Type: Article
Publication Date: 2020-11-08
Citations: 8
DOI: https://doi.org/10.1007/s00039-020-00553-1
Abstract Suppose that a binary operation $$\circ $$ <mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML"> <mml:mo>∘</mml:mo> </mml:math> on a finite set X is injective in each variable separately and also associative. It is easy to prove that $$(X,\circ )$$ <mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML"> <mml:mrow> <mml:mo>(</mml:mo> <mml:mi>X</mml:mi> <mml:mo>,</mml:mo> <mml:mo>∘</mml:mo> <mml:mo>)</mml:mo> </mml:mrow> </mml:math> must be a group. In this paper we examine what happens if one knows only that a positive proportion of the triples $$(x,y,z)\in X^3$$ <mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML"> <mml:mrow> <mml:mrow> <mml:mo>(</mml:mo> <mml:mi>x</mml:mi> <mml:mo>,</mml:mo> <mml:mi>y</mml:mi> <mml:mo>,</mml:mo> <mml:mi>z</mml:mi> <mml:mo>)</mml:mo> </mml:mrow> <mml:mo>∈</mml:mo> <mml:msup> <mml:mi>X</mml:mi> <mml:mn>3</mml:mn> </mml:msup> </mml:mrow> </mml:math> satisfy the equation $$x\circ (y\circ z)=(x\circ y)\circ z$$ <mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML"> <mml:mrow> <mml:mi>x</mml:mi> <mml:mo>∘</mml:mo> <mml:mo>(</mml:mo> <mml:mi>y</mml:mi> <mml:mo>∘</mml:mo> <mml:mi>z</mml:mi> <mml:mo>)</mml:mo> <mml:mo>=</mml:mo> <mml:mo>(</mml:mo> <mml:mi>x</mml:mi> <mml:mo>∘</mml:mo> <mml:mi>y</mml:mi> <mml:mo>)</mml:mo> <mml:mo>∘</mml:mo> <mml:mi>z</mml:mi> </mml:mrow> </mml:math> . 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. A recent result of Green shows that this metric approximation is necessary: it is not always possible to obtain a proportional-sized subset that agrees with part of the multiplication table of a group.