Splitting Extensions of Nonassociative Algebras and Modules with Metagroup Relations

Abstract

A class of nonassociative algebras is investigated with mild relations induced from metagroup structures. Modules over nonassociative algebras are studied. For a class of modules over nonassociative algebras, their extensions and splitting extensions are scrutinized. For this purpose tensor products of modules and induced modules over nonassociative algebras are investigated. Moreover, a developed cohomology theory on them is used.

1. Introduction

Extensions and splitting extensions of associative algebras play a very important role and have found many-sided applications (see, for example, refs. [1,2,3,4,5] and references therein). Studies of their structure are grounded on cohomology theory. In particular, great attention is paid to algebras with group identities. On the other hand, cohomology theory of associative algebras was investigated by Hochschild and other authors [6,7,8], but it is not applicable to nonassociative algebras. Cohomology theory of group algebras is an important and great part of algebraic topology. Nonassociative algebras with some identities in them, such as Cayley–Dickson algebras and their generalizations are wide spread not only in algebra but also in many-sided applications in physics, noncommutative geometry, quantum field theory, PDEs, and other sciences (see [9,10,11,12,13,14,15,16,17,18] and references therein). For example, the Klein–Gordon hyperbolic PDE of the second order with constant coefficients was solved by Dirac with the help of complexified quaternions [19]. Cayley–Dickson algebras were used for decompositions of higher-order PDEs into lower-order PDEs, which subsequently permitted integrating and analyzing them [15]. PDEs or their systems frequently possess groups of their symmetries [8]. Group algebras appearing over $\mathbf{C}$ in conjunction with Cayley–Dickson algebras lead to extensions that are generalized Cayley–Dickson algebras or even more general metagroup algebras. In their turn operator algebras over Cayley–Dickson algebras also induce the metagroup algebras. Besides algebras over $\mathbf{R}$ or $\mathbf{C}$, there are such algebras over other fields, that is important in non-archimedean quantum mechanics and quantum field theory. Analysis of PDEs and operators over Cayley–Dickson algebras provide generalized Cayley–Dickson algebras or metagroup algebras acting on modules of functions.

It is worth mentioning that in the 20th century it was demonstrated that a noncommutative geometry exists if there exists a corresponding quasi-group [20,21,22]. On the other hand, metagroups are quasigroups with some weak relations.

An extensive area of investigations of PDEs intersects with cohomologies and deformed cohomologies [8]. Therefore, it is important to develop this area over octonions, Cayley–Dickson algebras, and more general nonassociative algebras with metagroup relations. Examples of nonassociative algebras, modules, and homological complexes with metagroup relations are given in [12,13,23]. In these works, it is shown that generalized Cayley–Dickson algebras are particular cases of metagroup algebras. For related with them digital Hopf spaces cohomologies were investigated in [24].

In [14], smashed and twisted wreath products of metagroups or groups were studied. It permitted constructing wide families of metagroups even starting from groups. It also demonstrates that meta groups appear naturally in algebra and metagroup algebras compose an enormous class of nonassociative algebras.

This article is devoted to extensions and splitting extensions of nonassociative algebras. For this purpose cohomology theory on them was developed in [12]. In its turn, it demanded to impose some mild conditions on algebras to develop cohomology theory on them.

A class of nonassociative algebras is investigated with mild relations induced from metagroup structures. Modules over nonassociative metagroup algebras are studied. For a class of modules over nonassociative algebras their extensions, splitting extensions, and ideals are scrutinized in Section 3.

For this purpose tensor products of modules and induced modules over nonassociative metagroup algebras are investigated in Section 2. Smashly graded modules and algebras are studied. The smash gradation is considered over metagroups. The considered tensor products take into account smash nonassociative structures (see Theorem 1, Proposition 1, Remark 1). Moreover, a developed cohomology theory on them is used. Algebras satisfying these conditions are described (see Theorem 2). In Propositions 2 and 3 and Theorems 3 and 5, tensor products of morphisms are studied. The exactness of morphisms is investigated in Theorems 4 and 7 and Corollary 3. Generic morphisms are scrutinized in Theorem 6. Isomorphisms of such nonassociative algebras are studied in Proposition 4 and Corollary 6. Their meta-isomorphisms are investigated in Proposition 5 and Theorem 8.

The exactness of isomorphisms related with extensions and metagroup algebras is studied in Lemma 6 and Propositions 6 and 7 (see also Definitions 1 and 6). Extensions of smashly graded modules and algebras are investigated in Theorems 9 and 10 and Proposition 8.

Necessary definitions are provided in Appendix A (see also Formulas (A1)–(A19)).

All main results of this paper are obtained for the first time. They can be used for further studies of nonassociative algebras, their cohomologies, algebraic geometry, PDEs, their applications in the sciences, etc.

2. Modules over Nonassociative Algebras with Metagroup Relations

Definition 1.

Let $\mathcal{T}$ be a commutative associative unital ring, G be a metagroup and $A=\mathcal{T}\left[G\right]$ be a metagroup algebra. Let also B be a G-graded unital right A-module (see Definitions A2 and A3). Suppose in addition that

$\left(1\right)$ there exists a mapping $B\times B\ni (x,y)\mapsto xy\in B$ such that

$x(y+z)=xy+xz$ and $(y+z)x=yx+zx$ and $\left(bx\right)y=b\left(xy\right)$ and $\left(xb\right)y=x\left(by\right)$ and $\left(xy\right)b=x\left(yb\right)$

for all x, y, z in B, $b\in \mathcal{T}$;

$\left(2\right)$$\left(x_g{y}_h\right)z_s={\mathsf{t}}_3(g,h,s)x_g\left(y_h{z}_s\right)$ and $x_g{y}_h\in B_{gh}$

for every g, h, s in G, $x_g\in B_g$, $y_h\in B_h$, $z_s\in B_s$.

Then we call B a smashly G-graded algebra over A (or a smashly G-graded A-algebra). For short it will be written “an A-algebra” instead of “a smashly G-graded A-algebra”. The algebra B is called unital if and only if

$\left(3\right)$ B has a unit element $1=1_B$ such that $1_{B}x=x$ and $x{1}_B=x$ for each $x\in B$.

Suppose that X is a $G_1$-graded left $A_1$-module and Y is a $G_2$-graded left $A_2$-module, $A_j=\mathcal{T}\left[G_j\right]$ is a metagroup algebra for each $j\in \{1,2,3,\cdots \}$. Suppose also that $f:X\to Y$ is a map such that f is

$\left(4\right)$ a left $\mathcal{T}$-homomorphism and $f\left(X_g\right)\subseteq Y_{f_1\left(g\right)}$ for each $g\in G_1$, where $f_1:G_1\to G_2$ is a homomorphism of metagroups:

$\left(5\right)$$f_1\left(gh\right)=f_1\left(g\right)f_1\left(h\right)$ and $f_1(g\backslash h)=f_1\left(g\right)\backslash f_1\left(h\right)$ and

$f_1(g/h)=f_1\left(g\right)/f_1\left(h\right)$ for every g and h in G.

The map $f:X\to Y$ satisfying conditions $\left(4\right)$ and $\left(5\right)$ will be called a $(G_1,G_2)$-graded left $\mathcal{T}$-homomorphism of the left modules X and Y. If $\mathcal{T}$ is specified, it may be said a homomorphism instead of a $\mathcal{T}$-homomorphism. Symmetrically is defined a $(G_3,G_4)$-graded right homomorphism of a right $A_3$-module X and a right $A_4$-module Y. For a $(A_1,A_3)$-bimodule X and a $(A_2,A_4)$-bimodule Y if a map $f:X\to Y$ is $(G_1,G_2)$-graded left and $(G_3,G_4)$-graded right homomorphism, then f will be called a $((G_1,G_2),(G_3,G_4))$-graded homomorphism of bimodules X and Y.

Assume that X is a left $A_1$-module and Y is a left $A_2$-module and $f:X\to Y$ is a map, where $A_j=\mathcal{T}\left[G_j\right]$ for each j, such that

$\left(6\right)$$f:X\to Y$ is a left $\mathcal{T}$ homomorphism and $f(ax+by)=f_1\left(a\right)f\left(x\right)+f_1\left(b\right)f\left(y\right)$, where

$\left(7\right)$$f_1$ is a homomorphism from $A_1$ into $A_2$, that is

$f_1:A_1\to A_2$ and $f_1:G_1\to G_2$ and $f_1:\mathcal{T}\left[N\left(G_1\right)\right]\to \mathcal{T}\left[N\left(G_2\right)\right]$ and

$f_1\left(ab\right)=f_1\left(a\right)f_1\left(b\right)$ and $f_1(g\backslash h)=f_1\left(g\right)\backslash f_1\left(h\right)$ and

$f_1(g/h)=f_1\left(g\right)/f_1\left(h\right)$ and $f_1(pa+bs)=p{f}_1\left(a\right)+f_1\left(b\right)s$

for every g and h in $G_1$, $a$ and b in $A_1$, $p$ and s in $\mathcal{T}$, where $G_j$ is considered embedded into $\mathcal{T}\left[G_j\right]$ as $G_j{1}_{\mathcal{T}}$ identifying $g{1}_{\mathcal{T}}$ with g for simplicity of the notation, where $1_{\mathcal{T}}$ is the unit element of the ring $\mathcal{T}$, $g\in G_j$.

If f satisfies conditions $\left(6\right)$ and $\left(7\right)$, then f will be called an $(A_1,A_2)$-generic left homomorphism of left modules X and Y. For right modules an $(A_1,A_2)$-generic right homomorphism is defined analogously. If X is an $(A_1,A_3)$-bimodule and Y is an $(A_2,A_4)$-bimodule and f is an $(A_1,A_2)$-generic left and $(A_3,A_4)$-generic right homomorphism, then f will be called an $((A_1,A_2),(A_3,A_4))$-generic homomorphism of bimodules X and Y.

If additionally the homomorphism $f_1$ is bijective and surjective in $\left(7\right)$ and $f_1^{-1}:A_2\to A_1$ is the homomorphism, then $f_1$ is called an isomorphism of $A_1$ with $A_2$ (or automorphism if $A_1=A_2$).

In particular, if $G_1=G_2$, then “$(G_1,G_1)$-graded” or “$(A_1,A_1)$-generic” will be shortened to “$G_1$-graded” or “$A_1$-generic” correspondingly, etc. If $f_1$ is an automorphism of $G_1$ (or of $A_1$ correspondingly), then a $G_1$-graded (or an $A_1$-generic) left homomorphism from X into Y will be called $G_1$-exact (or $A_1$-exact correspondingly). Similarly $G_3$-exact or $A_3$-exact right homomorphisms of right modules and $(G_1,G_3)$-exact or $(A_1,A_3)$-exact homomorphisms of bimodules are defined.

If X and Y are A-algebras and f is a G-graded (or G-exact or A-generic or A-exact) homomorphism from X into Y considered as A-bimodules and in addition the following condition is satisfied

$\left(8\right)$$f\left(vx\right)=f\left(v\right)f\left(x\right)$ for each x and v in X,

then f will be called a G-graded (or G-exact or A-generic or A-exact correspondingly) homomorphism of the A-algebras.

We consider the Cartesian product $X\times Y$ of A-bimodules X and Y. Let $X{\times }_{A}Y$ be an A-bimodule generated from $X\times Y$ using finite additions of elements $(x,y)\in X\times Y$ and the left and right multiplications on elements $a\in A$ such that

$\left(9\right)$$(x,y)+(x_1,y_1)=(x+x_1,y+y_1)$ and

$\left(10\right)$$a(x,y)=(ax,ay)$ and $(x,y)a=(xa,ya)$ and

$\left(11\right)$$g(X_e,Y_e)=(X_g,Y_g)$ and $(X_e,Y_e)g=(X_g,Y_g)$ (see also $\left(A16\right)$) for each x and $x_1$ in X, $y$, and $y_1$ in Y, $a\in A$, $g\in G$.

Suppose that X, Y and Z are A-bimodules.

$\left(12\right)$ Let $\Lambda :X\times Y\to Z$ be a $\mathcal{T}\left[\mathcal{C}\right(G\left)\right]$-bilinear map. Let also Λ satisfy the following identities:

$\left(13\right)$$\Lambda (x_g{b}_h,y_s)={\mathsf{t}}_3(g,h,s)\Lambda (x_g,b_h{y}_s)$ and $\Lambda (cx,y)=c\Lambda (x,y)$ and

$\Lambda (x,yc)=\Lambda (x,y)c$ for each $c\in N\left(G\right)$, $x\in X$, $y\in Y$, g and h and s in G. If Λ fulfills conditions $\left(12\right)$ and $\left(13\right)$, then it will be said that the map Λ is G-balanced.

Definition 2.

Let X and Y be A-bimodules (see Definitions A3 and 1), where $A=\mathcal{T}\left[G\right]$ is the metagroup algebra. Let C be an A-bimodule supplied with a $\mathcal{T}\left[\mathcal{C}\right(G\left)\right]$-bilinear map $\xi :X\times Y\to C$ denoted by $\xi (x,y)=x\otimes y$ for each $x\in X$ and $y\in Y$ such that

$\left(14\right)$ C is generated by a set $\{x\otimes y:x\in X,y\in Y\}$ and

$\left(15\right)$ if $\Lambda :X\times Y\to Z$ is a G-balanced map of A-bimodules X, Y, and Z, and for each fixed $x\in X$ the map $\Lambda (x,·):Y\to Z$ and for each fixed $y\in Y$ the map $\Lambda (·,y):X\to Z$ are G-graded homomorphisms of A-bimodules, then there exists a G-graded homomorphism $\psi :C\to Z$ of A-bimodules such that $\psi (x\otimes y)=\Lambda (x,y)$ for each $x\in X$ and $y\in Y$.

If conditions $\left(14\right)$ and $\left(15\right)$ are satisfied, then the A-bimodule C is called a tensor product of X with Y over A and denoted by $X{\otimes }_{A}Y$.

Remark 1.

Definition 2 implies the following identities in $X{\otimes }_{A}Y$:

$\left(16\right)$$(cx+d{x}_1)\otimes y=c(x\otimes y)+d(x_1\otimes y)$;

$\left(17\right)$$x\otimes (yc+y_{1}d)=(x\otimes y)c+(x\otimes y_1)d$;

$\left(18\right)$$\left(x_g{b}_h\right)\otimes y_s={\mathsf{t}}_3(g,h,s)x_g\otimes \left(b_h{y}_s\right)$;

$\left(19\right)$$\left(b_h{x}_g\right)\otimes y_s={\mathsf{t}}_3(h,g,s)b_h(x_g\otimes y_s)$;

$\left(20\right)$$(x_g\otimes y_s)b_h={\mathsf{t}}_3(g,s,h)x_g\otimes \left(y_s{b}_h\right)$;

$\left(21\right)$$\gamma (x\otimes y)=(x\otimes y)\gamma$;

$\left(22\right)$$x\otimes 0=0$ and $0\otimes y=0$

for each x and $x_1$ in X; $y$ and $y_1$ in Y; c and d in $\mathcal{T}\left[N\right(G\left)\right]$, $g$ and h and s in G, $b_h\in A_h$, $x_g\in X_g$, $y_s\in Y_s$, $\gamma \in \mathcal{T}\left[\mathcal{C}\right(G\left)\right]$.

If $p:G\to H$ is a homomorphism of metagroups, then $p\left(G\right)$ is a submetagroup in H such that $p\left(N\right(G\left)\right)\subseteq N\left(p\right(G\left)\right)$ and $p\left(Com\right(G\left)\right)\subseteq Com\left(p\right(G\left)\right)$, hence $p\left(\mathsf{C}\right(G\left)\right)\subseteq \mathsf{C}\left(p\right(G\left)\right)$.

We remind the reader that a submetagroup Q in G is called normal if it satisfies the following conditions: $gQ=Qg$, $\left(hg\right)Q=h\left(gQ\right)$, $\left(hQ\right)g=h\left(Qg\right)$ and $Q\left(hg\right)=\left(Qh\right)g$ for each g and h in G. There exists a quotient metagroup $G/$·$/Q$ consisting of classes $gQ$ in G with $g\in G$, since

$\left[\right(\left(gQ\right)\left(hQ\right)\left)\right(sQ\left)\right]/\left[\right(gQ\left)\right(\left(hQ\right)\left(sQ\right)\left)\right]=\left(\right[\left(gh\right)s]/[g\left(hs\right)\left]\right)Q\in \mathsf{C}(G/$·$/Q)$

for each g, h, and s in G. Moreover, there exists a quotient homomorphism ${\pi }_{G,Q}$ from G onto $G/$·$/Q$. In particular, $p^{-1}\left(e_H\right)=:Ker\left(p\right)$ is a normal submetagroup in G, where $e_H$ denotes the unit element in H, such that the image $p\left(G\right)$ is isomorphic with $G/$·$/Ker\left(p\right)$.

As usually $i{d}_X$ denotes the identity homomorphism on X, $i{d}_X\left(x\right)=x$ for each $x\in X$. Note that there are natural embeddings of G and $\mathcal{T}$ in $A=\mathcal{T}\left[G\right]$ as $G1$ and $e\mathcal{T}$ correspondingly, where $1_{\mathcal{T}}=1$ is a unit of the ring $\mathcal{T}$ and $e=e_G$ is a unit of the metagroup G. In particular, there may be a case when G and $\mathcal{T}$ are contained in a nonassociative ring $\mathcal{R}$ such that $G\cap \mathcal{T}\ne \varnothing$. This may induce algebraic identities in the metagroup algebra.

Lemma 1.

Let X, Y, Z, and A be as in Definition 2. Let also ϕ and θ be G-graded homomorphisms from $X{\otimes }_{A}Y$ into Z such that $\varphi (x\otimes y)=\theta (x\otimes y)$ for each $x\in X$ and $y\in Y$. Then ϕ and θ coincide on $X{\otimes }_{A}Y$.

Proof. 

Certainly $ker(\varphi -\theta )$ is a submodule in $X{\otimes }_{A}Y$. From conditions $\left(14\right)$ and $\left(15\right)$ it follows that $\varphi =\theta$. □

Corollary 1.

The homomorphism ξ is uniquely defined by condition $\left(15\right)$.

Corollary 2.

If $C=X{\otimes }_{A}Y$ and $C^{\prime }=X{{\otimes }^{\prime }}_{A}Y$ are two tensor products of X and Y over A (see Definition 2), then there exists a unique G-graded isomorphism $\varphi :C\to C^{\prime }$ such that

$\left(23\right)$$\varphi (x\otimes y)=x{\otimes }^{\prime }y$ for each $x\in X$ and $y\in Y$, where $x\otimes y\in C$ and $x{\otimes }^{\prime }y\in C^{\prime }$.

Proof. 

An existence of $\varphi$ satisfying condition $\left(23\right)$ follows from identities $\left(16\right)$–$\left(22\right)$. Similarly, there exists a G-graded homomorphism $\eta :C^{\prime }\to C$ such that $\eta \left(x{\otimes }^{\prime }y\right)=x\otimes y$ for each $x\in X$ and $y\in Y$. In view of Lemma 1 $\varphi \eta =id$ and $\eta \varphi =id$, where $id$ denotes the identity map. □

Theorem 1.

Assume that X and Y are A-bimodules, where $A=\mathcal{T}\left[G\right]$ is a metagroup algebra. Then the tensor product $X{\otimes }_{A}Y$ exists and it is an A-bimodule.

Proof. 

We consider an A-bisubmodule H in the A-bimodule $K:=X{\times }_{A}Y$ (see Definition 1). Let H be generated by elements of the following types:

$\left(24\right)$$(cx+d{x}_1,y)-c(x,y)-d(x_1,y)$;

$\left(25\right)$$(x,yc+y_{1}d)-(x,y)c-(x,y_1)d$;

$\left(26\right)$$(x_g{b}_h,y_s)-({\mathsf{t}}_3(g,h,s)x_g,b_h{y}_s)$;

$\left(27\right)$$(b_h{x}_g,y_s)-{\mathsf{t}}_3(h,g,s)b_h(x_g,y_s)$;

$\left(28\right)$$(x_g,y_s)b_h-{\mathsf{t}}_3(g,s,h)(x_g,y_s{b}_h)$

$\left(29\right)$$\gamma (x,y)-(x,y)\gamma$;

$\left(30\right)$$(x,0)$ and $(0,y)$

for each x and $x_1$ in X; $y$ and $y_1$ in Y; c and d in $\mathcal{T}\left[N\right(G\left)\right]$, $g$ and h and s in G, $b_h\in A_h$, $x_g\in X_g$, $y_s\in Y_s$, $\gamma \in \mathcal{T}\left[\mathcal{C}\right(G\left)\right]$.

There exists the quotient module $Q:=K/$·$/H$ of K by H, since K and H have a structure of commutative groups relative to the addition. One can put $x\otimes y=(x,y)+H$. Note that all pairs $x\otimes y$ generate the quotient module Q, because all pairs $x\times y$ generate K. Let $\Lambda :X\times Y\to Z$ be a G-balanced map of A-bimodules X, Y and Z such that for each fixed $x\in X$ the map $\Lambda (x,·):Y\to Z$ and for each fixed $y\in Y$ the map $\Lambda (·,y):X\to Z$ are G-graded homomorphisms (see Definition 1). Evidently there exists a $\mathcal{T}\left[\mathcal{C}\right(G\left)\right]$-bilinear extension $\eta$ of $\Lambda$ on K, $\eta :K\to Z$. From the properties of the map $\Lambda$ it follows that all elements of the types $\left(24\right)$–$\left(30\right)$ belong to the kernel $ker\left(\eta \right)$ of $\eta$. This induces a homomorphism $\psi :Q\to Z$ such that $\psi \left(\right(x\otimes y\left)\right)=\psi \left(\right(x,y)+H)=\eta \left(\right(x,y\left)\right)=\Lambda (x,y)$ for each $x\in X$ and $y\in Y$. □

Proposition 1.

Let $X_1$, $X_2$, $Y_1$, and $Y_2$ be A-bimodules, where $A=\mathcal{T}\left[G\right]$ is a metagroup algebra. Let $\varphi :X_1\to X_2$ and $\psi :Y_1\to Y_2$ be their G-graded homomorphisms such that ${\varphi }_1\left(g\right)={\psi }_1\left({\theta }_1\left(g\right)\right)$ for each $g\in G$, where ${\theta }_1$ is an automorphism of G. Then there exists a unique G-graded homomorphism

$\left(31\right)$$\varphi \otimes \psi :X_1{\otimes }_A{Y}_1\to X_2{\otimes }_A{Y}_2$ such that $(\varphi \otimes \psi )(x\otimes y)=\varphi \left(x\right)\otimes \psi \left(y\right)$

for each $x\in X_1$ and $y\in Y_1$. Moreover, the following identities are accomplished:

$\left(32\right)$$\varphi \otimes (\psi a+\xi b)=(\varphi \otimes \psi )a+(\varphi \otimes \xi )b$,

$\left(33\right)$$(a\varphi +b\nu )\otimes \psi =a(\varphi \otimes \psi )+b(\nu \otimes \psi )$

for each a and b in $\mathcal{T}\left[N\right(G\left)\right]$, G-graded homomorphisms $\nu :X_1\to X_2$, $\xi :Y_1\to Y_2$ such that ${\varphi }_1\left(g\right)={\xi }_1\left({\theta }_1\left(g\right)\right)$ and ${\nu }_1\left(g\right)={\psi }_1\left({\theta }_1\left(g\right)\right)$ for each $g\in G$;

$\left(34\right)$$\left(\varphi a\right)\otimes \psi =\varphi \otimes \left(a\psi \right)$, for each a in $\mathcal{T}\left[N\right(G\left)\right]$;

$\left(35\right)$$\left(\gamma \varphi \right)\otimes \psi =\varphi \otimes \left(\psi \gamma \right)$ for each $\gamma \in \mathcal{T}\left[\mathcal{C}\right(G\left)\right]$;

$\left(36\right)$$i{d}_{X_1}\otimes i{d}_{Y_1}=i{d}_{X_1{\otimes }_A{Y}_1}$;

$\left(37\right)$$(\mu \otimes \eta )(\varphi \otimes \psi )=\left(\mu \varphi \right)\otimes \left(\eta \psi \right)$

for each G-graded homomorphisms $\mu :X_2\to X_3$, $\eta :Y_2\to Y_3$ of A-bimodules with ${\mu }_1\left(g\right)={\eta }_1\left({\theta }_2\left(g\right)\right)$ for each $g\in G$, where ${\theta }_2$ is an automorphism of G;

$\left(38\right)$$\varphi \otimes 0=0$ and $0\otimes \psi =0$.

Proof. 

The homomorphisms $\varphi$ and $\psi$ are G-graded. Therefore,

$\left(39\right)$$(\varphi \otimes \psi )(x_g\otimes y_h)\in X_{1,{\varphi }_1\left(g\right)}{\otimes }_{\mathcal{T}}{Y}_{1,{\psi }_1\left(h\right)}$

for each g and h in G, $x_g\in X_{1,g}$, $y_h\in Y_{1,h}$. On the other hand, by the conditions of this proposition ${\varphi }_1$ and ${\psi }_1$ are homomorphisms from G into G such that ${\varphi }_1\left(g\right)={\psi }_1\left({\theta }_1\left(g\right)\right)$ for each $g\in G$. Therefore, from $\left(A16\right)$ it follows that

$\left(40\right)$$X_{1,{\varphi }_1\left(g\right)}{\otimes }_{\mathcal{T}}{Y}_{1,{\psi }_1\left(h\right)}$ is isomorphic as the $\mathcal{T}$-bimodule with

$X_{1,s}{\otimes }_{\mathcal{T}}{Y}_{1,v}$ for each s in G and $v=s\backslash {\psi }_1\left({\theta }_1\left(g\right)h\right)$.

Thus the properties $\left(39\right)$ and $\left(40\right)$ and Definition 2 of $X_1{\otimes }_A{Y}_1$ imply that there exists a unique G-graded homomorphism $(\varphi \otimes \psi )$. Identities $\left(32\right)$–$\left(37\right)$ follow from Lemma 1, Theorem 1 and $\left(16\right)$–$\left(22\right)$. □

Theorem 2.

Assume that X and Y are A-bimodules, where $A=\mathcal{T}\left[G\right]$ is a metagroup algebra. Assume also that P is a quotient metagroup $P:=G/$·$/S$ of a metagroup G by a subgroup S such that $S\subseteq \mathcal{C}\left(G\right)$ and $X_g\ne X_e$ and $Y_g\ne Y_e$ for each $g\in G-S$. Then there exist $\mathcal{T}\left[S\right]$-bimodules $X_g^{\sigma }$ such that X considered as a $\mathcal{T}\left[S\right]$-bimodule is isomorphic with the direct sum ${⨁}_{g\in P}{X}_g^{\sigma }$ and ${\left(X{\otimes }_{A}Y\right)}_r^{\sigma }$ is isomorphic with ${⨁}_{h\in P,s\in P,hs=r}\left(X_h^{\sigma }{\otimes }_{\mathcal{T}\left[S\right]}{Y}_s^{\sigma }\right)$ for each $r\in P$.

Proof. 

Notice that S is a normal commutative subgroup in G, because S is contained in $\mathcal{C}\left(G\right)$. On the other hand, ${\mathsf{t}}_3({\gamma }_{1}a,{\gamma }_{2}b,{\gamma }_{3}c)={\mathsf{t}}_3(a,b,c)\in \mathcal{C}\left(G\right)$ for each a, b, and c in G, ${\gamma }_j$ in $\mathcal{C}\left(G\right)$, $j\in \{1,2,3\}$. Therefore, the quotient P of G by S is a metagroup, where elements of P are classes $bS$, $b\in G$. We denote by $\theta$ a quotient map from G onto P such that $\theta \left(b\right)=bS$ for each $b\in G$.

Apparently $\mathcal{T}\left[S\right]$ is an algebra over the ring $\mathcal{T}$ and $\mathcal{T}\left[S\right]$ also has a structure of an associative commutative unital ring, because S is a commutative group and $\mathcal{T}$ is the commutative associative unital ring, $1_{\mathcal{T}\left[S\right]}=1_{\mathcal{T}}{e}_G$ (see Definitions A1 and A2). We put $X_g^{\sigma }={\sum }_{a\in {\theta }^{-1}\left(g\right)}{X}_a$ (see also Definition A3). This implies that $X_g^{\sigma }$ is a $\mathcal{T}\left[S\right]$-bimodule for each $g\in P$, since ${\theta }^{-1}\left(g\right)=S$, while $X_a$ is the $\mathcal{T}$-bimodule. From $X_g\ne X_e$ for each $g\in G-S$ and $\left(A16\right)$–$\left(A19\right)$ it follows that for each $g\ne h$ in P the intersection $X_g^{\sigma }\cap X_h^{\sigma }$ is null. Therefore X, considered as the $\mathcal{T}\left[S\right]$-bimodule, is isomorphic with the direct sum ${⨁}_{g\in P}{X}_g^{\sigma }$ of $\mathcal{T}\left[S\right]$-bimodules. From Theorem 1 and Remark 1 it follows that $\mathcal{T}\left[S\right]$-bimodules ${\left(X{\otimes }_{A}Y\right)}_r^{\sigma }$ and ${⨁}_{h\in P,s\in P,hs=r}\left(X_h^{\sigma }{\otimes }_{\mathcal{T}\left[S\right]}{Y}_s^{\sigma }\right)$ are isomorphic for each $r\in P$. □

Proposition 2.

If the conditions of Proposition 1 are satisfied, ϕ and ψ are G-exact, then $\varphi \otimes \psi$ is G-exact.

Proof. 

Since ${\varphi }_1:G\to G$ is an automorphism and ${\varphi }_1\left(g\right)={\psi }_1\left({\theta }_1\left(g\right)\right)$ for each $g\in G$, then

$\left(41\right)$${\varphi }_1\times {\psi }_1:G\times G\to G\times G$ is an automorphism of the direct product $G\times G$ of the metagroup G.

From Theorem 1 and Definition A3 we deduce that

$\left(42\right)$${\left(X_j{\otimes }_A{Y}_j\right)}_g$ is isomorphic as the $\mathcal{T}$-bimodule with

${\sum }_{h\in G,s=h\backslash g}{X}_{j,h}{\otimes }_{\mathcal{T}}{Y}_{j,s}$ for each j.

Thus $\left(41\right)$, $\left(42\right)$, conditions $\left(A16\right)$, $\left(A17\right)$, $\left(14\right)$, and $\left(15\right)$ induce a bijective surjective map ${(\varphi \otimes \psi )}_1:G\to G$. Since ${\varphi }_1\times {\psi }_1:G\times G\to G\times G$ is an automorphism and $\varphi$ and $\psi$ also possess properties of $\mathcal{T}$-homomorphisms, then ${(\varphi \otimes \psi )}_1$ is an automorphism of G. From this and Proposition 1 it follows that $\varphi \otimes \psi$ is a G-exact homomorphism. □

Proposition 3.

Assume that $X_1$, $X_2$, $Y_1$, and $Y_2$ are A-bimodules, where $A=\mathcal{T}\left[G\right]$ is a metagroup algebra. Assume also that P is a quotient metagroup $P:=G/$·$/S$ of a metagroup G by a subgroup S such that $S\subseteq \mathcal{C}\left(G\right)$ and $X_{k,g}\ne X_{k,e}$ and $Y_{k,g}\ne Y_{k,e}$ for each $g\in G-S$, $k\in \{1,2\}$. Assume also that $\varphi :X_1\to X_2$ and $\psi :Y_1\to Y_2$ are their P-graded $\mathcal{T}\left[S\right]$-homomorphisms such that ${\varphi }_1\left(g\right)={\psi }_1\left({\theta }_1\left(g\right)\right)$ for each $g\in P$, where ${\theta }_1$ is an automorphism of P. Then a unique P-graded $\mathcal{T}\left[S\right]$-homomorphism exists

$\left(43\right)$$\varphi \otimes \psi :X_1{\otimes }_A{Y}_1\to X_2{\otimes }_A{Y}_2$ such that $(\varphi \otimes \psi )(x\otimes y)=\varphi \left(x\right)\otimes \psi \left(y\right)$

for each $x\in X_1$ and $y\in Y_1$. Moreover, if ϕ and ψ are P-exact, then $\varphi \otimes \psi$ is P-exact.

Proof. 

In view of Theorem 2 $X_1$, $X_2$, $Y_1$, $Y_2$ also have structures of P-graded $\mathcal{T}\left[S\right]$-bimodules. Note that $\mathcal{T}\left[S\right]X_{1,p}^{\sigma }=X_{1,p}^{\sigma }$ for each $p\in P$, since $\mathcal{T}\left[S\right]$ is a commutative associative unital ring. Therefore Theorem 1, Proposition 1 and 2 and Remark 1 imply the assertion of this proposition. □

Theorem 3.

Let X, Y, $X_1$, and $Y_1$ be A-bimodules, where $A=\mathcal{T}\left[G\right]$ is a metagroup algebra. Then there exist A-exact $\mathcal{T}$-isomorphisms

$\eta :X{\otimes }_A(Y\oplus Y_1)\to \left(X{\otimes }_{A}Y\right)\oplus \left(X{\otimes }_A{Y}_1\right)$ of A-bimodules and

$\rho :(X\oplus X_1){\otimes }_{A}Y\to \left(X{\otimes }_{A}Y\right)\oplus \left(X_1{\otimes }_{A}Y\right)$ of A-bimodules

such that

$\left(44\right)$$\eta (x\otimes (y,y_1))=(x\otimes y,x\otimes y_1)$ and

$\rho ((x,x_1)\otimes y)=(x\otimes y,x_1\otimes y)$ for each $x\in X$, $x_1\in X_1$, $y\in Y$ and $y_1\in Y_1$;

there exists a G-exact $\mathcal{T}$-isomorphism

$\nu :\left(X{\otimes }_A{X}_1\right){\otimes }_{A}Y\to X{\otimes }_A\left(X_1{\otimes }_{A}Y\right)$ of A-bimodules such that

$\left(45\right)$$\nu ((x_g\otimes x_{1,h})\otimes y_s)={\mathsf{t}}_3(g,h,s)x_g\otimes (x_{1,h}\otimes y_s)$ for each g, h, and s in G, $x_g\in X_g$, $x_{1,h}\in X_{1,h}$ and $y_s\in Y_s$;

$\left(46\right)$ for each g and h in G there exists an isomorphism of $\mathcal{T}$-bimodules ${\mu }_{g,h}:X_g{\otimes }_{\mathcal{T}}{Y}_h\to Y_g{\otimes }_{\mathcal{T}}{X}_h$;

$\left(47\right)$ there are A-exact $\mathcal{T}$-isomorphisms

$X{\otimes }_{A}A\cong X$ and $A{\otimes }_{A}X\cong X$ of A-bimodules.

Proof. 

Note that $X{\otimes }_A(Y_1\oplus Y_2)$ and $\left(X{\otimes }_A{Y}_1\right)\oplus \left(X{\otimes }_A{Y}_2\right)$ are isomorphic as $\mathcal{T}$-bimodules, because the ring $\mathcal{T}$ is associative and commutative. On the other hand, $\eta (\left(b{x}_e\right)\otimes (y,y_1))=b(x_e\otimes y,x_e\otimes y_1)$ and $\eta (x\otimes \left((y_e,y_{1,e})b\right))=(x\otimes y_e,x\otimes y_{1,e})b$ for each $b\in A$, $x\in X$, $x_e\in X_e$, $y\in Y$, $y_e\in Y_e$, $y_1\in Y_1$ and $y_{1,e}\in Y_{1,e}$. In view of Proposition 1 $X{\otimes }_A(Y_1\oplus Y_2)$ and $\left(X{\otimes }_A{Y}_1\right)\oplus \left(X{\otimes }_A{Y}_2\right)$ are isomorphic as A-bimodules. Using Definition A3, conditions $\left(14\right)$ and $\left(15\right)$ it is sufficient to take in $\left(6\right)$ and $\left(7\right)$ a homomorphism ${\eta }_1\left(a\right)=a$ for each $a\in A$, since $\left(b{x}_e\right)\otimes (y,y_1)=b(x_e\otimes (y,y_1))$ and $x\otimes \left((y_e,y_{1,e})b\right)=(x\otimes (y_e,y_{1,e}))b$ and $\left(x_{e}g\right)\otimes (y,y_1)=x_e\otimes (gy,g{y}_1)$ and $x\otimes (g{y}_e,g{y}_{1,e})=\left(xg\right)\otimes (y_e,y_{1,e})$ for each $g\in G$. Thus the isomorphism $\eta$ is A-exact. Symmetrically it is proved that the isomorphism $\rho$ is A-exact.

Since $\mathcal{T}\subset \mathcal{T}\left[\mathcal{C}\right(G\left)\right]\subset A$, then $\left(X{\otimes }_A{X}_1\right){\otimes }_{A}Y$ and $X{\otimes }_A\left(X_1{\otimes }_{A}Y\right)$ are isomorphic as $\mathcal{T}$-bimodules. Note that $\left(42\right)$ implies

$\left(48\right)$${\left(\left(X{\otimes }_A{X}_1\right){\otimes }_{A}Y\right)}_v={\sum }_{g\in G,h\in G}\left(X_g{\otimes }_{\mathcal{T}}{X}_{1,h}\right){\otimes }_{\mathcal{T}}{Y}_{\left(gh\right)\backslash v}$

for each v in G. We put ${\nu }_1\left(g\right)=g$ for each $g\in G$. Therefore, the latter formula and $\left(48\right)$ induce a G-exact isomorphism $\nu$ satisfying $\left(45\right)$.

Assertion $\left(46\right)$ follows from $\left(A16\right)$–$\left(A19\right)$ and the conditions of this proposition, since $X_e{\otimes }_{\mathcal{T}}{Y}_e$ and $Y_e{\otimes }_{\mathcal{T}}{X}_e$ are isomorphic $\mathcal{T}$-bimodules.

There exists a left and right $\mathcal{T}$-linear homomorphism $\xi$ from $X{\otimes }_{A}A$ onto X such that $\xi (x\otimes a)=xa$ for each $x\in X$ and $a\in A$, since A is the unital algebra and $x1=x$ for each $x\in X$, because for unital rings modules are supposed to be unital according to Definition A3. On the other side, there exists a homomorphism $\omega :X\to X{\otimes }_{A}A$ such that $\omega \left(x\right)=x\otimes 1$ for each $x\in X$. Therefore, $\omega \left(\xi \right(x\otimes a\left)\right)=\left(xa\right)\otimes 1=x\otimes a$ by $\left(18\right)$ and $\xi \left(\omega \right(x\left)\right)=x$ for each $x\in X$ and $a\in A$. It is sufficient to put ${\xi }_1\left(a\right)=a$ and ${\omega }_1\left(a\right)=a$ for each $a\in A$. Thus $\xi$ is A-exact. Similarly there exists an A-exact isomorphism $A{\otimes }_{A}X\cong X$ of A-bimodules. □

Theorem 4.

Assume that $X_1$, $X_2$, $X_3$, and Y are A-bimodules, where $A=\mathcal{T}\left[G\right]$ is a metagroup algebra. Assume also that $X_1{}_{\overrightarrow{f^1}}{X}_2{}_{\overrightarrow{f^2}}{X}_3\to 0$ is an exact sequence with G-graded homomorphisms $f^1$ and $f^2$. Then a sequence $X_1{⨂}_{A}Y{}_{\overrightarrow{s^1}}{X}_2{⨂}_{A}Y{}_{\overrightarrow{s^2}}{X}_3{⨂}_{A}Y\to 0$ is exact with G-graded homomorphisms $s^1$ and $s^2$ such that $s^j{|}_{X_j{⨂}_{\mathcal{T}}{Y}_e}=f^j\otimes i{d}_{Y_e}$ and $s_1^j=f_1^j$, where $j\in \{1,2\}$, $i{d}_{Y_e}\left(y_e\right)=y_e$ for each $y_e\in Y_e$.

Proof. 

Apparently the formulas $s^j{|}_{X_j{⨂}_{\mathcal{T}}{Y}_e}=f^j\otimes i{d}_{Y_e}$, $s_1^j=f_1^j$, $\left(A16\right)$–$\left(A19\right)$, $\left(4\right)$, $\left(5\right)$ and $\left(16\right)$–$\left(21\right)$ induce G-graded homomorphisms $s^j:X_j{⨂}_{A}Y\to X_{j+1}{⨂}_{A}Y$, where $j\in \{1,2\}$. Notice that $s^2\left(X_2{⨂}_{A}Y\right)$ contains all tensors of rank 1, that is $x_3\otimes y$ with $x_3\in X_3$ and $y\in Y$. Then $f_1^1\left(G\right)$ is a submetagroup in G isomorphic with $G/$·$/Ker\left(f_1^1\right)$ (see Remark 1). Evidently $f^2\circ f^1=0$ implies $s^2\circ s^1=0$, consequently, $Im\left(s^1\right)\subseteq Ker\left(s^2\right)$. Therefore we get

$Ker\left(f^2\right){⨂}_{A}Y=Im\left(f^1\right){⨂}_{A}Y\subseteq Im(f^1\otimes i{d}_Y)$,

where $Im\left(f^1\right)=f^1\left(X_1\right)$, $Ker\left(f^2\right)={\left(f^2\right)}^{-1}\left(0\right)$. For $X_2{⨂}_{A}Y$ considered as the $\mathcal{T}$-bimodule there exists a natural projection $\pi$ of $X_2{⨂}_{A}Y$ onto a quotient $\mathcal{T}$-bimodule $V:=\left(X_2{⨂}_{A}Y\right)/Im\left(s^1\right)$, because $Im\left(s^1\right)$ is a $\mathcal{T}$-bimodule contained in $X_2{⨂}_{A}Y$. Therefore the map $\Phi (x_3,y):=\pi ({\left(f^2\right)}^{-1}{x}_3\otimes y)$ is $\mathcal{T}$-bilinear from $X_3\times Y$ into V, where $x_3\in X_3$ and $y\in Y$. This induces a homomorphism $\mu$ from $X_3{⨂}_{A}Y$ onto V considered as $\mathcal{T}$-bimodules such that $\mu (x_3\otimes y)=\pi (\left({\left(f^2\right)}^{-1}{x}_3\right)\otimes y)$, hence $\mu (f^2\left(x_2\right)\otimes y)=\pi (x_2\otimes y)$ for each $x_2\in X_2$ and $y\in Y$. In view of Lemma 1 $\mu \circ s^2=\pi$. Therefore $Ker\left(s^2\right)\subseteq Ker\left(\pi \right)=Im\left(s^1\right)$. Thus $Ker\left(s^2\right)=Im\left(s^1\right)$. □

Corollary 3.

Suppose that $X_1$, $X_2$, $X_3$, and Y are A-bimodules, where $A=\mathcal{T}\left[G\right]$ is a metagroup algebra. Suppose also that $0\to X_1{}_{\overrightarrow{f^1}}{X}_2{}_{\overrightarrow{f^2}}{X}_3\to 0$ is a splitting exact sequence with G-exact homomorphisms $f^1$ and $f^2$. Then a sequence

$0\to X_1{⨂}_{A}Y{}_{\overrightarrow{s^1}}{X}_2{⨂}_{A}Y{}_{\overrightarrow{s^2}}{X}_3{⨂}_{A}Y\to 0$ is exact with G-exact homomorphisms $s^1$ and $s^2$ such that $s^1=f^1\otimes i{d}_Y$ and $s^2=f^2\otimes i{d}_Y$.

Proof. 

There exists a homomorphism $\nu :X_2\to X_1$ such that $\nu \circ f^1=i{d}_{X_1}$, because the sequence $0\to X_1{}_{\overrightarrow{f^1}}{X}_2{}_{\overrightarrow{f^2}}{X}_3\to 0$ splits. Therefore, the homomorphism $\nu$ is also G-exact such that ${\nu }_1={\left(f_1^1\right)}^{-1}$, since the homomorphism $f^j$ is G-exact for each j. In view of Theorem 4 the sequence $X_1{⨂}_{A}Y{}_{\overrightarrow{s^1}}{X}_2{⨂}_{A}Y{}_{\overrightarrow{s^2}}{X}_3{⨂}_{A}Y\to 0$ is exact with G-exact homomorphisms $s^1$ and $s^2$ and $(\nu \otimes i{d}_Y)(f^1\otimes i{d}_Y)=i{d}_{X_1{⨂}_{A}Y}$, consequently, $s^1$ is injective. □

Theorem 5.

Let X and Y be A-algebras, where $A=\mathcal{T}\left[G\right]$ is a metagroup algebra. Then on the A-bimodule $X{⨂}_{A}Y$ there exists a multiplication satisfying the following conditions:

$\left(49\right)$$(a{x}_1\otimes y_1)(x_2\otimes y_{2}b)=a\left(x_1{x}_2\right)\otimes \left(y_1{y}_2\right)b$

for each $x_1$ and $x_2$ in X, $y_1$ and $y_2$ in Y, $a$ and b in $\mathcal{T}\left[N\right(G\left)\right]$;

$\left(50\right)$$\left((x_{1,g_1}\otimes y_{1,h_1})(x_{2,g_2}\otimes y_{2,h_2})\right)(x_{3,g_3}\otimes y_{3,h_3})=$

${\mathsf{t}}_3(g_1,g_2,g_3)(x_{1,g_1}\otimes y_{1,h_1})\left((x_{2,g_2}\otimes y_{2,h_2})(x_{3,g_3}\otimes y_{3,h_3})\right){\mathsf{t}}_3(h_1,h_2,h_3)$

for each $x_{j,g_j}\in X_{g_j}$, $y_{j,h_j}\in Y_{h_j}$, $g_j$ and $h_j$ in G, $j\in \{1,2,3\}$;

$\left(51\right)$$1_X\otimes 1_Y=1_{X{\otimes }_{A}Y}$ for unital algebras X and Y.

Proof. 

By virtue of Proposition 1 $(a{L}_{x_{1,g_1}}\otimes L_{y_{1,h_1}})(x_2\otimes y_{2}b)=a\left(x_{1,g_1}{x}_2\right)\otimes \left(y_{1,h_1}{y}_2\right)b$ for each $x_{1,g_1}\in X_{g_1}$ and $x_2$ in X, $y_{1,h_1}\in Y_{h_1}$ and $y_2$ in Y, $a$ and b in $\mathcal{T}\left[N\right(G\left)\right]$, where $L_{x_1}{x}_2=x_1{x}_2$, $a{L}_{x_1}=\left(a{I}_X\right)L_{x_1}$, $L_{x_1}a=L_{x_1}\left(a{I}_X\right)$, $I_X{x}_1=x_1$ for each $x_1\in X$. Evidently, $L_{a{x}_1+x_{2}b}=a{L}_{x_1}+L_{x_2}b$ for each $x_1$ and $x_2$ in X, $a$ and b in $\mathcal{T}\left[N\right(G\left)\right]$. Therefore $a{L}_{x_{1,g_1}}\otimes L_{y_{1,h_1}}$ is a G-graded endomorphism of $X{⨂}_{A}Y$. This induces a $\mathcal{T}$-bilinear map $X\times Y\ni (x_1,y_1)\mapsto L_{x_1}\otimes L_{y_1}\in Ho{m}_{\mathcal{T}}(U,U)$, where $Ho{m}_{\mathcal{T}}(V,W)$ denotes the family of all $\mathcal{T}$-bilinear homomorphisms from V into W, where V and W are $\mathcal{T}$-bimodules, U is $X{⨂}_{A}Y$ considered as the $\mathcal{T}$-bimodule. Thus there exists a $\mathcal{T}$-bilinear homomorphism of $\mathcal{T}$-bimodules $f:U\to Ho{m}_{\mathcal{T}}(U,U)$ such that $f(x_1\otimes y_1)=L_{x_1}\otimes L_{y_1}$ for each $x_1\in X$ and $y_1\in Y$. We define a map $\mu :U\times U\to U$ by the following formula $\mu (u,v)=f\left(u\right)\left(v\right)$ for each u and v in U. The construction of $\mu$ and identities $\left(16\right)$–$\left(22\right)$ imply that $\mu$ is $\mathcal{T}$-bilinear and satisfies $\left(49\right)$ and $\left(50\right)$. Thus $\mu$ is the multiplication on $X{⨂}_{A}Y$. From Lemma 1 it follows that $1_X\otimes 1_Y$ is the unit element in $X{⨂}_{A}Y$, if X and Y are unital algebras. □

Corollary 4.

Let W, $X$, and Y be A-algebras, where $A=\mathcal{T}\left[G\right]$ is a metagroup algebra. Then

$\left(52\right)$ the A-algebra $(W\oplus X){⨂}_{A}Y$ is isomorphic with

$\left(W{⨂}_{A}Y\right)\oplus \left(X{⨂}_{A}Y\right)$;

$\left(53\right)$$(w_g\otimes x_h)\otimes y_s={\mathsf{t}}_3(g,h,s)w_g\otimes (x_h\otimes y_s)$ for every $w_g\in W_g$, $x_h\in X_h$, $y_s\in Y_s$, $g$, h, and s in G;

$\left(54\right)$ the $\mathcal{T}$-algebra $W_e{⨂}_{\mathcal{T}}{X}_e$ is isomorphic with $X_e{⨂}_{\mathcal{T}}{W}_e$;

$\left(55\right)$ the A-algebras $W{⨂}_{A}A$ and $A{⨂}_{A}W$ are isomorphic with W.

Proof. 

This follows from Theorems 3 and 5 and Lemma 1. □

Lemma 2.

Suppose that X and Y are unital A-algebras, where $A=\mathcal{T}\left[G\right]$ is a metagroup algebra. Suppose also that $f_X:X\to X{⨂}_{A}Y$ and $f_Y:Y\to X{⨂}_{A}Y$ are maps such that $f_X\left(x\right)=x\otimes 1_Y$ and $f_Y\left(y\right)=1_X\otimes y$ for each $x\in X$ and $y\in Y$. Then $f_X$ and $f_Y$ are A-homomorphisms such that $f_X\left(X\right)\cup f_Y\left(Y\right)$ generates $X{⨂}_{A}Y$ and $f_X\left(x\right)f_Y\left(y\right)=f_Y\left(y\right)f_X\left(x\right)$ for each $x\in X$ and $y\in Y$. Particularly, if $\mathcal{T}=F$ is a field, then $f_X$ and $f_Y$ are injective.

Proof. 

From $\left(49\right)$ and $\left(50\right)$ it follows that $f_X$ and $f_Y$ are A-homomorphisms such that $f_X\left(x\right)f_Y\left(y\right)=x\otimes y$ and $f_Y\left(y\right)f_X\left(x\right)=x\otimes y$ for each $x\in X$ and $y\in Y$, since $f_X\left(ax\right)=\left(ax\right)\otimes 1_Y=a(x\otimes 1_Y)$ and $f_X\left(xa\right)=\left(xa\right)\otimes 1_Y=x\otimes \left(1_{Y}a\right)=(x\otimes 1_Y)a$ for each $x\in X$ and $a\in A$. Therefore $f_X\left(X\right)\cup f_Y\left(Y\right)$ generates $X{⨂}_{A}Y$, because the set $\{x\otimes y:x\in X,y\in Y\}$ generates the tensor product $X{⨂}_{A}Y$ of algebras X and Y over A.

If $\mathcal{T}=F$ is a field, then the restriction of $f_X$ on $X_e$ is injective, $f_X:X_e\to X_e{⨂}_F{Y}_e$. On the other hand, $X_e{⨂}_F{Y}_e\subset X{⨂}_{A}Y$ and $f_X\left({\sum }_{g\in G}{x}_g\right)={\sum }_{g\in G}{f}_X\left(x_g\right)$ for each $x={\sum }_{g\in G}{x}_g$ in X, where $x_g\in X_g$ for each $g\in G$. For each $g\in G$ the F-bimodule $X_g$ is F-isomorphic with $X_{e}g$, consequently, the homomorphism $f_X:X\to X{⨂}_{A}Y$ is injective. □

Definition 3.

Assume that X is an A-algebra and $B\subseteq X$, where $A=\mathcal{T}\left[G\right]$ is a metagroup algebra. We put

$\left(56\right)$$Co{m}_X\left(B\right):=\{x\in X:\forall b\in B,xb=bx\}$;

$\left(57\right)$$N_{X,l}\left(B\right):=\{x\in X:\forall b\in B,\forall c\in B,\left(xb\right)c=x\left(bc\right)\}$;

$\left(58\right)$$N_{X,m}\left(B\right):=\{x\in X:\forall b\in B,\forall c\in B,\left(bx\right)c=b\left(xc\right)\}$;

$\left(59\right)$$N_{X,r}\left(B\right):=\{x\in X:\forall b\in B,\forall c\in B,\left(bc\right)x=b\left(cx\right)\}$;

$\left(60\right)$$N_X\left(B\right):=N_{X,l}\left(B\right)\cap N_{X,m}\left(B\right)\cap N_{X,r}\left(B\right)$ and

$\left(61\right)$${\mathsf{C}}_X\left(B\right):=Co{m}_X\left(B\right)\cap N_X\left(B\right)$.

Then $Co{m}_X\left(B\right)$, $N_X\left(B\right)$, and ${\mathsf{C}}_X\left(B\right)$ are called a commutant, a nucleus and a centralizer correspondingly of the algebra X relative to a subset B in X. Instead of $Co{m}_X\left(X\right)$, $N_X\left(X\right)$, or ${\mathsf{C}}_X\left(X\right)$ it will be also written shortly $Com\left(X\right)$, $N\left(X\right)$, or $\mathsf{C}\left(X\right)$ correspondingly. We put $B_e=B\cap X_e$ and $B_{\mathsf{C}}:=B\cap X_{\mathsf{C}}$, where $X_{\mathsf{C}}={\sum }_{g\in \mathsf{C}\left(G\right)}{X}_g$.

Lemma 3.

Let X and Y be A-algebras, where $A=\mathcal{T}\left[G\right]$ is a metagroup algebra. Let also $Y\subseteq X$, $B\subseteq X$, $D\subseteq X$ and $E\subset X$ such that $B_{e}A\subseteq B\subseteq B_e{⨂}_{\mathcal{T}}A$ and $D_{e}A\subseteq D\subseteq D_e{⨂}_{\mathcal{T}}A$. Then

$\left(62\right)$${\mathsf{C}}_X\left(B\right)$ is a $\mathcal{T}$-subalgebra in X;

$\left(63\right)$$N_X\left(B\right)$ is a $\mathcal{T}\left[N\right(G\left)\right]$-subalgebra in X;

$\left(64\right)$ if $D\subseteq B$, then ${\mathsf{C}}_X\left(B\right)\subseteq {\mathsf{C}}_X\left(D\right)$ and $N_X\left(B\right)\subseteq N_X\left(D\right)$;

$\left(65\right)$$B_{\mathsf{C}}\subseteq {\mathsf{C}}_X\left(D\right)$ if and only if $D_{\mathsf{C}}\subseteq {\mathsf{C}}_X\left(B\right)$;

$\left(66\right)$$Y\cap {\mathsf{C}}_X\left(Y\right)={\mathsf{C}}_Y\left(Y\right)$;

$\left(67\right)$$Co{m}_X\left(E\right)=X$ if and only if $E\subseteq Co{m}_X\left(X\right)$;

$\left(68\right)$ if $E\subseteq N_X\left(X\right)$, then $N_X\left(E\right)=X$ and

$\left(69\right)$ if $E\subseteq {\mathsf{C}}_X\left(X\right)$, then ${\mathsf{C}}_X\left(E\right)=X$.

Proof. 

From Formulas $\left(1\right)$, $\left(60\right)$, $\left(61\right)$, and $\left(A13\right)$–$\left(A19\right)$, it follows that $a_1{x}_1+a_2{x}_2\in {\mathsf{C}}_X\left(B\right)$ and $a_1\left(x_1{x}_2\right)\in {\mathsf{C}}_X\left(B\right)$ for each $x_1$ and $x_2$ in ${\mathsf{C}}_X\left(B\right)$, $a_1$ and $a_2$ in $\mathcal{T}$, since $B_{e}A\subseteq B\subseteq B_e{⨂}_{\mathcal{T}}A$. This implies $\left(62\right)$.

If $x_1$ and $x_2$ in $N_X\left(B\right)$, $a_1$ and $a_2$ in $\mathcal{T}\left[N\right(G\left)\right]$, then $a_1{x}_1+a_2{x}_2$, $a_1\left(x_1{x}_2\right)$ and $x_1{a}_1+x_2{a}_2$, $\left(x_1{x}_2\right)a_1$ belong to $N_X\left(B\right)$ by Formulas $\left(57\right)$–$\left(60\right)$, $\left(A2\right)$–$\left(A5\right)$ and Definition A2, since $B_{e}A\subseteq B\subseteq B_e{⨂}_{\mathcal{T}}A$ and $D_{e}A\subseteq D\subseteq D_e{⨂}_{\mathcal{T}}A$. This implies $\left(63\right)$. Then $\left(64\right)$ evidently follows from $\left(60\right)$ and $\left(61\right)$.

Let $B_{\mathsf{C}}\subseteq {\mathsf{C}}_X\left(D\right)$. If $b\in B_{\mathsf{C}}$, then there exist $g_1$,…, $g_n$ in $\mathsf{C}\left(G\right)$ such that $b=b_{g_1}+$⋯$+b_{g_n}$, since $B_{e}A\subseteq B\subseteq B_e{⨂}_{\mathcal{T}}A$, where $b_{g_i}\in B_{g_i}$ for each i, $B_g:=B\cap X_g$. Therefore $bc=cb$, $b\left(cd\right)=\left(bc\right)d$, $\left(cb\right)d=c\left(bd\right)$ and $\left(cd\right)b=c\left(db\right)$ for each c and d in D. From $B_{e}A\subseteq B\subseteq B_e{⨂}_{\mathcal{T}}A$ and $D_{e}A\subseteq D\subseteq D_e{⨂}_{\mathcal{T}}A$ it follows that

$spa{n}_{\mathcal{T}}{B}_{\mathsf{C}}D=spa{n}_{\mathcal{T}}B{D}_{\mathsf{C}}$ and $spa{n}_{\mathcal{T}}{D}_{\mathsf{C}}B=spa{n}_{\mathcal{T}}D{B}_{\mathsf{C}}$,

where $spa{n}_{\mathcal{T}}S$ denotes the family of all elements $v_1{s}_1+$⋯$+v_m{s}_m$ in X with $s_j\in S$ and $v_j\in \mathcal{T}$ for each j; $BD:=\{bd:b\in B,d\in D\}$. This together with $\left(A16\right)$–$\left(A19\right)$ and $\left(1\right)$ implies that $D_{\mathsf{C}}\subseteq {\mathsf{C}}_X\left(B\right)$. Symmetrically, $D_{\mathsf{C}}\subseteq {\mathsf{C}}_X\left(B\right)$ implies $B_{\mathsf{C}}\subseteq {\mathsf{C}}_X\left(D\right)$ and hence $\left(65\right)$.

From $\left(56\right)$–$\left(60\right)$ it follows that $Y\cap Co{m}_X\left(Y\right)=Co{m}_Y\left(Y\right)$ and $Y\cap N_X\left(Y\right)=N_Y\left(Y\right)$, consequently, $Y\cap {\mathsf{C}}_X\left(Y\right)={\mathsf{C}}_Y\left(Y\right)$ giving $\left(66\right)$.

The equality $Co{m}_X\left(E\right)=X$ is equivalent to $ax=xa$ for each $x\in X$ and $a\in E$. That is $Co{m}_X\left(E\right)=X$⇔$E\subseteq Co{m}_X\left(X\right)$. Thus assertion $\left(67\right)$ is proven.

Then $\left(68\right)$ follows from $\left(57\right)$–$\left(60\right)$. Finally $\left(61\right)$, $\left(67\right)$, and $\left(68\right)$ imply $\left(69\right)$. □

Theorem 6.

Let W, X, and Y be A-algebras, where $A=\mathcal{T}\left[G\right]$ is a metagroup algebra. Let $\xi :X\to W$ and $\eta :Y\to W$ be left and right A-generic homomorphisms correspondingly such that $\eta \left(Y_e\right)\subseteq N\left(W\right)$, $\xi \left(X_e\right)\subseteq N\left(W\right)$, and $\eta \left(Y_e\right)\subseteq Co{m}_{N\left(W\right)}\left(\xi \left(X_e\right)\right)$. Then there exists a unique homomorphism $\nu :X{⨂}_{A}Y\to W$ such that $\nu :X{⨂}_{\mathcal{T}}{Y}_e\to W$ and $\nu :X_e{⨂}_{\mathcal{T}}Y\to W$ are left and right A-generic homomorphisms of a left and right A-modules correspondingly and $\nu (x\otimes y)=\xi \left(x\right)\eta \left(y\right)$ for each $x\in X$ and $y\in Y$.

Proof. 

From the conditions of this proposition and Definition 1 it follows that

$\xi \left(\left(ab\right)x_e\right)={\xi }_1\left(a\right){\xi }_1\left(b\right)\xi \left(x_e\right)={\xi }_1\left(a\right)\xi \left(b{x}_e\right)$ and

$\eta \left(y_e\left(ab\right)\right)=\eta \left(y_e\right){\eta }_1\left(a\right){\eta }_1\left(b\right)=\eta \left(y_{e}a\right){\eta }_1\left(b\right)$

for each a and b in A, $x_e\in X_e$, $y_e\in Y_e$, consequently,

$\xi \left(\left(ab\right)x_e\right)\eta \left(y_e\right)={\xi }_1\left(a\right){\xi }_1\left(b\right)\xi \left(x_e\right)\eta \left(y_e\right)={\xi }_1\left(a\right)\xi \left(b{x}_e\right)\eta \left(y_e\right)$ and

$\xi \left(x_e\right)\eta \left(y_e\left(ab\right)\right)=\xi \left(x_e\right)\eta \left(y_e\right){\eta }_1\left(a\right){\eta }_1\left(b\right)=\xi \left(x_e\right)\eta \left(y_{e}a\right){\eta }_1\left(b\right)$.

Then we put $\nu (x\otimes y)=\xi \left(x\right)\eta \left(y\right)$ for each $x\in X$ and $y\in Y$. Therefore, we infer that

$\nu \left(\left(ab\right)(x_e\otimes y_e)\right)={\xi }_1\left(a\right)\nu (\left(b{x}_e\right)\otimes y_e)$ and

$\nu \left((x_e\otimes y_e)\left(ab\right)\right)=\nu (x_e\otimes \left(y_{e}a\right)){\eta }_1\left(b\right)$

for each a and b in A, $x_e\in X_e$, $y_e\in Y_e$. Then we deduce that

$\nu \left((\left(ab\right)x_{1,e}\otimes y_{1,e})(x_{2,e}\otimes y_{2,e})\right)={\xi }_1\left(a\right){\xi }_1\left(b\right)\nu \left((x_{1,e}\otimes y_{1,e})(x_{2,e}\otimes y_{2,e})\right)=$

${\xi }_1\left(a\right){\xi }_1\left(b\right)\nu (x_{1,e}{x}_{2,e}\otimes y_{1,e}{y}_{2,e})=$

${\xi }_1\left(a\right)\nu (\left(b{x}_{1,e}{x}_{2,e}\right)\otimes y_{1,e}{y}_{2,e})$ and

$\nu \left((x_{1,e}\otimes y_{1,e})(x_{2,e}\otimes y_{2,e}\left(ab\right))\right)=$

$\nu \left((x_{1,e}\otimes y_{1,e})(x_{2,e}\otimes y_{2,e})\right){\eta }_1\left(a\right){\eta }_1\left(b\right)=\nu (x_{1,e}{x}_{2,e}\otimes \left(y_{1,e}{y}_{2,e}a\right)){\eta }_1\left(b\right)$.

From $\mathcal{T}$-linearity of homomorphisms $\xi$ and $\eta$ it follows that $\nu$ is also $\mathcal{T}$-linear. The latter together with the identities given above lead to the conclusion that $\nu :X{⨂}_{\mathcal{T}}{Y}_e\to W$ and $\nu :X_e{⨂}_{\mathcal{T}}Y\to W$ are left and right A-generic homomorphisms of a left and right A-modules correspondingly. □

Corollary 5.

Let V, W, X, and Y be A-algebras, where $A=\mathcal{T}\left[G\right]$ is a metagroup algebra. Let $\xi :X\to V$ and $\eta :Y\to W$ be left and right A-generic homomorphisms correspondingly such that $\xi \left(X_e\right)\subseteq N\left(V\right)$, $\eta \left(Y_e\right)\subseteq N\left(W\right)$. Then there exists a left and right A-generic homomorphism $\xi \otimes \eta :X{⨂}_{A}Y\to V{⨂}_{A}W$. If $\mathcal{T}=F$ is field and homomorphisms ξ and η are injective, then the homomorphism $\xi \otimes \eta$ is also injective.

Proof. 

Note that there exists an isomorphism of $\mathcal{T}$-algebras:

$$N\left(V\underset{A}{⨂}W\right)\simeq \sum _{g\in G,h\in G,gh\in N\left(G\right)}\left(V_g\underset{\mathcal{T}}{⨂}{W}_h\right).$$

Therefore $N\left(V\right){⨂}_{\mathcal{T}}N\left(W\right)$ is contained in $N\left(V{⨂}_{A}W\right)$. This inclusion together with Lemma 2 and Theorems 5, 6 imply the assertion of this corollary. □

Proposition 4.

Assume that W, X, and Y are A-algebras, where $A=F\left[G\right]$ is a metagroup algebra over a field F. Then there exists an isomorphism of A-algebras $W\simeq \left(X{⨂}_{A}Y\right)$ if and only if the following conditions are fulfilled:

$\left(70\right)$ W contains A-subalgebras U and V such that there exist A-exact isomorphisms $\xi :X\to U$ and $\eta :Y\to V$;

$\left(71\right)$$V_e\subseteq Co{m}_{N\left(W\right)}\left(U_e\right)$;

$\left(72\right)$ there exist bases $\{u_i:i\in \Omega \}$ and $\{v_j:j\in {\rm Y}\}$ of the F-algebras $U_e$ and $V_e$ such that $\{\left(s{u}_i\right)\left((s\backslash e)v_j\right):i\in \Omega ,j\in {\rm Y},s\in S\}$ is a basis of $W_e$, where Ω and Υ are sets, where S is a coset of representatives of classes $q(G\cap F)$ in G, $q\in G$.

If $W_e$ is finite dimensional over F, then condition $\left(72\right)$ can be substituted with:

$\left(73\right)$$W_e$ as the F-algebra is generated by $U_e\cup V_e$ and $di{m}_F{W}_e=di{m}_F{U}_{e}di{m}_F{V}_{e}card(G/$·$/(F\cap G))$, where $di{m}_F{U}_e$ denotes a dimension of $U_e$ over the field F.

Proof. 

In view of Lemma 2 conditions $\left(70\right)$–$\left(73\right)$ are necessary (see also Remark 1). Vice versa if conditions $\left(70\right)$ and $\left(71\right)$ are satisfied, then there exists a homomorphism of A-algebras $\nu :X{⨂}_{A}Y\to W$ such that $\nu (x\otimes y)=\xi \left(x\right)\eta \left(y\right)$ for each $x\in X$ and $y\in Y$ by Theorem 6. From condition $\left(72\right)$ and Lemma 2 it follows that $\nu$ is an isomorphism.

If $di{m}_F{W}_e<{\aleph }_0$ and condition $\left(73\right)$ is fulfilled, then $\nu$ is surjective, since $\nu \left(X{⨂}_{A}Y\right)$ is a A-subalgebra in W containing $U\cup V$. By the conditions of this proposition

$G={\bigcup }_{s\in S}s(G\cap F)$ such that $s(G\cap F)\cap q(G\cap F)=\varnothing$

for each $s\ne q$ in S. Notice that

$di{m}_F{W}_e=di{m}_F{U}_{e}di{m}_F{V}_{e}card(G/·/(F\cap G))=$

$di{m}_F\left(X_e{⨂}_F{Y}_e\right)card(G/·/(F\cap G))=di{m}_F{\left(X{⨂}_{A}Y\right)}_e$ and $W={\sum }_{g\in G}{W}_g$

with $W_g={\sum }_{s\in G}\left(X_s{⨂}_F{Y}_{s\backslash g}\right)$, where $W_g$ is isomorphic with $W_{e}g$ and with $g{W}_e$ as the F-algebra. By virtue of Lemma 2 the homomorphism $\nu$ is injective. □

Corollary 6.

Let $G_1$ and $G_2$ be two metagroups and F be a field such that $card(G_j/$·$/(F\cap G_j))<{\aleph }_0$ for each $j\in \{1,2\}$. Let $G=G_1\times G_2$ be the direct product of metagroups and let X and Y be associative algebras over F. Then a metagroup algebra $A=F\left[G\right]$ is isomorphic with $A_1{⨂}_F{A}_2$, where $A_j=F\left[G_j\right]$ for each $j\in \{1,2\}$. Moreover, $\left(X{⨂}_{F}A\right){⨂}_{F}Y$ and $\left(X{⨂}_F{A}_1\right){⨂}_F\left(A_2{⨂}_{F}Y\right)$ can be supplied with A-exact F-isomorphic structures of A-algebras.

Proof. 

Since $ab=ba$ for each $a\in G_1$ and $b\in G_2$, then $A_{1,e}\subseteq Co{m}_{N\left(A\right)}\left(A_{2,e}\right)$. The union $A_1\cup A_2$ generates A as the F-algebra. In view of Proposition 4, A and $A_1{⨂}_F{A}_2$ are isomorphic as the F-algebras. We put $(a_1,b_1)((x\otimes (a_2,b_2))\otimes y)=(x\otimes (a_1{a}_2,b_1{b}_2))\otimes y$ and $((x\otimes (a_2,b_2))\otimes y)(a_1,b_1)=(x\otimes (a_2{a}_1,b_2{b}_1))\otimes y$ and $(a_1,b_1)((x\otimes a_2)\otimes (b_2\otimes y))=(x\otimes \left(a_1{a}_2\right))\otimes (\left(b_1{b}_2\right)\otimes y)$ and $((x\otimes a_2)\otimes (b_2\otimes y))(a_1,b_1)=(x\otimes \left(a_2{a}_1\right))\otimes (\left(b_2{b}_1\right)\otimes y)$ for each $a_j\in A_1$ and $b_j\in A_2$, $j\in \{1,2\}$, $x\in X$ and $y\in Y$. By F-linearity we extend the latter formulas from $(a_j,b_j)\in A$ on any $a\in A$, because A and $A_1{⨂}_F{A}_2$ are isomorphic as the F-algebras. This supplies $\left(X{⨂}_{F}A\right){⨂}_{F}Y$ and $\left(X{⨂}_F{A}_1\right){⨂}_F\left(A_2{⨂}_{F}Y\right)$ with A-algebras structures, which are A-exact isomorphic by Theorem 3, Proposition 4, and Corollary 4. □

Lemma 4.

Let X and Y be A-algebras, U be a right X-module, and V be a right Y-module, let also U and V be A-bimodules, where $A=\mathcal{T}\left[G\right]$ is a metagroup algebra. Then $U{⨂}_{A}V$ can be supplied with a structure of a right $X{⨂}_{A}Y$-module such that

$\left(74\right)$$(u\otimes v)(xa\otimes yb)=\left(ux\right)a\otimes \left(vy\right)b$ for each $u\in U$, $v\in V$, $x\in X$, $y\in Y$, $a$ and b in $N\left(A\right)$;

$\left(75\right)$$\left((u_{g_1}\otimes v_{h_1})(x_{2,g_2}\otimes y_{2,h_2})\right)(x_{3,g_3}\otimes y_{3,h_3})=$

$(u_{g_1}\otimes v_{h_1})((x_{2,g_2}\otimes y_{2,h_2})(x_{3,g_3}{\mathsf{t}}_3(g_1,g_2,g_3)\otimes y_{3,h_3}{\mathsf{t}}_3(h_1,h_2,h_3))$

for each $u_{g_1}\in U_{g_1}$, $v_{h_1}\in V_{h_1}$, $x_{j,g_j}\in X_{g_j}$, $y_{j,h_j}\in Y_{h_j}$, $g_j$ and $h_j$ in G, $j\in \{1,2,3\}$;

$\left(76\right)$ if the algebras X and Y are unital and the modules U and V are unital, then the right $X{⨂}_{A}Y$-module $U{⨂}_{A}V$ is also unital.

Proof. 

By virtue of Theorem 5 $X{⨂}_{A}Y$ is a A-algebra. In view of Proposition 1 $(R_a{R}_x\otimes R_b{R}_y)(u\otimes v)=\left(ux\right)a\otimes \left(vy\right)b$ for each $u\in U$, $v\in V$, $x\in X$, $y\in Y$, $a$ and b in $N\left(A\right)$, where $R_{x}u=ux$, $R_a{R}_{x}u=\left(ux\right)a=u\left(xa\right)=R_{xa}u$ for each $u\in U$, $x\in X$, $a\in N\left(A\right)$. This implies that $R_{a{x}_1+x_{2}b}=R_{x_1}{R}_a+R_b{R}_{x_2}$ for each $x_1$ and $x_2$ in X, $a$ and b in $N\left(A\right)$. Therefore $R_a{R}_{x_{1,g_1}}\otimes R_{y_{1,h_1}}$ is a G-graded $\mathcal{T}$-endomorphism of $U{⨂}_{A}V$ as the right A-module. This provides a $\mathcal{T}$-bilinear map $X\times Y\ni (x_1,y_1)\mapsto R_{x_1}\otimes R_{y_1}\in Ho{m}_{\mathcal{T}}(W,W)$, where $W=U{⨂}_{A}V$. Therefore there exists a $\mathcal{T}$-bilinear homomorphism of $\mathcal{T}$-bimodules $f:W\to Ho{m}_{\mathcal{T}}(W,W)$ such that $f(x_1\otimes y_1)=R_{x_1}\otimes R_{y_1}$ for each $x_1\in X$ and $y_1\in Y$. The latter property and Proposition 4 give $\left(74\right)$.

From Proposition 1 it follows that W is the A-bimodule. Then identity $\left(75\right)$ follows from $\left(74\right)$ and Remark 1. Thus identities $\left(74\right)$ and $\left(75\right)$ supply W with a right $X{⨂}_{A}Y$-module structure.

By virtue of Theorem 5 $1_X\otimes 1_Y$ is the unit element in $X{⨂}_{A}Y$, if X and Y are unital algebras. Therefore if modules U and V are unital, then W is also unital. □

Theorem 7.

Let $X_1$, $X_2$, $Y_1$, and $Y_2$ be A-bimodules and $X_1$, $X_2$ be right D-modules, $Y_1$ and $Y_2$ be right B-modules, where B and D are A-algebras, $A=\mathcal{T}\left[G\right]$ is a metagroup algebra, $\mathcal{T}$ is a commutative associative unital ring. Then

$\left(77\right)$ if $\xi \in Ho{m}_D(X_1,X_2)$ and $\eta \in Ho{m}_B(Y_1,Y_2)$, then

$\xi \otimes \eta \in Ho{m}_{D{⨂}_{A}B}(X_1{⨂}_A{Y}_1,X_2{\otimes }_A{Y}_2)$,

where $Ho{m}_D(X_1,X_2)$ denotes a $\mathcal{T}$-bimodule of all right D-homomorphism from $X_1$ into $X_2$ of right D-modules;

$\left(78\right)$ a map $(\xi ,\eta )\mapsto \xi \otimes \eta$ induces a $\mathcal{T}$-homomorphism of $\mathcal{T}$-bimodules:

$\omega :Ho{m}_D(X_1,X_2){⨂}_{\mathcal{T}}Ho{m}_B(Y_1,Y_2)\to Ho{m}_{D{⨂}_{A}B}(X_1{⨂}_A{Y}_1,X_2{\otimes }_A{Y}_2)$;

$\left(79\right)$$\omega :E_D\left(X_1\right){⨂}_{\mathcal{T}}{E}_B\left(Y_1\right)\to E_{D{⨂}_{A}B}\left(X_1{⨂}_A{Y}_1\right)$ is a $\mathcal{T}$-homomorphism of $\mathcal{T}$-algebras, where $E_D\left(X_1\right):=Ho{m}_D(X_1,X_1)$ and

$\left(80\right)$ if D is unital, then each $\xi \in Ho{m}_D(X_1,X_2)$ is A-exact.

Proof. 

Let $\xi \in Ho{m}_D(X_1,X_2)$ and $\eta \in Ho{m}_B(Y_1,Y_2)$. Then

$(\xi \otimes \eta )\left(\right(x\otimes y\left)\right(d\otimes b\left)\right)=\xi \left(xd\right)\otimes \eta \left(yb\right)$

$=\left(\xi \right(x\left)d\right)\otimes \left(\eta \right(y\left)b\right)=\left(\right(\xi \otimes \eta \left)\right(x\otimes y\left)\right)(d\otimes b)$

for each $x\in X_1$, $y\in Y_1$, $d\in D$, $b\in B$. From Lemmas 1 and 4 it follows that $\xi \otimes \eta \in Ho{m}_{D{⨂}_{A}B}(X_1{⨂}_A{Y}_1,X_2{\otimes }_A{Y}_2)$.

The map $(\xi ,\eta )\mapsto \xi \otimes \eta$ is $\mathcal{T}$-bilinear. In view of Proposition 1 the $\mathcal{T}$-homomorphism $\omega$ exists.

Let ${\xi }_j\in E_D\left(X_1\right)$ and ${\eta }_j\in E_B\left(Y_1\right)$ for each $j\in \{1,2\}$. Let also $\left({\xi }_j{\otimes }^{\prime }{\eta }_j\right)$ denotes a tensor of rank 1 in $E_D\left(X_1\right){⨂}_{\mathcal{T}}{E}_B\left(Y_1\right)$. Then the assertion $\left(79\right)$ follows from Theorem 5 and $\left(78\right)$, since

$\omega \left(\left({\xi }_1{\otimes }^{\prime }{\eta }_1\right)\left({\xi }_2{\otimes }^{\prime }{\eta }_2\right)\right)=\omega \left({\xi }_1{\xi }_2{\otimes }^{\prime }{\eta }_1{\eta }_2\right)=({\xi }_1{\xi }_2\otimes {\eta }_1{\eta }_2)=({\xi }_1\otimes {\eta }_1)({\xi }_2\otimes {\eta }_2)=\omega \left({\xi }_1{\otimes }^{\prime }{\eta }_1\right)\omega \left({\xi }_2{\otimes }^{\prime }{\eta }_2\right)$.

Note that if the algebra D is unital, then there exists an embedding of A into D as $A{1}_D$, where $1_D$ is a unit element in D. Therefore in this case each $\xi \in Ho{m}_D(X_1,X_2)$ is A-exact by $\left(6\right)$, $\left(7\right)$, $\left(A16\right)$. □

Lemma 5.

Let B and D be unital A-algebras, where $A=\mathcal{T}\left[G\right]$ is a metagroup algebra, $\mathcal{T}$ is a commutative associative unital ring. Let also X be a right B-module and Y be a $(B,D)$-bimodule. Then their tensor product $X{⨂}_{B}Y$ is a right D-module such that

$\left(81\right)$$(x_g\otimes y_h)d_s=x_g\otimes \left(y_h\left({\mathsf{t}}_3(g,h,s)d_s\right)\right)$ for each $x_g\in X_g$, $y_h\in Y_h$, $d_s\in D_s$, $g$, h, and s in G.

Proof. 

We put ${\Phi }_d(x,y)=x\otimes \left(yd\right)$ for each $x\in X$, $y\in Y$, $d\in D$. The latter formula defines a map ${\Phi }_d:X\times Y\to X{⨂}_{B}Y$. This map is $\mathcal{T}\left[\mathcal{C}\right(G\left)\right]$-bilinear and

$\left(82\right)$${\Phi }_{d_s}(x_g{b}_q,y_h)=\left(x_g{b}_q\right)\otimes \left(y_h{d}_s\right)=$

$x_g\otimes \left(\left(b_q{y}_h\right)[{\mathsf{t}}_3(g,q,hs)/{\mathsf{t}}_3(q,h,s)]d_s\right)={\Phi }_z(x_g,\left(b_q{y}_h\right))$ with

$z=[{\mathsf{t}}_3(g,q,hs)/{\mathsf{t}}_3(q,h,s)]d_s$ for each $x_g\in X_g$, $y_h\in Y_h$, $b_q\in B_q$, $d_s\in D_s$, $g$, h, q and s in G. Then we deduce that

$\left(83\right)$${\Phi }_{d\alpha +f\beta }={\Phi }_d\alpha +{\Phi }_f\beta$ and

$\left(84\right)$$x\otimes \left(\left(y_h{d}_s\right)f_q\right)=x\otimes \left(y_h\left({\mathsf{t}}_3(h,s,q)d_s{f}_q\right)\right)$

for each $\alpha$ and $\beta$ in $\mathcal{T}\left[\mathcal{C}\right(G\left)\right]$, d and f in D, $x\in X$, $y_h\in Y_h$, $d_s\in D_s$, $f_q\in D_q$, $h$, s, q in G. Note that ${\left(X{⨂}_{B}Y\right)}_v$ is a $\mathcal{T}$-bimodule generated by sums of elements of the form $\left(x_g{b}_q\right)\otimes y_h$ with $x_g\in X_g$, $b_q\in B_q$, $y_h\in Y_h$, $g$, q, and h in G such that $\left(gq\right)h=v$, where $v\in G$. Thus Formulas $\left(81\right)$–$\left(84\right)$ supply $X{⨂}_{B}Y$ with a structure of a right D-module. On the other hand, the metagroup algebra A has embeddings $A{1}_B$ in B and $A{1}_D$ in D, where $1_B$ denotes the unit element in B. Hence the tensor product $X{⨂}_{B}Y$ is G-graded by Theorem 1. □

Remark 2.

Similarly to Lemma 5, if X is a $(B,D)$-bimodule and Y is a left D-module, then $X{⨂}_{D}Y$ is a left B-module. Then if X is a $(B,D)$-bimodule and Y is a $(D,J)$-module, where B, D and J are unital A-algebras, then $X{⨂}_{D}Y$ is a $(B,J)$-bimodule such that

$\left(85\right)$$(\left(b_r{x}_g\right)\otimes y_h)j_s={\mathsf{t}}_3(r,g,h){\mathsf{t}}_3(r,gh,s){\mathsf{t}}_3(g,h,s)b_r(x_g\otimes \left(y_h{j}_s\right))$

for each $b_r\in B_r$, $x_g\in X_g$, $y_h\in Y_h$, $j_s\in J_s$, r, g, h, and s in G. Moreover, if $a\in \mathcal{T}\left[N\right(G\left)\right]$, $x\in X$, $y\in Y$, then

$\left(86\right)$$a(x\otimes y)=\left(ax\right)\otimes y$ and $(x\otimes y)a=x\otimes \left(ya\right)$.

In particular, if D is a subalgebra in B, then B can be considered as a $(D,B)$-bimodule. For a right D-module X then $X{⨂}_{D}B$ is a right B-module, which is called a module induced from X and it is denoted by $X^B$.

Definition 4.

Assume that B is a unital A-algebra, where $A=\mathcal{T}\left[G\right]$ is a metagroup algebra, $\mathcal{T}$ is a commutative associative unital ring. If X and Y are right B-modules and isomorphic as right $\mathcal{T}\left[\mathcal{C}\right(G\left)\right]$-modules and such that $X_e{⨂}_{\mathcal{T}}\mathcal{T}\left[\mathcal{C}\left(G\right)\right]$ and $Y_e{⨂}_{\mathcal{T}}\mathcal{T}\left[\mathcal{C}\left(G\right)\right]$ are isomorphic right $\mathcal{T}\left[\mathcal{C}\right(G\left)\right]$-modules, then we say that X and Y are meta-isomorphic. A meta-isomorphism for left or two-sided D-modules is similarly defined.

Proposition 5.

Suppose that B and D are unital A-algebras, where $A=\mathcal{T}\left[G\right]$ is a metagroup algebra over a commutative associative unital ring $\mathcal{T}$. Suppose also that X is a right B-module, Y is a $(B,D)$-bimodule, Z is a left D-module such that X, Y and Z also have structures of A-bimodules. Then $\left(X{⨂}_{B}Y\right){⨂}_{D}Z$ and $X{⨂}_B\left(Y{⨂}_{D}Z\right)$ are A-bimodules which are meta-isomorphic.

Proof. 

By virtue of Lemma 5 and Remark 2 $\left(X{⨂}_{B}Y\right){⨂}_{D}Z$ and $X{⨂}_B\left(Y{⨂}_{D}Z\right)$ are A-bimodules. From identities $\left(85\right)$, $\left(86\right)$ and the inclusion ${\mathsf{t}}_3(g,h,s)\in \mathcal{C}\left(G\right)$ for each g, h, and s in G it follows that $\left(X{⨂}_{B}Y\right){⨂}_{D}Z$ and $X{⨂}_B\left(Y{⨂}_{D}Z\right)$ are isomorphic as $\mathcal{T}\left[\mathcal{C}\right(G\left)\right]$-bimodules, also $\left(\left(X{⨂}_{B}Y\right){⨂}_{D}Z\right){⨂}_{\mathcal{T}}\mathcal{T}\left[\mathcal{C}\left(G\right)\right]$ and $\left(X{⨂}_B\left(Y{⨂}_{D}Z\right)\right){⨂}_{\mathcal{T}}\mathcal{T}\left[\mathcal{C}\left(G\right)\right]$ are isomorphic $\mathcal{T}\left[\mathcal{C}\right(G\left)\right]$-bimodules. □

Theorem 8.

Let B, C, and D be unital A-algebras, let D be a subalgebra in B and B be a subalgebra in C, where $A=\mathcal{T}\left[G\right]$ is a metagroup algebra, $\mathcal{T}$ is an associative commutative unital ring. Let also X and Y be right D-modules. Then

$\left(87\right)$${(X\oplus Y)}^B$ and $X^B\oplus Y^B$ are isomorphic as right B-modules;

$\left(88\right)$${\left(X^B\right)}^C$ and $X^C$ are meta-isomorphic right C-modules and

$\left(89\right)$$X^D$ is a right D-module meta-isomorphic with X.

Proof. 

Assertion $\left(87\right)$ follows from Lemma 5 and the proof is similar to that of Theorem 3 and Corollary 4, since $(x\oplus (yd\left)\right)b=\left(xb\right)\oplus \left(\right(yd\left)b\right)$ and $\left(\right(xd)\oplus y)b=\left(\right(xd\left)b\right)\oplus \left(yb\right)$ for each $x\in X$, $y\in Y$, $d\in D$ and $b\in B$.

In view of Proposition 5 and Remark 2

$\left(90\right)$$(x_g\otimes b_h)\otimes c_s=x_g\otimes (b_h\otimes c_s){\mathsf{t}}_3(g,h,s)$ and $(x_g\otimes \left(b_h{d}_r\right))\otimes c_s=x_g\otimes (\left(b_h{d}_r\right)\otimes c_s){\mathsf{t}}_3(g,hr,s)$ for each $x_g\in X_g$, $b_h\in B_h$, $d_r\in D_r$, $c_s\in C_s$, g, h, r, s in G. Consider the map $w:b_h\otimes c_s\mapsto 1\otimes \left(b_h{c}_s\right)\in 1\otimes C_{hs}$, consequently,

$w(\left(a_g{b}_h\right)\otimes c_s)=1\otimes \left(\left(a_g{b}_h\right)c_s\right)=1\otimes \left(a_g\left(b_h{c}_s\right){\mathsf{t}}_3(g,h,s)\right)=w(a_g\otimes \left(b_h{c}_s{\mathsf{t}}_3(g,h,s)\right))$

for each $a_g\in B_g$, $b_h\in B_h$, $c_s\in C_s$, $g$, h, and s in G. Therefore $B{\otimes }_{B}C$ is meta-isomorphic with C, because the algebras B and C are unital (see Definition 4). The latter meta-isomorphism and identities $\left(90\right)$ imply assertion $\left(88\right)$. Then $x\otimes \left(dv\right)=\left(x\right(dv\left)\right)\otimes 1$ for each $x\in X$, d and v in D; $(x_g\otimes d_h)\otimes v_s=x_g\otimes (d_h\otimes v_s){\mathsf{t}}_3(g,h,s)$ for each $x_g\in X_g$, $d_h\in D_h$, $v_s\in D_s$, g, h and s in G. On the other hand, $\left(x1\right)d=x\left(1d\right)=xd$ for each $x\in X$ and $d\in D$. Thus $X^D$ is a right D-module meta-isomorphic with X. □

3. Extensions of Nonassociative Algebras with Metagroup Relations

Definition 5.

Let B and D be A-algebras, where $A=\mathcal{T}\left[G\right]$ is a metagroup algebra and $\mathcal{T}$ is a commutative associative unital ring. Let also $f:B\to D$ be an A-exact homomorphism (see Definition 1). For a right D-module X we put $x_g{b}_h:=\xi \left(x_g\right)f\left(b_h\right)$ for each $x_g\in X_g$ and $b_h\in B_h$, g and h in G, where $\xi \left(x_g\right):=(\left(x_g\right)/g)f_1\left(g\right)$, $\left(x_g\right)/g=x_e$ with $x_{e}g=x_g$, $x_e\in X_e$, because $X_g=X_{e}g$. This provides on X a structure of a right B-module denoted by $X_B$, because $f_1:G\to G$ is an automorphism of the metagroup G. In particular, if B is a subalgebra in D and f is an embedding, then the correspondence $X\to X_B$ is called a restriction functor.

Lemma 6.

Let B, C, and D be unital A-algebras and let D be a subalgebra in B, where $A=\mathcal{T}\left[G\right]$ is a metagroup algebra and $\mathcal{T}$ is a commutative associative unital ring. Let also X be a right D-module and Y be a right B-module. Then there exist A-exact homomorphisms ${\nu }_X:X\to {\left(X^B\right)}_D$ of right D-modules and ${\mu }_Y:{\left(Y_D\right)}^B\to Y$ of right B-modules such that

$\left(91\right)$${\nu }_X\left(x\right)=x\otimes 1_B$ for each $x\in X$ and

$\left(92\right)$${\mu }_Y(y\otimes 1_B)=y$ for each $y\in Y$.

Moreover, if X is a $(C,D)$-bimodule, then ${\nu }_X$ is an A-exact homomorphism of left C-modules; if Y is a $(C,B)$-bimodule, then ${\mu }_Y$ is an A-exact homomorphism of left C-modules.

Proof. 

The A-algebras B, C, and D are unital, consequently, there are natural embeddings of A into them $A\hookrightarrow A{1}_B$, etc. Evidently formula $\left(91\right)$ defines a homomorphism of $\mathcal{T}$-modules. Then ${\nu }_X\left(xd\right)=\left(xd\right)\otimes 1_B=x\otimes d=(x\otimes 1_B)d={\nu }_X\left(x\right)d$ for each $x\in X$ and $d\in D$ by Lemma 5 and Theorem 8. Thus ${\nu }_X$ is an A-exact homomorphism of right D-modules.

For a $(C,D)$-bimodule X formula $\left(91\right)$ means that ${\nu }_X$ is an A-exact homomorphism of left C-modules.

Let $\eta :Y_D\times B\to Y$ be a mapping such that $\eta (y,b)=yb$ for each $y\in Y$ and $b\in B$. Apparently the mapping $\eta$ is $\mathcal{T}$-bilinear and G-balanced (see Definition 1). Therefore there exists a homomorphism of $\mathcal{T}$-modules ${\mu }_Y:Y_D{⨂}_{D}B\to Y$ such that ${\mu }_Y(y\otimes b)=yb$ for each $y\in Y$ and $b\in B$. By virtue of Lemma 5 and Remark 2 ${\mu }_Y$ is an A-exact homomorphism of right B-modules. If Y is a $(C,B)$-bimodule, then evidently ${\mu }_Y$ is an A-exact homomorphism of left C-modules, since ${\mu }_{Y,1}{{|}_{A{1}_C}=id|}_{A{1}_C}$. □

Definition 6.

Let X be a $(B,B)$-bimodule, where B is an A-algebra, $A=\mathcal{T}\left[G\right]$ (see Definitions A3 and A5). Let $X\times X\ni (u,v)\mapsto uv\in X$ be a mapping such that

$\left(93\right)$$u(v+w)=uv+uw$ and $(v+w)u=vu+wu$ and $\left(bu\right)v=b\left(uv\right)$ and $\left(ub\right)v=u\left(bv\right)$ and $\left(uv\right)b=u\left(vb\right)$

for all u, v, w in X, $b\in \mathcal{T}$;

$\left(94\right)$$\left(u_g{v}_h\right)w_s={\mathsf{t}}_3(g,h,s)u_g\left(v_h{w}_s\right)$ and $u_g{v}_h\in X_{gh}$ and

$\left(95\right)$$\left(x_s{v}_h\right)u_g={\mathsf{t}}_3(s,h,g)x_s\left(v_h{u}_g\right)$ and $\left(u_g{x}_s\right)v_h={\mathsf{t}}_3(g,s,h)u_g\left(x_s{v}_h\right)$ and

$\left(u_g{v}_h\right)x_s={\mathsf{t}}_3(g,h,s)u_g\left(v_h{x}_s\right)$ and $u_g{x}_s\in X_{gs}$ and $x_s{u}_g\in X_{sg}$ and

$\left(96\right)$$\left(x_s{y}_h\right)u_g={\mathsf{t}}_3(s,h,g)x_s\left(y_h{u}_g\right)$ and

$\left(u_g{x}_s\right)y_h={\mathsf{t}}_3(g,s,h)u_g\left(x_s{y}_h\right)$ and

$\left(y_h{u}_g\right)x_s={\mathsf{t}}_3(h,g,s)y_h\left(u_g{x}_s\right)$

for every g, h, s in G, $u_g\in X_g$, $v_h\in X_h$, $w_s\in X_s$, $x_s\in B_s$, $y_h\in B_h$. Then X will be called a multiplicative two-sided B-module.

We use the following notation for left ordered products: ${\left\{u_1\right\}}_{l\left(1\right)}=u_1$, ${\left\{u_1{u}_2\right\}}_{l\left(2\right)}=u_1{u}_2$, ${\{u_1\cdots u_{k+1}\}}_{l(k+1)}={\{u_1\cdots u_k\}}_{l\left(k\right)}{u}_{k+1}$ for each $k\ge 2$ and $u_1,\cdots ,u_{k+1}$ in X. If ${\{u_1\cdots u_k\}}_{l\left(k\right)}=0$ for each $u_1,\cdots ,u_k$ in X, where $k\ge 2$, then the multiplicative two-sided B-module X will be called k-nilpotent.

If M and X are two multiplicative two-sided B-modules and $\varphi :M\to X$ is an additive mapping $\varphi (u+v)=\varphi \left(u\right)+\varphi \left(v\right)$ which is left $\mathcal{T}$-homogeneous $\varphi \left(bu\right)=b\varphi \left(u\right)$ or right $\mathcal{T}$-homogeneous $\varphi \left(ub\right)=\varphi \left(u\right)b$ for each $b\in \mathcal{T}$, u and v in M, then ϕ is called left $\mathcal{T}$-linear or right $\mathcal{T}$-linear respectively. If ϕ is left and right $\mathcal{T}$-linear, then it is called $\mathcal{T}$-linear. If ϕ is $\mathcal{T}$-linear and $\varphi \left(u_g{v}_h\right)=\zeta (g,h)\varphi \left(u_g\right)\varphi \left(v_h\right)$ with $\zeta (g,h)\in \mathcal{T}\left[\mathcal{C}\right(G\left)\right]-\left\{0\right\}$ for each g and h in G, $u_g\in M_g$, $v_h\in M_h$, then ϕ will be called a metahomomorphism.

Remark 3.

A multiplication in a multiplicative two-sided B-module X induces a homomorphism $\mu \in Ho{m}_{\mathcal{T}}(X{\otimes }_{B}X,X)$ satisfying the following conditions:

$\left(97\right)$$\mu (u\otimes (v+w\left)\right)=\mu (u\otimes v)+\mu (u\otimes w)$ and

$\mu \left(\right(v+w)\otimes u)=\mu (v\otimes u)+\mu (w\otimes u)$ and

$\mu \left(\right(bu)\otimes v)=b\mu (u\otimes v)$ and $\mu \left(\right(ub)\otimes v)=\mu (u\otimes (bv\left)\right)$ and

$\mu (u\otimes (vb\left)\right)=\mu (u\otimes v)b$ for all u, v, w in X, $b\in \mathcal{T}$;

$\left(98\right)$$\mu (\mu (u_g\otimes v_h)\otimes w_s)=\mu (\left({\mathsf{t}}_3(g,h,s)u_g\right)\otimes \mu (v_h\otimes w_s))$ and

$\left(99\right)$$\mu (\left(x_s{v}_h\right)\otimes u_g)=\mu \left(\left({\mathsf{t}}_3(s,h,g)x_s\right)(v_h\otimes u_g)\right)$ and

$\mu (\left(u_g{x}_s\right)\otimes v_h)=\mu (\left({\mathsf{t}}_3(g,s,h)u_g\right)\otimes \left(x_s{v}_h\right))$ and

$\mu \left((u_g\otimes v_h)x_s\right)=\mu (\left({\mathsf{t}}_3(g,h,s)u_g\right)\otimes \left(v_h{x}_s\right))$

for every g, h, s in G, $u_g\in X_g$, $v_h\in X_h$, $w_s\in X_s$, $x_s\in B_s$.

Vice versa, each homomorphism satisfying conditions $\left(97\right)$–$\left(99\right)$ induces on the two-sided B-module X a multiplicative structure. As $B^e:=B{\otimes }_{\mathcal{T}}{B}^{op}$ usually denotes the enveloping algebra of B, $B^{op}$ notates the opposite algebra of B.

Next we take an A-algebra C and its A-subalgebra B and an ideal X in C, where $A=\mathcal{T}\left[G\right]$ is a metagroup algebra. Then a multiplication in C provides a structure of a two-sided B-module $X\times B\ni (u,b)\mapsto ub$, $B\times X\ni (b,u)\mapsto bu$ and a multiplication $(X\times X)\ni (u,v)\mapsto uv\in X$ in X for each u and v in X, $b\in B$. This construction makes of X a multiplicative two-sided B-module.

Proposition 6.

Assume that X is a multiplicative two-sided B-module with B being a unital A-algebra, where $A=\mathcal{T}\left[G\right]$ is a metagroup algebra. Put $X\uplus B$ to be $X\oplus B$ as an A-module and define a multiplication in it by:

$\left(100\right)$$(w,b)(v,a)=(wv+bv+wa,ba)$ for each w, v in X, $a$ and b in B. Then

$\left(101\right)$$X\uplus B$ is the A-algebra with a unit element $(0,1)$;

$\left(102\right)$$B^{\prime }:=\{(0,b):b\in B\}$ is a subalgebra in $X\uplus B$ and a mapping $(0,b)\mapsto b$ is an A-exact isomorphism of A-algebras $B^{\prime }$ with B;

$\left(103\right)$$X^{\prime }:=\{(w,0):w\in X\}$ is an ideal in $X\uplus B$ and a mapping $(w,0)\mapsto w$ is an A-exact isomorphism of multiplicative two-sided B-modules $X^{\prime }$ with X;

$\left(104\right)$$X\uplus B=X^{\prime }\oplus B^{\prime }$.

Proof. 

Since B is the A-algebra and $1\in B$ (see Definition 6), then X also has a structure of a two-sided A-module. Evidently, A is embedded into $X\uplus B$ as $(0,A)$. Therefore, the multiplication rule $\left(100\right)$ shows $X\uplus B$ is the A-algebra with the unit element $(0,1)$. Then from $\left(100\right)$ and $\left(101\right)$ assertions $\left(102\right)$ and $\left(103\right)$ evidently follow, since $(0,b)(0,a)=(0,ba)$; $(w,0)(v,a)=(wv+wa,0)\in X^{\prime }$ and $(w,b)(v,0)=(wv+bv,0)\in X^{\prime }$ for each w, v in X and a, b in B. The assertion $\left(104\right)$ follows from $\left(100\right)$, $\left(102\right)$, and $\left(103\right)$. □

Proposition 7.

Let C be an A-algebra and let B be its unital A-subalgebra, where $A=\mathcal{T}\left[G\right]$ is a metagroup algebra. Let X be an ideal in C and $C=X\oplus B$ as two-sided A-modules. Then X is a multiplicative two-sided B-module relative to operations induced from C and $C=X\uplus B$.

Proof. 

From the conditions of this lemma and Remark 3 it follows that X is a multiplicative two-sided B-module. A mapping $p:(w,b)\mapsto w+b$ for each $w\in X$ and $b\in B$ provides an isomorphism between two-sided A-modules $X\uplus B$ and C. Indeed, $C=X\oplus B$ and $p\left(\right(w,b\left)\right(v,a\left)\right)=p(wv+bv+wa,ba)=wv+bv+wa+ba=(w+b)(v+a)=p(w,b)p(v,a)$ for each w and v in X, a and b in B. Then $p\left((w_g,b_g)(v_h,a_h)\right)=(w_g+b_g)(v_h+a_h)=p(w_g,b_g)p(v_h,a_h)$ for each $w_g\in X_g$, $v_h\in X_h$, $b_g\in B_g$, $a_h\in B_h$, $g$, and h belonging to G. Therefore this mapping p is an isomorphism of C with $X\uplus B$. □

Definition 7.

If the conditions of Proposition 7 are satisfied, then C is called a splitting extension of X with the help of B.

If an algebra C over a metagroup algebra $A=\mathbf{F}\left[G\right]$ is such that $C_e$ is finite dimensional over a field $\mathbf{F}$, then it will be said that C is finite dimensional over A, where e is a unit element in G.

Remark 4.

A direct sum $X\oplus B$ of two A-algebras may usually be not a splitting extension, because B may usually be not a subalgebra in $X\oplus B$, since $1_{X\oplus B}\ne 1_B$ if $X\ne 0$.

Proposition 8.

Let C be an A-algebra finite dimensional over A, where $A=\mathbf{F}\left[G\right]$ is a metagroup algebra over a field $\mathbf{F}$.

$\left(105\right)$ If $B=C/J\left(C\right)$ is separable, then a radical $J\left(C\right)$ is a nilpotent multiplicative two-sided B-module and C is isomorphic with $J\left(C\right)\uplus B$.

$\left(106\right)$ If B is a semisimple A-algebra, X is a nilpotent multiplicative two-sided A-module and $C=X\uplus B$, then X is isomorphic with a radical $J\left(C\right)$ and B is isomorphic with $C/J\left(C\right)$.

Proof. 

This follows from Theorem 3 in [13] and Propositions 6 and 7 above. □

Theorem 9.

Let B and D be two unital A-algebras such that B and D also have structure of separable algebras over a field $\mathbf{F}$, where $A=\mathbf{F}\left[G\right]$ is a metagroup algebra. Let X and M be a multiplicative two-sided B-module and D-module respectively.

(i). 

If $X\uplus B$ is isomorphic with $M\uplus D$ as $\mathbf{F}$-algebras and two-sided A-modules, then there exists an A-exact isomorphism of A-algebras $\theta :B\to D$ and an A-exact isomorphism of two-sided A-modules $\psi :X\to M$ such that

$\left(107\right)$$\psi \left(uy\right)=\psi \left(u\right)\psi \left(y\right)$,

$\left(108\right)$$\psi \left(bu\right)=\theta \left(b\right)\psi \left(u\right)$,

$\left(109\right)$$\psi \left(ub\right)=\psi \left(u\right)\theta \left(b\right)$,

$\left(110\right)$$\left(\psi \left(v_g\right)\psi \left(u_h\right)\right)\psi \left(y_s\right)={\mathsf{t}}_3(g,h,s)\psi \left(v_g\right)\left(\psi \left(u_h\right)\psi \left(y_s\right)\right)$ and

$\left(111\right)$$\left(\theta \left(a_g\right)\theta \left(b_h\right)\right)\theta \left(c_s\right)={\mathsf{t}}_3(g,h,s)\theta \left(a_g\right)\left(\theta \left(b_h\right)\theta \left(c_s\right)\right)$

for each u and y in X and $b\in B$; $g$, h, and s in G; $v_g\in X_g$, $u_h\in X_h$, $y_s\in X_s$; $a_g\in B_g$, $b_h\in B_h$, $c_s\in B_s$.

(ii). 

If $\theta :B\to D$ is an A-exact isomorphism of A-algebras and $\psi :X\to M$ is an A-exact isomorphism of two-sided A-modules such that conditions $\left(107\right)$–$\left(111\right)$ are satisfied, then $X\uplus B$ is A-exact isomorphic with $M\uplus D$.

Proof. 

$\left(i\right)$. The metagroup algebra A is embedded into the unital A-algebras B and D as $A{1}_B$ and $A{1}_D$ correspondingly, where $1_B$ is the unit element in B. Let $f:X\uplus B\to M\uplus D$ be an isomorphism of $X\uplus B$ with $M\uplus D$ as $\mathbf{F}$-algebras and A-exact as two-sided A-modules. By virtue of Proposition 8 $f\left(X\right)=f\left(J\right(X\uplus B\left)\right)=J(M\uplus D)=M$ and $M\uplus D=J(M\uplus D)\oplus f\left(B\right)$. In view of Proposition 8 $(1-w)D=f\left(A\right)(1-w)$ for some $w\in M$ such that $(1-w)$ has a left inverse ${(1-w)}_l$ and a right inverse ${(1-w)}_r$, that is ${(1-w)}_l(1-w)=1$ and $(1-w){(1-w)}_r=1$.

On the other hand, we have

$X\uplus B={\sum }_{g\in G}{(X\uplus B)}_g$

(see Definition A3). Note that $X_e$ and $M_e$ is a multiplicative two-sided $B_e$-module and $D_e$-module respectively, since X, M, B and D are G-graded, $B_e$ and $D_e$ are F-algebras, $X_e$ and $M_e$ are F-bimodules. From Definition 6 it follows that $X_e$, $M_e$, $B_e$, and $D_e$ are associative relative to the multiplication and addition. In view of Proposition 7 ${(X\uplus B)}_e$ is isomorphic with ${(M\uplus D)}_e$ as $\mathbf{F}$-algebras. From Proposition 8 it follows that that there exists $v\in M_e$ with left and right invertible $1-v$ such that $(1-v)D_e=f_e\left(B_e\right)(1-v)$, where $f_e:{(X\uplus B)}_e\to {(M\uplus D)}_e$ denotes an isomorphism of the $\mathbf{F}$-algebras. From $f\left(gu\right)=gf\left(u\right)$ and $f\left(ug\right)=f\left(u\right)g$ for each $u\in {(X\uplus B)}_e$ and $g\in G$ it follows that $f_e$ is the restriction of f on ${(X\uplus B)}_e$. Since $v\in M_e$, then the left and right inverses of $1-v$ coincide, because $M_e$ is the associative multiplicative two-sided $B_e$-module. Therefore $\left({(1-v)}^{-1}g\right)(1-v)={(1-v)}^{-1}\left(g(1-v)\right)$ for each $g\in G$. We put $H={(1-v)}^{-1}G(1-v):=\{h:h={(1-v)}^{-1}g(1-v),g\in G\}$. Hence H and G are isomorphic metagroups. This implies that $(1-v)D{(1-v)}^{-1}$ is isomorphic with $f\left(B\right)$ as the $\mathbf{F}$-algebras and two-sided A-modules, since

$f\left(\left(a_g{b}_h\right)c_s\right)=\left(f\left(a_g\right)f\left(b_h\right)\right)f\left(c_s\right)={\mathsf{t}}_3(g,h,s)f\left(a_g\right)\left(f\left(b_h\right)f\left(c_s\right)\right)={\mathsf{t}}_3(g,h,s)f\left(a_g\left(b_h{c}_s\right)\right)$ for each $a_g\in B_g$, $b_h\in B_h$, $c_s\in B_s$ and g, h, s in G. Hence there exists and isomorphism $\varphi :X\uplus B\to M\uplus D$ as $\mathbf{F}$-algebras and as two-sided A-modules, $\varphi$ is A-exact such that $(1-v)D{(1-v)}^{-1}=\varphi \left(B\right)$, because $v\in M_e$ and $e\in \mathcal{C}\left(G\right)\subseteq \mathcal{C}\left(A\right)$.

We put a morphism $\theta :B\to D$ as algebras over $\mathbf{F}$ and two-sided A-modules to be $(1-v)\theta \left(b\right)=\varphi \left(b\right)(1-v)$ for each $b\in B$. Let a morphism $\psi :X\to M$ as $\mathbf{F}$-linear spaces and two-sided A-modules be given by the following equality $(1-v)\psi \left(u\right)=\varphi \left(u\right)(1-v)$ for each $u\in X$. Then $(1-v){(X\uplus B)}_e$ and ${(X\uplus B)}_e$ are isomorphic as $\mathbf{F}$-linear spaces, since

${(1-v)}^{-1}\left[(1-v){(X\uplus B)}_e\right]=\left[{(1-v)}^{-1}(1-v)\right]{(X\uplus B)}_e$.

Therefore $(1-v)(X\uplus B)$ and $X\uplus B$ are isomorphic as $\mathbf{F}$-linear spaces and as two-sided A-modules they are A-exact isomorphic, since $\left[(1-v){(X\uplus B)}_e\right]g=(1-v){(X\uplus B)}_g$ for each $g\in G$, because ${\mathsf{t}}_3(h,e,g)=e$ for each h and g in G. Hence $\psi$ is the isomorphism of X and M as $\mathbf{F}$-linear spaces and as two-sided A-modules $\psi$ is A-exact. Then

$(1-v)\psi \left(uy\right)=\left(\varphi \right(u\left)\varphi \right(y\left)\right)(1-v)$ and

$(1-v)\left(\psi \left(u\right)\psi \left(y\right)\right)=(1-v)\left(\psi \left(u\right){(1-v)}^{-1}\left((1-v)\psi \left(y\right)\right)\right)=\left(\varphi \left(u\right)\varphi \left(y\right)\right)(1-v)$

for each u and y in X. This implies Identity $\left(107\right)$. Then from

$\theta (\alpha a+\beta b)=\alpha {(1-v)}^{-1}\varphi \left(a\right)(1-v)+\beta {(1-v)}^{-1}\varphi \left(b\right)(1-v)$ and

$\theta \left(ab\right)={(1-v)}^{-1}\left(\varphi \left(a\right)\varphi \left(b\right)\right)(1-v)=\left({(1-v)}^{-1}\varphi \left(a\right)(1-v)\right)\left({(1-v)}^{-1}\varphi \left(b\right)(1-v)\right)=\theta \left(a\right)\theta \left(b\right)$

for each a and b in B and $\alpha$ and $\beta$ in $\mathbf{F}$ it follows that $\theta :B\to D$ is the isomorphism of them as the $\mathbf{F}$-algebras and as two-sided A-modules the isomorphism $\theta$ is A-exact, because $\varphi$ is A-exact and $v\in M_e$. Then

$\psi \left(bu\right)={(1-v)}^{-1}\varphi \left(bu\right)(1-v)=\left({(1-v)}^{-1}\varphi \left(b\right)(1-v)\right)\left({(1-v)}^{-1}\varphi \left(u\right)(1-v)\right)$

for each $b\in B$ and $u\in X$, consequently, Identity $\left(108\right)$ is satisfied. Similarly is verified Identity $\left(109\right)$.

For each $v_g$, $u_h$, $y_s$, $a_g$, $b_h$, $c_s$ satisfying the conditions of this theorem we infer that

$\left(\psi \left(v_g\right)\psi \left(u_h\right)\right)\psi \left(y_s\right)=\left[\left({(1-v)}^{-1}\varphi \left(v_g\right)(1-v)\right)\left({(1-v)}^{-1}\varphi \left(u_h\right)(1-v)\right)\right]$

$\left({(1-v)}^{-1}\varphi \left(y_s\right)(1-v)\right)={(1-v)}^{-1}\varphi \left(\left(v_g{u}_h\right)y_s\right)(1-v)$

$={\mathsf{t}}_3(g,h,s)\left({(1-v)}^{-1}\varphi \left(v_g\right)(1-v)\right)[\left({(1-v)}^{-1}\varphi \left(u_h\right)(1-v)\right)$

$\left({(1-v)}^{-1}\varphi \left(y_s\right)(1-v)\right)]={\mathsf{t}}_3(g,h,s)\psi \left(v_g\right)\left(\psi \left(u_h\right)\psi \left(y_s\right)\right)$ and

$\left(\theta \left(a_g\right)\theta \left(b_h\right)\right)\theta \left(c_s\right)=\left[\left({(1-v)}^{-1}\varphi \left(a_g\right)(1-v)\right)\left({(1-v)}^{-1}\varphi \left(b_h\right)(1-v)\right)\right]$

$\left({(1-v)}^{-1}\varphi \left(c_s\right)(1-v)\right)={(1-v)}^{-1}\varphi \left(\left(a_g{b}_h\right)c_s\right)(1-v)=$

${\mathsf{t}}_3(g,h,s)\left({(1-v)}^{-1}\varphi \left(a_g\right)(1-v)\right)[\left({(1-v)}^{-1}\varphi \left(b_h\right)(1-v)\right)$

$\left({(1-v)}^{-1}\varphi \left(c_s\right)(1-v)\right)]={\mathsf{t}}_3(g,h,s)\theta \left(a_g\right)\left(\theta \left(b_h\right)\theta \left(c_s\right)\right)$,

since $v\in M_e$. Thus identities $\left(110\right)$ and $\left(111\right)$ are satisfied.

$\left(ii\right)$. Vice versa if $\theta :B\to D$ is an A-exact isomorphism of A-algebras and $\psi :X\to M$ is an A-exact isomorphism of two-sided A-modules such that conditions $\left(107\right)$–$\left(111\right)$ are satisfied, then $\varphi (u\uplus b)=\psi \left(u\right)\uplus \theta \left(b\right)$ for each $u\in X$ and $b\in B$ provides an A-exact isomorphism of the A-algebras $X\uplus B$ and $M\uplus D$. □

Theorem 10.

Assume that B and D are A-algebras, where $A=\mathcal{T}\left[G\right]$ is a metagroup algebra, $\mathcal{T}$ is an associative commutative unital ring. Assume also that D is a subalgebra of B. Then the following conditions are equivalent:

$\left(112\right)$$B=D\oplus Y$, where Y is a $(D,D)$-bisubmodule in B,

$\left(113\right)$ for each A-algebra C and each $(C,D)$-bimodule X a homomorphism ${\nu }_X:X\to {\left(X^B\right)}_D$ is a splitting injective A-exact homomorphism of $(C,D)$-bimodules.

Proof. 

$\left(112\right)\Rightarrow \left(113\right)$. From $\left(112\right)$ it follows that there exists an A-exact homomorphism $f:B\to D$ such that ${f|}_D=i{d}_D$, where $i{d}_D\left(d\right)=d$ for each $d\in D$. Therefore a mapping $p:X\times B\to X$ such that $p(x,b)=xf\left(b\right)$ for each $x\in X$ and $b\in B$ is A-exact and G-balanced (see Definition 1). Thus there exists a homomorphism $w:X^B\to X$ such that $w(x\otimes b)=xf\left(b\right)$ for each $x\in X$ and $b\in B$. Evidently w is an A-exact homomorphism of left D-modules and an A-exact homomorphism of right D-modules, since f is the A-exact homomorphism of right D-modules. On the other hand, $1_B=1_D\in D$, consequently, $w{\nu }_X\left(x\right)=w(x\otimes 1_B)=x{1}_B=x$ for each $x\in X$ (see Lemma 6). This means that ${\nu }_X$ is an A-exact splitting injective homomorphism.

Vice versa $\left(113\right)$ implies $\left(112\right)$ by taking in particular $C=D$ and $X=D$ considered as $(D,D)$-bimodules. □

4. Conclusions

In this article new specific tensor products of nonassociative algebras and modules induced over nonassociative algebras with metagroup relations are investigated (see also Formulas $\left(1\right)$–$\left(113\right)$ above). In particular, their splitting extensions also are scrutinized. Their radicals and separability are studied. For this purpose, their cohomologies are used.

The obtained results can be used for further studies of the structure of nonassociative algebras, modules, and homological complexes over them, their tensor products, ideals, extensions, homomorphisms. On the other side, smashed twisted products and smashed twisted wreath products of metagroups or groups were described in [14]. They also provide tools for the construction of a wide class of nonassociative algebras, modules, and homological complexes over them with metagroup relations. With the help of the results presented above it also is possible to continue investigations of nonassociative generalized Cayley–Dickson algebras cohomologies, noncommutative geometry, algebraic geometry, operator theory, spectral theory, PDEs, their applications in the sciences, etc. [6,8,10,15,18,20,22,24,25,26]. It can also be applied in information technologies for the classification of flows of information [27,28].