It is known that a regular ring has stable range one if and only if it is unit regular. The purpose of this note is to give an independent and more elementary proof of this result.2000 Mathematics Subject Classification: 16E50, 16E65.1. Introduction. All rings considered in this note are associative with identity. A ring R is said to be (von Neumann) regular if, given any x ∈ R, there exists y ∈ R such that xyx = x. If, given any x ∈ R, there exists an invertible element u ∈ R such that xux = x, then R is said to be unit regular. A ring R is said to have stable range one if for any a, b ∈ R satisfying aR + bR = R, there exists y ∈ R such that a + by is right invertible. By Vaserstein [4, Theorem 1], this definition is left-right symmetric.It has been shown independently in [1,3] that a regular ring has stable range one if and only if it is unit regular (see also [2]). The aim of this note is to provide a rather straightforward and more elementary proof of this result.We need the following proposition.
Let G be a finite group and let T be a non-empty subset of G. For any positive integer k, let Tk={t1…tk∣t1,…,tk∈T}. The set T is called exhaustive if Tn=G for some positive integer n where the smallest positive integer n, if it exists, such that Tn=G is called the exhaustion number of T and is denoted by e(T). If Tk≠G for any positive integer k, then T
is a non-exhaustive subset and we write e(T)=∞. In this paper, we investigate the exhaustion numbers of subsets of the generalized quaternion group Q2n=⟨x, y∣x2n−1=1, x2n−2=y2, yx=x2n−1−1y⟩ where n≥3. We show that Q2n has no exhaustive subsets of size 2 and that the smallest positive integer k such that any subset T⊆Q2n of size greater than or equal to k is exhaustive is 2n−1+1. We also show that for any integer k∈{3,…,2n}, there exists an exhaustive subset T of Q2n such that |T|=k
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