Construction of 2-Gyrogroups in Which Every Proper Subgyrogroup Is Either a Cyclic or a Dihedral Group

Abstract

In this paper, a 2-gyrogroup $G(n)$ of order $2^n$, $n\ge 3$, is constructed in which every proper subgyrogroup is either a cyclic or a dihedral group. It is proved that the subgyrogroup lattice and normal subgyrogroup lattice of $G(n)$ are isomorphic to the subgroup lattice and normal subgroup lattice of the dihedral group of order $2^n$, which causes us to use the name dihedral gyrogroup for this class of gyrogroups of order $2^n$. Moreover, all proper subgyrogroups of $G(n)$ are subgroups.

1. Basic Concept and History

A pair $(G,\oplus )$ consisting of a nonempty set G and a binary operation $\oplus :G\times G⟶G$ is called a groupoid. A bijection $\varphi$ from the groupoid G to itself is called an automorphism of G if $\varphi (a\oplus b)=\varphi (a)\oplus \varphi (b)$, for all $a,b\in G$. The set of all automorphisms of G is denoted by $Aut(G)$. It is straightforward to check that $Aut(G)$ forms a group under composition of function.

A groupoid $(G,\oplus )$ is called a gyrogroup if the following conditions are satisfied:

• there exists an element $0\in G$ such that for all $x\in G$, $0\oplus x=x$;
• for each $a\in G$, there exists $b\in G$ such that $b\oplus a$$=0$;
• (gyroassociative law) there exists a function $gyr:G\times G⟶Aut(G)$ such that for every $a,b,c\in G$, $a\oplus (b\oplus c)=(a\oplus b)\oplus gyr[a,b]c$, where $gyr[a,b]c=gyr(a,b)(c)$;
• for each $a,b\in G$, $gyr[a,b]$=$gyr[a\oplus b,b]$.

The gyrogroup G is called gyrocommutative if and only if for all $a,b\in G$, $a\oplus b$=$gyr[a,b](b\oplus a)$. The function $gyr[a,b]$, $a,b\in G$, is called the gyroautomorphism generated by a and b. Note that the axioms of a gyrogroup apply their right counterpart. Suppose G is a group and $gyr[a,b]=1_G$, where $1_G$ denotes the trivial automorphism and $a,b$ are arbitrary elements of G. Then it is easy to see that G will have the structure of a gyrogroup. This shows that gyrogroup is a generalization of the classical notion of a group.

Abraham Ungar in his seminal paper [1] initiated a new approach to special theory of relativity and in [2], he employed the Thomas precession and its interplay with Einstein’s addition to formally defines gyrogroups. In 2001, Ungar published a book containing many results about gyrogroups and gyrovector spaces [2]. In [3], he applied gyrogroups and gyrovector spaces to introduce an interesting analogy between three models in hyperbolic geometry: the Poincar$\acute{\mathrm{e}}$ ball model, the Beltrami ball model and the PV space model. In [4], it is shown that the gyrocommutative gyrogroups and their resulting gyrovector spaces in hyperbolic geometry have the role as vector spaces in Euclidean geometry. In 2009, Ungar published another book in which he presented his idea that Thomas gyration turns Euclidean geometry into hyperbolic geometry, and what philosophical analogies the two geometries share [5]. Ungar published four other books [6,7,8,9] and some survey articles like [10] in which he explained the history and philosophy of his theory.

In this paper, we present an algebraic approach to constructing new gyrogroups. We will construct a gyrogroup $G(n),n\ge 3$, of order $2^n$ by considering a cyclic group of order $2^n$. It is proved that all proper subgyrogroups of $G(n)$ are either cyclic or dihedral groups. The structure of the subgyrogroup lattice of $G(n)$ will also be given.

2. Preliminaries

In this section we introduce some useful lemmas which are crucial in our main results. We refer to [11,12] for the main properties of the subgyrogroups, gyrogroup homomorphisms and quotient gyrogroup. Our calculations are done with the aid of GAP [13].

Throughout this paper $P(n)=\{0,1,2,\cdots ,2^{n-1}-1\}$, $H(n)=\{2^{n-1},2^{n-1}+1,\cdots ,2^n-1\}$ and $G(n)=P(n)\cup H(n)$, where $n\ge 3$ is a natural number. It is clear that $P(n)$ is a cyclic group under addition modulo $m=2^{n-1}$ and $H(n)=P(n)+m$. This shows that $G(n)=P(n)\cup (P(n)+m)$. Define the binary operation ⊕ on $G(n)$ as follows:

$$i\oplus j=\left\{\begin{array}{cc}\hfill t\hfill & \hfill (i,j)\in P(n)\times P(n)\hfill \\ \hfill t+m\hfill & \hfill (i,j)\in P(n)\times H(n)\hfill \\ \hfill s+m\hfill & \hfill (i,j)\in H(n)\times P(n)\hfill \\ \hfill k\hfill & \hfill (i,j)\in H(n)\times H(n)\hfill \end{array}\right.$$

and $t,s,k\in P(n)$ are the following non-negative integers:

$$\left\{\begin{array}{ccc}\hfill t& \hfill \equiv i+j\hfill & (modm)\\ \hfill s& \hfill \equiv i+(\frac{m}{2}-1)j\hfill & (modm)\\ \hfill k& \hfill \equiv (\frac{m}{2}+1)i+(\frac{m}{2}-1)j\hfill & (modm)\end{array}\right..$$

Suppose the greatest common divisor of positive integers r and s is denoted by $(r,s)$. It is obvious that the operation ⊕ is well-defined, and if $x\equiv y(modm)$ and $x,y$ are simultaneously in $P(n)$ or $H(n)$, then $x=y$.

Lemma 1.

Suppose $A:G(n)⟶G(n)$ is defined as:

$$A(i)=\left\{\begin{array}{cc}\hfill i\hfill & \hfill i\in P(n)\hfill \\ \hfill r+m\hfill & \hfill i\in H(n)\hfill \end{array}\right.$$

where $r\in P(n)$ and ${\displaystyle r\equiv i+\frac{m}{2}(modm)}$. Then A is an automorphism of $(G(n),\oplus )$.

Proof. 

It is clear that the mapping A is well defined, one to one and onto mapping on $G(n)$. To prove A is a homomorphism, we assume $i,j\in G(n)$ are arbitrary elements. If $i,j\in P(n)$, then it is obvious that $A(i\oplus j)=A(i)\oplus A(j)$. Our main proof will consider three separate cases as follows:

1.

$i\in P(n)$ and $j\in H(n)$. In this case, $i\oplus j=t_1+m\in H(n)$ such that $t_1\in P(n)$ and $t_1\equiv i+j(modm)$. Now by our assumption, $A(i)=i$, $A(j)=r_1+m$ and

$$A(i\oplus j)=A(t_1+m)=r_2+m,$$

(1)

where $r_1,r_2\in P(n)$, $r_1\equiv j+\frac{m}{2}(modm)$ and $r_2\equiv t_1+m+\frac{m}{2}(modm)$. By these equations,

$$A(i)\oplus A(j)=i\oplus (r_1+m)=t_2+m$$

(2)

such that $t_2\in P(n)$ and $t_2\equiv i+r_1+m(modm)$. Hence, $r_2$≡$t_1+\frac{m}{2}$≡$i+j+\frac{m}{2}$≡$i+r_1$≡$t_2(modm).$ Since $r_2,t_2\in P(n)$, $r_2=t_2$. We now apply Equations (1) and (2) to deduce that $A(i\oplus j)=A(i)\oplus A(j)$.

2.

$i\in H(n)$ and $j\in P(n)$. In this case, $i\oplus j=s_1+m\in H(n)$, where $s_1\in P(n)$ and $s_1\equiv i+(\frac{m}{2}-1)j(modm)$. By our assumption $A(i)=r_1+m$, $A(j)=j$ and

$$A(i\oplus j)=A(s_1+m)=r_2+m,$$

(3)

where $r_1,r_2\in P(n)$, $r_1\equiv i+\frac{m}{2}(modm)$ and $r_2\equiv s_1+m+\frac{m}{2}(modm).$ By these calculations, one can see that

$$A(i)\oplus A(j)=(r_1+m)\oplus j=s_2+m$$

(4)

such that $s_2\in P(n)$ and $s_2\equiv r_1+m+(\frac{m}{2}-1)j(modm)$. Hence, $r_2\equiv s_1+\frac{m}{2}$≡$i+(\frac{m}{2}-1)j+\frac{m}{2}$≡$r_1+(\frac{m}{2}-1)j$≡$s_2(modm).$ Since $r_2,s_2\in P(n)$, $r_2=s_2$, and by Equations (3) and (4), $A(i\oplus j)=A(i)\oplus A(j)$, as desired.

3.

$i,j\in H(n)$. By definition, $i\oplus j=k_1\in P(n)$, where $k_1\equiv (\frac{m}{2}+1)i+(\frac{m}{2}-1)j(modm)$. We now apply our assumption to deduce that $A(i)=r_1+m$, $A(j)=r_2+m$ and

$$A(i\oplus j)=A(k_1)=k_1$$

(5)

such that $r_1,r_2\in P(n)$, $r_1\equiv i+\frac{m}{2}(modm)$ and $r_2\equiv j+\frac{m}{2}(modm).$ By above calculations,

$$A(i)\oplus A(j)=(r_1+m)\oplus (r_2+m)=k_2$$

(6)

in which $k_2\in P(n)$ and $k_2\equiv (\frac{m}{2}+1)(r_1+m)+(\frac{m}{2}-1)(r_2+m)(modm)$. Hence, $k_2\equiv (\frac{m}{2}+1)r_1+(\frac{m}{2}-1)r_2$≡$(\frac{m}{2}+1)i+(\frac{m}{2}-1)j$≡$k_1(modm).$ Since $k_1,k_2\in P(n)$, again $k_1=k_2$ and by Equations (5) and (6), $A(i\oplus j)=A(i)\oplus A(j)$.

This completes our argument. □

Set $O_P=\{i\in P(n)\mid 2\nmid i\}$, $O_H=\{i\in H(n)\mid 2\nmid i\}$, $E_H=\{i\in H(n)\mid 2\mid i\}$ and $M=[O_P\times (O_H\cup E_H)]\bigcup [O_H\times (O_P\cup E_H)]\bigcup [E_H\times (O_P\cup O_H)]$. Define the map $gyr:G(n)\times G(n)⟶Aut(G(n),\oplus )$ as follows:

$$gyr(a,b)=gyr[a,b]=\left\{\begin{array}{cc}\hfill A\hfill & \hfill (a,b)\in M\hfill \\ \hfill I\hfill & \hfill \mathrm{otherwise}\hfill \end{array}\right.$$

where I is the identity automorphism of $G(n)$ and A is the automorphism defined in Lemma 1.

3. Main Results

The aim of this section is to construct finite gyrogroups of order $2^n$ by cyclic group ${\mathbb{Z}}_{2^n}$, for $n\ge 3$.

Theorem 1.

$(G(n),\oplus )$ is a gyrogroup.

Proof. 

By Lemma 1, $gyr[a,b]\in Aut(G(n),\oplus )$. By definition of ⊕, $0\oplus i=i$ and so 0 is the identity element of $G(n)$. It is easy to see that the inverse of each element $x\in G(n)$ can be computed by the following formula:

$$\ominus x=\left\{\begin{array}{cc}\hfill -x\hfill & \hfill x\in P(n)\hfill \\ \hfill x\hfill & \hfill x\in H(n)\hfill \end{array}\right.$$

in which $-x$ is the inverse of x in $P(n)$.

Now we will prove the loop property. Suppose $(a,b)$ is an arbitrary element of $G(n)\times G(n)$. We will have four separate cases as follows:

1.

$(a,b)\in P(n)\times P(n)$. In this case, $a\oplus b\in P(n)$. Clearly $(a\oplus b,b),(a,b)\notin M$. Thus, by definition of $gyr$, $gyr[a\oplus b,b]=I=gyr[a,b],$ as desired.

2.

$(a,b)\in H(n)\times H(n)$. In this case, $a\oplus b\in P(n)$ and

$$a\oplus b\equiv (\frac{m}{2}+1)a+(\frac{m}{2}-1)b\equiv (\frac{m}{2}+1)(a-b)(modm).$$

(7)

If $a\oplus b\notin O_P$, then by Equation (7), $(a\oplus b,b),(a,b)\notin M$. Hence $gyr[a\oplus b,b]=I=gyr[a,b].$ If $a\oplus b\in O_P$, then by Equation (7), $(a\oplus b,b),(a,b)\in M$. Hence for every $i\in G(n)$, if $i\in P(n)$, then $gyr[a\oplus b,b](i)=i=gyr[a,b](i).$ If $i\in H(n)$, then $gyr[a\oplus b,b](i)=r+m=gyr[a,b](i)$ such that $r\equiv i+\frac{m}{2}(modm)$.

3.

$(a,b)\in P(n)\times H(n)$. In this case, $a\oplus b\in H(n)$ and

$$a\oplus b\equiv a+b(modm).$$

(8)

If $a\oplus b\in O_H$, then by Equation (8), we have the following two subcases:

(a)

$a\notin O_P$ and $b\in O_H$. In this case, $(a\oplus b,b),(a,b)\notin M$ and by definition of $gyr$, $gyr[a\oplus b,b]=I=gyr[a,b]$.

(b)

$a\in O_P$ and $b\in E_H$. In this case, $(a\oplus b,b),(a,b)\in M$. For every $i\in G(n)$, if $i\in P(n)$, then $gyr[a\oplus b,b](i)=i=gyr[a,b](i).$ If $i\in H(n)$, then $gyr[a\oplus b,b](i)=r+m=gyr[a,b](i)$ such that $r\equiv i+\frac{m}{2}(modm)$.

If $a\oplus b\in E_H$, then by Equation (8), we have the following two subcases:

(a)

$a\notin O_P$ and $b\in E_H$. In this case $(a\oplus b,b),(a,b)\notin M$. The proof is now similar to the subcase (a).

(b)

$a\in O_P$ and $b\in O_H$. In this case $(a\oplus b,b),(a,b)\in M$. The proof of this subcase is similar to the subcase (b).

4.

$(a,b)\in H(n)\times P(n)$. In this case, $a\oplus b\in H(n)$ and $a\oplus b\equiv a+(\frac{m}{2}-1)b(modm).$ The proof of this case is similar to (3) and so it is omitted.

Therefore, the loop property is valid. Finally, we investigate the left gyroassociative law. To do this, we have four separate cases as follows:

1.

$(a,b)\in P(n)\times P(n)$. In this case $a\oplus b\in P(n)$ and $gyr[a,b]=I$. If $c\in P(n)$, then $b\oplus c\in P(n)$ and by definition of ⊕, we have:

$$\begin{array}{ccc}\hfill (a\oplus b)\oplus gyr[a,b](c)& =& \hfill (a\oplus b)\oplus c\equiv (a\oplus b)+c(modm)\hfill \\ & \equiv & \hfill (a+b)+c\equiv a\oplus (b\oplus c)(modm)\hfill \end{array}$$

This proves that $a\oplus (b\oplus c)=(a\oplus b)\oplus gyr[a,b](c)$. If $c\in H(n)$, then $b\oplus c\in H(n)$ and by definition of ⊕,

$$\begin{array}{ccc}\hfill (a\oplus b)\oplus gyr[a,b](c)& =& \hfill (a\oplus b)\oplus c\equiv (a\oplus b)+c(modm)\hfill \\ & \equiv & \hfill a+(b+c+m)\equiv a+(b\oplus c)(modm)\hfill \\ & \equiv & \hfill a+(b\oplus c)+m\equiv a\oplus (b\oplus c)(modm)\hfill \end{array}$$

Therefore, $a\oplus (b\oplus c)=(a\oplus b)\oplus gyr[a,b](c)$, which proves the left gyroassociative law for this case.

2.

$(a,b)\in H(n)\times H(n)$. In this case, $a\oplus b\in P(n)$ and

$$gyr[a,b]=\left\{\begin{array}{cc}\hfill A\hfill & \hfill (a,b)\in N\hfill \\ \hfill I\hfill & \hfill \mathrm{otherwise}\hfill \end{array}\right.$$

in which $N=(O_H\times E_H)\cup (E_H\times O_H)\subseteq M$. We consider two subcases as follows:

(a)

$(a,b)\notin N$. In this subcase, $gyr[a,b]=I$ and ($a,b\in O_H$ or $a,b\in E_H$). If $c\in P(n)$, then $b\oplus c\in H(n)$ and by definition of ⊕,

$$\begin{array}{ccc}\hfill (a\oplus b)\oplus gyr[a,b](c)& =& \hfill (a\oplus b)\oplus c\equiv (a\oplus b)+c(modm)\hfill \\ & \equiv & \hfill [(\frac{m}{2}+1)a+(\frac{m}{2}-1)b]+c(modm)\hfill \end{array}$$

(9)

and

$$\begin{array}{ccc}\hfill a\oplus (b\oplus c)& \equiv & \hfill (\frac{m}{2}+1)a+(\frac{m}{2}-1)(b\oplus c)(modm)\hfill \\ & \equiv & \hfill (\frac{m}{2}+1)a+(\frac{m}{2}-1)[b+(\frac{m}{2}-1)c+m](modm)\hfill \\ & \equiv & \hfill (\frac{m}{2}+1)a+[(\frac{m}{2}-1)b+c+(\frac{m}{4}-1)mc](modm)\hfill \\ & \equiv & \hfill (\frac{m}{2}+1)a+[(\frac{m}{2}-1)b+c](modm).\hfill \end{array}$$

(10)

By Equations (9) and (10), $a\oplus (b\oplus c)=(a\oplus b)\oplus gyr[a,b](c)$. If $c\in H(n)$, then $b\oplus c\in P(n)$ and by definition of ⊕,

$$\begin{array}{ccc}\hfill (a\oplus b)\oplus gyr[a,b](c)& =& \hfill (a\oplus b)\oplus c\hfill \\ & \equiv & \hfill (a\oplus b)+c+m(modm)\hfill \\ & \equiv & \hfill [(\frac{m}{2}+1)a+(\frac{m}{2}-1)b]+c(modm)\hfill \\ & \equiv & \hfill a-b+c+\frac{m}{2}(a+b)(modm)\hfill \\ & \equiv & \hfill a-b+c(modm)\hfill \end{array}$$

(11)

On the other hand,

$$\begin{array}{ccc}\hfill a\oplus (b\oplus c)& \equiv & \hfill a+(\frac{m}{2}-1)(b\oplus c)+m(modm)\hfill \\ & \equiv & \hfill a+(\frac{m}{2}-1)[(\frac{m}{2}+1)b+(\frac{m}{2}-1)c](modm)\hfill \\ & \equiv & \hfill a+[(\frac{m^2}{4}-1)b+{(\frac{m}{2}-1)}^{2}c](modm)\hfill \\ & \equiv & \hfill a-b+c+m[\frac{m}{4}b+(\frac{m}{4}-1)c](modm)\hfill \\ & \equiv & \hfill a-b+c(modm).\hfill \end{array}$$

(12)

By Equations (11) and (12), $a\oplus (b\oplus c)=(a\oplus b)\oplus gyr[a,b](c)$.

(b)

$(a,b)\in N$. In this subcase, $gyr[a,b]=A$ and one of $(a\in O_H,b\in E_H)$ or $(a\in E_H,b\in O_H)$ can be occurred. If $c\in P(n)$, then $b\oplus c\in H(n)$ and $gyr[a,b](c)=c$. A similar argument as (9) and (10) shows that $a\oplus (b\oplus c)=(a\oplus b)\oplus gyr[a,b](c)$. If $c\in H(n)$, then $b\oplus c\in P(n)$, $gyr[a,b](c)\in H(n)$ and $gyr[a,b](c)\equiv c+\frac{m}{2}(modm)$. Again by definition of ⊕,

$$\begin{array}{ccc}\hfill (a\oplus b)\oplus gyr[a,b](c)& \equiv & \hfill (a\oplus b)+gyr[a,b](c)+m(modm)\hfill \\ & \equiv & \hfill [(\frac{m}{2}+1)a+(\frac{m}{2}-1)b]+c+\frac{m}{2}(modm)\hfill \\ & \equiv & \hfill a-b+c(modm)\hfill \end{array}$$

(13)

By a similar argument as Equation (12), $a\oplus (b\oplus c)\equiv a-b+c(modm)$. By the last equation and Equation (13), $a\oplus (b\oplus c)=(a\oplus b)\oplus gyr[a,b](c)$.

3.

$(a,b)\in P(n)\times H(n)$. In this case, $a\oplus b\in H(n)$ and

$$gyr[a,b]=\left\{\begin{array}{cc}\hfill A\hfill & \hfill a\in O_P\hfill \\ \hfill I\hfill & \hfill a\notin O_P\hfill \end{array}\right..$$

We consider two subcases as follows:

(a)

$a\notin O_P$. In this subcase, $gyr[a,b]=I$ and if $c\in P(n)$, then $b\oplus c\in H(n)$. Now by definition ⊕,

$$\begin{array}{cc}\hfill (a\oplus b)\oplus gyr[a,b](c)& \hfill =(a\oplus b)\oplus c\hfill \\ \hfill & \hfill \equiv (a\oplus b)+(\frac{m}{2}-1)c+m(modm)\hfill \\ \hfill & \hfill \equiv (a+b+m)+(\frac{m}{2}-1)c(modm)\hfill \\ \hfill & \hfill \equiv a+[b+(\frac{m}{2}-1)c+m](modm)\hfill \\ \hfill & \hfill \equiv a+[b\oplus c]+m(modm)\hfill \\ \hfill & \hfill \equiv a\oplus (b\oplus c)(modm)\hfill \end{array}$$

(14)

By Equation (14), $a\oplus (b\oplus c)=(a\oplus b)\oplus gyr[a,b](c)$. If $c\in H(n)$, then $b\oplus c\in P(n)$ and by definition of ⊕,

$$\begin{array}{ccc}\hfill (a\oplus b)\oplus gyr[a,b](c)& =& \hfill (a\oplus b)\oplus c\hfill \\ & \equiv & \hfill (\frac{m}{2}+1)(a\oplus b)+(\frac{m}{2}-1)c(modm)\hfill \\ & \equiv & \hfill (\frac{m}{2}+1)(a+b+m)+(\frac{m}{2}-1)c(modm)\hfill \\ & \equiv & \hfill a+(\frac{m}{2}+1)b+(\frac{m}{2}-1)c+\frac{m}{2}a(modm)\hfill \\ & \equiv & \hfill a+(\frac{m}{2}+1)b+(\frac{m}{2}-1)c(modm)\hfill \\ & \equiv & \hfill a+(b\oplus c)(modm)\hfill \\ & \equiv & \hfill a\oplus (b\oplus c)(modm)\hfill \end{array}$$

By the last equalities, $a\oplus (b\oplus c)=(a\oplus b)\oplus gyr[a,b](c)$.

(b)

$a\in O_P$. In this subcase, $gyr[a,b]=A$. If $c\in P(n)$, then $b\oplus c\in H(n)$ and $gyr[a,b](c)=c$. A similar argument as Equation (14) shows that $a\oplus (b\oplus c)=(a\oplus b)\oplus gyr[a,b](c)$. If $c\in H(n)$, then $b\oplus c\in P(n)$, $gyr[a,b](c)\in H(n)$ and $gyr[a,b](c)\equiv c+\frac{m}{2}(modm)$. Apply again definition of ⊕ to deduce that

$$\begin{array}{cc}\hfill (a\oplus b)& \hfill \oplus gyr[a,b](c)\equiv (\frac{m}{2}+1)(a\oplus b)+(\frac{m}{2}-1)(gyr[a,b](c))(modm)\hfill \\ \hfill & \hfill \equiv (\frac{m}{2}+1)(a+b+m)+(\frac{m}{2}-1)(c+\frac{m}{2})(modm)\hfill \\ \hfill & \hfill \equiv a+(\frac{m}{2}+1)b+(\frac{m}{2}-1)c+\frac{m}{2}(a+\frac{m}{2}-1)(modm)\hfill \\ \hfill & \hfill \equiv a+(\frac{m}{2}+1)b+(\frac{m}{2}-1)c(modm)\hfill \\ \hfill & \hfill \equiv a+(b\oplus c)\equiv a\oplus (b\oplus c)(modm)\hfill \end{array}$$

By the last equalities, $a\oplus (b\oplus c)=(a\oplus b)\oplus gyr[a,b](c)$.

4.

$(a,b)\in H(n)\times P(n)$. In this case, $a\oplus b\in H(n)$ and $gyr[a,b]$=A, when $b\in O_P$; and I, otherwise. We now consider the following two subcases:

(a)

$b\notin O_P$. In this subcase, $gyr[a,b]=I$. If $c\in P(n)$, then $b\oplus c\in P(n)$ and by definition of ⊕,

$$\begin{array}{cc}\hfill (a\oplus b)\oplus gyr[a,b](c)& \hfill =(a\oplus b)\oplus c\hfill \\ \hfill & \hfill \equiv (a\oplus b)+(\frac{m}{2}-1)c+m(modm)\hfill \\ \hfill & \hfill \equiv [a+(\frac{m}{2}-1)b+m]+(\frac{m}{2}-1)c(modm)\hfill \\ \hfill & \hfill \equiv a+(\frac{m}{2}-1)(b+c)(modm)\hfill \\ \hfill & \hfill \equiv a+(\frac{m}{2}-1)(b\oplus c)+m(modm)\hfill \\ \hfill & \hfill \equiv a\oplus (b\oplus c)(modm)\hfill \end{array}$$

(15)

By (15), $a\oplus (b\oplus c)=(a\oplus b)\oplus gyr[a,b](c)$. If $c\in H(n)$, then $b\oplus c\in H(n)$ and we have:

$$\begin{array}{cc}\hfill (a\oplus b)& \hfill \oplus gyr[a,b](c)=(a\oplus b)\oplus c\hfill \\ \hfill & \hfill \equiv (\frac{m}{2}+1)(a\oplus b)+(\frac{m}{2}-1)c(modm)\hfill \\ \hfill & \hfill \equiv (\frac{m}{2}+1)[a+(\frac{m}{2}-1)b+m]+(\frac{m}{2}-1)c(modm)\hfill \\ \hfill & \hfill \equiv (\frac{m}{2}+1)a+(\frac{m}{2}-1)(b+c)+\frac{m}{2}(\frac{m}{2}-1)b(modm)\hfill \\ \hfill & \hfill \equiv (\frac{m}{2}+1)a+(\frac{m}{2}-1)(b+c+m)(modm)\hfill \\ \hfill & \hfill \equiv (\frac{m}{2}+1)a+(\frac{m}{2}-1)(b\oplus c)(modm)\hfill \\ \hfill & \hfill \equiv a\oplus (b\oplus c)(modm)\hfill \end{array}$$

The last equalities give $a\oplus (b\oplus c)=(a\oplus b)\oplus gyr[a,b](c)$.

(b)

$b\in O_P$. In this subcase, $gyr[a,b]=A$. If $c\in P(n)$, then $b\oplus c\in P(n)$ and $gyr[a,b](c)=c$. A similar argument as (15) shows that $a\oplus (b\oplus c)=(a\oplus b)\oplus gyr[a,b](c)$. If $c\in H(n)$, then $b\oplus c\in H(n)$, $gyr[a,b](c)\in H(n)$ and $gyr[a,b](c)\equiv c+\frac{m}{2}(modm)$. Therefore,

$$\begin{array}{cc}\hfill & \hfill (a\oplus b)\oplus gyr[a,b](c)\equiv (\frac{m}{2}+1)(a\oplus b)+(\frac{m}{2}-1)(gyr[a,b](c))(modm)\hfill \\ \hfill & \hfill \equiv (\frac{m}{2}+1)[a+(\frac{m}{2}-1)b+m]+(\frac{m}{2}-1)(c+\frac{m}{2})(modm)\hfill \\ \hfill & \hfill \equiv (\frac{m}{2}+1)a+(\frac{m}{2}-1)(b+c)+\frac{m}{2}(\frac{m}{2}-1)(b+1)(modm)\hfill \\ \hfill & \hfill \equiv (\frac{m}{2}+1)a+(\frac{m}{2}-1)(b+c+m)(modm)\hfill \\ \hfill & \hfill \equiv (\frac{m}{2}+1)a+(\frac{m}{2}-1)(b\oplus c)(modm)\hfill \\ \hfill & \hfill \equiv a\oplus (b\oplus c)(modm)\hfill \end{array}$$

The last equalities give $a\oplus (b\oplus c)=(a\oplus b)\oplus gyr[a,b](c)$.

Hence the result. □

Example 1.

In this example, we investigate the gyrogroup $G(3)$ of order 8 constructed by the cyclic group ${\mathbb{Z}}_8$. By definition, $G(3)=\{0,1,2,3,4,5,6,7\}$ and the binary operation ⊕ is defined as follows:

$$i\oplus j=\left\{\begin{array}{cc}\hfill t\hfill & \hfill (i,j)\in P(3)\times P(3)\hfill \\ \hfill t+4\hfill & \hfill (i,j)\in (P(3)\times H(3))\cup (H(3)\times P(3))\hfill \\ \hfill k\hfill & \hfill (i,j)\in H(3)\times H(3)\hfill \end{array}\right.$$

in which $P(3)=\{0,1,2,3\}$, $H(3)=\{4,5,6,7\}$ and $t,k\in P(3)$. Also,

$$\left\{\begin{array}{cc}\hfill t& \hfill \equiv i+j(mod4)\hfill \\ \hfill k& \hfill \equiv 3i+j(mod4)\hfill \end{array}\right.$$

Therefore, the addition table and the gyration table for $G(3)$ are presented in

Table 1. The addition and Gyration Tables of $G(3)$. (a) The addition table of $G(3)$. (b) The gyration table of $G(3)$ such that $A=(4,6)(5,7)$.

(a)
⊕	$\mathbf{0}$	$\mathbf{1}$	$\mathbf{2}$	$\mathbf{3}$	$\mathbf{4}$	$\mathbf{5}$	$\mathbf{6}$	$\mathbf{7}$
$\mathbf{0}$	0	1	2	3	4	5	6	7
$\mathbf{1}$	1	2	3	0	5	6	7	4
$\mathbf{2}$	2	3	0	1	6	7	4	5
$\mathbf{3}$	3	0	1	2	7	4	5	6
$\mathbf{4}$	4	5	6	7	0	1	2	3
$\mathbf{5}$	5	6	7	4	3	0	1	2
$\mathbf{6}$	6	7	4	5	2	3	0	1
$\mathbf{7}$	7	4	5	6	1	2	3	0
(b)
$\mathbf{gyr}$	$\mathbf{0}$	$\mathbf{1}$	$\mathbf{2}$	$\mathbf{3}$	$\mathbf{4}$	$\mathbf{5}$	$\mathbf{6}$	$\mathbf{7}$
$\mathbf{0}$	I	I	I	I	I	I	I	I
$\mathbf{1}$	I	I	I	I	A	A	A	A
$\mathbf{2}$	I	I	I	I	I	I	I	I
$\mathbf{3}$	I	I	I	I	A	A	A	A
$\mathbf{4}$	I	A	I	A	I	A	I	A
$\mathbf{5}$	I	A	I	A	A	I	A	I
$\mathbf{6}$	I	A	I	A	I	A	I	A
$\mathbf{7}$	I	A	I	A	A	I	A	I

in which A is the unique nonidentity gyroautomorphism of $G(3)$ given by $A=(4,6)(5,7)$. Also, the addition table and the gyration table for $G(4)$ are presented in

Table 2. The addition table of $G(4)$.

⊕	0	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15
$\mathbf{0}$	0	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15
$\mathbf{1}$	1	2	3	4	5	6	7	0	9	10	11	12	13	14	15	8
$\mathbf{2}$	2	3	4	5	6	7	0	1	10	11	12	13	14	15	8	9
$\mathbf{3}$	3	4	5	6	7	0	1	2	11	12	13	14	15	8	9	10
$\mathbf{4}$	4	5	6	7	0	1	2	3	12	13	14	15	8	9	10	11
$\mathbf{5}$	5	6	7	0	1	2	3	4	13	14	15	8	9	10	11	12
$\mathbf{6}$	6	7	0	1	2	3	4	5	14	15	8	9	10	11	12	13
$\mathbf{7}$	7	0	1	2	3	4	5	6	15	8	9	10	11	12	13	14
$\mathbf{8}$	8	11	14	9	12	15	10	13	0	3	6	1	4	7	2	5
$\mathbf{9}$	9	12	15	10	13	8	11	14	5	0	3	6	1	4	7	2
$\mathbf{10}$	10	13	8	11	14	9	12	15	2	5	0	3	6	1	4	7
$\mathbf{11}$	11	14	9	12	15	10	13	8	7	2	5	0	3	6	1	4
$\mathbf{12}$	12	15	10	13	8	11	14	9	4	7	2	5	0	3	6	1
$\mathbf{13}$	13	8	11	14	9	12	15	10	1	4	7	2	5	0	3	6
$\mathbf{14}$	14	9	12	15	10	13	8	11	6	1	4	7	2	5	0	3
$\mathbf{15}$	15	10	13	8	11	14	9	12	3	6	1	4	7	2	5	0

and

Table 3. The gyration table of $G(4)$ such that $A=(8,12)(9,13)(10,14)(11,15)$.

$\mathbf{gyr}$	0	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15
0	I	I	I	I	I	I	I	I	I	I	I	I	I	I	I	I
$\mathbf{1}$	I	I	I	I	I	I	I	I	A	A	A	A	A	A	A	A
$\mathbf{2}$	I	I	I	I	I	I	I	I	I	I	I	I	I	I	I	I
$\mathbf{3}$	I	I	I	I	I	I	I	I	A	A	A	A	A	A	A	A
$\mathbf{4}$	I	I	I	I	I	I	I	I	I	I	I	I	I	I	I	I
$\mathbf{5}$	I	I	I	I	I	I	I	I	A	A	A	A	A	A	A	A
$\mathbf{6}$	I	I	I	I	I	I	I	I	I	I	I	I	I	I	I	I
$\mathbf{7}$	I	I	I	I	I	I	I	I	A	A	A	A	A	A	A	A
$\mathbf{8}$	I	A	I	A	I	A	I	A	I	A	I	A	I	A	I	A
$\mathbf{9}$	I	A	I	A	I	A	I	A	I	A	I	A	I	A	I	A
$\mathbf{10}$	I	A	I	A	I	A	I	A	I	A	I	A	I	A	I	A
$\mathbf{11}$	I	A	I	A	I	A	I	A	I	A	I	A	I	A	I	A
$\mathbf{12}$	I	A	I	A	I	A	I	A	I	A	I	A	I	A	I	A
$\mathbf{13}$	I	A	I	A	I	A	I	A	I	A	I	A	I	A	I	A
$\mathbf{14}$	I	A	I	A	I	A	I	A	I	A	I	A	I	A	I	A
$\mathbf{15}$	I	A	I	A	I	A	I	A	I	A	I	A	I	A	I	A

in which A is the unique nonidentity gyroautomorphism of $G(4)$ given by $A=(8,12)(9,13)(10,14)(11,15)$.

Theorem 2.

The gyrogroup $(G(n),\oplus )$ is non-gyrocommutative.

Proof. 

We know $\left|G(n)\right|=2^n$, for $n\ge 3$. If $n=3$, then by Example 1, $gyr[1,5]=A$ and $1\oplus 5=6\ne 4=gyr[1,5](6)=gyr[1,5](5\oplus 1)$. Therefore, in this case $G(n)$ is non-gyrocommutative. For $n>3$, suppose that $G(n)$ is gyrocommutative. So, for every $a,b\in G(n)$, the following equality is satisfied:

$$a\oplus b=gyr[a,b](b\oplus a)$$

(16)

Suppose $a=2$ and $b=m+1$. By definition, $2\in P(n)$, $m+1\in H(n)$ and $gyr[2,m+1]=I$. Also, $2\oplus (m+1)=t+m$, $(m+1)\oplus 2=s+m$ such that $t\equiv 2+m+1\equiv 3(modm)$ and $s\equiv m+1+(\frac{m}{2}-1)2\equiv m-1(modm)$. Thus $2\oplus (m+1)=m+3$ and $(m+1)\oplus 2=2m-1$. By these relations and Equation (16), $2\oplus (m+1)=gyr[2,m+1]((m+1)\oplus 2)$=$(m+1)\oplus 2$. This proves that $m+3=2m-1$ and so $m=4$. Therefore, $2^{n-1}=2^2$ and hence $n=3$ which is a contradiction. □

Theorem 3.

Let $G(n)$ be the dihedral 2-gyrogroup of order $2^n$. B is a subgyrogroup of $G(n)$ if and only if it has one of the following forms:

1.

$B=G(n)$;

2.

B is a subgroup of $P(n)$;

3.

$B=\{0,i\}$, where $i\in H(n)$;

4.

There are integers r and s such that $1\le r\le n-2$, $0\le s\le 2^r-1$ and $B=\langle 2^r\rangle \cup \langle 2^r\rangle +(m+s)$, where $m=2^{n-1}$.

Proof. 

Suppose B is a proper subgyrogroup of $G(n)$. We first assume that $B\subseteq P(n)$. Since the restriction of ⊕ to B is the group addition of $P(n)$, B will be a subgroup of $P(n)$, as desired. We next assume that $B⊈P(n)$. Since $G(n)$ can be partitioned by $P(n)$ and $H(n)$, we can write the subgyrogroup B as $B_1\cup B_2$ in which $B_1\subseteq P(n)$ and $B_2\subseteq H(n)$. It is easy to see $B_1$ is a subgroup of $P(n)$. Suppose $B_1=\left\{0\right\}$ and $i,j\in B_2$. By an easy calculation one can see that $(\frac{m}{2}+1)i+(\frac{m}{2}-1)j\equiv 0$ ($modm$) if and only if $(\frac{m}{2}+1)i\equiv (1-\frac{m}{2})j+mj$ ($modm$) if and only if $(\frac{m}{2}+1)i\equiv (\frac{m}{2}+1)j$ ($modm$) if and only if $i\equiv j$ ($modm$). Since $i,j\in B_2\subseteq H(n)$, $i=j$. This shows that $B_2=\left\{i\right\}$ and $B=\{0,i\}$, for some $i\in H(n)$. Therefore, B has the form of $(3)$. Finally, suppose that $B_1=\langle 2^r\rangle$, $1\le r\le n-2$. We will prove that $B=\langle 2^r\rangle \cup \langle 2^r\rangle +(m+s)$, for some s, $0\le s\le 2^r-1$. Choose arbitrary elements $u,v\in B_2$. By definition $u\oplus v\in B_1$ and there exists an integer k such that $u\oplus v=\frac{m}{2}(u+v)+(u-v)+mk$ and so $u-v\in \langle 2^r\rangle$. Suppose x is the least element of $B_2$, then $x=m+s$ in which $0\le s\le 2^r-1$. If $j\in B_2$, then $j-x\in \langle 2^r\rangle$ and so $j\in \langle 2^r\rangle +x=\langle 2^r\rangle +(m+s)$. This shows that $B_2\subseteq \langle 2^r\rangle +(m+s)=B_1+(m+s)$. Next fix an element $b\in B_2$. Define two one to one mappings ${\eta }_1:B_1⟶B_2$ and ${\eta }_2:B_2⟶B_1$ by ${\eta }_1(c)=b\oplus c$ and ${\eta }_2(d)=b\oplus d$. This proves that $|B_1|=|B_2|$ and so $B_2=B_1+(m+s)$, as desired.

Conversely, it is obvious that $G(n)$ is a subgyrogroup of itself. Suppose H is a subgroup of $P(n)$, then by (Proposition 23) [14] it will be a subgyrogroup of $G(n)$. We now assume that $H=\{0,i\}$, where $i\in H(n)$. By definition of $G(n)$, $i\oplus i=0$ and by (Proposition 22) [14], H is a subgyrogroup of $G(n)$. Finally, we assume that $B=\langle 2^r\rangle \cup \langle 2^r\rangle +(m+s)$, where r and s are positive integers such that $1\le r\le n-2$ and $0\le s\le 2^r-1$. If $i,j\in \langle 2^r\rangle$ then $i\oplus j\in \langle 2^r\rangle \subseteq B$. If $i\in B_1$ and $j\in B_2$, then there are integers $k_1,k_2,k_3,k_4$ such that $i=2^r{k}_1$, $j=2^r{k}_2+(m+s)$ and $i\oplus j$=$i+j+2^r{k}_3+m$=$2^r{k}_1+2^r{k}_2+(m+s)+2^r{k}_3+m$= $2^r{k}_4+(m+s)$. Hence, $i\oplus j$=$2^r{k}_4+(m+s)\in B_2\subseteq B$. A similar calculations shows that if $i\in B_2$, $j\in B_1$ or $i,j\in B_2$ then $i\oplus j\in B$. This completes the proof. □

4. Concluding Remarks

Miller [15] characterized the non-abelian and non-dihedral finite group G with this property that all non-abelian subgroups of G are dihedral. In this paper, we continue an earlier work of some of us [16] to construct a 2-gyrogroup $G(n)$ of order $2^n$, $n\ge 3$, such that all non-abelian subgyrogroups are dihedral. The structure of the subgyrogroups and normal subgyrogroups are completely determined. As a consequence, the subgyrogroup lattice and the normal subgyrogroup lattices of $G(n)$ are isomorphic to the subgroup and normal subgroup lattices of $D_{2^n}$, respectively. Therefore, all non-abelian subgroups of G are dihedral. This is a research problem for future to characterize non-degenerate gyrogroups for which all subgyrogroups are subgroups and non-abelian subgroups of this gyrogroup are dihedral.

Suppose $\langle x_1,\cdots ,x_r\rangle$ denotes as usual the subgyrogroup generated by the set $\{x_1,\cdots ,x_r\}$. By Theorem 3, it is easy to prove that $\langle 1\rangle \cong Z_{2^{n-1}}$, $\langle 2\rangle \cong Z_{2^{n-2}}$, ⋯, $\langle 2^{n-2}\rangle \cong Z_2$, $\langle 2,m\rangle \cong \langle 2,m+1\rangle \cong D_m$, $\langle 4,m\rangle \cong \langle 4,m+1\rangle \cong \langle 4,m+2\rangle \cong \langle 4,m+3\rangle \cong D_{\frac{m}{2}}$, $\langle 2^{n-2},m\rangle \cong \langle 2^{n-2},m+1\rangle \cong \langle 2^{n-2},m+2\rangle \cong \cdots \cong \langle 2^{n-2},m+2^{n-2}-1\rangle \cong D_{\frac{m}{2^{n-3}}}=K_4$, $\langle m\rangle \cong \langle m+1\rangle \cong \langle m+2\rangle \cong \cdots \cong \langle 2m-2\rangle \cong \langle 2m-1\rangle \cong Z_2$. In Figure 1, the subgyrogroup and normal subgyrogroup lattices of $G(n)$ are depicted. In this figure, the circled subgyrogroups are denoted the normal subgyrogroups of $G(n)$. Because of this similarity between the dihedral group $D_{2^n}$ and the gyrogroup $G(n)$ of order $2^n$, we like to use the name dihedral gyrogroup of order $2^n$ for $G(n)$. Moreover, all proper subgyrogroups of $G(n)$, $n\ge 3$ are subgroups, see Figure 2.

We end this paper with the following question:

Question 1.

Is it possible to apply the same method for constructing a gyrogroup $H_{2n}$ of order $2n$, $n\ge 3$, in which the the subgyrogroup and normal subgyrogroup lattices of $H_{2n}$ are isomorphic to the subgroup and normal subgroup lattices of $D_{2n},$ respectively?

Note that for every integer n, $n\ge 3$, A is an automorphism of order two and so $\{I,A\}$ is a group of order 2.