On Non-abundance Of Certain Semigroups Of Finite Full Transformations

Let Sn- be the semigroup of all decreasing full transformations of Xn={1,2,..., n}. We write where A C Xn\{1}. Then Sn-(A) is a subsemigroup of Sn-. In this paper, we prove that the semigroup Sn-(A) is not abundant if A≠(?) and A≠{n}. Moreover, we consider a particular case which A={x ∈Xn:x is even} for n is an even number and determine its Green’s relations or their generalisations, regular subsemigroups and rank. The main result are given in following:In Section 2, we introduce the notions of the semigroups S-(A) and BSn-. We show that any element of BSn-is not regular unless the element is idempotent. We also trivial-ly obtain the formula for the number of nilpotent elements in BSn-and the order of BSn-Lemma 2.1. The semigroup Sn-(A) is R-trivial.Proposition 2.10. Suppose that n=2k. Then Furthermore,In Section 3, we characterise the Green’s relations on Sn-(A) (BSn-) and their gener-alisations, and show that Sn-(A) (BSn-(n> 4)) is a non-abundant semigroup in the sense of Fountain. To end this section we amend the defect R* V(?)≠(?)*(?)* (in BSn-), and modify (?)*,(?)* of BSn- to the relations (?)(u),(?)(u).Lemma 3.1.2. Let α,β be elements in Sn-(A).(1) For 2∈A, (a, β ∈(?)* if and only if im(a)\{1,2}= im(β)\{1,2}.(2) For 2(?)A, (a,β) ∈(?)* if and only if im(a)=im(β).Lemma 3.1.3.Let α,β be elements in Sn-(A).(1)For n∈A,(α,β)∈(?)* if and only if ker(α)|xn-1,=ker(β)+Xn-(2)For n(?)A,(α,β)∈(?)* if and only if ker(α)=ker(β).Theorem 3.1.4. Let Sn-(A)be as defined above.If A≠(?) and A≠{n},the semigroup Sn-(A)is not abundant.In Section 4,we specialise the regular part of BSn-.We obtain,perhaps less trivially, the formula for the number of the idempotents of BSn-.Then we determine the maximal subband of BSn-.Theorem 4.3. Let A={1)3,…,2k-1)for n=2k and suppose that(?)={{A1,A2,...,Ak}:A1={1},Ak=A,j∈A,Ai+1=Ai∪{j},j(?)Ai}. Suppose that where Ω∈(?).Then M is the maximal subband of BSn-.Conversely,every maximal subband of BSni is isomorphic to one constructed in this way.In Section 5,we concentrate on the rank of BSn-.We show that BSn- is generated by the collection of all indecomposable elements in BSn-.Proposition 5.4. The set of all indecomposable elements of BSn- is the minimum generating set of BSn-.Theorem 5.5. Suppose that n=2k.Then(1)r1(BSn-)=1；(2)r5(BSn-)=|BSn-|=[(2k-1)!!]2.