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

S S symmetry

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

Soheila Mahdavi 1,† , Ali Reza Ashraﬁ 1,*,† , Mohammad Ali Salahshour 2,† and Abraham Albert Ungar 3,†

1 Department of Pure Mathematics, Faculty of Mathematical Sciences, University of Kashan, Kashan 87317-53153, Iran; [email protected] 2 Department of Mathematics, Savadkooh Branch, Islamic Azad University, Savadkooh 47418-39959, Iran; [email protected] 3 Department of Mathematics, North Dakota State University, Fargo, ND 58108-6050, USA; [email protected] * Correspondence: ashraﬁ@kashanu.ac.ir † These authors contributed equally to this work.

Abstract: In this paper, a 2-gyrogroup G(n) of order 2n, n 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 2n, which causes us to use the name dihedral gyrogroup for this class of gyrogroups of order 2n. Moreover, all proper subgyrogroups of G(n) are subgroups.

Keywords: gyrogroup; gyroautomorphism; gyroassociative; gyrocommutative

MSC: Primary 20N05; Secondary 20F99; 20D99

Citation: Mahdavi, S.; Ashrafi, A.R.; Salahshour, M.A.; Ungar, A.A. Construction of 2-Gyrogroups in 1. Basic Concept and History Which Every Proper Subgyrogroup Is A pair (G, ) consisting of a nonempty set G and a binary operation : G G G Either a Cyclic or a Dihedral Group. ⊕ ⊕ × −→ Symmetry 2021, 13, 316. https:// is called a groupoid. A bijection φ from the groupoid G to itself is called an automorphism of G if φ(a b) = φ(a) φ(b), for all a, b G. The set of all automorphisms of G is denoted doi.org/10.3390/sym13020316 ⊕ ⊕ ∈ by Aut(G). It is straightforward to check that Aut(G) forms a group under composition Academic Editor: Antonio Masiello of function. A groupoid (G, ) is called a gyrogroup if the following conditions are satisfied: ⊕ Received: 8 January 2021 • there exists an element 0 G such that for all x G, 0 x = x; ∈ ∈ ⊕ Accepted: 8 February 2021 • for each a G, there exists b G such that b a = 0; ∈ ∈ ⊕ Published: 14 February 2021 • (gyroassociative law) there exists a function gyr : G G Aut(G) such that for every × −→ a, b, c G, a (b c) = (a b) gyr[a, b]c, where gyr[a, b]c = gyr(a, b)(c); ∈ ⊕ ⊕ ⊕ ⊕ Publisher’s Note: MDPI stays neutral • for each a, b G, gyr[a, b] = gyr[a b, b]. with regard to jurisdictional claims in ∈ ⊕ The gyrogroup G is called gyrocommutative if and only if for all a, b G, a b = published maps and institutional affil- ∈ ⊕ gyr[a, b](b a). The function gyr[a, b], a, b G, is called the gyroautomorphism generated iations. ⊕ ∈ by a and b. Note that the axioms of a gyrogroup apply their right counterpart. Suppose G is a group and gyr[a, b] = 1G, where 1G 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. Copyright: © 2021 by the authors. Abraham Ungar in his seminal paper [1] initiated a new approach to special theory of Licensee MDPI, Basel, Switzerland. relativity and in [2], he employed the Thomas precession and its interplay with Einstein’s This article is an open access article addition to formally defines gyrogroups. In 2001, Ungar published a book containing many distributed under the terms and conditions of the Creative Commons results about gyrogroups and gyrovector spaces [2]. In [3], he applied gyrogroups and Attribution (CC BY) license (https:// gyrovector spaces to introduce an interesting analogy between three models in hyperbolic creativecommons.org/licenses/by/ geometry: the Poincare´ ball model, the Beltrami ball model and the PV space model. In [4], 4.0/). it is shown that the gyrocommutative gyrogroups and their resulting gyrovector spaces

Symmetry 2021, 13, 316. https://doi.org/10.3390/sym13020316 https://www.mdpi.com/journal/symmetry Symmetry 2021, 13, 316 2 of 12

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 [6–9] 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 3, of order 2n by considering a cyclic group of order ≥ 2n. 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, , 2n 1 1 , H(n) = 2n 1, 2n 1 + 1, , 2n { ··· − − } { − − ··· − 1 and G(n) = P(n) H(n), where n 3 is a natural number. It is clear that P(n) is a } ∪ ≥ n 1 cyclic group under addition modulo m = 2 − and H(n) = P(n) + m. This shows that G(n) = P(n) (P(n) + m). Deﬁne the binary operation on G(n) as follows: ∪ ⊕  t (i, j) P(n) P(n)  ∈ × t + m (i, j) P(n) H(n) i j = ∈ × ⊕ s + m (i, j) H(n) P(n)  ∈ × k (i, j) H(n) H(n) ∈ × and t, s, k P(n) are the following non-negative integers: ∈  t i + j (mod m)  ≡  m s i + ( 1)j (mod m) ≡ 2 − .  m m k ( + 1)i + ( 1)j (mod m) ≡ 2 2 − 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 y (mod m) 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: −→ ( i i P(n) A(i) = ∈ r + m i H(n) ∈ m where r P(n) and r i + (mod m). Then A is an automorphism of (G(n), ). ∈ ≡ 2 ⊕

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 G(n) are arbitrary elements. If ∈ i, j P(n), then it is obvious that A(i j) = A(i) A(j). Our main proof will consider ∈ ⊕ ⊕ three separate cases as follows: 1. i P(n) and j H(n). In this case, i j = t + m H(n) such that t P(n) and ∈ ∈ ⊕ 1 ∈ 1 ∈ t i + j (mod m). Now by our assumption, A(i) = i, A(j) = r + m and 1 ≡ 1 A(i j) = A(t + m) = r + m, (1) ⊕ 1 2 Symmetry 2021, 13, 316 3 of 12

where r , r P(n), r j + m (mod m) and r t + m + m (mod m). By 1 2 ∈ 1 ≡ 2 2 ≡ 1 2 these equations, A(i) A(j) = i (r + m) = t + m (2) ⊕ ⊕ 1 2 such that t P(n) and t i + r + m (mod m). Hence, r t + m i + j + m 2 ∈ 2 ≡ 1 2 ≡ 1 2 ≡ 2 ≡ i + r t (mod m). Since r , t P(n), r = t . We now apply Equations (1) and (2) 1 ≡ 2 2 2 ∈ 2 2 to deduce that A(i j) = A(i) A(j). ⊕ ⊕ 2. i H(n) and j P(n). In this case, i j = s + m H(n), where s P(n) and ∈ ∈ ⊕ 1 ∈ 1 ∈ s i + ( m 1)j (mod m). By our assumption A(i) = r + m, A(j) = j and 1 ≡ 2 − 1 A(i j) = A(s + m) = r + m, (3) ⊕ 1 2 where r , r P(n), r i + m (mod m) and r s + m + m (mod m). By these 1 2 ∈ 1 ≡ 2 2 ≡ 1 2 calculations, one can see that

A(i) A(j) = (r + m) j = s + m (4) ⊕ 1 ⊕ 2 such that s P(n) and s r + m + ( m 1)j (mod m). Hence, r s + m 2 ∈ 2 ≡ 1 2 − 2 ≡ 1 2 ≡ i + ( m 1)j + m r + ( m 1)j s (mod m). Since r , s P(n), r = s , and by 2 − 2 ≡ 1 2 − ≡ 2 2 2 ∈ 2 2 Equations (3) and (4), A(i j) = A(i) A(j), as desired. ⊕ ⊕ 3. i, j H(n). By deﬁnition, i j = k P(n), where k ( m + 1)i + ( m 1)j ∈ ⊕ 1 ∈ 1 ≡ 2 2 − (mod m). We now apply our assumption to deduce that A(i) = r1 + m, A(j) = r2 + m and A(i j) = A(k ) = k (5) ⊕ 1 1 such that r , r P(n), r i + m (mod m) and r j + m (mod m). By 1 2 ∈ 1 ≡ 2 2 ≡ 2 above calculations,

A(i) A(j) = (r + m) (r + m) = k (6) ⊕ 1 ⊕ 2 2 in which k P(n) and k ( m + 1)(r + m) + ( m 1)(r + m)(mod m). Hence, 2 ∈ 2 ≡ 2 1 2 − 2 k ( m + 1)r + ( m 1)r ( m + 1)i + ( m 1)j k (mod m). Since k , k P(n), 2 ≡ 2 1 2 − 2 ≡ 2 2 − ≡ 1 1 2 ∈ again k = k and by Equations (5) and (6), A(i j) = A(i) A(j). 1 2 ⊕ ⊕ This completes our argument.

Set OP = i P(n) 2 - i , OH = i H(n) 2 - i , EH = i H(n) 2 i { ∈ | } { ∈ | } { ∈ | | } and M = [O (O E )] S[O (O E )] S[E (O O )]. Deﬁne the map P × H ∪ H H × P ∪ H H × P ∪ H gyr : G(n) G(n) Aut(G(n), ) as follows: × −→ ⊕ ( A (a, b) M gyr(a, b) = gyr[a, b] = ∈ I otherwise

where I is the identity automorphism of G(n) and A is the automorphism deﬁned in Lemma1.

3. Main Results The aim of this section is to construct ﬁnite gyrogroups of order 2n by cyclic group Z2n , for n 3. ≥ Theorem 1. (G(n), ) is a gyrogroup. ⊕ Proof. By Lemma1, gyr[a, b] Aut(G(n), ). By deﬁnition of , 0 i = i and so 0 is the ∈ ⊕ ⊕ ⊕ identity element of G(n). It is easy to see that the inverse of each element x G(n) can be ∈ computed by the following formula: ( x x P(n) x = − ∈ x x H(n) ∈ Symmetry 2021, 13, 316 4 of 12

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) G(n). We will have four separate cases as follows: × 1. (a, b) P(n) P(n). In this case, a b P(n). Clearly (a b, b), (a, b) M. Thus, ∈ × ⊕ ∈ ⊕ 6∈ by definition of gyr, gyr[a b, b] = I = gyr[a, b], as desired. ⊕ 2. (a, b) H(n) H(n). In this case, a b P(n) and ∈ × ⊕ ∈ m m m a b ( + 1)a + ( 1)b ( + 1)(a b)(mod m). (7) ⊕ ≡ 2 2 − ≡ 2 − If a b O , then by Equation (7), (a b, b), (a, b) M. Hence gyr[a b, b] = ⊕ 6∈ P ⊕ 6∈ ⊕ I = gyr[a, b]. If a b O , then by Equation (7), (a b, b), (a, b) M. Hence for ⊕ ∈ P ⊕ ∈ every i G(n), if i P(n), then gyr[a b, b](i) = i = gyr[a, b](i). If i H(n), then ∈ ∈ ⊕ ∈ gyr[a b, b](i) = r + m = gyr[a, b](i) such that r i + m (mod m). ⊕ ≡ 2 3. (a, b) P(n) H(n). In this case, a b H(n) and ∈ × ⊕ ∈ a b a + b (mod m). (8) ⊕ ≡ If a b O , then by Equation (8), we have the following two subcases: ⊕ ∈ H (a) a O and b O . In this case, (a b, b), (a, b) M and by definition of gyr, 6∈ P ∈ H ⊕ 6∈ gyr[a b, b] = I = gyr[a, b]. ⊕ (b) a O and b E . In this case, (a b, b), (a, b) M. For every i G(n), ∈ P ∈ H ⊕ ∈ ∈ if i P(n), then gyr[a b, b](i) = i = gyr[a, b](i). If i H(n), then gyr[a ∈ ⊕ ∈ ⊕ b, b](i) = r + m = gyr[a, b](i) such that r i + m (mod m). ≡ 2 If a b E , then by Equation (8), we have the following two subcases: ⊕ ∈ H (a) a O and b E . In this case (a b, b), (a, b) M. The proof is now similar 6∈ P ∈ H ⊕ 6∈ to the subcase (a). (b) a O and b O . In this case (a b, b), (a, b) M. The proof of this subcase ∈ P ∈ H ⊕ ∈ is similar to the subcase (b). 4. (a, b) H(n) P(n). In this case, a b H(n) and a b a + ( m 1)b (mod m). ∈ × ⊕ ∈ ⊕ ≡ 2 − 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) P(n) P(n). In this case a b P(n) and gyr[a, b] = I. If c P(n), then ∈ × ⊕ ∈ ∈ b c P(n) and by definition of , we have: ⊕ ∈ ⊕ (a b) gyr[a, b](c) = (a b) c (a b) + c (mod m) ⊕ ⊕ ⊕ ⊕ ≡ ⊕ (a + b) + c a (b c)(mod m) ≡ ≡ ⊕ ⊕ This proves that a (b c) = (a b) gyr[a, b](c). If c H(n), then b c H(n) ⊕ ⊕ ⊕ ⊕ ∈ ⊕ ∈ and by definition of , ⊕ (a b) gyr[a, b](c) = (a b) c (a b) + c (mod m) ⊕ ⊕ ⊕ ⊕ ≡ ⊕ a + (b + c + m) a + (b c)(mod m) ≡ ≡ ⊕ a + (b c) + m a (b c)(mod m) ≡ ⊕ ≡ ⊕ ⊕ Therefore, a (b c) = (a b) gyr[a, b](c), which proves the left gyroassociative ⊕ ⊕ ⊕ ⊕ law for this case. 2. (a, b) H(n) H(n). In this case, a b P(n) and ∈ × ⊕ ∈ ( A (a, b) N gyr[a, b] = ∈ I otherwise

in which N = (O E ) (E O ) M. We consider two subcases as follows: H × H ∪ H × H ⊆ Symmetry 2021, 13, 316 5 of 12

(a) (a, b) N. In this subcase, gyr[a, b] = I and (a, b O or a, b E ). If 6∈ ∈ H ∈ H c P(n), then b c H(n) and by definition of , ∈ ⊕ ∈ ⊕ (a b) gyr[a, b](c) = (a b) c (a b) + c (mod m) ⊕ ⊕ m⊕ ⊕ ≡ m ⊕ [( + 1)a + ( 1)b] + c (mod m) (9) ≡ 2 2 − and m m a (b c) ( + 1)a + ( 1)(b c)(mod m) ⊕ ⊕ ≡ 2 2 − ⊕ m m m ( + 1)a + ( 1)[b + ( 1)c + m](mod m) ≡ 2 2 − 2 − m m m ( + 1)a + [( 1)b + c + ( 1)mc](mod m) ≡ 2 2 − 4 − m m ( + 1)a + [( 1)b + c](mod m). (10) ≡ 2 2 − By Equations (9) and (10), a (b c) = (a b) gyr[a, b](c). If c H(n), ⊕ ⊕ ⊕ ⊕ ∈ then b c P(n) and by definition of , ⊕ ∈ ⊕ (a b) gyr[a, b](c) = (a b) c ⊕ ⊕ ⊕ ⊕ (a b) + c + m (mod m) ≡ m⊕ m [( + 1)a + ( 1)b] + c (mod m) ≡ 2 2 − m a b + c + (a + b)(mod m) ≡ − 2 a b + c (mod m) (11) ≡ − On the other hand, m a (b c) a + ( 1)(b c) + m (mod m) ⊕ ⊕ ≡ 2 − ⊕ m m m a + ( 1)[( + 1)b + ( 1)c](mod m) ≡ 2 − 2 2 − m2 m a + [( 1)b + ( 1)2c](mod m) ≡ 4 − 2 − m m a b + c + m[ b + ( 1)c](mod m) ≡ − 4 4 − a b + c (mod m). (12) ≡ − By Equations (11) and (12), a (b c) = (a b) gyr[a, b](c). ⊕ ⊕ ⊕ ⊕ (b) (a, b) N. In this subcase, gyr[a, b] = A and one of (a O , b E ) or ∈ ∈ H ∈ H (a E , b O ) can be occurred. If c P(n), then b c H(n) and ∈ H ∈ H ∈ ⊕ ∈ gyr[a, b](c) = c. A similar argument as (9) and (10) shows that a (b c) = ⊕ ⊕ (a b) gyr[a, b](c). If c H(n), then b c P(n), gyr[a, b](c) H(n) and ⊕ ⊕ ∈ ⊕ ∈ ∈ gyr[a, b](c) c + m (mod m). Again by definition of , ≡ 2 ⊕ (a b) gyr[a, b](c) (a b) + gyr[a, b](c) + m (mod m) ⊕ ⊕ ≡ m⊕ m m [( + 1)a + ( 1)b] + c + (mod m) ≡ 2 2 − 2 a b + c (mod m) (13) ≡ − By a similar argument as Equation (12), a (b c) a b + c (mod m). By ⊕ ⊕ ≡ − the last equation and Equation (13), a (b c) = (a b) gyr[a, b](c). ⊕ ⊕ ⊕ ⊕ Symmetry 2021, 13, 316 6 of 12

3. (a, b) P(n) H(n). In this case, a b H(n) and ∈ × ⊕ ∈ ( A a O gyr[a, b] = ∈ P . I a O 6∈ P We consider two subcases as follows: (a) a O . In this subcase, gyr[a, b] = I and if c P(n), then b c H(n). Now 6∈ P ∈ ⊕ ∈ by definition , ⊕ (a b) gyr[a, b](c) = (a b) c ⊕ ⊕ ⊕ ⊕ m (a b) + ( 1)c + m (mod m) ≡ ⊕ 2 − m (a + b + m) + ( 1)c (mod m) ≡ 2 − (14) m a + [b + ( 1)c + m](mod m) ≡ 2 − a + [b c] + m (mod m) ≡ ⊕ a (b c)(mod m) ≡ ⊕ ⊕ By Equation (14), a (b c) = (a b) gyr[a, b](c). If c H(n), then ⊕ ⊕ ⊕ ⊕ ∈ b c P(n) and by definition of , ⊕ ∈ ⊕ (a b) gyr[a, b](c) = (a b) c ⊕ ⊕ m⊕ ⊕ m ( + 1)(a b) + ( 1)c (mod m) ≡ 2 ⊕ 2 − m m ( + 1)(a + b + m) + ( 1)c (mod m) ≡ 2 2 − m m m a + ( + 1)b + ( 1)c + a (mod m) ≡ 2 2 − 2 m m a + ( + 1)b + ( 1)c (mod m) ≡ 2 2 − a + (b c)(mod m) ≡ ⊕ a (b c)(mod m) ≡ ⊕ ⊕ By the last equalities, a (b c) = (a b) gyr[a, b](c). ⊕ ⊕ ⊕ ⊕ (b) a O . In this subcase, gyr[a, b] = A. If c P(n), then b c H(n) and ∈ P ∈ ⊕ ∈ gyr[a, b](c) = c. A similar argument as Equation (14) shows that a (b c) = ⊕ ⊕ (a b) gyr[a, b](c). If c H(n), then b c P(n), gyr[a, b](c) H(n) and ⊕ ⊕ ∈ ⊕ ∈ ∈ gyr[a, b](c) c + m (mod m). Apply again definition of to deduce that ≡ 2 ⊕ m m (a b) gyr[a, b](c) ( + 1)(a b) + ( 1)(gyr[a, b](c))(mod m) ⊕ ⊕ ≡ 2 ⊕ 2 − m m m ( + 1)(a + b + m) + ( 1)(c + )(mod m) ≡ 2 2 − 2 m m m m a + ( + 1)b + ( 1)c + (a + 1)(mod m) ≡ 2 2 − 2 2 − m m a + ( + 1)b + ( 1)c (mod m) ≡ 2 2 − a + (b c) a (b c)(mod m) ≡ ⊕ ≡ ⊕ ⊕ By the last equalities, a (b c) = (a b) gyr[a, b](c). ⊕ ⊕ ⊕ ⊕ 4. (a, b) H(n) P(n). In this case, a b H(n) and gyr[a, b] = A, when b O ; and ∈ × ⊕ ∈ ∈ P I, otherwise. We now consider the following two subcases: (a) b O . In this subcase, gyr[a, b] = I. If c P(n), then b c P(n) and by 6∈ P ∈ ⊕ ∈ definition of , ⊕ Symmetry 2021, 13, 316 7 of 12

(a b) gyr[a, b](c) = (a b) c ⊕ ⊕ ⊕ ⊕ m (a b) + ( 1)c + m (mod m) ≡ ⊕ 2 − m m [a + ( 1)b + m] + ( 1)c (mod m) ≡ 2 − 2 − m (15) a + ( 1)(b + c)(mod m) ≡ 2 − m a + ( 1)(b c) + m (mod m) ≡ 2 − ⊕ a (b c)(mod m) ≡ ⊕ ⊕ By (15), a (b c) = (a b) gyr[a, b](c). If c H(n), then b c H(n) ⊕ ⊕ ⊕ ⊕ ∈ ⊕ ∈ and we have: (a b) gyr[a, b](c) = (a b) c ⊕ ⊕ ⊕ ⊕ m m ( + 1)(a b) + ( 1)c (mod m) ≡ 2 ⊕ 2 − m m m ( + 1)[a + ( 1)b + m] + ( 1)c (mod m) ≡ 2 2 − 2 − m m m m ( + 1)a + ( 1)(b + c) + ( 1)b (mod m) ≡ 2 2 − 2 2 − m m ( + 1)a + ( 1)(b + c + m)(mod m) ≡ 2 2 − m m ( + 1)a + ( 1)(b c)(mod m) ≡ 2 2 − ⊕ a (b c)(mod m) ≡ ⊕ ⊕ The last equalities give a (b c) = (a b) gyr[a, b](c). ⊕ ⊕ ⊕ ⊕ (b) b O . In this subcase, gyr[a, b] = A. If c P(n), then b c P(n) ∈ P ∈ ⊕ ∈ and gyr[a, b](c) = c. A similar argument as (15) shows that a (b c) = ⊕ ⊕ (a b) gyr[a, b](c). If c H(n), then b c H(n), gyr[a, b](c) H(n) and ⊕ ⊕ ∈ ⊕ ∈ ∈ gyr[a, b](c) c + m (mod m). Therefore, ≡ 2 m m (a b) gyr[a, b](c) ( + 1)(a b) + ( 1)(gyr[a, b](c))(mod m) ⊕ ⊕ ≡ 2 ⊕ 2 − m m m m ( + 1)[a + ( 1)b + m] + ( 1)(c + )(mod m) ≡ 2 2 − 2 − 2 m m m m ( + 1)a + ( 1)(b + c) + ( 1)(b + 1)(mod m) ≡ 2 2 − 2 2 − m m ( + 1)a + ( 1)(b + c + m)(mod m) ≡ 2 2 − m m ( + 1)a + ( 1)(b c)(mod m) ≡ 2 2 − ⊕ a (b c)(mod m) ≡ ⊕ ⊕ The last equalities give a (b c) = (a b) 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 Z8. By deﬁnition, G(3) = 0, 1, 2, 3, 4, 5, 6, 7 and the binary operation is deﬁned { } ⊕ as follows:  t (i, j) P(3) P(3)  ∈ × i j = t + 4 (i, j) (P(3) H(3)) (H(3) P(3)) ⊕  ∈ × ∪ × k (i, j) H(3) H(3) ∈ × Symmetry 2021, 13, 316 8 of 12

in which P(3) = 0, 1, 2, 3 ,H(3) = 4, 5, 6, 7 and t, k P(3). Also, { } { } ∈ ( t i + j (mod 4) ≡ k 3i + j (mod 4) ≡ Therefore, the addition table and the gyration table for G(3) are presented in Table1 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 Tables2 and3 in which A is the unique nonidentity gyroautomorphism of G(4) given by A = (8, 12)(9, 13)(10, 14)(11, 15).

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) 0 1 2 3 4 5 6 7 ⊕ 0 0 1 2 3 4 5 6 7 1 1 2 3 0 5 6 7 4 2 2 3 0 1 6 7 4 5 3 3 0 1 2 7 4 5 6 4 4 5 6 7 0 1 2 3 5 5 6 7 4 3 0 1 2 6 6 7 4 5 2 3 0 1 7 7 4 5 6 1 2 3 0 (b) gyr 0 1 2 3 4 5 6 7 0 IIIIIIII 1 IIIIAAAA 2 IIIIIIII 3 IIIIAAAA 4 IAIAIAIA 5 IAIAAIAI 6 IAIAIAIA 7 IAIAAIAI

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

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 ⊕ 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 1 1 2 3 4 5 6 7 0 9 10 11 12 13 14 15 8 2 2 3 4 5 6 7 0 1 10 11 12 13 14 15 8 9 3 3 4 5 6 7 0 1 2 11 12 13 14 15 8 9 10 4 4 5 6 7 0 1 2 3 12 13 14 15 8 9 10 11 5 5 6 7 0 1 2 3 4 13 14 15 8 9 10 11 12 6 6 7 0 1 2 3 4 5 14 15 8 9 10 11 12 13 7 7 0 1 2 3 4 5 6 15 8 9 10 11 12 13 14 8 8 11 14 9 12 15 10 13 0 3 6 1 4 7 2 5 9 9 12 15 10 13 8 11 14 5 0 3 6 1 4 7 2 10 10 13 8 11 14 9 12 15 2 5 0 3 6 1 4 7 11 11 14 9 12 15 10 13 8 7 2 5 0 3 6 1 4 12 12 15 10 13 8 11 14 9 4 7 2 5 0 3 6 1 13 13 8 11 14 9 12 15 10 1 4 7 2 5 0 3 6 14 14 9 12 15 10 13 8 11 6 1 4 7 2 5 0 3 15 15 10 13 8 11 14 9 12 3 6 1 4 7 2 5 0 Symmetry 2021, 13, 316 9 of 12

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

gyr 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 0 IIIIIIIIIIIIIIII 1 IIIIIIIIAAAAAAAA 2 IIIIIIIIIIIIIIII 3 IIIIIIIIAAAAAAAA 4 IIIIIIIIIIIIIIII 5 IIIIIIIIAAAAAAAA 6 IIIIIIIIIIIIIIII 7 IIIIIIIIAAAAAAAA 8 IAIAIAIAIAIAIAIA 9 IAIAIAIAIAIAIAIA 10 IAIAIAIAIAIAIAIA 11 IAIAIAIAIAIAIAIA 12 IAIAIAIAIAIAIAIA 13 IAIAIAIAIAIAIAIA 14 IAIAIAIAIAIAIAIA 15 IAIAIAIAIAIAIAIA

Theorem 2. The gyrogroup (G(n), ) is non-gyrocommutative. ⊕ Proof. We know G(n) = 2n, for n 3. If n = 3, then by Example1, gyr[1, 5] = A | | ≥ and 1 5 = 6 = 4 = gyr[1, 5](6) = gyr[1, 5](5 1). Therefore, in this case G(n) is ⊕ 6 ⊕ non-gyrocommutative. For n > 3, suppose that G(n) is gyrocommutative. So, for every a, b G(n), the following equality is satisﬁed: ∈ a b = gyr[a, b](b a) (16) ⊕ ⊕ Suppose a = 2 and b = m + 1. By deﬁnition, 2 P(n), m + 1 H(n) and gyr[2, m + ∈ ∈ 1] = I. Also, 2 (m + 1) = t + m, (m + 1) 2 = s + m such that t 2 + m + 1 3 ⊕ ⊕ ≡ ≡ (mod m) and s m + 1 + ( m 1)2 m 1 (mod m). Thus 2 (m + 1) = m + 3 and ≡ 2 − ≡ − ⊕ (m + 1) 2 = 2m 1. By these relations and Equation (16), 2 (m + 1) = gyr[2, m + ⊕ − ⊕ 1]((m + 1) 2) = (m + 1) 2. This proves that m + 3 = 2m 1 and so m = 4. Therefore, n 1 2 ⊕ ⊕ − 2 − = 2 and hence n = 3 which is a contradiction.

Theorem 3. Let G(n) be the dihedral 2-gyrogroup of order 2n. 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 H(n); { } ∈ 4. There are integers r and s such that 1 r n 2, 0 s 2r 1 and B = 2r 2r + n 1 ≤ ≤ − ≤ ≤ − h i ∪ h i (m + s), where m = 2 − .

Proof. Suppose B is a proper subgyrogroup of G(n). We ﬁrst assume that B 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), 6⊆ we can write the subgyrogroup B as B B in which B P(n) and B H(n). It is easy 1 ∪ 2 1 ⊆ 2 ⊆ to see B is a subgroup of P(n). Suppose B = 0 and i, j B . By an easy calculation 1 1 { } ∈ 2 one can see that ( m + 1)i + ( m 1)j 0 (mod m) if and only if ( m + 1)i (1 m )j + mj 2 2 − ≡ 2 ≡ − 2 (mod m) if and only if ( m + 1)i ( m + 1)j (mod m) if and only if i j (mod m). Since 2 ≡ 2 ≡ i, j B H(n), i = j. This shows that B = i and B = 0, i , for some i H(n). ∈ 2 ⊆ 2 { } { } ∈ Therefore, B has the form of (3). Finally, suppose that B = 2r , 1 r n 2. We 1 h i ≤ ≤ − will prove that B = 2r 2r + (m + s), for some s, 0 s 2r 1. Choose arbitrary h i ∪ h i ≤ ≤ − elements u, v B . By deﬁnition u v B and there exists an integer k such that ∈ 2 ⊕ ∈ 1 Symmetry 2021, 13, 316 10 of 12

u v = m (u + v) + (u v) + mk and so u v 2r . Suppose x is the least element ⊕ 2 − − ∈ h i of B , then x = m + s in which 0 s 2r 1. If j B , then j x 2r and so 2 ≤ ≤ − ∈ 2 − ∈ h i j 2r + x = 2r + (m + s). This shows that B 2r + (m + s) = B + (m + s). Next ∈ h i h i 2 ⊆ h i 1 fix an element b B . Define two one to one mappings η : B B and η : B B ∈ 2 1 1 −→ 2 2 2 −→ 1 by η (c) = b c and η (d) = b d. This proves that B = B and so B = B + (m + s), 1 ⊕ 2 ⊕ | 1| | 2| 2 1 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 H(n). By definition of G(n), i i = 0 and by (Proposi- { } ∈ ⊕ tion 22) [14], H is a subgyrogroup of G(n). Finally, we assume that B = 2r 2r + (m + s), h i ∪ h i where r and s are positive integers such that 1 r n 2 and 0 s 2r 1. If i, j 2r ≤ ≤ − ≤ ≤ − ∈ h i then i j 2r B. If i B and j B , then there are integers k , k , k , k such that ⊕ ∈ h i ⊆ ∈ 1 ∈ 2 1 2 3 4 i = 2rk , j = 2rk + (m + s) and i j = i + j + 2rk + m = 2rk + 2rk + (m + s) + 2rk + m 1 2 ⊕ 3 1 2 3 = 2rk + (m + s). Hence, i j = 2rk + (m + s) B B. A similar calculations shows 4 ⊕ 4 ∈ 2 ⊆ that if i B , j B or i, j B then i j B. This completes the proof. ∈ 2 ∈ 1 ∈ 2 ⊕ ∈ 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 2n, n 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 D2n , 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 x1, , xr denotes as usual the subgyrogroup generated by the set x1, , xr . h ··· i n 2 { ··· } By Theorem3, it is easy to prove that 1 = Z n 1 , 2 = Z n 2 , , 2 = Z , 2, m = ∼ 2 − ∼ 2 − − ∼ 2 ∼ h i h i ··· h i n h2 i 2, m + 1 = Dm, 4, m = 4, m + 1 = 4, m + 2 = 4, m + 3 = D m , 2 − , m = ∼ ∼ ∼ ∼ ∼ 2 ∼ h n 2 i hn 2 i h i h n 2 i n h2 i h i 2 − , m + 1 = 2 − , m + 2 = = 2 − , m + 2 − 1 = D m = K4, m = ∼ ∼ ∼ ∼ n 3 ∼ h i h i ··· h − i 2 − h i m + 1 = m + 2 = = 2m 2 = 2m 1 = Z . In Figure1, the subgyrogroup and h i ∼ h i ∼ ··· ∼ h − i ∼ h − i ∼ 2 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 n dihedral group D2n and the gyrogroup G(n) of order 2 , we like to use the name dihedral gyrogroup of order 2n for G(n). Moreover, all proper subgyrogroups of G(n), n 3 are ≥ subgroups, see Figure2. We end this paper with the following question:

Question 1. Is it possible to apply the same method for constructing a gyrogroup H2n of order 2n, n 3, in which the the subgyrogroup and normal subgyrogroup lattices of H are isomorphic to ≥ 2n the subgroup and normal subgroup lattices of D2n, respectively? Version January 29, 2021 submitted to Journal Not Speciﬁed 9 of 11

m =Versionm + January1 = 29,m 2021+ 2 submitted= to=Journal2m Not2 Specified= 2m 1 = Z . In Figure 1, the subgyrogroup and normal subgyrogroup9 of 11 ⟨ ⟩ ∼ ⟨ ⟩ ∼ ⟨ ⟩ ∼ ··· ∼ ⟨ − ⟩ ∼ ⟨ − ⟩ ∼ 2 lattices of G(n) are depicted. In this figure, the circled subgyrogroups are denoted the normal subgyrogroups of G(n). n Because of this similarity between the dihedral group D2n and the gyrogroup G(n) of order 2 , we like to use the name n dihedralm gyrogroup= m + 1 = ofm order+ 2 2= for=G(2nm). Moreover,2 = 2m all1 proper= Z . In subgyrogroups Figure 1, the subgyrogroup of G(n), n and3 are normal subgroups, subgyrogroup see Figure ⟨ ⟩ ∼ ⟨ ⟩ ∼ ⟨ ⟩ ∼ ··· ∼ ⟨ − ⟩ ∼ ⟨ − ⟩ ∼ 2 ≥ 2. lattices of G(n) are depicted. In this figure, the circled subgyrogroups are denoted the normal subgyrogroups of G(n). n WeBecause end thisof this paper similarity by the between following the question:dihedral group D2n and the gyrogroup G(n) of order 2 , we like to use the name dihedral gyrogroup of order 2n for G(n). Moreover, all proper subgyrogroups of G(n), n 3 are subgroups, see Figure ≥ Question2. 1. Is it possible to apply the same method for constructing a gyrogroup H of order 2n, n 3, in which the the 2n ≥ subgyrogroupWe end and this normal paper subgyrogroup by the following lattices question: of H2n are isomorphic to the subgroup and normal subgroup lattices of D2n, respectively.? Question 1. Is it possible to apply the same method for constructing a gyrogroup H of order 2n, n 3, in which the the 2n ≥ Symmetrysubgyrogroup2021, 13, and 316 normal subgyrogroup lattices of H2n are isomorphic to the subgroup and normal subgroup lattices11 of of 12 D2n, respectively.? G = 1, m ⟨ ⟩

G = 1, m ⟨ ⟩ 2, m + 1 1 2, m ⟨ ⟩ ⟨ ⟩ ⟨ ⟩

2, m + 1 1 2, m ⟨ ⟩ ⟨ ⟩ ⟨ ⟩

4, m + 3 4, m + 1 2 4, m 4, m + 2 ⟨ ⟩ ⟨ ⟩ ⟨ ⟩ ⟨ ⟩ ⟨ ⟩ 4, m + 3 4, m + 1 2 4, m 4, m + 2 ⟨ ⟩ ⟨ ⟩ ⟨ ⟩ ⟨ ⟩ ⟨ ⟩ 8, m + 7 8, m + 5 8, m + 3 8, m + 1 4 8, m 8, m + 2 8, m + 4 8, m + 6 ⟨ ⟩ ⟨ ⟩ ⟨ ⟩ ⟨ ⟩ ⟨ ⟩ ⟨ ⟩ ⟨ ⟩ ⟨ ⟩ ⟨ ⟩ 8, m + 7 8, m + 5 8, m + 3 8, m + 1 4 8, m 8, m + 2 8, m + 4 8, m + 6 ⟨ ⟩ ⟨ ⟩ ⟨ ⟩ ⟨ ⟩ ⟨ ⟩ ⟨ ⟩ ⟨ ⟩ ⟨ ⟩ ⟨ ⟩

2m 1 m + 3 m + 1 2n 2 m m + 2 ⟨ − ⟩ ⟨ ⟩ ⟨ ⟩ − 2m 2 ⟨ n 2⟩ ⟨ ⟩ ⟨ ⟩ ⟨ − ⟩ 2m 1 m + 3 m + 1 2 − m m + 2 2m 2 ⟨ − ⟩ ⟨ ⟩ ⟨ ⟩ ⟨ ⟩ ⟨ ⟩ ⟨ ⟩ ⟨ − ⟩

0 ⟨ 0⟩ ⟨ ⟩

Figure 1. The subgyrogroup andFigure normal 1. The subgyrogroup subgyrogroup lattices and normal of G(n subgyrogroup). lattices of G(n). Figure 1. The subgyrogroup and normal subgyrogroup lattices of G(n).

G(4) D G(4) D1616

Z8 Z8 G(3G)(3) D8D8 DD8 8

Z D D D D Z4 4 D4 4 D4 4 D44 4D4

Z2 D2 D2 D2 D2 D2 D2 D2 D2 Z2 D2 D2 D2 D2 D2 D2 D2 D2

0 0 Figure 2. The subgyrogroup lattice of G(3), G(4), D8 and D16 in one frame. Figure 2. The subgyrogroup latticeFigure of 2.GThe(3), subgyrogroupG(4), D8 and latticeD16 in of G one(3) frame., G(4), D8 and D16 in one frame.

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

Author Contributions: S.M., A.R.A., M.A.S., and A.A.U. contributed equally to this article. All authors have read and agreed to the published version of the manuscript. Funding: This research received no external funding. Institutional Review Board Statement: Not applicable. Symmetry 2021, 13, 316 12 of 12

Informed Consent Statement: Not applicable. Data Availability Statement: Not applicable. Conﬂicts of Interest: The authors declare no conﬂict of interest.

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