About derivations on G-algebras

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.

KEYWORDS:

• G-algebras
• normal subsets
• derivations on G-algebra

1. Introduction

Since 1966 many algebraic structures are introduced such as BCK-algebras, BCH-algebra [1], BCI-algebra [2] and B-algebra [3–5].

In 2012, R. K. Bandru and N. Rafi introduced in [6] a new notion called G-algebra. This notion played an important role in algebra and has many applications [7, 8].

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 ${\mathrm{\Omega }}_X$ is the set of derivations on a finite G-algebra X, then the order of ${\mathrm{\Omega }}_X$ divides the order of X. As consequence, we show that:

• If X is a G-algebra of odd order, then ${\mathrm{\Omega }}_X=\left\{I\right\}$.

• Let $(X,\star ,e)$ be a G-algebra. Then $({\mathrm{\Omega }}_X,\circ )$ is abelian, and has a structure of G-algebra with the same operation.

• The function $d_a\left(x\right)=x\star a$ is a derivation for all $a\in X$ 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 $d_a:X⟶X$ defined by $d_a\left(b\right)=a\star b$ is not a derivation on X, then $|{\mathrm{\Omega }}_X|<|X|$.

• 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:

1. there exists an element $e\in X$ such that $x\star x=e$ for all $x\in X$.

2. $x\star (x\star y)=y$ for all $x,y\in X$.

Example 2.2

Let F be a field and $X\left(F\right)=F-\left\{0\right\}$. Consider the binary operation ⋆ on $X\left(F\right)$ defined by $x\star y=x{y}^{-1}$. Then $\left(X\right(F),\star ,1)$ is a G-algebra. In fact, for all x, y and z in $X\left(F\right)$, $x\star x=x{x}^{-1}=1$ and $x\star (x\star y)=x\star \left(x{y}^{-1}\right)=x\star \left(x{y}^{-1}\right)=x(x{y}^{-1}{)}^{-1}=y.$

Proposition 2.3

Let $(X,\star ,0)$ be a G-algebra, then the following conditions hold:

1. $x\star e=x$ for all $x\in X$.

2. $x\star y=e,$ then x = y.

3. if $a\star x=a\star y,$ then x = y.

Definition 2.4

[7,8]

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 $x\star y\in S$ for all $x,y\in S$.

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 $x\star y\in N$ and $z\star t\in N$, we have $(x\star z)\star (y\star t)\in N.$

Example 2.6

Let F be a field. Consider the G-algebra $X\left(F\right)$ mentioned in Example 2.2. Then $\{1,-1\}$ is a proper normal subset of $X\left(F\right)$.

Clearly, every normal subset of X is a G-subalgebra but the converse is not true in general.

Proposition 2.7

Let $(X,\star ,e)$ be a G-algebra. Then

1. every normal subset contains e,

2. X and $\left\{e\right\}$ are normal subset of X,

3. 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 $x\star y=e$ and $z\star t=e$. From Proposition 2.3 (2), we obtain that x = y and z = t. Therefore $x\ast z=y\star t$, and hence $(x\ast z)\star (y\star t)=e.$ This shows that $\left\{e\right\}$ 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 $\left\{e\right\}$.

Example 2.9

Let $X=\{e,a,b,c\}$ a set and ⋆ is the binary operation on X defined as follows: $\begin{array}{|ccccc|}\hline \star & \mathrm{e}& \mathrm{a}& \mathrm{b}& \mathrm{c}\\ \mathrm{e}& \mathrm{e}& \mathrm{a}& \mathrm{b}& \mathrm{c}\\ \mathrm{a}& \mathrm{a}& \mathrm{e}& \mathrm{b}& \mathrm{c}\\ \mathrm{b}& \mathrm{b}& \mathrm{a}& \mathrm{e}& \mathrm{c}\\ \mathrm{c}& \mathrm{c}& \mathrm{a}& \mathrm{b}& \mathrm{e}\\ \hline\end{array}$ We know that $(X,e,\star )$ is a G-algebra. Let $N_1=\{e,a\}$, $N_2=\{e,a,b\}$ be the unique proper G-sub-algebras of X. In addition, the set $N_1$ is the only proper normal subset of X.

Definition 2.10

If $S_1$ and $S_2$ are two G-algebras of X, we define their product $S_1\star S_2$ by $S_1\star S_2=\{x\in X\mid x\star y\in S_1\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{s}\mathrm{o}\mathrm{m}\mathrm{e}y\in S_2\}.$

Lemma 2.11

As in the previous definition, if $S_1$ is a normal subset and $S_2$ is a G-subalgebra, then $S_1\star S_2$ is a G-subalgebra of X.

Proof.

Assume that $S_1$ is a normal subset and $S_2$ is a G-subalgebra. It is clear that $e\in S_1\star S_2,$ so $S_1\star S_2\ne \mathrm{\varnothing }.$ Let x and $x^{\prime }$ be two elements in $S_1\star S_2$. By definition, there exist y and $y^{\prime }$ in $S_2$ such that $x\star y\in S_1$ and $x^{\prime }\star y^{\prime }\in S_1$. Since $S_1$ is normal, we get (1) $(x\star x^{\prime })\star (y\star y^{\prime })\in S_1.$(1) As $S_2$ is G-subalgebra, $y\star y^{\prime }\in S_2$, and from (1(1) $(x\star x^{\prime })\star (y\star y^{\prime })\in S_1.$(1) ), we obtain $x\star x^{\prime }\in S_1\star S_2$.

3. Lagrange theorem for G-algebras

Let N be a normal subset of a G-algebra $(X,e,\star )$. Consider the relation $\mathcal{R}$ on X defined as follows: $x\mathcal{R}y⟺x\star y\in N.$ As N is normal subset of X, $\mathcal{R}$ is an equivalent relation. For all $x\in X$, denote $[x{]}_N$ by the equivalence class of x under this relation. That means $[x{]}_N=\{z\in X\mid x\star z\in N\}.$ The set X/N of the equivalence classes of this relation form a G-algebra under the following binary operation: $[x{]}_N\star [y{]}_N=[x\star y{]}_N.$

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 $|X/N|=n$ and $[x_i{]}_N$ be the distinct classes. We know that $|X|=\sum _{i=1}^n|[x_i{]}_N|.$ Consider the function $f_x:[x{]}_N⟶N$ defined as $f_x\left(z\right)=x\star z$.

This function is a bijection. In fact $f_x\left(z_1\right)=f_x\left(z_2\right)$ means $x\star z_1=x\star z_2$. Then $z_1=z_2$. In addition for all $y\in N$, $x\star y\in [x{]}_N$. Also $f_x(x\star y)=x\star (x\star y)=y\in N.$ As a consequence, we obtain $|[x_i{]}_n|=|N|$ for all i = 1, 2, …, n. Then $|X|=|X/N|\cdot |N|,$ and this proves the theorem.

Remark 3.3

The set $N_2$ in Example 2.9 is not a normal subset because its order does not divide the order of X. Otherwise the order of $N_1$ divides that of X.

Definition 3.4

Let $(X,\star ,e)$ and $(X^{\prime },{\star }^{\prime },e^{\prime })$ be two G-algebras. A mapping $\phi :X⟶X^{\prime }$ is called a homomorphism of G-algebras, if $\psi (x\star y)=\psi \left(x\right){\star }^{\prime }\psi \left(y\right),$ for all $x,y\in X$.

Example 3.5

If N is a normal subset of a G-algebra X, then the natural mapping ${\pi }_N:X⟶X/N$ defined as ${\pi }_N\left(x\right)=[x{]}_N$ is a homomorphism of G-algebras.

Definition 3.6

Let $\phi :X⟶X^{\prime }$ be a homomorphism of G-algebras. The set $\{x\in X\mid \phi (x)=e^{\prime }\}$ is called the kernel of φ, denoted by $\ker \phi$. for all $x,y\in X$. We define the image of φ by the set $\left\{\phi \right(x)\mid x\in X\}$, and denote by $\mathrm{\Im }\left(\phi \right)$

Theorem 3.7

Let $\psi :X⟶X^{\prime }$ be a homomorphism of G-algebras. Then:

1. $\psi \left(e\right)=e^{\prime }$.

2. $\ker \left(\psi \right)$ is a normal subset of X.

3. If S is a G-subalgebra of X, then its direct image $\psi \left(S\right)$ is a G-subalgebra of $X^{\prime }$.

4. If $N^{\prime }$ is a normal subset (resp. G-subalgebra) of $X^{\prime },$ then its inverse image ${\psi }^{-1}\left(N^{\prime }\right)$ 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 $\psi :X⟶X^{\prime }$ of G-algebra is called:

1. A monomorphism if ψ is one-to-one.

2. An epimorphism if ψ is onto.

3. An isomorphism if ψ is one-to-one and onto.

Lemma 3.9

A homomorphism $\psi :X⟶X^{\prime }$ of G-algebra is a monomorphism if and only if $\ker \left(\psi \right)=\left\{e\right\}$.

Proof.

If x is an element in $\ker \psi$, then $\psi \left(x\right)=e^{\prime }=\psi \left(e\right)$. Hence if ψ is a monomorphism, we obtain directly $\ker \psi =\left\{e\right\}$. Reciprocally, if $\ker \psi =\left\{e\right\}$, then for all x and y in X such that $\psi \left(x\right)=\psi \left(y\right)$, we have $\psi \left(x\right)\star \psi \left(y\right)=e^{\prime }$. Since ψ is a homomorphism, we get $\psi (x\star y)=e^{\prime }$, and hence $x\star y\in \ker \psi =\left\{e\right\}$. So $x\star y=e$. Hence x = y, and this shows that ψ is a monomorphism.

4. Derivation on G-algebras

Definition 4.1

Let $(X,\star ,e)$ be a G-algebra. A G-derivation on X is a mapping $d:X⟶X$ such that $d(x\star y)=d\left(x\right)\star y=x\star d\left(y\right)$ for all $x,y\in X$. We denote ${\mathrm{\Omega }}_X$ be the set of all G-derivation on X.

Remark 4.2

As $x=x\star e$ for all $x\in X$, every G-derivation d can be determined by $d\left(e\right)$ as follows: $d\left(x\right)=x\star d\left(e\right),$ for all $x\in X$.

Lemma 4.3

Let $a\in X$ and $d:X⟶X$ be a function defined as $d\left(x\right)=x\star a,$ for all $x\in X$. If d is a derivation, then $e\star a=a$.

Proof.

By definition of d, we have (2) $d\left(e\right)=e\star a.$(2) If d is a derivation, then (3) $d\left(e\right)=d(e\star e)=e\star d\left(e\right)=e\star (e\star a)=a.$(3) From (2(2) $d\left(e\right)=e\star a.$(2) ) and (3(3) $d\left(e\right)=d(e\star e)=e\star d\left(e\right)=e\star (e\star a)=a.$(3) ), we obtain $e\star a=a$.

Proposition 4.4

Let $(X,\star ,e)$ be a G-algebra.

1. The identity map I is a G-derivation on X.

2. For all $d_1,d_2\in {\mathrm{\Omega }}_X$, $d_1\circ d_2\in {\mathrm{\Omega }}_X$.

3. For all $d_1,d_2\in {\mathrm{\Omega }}_X$, $d_1\circ d_2=d_2\circ d_1$.

4. For all $d\in {\mathrm{\Omega }}_X,$ there exist $d^{-1}\in {\mathrm{\Omega }}_X$ such that $d\circ d^{-1}=d^{-1}\circ d=I$.

Proof.

The first three properties are proved in [8]. To prove (4), we need first to show that every derivation on X is a bijection.

1. Let x and y in X such that $d\left(x\right)=d\left(y\right)$. Then $d\left(x\right)\star d\left(y\right)=e$ and $d\left(y\right)\star d\left(x\right)=e$. That means $d^2\left(x\right)\star y=e$ and $d^2\left(y\right)\star x=e$. Hence, $d^2\left(x\right)=y$ and $d^2\left(y\right)=x$. Therefore $d\left(d\right(x\left)\right)=y$ and $d\left(d\right(y\left)\right)=x$. Since $d\left(x\right)=d\left(y\right)$, we get x = y.

2. If we take in the previous item x = y, we obtain that $x=d\left(d\right(x\left)\right)$ for all $x\in X$

Hence d is a bijection and $d^{-1}=d$.

Theorem 4.5

Let $(X,\star ,e)$ be a G-algebra. Then $({\mathrm{\Omega }}_X,\circ )$ is abelian, and has a structure of G-algebra with the same operation.

Proof.

The previous proposition shows that ${\mathrm{\Omega }}_X$ is an abelian group. We have $d\circ d=I$ for all $d\in {\mathrm{\Omega }}_X$, and by the associativity of the composite operation, we can show that $d_1\circ (d_1\circ d_2)=d_2$ for all $d_1$ and $d_2$ in ${\mathrm{\Omega }}_X$.

Corollary 4.6

Every G-subalgebra of ${\mathrm{\Omega }}_X$ is a normal.

Consider the map ${\omega }_X:{\mathrm{\Omega }}_X⟶X$ defined by: ${\omega }_X\left(d\right)=d\left(e\right).$

Lemma 4.7

Let X be a G-algebra. Then ${\omega }_X$ is a monomorphism of G-algebras.

Proof.

Let $d_1$ and $d_2$ be G-derivations on X. We have $\begin{array}{rl}{\omega }_X(d_1\circ d_2)& =(d_1\circ d_2)\left(e\right)\\ & =(d_1\circ d_2)(e\star e)\\ & =d_1\left(e\right)\star d_2\left(e\right)\\ & ={\omega }_X\left(d_1\right)\star {\omega }_X\left(d_2\right).\end{array}$ Hence, ${\omega }_X$ is a homomorphism of G-algebra. Let $d\in \ker {\omega }_X$. Then $d\left(e\right)=e$, so for all $x\in X$, we have $\begin{array}{rl}d\left(x\right)& =d(x\star e)\\ & =x\star d\left(e\right)\\ & =x\star e\\ & =x.\end{array}$ So d = I and $\ker {\omega }_X=\left\{I\right\}$. Hence, ${\omega }_X$ is one-to-one.

Proposition 4.8

$\omega \left({\mathrm{\Omega }}_X\right)$ is a normal subset of X.

Proof.

Let $x\star y$ and $z\star t$ be two elements in $\omega \left({\mathrm{\Omega }}_X\right)$. Then there exist G-derivations $d_1$ and $d_2$ on X such that $x\star y=d_1\left(e\right)$ and $z\star t=d_2\left(e\right)$. Since $d_i^2=I$ for i = 1, 2, $d_1(x\star y)=e$ and $d_2(z\star t)=e$. This implies that $d_1\left(x\right)\star y=e$ and $d_2\left(z\right)\star t=e$. So $d_1\left(x\right)=y$ and $d_2\left(z\right)=t.$ Therefore, (4) $d_1\left(x\right)\star d_2\left(z\right)=y\star t.$(4) Consider the G-derivation $d=d_1\circ d_2$. Then from (4(4) $d_1\left(x\right)\star d_2\left(z\right)=y\star t.$(4) ) $d(x\star z)=y\star t.$ So $d(x\star z)\star (y\star t)=e.$ This shows that the derivation d satisfies the equality $d\left(\right(x\star z)\star (y\star t\left)\right)=e.$ Since $d^2=I$, we deduce from the previous equality that (5) $(x\star z)\star (y\star t)=d\left(e\right).$(5) Finally, we conclude that $(x\star z)\star (y\star t)\in \omega \left({\mathrm{\Omega }}_X\right)$ for all $x\star y$ and $z\star t$ in $\omega \left({\mathrm{\Omega }}_X\right)$. Hence, the set $\omega \left({\mathrm{\Omega }}_X\right)$ is normal in X.

Corollary 4.9

Let X be a finite G-algebra. Then the order of ${\mathrm{\Omega }}_X$ divides the order of X.

Proof.

Since ${\omega }_X\left({\mathrm{\Omega }}_X\right)$ is a normal subset of X, we get by the Lagrange theorem, $|{\omega }_X\left({\mathrm{\Omega }}_X\right)|$ divides $|X|$. Apply the first isomorphism theorem to the monomorphism ${\omega }_X$, we obtain ${\mathrm{\Omega }}_X\backsimeq {\omega }_X\left({\mathrm{\Omega }}_X\right).$ Hence, the order of ${\mathrm{\Omega }}_X$ equals the order of ${\omega }_X\left({\mathrm{\Omega }}_X\right).$ Consequently, the order of ${\mathrm{\Omega }}_X$ divides the order of X.

Example 4.10

Consider the G-algebra $X=\{e,a,b,c\}$ with the binary operation ⋆ defined as follows: $\begin{array}{|lcccc|}\hline \star & \mathrm{e}& \mathrm{a}& \mathrm{b}& \mathrm{c}\\ \mathrm{e}& \mathrm{e}& \mathrm{c}& \mathrm{b}& \mathrm{a}\\ \mathrm{a}& \mathrm{a}& \mathrm{e}& \mathrm{c}& \mathrm{b}\\ \mathrm{b}& \mathrm{b}& \mathrm{a}& \mathrm{e}& \mathrm{c}\\ \mathrm{c}& \mathrm{c}& \mathrm{b}& \mathrm{a}& \mathrm{e}\\ \hline\end{array}$

From Remark 4.2 and Lemma 4.3, we know that every derivation on X can be written as $d\left(x\right)=x\star z$ for some $z\in X$ satisfying $e\star z=z$. 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 $d\left(x\right)=x\star b$ is already a G-derivation.

Finally, ${\mathrm{\Omega }}_X=\{d_1,d_2\}$ and ${\omega }_X\left({\mathrm{\Omega }}_X\right)=\{e,b\}$. In this example we remark that the order of ${\mathrm{\Omega }}_X$ divides the order of X and ${\omega }_X\left({\mathrm{\Omega }}_X\right)=\{e,b\}$ is a normal subset of X.

Example 4.11

Consider the G-algebra $X=\{e,a,b,c,d\}$ presented as follows: $\begin{array}{|lccccc|}\hline \star & \mathrm{e}& \mathrm{a}& \mathrm{b}& \mathrm{c}& \mathrm{d}\\ \mathrm{e}& \mathrm{e}& \mathrm{a}& \mathrm{b}& \mathrm{c}& \mathrm{d}\\ \mathrm{a}& \mathrm{a}& \mathrm{e}& \mathrm{b}& \mathrm{c}& \mathrm{d}\\ \mathrm{b}& \mathrm{b}& \mathrm{a}& \mathrm{e}& \mathrm{c}& \mathrm{d}\\ \mathrm{c}& \mathrm{c}& \mathrm{a}& \mathrm{b}& \mathrm{e}& \mathrm{d}\\ \mathrm{d}& \mathrm{d}& \mathrm{a}& \mathrm{b}& \mathrm{c}& \mathrm{e}\\ \hline\end{array}$

We know that every derivation d on X is determined by the value of d at e. Consider the five functions $d_i:X⟶X$ for i = 1, 2, 3, 4, 5 defined as follows: $d_1\left(x\right)=x\star e=x,$ $d_2\left(x\right)=x\star a,$ $d_3\left(x\right)=x\star b,$ $d_4\left(x\right)=x\star c$ and $d_5\left(x\right)=x\star d.$

Clearly $d_1=I$ is a G-derivation, but $d_2$ is not a G-derivation, since $d_2(b\star c)=d_2\left(c\right)\ne d_2\left(b\right)\star c$.

Applying Corollary 4.9, we obtain that the order of ${\mathrm{\Omega }}_X$ equals 1 or 5. Since $d_2$ is not a derivation, we have necessarily $|{\mathrm{\Omega }}_X|=1$.

Proposition 4.12

Let p be a prime number and $X_n=\{0,1,2,\dots ,p\}$ the G-algebra with the operation defined as follows: $x\star y=e$ if $x=y,$ and otherwise: $x\star y=\left\{\begin{array}{l}x\mathrm{i}\mathrm{f}y=e,\\ y\mathrm{i}\mathrm{f}y\ne e.\end{array}\right.$ Then X has only the identity derivation, more precisely ${\mathrm{\Omega }}_{X_p}=\left\{I\right\}.$

Proposition 4.8 can be generalized as follows:

Proposition 4.13

Let S be a subset of ${\mathrm{\Omega }}_X$ closed under composition. Then ${\omega }_X\left(S\right)$ 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 ${\mathrm{\Omega }}_X=\left\{I\right\}$

Proof.

Assume the contrary. That means $|{\mathrm{\Omega }}_X|>1$. Then there exists a G-derivation $d\ne I$. As $d^2=I$ and $d\circ I=I\circ d=d$, the subset $S=\{I,d\}$ is normal in ${\mathrm{\Omega }}_X$. Form Proposition 4.13, the subset $\{e,d(e\left)\right\}$ is normal in X. The fact $d\ne e$ implies that $d\left(e\right)\ne e$. Using Theorem 3.2, we obtain the result.

Corollary 4.15

Let X be a finite G-algebra. Then $|X|=\left\{\begin{array}{l}oddnumberif|{\mathrm{\Omega }}_X|=1,\\ evennumberif|{\mathrm{\Omega }}_X|\ge 2.\end{array}\right.$

The following example shows that if X is an infinite G-algebra, then ${\mathrm{\Omega }}_{X\left(F\right)}$ can be finite.

Example 4.16

Let F be a field and $X=X\left(F\right)$. Take an arbitrary derivation d on X. Then $d\left(x\right)=x\star a=x{a}^{-1}$ for some $a\in X$.

For all x, y in X, $d(x\star y)=x\star d\left(y\right)$. That means $\left(x{y}^{-1}\right)a^{-1}=x(y{a}^{-1}{)}^{-1}=x(y^{-1}a).$ So d is a derivation if and only if $a=a^{-1}$. As F is a field, this equality holds only in the case when a = 1 or a = −1. Then ${\mathrm{\Omega }}_{X\left(F\right)}=\{I,-I\}\mathrm{a}\mathrm{n}\mathrm{d}\omega \left({\mathrm{\Omega }}_{X\left(F\right)}\right)=\{-1,1\}.$

Theorem 4.17

Let X be a G-algebra. Then the following statements are equivalent:

1. The function $d_a\left(x\right)=x\star a$ is a derivation for all $a\in X$.

2. The operation ⋆ is commutative and associative on X (i.e. X is an abelian group).

Proof.

Assume that the function $d_a\left(x\right)=x\star a$ is a derivation for all $a\in X$. Applying Proposition 2.3 and Lemma 4.3 we obtain (6) $a\star e=e\star a=a\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{a}\mathrm{l}\mathrm{l}a\in X.$(6) Let b, c be elements in X. As $d_c$ is a derivation, $d_c(a\star b)=a\star d_c\left(b\right);$ therefore, (7) $(a\star b)\star c=a\star (b\star c)\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{a}\mathrm{l}\mathrm{l}a,b,c\in X.$(7) In addition, the fact $d_a$ is a derivation, $d_a(b\star c)=d_a\left(b\right)\star c$. Then (8) $(b\star c)\star a=(b\star a)\star c.$(8) Replacing b by e in the last equality, we obtain that (9) $(e\star c)\star a=(e\star a)\star c.$(9) Using (6(6) $a\star e=e\star a=a\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{a}\mathrm{l}\mathrm{l}a\in X.$(6) ), we get from (9(9) $(e\star c)\star a=(e\star a)\star c.$(9) ): (10) $c\star a=a\star c\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{a}\mathrm{l}\mathrm{l}a,c\in X.$(10) The properties (6(6) $a\star e=e\star a=a\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{a}\mathrm{l}\mathrm{l}a\in X.$(6) ), (7(7) $(a\star b)\star c=a\star (b\star c)\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{a}\mathrm{l}\mathrm{l}a,b,c\in X.$(7) ) and (10(10) $c\star a=a\star c\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{a}\mathrm{l}\mathrm{l}a,c\in X.$(10) ) show that X is an abelian group. Reciprocally, suppose that X is an abelian group. We will show that $d_a$ is a derivation for all $a\in X$. Let x, y be elements in X. As the operation ⋆ is commutative and associative, $\begin{array}{rl}{d}_a(x\star y)& =(x\star y)\star a=x\star (y\star a)\\ & =(x\star a)\star y=x\star d_a\left(y\right).\end{array}$ So $d_a(x\star y)=d_a\left(x\right)\star y=x\star d_a\left(y\right).$

Corollary 4.18

Let X be a finite G-algebra. If there is an element a such that $d_a$ is not a derivation on X (i.e. G is not a group), then $|{\mathrm{\Omega }}_X|<|X|$.

Corollary 4.19

Let X be a G-algebra of prime order. If there is an element a such that $d_a$ is not a derivation on X, then ${\mathrm{\Omega }}_X=\left\{I\right\}$.

Corollary 4.20

Let N be a normal subset of X of order 2. Then ${\mathrm{\Omega }}_X=\{e,d(e\left)\right\}$ for some derivation d.

Let $a\ne e$ be the second element of N. $d\left(x\right)=x\star a.$ Then $d\left(a\right)=e$ and $d\left(e\right)=e\star a=a.$

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.