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Article

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

by
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
2
Department of Mathematics, Savadkooh Branch, Islamic Azad University, Savadkooh 47418-39959, Iran
3
Department of Mathematics, North Dakota State University, Fargo, ND 58108-6050, USA
*
Author to whom correspondence should be addressed.
These authors contributed equally to this work.
Symmetry 2021, 13(2), 316; https://doi.org/10.3390/sym13020316
Submission received: 8 January 2021 / Revised: 5 February 2021 / Accepted: 8 February 2021 / Published: 14 February 2021
(This article belongs to the Special Issue Symmetry and Geometry in Physics)

Abstract

:
In this paper, a 2-gyrogroup G ( n ) of order 2 n , 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 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.
MSC:
Primary 20N05; Secondary 20F99; 20D99

1. Basic Concept and History

A pair ( G , ) consisting of a nonempty set G and a binary operation : G × G G 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 by A u t ( G ) . It is straightforward to check that A u t ( G ) forms a group under composition of function.
A groupoid ( G , ) is called a gyrogroup if the following conditions are satisfied:
  • there exists an element 0 G such that for all x G , 0 x = x ;
  • for each a G , there exists b G such that b a = 0 ;
  • (gyroassociative law) there exists a function g y r : G × G A u t ( G ) such that for every a , b , c G , a ( b c ) = ( a b ) g y r [ a , b ] c , where g y r [ a , b ] c = g y r ( a , b ) ( c ) ;
  • for each a , b G , g y r [ a , b ] = g y r [ a b , b ] .
The gyrogroup G is called gyrocommutative if and only if for all a , b G , a b = g y r [ a , b ] ( b a ) . The function g y r [ a , b ] , a , b 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 g y r [ 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 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 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 , , 2 n 1 1 } , H ( n ) = { 2 n 1 , 2 n 1 + 1 , , 2 n 1 } and G ( n ) = P ( n ) H ( n ) , where n 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 ) ( P ( n ) + m ) . Define the binary operation ⊕ on G ( n ) as follows:
i j = t ( i , j ) P ( n ) × P ( n ) t + m ( i , j ) P ( n ) × H ( n ) 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 ) s i + ( m 2 1 ) j ( mod m ) k ( m 2 + 1 ) i + ( m 2 1 ) j ( mod m ) .
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:
A ( i ) = i i P ( n ) r + m i H ( n )
where r P ( n ) and r i + m 2 ( mod m ) . Then A is an automorphism of ( G ( n ) , ) .
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 1 + m H ( n ) such that t 1 P ( n ) and t 1 i + j ( mod m ) . Now by our assumption, A ( i ) = i , A ( j ) = r 1 + m and
A ( i j ) = A ( t 1 + m ) = r 2 + m ,
where r 1 , r 2 P ( n ) , r 1 j + m 2 ( mod m ) and r 2 t 1 + m + m 2 ( mod m ) . By these equations,
A ( i ) A ( j ) = i ( r 1 + m ) = t 2 + m
such that t 2 P ( n ) and t 2 i + r 1 + m ( mod m ) . Hence, r 2 t 1 + m 2 i + j + m 2 i + r 1 t 2 ( mod m ) . Since r 2 , t 2 P ( n ) , r 2 = t 2 . We now apply Equations (1) and (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 1 + m H ( n ) , where s 1 P ( n ) and s 1 i + ( m 2 1 ) j ( mod m ) . By our assumption A ( i ) = r 1 + m , A ( j ) = j and
A ( i j ) = A ( s 1 + m ) = r 2 + m ,
where r 1 , r 2 P ( n ) , r 1 i + m 2 ( mod m ) and r 2 s 1 + m + m 2 ( mod m ) . By these calculations, one can see that
A ( i ) A ( j ) = ( r 1 + m ) j = s 2 + m
such that s 2 P ( n ) and s 2 r 1 + m + ( m 2 1 ) j ( mod m ) . Hence, r 2 s 1 + m 2 i + ( m 2 1 ) j + m 2 r 1 + ( m 2 1 ) j s 2 ( mod m ) . Since r 2 , s 2 P ( n ) , r 2 = s 2 , and by Equations (3) and (4), A ( i j ) = A ( i ) A ( j ) , as desired.
3.
i , j H ( n ) . By definition, i j = k 1 P ( n ) , where k 1 ( m 2 + 1 ) i + ( m 2 1 ) j ( mod m ) . We now apply our assumption to deduce that A ( i ) = r 1 + m , A ( j ) = r 2 + m and
A ( i j ) = A ( k 1 ) = k 1
such that r 1 , r 2 P ( n ) , r 1 i + m 2 ( mod m ) and r 2 j + m 2 ( mod m ) . By above calculations,
A ( i ) A ( j ) = ( r 1 + m ) ( r 2 + m ) = k 2
in which k 2 P ( n ) and k 2 ( m 2 + 1 ) ( r 1 + m ) + ( m 2 1 ) ( r 2 + m ) ( mod m ) . Hence, k 2 ( m 2 + 1 ) r 1 + ( m 2 1 ) r 2 ( m 2 + 1 ) i + ( m 2 1 ) j k 1 ( mod m ) . Since k 1 , k 2 P ( n ) , again k 1 = k 2 and by Equations (5) and (6), A ( i j ) = A ( i ) A ( j ) .
This completes our argument. □
Set O P = { i P ( n ) 2 i } , O H = { i H ( n ) 2 i } , E H = { i H ( n ) 2 i } and M = [ O P × ( O H E H ) ] [ O H × ( O P E H ) ] [ E H × ( O P O H ) ] . Define the map g y r : G ( n ) × G ( n ) A u t ( G ( n ) , ) as follows:
g y r ( a , b ) = g y r [ a , b ] = A ( a , b ) M I otherwise
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 Z 2 n , for n 3 .
Theorem 1.
( G ( n ) , ) is a gyrogroup.
Proof. 
By Lemma 1, g y r [ a , b ] A u t ( 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 x P ( n ) x x H ( n )
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, by definition of g y r , g y r [ a b , b ] = I = g y r [ a , b ] , as desired.
2.
( a , b ) H ( n ) × H ( n ) . In this case, a b P ( n ) and
a b ( m 2 + 1 ) a + ( m 2 1 ) b ( m 2 + 1 ) ( a b ) ( mod m ) .
If a b O P , then by Equation (7), ( a b , b ) , ( a , b ) M . Hence g y r [ a b , b ] = I = g y r [ a , b ] . If a b O P , then by Equation (7), ( a b , b ) , ( a , b ) M . Hence for every i G ( n ) , if i P ( n ) , then g y r [ a b , b ] ( i ) = i = g y r [ a , b ] ( i ) . If i H ( n ) , then g y r [ a b , b ] ( i ) = r + m = g y r [ a , b ] ( i ) such that r i + m 2 ( mod m ) .
3.
( a , b ) P ( n ) × H ( n ) . In this case, a b H ( n ) and
a b a + b ( mod m ) .
If a b O H , then by Equation (8), we have the following two subcases:
(a)
a O P and b O H . In this case, ( a b , b ) , ( a , b ) M and by definition of g y r , g y r [ a b , b ] = I = g y r [ a , b ] .
(b)
a O P and b E H . In this case, ( a b , b ) , ( a , b ) M . For every i G ( n ) , if i P ( n ) , then g y r [ a b , b ] ( i ) = i = g y r [ a , b ] ( i ) . If i H ( n ) , then g y r [ a b , b ] ( i ) = r + m = g y r [ a , b ] ( i ) such that r i + m 2 ( mod m ) .
If a b E H , then by Equation (8), we have the following two subcases:
(a)
a O P and b E H . In this case ( a b , b ) , ( a , b ) M . The proof is now similar to the subcase (a).
(b)
a O P and b O H . In this case ( a b , b ) , ( a , b ) M . The proof of this subcase 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 2 1 ) b ( mod m ) . 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 g y r [ a , b ] = I . If c P ( n ) , then b c P ( n ) and by definition of ⊕, we have:
( a b ) g y r [ 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 ) g y r [ a , b ] ( c ) . If c H ( n ) , then b c H ( n ) and by definition of ⊕,
( a b ) g y r [ 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 ) g y r [ 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
g y r [ a , b ] = A ( a , b ) N I otherwise
in which N = ( O H × E H ) ( E H × O H ) M . We consider two subcases as follows:
(a)
( a , b ) N . In this subcase, g y r [ a , b ] = I and ( a , b O H or a , b E H ). If c P ( n ) , then b c H ( n ) and by definition of ⊕,
( a b ) g y r [ a , b ] ( c ) = ( a b ) c ( a b ) + c ( mod m ) [ ( m 2 + 1 ) a + ( m 2 1 ) b ] + c ( mod m )
and
a ( b c ) ( m 2 + 1 ) a + ( m 2 1 ) ( b c ) ( mod m ) ( m 2 + 1 ) a + ( m 2 1 ) [ b + ( m 2 1 ) c + m ] ( mod m ) ( m 2 + 1 ) a + [ ( m 2 1 ) b + c + ( m 4 1 ) m c ] ( mod m ) ( m 2 + 1 ) a + [ ( m 2 1 ) b + c ] ( mod m ) .
By Equations (9) and (10), a ( b c ) = ( a b ) g y r [ a , b ] ( c ) . If c H ( n ) , then b c P ( n ) and by definition of ⊕,
( a b ) g y r [ a , b ] ( c ) = ( a b ) c ( a b ) + c + m ( mod m ) [ ( m 2 + 1 ) a + ( m 2 1 ) b ] + c ( mod m ) a b + c + m 2 ( a + b ) ( mod m ) a b + c ( mod m )
On the other hand,
a ( b c ) a + ( m 2 1 ) ( b c ) + m ( mod m ) a + ( m 2 1 ) [ ( m 2 + 1 ) b + ( m 2 1 ) c ] ( mod m ) a + [ ( m 2 4 1 ) b + ( m 2 1 ) 2 c ] ( mod m ) a b + c + m [ m 4 b + ( m 4 1 ) c ] ( mod m ) a b + c ( mod m ) .
By Equations (11) and (12), a ( b c ) = ( a b ) g y r [ a , b ] ( c ) .
(b)
( a , b ) N . In this subcase, g y r [ a , b ] = A and one of ( a O H , b E H ) or ( a E H , b O H ) can be occurred. If c P ( n ) , then b c H ( n ) and g y r [ a , b ] ( c ) = c . A similar argument as (9) and (10) shows that a ( b c ) = ( a b ) g y r [ a , b ] ( c ) . If c H ( n ) , then b c P ( n ) , g y r [ a , b ] ( c ) H ( n ) and g y r [ a , b ] ( c ) c + m 2 ( mod m ) . Again by definition of ⊕,
( a b ) g y r [ a , b ] ( c ) ( a b ) + g y r [ a , b ] ( c ) + m ( mod m ) [ ( m 2 + 1 ) a + ( m 2 1 ) b ] + c + m 2 ( mod m ) a b + c ( mod m )
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 ) g y r [ a , b ] ( c ) .
3.
( a , b ) P ( n ) × H ( n ) . In this case, a b H ( n ) and
g y r [ a , b ] = A a O P I a O P .
We consider two subcases as follows:
(a)
a O P . In this subcase, g y r [ a , b ] = I and if c P ( n ) , then b c H ( n ) . Now by definition ⊕,
( a b ) g y r [ a , b ] ( c ) = ( a b ) c ( a b ) + ( m 2 1 ) c + m ( mod m ) ( a + b + m ) + ( m 2 1 ) c ( mod m ) a + [ b + ( m 2 1 ) c + m ] ( mod m ) a + [ b c ] + m ( mod m ) a ( b c ) ( mod m )
By Equation (14), a ( b c ) = ( a b ) g y r [ a , b ] ( c ) . If c H ( n ) , then b c P ( n ) and by definition of ⊕,
( a b ) g y r [ a , b ] ( c ) = ( a b ) c ( m 2 + 1 ) ( a b ) + ( m 2 1 ) c ( mod m ) ( m 2 + 1 ) ( a + b + m ) + ( m 2 1 ) c ( mod m ) a + ( m 2 + 1 ) b + ( m 2 1 ) c + m 2 a ( mod m ) a + ( m 2 + 1 ) b + ( m 2 1 ) c ( mod m ) a + ( b c ) ( mod m ) a ( b c ) ( mod m )
By the last equalities, a ( b c ) = ( a b ) g y r [ a , b ] ( c ) .
(b)
a O P . In this subcase, g y r [ a , b ] = A . If c P ( n ) , then b c H ( n ) and g y r [ a , b ] ( c ) = c . A similar argument as Equation (14) shows that a ( b c ) = ( a b ) g y r [ a , b ] ( c ) . If c H ( n ) , then b c P ( n ) , g y r [ a , b ] ( c ) H ( n ) and g y r [ a , b ] ( c ) c + m 2 ( mod m ) . Apply again definition of ⊕ to deduce that
( a b ) g y r [ a , b ] ( c ) ( m 2 + 1 ) ( a b ) + ( m 2 1 ) ( g y r [ a , b ] ( c ) ) ( mod m ) ( m 2 + 1 ) ( a + b + m ) + ( m 2 1 ) ( c + m 2 ) ( mod m ) a + ( m 2 + 1 ) b + ( m 2 1 ) c + m 2 ( a + m 2 1 ) ( mod m ) a + ( m 2 + 1 ) b + ( m 2 1 ) c ( mod m ) a + ( b c ) a ( b c ) ( mod m )
By the last equalities, a ( b c ) = ( a b ) g y r [ a , b ] ( c ) .
4.
( a , b ) H ( n ) × P ( n ) . In this case, a b H ( n ) and g y r [ a , b ] =A, when b O P ; and I, otherwise. We now consider the following two subcases:
(a)
b O P . In this subcase, g y r [ a , b ] = I . If c P ( n ) , then b c P ( n ) and by definition of ⊕,
( a b ) g y r [ a , b ] ( c ) = ( a b ) c ( a b ) + ( m 2 1 ) c + m ( mod m ) [ a + ( m 2 1 ) b + m ] + ( m 2 1 ) c ( mod m ) a + ( m 2 1 ) ( b + c ) ( mod m ) a + ( m 2 1 ) ( b c ) + m ( mod m ) a ( b c ) ( mod m )
By (15), a ( b c ) = ( a b ) g y r [ a , b ] ( c ) . If c H ( n ) , then b c H ( n ) and we have:
( a b ) g y r [ a , b ] ( c ) = ( a b ) c ( m 2 + 1 ) ( a b ) + ( m 2 1 ) c ( mod m ) ( m 2 + 1 ) [ a + ( m 2 1 ) b + m ] + ( m 2 1 ) c ( mod m ) ( m 2 + 1 ) a + ( m 2 1 ) ( b + c ) + m 2 ( m 2 1 ) b ( mod m ) ( m 2 + 1 ) a + ( m 2 1 ) ( b + c + m ) ( mod m ) ( m 2 + 1 ) a + ( m 2 1 ) ( b c ) ( mod m ) a ( b c ) ( mod m )
The last equalities give a ( b c ) = ( a b ) g y r [ a , b ] ( c ) .
(b)
b O P . In this subcase, g y r [ a , b ] = A . If c P ( n ) , then b c P ( n ) and g y r [ a , b ] ( c ) = c . A similar argument as (15) shows that a ( b c ) = ( a b ) g y r [ a , b ] ( c ) . If c H ( n ) , then b c H ( n ) , g y r [ a , b ] ( c ) H ( n ) and g y r [ a , b ] ( c ) c + m 2 ( mod m ) . Therefore,
( a b ) g y r [ a , b ] ( c ) ( m 2 + 1 ) ( a b ) + ( m 2 1 ) ( g y r [ a , b ] ( c ) ) ( mod m ) ( m 2 + 1 ) [ a + ( m 2 1 ) b + m ] + ( m 2 1 ) ( c + m 2 ) ( mod m ) ( m 2 + 1 ) a + ( m 2 1 ) ( b + c ) + m 2 ( m 2 1 ) ( b + 1 ) ( mod m ) ( m 2 + 1 ) a + ( m 2 1 ) ( b + c + m ) ( mod m ) ( m 2 + 1 ) a + ( m 2 1 ) ( b c ) ( mod m ) a ( b c ) ( mod m )
The last equalities give a ( b c ) = ( a b ) g y r [ 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 Z 8 . By definition, G ( 3 ) = { 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 } and the binary operation ⊕ is defined as follows:
i j = t ( i , j ) P ( 3 ) × P ( 3 ) t + 4 ( i , j ) ( P ( 3 ) × H ( 3 ) ) ( H ( 3 ) × P ( 3 ) ) k ( i , j ) H ( 3 ) × H ( 3 )
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 3 i + j ( mod 4 )
Therefore, the addition table and the gyration table for G ( 3 ) are presented in Table 1 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 and Table 3 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 ) , ) is non-gyrocommutative.
Proof. 
We know | G ( n ) | = 2 n , for n 3 . If n = 3 , then by Example 1, g y r [ 1 , 5 ] = A and 1 5 = 6 4 = g y r [ 1 , 5 ] ( 6 ) = g y r [ 1 , 5 ] ( 5 1 ) . Therefore, in this case G ( n ) is 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 = g y r [ a , b ] ( b a )
Suppose a = 2 and b = m + 1 . By definition, 2 P ( n ) , m + 1 H ( n ) and g y r [ 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 2 1 ) 2 m 1 ( mod m ) . Thus 2 ( m + 1 ) = m + 3 and ( m + 1 ) 2 = 2 m 1 . By these relations and Equation (16), 2 ( m + 1 ) = g y r [ 2 , m + 1 ] ( ( m + 1 ) 2 ) = ( m + 1 ) 2 . This proves that m + 3 = 2 m 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 H ( n ) ;
4.
There are integers r and s such that 1 r n 2 , 0 s 2 r 1 and B = 2 r 2 r + ( m + s ) , where m = 2 n 1 .
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 ) , we can write the subgyrogroup B as B 1 B 2 in which B 1 P ( n ) and B 2 H ( n ) . It is easy to see B 1 is a subgroup of P ( n ) . Suppose B 1 = { 0 } and i , j B 2 . By an easy calculation one can see that ( m 2 + 1 ) i + ( m 2 1 ) j 0 ( m o d m ) if and only if ( m 2 + 1 ) i ( 1 m 2 ) j + m j ( m o d m ) if and only if ( m 2 + 1 ) i ( m 2 + 1 ) j ( m o d m ) if and only if i j ( m o d m ). Since i , j B 2 H ( n ) , i = j . This shows that B 2 = { i } and B = { 0 , i } , for some i H ( n ) . Therefore, B has the form of ( 3 ) . Finally, suppose that B 1 = 2 r , 1 r n 2 . We will prove that B = 2 r 2 r + ( m + s ) , for some s, 0 s 2 r 1 . Choose arbitrary elements u , v B 2 . By definition u v B 1 and there exists an integer k such that u v = m 2 ( u + v ) + ( u v ) + m k and so u v 2 r . Suppose x is the least element of B 2 , then x = m + s in which 0 s 2 r 1 . If j B 2 , then j x 2 r and so j 2 r + x = 2 r + ( m + s ) . This shows that B 2 2 r + ( m + s ) = B 1 + ( m + s ) . Next fix an element b B 2 . Define two one to one mappings η 1 : B 1 B 2 and η 2 : B 2 B 1 by η 1 ( c ) = b c and η 2 ( d ) = b 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 H ( n ) . By definition of G ( n ) , i i = 0 and by (Proposition 22) [14], H is a subgyrogroup of G ( n ) . Finally, we assume that B = 2 r 2 r + ( m + s ) , where r and s are positive integers such that 1 r n 2 and 0 s 2 r 1 . If i , j 2 r then i j 2 r B . If i B 1 and j 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 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 j = 2 r k 4 + ( m + s ) B 2 B . A similar calculations shows that if i B 2 , j B 1 or i , j B 2 then i j 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 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 x 1 , , x r denotes as usual the subgyrogroup generated by the set { x 1 , , x r } . By Theorem 3, it is easy to prove that 1 Z 2 n 1 , 2 Z 2 n 2 , ⋯, 2 n 2 Z 2 , 2 , m 2 , m + 1 D m , 4 , m 4 , m + 1 4 , m + 2 4 , m + 3 D m 2 , 2 n 2 , m 2 n 2 , m + 1 2 n 2 , m + 2 2 n 2 , m + 2 n 2 1 D m 2 n 3 = K 4 , m m + 1 m + 2 2 m 2 2 m 1 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 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 2 n of order 2 n , n 3 , in which the the subgyrogroup and normal subgyrogroup lattices of H 2 n are isomorphic to the subgroup and normal subgroup lattices of D 2 n , respectively?
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.

Informed Consent Statement

Not applicable.

Data Availability Statement

Not applicable.

Conflicts of Interest

The authors declare no conflict of interest.

References

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Figure 1. The subgyrogroup and normal subgyrogroup lattices of G ( n ) .
Figure 1. The subgyrogroup and normal subgyrogroup lattices of G ( n ) .
Symmetry 13 00316 g001
Figure 2. The subgyrogroup lattice of G ( 3 ) , G ( 4 ) , D 8 and D 16 in one frame.
Figure 2. The subgyrogroup lattice of G ( 3 ) , G ( 4 ) , D 8 and D 16 in one frame.
Symmetry 13 00316 g002
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 ) .
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 01234567
1 12305674
2 23016745
3 30127456
4 45670123
5 56743012
6 67452301
7 74561230
(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 ) .
Table 2. The addition table of G ( 4 ) .
0123456789101112131415
0 0123456789101112131415
1 1234567091011121314158
2 2345670110111213141589
3 3456701211121314158910
4 4567012312131415891011
5 5670123413141589101112
6 6701234514158910111213
7 7012345615891011121314
8 8111491215101303614725
9 9121510138111450361472
10 1013811149121525036147
11 1114912151013872503614
12 1215101381114947250361
13 1381114912151014725036
14 1491215101381161472503
15 1510138111491236147250
Table 3. The gyration table of G ( 4 ) such that A = ( 8 , 12 ) ( 9 , 13 ) ( 10 , 14 ) ( 11 , 15 ) .
Table 3. The gyration table of G ( 4 ) such that A = ( 8 , 12 ) ( 9 , 13 ) ( 10 , 14 ) ( 11 , 15 ) .
gyr 0123456789101112131415
0IIIIIIIIIIIIIIII
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
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Mahdavi, S.; Ashrafi, A.R.; Salahshour, M.A.; Ungar, A.A. Construction of 2-Gyrogroups in Which Every Proper Subgyrogroup Is Either a Cyclic or a Dihedral Group. Symmetry 2021, 13, 316. https://doi.org/10.3390/sym13020316

AMA Style

Mahdavi S, Ashrafi AR, Salahshour MA, Ungar AA. Construction of 2-Gyrogroups in Which Every Proper Subgyrogroup Is Either a Cyclic or a Dihedral Group. Symmetry. 2021; 13(2):316. https://doi.org/10.3390/sym13020316

Chicago/Turabian Style

Mahdavi, Soheila, Ali Reza Ashrafi, Mohammad Ali Salahshour, and Abraham Albert Ungar. 2021. "Construction of 2-Gyrogroups in Which Every Proper Subgyrogroup Is Either a Cyclic or a Dihedral Group" Symmetry 13, no. 2: 316. https://doi.org/10.3390/sym13020316

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