MATHEMATICS

I. S. Ponizovskii

INVERSE SEMIGROUPS WITH A FINITE NUMBER OF IDEMPOTENTS

(Presented by Academician A. I. Mal'cev, 12 XII 1961)

Let \(\mathfrak K\) denote the class of inverse semigroups with a finite number of idempotents. To simplify the formulations we shall assume that every semigroup in \(\mathfrak K\) contains a zero. For semigroups of the class \(\mathfrak K\) we shall briefly describe: their structure, the automorphism group, the semigroup of related one-sided shifts \((^1)\), homomorphisms, and the construction procedure. The latter, together with the isomorphism theorem (Theorem 2), also gives at the same time a description of semigroups of the class \(\mathfrak K\).

Let \(\mathfrak E\) denote the class of completely simple semigroups whose defining matrix is finite and may be chosen to be the identity matrix. Everywhere below \(\mathfrak A\) denotes a semigroup with zero \(0\). We shall write \(A \to B\) if \(B\) follows from \(A\).

1. The foundation of the theory set forth is

Theorem 1. \(\mathfrak A\) belongs to the class \(\mathfrak K\) if and only if \(\mathfrak A\) has a finite ideal composition series

\[ \mathfrak A=\mathfrak A_1 \supset \mathfrak A_2 \supset \cdots \supset \mathfrak A_n \supset \mathfrak A_{n+1}=0, \tag{1} \]

all of whose factors belong to \(\mathfrak E\). In particular, \(\mathfrak E \subset \mathfrak K\).

1, A. Everywhere (1) will denote an ideal composition series of \(\mathfrak A\); the factors of (1) will be denoted by \(\overline{\mathfrak A}_i\) \((i=1,2,\ldots,n)\), \(\overline{\mathfrak A}_i=\mathfrak A_i/\mathfrak A_{i+1}\).

1. Let \(P\) be a field, and let \(P(\mathfrak B)\) be the semigroup algebra of the semigroup \(\mathfrak B\) over \(P\)*.

To the series (1) there corresponds the ideal series \(P(\mathfrak A)\)

\[ P(\mathfrak A)=P(\mathfrak A_1)\supset P(\mathfrak A_2)\supset \cdots \supset P(\mathfrak A_n)\supset P(\mathfrak A_{n+1})=0. \tag{2} \]

Here
\[ P(\mathfrak A_i)/P(\mathfrak A_{i+1}) \simeq P(\overline{\mathfrak A}_i) \quad (i=1,2,\ldots,n). \]
Let \(\alpha_i\) denote the ideal homomorphism of \(\mathfrak A\) generated by \(\mathfrak A_i\), a term of (1); the induced homomorphism of \(P(\mathfrak A)\) by \(\alpha_i\) will also be denoted by \(\alpha_i\). Then
\[ P(\overline{\mathfrak A}_i)=\alpha_{i+1}P(\mathfrak A_i)\quad (i=1,2,\ldots,n). \]
Let \(E_i\) denote the sum in \(P(\mathfrak A)\) of all idempotents from
\[ \mathfrak A_i \setminus \mathfrak A_{i+1}\quad (i=1,2,\ldots,n). \]
Let now \(k\le i\le n\). Put
\[ x\in \mathfrak A_k\setminus \mathfrak A_{k+1}\to \varphi_{ik}x=\alpha_{i+1}(xE_i). \]

Then \(\varphi_{ik}\) is a mapping of \(\mathfrak A_k\setminus \mathfrak A_{k+1}\) into \(P(\overline{\mathfrak A}_i)\). If we assume that \(\alpha_{i+1}y=y\) for \(y\in \mathfrak A_i\setminus \mathfrak A_{i+1}\) (as is done, for example, in (2)), then \(\varphi_{ii}\) is simply the embedding of \(\mathfrak A_i\setminus \mathfrak A_{i+1}\) into \(P(\overline{\mathfrak A}_i)\). We assemble the mappings \(\varphi_{ik}\) \((k\le i)\) into the matrix

\[ \Phi= \begin{pmatrix} \varphi_{11} & 0 & \cdots & 0\\ \varphi_{21} & \varphi_{22} & \cdots & 0\\ \cdots & \cdots & \cdots & \cdots\\ \varphi_{n1} & \varphi_{n2} & \cdots & \varphi_{nn} \end{pmatrix}. \]

* That is, a basis of \(P(\mathfrak B)\) consists of all nonzero elements of \(\mathfrak B\), which are multiplied in the same way as in \(\mathfrak B\), and the zero of \(P(\mathfrak B)\) is identified with the zero of \(\mathfrak B\) (if \(\mathfrak B\) has a zero).

It turns out that specifying the factors \(\overline{\mathfrak A}_i\) \((i=1,2,\ldots,n)\) of the series (1) and the matrix \(\Phi\) completely determines \(\mathfrak A\). Therefore we shall write
\[ \mathfrak A=(\overline{\mathfrak A}_1,\ldots,\overline{\mathfrak A}_n,\Phi). \]

1. Let us show how the question of isomorphism of semigroups from \(\mathfrak K\) is solved. A necessary condition for isomorphism is equality of the lengths of the ideal composition series of the semigroups. Therefore, considering \(\mathfrak A,\mathfrak B\in\mathfrak K\), we may assume:
   \[ \mathfrak A=(\overline{\mathfrak A}_1,\ldots,\overline{\mathfrak A}_n,\Phi),\qquad \mathfrak B=(\overline{\mathfrak B}_1,\ldots,\overline{\mathfrak B}_n,\Psi). \]

Denote by \(L=(\lambda_{ip})\) an \(n\times n\)-matrix in each row and each column of which there is exactly one nonzero element. Moreover, if \(\lambda_{ip}\ne0\), then \(\lambda_{ip}\) is an isomorphism of \(\overline{\mathfrak A}_i\) onto \(\overline{\mathfrak B}_p\), extended linearly to an isomorphism of \(P(\overline{\mathfrak A}_i)\) onto \(P(\overline{\mathfrak B}_p)\). In this case \(L\Phi\) and \(\Psi L\) have a natural meaning.

Theorem 2. For \(\mathfrak A\) and \(\mathfrak B\) to be isomorphic it is necessary and sufficient that there exist a matrix \(L\) such that
\[ L\Phi=\Psi L. \]

1. Let us give a vector interpretation of the semigroup \(\mathfrak A\in\mathfrak K\), \(\mathfrak A=(\overline{\mathfrak A}_1,\ldots,\overline{\mathfrak A}_n,\Phi)\). Denote by \(V(\mathfrak A)\) the totality of all \(n\)-dimensional vectors of the form
   \[ \overline{x}_i=(0,\ldots,0,x_i,\ldots,\varphi_{ki}x_i,\ldots), \]
   where \(x_i\in\mathfrak A_i\setminus\mathfrak A_{i+1}\), \(\{\overline{x}_i\}_p=0\) for \(p\le i\), \(\{\overline{x}_i\}_p=\varphi_{pi}x_i\) for \(p\ge i\).* The zero standing in the place of the \(k\)-th component \(x_i\) is regarded as the zero of \(P(\overline{\mathfrak A}_k)\). Then
   \[ \overline{x}\in V(\mathfrak A)\to \{\overline{x}\}_k\in P(\overline{\mathfrak A}_k). \]
   This makes it possible to define multiplication in \(V(\mathfrak A)\):
   \[ \overline{x},\overline{y}\in V(\mathfrak A)\to \{\overline{x}\cdot\overline{y}\}_k=\{\overline{x}\}_k\cdot\{\overline{y}\}_k. \tag{3} \]

Define also a mapping \(\omega\) of \(V(\mathfrak A)\) onto \(\mathfrak A\), putting:
\[ \omega\overline{x}_i=x_i. \tag{4} \]

Theorem 3. With respect to the multiplication (3), \(V(\mathfrak A)\) forms a semigroup isomorphic to \(\mathfrak A\). An isomorphism of \(V(\mathfrak A)\) onto \(\mathfrak A\) may be given by the mapping \(\omega\).

1. Let us describe the automorphisms of \(\mathfrak A\in\mathfrak K\), \(\mathfrak A=(\overline{\mathfrak A}_1,\ldots,\overline{\mathfrak A}_n,\Phi)\). For this purpose we replace in item 3 the semigroup \(\mathfrak B\) by \(\mathfrak A=(\overline{\mathfrak A}_1,\ldots,\overline{\mathfrak A}_n,\Phi)\). Then the nonzero element \(\lambda_{ip}\) of the matrix \(L\) will be an isomorphism of \(\overline{\mathfrak A}_i\) onto \(\overline{\mathfrak A}_p\). Define a mapping \(A\) of the semigroup \(\mathfrak A\), putting for \(x\in\mathfrak A\)
   \[ Ax=\omega(L\cdot\overline{x}). \tag{5} \]

Here \(L\cdot\overline{x}\) is the usual product of a matrix and a vector, and \(\omega\) is defined by (4).

Theorem 4. The automorphism group of \(\mathfrak A\) is isomorphic to the multiplicative group \(\mathfrak L\) of all matrices \(L\) such that \(L\Phi=\Phi L\). Every automorphism \(A\) of the semigroup \(\mathfrak A\) has the form (5), where \(L\in\mathfrak L\), and conversely: if \(L\in\mathfrak L\), then the mapping \(A\), defined by (5), is an automorphism of \(\mathfrak A\).

1. Let us call \(\delta\) a two-sided operator in the semigroup \(\mathfrak B\) if for all \(x\in\mathfrak B\) there are defined \(\delta x,x\delta\in\mathfrak B\) (not necessarily \(\delta x=x\delta\)). Equality and multiplication of two-sided operators in \(\mathfrak B\) are defined naturally. By \(T\mathfrak B\) we denote the semigroup of all such two-sided operators \(\delta\) in \(\mathfrak B\), so that
   \[ x,y\in\mathfrak B\to \delta(xy)=(\delta x)y,\qquad (x\delta)y=x(\delta y),\qquad (xy)\delta=x(y\delta). \]
   \(T\mathfrak B\) is nothing other than the semigroup of pairs of linked one-sided translations of \(\mathfrak B\) ([1], p. 173). If \(\mathfrak B\in\mathfrak K\), then one may regard \(\mathfrak B\) as a two-sided ideal of \(T\mathfrak B\). Note the proposition:

Theorem 5. \(\mathfrak A\in\mathfrak K\to T\mathfrak A\in\mathfrak K\).

\[ \text{* By } \{x\}_p \text{ is denoted the } p\text{-th component of the vector } \overline{x}. \]

1. We shall call a semigroup \(\mathfrak S\) an extension of a semigroup \(\mathfrak X\) by means of a semigroup \(\mathfrak D\), if \(\mathfrak X\) is a two-sided ideal of \(\mathfrak S\) and \(\mathfrak S/\mathfrak X \simeq \mathfrak D\). For the construction of semigroups of the class \(\mathfrak K\) it is enough to be able to construct extensions of any concrete semigroup from \(\mathfrak K\) by means of a semigroup of the class \(\mathfrak C\) (this follows from Theorem 1). These extensions are described by

Theorem 6\(^*\). Let \(\mathfrak A \in \mathfrak K\), \(\mathfrak B \in \mathfrak C\); \(\theta\) be the zero of \(\mathfrak B\); \(\Omega=\mathfrak B\setminus \theta\). As noted in item 6, we may regard \(\mathfrak A\) as a two-sided ideal of \(T\mathfrak A\). Suppose further that for a mapping \(\tau\) one has: a) either \(\tau\) is a homomorphism of \(\mathfrak B\) into some semigroup \(\mathfrak A_k\) such that \(\tau\mathfrak B\) can be equal to the zero of \(\mathfrak A_k\) only for \(k=n\); b) or \(\tau\) is a homomorphism of \(\mathfrak B\) into \(T\mathfrak A/\mathfrak A\), mapping the zero of \(\mathfrak B\) to the zero of \(T\mathfrak A/\mathfrak A\).

Denoting by \(\alpha\) the ideal homomorphism of \(T\mathfrak A\) generated by \(\mathfrak A\), and by \(\alpha_{k+1}\) the same as in item 2, put, in case a), \(\rho=\alpha_{k+1}^{-1}\tau\), and, in case b), \(\rho=\alpha^{-1}\tau\). We now define in \(\mathfrak S=\mathfrak A\cup\mathfrak C\) the multiplication “\(\circ\)” by setting:

\[ \begin{gathered} x,y\in\mathfrak A \to x\circ y=xy;\qquad x,y,xy\in\mathfrak C \to x\circ y=xy;\\ x,y\in\mathfrak C,\quad xy\notin\mathfrak C \to x\circ y=\rho x\cdot \rho y\quad **;\\ x\in\mathfrak C,\quad y\in\mathfrak A \to x\circ y=\rho x\cdot y,\qquad yx=y\cdot \rho x. \end{gathered} \tag{6} \]

Then \(\mathfrak S\), with respect to the multiplication “\(\circ\),” is a semigroup—an extension of \(\mathfrak A\) by means of \(\mathfrak B\). Conversely, every extension of \(\mathfrak A\) by means of \(\mathfrak B\) can be obtained by the method described in the theorem.

7, A. If \(\mathfrak S\) is an extension of \(\mathfrak A\) by means of \(\mathfrak B\) and the multiplication in \(\mathfrak S\) is defined by formulas (6), then we shall write \(\mathfrak S=(\mathfrak B,\mathfrak A,\rho)\).

7, B. The homomorphisms \(\mathfrak B,\mathfrak B\in\mathfrak C\) are described by the method \((^3)\). Further, for \(\mathfrak B\in\mathfrak C\) it is easy to construct \(T\mathfrak B\) ((1), Theorem 6). Therefore we shall have the possibility of an unrestricted application of the construction of Theorem 6 if we show how, knowing \(T\mathfrak B\) and \(T\mathfrak A\), one can construct \(T\mathfrak S\), \(\mathfrak S=(\mathfrak B,\mathfrak A,\rho)\). This construction is given by

Theorem 7. Let \(\mathfrak A,\mathfrak B,\mathfrak C\) be the same as in Theorem 6, and let \(\mathfrak S=(\mathfrak B,\mathfrak A,\rho)\). Suppose further that \(\delta_1\in T\mathfrak B\), \(\delta_2\in T\mathfrak A\), and that for \(x\in\mathfrak B\) the following conditions are satisfied:

\[ \delta_1x\in\mathfrak C \to \delta_2\cdot \rho x=\rho(\delta_1x);\qquad x\delta_1\in\mathfrak C \to \rho x\cdot \delta_2=\rho(x\delta_1), \]

\[ \delta_1x\notin\mathfrak C \to \delta_2\cdot \rho x\in\mathfrak A;\qquad x\delta_1\notin\mathfrak C \to \rho x\cdot \delta_2\in\mathfrak A. \]

Define in \(\mathfrak S\) a two-sided operator \(\delta\), setting:

\[ x\in\mathfrak A \to \delta x=\delta_2x;\qquad x\delta=x\delta_2; \]

\[ x,\delta_1x\in\mathfrak C \to \delta x=\delta_1x;\qquad x,x\delta_1\in\mathfrak C \to x\delta=x\delta_1; \]

\[ x\in\mathfrak C,\ \delta_1x\notin\mathfrak C \to \delta x=\delta_2\cdot \rho x;\qquad x\in\mathfrak C,\ x\delta_1\notin\mathfrak C \to x\delta=\rho x\cdot \delta_2. \]

Then \(\delta\in T\mathfrak S\), and every element \(\delta\) of \(T\mathfrak S\) can be obtained by the method described in the theorem.

1. We describe, in conclusion, the homomorphisms of semigroups of the class \(\mathfrak K\). It suffices to show how homomorphisms \(\mathfrak A\in\mathfrak K\) are constructed from homomorphisms \(\mathfrak A_i\) \((i=1,2,\ldots,n)\).

If \(\mu,\nu\) are homomorphisms of the semigroup \(\mathfrak S\) and \(\mu=\lambda\nu\), where \(\lambda\) is an isomorphism of \(\nu\mathfrak S\), then we shall write \(\mu\sim\nu\). The greatest common right divisor \((^2)\) of the homomorphisms \(\mu_1,\ldots,\mu_m\) of the semigroup \(\mathfrak S\) will be denoted by \(o.\ \mathrm{n}.\ \mathrm{d.}\ \mu_i\) \((i=1,2,\ldots,m)\).

* The construction of Theorem 6 is based on a construction proposed by Clifford \((^1)\).

** \(xy\notin\mathfrak C\), i.e. \(xy=\theta\) in the semigroup \(\mathfrak B\).

Theorem 8. Let \(\mathfrak A \in K\), and let \(\bar{\mu}_i\) be a homomorphism of \(\bar{\mathfrak A}_i\). In the notation of §2, for \(x \in \mathfrak A\) put

\[ \mu_i x = \bar{\mu}_i \alpha_{i+1}(xE_i). \tag{7} \]

Then: a) \(\mu_i\) is a homomorphism of \(\mathfrak A\); b) if \(\mu\) is a homomorphism of \(\mathfrak A\), then there exist homomorphisms \(\bar{\mu}_i\) of the semigroups \(\bar{\mathfrak A}_i\) \((i = 1, 2, \ldots, n)\) such that \(\mu \sim\) o. n. d. \(\mu_i\) \((i = 1, 2, \ldots, n)\), where \(\mu_i\) is obtained from \(\bar{\mu}_i\) by (7).

Received
27 XI 1961

CITED LITERATURE

1. A. H. Clifford, Trans. Am. Math. Soc., 68, No. 2, 165 (1950).
2. E. S. Lyapin, Semigroups, 1960.
3. L. M. Gluskin, Scientific Notes of the Kharkov Pedagogical Institute, 18, 41 (1956).