# The Regular Part of a Semigroup of Full Transformations with Restricted Range: Maximal Inverse Subsemigroups and Maximal Regular Subsemigroups of Its Ideals.

1. IntroductionLet T(X) be the set of all full transformations from a nonempty set X into itself. It is well-known that T(X) is a regular semigroup under composition of functions; see [1, p. 63]. Moreover, every semigroup can be embedded in T(X) for some nonempty set X; see [1, p. 7]. In [2, 3], Schein posed the problem to determine all maximal inverse subsemigroups of T(X); this problem is still unsolved. In 1976, Nichols [4] characterized one class of maximal inverse subsemigroup of T(X). Later in 1978, Reilly [5] has generalized Nichols construction and obtained a much wider class of maximal inverse subsemigroups of T(X). In 1999, Yang [6] described all of the maximal inverse subsemigroups of the finite symmetric inverse semigroup. Later in 2001, Yang [7] obtained the maximal subsemigroups of the finite singular transformation semigroups. In 2002, You [8] determined all the maximal regular subsemigroups of all ideals of the finite full transformation semigroup. In 2004, H. B. Yang and X. L. Yang [9] completely described the maximal subsemigroups of ideals of the finite full transformation semigroup. In 2014, Zhao et al. [10] showed that any maximal regular subsemigroup of ideals of the finite full transformation semigroup is idempotent generated. After that in 2015, East et al. [11] classified the maximal subsemigroups of T(X) when X is an infinite set containing certain subgroups of the symmetric group on X.

Let Y be a fixed nonempty subset of X and let T(X, Y) be the subsemigroup of T(X) of all elements with ranges contained in Y. In 1975, Symons [12] introduced and studied the semigroup T(X, Y). He described all the automorphisms of T(X, Y) and also determined the isomorphism theorem for two semigroups of type T(X, Y). In 2005, Nenthein et al. [13] characterized the regular elements of T(X, Y). In general, T(X, Y) is not a regular semigroup. In 2008, Sanwong and Sommanee [14] defined

F(X, Y) = {[alpha] [member of] T(X, Y): X[alpha] = Y[alpha]} (1)

and showed that F(X, Y) is the largest regular subsemigroup of T(X, Y). Obviously, F(X, Y) = T(X) when X = 7. Hence, we may regard F(X, Y) as a generalization of T(X). The authors also characterized Green's relations on T(X, Y) and gave one class of maximal inverse subsemigroups of T(X, Y). Later in 2009, Sanwong et al. [15] described all maximal and minimal congruences on T(X, Y). In 2011, Mendes-Gonfalves and Sullivan [16] obtained all the ideals of T(X, Y). In the same year, Sanwong [17] described Green's relations and ideals and all maximal regular subsemigroups of F(X, Y). The author also proved that every regular semigroup S can be embedded in F([S.sup.1],S). After that, in 2013, Sommanee and Sanwong [18] computed the rank of F(X, Y) when X is a finite set. They also obtained the rank and the idempotent rank of its ideals when X is finite. In 2014, Fernandes and Sanwong [19] calculated the rank of T(X, Y) when X is finite. In 2016, L. Sun and J. Sun [20] characterised the natural partial order on T(X, Y). In the same year, Tinpun and Koppitz [21] determined the relative rank of T(X, Y) modulo the semigroup S(X, Y) of all extensions of the bijections on Y when X is finite.

Let X be a chain and OT(X) the full order-preserving transformation semigroup on X. In 2011, Dimitrova and Koppitz [22] characterized the maximal regular subsemigroups of ideals of OT(X) when X is a finite chain. In 2015, Sommanee and Sanwong [23] investigated the regularity and Green's relations of the order-preserving transformation semigroup

OF(X, Y) = F(X, Y) [intersection] OT(X) = {[alpha] [member of] F(X, Y) : [alpha] is order-preserving}. (2)

They also proved that OF(X, Y) is idempotent generated when Y is finite.

Let V be a vector space and let T(V) denote the linear transformation semigroup from V into V. For a fixed subspace W of V, let T(V, W) be the semigroup consisting of all linear transformations from V into W. Recently in 2017, Sommanee and Sangkhanan [24] determined all the maximal regular subsemigroups of the semigroup

Q = {[alpha] [member of] T(V, W) : V[alpha] = W[alpha]}, (3)

when W is a finite dimensional subspace of V over a finite field.

In this paper, we describe the maximal inverse subsemigroups of F(X, Y) and completely determine all the maximal regular subsemigroups of its ideals.

2. Preliminaries and Notations

For all undefined notions, the reader is referred to [1].

For any set A, [absolute value of A] means the cardinality of the set A. If A is a subset of a semigroup S, then <A> denotes the subsemigroup of S generated by A. An element e of a semigroup S is called idempotent if [e.sup.2] = e. If U is a subset of a semigroup S, then E (U) denotes the set of all idempotents in the set U. An element a of a semigroup S is called regular if there exists x [member of] S such that a = axa. The semigroup S is called regular if all its elements are regular. If a is an element of a semigroup S, we say that a' is an inverse of a if a = aa'a and a' = a'aa'. We denote the set of inverses of an element a by V (a). A semigroup S is called an inverse semigroup if every element a in S has a unique inverse [a.sup.-1] in S; it is equivalent to Sbeing regular and idempotent elements commute. A proper (regular, inverse) subsemigroup M of a semigroup S is a maximal (regular, inverse) subsemigroup of S if whenever M [subset or equal to] T [subset or equal to] S for some a (regular, inverse) subsemigroup T of S, then M = T or T = S.

The Green's relations L, R, H, D, and J on a semigroup S are defined as follows. For a, b [member of] S,

(1) aLb if and only if [S.sub.1]a = [S.sub.1]b;

(2) aRb if and only if a[S.sub.1] = b[S.sub.1];

(3) aJb if and only if [S.sub.1]a[S.sub.1] = [S.sub.1]b[S.sub.1];

(4) H = L [intersection] R and D = L [??] R.

For each a [member of] S, we denote L-class, R-class, H-class, D-class, and J-class containing a by [L.sub.a], [R.sub.a], [H.sub.a], [D.sub.a], and [J.sub.a], respectively.

Lemma 1 (see [5, Lemma 1]). Let M be a maximal inverse subsemigroup of a semigroup S. If M contains a minimum idempotent e, then [H.sub.e] [subset or equal to] M.

Lemma 2. Let S be any inverse semigroup. If e is a minimum idempotent in S, then ea = ae for all a [member of] S.

Proof. Assume that e is a minimum idempotent in S. Then e [less than or equal to] f for all f [member of] E(S); that is, e = ef = fe. Let a [member of] S. Then [aea.sup.-1], [a.sup.-1] ea [member of] E(S) and ea = e([aea.sup.-1])a = ([aea.sup.-1])ea = a([ea.sup.-1]ea) = ae.

Throughout the paper we assume that Y is a nonempty subset of a set X. Let T(X, Y) = {[alpha] [member of] T(X) : X[alpha] [subset or equal to] Y}, where X[alpha] denotes the image of [alpha]. Define

F(X, Y) = {[alpha] [member of] T(X, Y): X[alpha] = Y[alpha]}. (4)

It is known that F(X, Y) is the largest regular subsemigroup of T(X, Y).

If [alpha] [member of] T(X) and x [member of] X, then the image of x under [alpha] is written as x[alpha]. The set of all inverse images of x under [alpha] is denoted by x[alpha].sup-1].

Lemma 3 (see [17, Lemma 1]). Let [alpha] [member of] T(X, Y). Then [alpha] [member of] F(X, Y) if and only if a[alpha].sup.-1] [intersection] Y [not equal to] 0 for all a [member of] X[alpha].

In [17], the author gave a complete description of Green's relations on F(X, Y) as the following lemma.

Lemma 4. Let [alpha], [beta] [member of] F(X, Y). Then

(1) [alpha]L[beta] if and only if X[alpha] = X[beta];

(2) [alpha]R[beta] if and only if [[pi].sub.[alpha]] = [[pi].sub.[beta]], where [[pi].sub.[alpha]] = {x[alpha].sup.-1] : x [member of] X[alpha]};

(3) [alpha]D[beta] if and only if [absolute value of X[alpha]] = [absolute value of X[beta]];

(4) D = J.

In addition, [alpha]H[beta] if and only if X[alpha] = X[beta] and [[pi].sub.[alpha]] = [[pi].sub.[beta]].

Lemma 5. F(X, Y) is an inverse semigroup if and only if [absolute value of Y] = 1.

Proof. If [absolute value of Y] = 1, then [absolute value of F(X, Y)] = 1 and so it is inverse. Conversely, assume that F(X, Y) is an inverse semigroup. Let a and b be elements in Y. Then there exist constant maps [X.sub.a] and [X.sub.b] with ranges {a} and {b}, respectively. Thus they are idempotents in F(X, Y) such that [X.sub.a] = [X.sub.b][X.sub.a] = [X.sub.a][X.sub.b] = [X.sub.b]; it follows that a = b. Hence, [absolute value of Y] = 1.

Proposition 6. M is a maximal inverse subsemigroup of T(X, Y) if and only if M is a maximal inverse subsemigroup of F(X, Y).

Proof. One direction is clear. Indeed, each inverse subsemigroup of T(X, Y) is contained in F(X, Y), the largest regular subsemigroup of T(X, Y). Conversely, assume that M is a maximal inverse subsemigroup of F(X, Y). Then M is an inverse subsemigroup of T(X, Y). Let V be an inverse subsemigroup of T(X, Y) such that M [subset or equal to] V [??] T(X, Y). Since F(X, Y) is the largest regular subsemigroup of T(X, Y), it follows that M [subset or equal to] V [subset or equal to] F(X, Y). We get that M = V or V = F(X, Y) by the maximality of M in F(X, Y). If F(X, Y) = V, then F(X, Y) is an inverse semigroup and so [absolute value of Y] = 1 by Lemma 5; this implies that T(X, Y) = F(X, Y). Therefore, M is a maximal inverse subsemigroup of T(X, Y).

For each a [member of] Y, let

[F.sub.a] = {[alpha] [member of] F(X, Y) : a[alpha] = a, [alpha] is injective on X \ a[[alpha].sup.-1]}. (5)

The authors [14] showed that [F.sub.a] is a maximal inverse subsemigroup of T(X, Y). Thus by Proposition 6, we have that [F.sub.a] is also a maximal inverse subsemigroup of F(X, Y).

3. Maximal Inverse Subsemigroups

In this section, we write G(A) for the permutation group on a set A.

Let e be any idempotent in T(X). In 1978, Reilly defined [I.sub.e] to be the set of those elements [alpha] [member of] T(X) which satisfy the following three conditions:

(1) The restriction of [alpha] to Xe is a permutation of Xe; that is, [[alpha]|.sub.Xe] [member of] G (Xe).

(2) For x [member of] X, if x[alpha] [not member of] Xe, then [absolute value of (x[alpha])[[alpha].sup.-1]] = 1.

(3) [alpha]e = e[alpha].

Lemma 7 (see [5, Theorem 2]). Let e be an idempotent in T(X) and let [I.sub.e] be defined as above. Then [I.sub.e] is a maximal inverse subsemigroup of T(X) with minimum idempotent e.

Notice that if e is a constant map with range {a} [subset or equal to] X, then [I.sub.e] = [I.sub.a], where

[I.sub.a] = {[alpha] [member of] T(X) : a[alpha] = a, [alpha] is injective on X \ a[[alpha].sup.-1]}; (6)

see [4, 5] for more details.

Now, let e be any idempotent in F(X, Y) and define

[F.sub.e] = [I.sub.e] [intersection] F(X, Y). (7)

Since e [member of] [I.sub.e] [intersection] F(X, Y) = [F.sub.e], we obtain [F.sub.e] [not equal to] 0 and so [F.sub.e] is a subsemigroup of F(X, Y). We note that if X = Y, then F(X, Y) = T(X) and [F.sub.e] = [I.sub.e] [intersection] F(X, Y) = [I.sub.e] [intersection] T(X) = [I.sub.e].

Here, we aim to prove that [F.sub.e] is the maximal inverse subsemigroups of F(X, Y) with minimum idempotent e.

To prove our main result we need the following five lemmas.

Lemma 8. Let e be any idempotent in F(X,Y) and [alpha] [member of] F(X, Y). Then [alpha] [member of] [F.sub.e] if and only if

(1) [[alpha]|.sub.Xe] [member of] G (Xe),

(2) for y [member of] Y, if y[alpha] [??] Xe, then [absolute value of (y[alpha])[[alpha].sub.- 1]] = 1,

(3) [alpha]e = e[alpha].

Proof. One direction is clear. Conversely, assume that conditions (1), (2), and (3) hold. We prove that [alpha] [member of] [I.sub.e]. Suppose that x [member of] X and x[alpha] [not member of] Xe. Since [alpha] [member of] F(X, Y), it follows that x[alpha] = y[alpha] for some y [member of] Y. By condition (2), we get that [absolute value of (y[alpha])[[alpha].sup.-1]] = 1. Thus [absolute value of (x[alpha])[[alpha].sup.-1]] = 1 and so [alpha] [member of] [I.sub.e].

Lemma 9. Let e be any idempotent in F(X, Y) and [alpha] [member of] F(X, Y). Then condition (2) in Lemma 8 is equivalent to the statement

(X \ Y) [alpha] [subset or equal to] Xe, [alpha] is injective on Y \ (Xe) [[alpha].sup.-1]. (8)

Proof. Assume that, for y [member of] Y, if y[alpha] [not member of] Xe, then [absolute value of (y[alpha])[[alpha.sup.-1]] = 1. We prove (X\Y)[alpha] [subset or equal to] Xe. Let x[alpha] [member of] (X\Y)[alpha] for some x [member of] X\Y. Since (X\Y) [alpha] [subset or equal to] X[alpha] = Y[alpha], there exists y [member of] Y such that x[alpha] = y[alpha]. Thus x, y [member of] (y[alpha])[[alpha].sup.-1]. If y[alpha] [not member of] Xe, then [absolute value of y[alpha])[[alpha].sup.-1]] = 1 by the assumption. Whence x = y, this is a contradiction. Hence y[alpha] [member of] Xe; that is, x[alpha] [member of] Xe. Next, we prove [alpha] is injective on Y\(Xe)[[alpha].sup.-1]. Let w, z [member of] Y \ (Xe)[[alpha].sup.-1] be such that w[alpha] = z[alpha]. Then w [member of] Y and w[alpha] [not member of] Xe. Since w, z [member of] (w[alpha])[[alpha].sup.-1], we obtain w = z by our assumption.

Conversely, suppose that (X \ Y)[alpha] [subset or equal to] Xe and [alpha] is injective on Y \ (Xe)[[alpha].sup.-1]. Let y [member of] Y and y[alpha] [not member of] Xe. To show that [absolute value of (y[alpha])[[alpha].sup.-1] = 1, we let u, v [member of] (y[alpha])[[alpha].sup.-1]. Then u[alpha] = v[alpha] = y[alpha] [not member of] Xe; this implies that u, v [not member of] (Xe)[[alpha].sup.-1]. Since (X\Y)[alpha] [subset or equal to] Xe, we obtain u, v [member of] Y. Thus u, v [member of] Y\ (Xe)[[alpha].sup.-1] and so u = v since [alpha] is injective on Y\(Xe)[[alpha].sup.11].

Lemma 10. Let [alpha] be any element in T(X) and let e be an idempotent in T(X). If there is [alpha]' [member of] V([alpha]) such that e = e[alpha]'[alpha] = [alpha][alpha]'e, e[alpha] = [alpha]e and e[alpha]' = [alpha]e, then [[alpha]|.sub.Xe] [member of] G (Xe).

Proof. Assume that the conditions hold. Since

Xe = X (e[alpha]'[alpha]) = X (e[alpha]')[alpha] = X ([alpha]'e)[alpha] [subset or equal to] (Xe) [alpha] = X (e[alpha]) = X ([alpha]e) [subset or equal to] Xe, (9)

it follows that (Xe)[alpha] = Xe; that is, [alpha] maps Xe onto Xe. We prove that [[alpha]|.sub.Xe] is injective. Let x, y [member of] Xe be such that x[alpha] = y[alpha]. Then

x = xe = x ([alpha][alpha]'e) = (x[alpha])[alpha]'e = (y[alpha]) [alpha]'e = y ([alpha][alpha]'e) = ye = y. (10)

Hence [[alpha]|.sub.Xe] : Xe [righht arrow] Xe is bijective.

Lemma 11. Let e be an idempotent in F(X, Y). Then [F.sub.e] is an inverse subsemigroup of F(X, Y) with minimum idempotent e.

Proof. Let [alpha] be any element in [F.sub.e]. Then [alpha] [member of] [I.sub.e]. Since [I.sub.e] is an inverse semigroup, there exists [beta] [member of] [I.sub.e] such that [alpha] = [alpha][[beta][alpha] and [beta] = [beta][alpha][beta]. We prove [beta] [member of] F(X, Y). Since [beta] = [beta][alpha][beta], we obtain X[beta] = X[beta][alpha][beta] [subset or equal to] X[alpha][beta] [subset or equal to] Y[beta] [subset or equal to] X[beta]; that is, X[beta] = Y[beta]. To show X[beta] [subset or equal to] Y, we let x[beta] [member of] X[beta] for some [member of] X. Since [alpha] [member of] [F.sub.e], (X\Y) [alpha] [subset or equal to] Xe by Lemma 9. If x[beta] [member of] X\Y, then (x[beta])[alpha] [member of] (X\Y) [alpha] [subset or equal to] Xe. It follows that x[beta] = (x[beta][alpha])[beta] [member of] (Xe)[beta] = Xe [subset or equal to] Y; this is a contradiction. Hence, x [beta] [member of] Y and so X[beta] [subset or equal to] Y. Thus, [beta] [member of] [I.sub.e] [intersection] F(X, Y) = [F.sub.e]. Whence [F.sub.e] is a regular subsemigroup of [I.sub.e], it follows from Lemma 7 that [F.sub.e] is an inverse subsemigroup of F(X, Y) and e is the minimum idempotent in [F.sub.e].

Lemma 12. Let e be an idempotent in F(X, Y). If M is any inverse subsemigroup of F(X, Y) such that [F.sub.e] [subset or equal to] M, then E (M) [subset or equal to] [F.sub.e].

Proof. Assume that M is an inverse subsemigroup of F(X, Y) containing [F.sub.e]. Let f be any idempotent in M. Since e, f are idempotents in M, we get ef = fe. We prove that Xe [subset or equal to] Xf. Suppose that Xe [not subset or equal to] Xf. Then Xef [subset or equal to] Xe [intersection] Xf [??] Xe, so there exists z [member of] Xe\ Xef. We see that zf [member of] Xef and zf [not equal to] z. Now, we have z, zef [member of] Xe such that z [not equal to] zef. Let [sigma] [member of] G (Xe) be such that z[sigma] = zef and (zef) [sigma] = z. We define [alpha] [member of] T(X) by

x[alpha] = (xe) [sigma] [for all]x [member of] X. (11)

We see that X[alpha] = (Xe) [sigma] = (Ye) [sigma] = Y[alpha] and X[alpha] = (Xe)[sigma] = Xe [subset or equal to] Y; we obtain [alpha] [member of] F(X, Y). We prove [[alpha]|.sub.Xe] [member of] G (Xe). Let x, y [member of] X be such thaT(Xe) [alpha] = (ye)[alpha]. Then ((xe)e) [sigma] = ((ye)e) [sigma]; that is, (xe) [sigma] = (ye) [sigma]. Thus xe = ye since [sigma] is injective on Xe. Since (Xe) [sigma] = ((Xe)e) [sigma] = (Xe)[sigma] = Xe, [alpha] is surjective on Xe. Hence, [[alpha]|.sub.Xe] [member of] G (Xe). Since y[alpha] = (ye) [sigma] [member of] Xe for all y [member of] Y, whence condition (2) in Lemma 8 holds. To prove [alpha]e = e[alpha]. Let x [member of] X. Since (xe) [sigma] [member of] Xe and e is an idempotent, we get that x[alpha]e = ((xe)[sigma])e = (xe)[sigma] = ((xe)e)[sigma] = xe[alpha]. So, [alpha] [member of] [F.sub.e]. By Lemma 11, we have [F.sub.e] as an inverse subsemigroup of F(X, Y). Then there exists [[alpha].sup.-1] [member of] V([alpha]) [intersection] [F.sub.e] and [[sigma].sub.-1] = [[alpha].sup.-l]|.sub.Xe] [member of] G (Xe). Since zef = zfe [member of] Xe, we obtain (zef) [[alpha].sup.-1] = (zef) [[sigma].sup.-1] = z. It is clear that [[alpha].sup.-1] f[alpha] is an idempotent in M. We see that

[mathematical expression not reproducible]. (12)

Since z [not equal to] zf, this implies that f ([[alpha].sup.-1] f[alpha]) [not equal to] ([[alpha].sup.-1] f[alpha])f which is a contradiction. Therefore, Xe [subset or equal to] Xf. Hence ef = e and so e = ef = fe. It is easy to see that [f|.sub.Xe] = [id.sub.Xe], the identity map on Xe, which implies that [f|.sub.Xe] [member of] G (Xe). Thus, conditions (1) and (3) in Lemma 8 are satisfied for f.

In order to verify condition (2) in Lemma 8 for f to be an element of [F.sub.e], let y [member of] Y be such that yf [not member of] Xe. We show that [absolute value of (yf)[f.sup.-1]] = 1. Let u, v [member of] (yf) [f.sup.-1]. Then uf = vf = yf [not member of] Xe.

Suppose that u [not equal to] v. Assume that v [member of] Y. We define d [member of] T(X) by

[mathematical expression not reproducible]. (13)

Obviously, Xd [subset or equal to] Y and it is easy to verify that (xe)d = xe = x(de) for all x [member of] X. Hence, (Xe)d = Xe and ed = de. It is easy to show that [d|.sub.Xe] is injective, whence [d|.sub.Xe] [member of] G(Xe). We prove Xd [subset or equal to] Yd. Let xd [member of] Xd for some x [member of] X. If x [member of] Y\{u}, then xd [member of] Yd. But if x [member of] {u} [union] (X\Y), then xd = xe = (xe)d [member of] Yd. Thus, Xd [subset or equal to] Yd and so d [member of] F(X, Y). We prove d satisfies the condition in Lemma 9. By the definition of d, we have (X\Y) d [subset or equal to] Xe. Let a, b [member of]Y\(Xe) [d.sup.-1]. Then a, b [member of] Y and ad, bd [not member of] Xe, which implies that a, b [member of] Y\{u}. So, if ad = bd, then a = b. Hence, d is injective on Y\(Xe)[d.sup.-1] and d [member of] [F.sub.e]. Next, we prove d is an idempotent. For x [member of] Y\{u}, (xd)d = xd. If x [member of] {u} u(X\Y), then (xd)d = (xe)d = xe = xd. Thus, [d.sup.2] = d. Since d and f are idempotents in M, df = fd. However,

udf = (ue) f = ue [member of] Xe, ufd = (vf) d = vdf = vf [not member of] Xe, (14)

a contradiction. Hence, v [member of] X\Y. Define g [member of] T(X) by

[mathematical expression not reproducible]. (15)

Similarly, by using the same argument as in the proof above for d, we can show that g is an idempotent in [F.sub.e]. Since f, g [member of] E(M), we obtain fg = gf. We see that

vfg = vf [not member of] Xe but vgf = (ve) f ve [member of] Xe, (16)

and this is a contradiction. Thus, u = v and so condition (2) in Lemma 8 is also satisfied. Whence f [member of] [F.sub.e], therefore, E(M) [subset or equal to] [F.sub.e] as required.

Now, we are ready to prove our main result (Theorem 13).

Theorem 13. Let e be an idempotent in F(X, Y). Then [F.sub.e] is a maximal inverse subsemigroup of F(X, Y) with minimum idempotent e.

Proof. From Lemma 11, [F.sub.e] is an inverse subsemigroup of F(X, Y) with minimum idempotent e. To prove the maximality, let M be any inverse semigroup of F(X, Y) such that [F.sub.e] [subset or equal to] M [??] F(X, Y). Then by Lemma 12, we have E(M) [subset or equal to] [F.sub.e], which implies that e is the minimum idempotent in M. We prove M = [F.sub.e] by letting [alpha] [member of] M. Since [alpha], [[alpha].sup.-1] [member of] M, we obtain e[alpha] = [alpha]e and e[[alpha].sup.-1] = [[alpha].sup.-1]e by Lemma 2. And since [alpha][[alpha].sup.-1] and [[alpha].sup.-1][alpha] are idempotents in M, e = [alpha][[alpha].sup.-1] [member of] and e = e[[alpha].sup.-1] [alpha]. Then by Lemma 10, we get that [[alpha]|.sub.Xe] [member of] G(Xe). Thus, conditions (1) and (3) in Lemma 8 are satisfied for a to be an element of [F.sub.e]. Next, we prove that [alpha] satisfies condition (2) in Lemma 8.

Let y [member of] Y be such that y[alpha] [not member of] Xe. Let u, v [member of] (y[alpha])[[alpha].sup.-1]. Then u[alpha] = v[alpha] = y[alpha] [not member of] Xe; this implies that u[alpha][[alpha].sup.-1] = v[alpha][[alpha].sup.-1] = y[alpha][[alpha].sup.-1] [not member of] Xe. Now, we have [alpha][[alpha].sup.-1] [member of] E([F.sub.e]),y [member of] Y and y([alpha][[alpha].sup.-1]) [not member of] Xe. Since u, v [member of] [y([alpha][[alpha].sup.-1])][([alpha][[alpha].sup.-1]).sup.-1] we obtain that u = v by using condition (2) in Lemma 8 for [alpha][[alpha].sup.-1] [member of] [F.sub.e]. Hence, [alpha] [member of] [F.sub.e] and so M [subset or equal to] [F.sub.e]. Therefore, M = [F.sub.e] and [F.sub.e] is a maximal inverse subsemigroup of F(X, Y).

Corollary 14. Let e be an idempotent in F(X, Y). Then [F.sub.e] is a maximal inverse subsemigroup of T(X, Y).

Proof. It follows directly from Theorem 13 and Proposition 6.

Remark 15. [F.sub.e] [not equal to] [F.sub.f] for all distinct idempotents e, f [member of] F(X, Y). In fact, if there exist idempotents e and f in F(X, Y) such that [F.sub.e] = [F.sub.f], then f [member of] [F.sub.e] and e [member of] [F.sub.f] such that e and f are minimum idempotents of [F.sub.e] and [F.sub.f], respectively. Thus, we get e [less than or equal to] f and f [less than or equal to] e; it follows that e = f.

Corollary 16. Let e be an idempotent in F(X, Y) such that Xe = Y. Then [F.sub.e] = [H.sub.e] is a maximal inverse subsemigroup of F(X, Y).

Proof. Since [F.sub.e] is a maximal inverse subsemigroup of F(X, Y) with the minimum idempotent e, we obtain that [H.sub.e] [subset or equal to] [F.sub.e] by Lemma 1. To show that [F.sub.e] [subset or equal to] [H.sub.e], let [alpha] [member of] [F.sub.e]. Then [[alpha]|.sub.Xe] [member of] G(Xe) and [alpha]e = e[alpha]. Since (Xe)[alpha] = Xe, it follows that X[alpha] = Y[alpha] = (Xe)[alpha] = Xe. We prove that [[pi].sub.a] = [[pi].sub.e]. Let x, y [member of] X such that x[alpha] = y[alpha]. Then (xe)[alpha] = (x[alpha])e = (y[alpha])e = (ye)[alpha]; this implies that xe = ye since [[alpha]|.sub.Xe] is injective. Hence [[pi].sub.a] [subset or equal to] [[pi].sub.e]. On the other hand, let xe = ye. Then (x[alpha])e = (xe)[alpha] = (ye)[alpha] = (y[alpha])e. Since x[alpha], y[alpha] [member of] Y = Xe and e is an idempotent, we obtain x[alpha] = y[alpha]. Hence [[pi].sub.e] [subset or equal to] [[pi].sub.[alpha]] and [[pi].sub.[alpha]] = [[pi].sub.e]. Then by Lemma 4, we get that [alpha] [member of] [H.sub.e]. Thus, [F.sub.e] [subset or equal to] [H.sub.e]. Therefore, [H.sub.e] = [F.sub.e] is the maximal inverse subsemigroup of F(X, Y).

Notice that if e is a constant map with range {a} [subset or equal to] Y, then [I.sub.e] = [I.sub.a] and

[F.sub.e] = F(X, Y) [intersection] [I.sub.a] = {[alpha] [member of] F(X, Y) : a[alpha] = a, [alpha] is injective on X\a[[alpha].sup.-1]}; (17)

that is, [F.sub.e] = [F.sub.a]. Thus, we get the following corollary which appeared in [14, Theorem 4.3].

Corollary 17. If e [member of] F((X, Y) is a constant map with range {a} [subset or equal to] Y, then [F.sub.e] = [F.sub.a] is a maximal inverse subsemigroup of F(X, Y) and T(X, Y).

Next, we count the number of elements in [F.sub.e]. Let e be any idempotent in F(X, Y). We know that [H.sub.e] [congruent to] G(Xe); see [17, Section 4]. Thus, if Xe = Y, then [F.sub.e] = [H.sub.e] [congruent to] G(Xe) = G(Y) by Corollary 16. Hence, [absolute value of [F.sub.e]] = [absolute value of G(Y)] when Xe = Y. In what follows, we assume that [X.sub.e] [??] Y.

Recall that the number of ways that r objects can be chosen from n distinct objects written as [mathematical expression not reproducible] is given by

[mathematical expression not reproducible]. (18)

Let A and B be (possibly empty) sets and let PI(A, B) be the set of all partial injective maps from A into B. We note that PI(A, B) contains the empty map. Moreover, if A = 0 or B = 0, then [absolute value of PI(A, B)] = 1. And if A and B are finite sets such that [absolute value of A] = m and [absolute value of B] = n, then we shall write PI(m, n) instead of PI(A, B). By a consideration of the cardinality of a domain of partial injective maps, we can verify that

[mathematical expression not reproducible]. (19)

Since Xe [??] Y, there exists x [member of] Y\Xe such that xe [member of] Xe. We see that x [member of] ((xe)[e.sup.-1] \ {xe}) [intersection] Y. For each y [member of] Xe, we define

A(y) = ([ye.sup.-1] \ {y}) [intersection] Y, B(y) = [ye.sup.-1] [intersection] (X\Y). (20)

Then [ye.sub.-1] = {y} [union] A(y) [union] B(y) is a disjoint union such that [ye.sup.-1] [intersection] Xe = {y} and A(y) [intersection] Xe = 0 = B(y) [intersection] Xe. Moreover, A(y) [not equal to] 0 for some y [member of] Xe since Xe [??] Y. Let

A = {[y.sub.i] [member of] Xe: A([y.sub.i]) [not equal to] 0, i [member of] I}, B = {[y.sub.j] [member of] Xe: B([y.sub.j]) [not equal to] 0, j [member of] J}. (21)

Then [y.sub.i]e = [y.sub.i] and [y.sub.i][e.sup.-1] [intersection] Xe = {[y.sub.i]} for all i [member of] I. It is easy to see that

[mathematical expression not reproducible]. (22)

Let [alpha] be any element in [F.sub.e]. Then [alpha]e = e[alpha] and [[alpha]|.sub.Xe] = [sigma] for some [sigma] [member of] G(Xe). For each x [member of] X \ Xe, x [member of] A([y.sub.i]) [union] B([y.sub.j]) for some [y.sub.i] [member of] A and [y.sub.j] [member of] B. If x [member of] A([y.sub.i]), then xe = [y.sub.i] [member of] Xe. Since [alpha]e = e[alpha], we obtain (x[alpha])e = (xe)[alpha] = [y.sub.i][alpha] = [y.sub.i][sigma]; that is, x[alpha] [member of] ([y.sub.i][sigma])[e.sup.-1] = {[y.sub.i][sigma]} [union] A([y.sub.i][sigma]) [union] B([y.sub.i][sigma]). Since x[alpha] [member of] Y and B([y.sub.i][sigma]) [subset or equal to] X\Y, we obtain that x[alpha] [member of] {[y.sub.i][sigma]} [union] A([y.sub.i][sigma]). If x [member of] B([y.sub.j]) [subset or equal to] X\Y, then x[alpha] [member of] (X\Y)[alpha] [subset or equal to] Xe by Lemma 9. Since e is an idempotent, we get x[alpha] = (x[alpha])e = (xe)[alpha] = [y.sub.j][alpha] = [y.sub.j][sigma]. By condition (2) in Lemma 8, we see that [alpha] is injective on all elements of A([y.sub.i]) which are not mapped into {[y.sub.i][sigma]} for all i [member of] I. Thus for each [mathematical expression not reproducible]) corresponds with an element [[theta].sub.i] of PI(A([y.sub.i]), A([y.sub.i][sigma])); that is, for u [member of] A([y.sub.i]),

[mathematical expression not reproducible]. (23)

We have shown the following. Given [alpha] [member of] [F.sub.e]. Then [alpha] is a union of a permutation [sigma] [member of] G(Xe), a union of partial injections [[theta].sub.i] : A([y.sub.i]) [[right arrow] A([y.sub.i][sigma]) for each [y.sub.i] [member of] A, and a union of functions from B([y.sub.j]) into {[y.sub.j][sigma]} for each [y.sub.j] [member of] B.

Theorem 18. Let e be any idempotent in F(X, Y) with Xe [??] Y. Let G(Xe) = {[[sigma].sub.k] : k [member of] [LAMBDA]} and A = {[y.sub.i] [member of] Xe : A([y.sub.i]) [not equal to] 0, i [member of] I}. Then there is a one-to-one correspondence between [F.sub.e] and [[universal].sub.k[member of][LAMBDA]] [{[[sigma].sub.k]} x [[PI].sub.i[member of]I] PI(A([y.sub.i]), A([y.sub.i][sigma].sub.k]))].

Proof. For each [alpha] [member of] [F.sub.e], we have [[alpha]|.sub.Xe] = [[sigma].sub.k] for some k [member of] [LAMBDA] and partial injections [[theta].sub.i]: A([y.sub.i]) [right arrow] A([y.sub.i][[sigma].sub.k]) for all [y.sub.i] [member of] A. We define

[mathematical expression not reproducible], (24)

by [alpha][PSI] = ([[sigma].sub.k], [([[theta].sub.i]).sub.i[member of]I]) for all [alpha] [member of] [F.sub.e]. We verify that [PSI] is injective. Let [alpha], [beta] [member of] [F.sub.e] be such that [alpha][PSI] = [beta][PSI]. Then [mathematical expression not reproducible]. For x [member of] X\Y, x [member of] B([y.sub.j]) for some [y.sub.j] [member of] B = {[y.sub.j] [member of] [X.sub.e] : B([y.sub.j]) [not equal to] 0, j [member of] J}. We get that x[alpha] = [y.sub.j][[sigma].sub.k] = [y.sub.j][[sigma].sub.l] = x[beta]. Whence ([[theta].sub.i]) [member of] [[pi].sub.i[member of]I] PI(A([y.sub.i]), A([y.sub.i][[sigma].sub.k])). We define [alpha] [member of] T(X, Y) which is determined by [[sigma].sub.k] and [[theta].sub.i] for all i [member of] I as follows. For x [member of] X,

[mathematical expression not reproducible]. (25)

We see that, for each x [member of] X\Y, x [member of] B([y.sub.j]) for some [y.sub.j] [member of] B. Then [y.sub.j] [member of] Xe [subset or equal to] Y and x[alpha] = [y.sub.j][[sigma].sub.k] = [y.sub.j][alpha]. So, X[alpha] [subset or equal to] Y[alpha], that is [alpha] [member of] F(X,Y). By the construction of [alpha], conditions (1) and (2) in Lemma 8 are satisfied for [alpha] to be an element of [F.sub.e]. To prove [alpha]e = e[alpha], we let x [member of] X. If x [member of] Xe, then (x[alpha])e = (x[[sigma].sub.k])e = x[sigma].sub.k] = x[alpha] = (xe)[alpha], since x, x[[sigma].sub.k] [member of] Xe and e is an idempotent. If x [member of] dom([[theta].sub.i]) [not equal to] 0, then x [member of] A([y.sub.i]); it follows that xe = [y.sub.i] and x[[theta].sub.i] [member of] A([y.sub.i])[[theta].sub.i] [subset or equal to] A([y.sub.i][[sigma].sub.k]). Thus, (x[[theta].sub.i]e = [y.sub.i][[sigma].sub.k] and (x[alpha])e = (x[[theta].sub.i])e = [y.sub.i][[sigma].sub.k] = [y.sub.i][alpha] = (xe)[alpha] since [y.sub.i] [member of] Xe. If x [member of] A([y.sub.i]) and (x [not equal to] dom([[theta].sub.i]) or [[theta].sub.t] is the empty map), then xe = [y.sub.i] and (x[alpha])e = ([y.sub.i][[sigma].sub.k])e = [y.sub.i][[sigma].sub.k] = [y.sub.i][alpha] = (xe)[alpha]. And if x [member of] B([y.sub.j]), then xe = [y.sub.j] and (x[alpha])e = ([y.sub.j][[sigma].sub.k])e = [y.sub.j][[sigma].sub.k] = [y.sub.j][alpha] = (xe)[alpha], since [y.sub.j] [member of] Xe. Hence, [alpha]e = e[alpha]. Whence [alpha] [member of] [F.sub.e] and [alpha][PSI] = ([[sigma].sub.k], [([[theta.sub.i]).sub.i[member of]I]), therefore, [PSI] is a bijection.

The following corollary is a straightforward consequence of Theorem 18.

Corollary 19. Let e be an idempotent in F(X, Y) such that Xe [??] Y and [absolute value of Xe] = p [member of] [Z.sup.+]. Let A = {[y.sub.i] [member of] Xe : A([y.sub.i] [not equal to] 0, 1 [less than or equal to] i [less than or equal to] t [less than or equal to] p}. For each [y.sub.i] [member of] A, we let [absolute value of A([y.sub.i])] = [m.sub.i] [greater than or equal to] 1 and [absolute value of A([y.sub.i][[sigma].sub.k])] = [n.sub.i,k] [greater than or equal to] 0(1[less than or equal to] i [less than or equal to] t, 1 [less than or equal to] k [less than or equal to] p!). Then

[mathematical expression not reproducible]. (26)

It is known that if X = Y, then F(X, Y) = T(X) and [F.sub.e] = [I.sub.e]. Then by Corollary 19, we have the following corollary.

Corollary 20. Let e be an idempotent in T(X) such that [X.sub.e] [??] X and [absolute value of Xe] = p [member of] [Z.sup.+]. Let A = {[y.sub.i] [member of] Xe : A([y.sub.i]) [not equal to] 0, 1 [less than or equal to] i [less than or equal to] t [less than or equal to] p}. For each [y.sub.i] [member of] A, we let [absolute value of A([y.sub.i])] = [m.sub.i] [greater than or equal to] 1 and [absolute value of A([y.sub.i] [[sigma.sub.k])] = [n.sub.i,k] [greater than or equal to] 0(1[less than or equal to] i[less than or equal to]t, 1 [less than or equal to] k [less than or equal to] p!). Then

[mathematical expression not reproducible]. (27)

If e is a constant map with range {a} [subset or equal to] Y, then we get the following corollary which appeared in [14, Theorem 4.5].

Corollary 21. Let Y be a finite subset of X with [absolute value of Y] = r [greater than or equal to] 2 and e a constant map in F(X, Y) with range {a} [subset or equal to] Y; then [mathematical expression not reproducible].

Proof. We note that [F.sub.e] = [F.sub.a] by Corollary 17. Since Xe = {a}, we obtain [absolute value of Xe] = 1 and there is a unique A(a) = Y \ {a}; that is, [absolute value of A(a)] = r - 1. By Corollary 19, we have

[mathematical expression not reproducible], (28)

where

[mathematical expression not reproducible]. (29)

Therefore, [mathematical expression not reproducible].

4. Maximal Regular of Ideals

Throughout this section, X is a finite set with n elements and Y a nonempty subset of X with r [less than or equal to] 2 elements. For convenience, we write [F.sub.n,r] for F(X, Y).

For each [alpha] [member of] [F.sub.n,r] with [absolute value of X[alpha]] = k [member of] [Z.sup.+], it follows from Lemma 3 that we can write

[mathematical expression not reproducible], (30)

where [a.sub.1], [a.sub.2], ..., [a.sub.k] [member of] Y and {[A.sub.1], [A.sub.2], ..., [A.sub.k]} is a partition of X such that [A.sub.i] [intersection] Y [not equal to] 0 for all 1 [less than or equal to] i [less than or equal to] k.

For 1 [less than or equal to] k [less than or equal to] r, we define J (F; k) = [[alpha] [member of] [F.sub.n,r] : [absolute value X[alpha]] = k}. Then J(F; k) is a J-class of the semigroup [F.sub.n,r]. Let Q(F; k) = J(F; 1) [union] J(F; 2) [union] ... [union] J(F; k) = [[alpha] [member of] [F.sub.n,r] : [absolute value of X[alpha]] [less than or equal to] k}, where 1 [less than or equal to] k [less than or equal to] r. Then Q(F; k) is an ideal of [F.sub.n,r] and it is a regular subsemigroup of [F.sup.n,r]; see [17, Lemma 7].

Recall that the principal factor of [F.sub.n,r] is the Rees quotient

[P.sub.k] = Q(F; k)/Q(F; k - 1), (31)

where 2 [less than or equal to] k [less than or equal to] r. It is usually convenient to think of it as J(F; k) [union] {0}, and the product of two elements of [P.sub.k] is taken to be zero if it falls in Q(F; k-1). Since [P.sub.k] is finite and it is not a zero semigroup, we obtain that [P.sub.k] is a completely 0-simple semigroup; see [18, p. 233] for more details.

We need the following lemmas for proof of our main result.

Lemma 22 (see [1, p. 98]). Let S be a completely 0-simple semigroup. If a, b [member of] S and ab [not equal to] 0, then ab [member of] [R.sub.a] [intersection] [L.sub.b].

Lemma 23 (see [1, Proposition 2.3.7]). Let a, b be elements in D-class D. Then ab [member of] [R.sub.a] [intersection] [L.sub.b] if and only if [L.sub.a] [intersection] [R.sub.b] contains an idempotent.

Lemma 24. Let a be any element in a semigroup S and T a regular subsemigroup of S.

(1) If [L.sub.a] [intersection] T [not equal to] 0, then [L.sub.a] [intersection] T contains an idempotent.

(2) If [R.sub.a] [intersection] T [not equal to] 0, then [R.sub.a] [intersection] T contains an idempotent.

Proof. (1) Assume that [L.sub.a] [intersection] T [not equal to] 0. Then there exists b [member of] [L.sub.a] [intersection] T. So, bLa and b = bb'b for some b' [member of] T. We get that b'b is an idempotent such that b'b [member of] T and b'bLbLa. Therefore, b'b is an idempotent in [L.sub.a] [intersection] T.

(2) If there is b [member of] [R.sub.a] [intersection] T, then we can write b = bb'b for some b' [member of] T and show that bb' is an idempotent in [R.sub.a] [intersection] T.

Lemma 25 (see [18, Lemma 3.5]). For 1 [less than or equal to] k [less than or equal to] r-1, Q(F; k) = <J(F; k)>.

Lemma 26 (see [18, Lemma 4.2]). If [alpha] [member of] J(F; k), then a can be written as a finite product of idempotents in J(F; k).

The following fact can be obtained from Lemmas 25 and 26 immediately.

Lemma 27. For 1 [less than or equal to] k [less than or equal to] r-1, Q(F; k) = <E(J(F; k))>.

Lemma 28. Let 1 [less than or equal to] k [less than or equal to] r-1 and [alpha] [member of] J(F; k). If [beta] [member of] [L.sub.a], then [beta] = [gamma][alpha] for some [gamma] [member of] J(F; k) such that [L.sub.[gamma]] [not equal to] [L.sub.a].

Proof. Suppose that [beta] [member of] [L.sub.a]. Thus X[alpha] = X[beta]; we write

[mathematical expression not reproducible], (32)

where [a.sub.1], [a.sub.2], ..., [a.sub.k] [member of] Y and [A.sub.i] [intersection] Y [not equal to] 0 [not equal to] [B.sub.i] [intersection] Y for all 1 [less than or equal to] i [less than or equal to] k. Since X[alpha] [not equal to] Y, there is y [member of] Y X[alpha] such that y [member of] [A.sub.j] for some j [member of] {1, 2, ..., k}. We choose [mathematical expression not reproducible] for all i [member of] {1, 2, ..., j-1, j + 1, ... k} and define

[mathematical expression not reproducible]. (33)

Then [gamma] [member of] J(F; k) such that X[gamma] [not equal to] X[alpha] and [beta] = [gamma][alpha].

Lemma 29. Let 2 [less than or equal to] k [less than or equal to] r-1 and [alpha] [member of] J(F; k). If [beta] [member of] [R.sub.[alpha]], then [beta] = [alpha][gamma] for some [gamma] [member of] J(F; k) such that [R.sub.[gamma]] [not equal to] [R.sub.[alpha]].

Proof. Assume that [beta] [member of] [R.sub.[alpha]]. Thus [[pi].sub.[alpha]] = [[pi].sub.[beta]]; we write

[mathematical expression not reproducible], (34)

where [a.sub.i], [b.sub.i] [member of] Y and [A.sub.i] [intersection] Y [not equal to] 0 for all 1 [less than or equal to] i [less than or equal to] k. Define

[mathematical expression not reproducible]. (35)

It is clear that [gamma] [member of] J(F; k). If [gamma] [not member of] [R.sub.[alpha]], then [R.sub.[gamma]] [not equal to] [R.sub.[alpha]] and [beta] = [alpha][gamma]. If such [gamma] [member of] [R.sub.[alpha]],we define

[mathematical expression not reproducible], (36)

where y [member of] Y \ X[alpha]. So, [gamma]' [member of] J(F; k) such that [R.sub.[gamma]'] [not equal to] [R.sub.[gamma]] = [R.sub.[alpha]] and [beta] = [alpha][gamma]'.

For the case k = 1, the ideal Q(F; 1) = J(F; 1) consisting of all constant maps [X.sub.a] with range {a} [subset or equal to] Y. It is easy to verify that J(F; 1) \ {[X.sub.a]} is a maximal regular subsemigroup of Q(F; 1) for all a [member of] Y.

Lemma 30. For 2 [less than or equal to] k [less than or equal to] r-1, M = Q(F; k-1) [union] (J(F; k) \ [L.sub.[alpha]]) is a maximal regular subsemigroup of Q(F; k) for all [alpha] [member of] J(F; k).

Proof. Let [alpha] be any element in J(F; k). Let [beta], [gamma] ]member of] J(F; k) \ [L.sub.[alpha]]. If [beta][gamma] [member of] Q(F; k-1), then [beta][gamma] [member of] M. If [beta][gamma] [not member of] Q(F; k-1), then [beta][gamma] [not equal to] 0 in [P.sub.k] = Q(F; k)/Q(F; k-1); that is, [beta][gamma] [member of] [R.sub.[beta]] [intersection] [L.sub.[gamma]] by Lemma 22. Thus, [beta][gamma]] [member of] [L.sub.[gamma]] [not equal to] [L.sub.[alpha]] and so [beta][gamma]] [member of] M. Hence, M is a subsemigroup of Q(F; k). We prove M is regular. Let [lambda] [member of] M. If [lambda] [member of] Q(F; k-1), then [lambda] is a regular element in Q(F; k-1). If [lambda] [member of] J(F; k) \ [L.sub.[alpha]], we write

[mathematical expression not reproducible], (37)

where [a.sub.i], [a.sub.2], ..., [a.sub.k] [member of] Y and [A.sub.i] [intersection] Y [not equal to] 0 for all 1 [less than or equal to] i [less than or equal to] k. Since X[alpha] [not equal to] Y, there exists y [member of] Y \ X[alpha] such that y [member of] [A.sub.j] for some j [member of] {1, 2, ..., k}. We choose [mathematical expression not reproducible] and define

[mathematical expression not reproducible]. (38)

Then [mu] [member of] J(F; k), X[mu] [not equal to] X[alpha], and [lambda] = [lambda][mu][lambda]. Hence, M is a regular semigroup. Next, we prove M is maximal. Let S be a regular subsemigroup of Q(F; k) such that M [??] S [subset or equal to] Q(F; k). Then there exists [xi] [member of] S \ M; that is, [xi] [member of] [L.sub.[alpha]] [intersection] S. Let [tau] be any element in [L.sub.[alpha]] = [L.sub.[xi]]. Then there is [gamma] [member of] J(F; k) such that [gamma] [not member of] [L.sub.[xi]] and [tau] = [gamma] [xi], by Lemma 28; that is, [gamma] [member of] M [subset or equal to] S. Thus, [tau] = [gamma][xi] [member of] S. Hence, [L.sub.[alpha]] [subset or equal to] S and therefore S = Q(F; k).

Lemma 31. For 2 [less than or equal to] k [less than or equal to] r-1, M = Q(F; k-1) [union] (J(F; k)\[R.sub.[alpha]]) is a maximal regular subsemigroup of Q(F; k) for all [alpha] [member of] J(F; k).

Proof. Let [alpha] be any element in J(F; k). Let [beta], [gamma] [member of] J(F; k)\[R.sub.[alpha]]. If [beta] [gamma] [member of] Q(F; k-1), then [beta] [gamma] [member of] M. If [beta] [gamma] [not member of] Q(F; k-1), then [beta] [gamma] [not equal to] 0 in [P.sub.k] and so [beta] [gamma] [member of] [R.sub.[beta]] [intersection] [L.sub.[gamma]] by Lemma 22. Thus, [beta] [gamma] [member of] [R.sub.[beta]] [not equal to] [R.sub.[alpha]]. So, [beta] [gamma] [member of] M. Hence, M is a subsemigroup of Q(F; k). We prove M is regular. Let [lambda] [member of] M. If [lambda] [member of] Q(F; k-1), then [lambda] is a regular element in Q(F; k-1). If [lambda] [member of] J(F; k) \[R.sub.[alpha]], we write

[mathematical expression not reproducible], (39)

where [a.sub.1], [a.sub.2], ..., [a.sub.k] [member of] Y and [A.sub.i] [intersection] Y [not equal to] 0 for all 1 [less than or equal to] i [less than or equal to] k. We choose [mathematical expression not reproducible] for al 1 [less than or equal to] i [less than or equal to] k and define

[mathematical expression not reproducible]. (40)

Then [lambda]' [member of] J(F; k) and [lambda] = [lambda][lambda]'[lambda]. If [lambda]' [not member of] [R.sub.[alpha]], then [lambda]' is a regular element in M. If [lambda]' [member of] [R.sub.[alpha]], we define

[mathematical expression not reproducible], (41)

where y [member of] Y \ X[lambda] since k < r. We obtain [lambda]" [not equal to] [R.sub.[lambda]'] = [R.sub.[alpha]] and [lambda] = [lambda][lambda]"[lambda]. Hence, M is a regular semigroup. For a maximality of M, we let S be a regular subsemigroup of Q(F; k) such that M [??] S [subset or equal to] Q(F; k). Then there exists [xi] [member of] S\M; that is, [xi] [member of] [R.sub.[alpha]] [intersection] S. Let [tau] be any element in [R.sub.[alpha]] = [R.sub.[xi]]. Then there is [gamma] [member of] J(F; k) such that [gamma] [not member of] [R.sub.[xi]] = [R.sub.[alpha]] and [tau] = [xi][gamma] by Lemma 29; that is, [gamma] [member of] M [subset or equal to] S. Thus, [tau] [member of] S and so [R.sub.[alpha]] [subset or equal to] S. Therefore, S = Q(F; k).

Theorem 32. Each maximal regular subsemigroup of Q(F; k), 2 [less than or equal to] k [less than or equal to] r-1 must be one of the following forms:

(1) Q(F; k-1) [union] (J(F; k)\[L.sub.[alpha]]);

(2) Q(F; k-1) [union] (J(F; k)\[R.sub.[beta]]),

where [alpha], [beta] [member of] J(F; k), [L.sub.[alpha]] is the L-class containing [alpha] in J(F; k), and [R.sub.[beta]] is the R-class containing [beta] in J(F; k).

Proof. By Lemmas 30 and 31, both (1) and (2) are maximal regular subsemigroups of Q(F; k).

On the other hand, let M be an arbitrary maximal regular subsemigroup of Q(F; k). It is easy to see that M [union] Q(F; k-1) is a regular subsemigroup of Q(F; k) such that M [subset or equal to] M [union] Q(F; k-1) [subset or equal to] Q(F; k). Then M = M [union] Q(F; k-1) or M [union] Q(F; k-1) = Q(F; k) by the maximality of M. If M [union] Q(F; k-1) = Q(F; k), then J(F; k) = M [intersection] J(F; k) [subset or equal to] M. We obtain M [subset or equal to] Q(F; k) = <J(F-,k)> [subset or equal to] M by Lemma 25; this implies that M = Q(F; k), a contradiction. Whence M = M [union] Q(F; k-1) and so Q(F; k-1) [subset or equal to] M, assume that [L.sub.[alpha]] [intersection] M [not equal to] 0 and [R.sub.[alpha]] [intersection] M [not equal to] 0 for all [alpha] [member of] J(F; k). We first prove that E(J(F; k)) [subset or equal to] M. Let e be an idempotent in J(F; k). Thus, [L.sub.e] [intersection] M [not equal to] 0 [not equal to] [R.sub.e] [intersection] M. By Lemma 24, [L.sub.e] [intersection] M and [R.sub.e] [intersection] M contain idempotents, say f and g respectively. So, f Le and gRe. Since [L.sub.f] [intersection] [R.sub.g] = [L.sub.e] [intersection] [R.sub.e] contains the idempotent e, fg [member of] [R.sub.f] [intersection] [L.sub.g] [subset or equal to] J(F; k) by Lemma 23. Since fg [member of] M and M is regular, fg = (fg) h (fg) for some h [member of] M. We see that f(ghf)g = fg [member of] J(F; k), which implies that ghf [member of] J(F; k); that is, (gh)f [not equal to] 0 in [P.sub.k]. Then by Lemma 22, we get (gh)f [member of] [R.sub.gh] [intersection] [L.sub.f]. Also, we have g(hf) [member of] [R.sub.g] [intersection] [L.sub.hf]. Whence ghf [member of] [L.sub.f] [intersection] [R.sub.g] = [L.sub.e] [intersection] [R.sub.e] = [H.sub.e], the group H-class contains e. So, there exists a positive integer t such that e = [(ghf).sup.t] [member of] M since g, h, f [member of] M. Therefore, E(J(F; k)) [subset or equal to] M. It follows from Lemma 27 that Q(F; k) = <E(J(F; k))> [subset of equal to] M [subset or equal to] Q(F; k); hence M = Q(F; k), a contradiction. Therefore, [L.sub.[alpha]] [intersection] M = 0 for some [alpha] [member of](F; k) or [R.sub.[beta]] [intersection] M = 0 for some [beta] [member of] J(F; k). If M [intersection] [L.sub.[alpha]] = [empty of] for some [alpha] [member of] J(F; k), then

M [subset or equal to] Q (F; k-1) [union] (J(F; k)\[L.sub.[alpha]]) [??] Q (F; k). (42)

Thus, M = Q(F; k-1) [union] (J(F; k) \ [L.sub.[alpha]]) by the maximality of M. If M [intersection] [R.sub.[beta]] = 0 for some [beta] [member of] J(F; k), then

M [subset or equal to] Q (F; k-1)[union](J(F; k)\[R.sub.[beta]]) [??]Q (F; k). (43)

Thus, M = Q(F; k-1) [union] (J(F; k) \ [R.sub.[beta]]) by the maximality of M.

Notice that the number of L-classes in J(F; k) equals ([??]) and the number of R-classes in J(F; k) equals S(r, k) [k.sup.n-r] where S(r, k) is the Stirling number of the second kind; see [18, Section 2] for details. Therefore, the number of maximal regular subsemigroups of Q(F; k) is equal to ([??]) + S(r, k) [k.sup.n-r] when 2 [less than or equal to] k [less than or equal to] r-1.

We shall normally write [T.sub.n] instead of T(X).For 1 [less than or equal to]k [less than or equal to] n, define

[J.sub.k] = {[alpha] [member of] [T.sub.n] : [absolute value of X[alpha]] = k}], K (n, k) = {[alpha] [member of] [T.sub.n] : [absolute value of X[alpha]] [less than or equal to] k}. (44)

It is well-known that K(n, k) = [J.sub.1] [union] [J.sub.2] [union] ... [union] [J.sub.k] and K(n, k) is an ideal of [T.sub.n] for all 1 [less than or equal to] k [less than or equal to] n. If n = r, then [F.sub.n,r] = [F.sub.n,n] = [T.sub.n], J(F; k) = [J.sub.k], and Q(F; k) = K(n, k). Therefore, we establish the following corollary which first appeared in [8, Theorem 2].

Corollary 33. Each maximal regular subsemigroup of K(n, k), 2 [less than or equal to] k [less than or equal to] n-1 must be one of the following forms:

(1) K(n, k-1) [union] ([J.sub.k]\[L.sub.[alpha]]);

(2) K(n, k-1) [union] ([J.sub.k]\[R.sub.[beta]]),

where [alpha], [beta] [member of] [J.sub.k], [L.sub.[alpha]] is the L-class containing [alpha] in [J.sub.k] of [T.sub.n], and [R.sub.[beta]] is the R-class containing [beta] in [J.sub.k] of [T.sub.n].

We note that if n = r, then we immediately obtain that the number of maximal regular subsemigroups of K(n, k) is equal to [mathematical expression not reproducible].

https://doi.org/10.1155/2018/2154745

Conflicts of Interest

The author declares that they have no conflicts of interest.

Acknowledgments

Financial support from the Coordinating Center for Thai Government Science and Technology Scholarship Students (CSTS), National Science and Technology Development Agency (NSTDA), is acknowledged.

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Worachead Sommanee (iD)

Department of Mathematics and Statistics, Faculty of Science and Technology, Chiang Mai Rajabhat University, Chiang Mai 50300, Thailand

Correspondence should be addressed to Worachead Sommanee; worachead_som@cmru.ac.th

Received 9 March 2018; Revised 25 March 2018; Accepted 1 April 2018; Published 7 May 2018

Academic Editor: Pentti Haukkanen

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Title Annotation: | Research Article |
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Author: | Sommanee, Worachead |

Publication: | International Journal of Mathematics and Mathematical Sciences |

Date: | Jan 1, 2018 |

Words: | 10583 |

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