Abstract
Suppose that a binary operation on a finite set is injective in each variable separately and also associative. It is easy to prove that must be a group. In this paper we examine what happens if one knows only that a positive proportion of the triples satisfy the equation . Other results in additive combinatorics would lead one to expect that there must be an underlying "group-like" structure that is responsible for the large number of associative triples. We prove that this is indeed the case: there must be a proportional-sized subset of the multiplication table that approximately agrees with part of the multiplication table of a metric group. We also present an example that suggests that our result cannot be strengthened to yield a dense subset that agrees with part of the multiplication table of a group.
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