Let S be a semigroup and T be a subsemigroup of finite index in S (that is, the set S \ T is finite). The subsemigroup T is also called a large subsemigroup of S. It is well known that if T has a finite complete rewriting system, then so does S. In this paper, we will prove the converse, that is, if S has a finite complete rewriting system, then so does T . Our proof is purely combinatorial and also constructive.
Introduction.Let S be a semigroup and T be a subsemigroup of finite index in S (that is, the set S \ T is finite). Then T is called a large subsemigroup of S, and S is called a small extension of T .In [4], Ruškuc asked if S is a small extension of T , whether S has a finite complete rewriting system if and only if T has a finite complete rewriting system (see [4, Problem 11.1 (iii)] and [6, Remark 4.2]). This problem was partially solved by Wang in [5, Theorem 1], who proved that if T has a finite complete rewriting system, then so does S. However it is still not known whether T has a finite complete rewriting system or not, when S has a finite complete rewriting system. In this paper we shall prove that this is true, i.e., we shall prove the following: Theorem 1.1. Suppose S is a small extension of T . If S has a finite complete rewriting system, then so does T . By Theorem 1.1 and the result of Wang [5, Theorem 1], we have completely answered the problem posed by Ruškuc (see [4, Problem 11.1 (iii)]).Corollary 1.2. Suppose S is a small extension of T . Then S has a finite complete rewriting system if and only if T has a finite complete rewriting system.Let A be a non-empty set. This set A is called the alphabet and the elements of A are called letters. We shall denote the free semigroup and free monoid on A by A + and A * , respectively.The elements of A + and A * are called words. Note that A * = A + ∪ {ǫ}, where ǫ is the empty 2010 Mathematics Subject Classification. Primary 20M05, 20M35, 68Q42.
