On subpolygroup commutativity degree of finite polygroups

1.   Introduction

Probability in finite groups has grabbed the interest of algebraists in the last few years. One of the concepts that have been studied is the probability that two group elements of a finite group $G$ commute ^[17], denoted by $d(G)$. Another concept ^[10] is the relative commutativity degree of a subgroup $H$ of a finite group $G$, denoted by $d(H, G)$. The latter measures the probability that an element of the subgroup $H$ commutes with an element of the group $G$. Furthermore, subgroup commutativity degree of a finite group $G$ ^[21] measures the probability of two random subgroups of $G$ commuting. Other related work can be found in ^[7,12,13], and a survey on statistical group theory can be found in ^[8]. On the other hand, hypergroup theory, a generalization of group theory, is a field that was introduced by Marty ^[14] in 1934. Special classes of hypergroups are canonical hypergroups, introduced in 1970 ^[15,16], and quasi-canonical hypergroups, introduced in 1981 ^[3,4]. The latter was studied by Comer ^[5] in 1984 under the name polygroup. For details about hyperstructure theory and its applications, we refer to ^[6,9]. Researchers involved in this field try to check the validity of the known results in group theory for hypergroups. Indeed, some generalizations have been accomplished, but the fact that the class of hypergroups is much larger than that of groups makes it more difficult to generalize many concepts Recently, there has been a growing interest in the use of probability in finite polygroup theory. Some related concepts were introduced such as the commutativity degree of finite polygroups, and some related work can be found in ^[18,19,20].

Inspired by the subgroup commutativity degree of a finite group $G$, our paper generalizes this concept to finite polygroups, and it is organized as follows: After an introduction, Section 2 presents some results related to the subgroup commutativity degree of finite groups. Then, Section 3 presents some results on the subpolygroups lattice of a particular class of polygroups. Finally, Section 4 defines the subpolygroup commutativity degree of a finite polygroup and presents some related results and examples by using the subpolygroup commutativity table. Moreover, it considers a special class of polygroups and finds an explicit formula for the subpolygroup commutativity degree of these.

2.   Subgroup commutativity degrees of finite groups

In ^[21], Tarnauceanu defined the subgroup commutativity degree of finite groups and found explicit formulas for the subgroup commutativity degrees of some special finite groups. In this section, we present some of their results and discuss some other related results.

Definition 2.1. Let $(G, \cdot)$ be a finite group and $L(G)$ be the set of all subgroups of $G$. Then, the subgroup commutativity degree of $G$ is defined as follows:

Remark 1. If all subgroups of $G$ are normal, for instance, if $G$ is abelian or if $G$ is the quaternion group $Q_8$, then $sd(G) = 1$.

Remark 2. Let $(G, \cdot)$ be a finite group. Then, $0 < sd(G)\leq 1$.

Definition 2.2. Let $k$ be a positive integer and $(G, \cdot)$ be a finite group with distinct subgroups $H_1, \ldots, H_k$. Then, the subgroup commutativity table of $G$ is defined in Table 1.

Here, for all $1\leq i, j\leq k$, $H_{ij} = \begin{cases} 1 & { if\ H_i\cdot H_j = H_j\cdot H_i, }\\ 0 & { otherwise.} \end{cases}$

Remark 3. Let $(G, \cdot)$ be a finite group with subgroup commutativity table $(H_{ij})$. Then,

Example 1. Let $S_3$ be the symmetric group on three letters. Then, the subgroup commutativity table of $S_3$ is given in Table 2.

It is clear that $sd(S_3) = \frac{5}{6}$.

Proposition 2.1. Let $(G, \cdot)$ be a finite group. Then, $sd(G) = 1$, or $sd(G)\geq \frac{9}{|L(G)|^2}$.

Proof. Let $G$ be a finite group with identity $e$ and $M(G) = \{(H, K)\in L(G)^2:HK = KH\}$. We consider the following cases.

If $G$ is the trivial group, or $G$ has no proper non-trivial subgroups, then $sd(G) = 1$. If $G$ has a proper non-trivial subgroup $H$, then $(\{e\}, \{e\})$, $(\{e\}, H)$, $(\{e\}, G)$, $(H, \{e\})$, $(H, H)$, $(H, G)$, $(G, \{e\})$, $(G, H)$, and $(G, G)$ are all in $M(G)$. Thus, $|M(G)|\geq 9$, and hence, $sd(G) = \frac{|M(G)|}{|L(G)|^2}\geq \frac{9}{|L(G)|^2}$. □

Example 2. Let $S_3$ be the symmetric group on three letters. Then, $sd(S_3) = \frac{5}{6}\geq \frac{9}{36} = \frac{9}{|L(S_3)|^2}$.

Proposition 2.2. Let $(G, \cdot)$ be a finite group and $R$ be the relation on $G$ defined as follows:

Then, $R$ is a reflexive and symmetric relation.

Proof. The proof is straightforward by using the subgroup commutativity table of $(G, \cdot)$. □

Theorem 2.1. ^[21] Let $G, G^\prime$ be finite groups with coprime orders. Then,

The dihedral group $D_n$ is the symmetry group of a regular polygon with $n$ sides. and it has the order $2n$. The generalized quaternion group $Q_{2^m}$ can be expressed via the following presentation:

The quasi-dihedral group $S_{2^m}$ with $m\geq 4$ can be expressed via the following presentation:

In ^[21], Tarnauceanu found explicit formulas for the subgroup commutativity degrees of some finite groups such as the dihedral group, quasi-dihedral group, and generalized quaternion group. Let $n\geq 2$ be a positive integer and $n = p_1^{\alpha_1}\ldots p_k^{\alpha_k}$ be the decomposition of $n$ as a product of prime factors. Then, $\tau(n)$, $\sigma(n)$ are the number and the sum of all divisors of $n$, respectivelym and

Theorem 2.2. ^[21] Let $D_n$ be as above and $n = 2^\alpha n^\prime$ with $n^\prime$ odd. Thenm

Example 3. Let $D_4, D_6$ be the symmetry groups of regular polygons with $4, 6$ sides, respectively. Then, $sd(D_4) = \frac{22}{25}$ and $sd(D_6) = \frac{101}{128}$.

Theorem 2.3. ^[21] Let $D_{2^{m-1}}$, $Q_{2^m}$, and $S_{2^m}$ be as above. Then,

In particular, $\lim_{m\rightarrow \infty}sd(D_{2^{m-1}}) = \lim_{m\rightarrow \infty}sd(Q_{2^m}) = \lim_{m\rightarrow \infty}sd(S_{2^m}) = 0$.

3.   Subpolygroup lattice of polygroups

In this section, we present some basic results and examples related to polygroup theory that are used throughout the paper. Moreover, we study the subpolygroup lattice for a particular class of polygroups.

3.1.   Polygroups

Let $P$ be a non-empty set and $\mathcal{P}^{\ast}(P)$ be the family of all non-empty subsets of $P$. A binary hyperoperation on $P$ is a mapping $\circ: P \times P \rightarrow \mathcal{P}^{\ast}(P)$. The couple $(P, \circ)$ is called a hypergroupoid.

In the above definition, if $H$ and $K$ are two non-empty subsets of $P$ and $p\in P$, then we define:

Definition 3.1. ^[5] A polygroup is a system $\langle P, \cdot, e, ^{-1}\rangle$, where $(P, \cdot)$ is a hypergroupoid, $e\in P$, $^{-1}:P\rightarrow P$ is a unary operation on $P$, and the following axioms hold for all $x, y, z\in P$:

(1) $(x\cdot y)\cdot z = x\cdot (y\cdot z)$,

(2) $e\cdot x = x\cdot e = \{x\}$,

(3) $x\in y\cdot z$ implies $y\in x\cdot z^{-1}$ and $z\in y^{-1}\cdot x$.

For simplicity, we write $x$ instead of $\{x\}$ for all $x$ in the polygroup $\langle P, \cdot, e, ^{-1}\rangle$.

A canonical hypergroup $\langle P, \cdot, e, ^{-1}\rangle$ is a commutative polygroup, i.e., $x\cdot y = y\cdot x$ for all $x, y\in P$. For more details, we refer to ^[15,16].

Example 4. Let $P = \{m_0, m_1, m_2\}$ and $(P, \circ)$ be defined by Table 3.

The identity under $\circ$ is $n_0$, and $x^{-1} = x$ for all $x\in P$. Then, $\langle P, \circ, m_0, ^{-1}\rangle$ is a canonical hypergroup.

Example 5. ^[11] Let $P^\prime = \{n_0, n_1, n_2, n_3\}$ and $(P^\prime, \cdot)$ be defined by Table 4.

Then, $\langle P^\prime, \cdot, n_0, ^{-1}\rangle$ is a non-canonical hypergroup.

Remark 4. Every group is a polygroup.

Let $\langle P, \circ, e, ^{-1}\rangle$ be a polygroup and $H\subseteq P$. Then, $H$ is a subpolygroup of $P$ if for all $x, y\in H$, we have $x\circ y\subseteq H$ and $x^{-1}\in H$.

Example 6. Let $\langle P^\prime, \cdot, n_0, ^{-1}\rangle$ be the polygroup in Example 5. Then, $\{n_0\}$ and $P^\prime$ are the only subpolygroups of $P^\prime$, i.e., $P^\prime$ has no non-trivial proper subpolygroups.

In ^[11], Jafarpour et al. described a method to get a polygroup from a group. Let $(G, \cdot)$ be a group, $a\notin G$, and $P_G = G\cup \{a\}$. Define "$\circ$" on $P_G$ as follows:

$(1)$ $a\circ a = e$;

$(2)$ $e\circ x = x\circ e = x$ for all $x\in P_G$;

$(3)$ $a\circ x = x\circ a = x$ for all $x\in P_G-\{e, a\}$;

$(4)$ $x\circ y = x\cdot y$ for all $x, y\in G$ with $y\neq x^{-1}$;

$(5)$ $x\circ x^{-1} = \{e, a\}$ for all $x\in P_G-\{e, a\}$.

Proposition 3.1. ^[11] If $(G, \cdot)$ is a group, then $\langle P_G, \circ, e, ^{-1}\rangle$ is a polygroup where $e$ and $^{-1}$ are the identity and inversion operations of $G$, respectively.

Example 7. Let $(\mathbb{Z}_3, +)$ be the group of integers modulo $3$ under standard addition modulo $3$. Then. $\langle P_{\mathbb{Z}_3}, \circ, 0, ^{-1})$ is a polygroup, and it is given by Table 5.

Definition 3.2. ^[9] Let $\langle P_1, \cdot_1, e_1, ^{-1}\rangle$, $\langle P_2, \cdot_2, e_2, ^{-1}\rangle$ be polygroups and $\psi:P_1\rightarrow P_2$ be a function. Then,

(1) $\psi$ is a homomorphism if $\psi(x\cdot_1 y)\subseteq \psi(x)\cdot_2 \psi(y)$ for all $x, y\in P_1$, and $\psi$ is a strong homomorphism if all these containments are equalities.

(2) $\psi$ is an isomorphism if it is a bijective strong homomorphism. In this case, we say that $P_1$ and $P_2$ are isomorphic polygroups.

3.2.   Subpolygroup lattice of $P_G$

We classify the subpolygroup lattice of $P_G$ in relation to the subgroup lattice of $G$.

Theorem 3.1. Let $(G, \cdot)$ be a group, $P_G$ be its associated polygroup, and $\emptyset\neq N\subseteq P_G$. Then, $N$ is a subpolygroup of $P_G$ if and only if $N = \{e\}$ or $N = P_S$ for some subgroup $S$ of $G$.

Proof. Let $S$ be a subgroup of $G$. Having $e\in S$ implies that $e\in P_S$, and hence, $P_S\neq\emptyset$. Let $x\in P_S$. Then,

For $x, y\in P_S$, we have

Thus, $P_S$ is a subpolygroup of $P_G$.

Conversely, let $N\neq\{e\}$ be a subpolygroup of $P_G$. Then, there exists $x\neq e\in N$. Since $N$ is a subpolygroup of $P_G$, it follows that $x^{-1}\in N$, and hence, $x\circ x^{-1} = \begin{cases} \{e, a\}, & \hbox{ if} \ x\neq a, \\ \{a\}, & \hbox{ otherwise, } \end{cases}\subseteq N$. Having $a\in N$ implies that we can write $N = S\cup \{a\}$ (with $a\notin S$). We need to show that $S$ is a subgroup of $G$. Let $x\in S$. Then, $x\neq a\in N$, and hence, $x^{-1}\neq a\in N$. Thus, $x^{-1}\in S$. Let $x, y\in S$. Then, $x\circ y = \begin{cases} xy\neq a, & \hbox{ if}\ y\neq x^{-1}, \\ \{e, a\}, & \hbox{ otherwise, } \end{cases}\subseteq N$. Thus, $xy\in S$. □

Corollary 3.1. Let $n$ be a positive integer, $(\mathbb{Z}_n, +)$ be the group of integers modulo $n$ under standard addition of integers modulo $n$, and $S$ be a subpolygroup of the polygroup $P_{\mathbb{Z}_n}$. Then, $S = \{0\}$ or $S = P_{\langle k\rangle }$ for integers $k$ that are divisors of $n$.

Theorem 3.1 is important to construct the subpolygroup lattice of the associated polygroup $P_G$.

Example 8. Let $(\mathbb{Z}_6, +)$ be the group of integers modulo $6$ under standard addition modulo $6$ and $P_{\mathbb{Z}_6}$ be its associated polygroup. Then, the subgroup lattice of $\mathbb{Z}_6$ is presented in Figure 1, and the subpolygroup lattice of $P_{\mathbb{Z}_6}$ is presented in Figure 2.

Notation 1. For a group $G$ and a polygroup $P$, $L(G), L(P)$ denote the sets of subgroups of $G$ and subpolygroups of $P$, respectively, and $|L(G)|, |L(P)|$ are their cardinalities.

Corollary 3.2. Let $(G, \cdot)$ be a finite group and $\langle P_G, \circ, e, ^{-1}\rangle$ its associated polygroup. Then, $|L(P_G)| = |L(G)|+1$.

Lemma 3.1. Let $(G, \cdot)$ be a group with subgroups $H$ and $K$ and $\langle P_G, \circ, e, ^{-1}\rangle$ be the associated polygroup. Then, $HK = KH$ if and only if $P_H\circ P_K = P_K\circ P_H$.

Proof. Let $HK = KH$. If $P_H\circ P_K = (H\cup\{a\})\circ(K\cup\{a\}) = HK\cup\{a\}$ and $HK = KH$ then $P_H\circ P_K = KH\cup\{a\} = P_K\circ P_H$. Similarly, if $P_H\circ P_K = P_K\circ P_H$, then $HK = KH$. □

Corollary 3.3. Let $(G, \cdot)$ be a group with subgroups $H$ and $K$ and $\langle P_G, \circ, e, ^{-1}\rangle$ be the associated polygroup. Then, $HK\in L(G)$ if and only if $P_H\circ P_K\in L(P_G)$.

Proof. This follows from Lemma 3.1. □

A lattice $L$ is called modular if, for any $x, y, z\in L$ with $x\leq y$, $x\vee (y\wedge z) = y\wedge (x\vee z)$. For more details about lattice theory, we refer to ^[2]. We prove that under a certain condition, the lattice subpolygroup of the associated polygroup is modular.

Lemma 3.2. Let $\left(G, \cdot \right)$ be a group and $\langle P_G, \circ, e, ^{-1}\rangle$ be its associated polygroup. If $HK = KH$ for all subgroups $H, K$ of $G$, then $\left(L(P_{G}\right), \wedge, \vee)$ is a lattice associated to $P_{G}$. Here,

Proof. It is clear that $P_{H_{i}}\cap P_{H_{j}}\in L(P_{G})$. Corollary 3.3 implies that $H_{i}H_{j}\in L(G)$ if and only if $P_{H_{i}}\circ P_{H_{j}}\in L(P_{G})$. We need to prove that $P_{H_{i}}\cup P_{H_{j}}\subseteq P_{H_{i}}\circ P_{H_{j}}.$ Let $x\in P_{H_{i}}\cup P_{H_{j}}$. Without loss of generality, we suppose that $x\in P_{H_{i}}$. Having $e\in P_{H_{j}}$, for any $j\in \{1, ..., |L(G)|\}$ implies that $x = x\circ e\subseteq P_{H_{i}}\circ P_{H_{j}}$. Now, we show that $P_{H_{i}}\circ P_{H_{j}}$ is the smallest polygroup which contains the subpolygroups $P_{H_{i}}$ and $P_{H_{j}}$. Let $P_{H_{k}}\in L(P_{G})$ such that $P_{H_{i}}\subseteq P_{H_{k}}$ and $P_{H_{j}}\subseteq P_{H_{k}}$. So, $P_{H_{i}}\circ P_{H_{j}}\subseteq P_{H_{k}}\circ P_{H_{k}} = P_{H_{k}}$. Therefore, $\left(L(P_{G}\right), \wedge, \vee)$ is a lattice associated to $P_{G}$. □

Theorem 3.2. Let $\left(G, \cdot \right)$ be a group and $\langle P_G, \circ, e, ^{-1}\rangle$ its associated polygroup. If $HK = KH$ for all subgroups $H, K$ of $G$ then $\left(L(P_{G}\right), \wedge, \vee)$ is a modular lattice associated to $P_{G}$.

Proof. Let $P_{H_{i}},$ $P_{H_{j}},$ $P_{H_{k}}\in L(P_{G})$, where $i$, $j$, $k\in \{1, ..., |L(G)|\}$ such that $P_{H_{j}}\subseteq P_{H_{i}}$. Any lattice satisfies the modularity inequality:

We show that $P_{H_{i}}\wedge (P_{H_{j}}\vee P_{H_{k}})\subseteq P_{H_{j}}\vee \left(P_{H_{i}}\wedge P_{H_{j}}\right)$. Let $x\in P_{H_{i}}\wedge \left(P_{H_{j}}\vee P_{H_{k}}\right)$. Having $x\in P_{H_{i}}$ and $x\in$ $P_{H_{j}}\circ P_{H_{k}}$ implies that there exist $y\in P_{H_{j}}$, $z\in P_{H_{k}}$ such that $x\in y\circ z$. The latter implies that $z\in y^{-1}\circ x$. Having $y^{-1}\in P_{H_{j}}$ implies that

Having $z\in P_{H_{k}}$ and $z\in P_{H_{i}}$ implies that $x\in y\circ z\subseteq P_{H_{j}}\circ \left(P_{H_{i}}\cap P_{H_{j}}\right)$, and hence, $P_{H_{i}}\wedge (P_{H_{j}}\vee P_{H_{k}})\subseteq P_{H_{j}}\vee \left(P_{H_{i}}\wedge P_{H_{j}}\right)$ for any $P_{H_{i}}, P_{H_{j}}, P_{H_{k}}\in L(P_{G})$. □

Lemma 3.3. Let $G, G^\prime$ be any finite non-trivial groups with identities $e, e^\prime$, respectively. Introduce elements $a, b$ such that $\{a, b\}\cap (G\cup G^\prime) = \emptyset$ and $P_G = G\cup \{a\}$, $P_{G^\prime} = G^\prime\cup \{b\}$. Then, $A_1 = \{(e, e^\prime)\}$, $A_2 = \{(e, e^\prime), (a, b)\}$, $A_3 = \{e\}\times P_{S^\prime}$ (where $S^\prime$ is a subgroup of $G^\prime$), $A_4 = P_S\times \{e^\prime\}$ (where $S$ is a subgroup of $G$), $A_5 = P_S\times P_{S^\prime}$ (where $S, S^\prime$ are subgroups of $G, G^\prime$ respectively) are subpolygroups of $P_G\times P_{G^\prime}$.

Proof. The proof is straightforward. □

In group theory, it is well known that if $G$ and $G^\prime$ are finite groups with coprime orders and $A$ is a subgroup of $G\times G^\prime$ then there exist subgroups $S, S^\prime$ of $G, G^\prime$ respectively such that $A = S\times S^\prime$. This fact from group theory may not hold for polygroups. We illustrate this remark via Example 9.

Example 9. Let $G, G^\prime$ be any non-trivial groups with identities $e, e^\prime$ respectively. Introduce elements $a, b$ such that $\{a, b\}\cap (G\cup G^\prime) = \emptyset$ and $P_G = G\cup \{a\}$, $P_{G^\prime} = G^\prime\cup \{b\}$. Then, $M = \{(e, e^\prime), (a, b)\}$, represented by Table 6, is a subpolygroup of $P_G\times P_{G^\prime}$.

It is clear that $M$ can not be written as a Cartesian product of two subpolygroups.

4.   Subpolygroup commutativity degree of finite polygroups

In this section, inspired by the definition of the subgroup commutativity degree of finite groups ^[21], we define the subpolygroup commutativity degree of finite polygroups. First, we present some general results. Then, we make a complete study on special polygroups that are associated to finite groups. We find an explicit formula for the subpolygroup commutativity degree of these polygroups, and we present some interesting results.

Definition 4.1. ^[1] Let $\langle P, \cdot, e, ^{-1}\rangle$ be a finite polygroup. Then,

It is clear that $0 < sd(P)\leq 1$.

Remark 5. Let $\langle P, \cdot, e, ^{-1}\rangle$ be a finite polygroup. Then, $sd(P) = 1$ if $P$ is commutative, or every two subpolygroups commute.

Example 10. Let $\langle P^\prime, \cdot, n_0, ^{-1}\rangle$ be the non-canonical hypergroup in Example 5. Since $\{n_0\}, P^\prime$ are the only subpolygroups of $P^\prime$, it follows that $sd(P^\prime) = 1$.

Proposition 4.1. Let $\langle P, \cdot, e, ^{-1}\rangle$ be a polygroup and $H, K$ be subpolygroups of $P$. If $H\cdot K = K\cdot H$, then $H\cdot K\in L(P)$.

Proof. The proof is straightforward. □

Proposition 4.2. Let $\langle P, \cdot, e, ^{-1}\rangle$ be a finite polygroup. Then, $sd(P) = 1$ or $sd(P)\geq \frac{9}{|L(P)|^2}$.

Proof. The proof is similar to that of Proposition 2.1. □

Corollary 4.1. Let $\langle P, \cdot, e, ^{-1}\rangle$ be a finite polygroup. Then, the following statements hold.

(1) If $|L(P)|\leq 3$, then $sd(P) = 1$.

(2) If $|L(P)|\leq 9$, then $sd(P)\geq \frac{1}{2}$.

Proof. This follows from Proposition 4.2. □

Proposition 4.3. Let $\langle P_1, \cdot_1, e_1, ^{-1}\rangle$ and $\langle P_2, \cdot_2, e_2, ^{-1}\rangle$ be isomorphic finite polygroups. Then, $sd(P_1) = sd(P_2)$.

Proof. The proof is straightforward. □

Remark 6. The converse of Proposition 4.3 may not hold. The polygroups in Examples 4 and 5 have the same subpolygroup commutativity degree (which is equal to $1$), but they are non-isomorphic polygroups.

Definition 4.2. Let $\langle P, \cdot, e, ^{-1}\rangle$ be a finite polygroup with distinct subpolygroups $H_1, \ldots, H_k$ where $k$ is a positive integer. Then, the subpolygroup commutativity table of $P$ is defined via Table 7.

Here for all $1\leq i, j\leq k$, $H_{ij} = \begin{cases} 1, & { if \ H_i\cdot H_j = H_j\cdot H_i }, \\ 0, & { otherwise.} \end{cases}$

Remark 7. Let $\langle P, \cdot, e, ^{-1}\rangle$ be a finite polygroup with subpolygroup commutativity table $(H_{ij})$. Then,

Example 11. Let $S_3$ be the symmetric group on three letters and $P_{S_3}$ be its associated polygroup. Then, the subpolygroup commutativity table of $P_{S_3}$ is given in Table 8.

It is clear that $sd(P_{S_3}) = \frac{43}{49}$.

Proposition 4.4. Let $\langle P, \cdot, e, ^{-1}\rangle$ be a finite polygroup. Then, $sd(P)\geq \frac{5|L(P)|-6}{|L(P)|^2}$.

Proof. Using the subpolygroup commutativity table, the first row and column, the last row and column and the diagonal are entirely $1$'s. So, we have at least $5|L(P)|-6$ of $1$'s after removing the repetitive $1$'s. Therefore, $sd(P)\geq \frac{5|L(P)|-6}{|L(P)|^2}$. □

Next, we find a formula for the subpolygroup commutativity degree of the polygroup $P_G$ associated to a finite group $G$.

Theorem 4.1. Let $(G, \cdot)$ be a finite group and $P_G$ be its associated polygroup. Then,

Proof. We have $sd(P_G) = \frac{|M(P_G)|}{|L(P_G)|^2}$ where $M(P_G) = \{(A, B)\in L^2(P_G):A\circ B = B\circ A \}$ and $|L(P_G)| = |L(G)|+1$. Lemma 3.1 implies that $M(P_G) = M_0\cup M_1\cup M_2$ where $M_0 = \{(A, \{e\}): A\in L(P_G)\}$, $M_1 = \{(\{e\}, A): A\in L(P_G)\}$ and $M_2 = \{(P_H, P_K):HK = KH\}$. Having $|M_0| = |M_1| = |L(G)|+1$, $M_0\cap M_1 = \{(e, e)\}$, $M_0\cap M_2 = M_1\cap M_2 = \emptyset$, and $|M_2| = |L(G)|^2sd(G)$ imply that

□

Example 12. Let $S_3$ be the symmetric group on three letters and $P_{S_3}$ be its associated polygroup. Having $sd(S_3) = \frac{5}{6}$ and $|L(S_3)| = 6$ implies that $sd(P_{S_3}) = \frac{30+2(6)+1}{7^2} = \frac{43}{49}$.

Corollary 4.2. Let $n = 2^\alpha n^\prime$ with $n^\prime$ odd, $D_n$ be the dihedral group, and $P_{D_n}$ be its associated polygroup. Then,

Proof. The proof follows from Theorems 2.2 and 4.1. □

Corollary 4.3. Let $m$ be a positive integer, and $P_{D_{2^{m-1}}}$, $P_{Q_{2^m}}$, $P_{S_{2^m}}$ be the associated polygroups of $D_{2^{m-1}}$, $Q_{2^m}$, and $S_{2^m}$, respectively. Then,

In particular, $\lim_{m\rightarrow \infty}sd(P_{D_{2^{m-1}}}) = \lim_{m\rightarrow \infty}sd(P_{Q_{2^m}}) = \lim_{m\rightarrow \infty}sd(P_{S_{2^m}}) = 0$.

Proof. The proof follows from Theorems 2.3 and 4.1. □

Proposition 4.5. Let $(G, \cdot)$ be a finite group and $P_G$ be its associated polygroup. Then, $sd(P_G)\geq sd(G)$. Moreover, the equality holds if and only if $sd(G) = 1$.

Proof. Having

implies that $sd(P_G)\geq sd(G).$ □

Proposition 4.6. Let $(G, \cdot)$ be a finite group and $P_G$ be its associated polygroup. Then, $sd(P_G)\leq \frac{sd(G)+1}{2}$.

Proof. If $|L(G)|\leq 3$, then $sd(G) = sd(P_G) = 1$, and the inequality holds. If $|L(G)|\rangle 3$, then $|L(G)|^2$$-2|L(G)|-1\rangle 0$. Having $|L(G)|^2-2|L(G)|-1\geq 0$ implies that

and hence, $sd(P_G)\leq \frac{sd(G)+1}{2}$. □

5.   Conclusions

This paper dealt with polygroup probability by introducing the subpolygroup commutativity degree of finite polygroups. Some basic results were elaborated, and some examples were presented. Moreover, an explicit formula for the subpolygroup commutativity degree of a particular class of finite polygroups was derived and applied to the polygroups associated to the dihedral group, to the quasi-dihedral group and to the generalized quaternion groups.

For future research, we raise some open problems.

$(1)$ Find an explicit formula for the subpolygroup commutativity degree of other classes of finite polygroups.

$(2)$ Given finite polygroups $P_1, \ldots, P_n$, find a necessary and sufficient condition so that $sd(P_1\times\ldots\times P_n) = sd(P_1)\ldots sd(P_n)$.

$(3)$ For a finite polygroup $\langle P, \cdot, e, ^{-1}\rangle$, find a relationship between its commutativity degree and its subpolygroup commutativity degree.

Use of AI tools declaration

The authors declare they have not used AI tools in the creation of this article.

Acknowledgments

The APC was supported by the Project for the Development of the Organization "DZRO Military autonomous and robotic systems-VAROPS".

The second author thanks the Ministry of Defence of the Czech Republic for the support. The authors are grateful for the referees valuable comments and suggestions that improved the quality of the manuscript.

Conflict of interest

The authors declare no conflict of interest.