Citation Context

...∗-regular semigroup, (a, b) ∈ S and p be the smallest positive integers such that (a, b)p ∈ S. Then (a, b)p ∈ L∗(a,b) ∩R∗(a,b) = H∗(a,b). 2. MAIN RESULTS Let V be the set of all inverse elements of S =-=[3]-=-, E is the set of all idempotent of S. Here we get a good result. Theorem 1. Let (a, b) and (x, y) be element of a Π∗-regular semigroup S. Then (1)(a, b)L∗(x, y)⇔ ∃(a, b)′, (x, y)′ ∈ V, (a, b)′(a, b)m...

Citation Context

... (x, y)(a, b)m [1]. In this paper, we consider some special case of Π∗-regular semigroups and completely Π∗-regular semigroups. Remark The marks we don’t illustrate in this paper please see reference =-=[2]-=-. 1 The Subclasses of Characterization on Π∗-Regular Semigroups Now let S be a Π∗-regular semigroup. Define the equivalence relations L∗, R∗, H∗, J∗ on S by (a, b)L∗(x, y)⇔ S(a, b)m = S(x, y)n (a, b)R...

Citation Context

... is Π∗-regular. A semigroup S is completely Π∗-regular, if S is Π∗-regular, for any (a, b) ∈ S and element (a, b) is a regular, there exist m ∈ Z+, (x, y) ∈ S, such that (a, b)m(x, y) = (x, y)(a, b)m =-=[1]-=-. In this paper, we consider some special case of Π∗-regular semigroups and completely Π∗-regular semigroups. Remark The marks we don’t illustrate in this paper please see reference [2]. 1 The Subclas...

Citation Context

... v) ∈ H∗(e,e) and p is the smallest positive integer such that (u, v)p ∈ S, then (u, v)q ∈ G(e,e) for every q ≥ p. Proof. Let (a, b) ∈ G(e,e) and let (s, t) be an inverse element for (a, b) in G(e,e) =-=[4]-=-. Since (s, t)(a, b) = (e, e) = (a, b)(s, t), we obtain that (s, t) ∈ V , so by theorem 1 (a, b) ∈ H∗(e,e). Hence G(e,e) ⊆ H∗(e,e). Assume (u, v) ∈ H∗(e,e) and let p be the smallest positive integer s...

Citation Context

...(a,b). Proof. Let p and q be the smallest positive integers such that ((a, b)(x, y)) p , ((x, y)(a, b)) q ∈ S. Then ((a, b)(x, y)) p ∈ G(e,e) ⊆ J∗(e,e), ((x, y)(a, b)) q ∈ G(f,f) ⊆ J∗(f,f). Whence by =-=[5]-=- we obtain ((a, b)(x, y)) p+m ∈ G(e,e) ⊆ J(e,e), ((x, y)(a, b)) q+n ∈ G(f,f) ⊆ J∗(f,f). For every p, q ≥ 0, so ((a, b)(x, y)) m J∗((a, b)(x, y))p+m, ((x, y)(a, b)) n J∗((x, y)(a, b))q+n. 4 Luo, X./Pro...