” by Jean-Louis et al. [12], and we hope that our work will be considered as a complementary to the results presented there.

The paper is arranged as follows. In Sect. 2, we present the definitions of a Riordan array, the fundamental theorem of Riordan arrays and the Riordan group, together with the Riordan subgroups that have been found so far. We define H[r, s, p], a family of Riordan subgroups, based on a collection of isomorphic Riordan subgroups, as shown in Sect. 3. We present conditions for involutions, and pseudo-involutions of this family. We also present common Riordan elements among those subgroups that form abelian Riordan subgroups. Properties of Riordan subgroups that do not belong in H[r, s, p] are presented in Sect. 4, where we show alternative ways of expressing the Riordan group as a semi-direct product, and some general forms of Stabilizers. Finally, in Sect. 5, we suggest a number of open problems that might be of interest, and worth further exploration.

2 Riordan arrays and their algebraic structure

Let $\mathbb {F}=\mathbb {K}[[z]]$ be the ring of formal power series (fps), with coefficients in $\mathbb {K}$, where $\mathbb {K}$ is the field $\mathbb {R}$ or $\mathbb {C}$. If $f(z)= f_0+f_1z+f_2z^2+f_3z^3+\cdots \in \mathbb {F},$ then the order of f(z) is defined to be the lowest index k for which the coefficient $f_k\ne 0$, while the set of fps of order k is denoted by $\mathbb {F}_k$ [22].

Definition 1

[17] An Ordinary Riordan array is a lower triangular infinite matrix R, constructed by two fps

$$\begin{aligned} g(z)=\sum \limits _{n=0}^\infty g_n z^{n} ,\quad \text {and} \quad f(z)=\sum \limits _{n=1}^\infty f_n z^{n}, \end{aligned}$$

where $g(z) \in \mathbb {F}_0$, and $f(z) \in \mathbb {F}_1$, in such a way that the generating function (gf) of the $k{\text {th}}$ column is $g(z)(f(z))^k$, for all $k \ge 0$. We say that R is a Riordan array or Riordan matrix and we write $R=(g(z),f(z))$.

Example 1

Expressing Pascal’s triangle as a Riordan array, we get

$$\begin{aligned} P=(g(z),f(z))=\left( \frac{1}{1-z}, \frac{z}{1-z} \right) = \begin{bmatrix} 1 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} \cdots \\ 1 &{} 1 &{} 0 &{} 0 &{} 0 &{} 0 &{} \cdots \\ 1 &{} 2 &{} 1 &{} 0 &{} 0 &{} 0 &{} \cdots \\ 1 &{} 3 &{} 3 &{} 1 &{} 0 &{} 0 &{} \cdots \\ 1 &{} 4 &{} 6 &{} 4 &{} 1 &{} 0 &{}\cdots \\ 1 &{} 5 &{} 10 &{} 10 &{} 5&{} 1&{} \cdots \\ \vdots &{} \vdots &{} \vdots &{} \vdots &{} \vdots &{} \vdots &{}\ddots \\ \end{bmatrix}, \end{aligned}$$

where the function $g(z)=\frac{1}{1-z}$ generates the all ones sequence [A000012 in OEIS [21], and it is multiplied by z in order to have $f(z) \in \mathbb {F}_1$.

Example 2

The Riordan array $J=\left( \frac{1-\sqrt{1-4z}}{2z}, \frac{z}{1-z-z^2} \right)$ gives rise to the Riordan matrix

$$\begin{aligned} \begin{bmatrix} 1 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} \cdots \\ 1 &{} 1 &{} 0 &{} 0 &{} 0 &{} 0 &{} \cdots \\ 2 &{} 2 &{} 1 &{} 0 &{} 0 &{} 0 &{} \cdots \\ 5 &{} 5 &{} 3 &{} 1 &{} 0 &{} 0 &{} \cdots \\ 14 &{} 12 &{} 9 &{} 4 &{} 1 &{} 0 &{}\cdots \\ 42 &{} 31 &{} 24 &{} 14 &{} 5 &{} 1&{} \cdots \\ \vdots &{} \vdots &{} \vdots &{} \vdots &{} \vdots &{} \vdots &{}\ddots \\ \end{bmatrix} \end{aligned}$$

where $\frac{1-\sqrt{1-4z}}{2z}$ is the generating function of the Catalan numbers [A000108 in OEIS [21]], and the function $\frac{z}{1-z-z^2}$ is expressed as a series which its coefficients are the Fibonacci numbers $0,1,1,2,3,5,8,13,\dots$ [A000045 in OEIS [21]].

2.1 The Riordan group

Theorem 1

(Fundamental theorem of Riordan arrays)[20] Let $R=(g(z),f(z))$ be a Riordan matrix, and let P, Q be two column vectors, where their gfs are A(z) and B(z), respectively, such that

$$\begin{aligned} R*P=Q. \end{aligned}$$

This relation holds if and only if the following relation among the gfs is true

$$\begin{aligned} (g(z),f(z))*A(z)= & {} B(z) \\ \Leftrightarrow g(z)A(f(z))= & {} B(z). \end{aligned}$$

Now, the operation $*$ of the Riordan group, combining the gfs of the matrices, is defined as follows. Suppose that we have two Riordan matrices $\varLambda =(g(z),f(z))$ and $K=(h(z),k(z))$, then we define the product $\varLambda *K$ as

$$\begin{aligned} \varLambda *K\,= & {} \,(g(z)h(f(z)),k(f(z))). \end{aligned}$$

This product can be shown to be associative, with $I=(1,z)$ being the identity element. Additionally, the inverse element is given by:

$$\begin{aligned} (g(z),f(z))^{-1}=\left( \frac{1}{g(\bar{f}(z))},\bar{f}(z)\right) , \end{aligned}$$

where $\bar{f}$ is the compositional inverse.

Definition 2

[17] The set $\mathcal {R}$ of all Riordan arrays together with the above operation form the Riordan group, $\langle \mathcal {R},*\rangle$.

The order of this group is infinite; however, if we restrict all entries to be integers, then any element of finite order must have order 1 or 2 [18].

Definition 3

[16, 18] Let $\varLambda =(g(z), f(z))$ be a Riordan matrix. If $\varLambda *\varLambda =I$, then $\varLambda$ is called involution, i.e., Riordan elements of order 2 are called involutions.

Example 3

The Riordan array

$$\begin{aligned} \left( \frac{1}{1-z}, - \frac{z}{1-z} \right) = \begin{bmatrix} 1 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} \cdots \\ 1 &{} -1 &{} 0 &{} 0 &{} 0 &{} 0 &{} \cdots \\ 1 &{} -2 &{} 1 &{} 0 &{} 0 &{} 0 &{} \cdots \\ 1 &{} -3 &{} 3 &{} -1 &{} 0 &{} 0 &{} \cdots \\ 1 &{} -4 &{} 6 &{} -4 &{} 1 &{} 0 &{}\cdots \\ 1 &{} -5 &{} 10 &{} -10 &{} 5&{} -1&{} \cdots \\ \vdots &{} \vdots &{} \vdots &{} \vdots &{} \vdots &{} \vdots &{}\ddots \\ \end{bmatrix} \end{aligned}$$

is an involution, as

$$\begin{aligned} \left( \frac{1}{1-z}, - \frac{z}{1-z} \right) * \left( \frac{1}{1-z}, - \frac{z}{1-z} \right)= & {} \left( \frac{1}{1-z} \frac{1}{1+\frac{z}{1-z}}, \frac{\frac{z}{1-z}}{1+ \frac{z}{1-z}} \right) \\= & {} (1,z) . \end{aligned}$$

Definition 4

[10, 18] Let K be a Riordan matrix and $M=(1,-z)$. If $K *M$ is an involution, then we call K a pseudo-involution, i.e. $(K *M)^2=I$, i.e. $K *M$ has order 2.

Proposition 1

[3] If the Riordan element $K=(g(z), f(z))$ is a pseudo-involution, then we have that $-f(-f(z))=z,$ and $g(z)=\frac{1}{g(-f(z))}.$

Example 4

An example of a pseudo-involution is the Riordan array which is known as Nkwanta’s RNA triangle [3]

$$\begin{aligned} N=(g(z), zg(z))= \begin{bmatrix} 1 &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} \cdots \\ 1 &{} 1 &{} 0 &{} 0 &{} 0 &{} 0 &{} \cdots \\ 1 &{} 2 &{} 1 &{} 0 &{} 0 &{} 0 &{} \cdots \\ 2 &{} 3 &{} 3 &{} 1 &{} 0 &{} 0 &{} \cdots \\ 4 &{} 6 &{} 6 &{} 4 &{} 1 &{} 0 &{}\cdots \\ 8 &{} 13 &{} 13 &{} 10 &{} 5&{} 1&{} \cdots \\ \vdots &{} \vdots &{} \vdots &{} \vdots &{} \vdots &{} \vdots &{}\ddots \\ \end{bmatrix} \end{aligned}$$

where $g(z)= \frac{1-z-z^2-\sqrt{1-2z-z^2-2z^3+z^4}}{2z^2}$ is the generating function which arises by enumerating secondary structures of RNA molecules [A004148 in OEIS [21]].

2.2 Riordan subgroups

Many of the Riordan subgroups that have been found so far [4, 12, 15, 17, 19, 20] are summarized in Table 1.

Table 1 Riordan subgroups

Additionally, in [14], Luzon et al. presented a general form of Riordan subgroups, as a family of subgroups

$$\begin{aligned} H[r,s]=\left\{ \left( \left( \frac{f(z)}{z}\right) ^r (f'(z))^s, f(z) \right) \Bigg | f(z) \in \mathbb {F}_{1} \right\} . \end{aligned}$$

(1)

3 Properties of the Riordan subgroups

3.1 A Riordan family of subgroups

He [8, 9] and Jean-Louis and Nkwanta [12] have shown that the Associated, the Bell, the Derivative, the Power-Bell, and the Hitting time subgroups are isomorphic to $\mathbb {F}_1[[z]]= \lbrace f(z) \mid f(z)=f_1z+f_2z^2+\ldots \rbrace ,$ which is the group of fps, under composition. We prove that the Stochastic subgroup also belongs to this isomorphic collection of Riordan subgroups.

Proposition 2

The Associated and the Stochastic Riordan subgroups are isomorphic.

Proof

Let

$$\begin{aligned} \phi : (1,f(z)) \rightarrow \left( \frac{f(z)-1}{z-1}, f(z) \right) \end{aligned}$$

be a mapping between the two subgroups, and suppose that (1, f(z)) and (1, h(z)) are two elements of the Associated subgroup. Then, we have

$$\begin{aligned} \phi ((1,f(z))*(1,h(z)))= & {} \phi (1,h(f(z))) \\= & {} \left( \frac{h(f(z))-1}{z-1},h(f(z))\right) \\= & {} \left( \frac{f(z)-1}{z-1},f(z)\right) *\left( \frac{h(z)-1}{z-1},h(z)\right) \\= & {} \phi (1,f(z))*\phi (1,h(z)), \end{aligned}$$

which means that $\phi$ is a homomorphism. We also have that

$$\begin{aligned} \text {Ker}(\phi )= & {} \lbrace (1,f(z)) \in \text { Associated } \mid \phi (1,f(z))=(1,z)\rbrace , \end{aligned}$$

which leads us to the equation

$$\begin{aligned} \left( \frac{f(z)-1}{z-1}, f(z)\right)= & {} (1,z), \end{aligned}$$

and to the simultaneous equations

$$\begin{aligned} \frac{f(z)-1}{z-1}=1 \text { ; } f(z)=z. \end{aligned}$$

Now, if $f(z)\ne z$, then $\frac{f(z)-1}{z-1} \notin \mathbb {F}_0$. Therefore, the only solution is $f(z)=z$. Hence, $\phi$ is an injection. It is also clear from the Stochastic subgroup as shown in Table 1 that $\phi$ is onto. Hence, $\phi$ is an isomorphism. $\square$

Considering the family of subgroups as defined in Eq. 1 in Sect. 2.2, which contains five isomorphic Riordan subgroups, we extend the definition by adding one extra parameter, which corresponds to the pth power of the first gf of the Stochastic subgroup, as follows.

$$\begin{aligned} H[r,s,p]{=}\left\{ \left( \left( \frac{f(z)}{z}\right) ^r (f'(z))^s \left( \frac{f(z)-1}{z-1} \right) ^{p}, f(z) \right) \Bigg | (r,s,p)\in \mathbb {Q}^3, f(z) \in \mathbb {F}_{1} \right\} . \end{aligned}$$

(2)

In order to avoid any possible confusion, we owe to explain the terms that we are going to use from now on, which are based on this family of subgroups. So, the Greek letters $\rho , \sigma$, and $\pi$ instead of the Latin r, s, and p will be used for fixed parameters, and the index f will be used to declare the dependence of the algebraic structure that we are referring to, from a function $f(z) \in \mathbb {F}_1$. Hence, we have that

$$\begin{aligned} H_{f}[\rho ,\sigma ,\pi ] \in H[\rho ,\sigma ,\pi ] \subseteq H[r,s,p], \end{aligned}$$

where $H[\rho ,\sigma ,\pi ]$ is a Riordan subgroup, and $H_{f}[\rho ,\sigma ,\pi ]$ is a Riordan element.

Now, the six isomorphic Riordan subgroups can be defined by $H[\rho ,\sigma ,\pi ]$ and are shown in Table 2.

Table 2 Riordan subgroups of H[r, s, p]

Proposition 3

The Riordan family H[r, s, p] represents a subgroup of the Riordan group, for each triple $(r,s,p) \in \mathbb {Q}^3$.

Proof

Let

$$\begin{aligned} H_{f_1}[\rho ,\sigma ,\pi ]=\left( \bigg (\frac{f_1(z)}{z}\bigg )^{\rho } (f'_1(z))^{\sigma } \left( \frac{f_1(z)-1}{z-1} \right) ^{\pi },f_1(z)\right) \end{aligned}$$

and

$$\begin{aligned} H_{f_2}[\rho ,\sigma ,\pi ]=\left( \left( \frac{f_2(z)}{z}\right) ^{\rho } (f'_2(z))^{\sigma } \left( \frac{f_2(z)-1}{z-1} \right) ^{\pi },f_2(z)\right) \end{aligned}$$

be two elements of H[r, s, p]. We have that

$$\begin{aligned}&H_{f_1}[\rho ,\sigma ,\pi ] *H_{f_2}[\rho ,\sigma ,\pi ] \\&\quad = \left( \left( \frac{f_1(z)}{z}\right) ^{\rho } (f'_1(z))^{\sigma } \left( \frac{f_1(z)-1}{z-1} \right) ^{\pi },f_1(z)\right) \\&\qquad *\left( \left( \frac{f_2(z)}{z}\right) ^{\rho } (f'_2(z))^{\sigma } \left( \frac{f_2(z)-1}{z-1} \right) ^{\pi },f_2(z)\right) \\&\quad =\left( \left( \frac{f_2(f_1(z))}{z}\right) ^{\rho } (f'_1(z))^{\sigma } (f'_2(f_1(z)))^{\sigma } \left( \frac{f_2(f_1(z))-1}{z-1}\right) ^{\pi } ,f_2(f_1(z))\right) \\&\quad =\left( \left( \frac{f_2(f_1(z))}{z}\right) ^{\rho } (d(f_2(f_1(z))))^{\sigma } \left( \frac{f_2(f_1(z))-1}{z-1}\right) ^{\pi } ,f_2(f_1(z))\right) , \end{aligned}$$

where $d(f_2(f_1(z)))$ is the derivative of $f_2(f_1(z))$. This shows closure.

Now, the inverse of an element $H_f[\rho , \sigma , \pi ]$ will be

$$\begin{aligned} H^{-1}_f[\rho , \sigma , \pi ]= & {} \left( \left( \frac{f(z)}{z}\right) ^{\rho } (f'(z))^{\sigma } \left( \frac{f(z)-1}{z-1} \right) ^{\pi } ,f(z)\right) ^{-1}\\= & {} \left( \frac{1}{\left( \frac{z}{\bar{f}(z)}\right) ^{\rho } (f'(\bar{f}(z)))^{\sigma } \left( \frac{z-1}{\bar{f}(z)-1}\right) ^{\pi }} ,\bar{f}(z)\right) \\= & {} \left( \left( \frac{\bar{f}(z)}{z}\right) ^{\rho } \left( \frac{1}{f'(\bar{f}(z))}\right) ^{\sigma } \left( \frac{\bar{f}(z)-1}{z-1} \right) ^{\pi },\bar{f}(z)\right) . \end{aligned}$$

As shown in [1], Section 6.4, p. 170, we get that

$$\begin{aligned} (\bar{f})'(z)\,= & {} \,\frac{1}{f'(\bar{f}(z))}. \end{aligned}$$

(3)

So, $H^{-1}_f[\rho , \sigma , \pi ]=\left( \left( \frac{\bar{f}(z)}{z}\right) ^{\rho } \left( \bar{f}'(z)\right) ^{\sigma } \left( \frac{\bar{f}(z)-1}{z-1} \right) ^{\pi },\bar{f}(z)\right)$.

Hence, for every triple $(\rho , \sigma , \pi ) \in \mathbb {Q}^3$, H[r, s, p] is a Riordan subgroup. $\square$

Using the general form of a Riordan subgroup of H[r, s, p], we now present the following results.

Proposition 4

An arbitrary Riordan element $H_f[\rho , \sigma , \pi ] \in H[r,s,p]$ is an involution if and only if $f(z)=\bar{f}(z)$.

Proof

Let $H_f [ \rho ,\sigma , \pi ]= \left( \left( \frac{f(z)}{z} \right) ^{\rho } \left( f'(z)\right) ^{\sigma } \left( \frac{f(z)-1}{z-1}\right) ^{\pi }, f(z)\right)$ and $f(z)=\bar{f}(z)$. Then, we have that

$$\begin{aligned}&H_f [ \rho ,\sigma , \pi ] *H_f [ \rho ,\sigma , \pi ]\\&\quad =\,\left( \left( \frac{f(z)}{z} \right) ^{\rho } \left( f'(z)\right) ^{\sigma } \left( \frac{f(z)-1}{z-1} \right) ^{\pi } \left( \frac{z}{f(z)} \right) ^{\rho } \left( \frac{1}{f'(z)}\right) ^{\sigma } \left( \frac{z-1}{f(z)-1} \right) ^{\pi }, z \right) \\&\quad =\,(1,z). \end{aligned}$$

Hence, $H_f [ \rho ,\sigma , \pi ]$ is an involution.

Now, let us assume that $H_f[ \rho ,\sigma , \pi ]$ is an involution. So, we have that

$$\begin{aligned} H_f [ \rho ,\sigma , \pi ] *H_f [ \rho ,\sigma , \pi ] = (1,z), \end{aligned}$$

and the equation

$$\begin{aligned} \left( \left( \frac{f(f(z))}{z} \right) ^{\rho } \left( (f(f(z)))' \right) ^{\sigma } \left( \frac{f(f(z))-1}{z-1} \right) ^{\pi }, f(f(z)) \right) = (1,z) \end{aligned}$$

which is satisfied if $f(z)=\bar{f}(z)$. $\square$

Proposition 5

Let $A=(g(z),f(z))$ and $B=(h(z),f(z))$ be two Riordan involutions such that $A,B \in H[r,s,p]$, where $g(z)\ne h(z)$ and f(z) is fixed, then

$$\begin{aligned} A*B=(g(z),f(z))*(h(z),f(z))=\left( \frac{g(z)}{h(z)},z\right) . \end{aligned}$$

Proof

We have that

$$\begin{aligned} (g(z), f(z))*(h(z),f(z))= & {} (g(z)h(f(z)),z), \end{aligned}$$

(4)

as $f=\bar{f}$, by Proposition 4. The Riordan element (h(z), f(z)) satisfies the equation $(h(z), f(z))*(h(z), f(z))=(1,z)$. Hence, we have that $h(z)h(f(z))=1 \Rightarrow h(f(z))=\frac{1}{h(z)}$, and the right-hand side of Eq. 4 becomes $\left( \frac{g(z)}{h(z)}, z \right)$. $\square$

Proposition 6

An element $H_f [ \rho ,\sigma , \pi ] \in H[r,s,p]$ is a pseudo-involution if $-f(-f(z))=z$ and $\pi =0$.

Proof

A pseudo-involution of H[r, s, p] needs to satisfy the following

$$\begin{aligned} H_{f^-} [ \rho ,\sigma , \pi ]= & {} H_f [ \rho ,\sigma , \pi ] *(1,-z) \\= & {} \left( \left( \frac{f(z)}{z} \right) ^{\rho } \left( f'(z)\right) ^{\sigma } \left( \frac{f(z)-1}{z-1}\right) ^{\pi }, f(z)\right) *(1,-z) \\= & {} \left( \left( \frac{f(z)}{z} \right) ^{\rho } \left( f'(z)\right) ^{\sigma } \left( \frac{f(z)-1}{z-1}\right) ^{\pi }, -f(z)\right) \end{aligned}$$

Now, $H_{f^-} [ \rho ,\sigma , \pi ] *H_{f^-} [ \rho ,\sigma , \pi ]$ becomes

$$\begin{aligned}&\left( \left( \frac{f(z)}{z} \right) ^{\rho } \left( f'(z)\right) ^{\sigma } \left( \frac{f(z)-1}{z-1}\right) ^{\pi } \left( \frac{f(-f(z))}{-f(z)} \right) ^{\rho } \left( f'(-f(z))\right) ^{\sigma } \right. \nonumber \\&\qquad \left. \left( \frac{f(-f(z))-1}{-f(z)-1}\right) ^{\pi }, -f(-f(z))\right) \end{aligned}$$

(5)

For $-f(-f(z))=z$, Eq. 5 becomes

$$\begin{aligned}&\left( \left( \frac{f(z)}{z} \right) ^{\rho } \left( \frac{-z}{-f(z)} \right) ^{\rho } \left( f'(z)\right) ^{\sigma } \left( f'(-f(z)) \right) ^{\sigma } \left( \frac{f(z)-1}{z-1} \right) ^{\pi } \left( \frac{-z-1}{-f(z)-1} \right) ^{\pi }, z \right) \nonumber \\&\quad = \left( \left( f'(z) f'(-f(z))\right) ^{ \sigma } \left( \frac{(-z-1)(f(z)-1)}{(-f(z)-1)(z-1)}\right) ^{ \pi }, z \right) \end{aligned}$$

(6)

Differentiating the expression $f(-f(z))=-z$, we get

$$\begin{aligned}{}[f(-f(z))]' = (-z)' \Rightarrow f'(-f(z))= \frac{1}{f'(z)} \end{aligned}$$

Substituting this result into Eq. 6, we get

$$\begin{aligned} \left( \left( \frac{(z+1)(f(z)-1)}{(f(z)+1)(z-1)}\right) ^{\pi }, z \right) \end{aligned}$$

The fraction $\frac{(z+1)(f(z)-1)}{(f(z)+1)(z-1)} = 1$, only for $f(z)= z$, hence $\pi =0$. $\square$

Proposition 7

All Riordan subgroups $H[\rho , \sigma , \pi ]$ for distinct triples $(\rho , \sigma , \pi ) \in \mathbb {Q}^3$ are isomorphic.

Proof

Let $H[\rho ,\sigma ,\pi ]=\left( \left( \frac{f(z)}{z}\right) ^{\rho } \left( f'(z) \right) ^{\sigma } \left( \frac{f(z)-1}{z-1} \right) ^{\pi }, f(z) \right)$, and $H[\rho ',\sigma ',\pi ']=\left( \left( \frac{f(z)}{z}\right) ^{\rho '} \left( f'(z) \right) ^{\sigma '} \left( \frac{f(z)-1}{z-1} \right) ^{\pi '}, f(z) \right)$ be two arbitrary Riordan subgroups of H[r, s, p]. We will prove that there is a mapping between these two subgroups, which is an isomorphism. Now, let $\psi$ be a mapping between the Associated subgroup, and $H[\rho ,\sigma ,\pi ]$, such that

$$\begin{aligned} \psi : (1,f(z)) \rightarrow H[\rho ,\sigma ,\pi ]. \end{aligned}$$

Suppose that (1, f(z)) and (1, h(z)) are two Riordan elements of the Associated subgroup. Then, we have that

$$\begin{aligned} \psi \left( (1,f(z))*(1,h(z)) \right)= & {} \,\left( \left( \frac{h(f(z))}{z} \right) ^{\rho } [(h(f(z)))']^{\sigma } \left( \frac{h(f(z))-1}{z-1}\right) ^{\pi },h(f(z)) \right) \\= & {} \,\left( \left( \frac{f(z)}{z} \right) ^{\rho } ((f'(z))^{\sigma } \left( \frac{f(z)-1}{z-1}\right) ^{\pi },f(z) \right) \\&\quad *\left( \left( \frac{h(z)}{z} \right) ^{\rho } (h'(z))^{\sigma } \left( \frac{h(z)-1}{z-1}\right) ^{\pi },h(z) \right) \\= & {} \,\psi ( 1,f(z) )*\psi (1,h(z)). \end{aligned}$$

Hence, $\psi$ is a homomorphism.

The homomorphism $\psi$ is also an epimorphism as $Im(\psi )= H[\rho ,\sigma ,\pi ]$, and it is an injection as $Ker(\psi ) = \left\{ (1,f(z)) \mid \psi (1,f(z))=\textit{1} \right\}$, which means that

$$\begin{aligned} \left( \bigg (\frac{f(z)}{z}\bigg )^{\rho } (f'(z))^{\sigma } \bigg (\frac{f(z)-1}{z-1} \bigg )^{\pi }, f(z) \right)= & {} (1,z). \end{aligned}$$

(7)

This is true only for $f(z)=z$. So, $Ker(\psi )=((1,z)\mid z \in \mathbb {C})$. Hence, the Associated subgroup is isomorphic to the arbitrary element $H[\rho ,\sigma ,\pi ]$ of H[r, s, p]. Now, using the inverse mapping of a similar isomorphism from the Associated subgroup to the arbitrary Riordan subgroup $H[\rho ',\sigma ',\pi ']=\left( \left( \frac{f(z)}{z}\right) ^{\rho '} \left( f'(z) \right) ^{\sigma '} \left( \frac{f(z)-1}{z-1} \right) ^{\pi '}, f(z) \right)$, we get the following commutative diagram.

where $\psi '$ is defined as

$$\begin{aligned} \psi '^{-1}: H[0,0,0] \rightarrow H[\rho ',\sigma ',\pi '] , \end{aligned}$$

and we finally have $\theta = \psi \circ \psi '^{-1}$. Hence, any two subgroups of the form H[r, s, p] are isomorphic. $\square$

3.2 Some intersections in H[r, s, p]

Proposition 8

The intersection of the subgroups H[1, 0, 0] and $H[-1,1,0]$ is given by the Riordan subgroup

$$\begin{aligned} P^c = \left\{ \left( \frac{1}{1-cz},\frac{z}{1-cz}\right) \Bigg | c \in \mathbb {R} \right\} . \end{aligned}$$

Proof

Solving the differential equation $\frac{f(z)}{z}=\frac{z f'(z)}{f(z)}$, we get that $f(z)=\frac{z}{1-cz}$. The intersection $P^c$ is a Riordan subgroup, as it satisfies closure and its inverse is also of the form of elements of this set. $\square$

Proposition 9

All non-identity elements in $P^c$ are pseudo-involutions.

Proof

We have

$$\begin{aligned} \left( \frac{1}{1-cz},\frac{z}{1-cz}\right) *(1,-z) = \left( \frac{1}{1-cz},-\frac{z}{1-cz} \right) , \end{aligned}$$

and

$$\begin{aligned} \left( \frac{1}{1-cz},-\frac{z}{1-cz} \right) *\left( \frac{1}{1-cz},-\frac{z}{1-cz} \right) = (1,z). \end{aligned}$$

$\square$

Corollary 1

The Riordan subgroup $P^c$ contains only trivial involutions.

Proposition 10

The Riordan subgroup $P^c$ is abelian.

Proof

We note that $P_c$ is the $c\text {th}$ power of the Pascal’s triangle; hence, for the two Riordan elements of $P^c$, $a=\left( \frac{1}{1-c_{1}z},\frac{z}{1-c_{1}z}\right)$ and $b=\left( \frac{1}{1-c_{2}z},\frac{z}{1-c_{2}z}\right)$, where $c_{1},c_{2}\in \mathbb {R}$, we have that

$$\begin{aligned} a*b= & {} \left( \frac{1}{1-(c_{1}+c_{2})z},\frac{z}{1-(c_{1}+c_{2})z)} \right) . \end{aligned}$$

(8)

By Eq. 8, $P^c$ satisfies commutativity, as $a*b= b*a$. Hence, $P^c$ is abelian. $\square$

This Riordan subgroup was firstly described in [3] as a class of generalized Pascal’s triangles and later on, in [12] as a larger subset of Riordan matrices. Additionally, Cheon et al. have shown in Lemma 4.1 in [5] that $P^c$ belongs to a family of cyclic subgroups, the intersection of the family of Power-Bell subgroups, and the Hitting time subgroup, while another reference of this can be found in Proposition 5 of [13], where it is written in its $T(d \mid h)$ form.

Proposition 11

The intersection of the subgroups H[0, 0, 1] and $|H[-1,1,0]$ is given by the subgroup

$$\begin{aligned} P_{c,c+1} = \left\{ \left( \frac{1}{1+cz},\frac{z(1+c)}{1+cz} \right) \Bigg | c\ne -1 \right\} . \end{aligned}$$

Proof

The differential equation $\frac{f(z)-1}{z-1}=\frac{zf'(z)}{f(z)}$ leads us to the solution $f(z)=\frac{z}{z(1-e^k)+e^k}$, where $k\ne 0$. We set $e^k= \frac{1}{c+1},$ where c is the constant, and we get $f(z)=\frac{z(c+1)}{1+cz}$.

Now, let $k=\,\left( \frac{1}{1+c_{1}z},\frac{z(1+c_1)}{1+c_{1}z}\right)$ and $m=\,\left( \frac{1}{1+c_{2}z},\frac{z(1+c_2)}{1+c_{2}z}\right)$ be two elements of $P_{c,c+1}$, where $c_{1},c_{2}\in \mathbb {R}$. Then,

$$\begin{aligned} k *m\,= & {} \,\left( \frac{1}{1+z(c_1+c_2+c_1c_2)}, \frac{z(1+c_1)(1+c_2)}{1+z(c_1+c_2+c_1c_2)}\right) \nonumber \\= & {} \,\left( \frac{1}{1+z(c_1+c_2+c_1c_2)}, \frac{z(1+c_1+c_2+c_1c_2)}{1+z(c_1+c_2+c_1c_2)}\right) , \end{aligned}$$

(9)

and for $A=c_1+c_2+c_1c_2$, Eq. 9 becomes $\left( \frac{1}{1+zA}, \frac{z(1+A)}{1+zA}\right)$. This proves that $P_{c,c+1}$ is closed under multiplication. The multiplicative inverse $\bar{f}(z)$ is

$$\begin{aligned} (f\circ \bar{f})(z)=z\Rightarrow & {} \frac{\bar{f}(z)}{1+c\bar{f}(z)}(1+c)=z\\\Rightarrow & {} \bar{f}(z)=\frac{z}{1+c-zc}. \end{aligned}$$

By the definition of the inverse, we have:

$$\begin{aligned} \left( \frac{1}{1+cz}, \frac{z}{1+cz}(1+c)\right) ^{-1}= \left( \frac{1+c}{1+c-cz},\frac{z}{1+c-cz} \right) , \end{aligned}$$

which for $\frac{c}{1+c}=K$ can also be transformed into an element of $P_{c,c+1}$. Hence, we proved that $P_{c,c+1}$ is a Riordan subgroup. $\square$

Proposition 12

The Riordan subgroup $P_{c,c+1}$ is abelian.

Proof

From Eq. 9, it is clear that $k *m = m *k$. Hence, $P_{c,c+1}$ satisfies commutativity.

$\square$

Proposition 13

The only element of $P_{c,c+1}$ which is a non-trivial involution is $\left( \frac{1}{1-2z}, -\frac{z}{1-2z}\right)$.

Proof

Let us try first to find any possible involutions in this subgroup

$$\begin{aligned}&\left( \frac{1}{1+cz},\frac{z}{1+cz}(c+1)\right) *\left( \frac{1}{1+cz},\frac{z}{1+cz}(c+1)\right) \\&\quad =\,\left( \frac{1}{1+c(c+2)z},\frac{z}{1+c(c+2)z}(c+1)^2\right) \end{aligned}$$

which for $c=0$, collapses to the identity, while for $c=-2$, we get $\left( \frac{1}{1-2z}, - \frac{z}{1-2z}\right)$. $\square$

Proposition 14

The Riordan subgroup $P_{c,c+1}$ does not contain non-trivial pseudo-involutions.

Proof

For the pseudo-involutions of $P_{c,c+1}$, we have

$$\begin{aligned} \left( \frac{1}{1+cz}, \frac{z}{1+cz}(c+1)\right) *(1,-z)=\left( \frac{1}{1+cz},-\frac{z}{1+cz}(c+1)\right) , \end{aligned}$$

and then,

$$\begin{aligned}&\left( \frac{1}{1+cz},-\frac{z}{1+cz}(c+1)\right) *\left( \frac{1}{1+cz},-\frac{z}{1+cz}(c+1)\right) \\&\quad =\left( \frac{1}{1-c^2z},\frac{z}{1-c^2z}(c+1)^2\right) , \end{aligned}$$

which for $c=0$, we get the identity element. $\square$

Proposition 15

• (a) The intersection of the subgroups H[0, 0, 0] and $H[-1,1,0]$ is the subgroup

  $$\begin{aligned} \textit{1}_c = \left\{ (1,cz) \mid c \in \mathbb {C} \right\} . \end{aligned}$$

• (b) The intersection of the subgroups H[0, 1, 0] and H[1, 0, 0] is the subgroup

  $$\begin{aligned} \textit{1}_{c,c} = \left\{ (c,cz) \mid c \in \mathbb {C} \right\} . \end{aligned}$$

Proof

Both of these intersections lead us to the differential equation

$$\begin{aligned} f'(z)-\frac{f(z)}{z}=0, \end{aligned}$$

where its solutions are given by $f(z)=cz$, where c is a constant. These solutions form the subsets $\textit{1}_c$ and $\textit{1}_{c,c}$, respectively, which can be easily shown to be Riordan subgroups. $\square$

Both of these Riordan subgroups can be thought of as c-extensions [19] of the unitary Riordan subgroup $I = \lbrace (1,z) \rbrace$.

Proposition 16

The Riordan subgroups $1_c$ and $1_{c,c}$ are abelian.

Proof

Let $(c_1,c_1z), (c_2,c_2z) \in 1_{c,c}$. Then,

$$\begin{aligned} (c_1,c_1z)*(c_2,c_2z)=(c_1c_2,c_1c_2z)=(c_2,c_2z) *(c_1,c_1z), \end{aligned}$$

Similarly shown for $1_c$. $\square$

Proposition 17

The only involutions in $1_c$ and $1_{c,c}$ except for the identity are $(1,-z)$ and $(-1,-z)$, respectively.

Proof

Left as exercise to the reader. $\square$

Now, let us focus on the products of the pairs of subgroups, from which $1_c$ and $1_{c,c}$ originated.

Proposition 18

Let $A=\left( 1,f(z)\right) \in H[0,0,0]$ , $T=\left( \frac{zf'(z)}{f(z)},f(z)\right) \in H[-1,1,0]$, $D=\left( f'(z),f(z)\right) \in H[0,1,0]$, $B=\left( \frac{f(z)}{z},f(z)\right) \in H[1,0,0]$, where $f=\bar{f}$, and f is fixed, we have that

$$\begin{aligned} A*T\,= & {} \,B*D\,=\,\left( \frac{f(z)}{zf'(z)},z\right) \quad \text {and} \end{aligned}$$

(10)

$$\begin{aligned} T*A\,= & {} \,D*B\,=\,\left( \frac{zf'(z)}{f(z)},z\right) . \end{aligned}$$

(11)

Proof

It is clear from Proposition 5. $\square$

Usually, matrices do not satisfy commutativity and Riordan arrays cannot be an exception, in general. We also notice that although subgroups such as $1_c$ and $1_{c,c}$ are abelian, commutativity cannot be inherited to any c-extension form of a commutative Riordan subgroup. A counterexample is the Appell subgroup which is abelian, while its c-Appell extension $\lbrace (g(z),cz) | g(z) \in \mathbb {F}_0 \rbrace$ is not.

The Power-Bell subgroup

$$\begin{aligned} \text {Power-Bell}\left( n\right) =\left\{ \Bigg (\left( \frac{f(z)}{z}\right) ^n,f(z)\Bigg ) \Bigg | n \in \mathbb {Z} \right\} , \end{aligned}$$

represents a whole family of Riordan subgroups, for any value of $n,\ldots ,$ where $n\in \mathbb {Z}$ [12], while for the trivial cases of $n=0$ and $n=1$, it collapses to the Associated and the Bell subgroups, respectively. Common Riordan elements with other subgroups can be found in some of the following cases.

Proposition 19

[5] The intersections of the subgroups H[n, 0, 0] and $H[-1,1,0]$ for the same arbitrary function f are given by the Riordan subgroup of the general form

$$\begin{aligned} B_n= & {} \left\{ \left( \frac{1}{1-cz^n}, \frac{z}{\root n \of {1-cz^n}} \right) \Bigg | c\in \mathbb {C}\right\} \end{aligned}$$

(12)

which contains the cyclic Riordan subgroup generated by the element

$$\begin{aligned} \left( \frac{1}{1-z^n}, \frac{z}{\root n \of {1-z^n}}\right) . \end{aligned}$$

For $n=1$, see Proposition 8.

Proposition 20

For $n \in \mathbb {Z}$, we have that

$$\begin{aligned} H[n+1,0,0] \cap H[0,1,0]\,= & {} \,\left( \frac{f(z)}{z}, z \right) *H[n,0,0] \cap H[-1,1,0]. \end{aligned}$$

(13)

Proof

We observe that the terms $\frac{f(z)}{z}$ and $f'(z)$ are contained in both sides of Eq. 13. Solving the LHS, we get the differential equation

$$\begin{aligned} \left( \frac{f(z)}{z} \right) ^{n+1} = f'(z), \end{aligned}$$

(14)

and the solution $f(z)=\frac{z}{(1-cz)^{\frac{1}{n}}}$, where c is the constant. Hence,

$$\begin{aligned} H[n+1,0,0] \cap H[0,1,0] = \left( \frac{1}{(1-cz^n)^{\frac{1}{n}+1}}, \frac{z}{(1-cz^n)^{\frac{1}{n}}} \right) \end{aligned}$$

For the intersection of the right-hand side of Eq. 13, we also get Eq. 14; hence,

$$\begin{aligned} H[n,0,0] \cap H[-1,1,0] = B_n = \left( \frac{1}{1-cz^n}, \frac{z}{(1-cz^n)^{\frac{1}{n}}} \right) . \end{aligned}$$

$\square$

Proposition 20 can also be extended, by using “Power-Derivative” and “Power-Hitting time” subgroups, as follows.

Corollary 2

For $m, k \in \mathbb {Z}$, we have that

$$\begin{aligned} H[m+k,0,0] \cap H[0,k,0]= & {} \left( \frac{f(z)}{z}, z \right) *H[m,0,0] \cap H[-k,k,0]. \end{aligned}$$

(15)

In Table 3, we have gathered all the possible intersections of the Riordan subgroups that we mentioned in this section.

Table 3 Intersections of Riordan subgroups

4 Relationships of H[r, s, p] and other Riordan subgroups

In the current section, we present some of the Riordan subgroups which cannot be expressed in terms of H[r, s, p], some of their algebraic properties, their connections with H[r, s, p], and any possible relationship among them.

4.1 The Appell subgroup

For every G(z) that can be written as $\frac{g(z)}{k(z)}$, where $g(z), k(z) \in \mathbb {F}_0$ and $k(z) \ne 1$, the subgroup $\lbrace (G(z),z) | G(z) \in \mathbb {F}_0 \rbrace$ is a subgroup of the Appell subgroup and it is normal. Using this form of the Appell subgroup, we present new semi-direct products of the Riordan group, as shown in Table 4.

Table 4 Semi-direct products

We note that we are also able to express the Riordan group as a semi-direct product of the Appell and the Stabilizer subgroups, as $\left( \frac{h(f(z))g(z)}{h(z)},z\right) \ltimes \left( \frac{h(z)}{h(f(z))}, f(z) \right)$. The Stabilizer subgroup is not included in the collection of isomorphic subgroups, because of the existence of the arbitrary function h(z). Nevertheless, by Group Theory, we have the following lemma.

Lemma 1

Let N be a normal subgroup of the group G, and let A and B be two subgroups of G, such that $G=N \ltimes A$, and $G=N \ltimes B$, then $A \backsimeq B$.

Proof

Applying the Second Isomorphism Theorem [2, Thm. 34.5, p. 308–309], we have that

$$\begin{aligned} NA \bigg / N \backsimeq A \bigg /( N \cap A )\quad \text {, and }\quad NB \bigg / N \backsimeq B \bigg / ( N \cap B ) \end{aligned}$$

(16)

By the definition of a semi-direct product, we have $N \cap A = N \cap B= I$, where I is the unitary subgroup, and $NA=NB=G$. Hence, Eq. 16 becomes

$$\begin{aligned} G / N \backsimeq A \quad \text {, and }\quad G / N \backsimeq B \end{aligned}$$

Hence, $A \backsimeq B$. $\square$

In our case, as the Stabilizer subgroup can be used in a semi-direct product, using this lemma, we get the following corollary.

Corollary 3

The Stabilizer subgroup is isomorphic to any subgroup in H[r, s, p].

Moreover, Appell matrices have already appeared in Proposition 5. We have the following corollary.

Corollary 4

Elements of the Appell subgroup can be expressed as products of two involutions of H[r, s, p].

4.2 The Checkerboard subgroup

Although the Checkerboard subgroup is contained in the Cheon subgroup [4], the set of elements $(g_{e}(z),f_{o}(z))$, where $g_e$ is an even and $f_o$ is an odd function, are elements of other Riordan subgroups.

Proposition 21

Riordan elements of H[r, s, p] are also elements of the Checkerboard subgroup, if and only if f is odd, and $p=0$.

Proof

The general element $H_f[ \rho , \sigma , \pi ]=\left( \left( \frac{f(z)}{z} \right) ^{ \rho } (f'(z))^{ \sigma } \left( \frac{f(z)-1}{z-1} \right) ^{ \pi } , f(z) \right)$ belongs to the Checkerboard subgroup if and only if f(z) is an odd function, and $\left( \frac{f(z)}{z} \right) ^{ \rho } (f'(z))^{ \sigma } \left( \frac{f(z)-1}{z-1} \right) ^{ \pi }$ is an even function. For f :  odd, we have that $\left( \frac{f(z)}{z} \right) ^{ \rho } (f'(z))^{ \sigma }$ is even for every value of $\rho , \sigma \in \mathbb {Q}$. Nevertheless, this condition is not satisfied for the term $\frac{f(z)-1}{z-1}$, as for $f(-z)=-f(z)$, we get that $\frac{f(z)-1}{z-1} \ne \frac{f(-z)-1}{-z-1}$; hence, it is not even. So, the first generating function of $H_f[ \rho , \sigma , \pi ]$ can be even only for $p=0$. $\square$

From Propositions 6 and 21, we come to the conclusion that every power-Stochastic subgroup which corresponds to the term $\left( \frac{f(z)-1}{z-1} \right) ^p$ remains unaffected by the parity of the function f(z). That leads us to the following corollary.

Corollary 5

The Checkerboard subgroup and every subgroup of the form $H[\rho ,\sigma ,\pi ]$, where $\pi \ne 0$, have no non-trivial common Riordan elements.

Elements of the Appell subgroup, where g is an even function also belong to the Checkerboard subgroup. More precisely, we have the corollary below.

Corollary 6

The set of Riordan elements $\hbox {App}_{e}=\lbrace (g_{e}(z),z) \rbrace$, where $g_{e}$ is even, is an abelian and normal subgroup of the Appell subgroup.

Similarly, elements of the Associated subgroup, where f is an odd function, also belong to the Checkerboard subgroup. Hence, we have the following proposition.

Proposition 22

The Checkerboard subgroup can be written as a semi-direct product of the subgroups $\hbox {App}_{e}=\lbrace (g_{e}(z),z)| g_{e}: \text { even} \rbrace$ and $\hbox {Assoc}_{o}=\lbrace (1,f_o(z)) | f_{o}: \text { odd} \rbrace$.

Proof

First, we have that $\hbox {App}_{e} \cap \hbox {Assoc}_{o} = I$. It suffices to show that $\hbox {App}_{e}$ is a normal subgroup of the Checkerboard subgroup. Let (e(z), o(z)) be an arbitrary element of the Checkerboard subgroup, where e(z) and o(z) are even and odd functions, respectively. Then, the inverse of this element will be

$$\begin{aligned} (e(z),o(z))^{-1} = \left( \frac{1}{e(\bar{o}(z))}, \bar{o}(z) \right) \end{aligned}$$

So, we have that

$$\begin{aligned} (e(z),o(z))^{-1} *\hbox {App}_{e} *(e(z),o(z))\,= & {} \,\left( \frac{1}{e(\bar{o}(z))}, \bar{o}(z) \right) *(g_e(z),z) *(e(z),o(z)) \\= & {} \,\left( \frac{g_e(\bar{o}(z))}{e(\bar{o}(z))} e(\bar{o}(z)), o(\bar{o}(z)) \right) \\= & {} \,(g_e(\bar{o}(z)), z), \end{aligned}$$

and $\bar{o}(z)$ has to be an odd function, so $g_e(\bar{o}(z))$ is an even function, which means that

$$\begin{aligned} (e(z),o(z))^{-1} *\hbox {App}_{e} *(e(z),o(z)) \in \hbox {App}_{e}. \end{aligned}$$

Hence, $(g_o(z), f_e(z))=(g_{e}(z),z) \ltimes (1,f_o(z))$ $\square$

4.3 The Stabilizer subgroup

The Stabilizer subgroup $S_h= \lbrace (g(z),f(z)) \vert (g(z),f(z)) *h(z)=h(z) \rbrace$ [11] is defined as a subgroup which stabilizes a column vector h(z), according to Theorem 1 as follows

$$\begin{aligned} (g(z),f(z)) *h(z)= & {} h(z)\\ \Leftrightarrow g(z)h(f(z))= & {} h(z). \end{aligned}$$

The Associated and the Stochastic subgroups are the stabilizers of the column vectors $(1,0,0,\ldots )^{T}$ and $(1,1,1,\ldots )^{T}$, respectively [19]. Jean-Louis and Nkwanta questioned the existence of other column vectors which are stabilized by any other Riordan subgroup [12], while He added that not all subgroups are stabilizers [11]. We present some forms of stabilizers for appropriate values of the parameters of H[r, s, p].

Proposition 23

Let H[r, s, p] be a family of Riordan subgroups.

• a. Riordan arrays of the subfamily H[r, 0, p] are stabilizers of column vectors of the form $\frac{1}{z^r(1-z)^p}$, where p is even.

• b. Riordan arrays of the subfamily $H[-s,s,p]$ are stabilizers of column vectors of the form $\frac{z^s}{(z-1)^p}$, where p is even.

Proof

• a. We have

  $$\begin{aligned} H[r,0,p] *\frac{1}{z^r(1-z)^{p}} =\,\frac{1}{z^r}\left( \frac{f(z)-1}{z-1}\right) ^{p}\frac{1}{(1-f(z))^{p}}. \end{aligned}$$

  For $p=2k$, we get

  $$\begin{aligned} \frac{1}{z^r}\left( \frac{f(z)-1}{z-1}\right) ^{2k}\frac{1}{(1-f(z))^{2k}}= & {} \,\frac{1}{z^r(z-1)^{2k}}. \end{aligned}$$

  Hence, p must be even.

• b. Similarly, we have that

  $$\begin{aligned} H[-s,s,p] *\frac{1}{((lnz)')^s(1-z)^{p}}= & {} \frac{z^s(f(z)-1)^p}{(z-1)^{p}(1-f(z))^p}. \end{aligned}$$

  For $p=2k$, we get

  $$\begin{aligned} \frac{z^s(f(z)-1)^{2k}}{(z-1)^{2k}(1-f(z))^{2k}}= & {} \frac{z^s}{(z-1)^{2k}}. \end{aligned}$$

  Hence, we proved that

  $$\begin{aligned} H[-s,s,2k] *\frac{1}{((lnz)')^s(1-z)^{2k}}=\frac{z^s}{(z-1)^{2k}}. \end{aligned}$$

$\square$

4.4 Relationships of non-H[r, s, p] subgroups

For elements of the Checkerboard subgroup which can be expressed in a stabilizer form, we have the following proposition.

Proposition 24

A Stabilizer element $\left( \frac{h(z)}{h(f(z))},f(z)\right)$ is contained in the Checkerboard subgroup, if and only if f is an odd function and h is either odd or even.

Proof

Suppose that f is an odd function, i.e., $f(z)=-f(-z)$, then we have two cases.

If h is also odd, we have $h(z)=-h(-z)$. Then, for $h\circ f$, we have that

$$\begin{aligned} h(f(z))=h(-f(-z))=-h(f(-z)). \end{aligned}$$

Hence, $h\circ f$ is odd and $\frac{h(z)}{h(f(z))}$ is then even as the quotient of two odd functions.

Similarly, if h is even, then the composition $h\circ f$ is even and the quotient $\frac{h(z)}{h(f(z))}$ is also even. In both cases, $\left( \frac{h(z)}{h(f(z))},f(z)\right)$ is contained in the Checkerboard subgroup.

Now suppose that $\left( \frac{h(z)}{h(f(z))},f(z)\right)$ is contained in the Checkerboard subgroup, it can be easily proven that f has to be an odd function and h has to be an odd or even function. $\square$

Corollary 7

The intersection of the Stabilizer and Checkerboard subgroups gives rise to the Riordan subgroup

$$\begin{aligned} \hbox {Stab}\cap \hbox {Checkb}\,= & {} \,\left\{ \left( \frac{h_{e}(z)}{h_{e}(f_o(z))},f_o(z)\right) \quad \bigg | f_o:\text {odd,}h_e: \text {even} \right\} \nonumber \\&\cup \left\{ \left( \frac{h_{o}(z)}{h_{o}(f_o(z))},f_o(z)\right) \quad \Bigg | f_o:\text {odd, }h_o: \text {odd} \right\} \nonumber \\= & {} \hbox {Stab}(h_e,f_o) \cup \hbox {Stab}(h_o,f_o). \end{aligned}$$

(17)

We observe that this new-formed subgroup, contrary to other Riordan subgroups which came as intersections of already known subgroups, is not abelian. Nevertheless, Eq. 17 allows us to characterize Riordan subgroups according to the form of their h function. Hence, we have two main categories of subgroups of $Stab\cap Checkb$, let us denote them as $(h_e,f_o)$ and $(h_o,f_o)$.

5 Summary

By defining H[r, s, p], we have shown that

$$\begin{aligned} \mathbb {F}_1[[z]]\simeq & {} \text {Associated}\simeq \text {Bell} \simeq \text {Derivative} \simeq \text {Stochastic} \simeq \text {Hitting time} \\\simeq & {} \text {Power-Bell} \simeq \text {Stabilizer} \simeq H[\rho , \sigma , \pi ], \end{aligned}$$

where $\mathbb {F}_1[[z]]$ is the group of fps under composition, and $(\rho , \sigma , \pi ) \in \mathbb {Q}^3$. Additionally, some intersections of the Riordan subgroups are abelian. As a result, we have partially answered the question which was proposed by Jean-Louis and Nkwanta [12], regarding the existence of other commutative subgroups. We have also presented new semi-direct products of the Riordan group, stabilizers of H[r, s, p], and conditions for the involutions and the pseudo-involutions of this family of Riordan subgroups.

The diagram in Fig. 1 contains the known Riordan subgroups that we investigated and the new Riordan subgroups that we found in this investigation.

Fig. 1

Diagram of Riordan subgroups

5.1 Future work and some open problems

Most of our work is related, but not limited to the algebraic structures of the Ordinary Proper Riordan arrays. Hence, part of our current and future work is to generalize our findings to other kinds of Riordan arrays [6, 7], explore the behavior of Riordan sets and subgroups, and expand Riordan group theory.

During our study, we found some intriguing questions. Some of them are still under investigation, while others may not have a clear answer. We list them in this subsection as open problems.

• 1. By Proposition 1, we get necessary conditions for pseudo-involutions. Are there any sufficient conditions?

• 2. The Riordan subgroups

  $$\begin{aligned} P^c {=} \left\{ \left( \frac{1}{1-cz},\frac{z}{1-cz}\right) \Bigg | n {\in } \mathbb {C} \right\} \text {, and } B_n {=} \left\{ \left( \frac{1}{1-cz^n}, \frac{z}{\root n \of {1-cz^n}} \right) \Bigg | c,n \in \mathbb {C}\right\} \end{aligned}$$

  are cyclic. Is there any other significant cyclic subgroup in the Riordan group?

• 3. Can we construct a semi-direct product for the Riordan group, without using the Appell subgroup? Equivalently, is there any other normal Riordan subgroup, except for the Appell which can be used for that purpose?

• 4. Is there any non-trivial $h(z) \in \mathbb {F}_k$, for $k \in \mathbb {N}$, which enables us to write the Derivative subgroup $(f'(z), f(z))$ in a stabilizer form, as

  $$\begin{aligned} f'(z)= & {} \frac{h(z)}{h(f(z))} , \end{aligned}$$

  where $f\in \mathbb {F}_1$?