A Pigeonhole Proof of Kaplansky's Theorem

Abstract

Proof of Lemma 1: Suppose b has distinct right inverses a1, a2,..., aN. Then a1 a,, a2 a1,. . ., aN a1 are N distinct solutions of bx = 0. We will show that the set {1 a1b, 1 a2b, ...,1 aNb} contains at least one additional solution of bx = O. Clearly all of the elements of this set are solutions. If there were not a new solution in this set, then for each j there is a k so that 1 ajb = ak a1. However, 1 ajb cannot equal a1 a1 = 0, because then a1 would be a left inverse for b, and in this case it is easy to see that b has only one right inverse, a contradiction. Thus, since there are N (1 ajb)'s (the pigeons) and only (N 1) acceptable (ak al)'s (the pigeon-holes), by the pigeon-hole principle we must have that for some m = n, 1 amb = 1anb. But then amb = anb, and multiplying this on the right by a1, we get am = an, a contradiction.