1 INTRODUCTION

Let \(R\) be a linearly ordered ring. Consider the subsemigroup \(G_{n}(R)\) in the group \(\mathrm{GL}_{n}(R)\) that consists of all matrices with nonnegative coefficients.

Mikhalev and Shatalova [1] described all automorphisms of the semigroup \(G_{n}(R)\) in the case where \(R\) is a linearly ordered skewfield and and \(n\geqslant 2\). Bunina and Mikhalev [2] described all automorphisms of \(G_{n}(R)\) in the case where \(R\) is an arbitrary linearly ordered associative ring with \(1/2\) and \(n\geqslant 3\). Bunina and Mikhalev [3] found necessary and sufficient conditions for these semigroups to be elementary equivalent. Bunina and Semenov [4] described automorphisms of the semigroups of invertible nonnegative matrices of order higher than two over commutative partially ordered rings with 2 invertible, and in [5] they found necessary and sufficient conditions for their elementary equivalence. Bunina [6] described automorphisms of the semigroup \(G_{2}(R)\) assuming that \(R\) is a partially ordered commutative ring with \(1/2\) generated by its invertible elements. Semenov [7] described automorphisms of the semigroup of nonnegative matrices over the ring of integers.

A less ordinary problem is the description of all endomorphisms of a given semigroup. In [8], endomorphisms of the semigroup \(G_{n}(R)\) are described in the case where \(R\) is a linearly ordered commutative ring.

In the present paper, we describe all endomorphisms of the semigroup of nonnegative matrices over a linearly ordered, not necessarily commutative, ring with \(n>2\). The main meaning of the proved theorem is that, for a sufficiently large image, every endomorphism is standard, as in the case of automorphisms.

2 BASIC NOTIONS

Let \(R\) be a linearly ordered ring. By \(G_{n}(R)\), we denote the subsemigroup of the group \(\mathrm{GL}_{n}(R)\) consisting of all matrices with nonnegative coefficients.

A ring \(R\) is called linearly ordered if there is a partial order relation \(\leqslant\) on \(R\) satisfying the following conditions:

(1) \(\forall x,y\in R(x\leqslant y\vee y\leqslant x)\);

(2) \(\forall x,y,z\in R(x\leqslant y\Rightarrow x+z\leqslant y+z)\);

(3) \(\forall x,y\in R(0\leqslant x\land 0\leqslant y\Rightarrow 0\leqslant xy)\).

We denote positive and negative elements of the ring \(R\) by \(R_{+}\) and \(R_{-}\), respectively.

Let \(e=I_{n}\) be the \(n\times n\) unit matrix; \(\Gamma_{n}(R)\) is the group consisting of all invertible elements from \(G_{n}(R)\); \(\Sigma_{n}\) is a symmetric group of order \(n\); \(\sigma_{s}\) is the permutation matrix (i.e., the matrix \((\delta_{is(j)})\), where \(\delta_{is(j)}\) is the Kronecker symbol and \(s\in\Sigma_{n}\)); and \(S_{n}\) is the group consisting of all matrices \(\sigma_{s}\). By \(N_{S}\), we denote a normal subgroup of \(S_{n}\); \(\textrm{diag}[\alpha_{1},\ldots,\alpha_{n}]\) is a diagonal matrix with elements \(\alpha_{1},\dots,\alpha_{n}\) on its diagonal, where \(\alpha_{1},\ldots,\alpha_{n}\in R^{*}_{+}\). We denote by \(D_{n}(R)\) the group of all invertible diagonal matrices from \(G_{n}(R)\), and \(N_{D}\) is a subgroup in the group \(D_{n}(R)\).

By \(t_{ij}(x)\) we denote the matrix \(e+xE_{i,j}\), and \(t_{ij}\) denotes \(t_{ij}(1)\). Let \(\mathbf{P}\) be the subsemigroup of \(G_{n}(R)\) generated by all matrices \(\sigma\) (\(\sigma\in S_{n}\)), \(t_{ij}(x)\) (\(x\in R_{+},i\neq j\)), and \(\textrm{diag}[\alpha_{1},\dots,\alpha_{n}]\in D_{n}(R)\).

We call the above matrices elementary.

Two matrices \(A,B\in G_{n}(R)\) are called \(\mathcal{P}\)-equivalent (see [5]), if there exist matrices \(A_{j}\in G_{n}(R)\), \(j=0,\dots,k\), such that \(A=A_{0}\), \(B=A_{k}\), and matrices \(P_{i},\widetilde{P}_{i},Q_{i},\widetilde{Q}_{i}\in\mathbf{P}\), \(i=0,\dots,k-1\), such that \(P_{i}A_{i}\widetilde{P}_{i}=Q_{i}A_{i+1}\widetilde{Q}_{i}\).

By \(GE_{n}^{+}(R)\), we denote the subsemigroup in \(G_{n}(R)\) generated by all matrices that are \(\mathcal{P}\)–equivalent to matrices of \(\mathbf{P}\).

If \(G\) is some semigroup (for example, \(G=R_{+}^{*}\), \(G_{n}(R)\), \(GE_{n}^{+}(R)\)), then a homomorphism \({\lambda\colon G\to G}\) is called a central homomorphism of \(G\) in the case if \({\lambda(G)\subset Z(G)}\). The mapping \(\Omega\colon X\mapsto\) \(\lambda(X)\cdot X\), where \(\lambda(\cdot)\) is a central homomorphism, is called a central homothety.

For each matrix \(M\in\Gamma_{n}(R)\), let \(\Phi_{M}\) denote an automorphism of the semigroup \(G_{n}(F)\) such that \(\Phi_{M}(X)=MXM^{-1}\) for all \(X\in G_{n}(F)\) (inner automorphism).

For every \(y(\cdot)\in\textrm{End}(R_{+})\), we denote by \(\Phi^{y}\) the endomorphism of \(G_{n}(R)\) such that \(\Phi^{y}(X)=\Phi^{y}((x_{ij}))=(y(x_{ij}))\) for all \(X=(x_{ij})\in G_{n}(R)\) (ring endomorphism).

Let us take the ring endomorphism \(\Phi^{y}\). If we have some homomorphism \(\lambda\colon G_{n}(R)\to Z_{+}^{*}(y(R_{+}))\), then the mapping \(\Omega^{y}\colon X\mapsto\lambda(X)\cdot\Phi^{y}(X)\) is called central-ring endomorphism.

3 ACTION OF THE ENDOMORPHISM ON ELEMENTARY MATRICES

Let some endomorphism \(\Phi\) be given. Let us show how this endomorphism acts on elementary matrices with nonnegative elements.

We estimate how the image of the endomorphism \(\Phi\) is related to the image of the matrices \(t_{ij}(x)\).

Lemma 1. If \(\Phi(t_{ij})=e\) for at least one pair \(i\neq j\), then \(\Phi(t_{kl}(x))=e\) for any pair \(k\neq l\) and for any \(x>0\).

Proof. Let \(\Phi(t_{ij}(x))=A\). Consider fixed \(k\) and \(l\). Let the elements \(t_{ij}(x)\) and \(t_{ik}(x)\) be conjugate to each other by the permutation matrix \(\sigma\), that is, the equality \(\sigma t_{ij}(x)\sigma^{-1}=t_{kl}(x)\) holds. Due to the fact that all \(t_{ij}\) are conjugate to each other and \(\Phi(t_{ij})=e\), we have \(\Phi(t_{kl})=e\) for any \(k,l\). Now, from the relation \(t_{ij}(x)t_{jk}(1)=t_{jk}(1)t_{ij}(x)t_{ik}(x)\), we obtain

$$\Phi(t_{ij}(x)t_{jk}(1))=\Phi(t_{jk}(1)t_{ij}(x)t_{ik}(x)),$$
$$\Phi(t_{ij}(x))\Phi(t_{jk}(1))=\Phi(t_{jk}(1))\Phi(t_{ij}(x))\Phi(t_{ik}(x)),$$
$$Ae=eA\Phi(\sigma t_{ij}(x)\sigma^{-1}),$$
$$A=ABAB^{-1},$$
$$e=A.$$

This means that \(\Phi(t_{ij}(x))=e\).

Lemma 2. Let the endomorphism \(\Phi\) be such that there exist \(i\), \(j\) such that \(\Phi(t_{ij})=e\). Then the image under the endomorphism of the semigroup \(GE_{n}^{+}\) is a subgroup of \(\Gamma_{n}\) .

Proof. According to Lemma 1, all \(t_{ij}(x)\) go into \(e\). In this case, \(\Phi(GE_{n}^{+})=\Phi({\mathbf{P}})\). Indeed, let the matrix \(M\) be \(\mathcal{P}\)-equivalent to the matrix \(M_{1}\in{\mathbf{P}}\). Then the transition from \(\Phi(M_{1})\) to \(\Phi(M)\) can be made by using invertible matrices. Hence, \(\Phi(M)\in\Phi(P)\). In turn, \(\Phi(P)\) lies in \(\Gamma_{n}\).

As the kernel of endomorphism of semigroups with identity, we understand the complete preimage of unity. It is easy to verify that the kernel of the endomorphism \(\Phi\) bounded on \(\Gamma_{n}\) is a normal subgroup of the group \(\Gamma_{n}\). Thus, in the case where \(\Phi(t_{ij})=e\), knowing the description of all normal subgroups in \(\Gamma_{n}\), we can understand what are the images of the semigroup \(GE_{n}^{+}(R)\).

Classification of normal subgroups of the group \(\Gamma_{n}\). Let us proceed to the description of all normal subgroups of the group \(\Gamma_{n}\). It is easy to see that this group consists of all monomial matrices; i.e., \(\Gamma_{n}(R)=D_{n}(R)\cdot S_{n}\).

In fact, if a matrix with nonnegative elements has at least two nonzero elements in its row, then its inverse matrix cannot consist only of nonnegative elements, since in this case the product of these two matrices gives a nonzero element outside the diagonal.

Since \(\Gamma_{n}(R)\) is a group and \(\textrm{Ker}\Phi\cap\Gamma_{n}(R)\triangleleft\Gamma_{n}(R)\), in order to describe the action of \(\Phi\) on the subgroup \(\Gamma_{n}(R)\) it suffices to describe the normal subgroups of the group of monomial matrices.

It is also easy to observe that the normal subgroup of the group \(\Gamma_{n}(R)\) is either a subgroup of diagonal matrices or a semidirect product of the groups \(N_{S}\) and \(N_{D}\).

Kernel triviality on the group \(\Gamma_{n}\). Let us consider how the kernel of endomorphism on monomial matrices and the image of the element \(t_{ij}\) are related. As stated earlier, \(N_{S}\) is a normal subgroup of the group \(S_{n}\). Nontrivial normal subgroups are the subgroup of even permutations for \(n\geqslant 3\) and the Klein subgroup for \(n=4\). We denote the group of permutation matrices corresponding to even permutations by \(A_{n}\) and the subgroup of permutation matrices of the group \(S_{4}\) corresponding to the Klein group by \(V_{4}\).

Lemma 3. Let \(N_{S}=S_{n}\) or \(N_{S}=A_{n}\). Then \(\Phi(t_{ij}(x))=e\).

Proof. Let \(\Phi(t_{12})=A\). Then \(\Phi(t_{23})=\Phi(\sigma_{(123)}t_{12}\sigma_{(123)^{-1}})=\Phi(t_{12})=A\). We now use the relation \(t_{13}t_{23}t_{12}=t_{12}t_{23}\). Since the matrix \(A\) is invertible in \(\mathrm{GL}_{n}(R)\), it follows that \(\Phi(t_{13})=e\). Given that all matrices \(t_{ij}\) are conjugate by permutation matrices, we have \(\Phi(t_{ij})=e\); consequently, \(\Phi(t_{ij}(x))=e\). Therefore, by Lemma 2, the image of \(G_{n}(R)\) is a group.

Lemma 4. Let \(N_{S}=V_{4}\). Then \(\Phi(t_{ij}(x))=e\).

Proof. Since the element \(\sigma_{(12)(34)}\) lies in the subgroup \(V_{4}\), we get that, for some \(a,b,c\), and \(d\), the kernel of \(\Phi\) contains a matrix \(A\) of the form \(A=\sigma_{(12)(34)}\alpha\), where \(\alpha=\textrm{diag}[a,b,c,d]\).

The kernel also contains \(A^{-1}\) and all matrices conjugate to \(A\). That is, the matrix \(B=[\alpha_{1}=2]A\left[\alpha_{1}=\frac{1}{2}\right]=[\alpha_{1}=2]\sigma_{(12)(34)}\alpha\left[\alpha_{1}=\frac{1}{2}\right]=\sigma_{(12)(34)}\alpha\left[\alpha_{1}=\frac{1}{2},\alpha_{2}=2\right]\) lies in the kernel.

We obtain that the diagonal matrix \(\beta=\textrm{diag}\left[\frac{1}{2},2,1,1\right]=A^{-1}B\) also lies in the kernel.

Using the expression \(t_{23}^{2}=t_{23}(2)=\beta t_{23}\beta^{-1}\), we conclude that \(\Phi(t_{23})=\Phi(t_{23})^{2}\); i.e., \(\Phi(t_{23})=e\). According to Lemma 1, we have \(\Phi(t_{ij}(x))=e\).

It follows from Lemmas 3 and 4 that, if any of the elements \(t_{ij}\) does not lie in the kernel of the endomorphism \(\Phi\), then \(\textrm{Ker}\Phi\cap S_{n}=e\).

Action of the endomorphism on substitution matrices. Consider the restriction \(\varphi\) of the endomorphism \(\Phi\) of the semigroup \(G_{n}(R)\) to the group \(\Gamma_{n}(R)\). It is an endomorphism of the group. Let \(N=N_{S}\rightthreetimes N_{D}=\textrm{Ker}\varphi\). Since we have described the normal subgroups of \(\Gamma_{n}(R)\), we can describe endomorphisms of \(\Gamma_{n}(R)\) as well.

Lemma 5. Let \(N_{S}=\{e\}\). Then there exists an automorphism \(\psi\in\textrm{Aut}(S_{n})\) such that, for each \(\sigma\in S_{n}\), we have \(\Phi(\sigma)=\alpha_{\sigma}\psi(\sigma),\) where \(\alpha_{\sigma}\in D_{n}(R)\).

Proof. Let \(\Phi(\sigma)=\alpha_{\sigma}\psi(\sigma)\), where \(\psi\) is some self-mapping of \(S_{n}\). We show that \(\psi\) is an automorphism of the group \(S_{n}\).

Indeed, the following chain of equalities holds for all \(\sigma_{1},\sigma_{2}\in S_{n}\):

$$\alpha_{\sigma_{1}\cdot\sigma_{2}}\cdot\psi(\sigma_{1}\sigma_{2})=\Phi(\sigma_{1}\cdot\sigma_{2})=\Phi(\sigma_{1})\cdot\Phi(\sigma_{2})=\alpha_{\sigma_{1}}\cdot\psi(\sigma_{1})\cdot\alpha_{\sigma_{2}}\cdot\psi(\sigma_{2})=\alpha_{\sigma_{1}}\alpha_{\sigma_{2}}^{\prime}\cdot\psi(\sigma_{1})\psi(\sigma_{2}).$$

This means that \(\psi(\sigma_{1})\psi(\sigma_{2})=\psi(\sigma_{1}\sigma_{2})\).

If \(\sigma_{1},\sigma_{2}\in S_{n}\), \(\sigma_{1}\neq\sigma_{2}\), and \(\sigma=\psi(\sigma_{1})=\psi(\sigma_{2})\), then \(\Phi(\sigma_{1}\cdot\sigma_{2}^{-1})=\alpha_{\sigma_{1}}\sigma\sigma^{-1}\alpha_{\sigma_{2}}^{-1}=\alpha_{\sigma_{1}}\alpha_{\sigma_{2}}^{-1}\in D_{n}(R).\)

Thus, we have \(\Phi(\rho)\in D_{n}(R)\) for some \(\rho\neq e\), which cannot be the case since the element \(\rho\) has finite order. It follows from all what has been said that \(\psi\) is an automorphism of the group \(S_{n}\).

Using the following lemma, we can prove the first part of the main theorem.

Lemma 6. Let \(N_{S}=\{e\}\) and \(\Phi(t_{ij})\neq e\) for some \(i,j\) . Then there exists an inner automorphism \(\Phi_{M}\) of the semigroup \(G_{n}(R)\) such that \(\Phi_{M}\Phi(\sigma)=\sigma\) for any \(\sigma\in S_{n}\) .

Proof. The proof is similar to the proof of Lemma 8 from [8, pp. 595–596].

Action of the endomorphism \(\Phi^{\prime}\) on diagonal matrices. In what follows, we will consider only endomorphisms \(\Phi\) not transforming the elements \(t_{ij}\) to unity.

According to the previous lemma, for such an endomorphism \(\Phi\) there exists an endomorphism \(\Phi^{\prime}\) such that \(\Phi^{\prime}=\Phi_{M}\circ\Phi\) and \(\Phi^{\prime}(\sigma)=\sigma\) for any \(\sigma\in S_{n}\). Further we will consider this endomorphism \(\Phi^{\prime}\).

Lemma 7. Let \(\Phi^{\prime}\) be an endomorphism of the semigroup \(G_{n}(R)\) such that \(\Phi^{\prime}(\sigma)=\sigma\) for any \(\sigma\in S_{n}\) . Then

(1) \(\Phi^{\prime}(D_{n}(R))\subset D_{n}(R)\);

(2) \(\Phi^{\prime}(\textrm{diag}[x,1,\dots,1])=\textrm{diag}[y,z,\dots,z]\).

Proof. We denote by \([\alpha_{i}=x]\) the diagonal matrix whose diagonal contains the element \(x\) at the position \(i\) and units \(x\) on the rest of the diagonal. Consider the matrix \([\alpha_{1}=x]=\textrm{diag}[x,1,\dots,1]\). Let \(\Phi^{\prime}([\alpha_{1}=x])=D\sigma\), where \(D\) is some diagonal matrix. The matrix \([\alpha_{1}=x]\) commutes with all the permutations that preserve \(1\) in place. Hence, its image will commute with all the images of these permutations. Whence we can conclude that \(\sigma=e\) and \(D=\textrm{diag}[y,z,\dots,z]\). A similar conclusion can be made regarding a matrix with a non-unit element in any (not necessarily first) place. These matrices generates the group of all diagonal matrices. Hence, the image of any diagonal matrix is diagonal.

We prove a lemma that defines the image of all diagonal matrices.

Lemma 8. Let the endomorphism \(\Phi^{\prime}\) transform all diagonal matrices to scalar ones. Then all the matrices \(t_{ij}\) lie in its kernel.

Proof. Consider the matrix \([\alpha_{1}=2]=\textrm{diag}[2,1,\ldots,1]\). Let the image of the matrix \([\alpha_{1}=2]\) be the matrix \(ae\). Then we obtain the chains of equalities

$$\Phi^{\prime}(t_{12})=\Phi^{\prime}\left(\left[\alpha_{3}=2]t_{12}[\alpha_{3}=\frac{1}{2}\right]\right)=\Phi^{\prime}([\alpha_{3}=2])\Phi^{\prime}(t_{12})\Phi^{\prime}\left(\left[\alpha_{3}=\frac{1}{2}\right]\right)$$
$${}=ae\Phi^{\prime}(t_{12})a^{-1}e=\Phi^{\prime}([\alpha_{1}=2])\Phi^{\prime}(t_{12})\Phi^{\prime}\left(\left[\alpha_{1}=\frac{1}{2}\right]\right)$$
$${}=\Phi^{\prime}\left([\alpha_{1}=2]t_{12}\left[\alpha_{1}=\frac{1}{2}\right]\right)=\Phi^{\prime}(t^{2}_{12})=(\Phi^{\prime}(t_{12}))^{2}.$$

Therefore, \(\Phi^{\prime}(t_{12})=e\). According to Lemma 1, all the matrices \(t_{ij}\) lie in the kernel.

We formally obtained that, in the case \(\Phi^{\prime}(t_{ij})\neq e\) under consideration, the image of diagonal matrices cannot consist only of scalar matrices.

Construction of an endomorphism of the semiring of nonnegative elements.

Lemma 9. If an endomorphism \(\Phi^{\prime}\in\textrm{End}(G_{n}(R))\) be such that \(\Phi^{\prime}(\sigma)=\sigma\) for any \(\sigma\in S_{n}\), then, for all \(X\in G_{2}(R)\),

$$\Phi^{\prime}\begin{pmatrix}X&\\ &I_{n-2}\end{pmatrix}=\begin{pmatrix}Y&\\ &aI_{n-2}\end{pmatrix},$$

where \(Y\in G_{2}(R)\), \(a\in R^{*}\).

Proof. Note that this matrix commutes with the permutation matrices preserving \(1\) and \(2\). The original matrix also commutes with all diagonal matrices of the form \(\textrm{diag}[1,\dots,t,\dots,1]\) and there exists \(t\) such that its image does not lie at the center. This immediately implies the assertion of the lemma.

Lemma 10. Let \(\Phi^{\prime}\) be an endomorphism of the semigroup \(G_{n}(R)\) such that \(\Phi^{\prime}(\sigma)=\sigma\) for any \(\sigma\in S_{n}\) and \(\Phi^{\prime}(t_{ij})\neq e\). Then \(\Phi^{\prime}(t_{12})=t_{12}\).

Proof. Note that \(\Phi^{\prime}(t_{12})\), \(\Phi^{\prime}(t_{13})\), and \(\Phi^{\prime}(t_{23})\) are related by the following relations:

(1) \(\Phi^{\prime}(t_{13})=\Phi^{\prime}(\sigma_{(23)}t_{12}\sigma_{(23)})=\sigma_{(23)}\Phi^{\prime}(t_{12})\sigma_{(23)}\);

(2) \(\Phi^{\prime}(t_{23})=\Phi^{\prime}(\sigma_{(123)}t_{12}\sigma_{(321)})=\sigma_{(123)}\Phi^{\prime}(t_{12})\sigma_{(321)}\).

We consider only the first three columns and three rows. According to Lemma 9, we obtain for the first three columns and rows that

$$\Phi^{\prime}(t_{12})=\begin{pmatrix}x&y&\\ z&t&\\ &&a\end{pmatrix},\Phi^{\prime}(t_{13})=\begin{pmatrix}x&&y\\ &a&\\ z&&t\end{pmatrix},\Phi^{\prime}(t_{23})=\begin{pmatrix}a&&\\ &x&y\\ &z&t\end{pmatrix}.$$

For the above elements, we have \(t_{12}t_{13}=t_{13}t_{12}\). Equating the values in the cell \((3,2)\), we find that \(zy=0\). That is, either \(z=0\) or \(y=0\). Therefore, we have \(t\neq 0\) and \(x\neq 0\) since, otherwise, the matrix \(\Phi^{\prime}(t_{12})\) would contain a column or a row of zeros, i.e., it would be degenerate.

In view of the equality \(t_{12}t_{23}=t_{23}t_{12}t_{23}\), we have

$$\begin{pmatrix}xa&yx&y^{2}\\ za&tx&ty\\ &az&at\end{pmatrix}=\begin{pmatrix}ax^{2}&aya&axy\\ xzx+yaz&xta&xzy+yat\\ z^{2}x+taz&zta&z^{2}y+tat\end{pmatrix}=\begin{pmatrix}ax^{2}&aya&axy\\ xzx&xta&yat\\ z^{2}x+taz&zta&tat\end{pmatrix}.$$

From the equality of the elements at the position \((3,3)\), we obtain \(at=tat\). The nondegeneracy of \(t\) and \(a\) implies \(t=1\).

We consider the elements located at the position \((3,1)\). We obtain \(z^{2}x+az=0\) or \(z^{2}x+zx=zx(z+1)=0\). Since \(x\neq 0\) and \(z\geqslant 0\), the equality \(zx(z+1)=0\) is possible only for \(z=0\).

The equality \(t_{12}t_{23}=t_{23}t_{12}t_{23}\) in matrix form will look as follows:

$$\begin{pmatrix}xa&yx&y^{2}\\ &x&y\\ &&a\end{pmatrix}=\begin{pmatrix}ax^{2}&aya&axy\\ &xa&ya\\ &&a\end{pmatrix}.$$

It follows from the equality \(xa=x\) at the position \((2,2)\) and \(x\neq 0\) that \(a=1\). By virtue of the equality \(xa=ax^{2}\) at the position \((1,1)\), we have \(x=1\). That is, \(y^{2}=y\) is written at the position \((1,3)\). This means that \(y=1\) or \(y=0\). In the case \(y=0\), we obtain that \(\Phi^{\prime}(t_{12})=e\), i.e., \(y=1\) and \(\Phi^{\prime}(t_{12})=t_{12}\).

Lemma 11. Let an endomorphism \(\Phi^{\prime}\in\textrm{End}(G_{n}(R))\) be such that \(\Phi^{\prime}(\sigma)=\sigma\) for any \(\sigma\in S_{n}\). Then \(\Phi(t_{12}(x))=t_{12}(b(x))\), where \(b\) is an endomorphism of the semiring \(R_{+}\) .

Proof. The proof is similar to the proof from [2, pp. 1877–1879].

Theorem. Let \(\Phi\) be an endomorphism of the semigroup \(G_{n}(R)\) for \(n\geqslant 3\) , where \(R\) is a linearly ordered, not necessarily commutative, ring with 1/2. Then one of the two conditions is fulfilled:

(1) the image of the semigroup \(GE_{n}^{+}(R)\) is some subgroup of the group \(\Gamma_{n}(R)\); in this case, all the matrices \(t_{ij}(x)\) are transformed into the identity one;

(2) there exist \(M\in\Gamma_{n}(R)\) and a central-ring endomorphism \(\Omega^{b}\) such that \(\Phi\) coincides with \({\Phi_{M}\circ\Omega^{b}}\) on the semigroup \(GE_{n}^{+}(R)\) .

Proof. To proof the first case, we consider \(N_{S}\). If \(N_{S}\) is a nontrivial subgroup, then, according to Lemmas 3 and 4, we obtain that \(\Phi(t_{ij})=e\).

In the case where \(n=6\), \(N_{S}=\{e\}\), and the automorphism \(\psi\colon S_{n}\to S_{n}\) is not standard, we obtain as a consequence of Lemma 6 that \(\Phi(t_{ij})=e\).

In both cases, according to Lemma 2, the image of the whole semigroup \(GE_{n}^{+}(R)\) is some subgroup of the group \(\Gamma_{n}(R)\); in this case, all the matrices \(t_{ij}(x)\) become the identity one.

It remains to consider the case where \(\Phi(t_{ij})\neq e\).

According to Lemma 6, there exists an inner automorphism \(\Phi_{M}\) of the semigroup \(G_{n}(R)\) such that \(\Phi_{M}\Phi(\sigma)=\sigma\) for any \(\sigma\in S_{n}\).

By Lemma 11, there exists an endomorphism \(b\) of the semiring \(R_{+}\) such that \(\Phi(t_{ij}(x))=t_{ij}(b(x))\). Note that the endomorphism \(\Phi^{\prime\prime}=\Phi_{M}\circ\Phi^{b}\) coincides with the endomorphism \(\Phi\) on the semigroup generated by the permutation matrices and \(t_{ij}(x)\).

Define \(\Phi^{\prime}=(\Phi_{M})^{-1}\circ\Phi\). Then the image of the diagonal matrix is as follows: \(\Phi^{\prime}([\alpha_{1}=x])=\Phi^{\prime}(\textrm{diag}[x,1,\ldots,1])=\textrm{diag}[\xi(x),\gamma(x),\dots,\gamma(x)].\)

Note that the matrix \([\alpha_{1}=x]\) commutes with any matrix of the form \(\begin{pmatrix}1&\\ &A_{n-1}\end{pmatrix}\). This implies that \(\gamma(x)\) commutes with \(b(R_{+})\); i.e., \(\gamma(x)\in Z_{+}^{*}(R_{+})\).

As known, \(\Phi^{b}([\alpha_{1}=x])=\textrm{diag}[b(x),1,\ldots,1]\). Using the relations \([\alpha_{1}=x]t_{12}(1)[\alpha_{1}=x^{-1}]=t_{12}(x)\) and \(\Phi^{\prime}(t_{12}(x))=\Phi^{b}(t_{12}(x))\), we obtain \(\xi(x)(\gamma(x))^{-1}=b(x)\). Then we have \(\Phi^{\prime}([\alpha_{1}=x])=\gamma(x)\Phi^{b}([\alpha_{1}=x])\). Note that the mapping \(\gamma(\cdot)\colon R_{+}\to R_{+}\) is multiplicative.

If \(\alpha=\textrm{diag}[x_{1},\ldots,x_{n}]\in D_{n}(R)\), then \(\Phi^{\prime}(\alpha)=\gamma(x_{1}\ldots x_{n})\Phi^{b}(\alpha)\).

It is clear that any matrix \(A\in\mathbf{P}\) can be represented as \(A=\textrm{diag}[\alpha_{1},\dots,\alpha_{n}]A_{1}\ldots A_{k},\) where \(\alpha_{1},\ldots,\alpha_{n}\in R_{+}\) and \(A_{1},\ldots,A_{k}\in\{\sigma,\ t_{ij}(x)|\sigma\in S_{n},\ x\in R_{+},\ i,j=1,\dots,n,\ i\neq j\}\). Then \(\Phi^{\prime}(A)=\gamma(\alpha_{1}\dots\alpha_{n})\cdot\Phi^{b}(A).\)

Now we introduce the mapping \(\overline{\gamma}(\cdot)\colon\mathbf{P}\to R_{+}\) using the following rule: if \(A\in\mathbf{P}\) and \(A=\textrm{diag}[\alpha_{1},\dots,\alpha_{n}]A_{1}\dots A_{k}\), where \(A_{1},\dots,A_{k}\in\{\sigma,\ t_{ij}(x)|\sigma\in S_{n},\ x\in R_{+},\ i,j=1,\dots,n,\ i\neq j\}\), then \(\overline{\gamma}(A)=\gamma(\alpha_{1}\ldots\alpha_{n})\). Since \(\overline{\gamma}(AA^{\prime})\Phi^{b}(AA^{\prime})=\Phi^{\prime}(AA^{\prime})=\Phi^{\prime}(A)\Phi^{\prime}(A^{\prime})=\overline{\gamma}(A)\Phi^{b}(A)\cdot\overline{\gamma}(A^{\prime})\times\Phi^{b}(A^{\prime})=\overline{\gamma}(A)\overline{\gamma}(A^{\prime})\Phi^{\prime}(AA^{\prime}),\) we find that \(\overline{\gamma}\) is a homomorphism \(\mathbf{P}\to R_{+}\).

Now we see that the endomorphism \(\Phi\) coincides with the endomorphism \(\Phi_{M}\circ\Omega^{b}\) on the semigroup \(\mathbf{P}\), where \(\Omega^{b}(A)=\overline{\gamma}(A)\cdot\Phi^{b}(A)\) for all \(A\in\mathbf{P}\).

Let \(B\in GE_{n}^{+}(R)\). Then the matrix \(B\) is \(\mathcal{P}\)-equivalent to some matrix \(A\in\mathbf{P}\). It is easy to see that we can extend the mapping \(\overline{\gamma}(\cdot)\colon\mathbf{P}\to R_{+}^{*}\) to a certain mapping \(\lambda(\cdot)\colon GE_{n}^{+}(R)\to R_{+}^{*}\), such that \(\Phi^{\prime}(B)=\lambda(B)\cdot\Phi^{b}(B)\) for each \(B\in GE_{n}^{+}(R)\). As a result, we obtain \(\Phi(B)=\Phi_{M}(B)\Omega^{b}(B)\), where \(\Omega^{b}\colon B\mapsto\lambda(B)\cdot\Phi^{b}(B)\). The theorem is proved.

The author would like to thank the referee for their attentive attitude to the paper and valuable comments.