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ABSTRACT
This paper presents several results from the study of certain properties of G-algebras. More precisely, we will show that, if N is a normal subset of a finite G-algebra X, then the order of N divides the order of X. In addition, we will give an important result about the set of derivations on a finite G-algebra X, and give a condition on this set to determine the commutativity and the associativity of X.
1. Introduction
Since 1966 many algebraic structures are introduced such as BCK-algebras, BCH-algebra [Citation1], BCI-algebra [Citation2] and B-algebra [Citation3–5].
In 2012, R. K. Bandru and N. Rafi introduced in [Citation6] a new notion called G-algebra. This notion played an important role in algebra and has many applications [Citation7, Citation8].
Our aim in this work is to study the number of derivations on G-algebras. We start with definitions and propositions on G-algebra, then we apply the Lagrange theorem on G-algebras to prove that if is the set of derivations on a finite G-algebra X, then the order of
divides the order of X. As consequence, we show that:
If X is a G-algebra of odd order, then
.
Let
be a G-algebra. Then
is abelian, and has a structure of G-algebra with the same operation.
The function
is a derivation for all
if and only if the operation ⋆ is commutative and associative on X (i. e. X is an abelian group).
If there is an element a such that the function
defined by
is not a derivation on X, then
.
Every G-algebra of order 2 is an abelian group.
Many other corollaries are mentioned in this paper and have interesting applications in different examples of G-algebras.
2. Preliminaries
Let X be a non-empty set throughout a binary operation ⋆. Let 0 be a fixed element in X.
Definition 2.1
A G-algebra is a non-empty set X throughout a binary operation ⋆ such that the following conditions are satisfied:
there exists an element
such that
for all
.
for all
.
Example 2.2
Let F be a field and . Consider the binary operation ⋆ on
defined by
. Then
is a G-algebra. In fact, for all x, y and z in
,
and
Proposition 2.3
Let be a G-algebra, then the following conditions hold:
for all
.
then x = y.
if
then x = y.
Definition 2.4
[Citation7,Citation8]
Let S be a non-empty subset of a G-algebra X. We say that S is a G-subalgebra of X if it is closed under ⋆, that means for all
.
Definition 2.5
Let N be a non-empty subset of a G-algebra X. We say that N is a normal subset of X if for all x, y, z and t in X such that and
, we have
Example 2.6
Let F be a field. Consider the G-algebra mentioned in Example 2.2. Then
is a proper normal subset of
.
Clearly, every normal subset of X is a G-subalgebra but the converse is not true in general.
Proposition 2.7
Let be a G-algebra. Then
every normal subset contains e,
X and
are normal subset of X,
the intersection of normal subsets (resp. G-subalgebras) of X is a normal subset (resp. G-subalgebra) of X.
Proof.
It is clear that X is a normal subset of X. Let x, y, z and t be elements in X such that and
. From Proposition 2.3 (2), we obtain that x = y and z = t. Therefore
, and hence
This shows that
is a normal subset of X. The proof of (3) is trivial.
Definition 2.8
A G-subalgebra (resp. a normal subset) of an G-algebra X is called proper, if it is not equal either to X nor .
Example 2.9
Let a set and ⋆ is the binary operation on X defined as follows:
We know that
is a G-algebra. Let
,
be the unique proper G-sub-algebras of X. In addition, the set
is the only proper normal subset of X.
Definition 2.10
If and
are two G-algebras of X, we define their product
by
Lemma 2.11
As in the previous definition, if is a normal subset and
is a G-subalgebra, then
is a G-subalgebra of X.
Proof.
Assume that is a normal subset and
is a G-subalgebra. It is clear that
so
Let x and
be two elements in
. By definition, there exist y and
in
such that
and
. Since
is normal, we get
(1)
(1) As
is G-subalgebra,
, and from (Equation1
(1)
(1) ), we obtain
.
3. Lagrange theorem for G-algebras
Let N be a normal subset of a G-algebra . Consider the relation
on X defined as follows:
As N is normal subset of X,
is an equivalent relation. For all
, denote
by the equivalence class of x under this relation. That means
The set X/N of the equivalence classes of this relation form a G-algebra under the following binary operation:
Definition 3.1
The G-algebra G/N obtained from the preceding construction is called the factor G-algebra (or quotient G-algebra) of X by N.
The following theorem is an analogue of the Lagrange theorem.
Theorem 3.2
If X is a finite G-algebra, then for every normal subset N of X, the order of N divides the order of X.
Proof.
Let and
be the distinct classes. We know that
Consider the function
defined as
.
This function is a bijection. In fact means
. Then
. In addition for all
,
. Also
As a consequence, we obtain
for all i = 1, 2, …, n. Then
and this proves the theorem.
Remark 3.3
The set in Example 2.9 is not a normal subset because its order does not divide the order of X. Otherwise the order of
divides that of X.
Definition 3.4
Let and
be two G-algebras. A mapping
is called a homomorphism of G-algebras, if
for all
.
Example 3.5
If N is a normal subset of a G-algebra X, then the natural mapping defined as
is a homomorphism of G-algebras.
Definition 3.6
Let be a homomorphism of G-algebras. The set
is called the kernel of φ, denoted by
. for all
. We define the image of φ by the set
, and denote by
Theorem 3.7
Let be a homomorphism of G-algebras. Then:
.
is a normal subset of X.
If S is a G-subalgebra of X, then its direct image
is a G-subalgebra of
.
If
is a normal subset (resp. G-subalgebra) of
then its inverse image
is a normal subset (resp. G-subalgebra) of X.
Proof.
The proof of this theorem is easy and can be obtained directly by using the usual ideas in theory of groups.
Definition 3.8
A homomorphism of G-algebra is called:
A monomorphism if ψ is one-to-one.
An epimorphism if ψ is onto.
An isomorphism if ψ is one-to-one and onto.
Lemma 3.9
A homomorphism of G-algebra is a monomorphism if and only if
.
Proof.
If x is an element in , then
. Hence if ψ is a monomorphism, we obtain directly
. Reciprocally, if
, then for all x and y in X such that
, we have
. Since ψ is a homomorphism, we get
, and hence
. So
. Hence x = y, and this shows that ψ is a monomorphism.
4. Derivation on G-algebras
Definition 4.1
Let be a G-algebra. A G-derivation on X is a mapping
such that
for all
. We denote
be the set of all G-derivation on X.
Remark 4.2
As for all
, every G-derivation d can be determined by
as follows:
for all
.
Lemma 4.3
Let and
be a function defined as
for all
. If d is a derivation, then
.
Proof.
By definition of d, we have
(2)
(2) If d is a derivation, then
(3)
(3) From (Equation2
(2)
(2) ) and (Equation3
(3)
(3) ), we obtain
.
Proposition 4.4
Let be a G-algebra.
The identity map I is a G-derivation on X.
For all
,
.
For all
,
.
For all
there exist
such that
.
Proof.
The first three properties are proved in [Citation8]. To prove (4), we need first to show that every derivation on X is a bijection.
Let x and y in X such that
. Then
and
. That means
and
. Hence,
and
. Therefore
and
. Since
, we get x = y.
If we take in the previous item x = y, we obtain that
for all
Hence d is a bijection and .
Theorem 4.5
Let be a G-algebra. Then
is abelian, and has a structure of G-algebra with the same operation.
Proof.
The previous proposition shows that is an abelian group. We have
for all
, and by the associativity of the composite operation, we can show that
for all
and
in
.
Corollary 4.6
Every G-subalgebra of is a normal.
Consider the map defined by:
Lemma 4.7
Let X be a G-algebra. Then is a monomorphism of G-algebras.
Proof.
Let and
be G-derivations on X. We have
Hence,
is a homomorphism of G-algebra. Let
. Then
, so for all
, we have
So d = I and
. Hence,
is one-to-one.
Proposition 4.8
is a normal subset of X.
Proof.
Let and
be two elements in
. Then there exist G-derivations
and
on X such that
and
. Since
for i = 1, 2,
and
. This implies that
and
. So
and
Therefore,
(4)
(4) Consider the G-derivation
. Then from (Equation4
(4)
(4) )
So
This shows that the derivation d satisfies the equality
Since
, we deduce from the previous equality that
(5)
(5) Finally, we conclude that
for all
and
in
. Hence, the set
is normal in X.
Corollary 4.9
Let X be a finite G-algebra. Then the order of divides the order of X.
Proof.
Since is a normal subset of X, we get by the Lagrange theorem,
divides
. Apply the first isomorphism theorem to the monomorphism
, we obtain
Hence, the order of
equals the order of
Consequently, the order of
divides the order of X.
Example 4.10
Consider the G-algebra with the binary operation ⋆ defined as follows:
From Remark 4.2 and Lemma 4.3, we know that every derivation on X can be written as for some
satisfying
. By the definition of the operation ⋆, we can remark that z must be equal to e or z. If z = e, then d is the identity derivation, and if z = b, then we can check that the function
is already a G-derivation.
Finally, and
. In this example we remark that the order of
divides the order of X and
is a normal subset of X.
Example 4.11
Consider the G-algebra presented as follows:
We know that every derivation d on X is determined by the value of d at e. Consider the five functions for i = 1, 2, 3, 4, 5 defined as follows:
and
Clearly is a G-derivation, but
is not a G-derivation, since
.
Applying Corollary 4.9, we obtain that the order of equals 1 or 5. Since
is not a derivation, we have necessarily
.
Proposition 4.12
Let p be a prime number and the G-algebra with the operation defined as follows:
if
and otherwise:
Then X has only the identity derivation, more precisely
Proposition 4.8 can be generalized as follows:
Proposition 4.13
Let S be a subset of closed under composition. Then
is a normal subset of X.
Proof.
The proof of this proposition is the same as that of Proposition 4.8.
Theorem 4.14
If X is a G-algebra of odd order, then
Proof.
Assume the contrary. That means . Then there exists a G-derivation
. As
and
, the subset
is normal in
. Form Proposition 4.13, the subset
is normal in X. The fact
implies that
. Using Theorem 3.2, we obtain the result.
Corollary 4.15
Let X be a finite G-algebra. Then
The following example shows that if X is an infinite G-algebra, then can be finite.
Example 4.16
Let F be a field and . Take an arbitrary derivation d on X. Then
for some
.
For all x, y in X, . That means
So d is a derivation if and only if
. As F is a field, this equality holds only in the case when a = 1 or a = −1. Then
Theorem 4.17
Let X be a G-algebra. Then the following statements are equivalent:
The function
is a derivation for all
.
The operation ⋆ is commutative and associative on X (i.e. X is an abelian group).
Proof.
Assume that the function is a derivation for all
. Applying Proposition 2.3 and Lemma 4.3 we obtain
(6)
(6) Let b, c be elements in X. As
is a derivation,
therefore,
(7)
(7) In addition, the fact
is a derivation,
. Then
(8)
(8) Replacing b by e in the last equality, we obtain that
(9)
(9) Using (Equation6
(6)
(6) ), we get from (Equation9
(9)
(9) ):
(10)
(10) The properties (Equation6
(6)
(6) ), (Equation7
(7)
(7) ) and (Equation10
(10)
(10) ) show that X is an abelian group. Reciprocally, suppose that X is an abelian group. We will show that
is a derivation for all
. Let x, y be elements in X. As the operation ⋆ is commutative and associative,
So
Corollary 4.18
Let X be a finite G-algebra. If there is an element a such that is not a derivation on X (i.e. G is not a group), then
.
Corollary 4.19
Let X be a G-algebra of prime order. If there is an element a such that is not a derivation on X, then
.
Corollary 4.20
Let N be a normal subset of X of order 2. Then for some derivation d.
Let be the second element of N.
Then
and
Corollary 4.21
Every G-algebra of order 2 is an abelian group.
Disclosure statement
No potential conflict of interest was reported by the authors.
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