AbstractSuppose that a binary operation $$\circ $$
∘
on a finite set X is injective in each variable separately and also associative. It is easy to prove that $$(X,\circ )$$
(
X
,
∘
)
must be a group. In this paper we examine what happens if one knows only that a positive proportion of the triples $$(x,y,z)\in X^3$$
(
x
,
y
,
z
)
∈
X
3
satisfy the equation $$x\circ (y\circ z)=(x\circ y)\circ z$$
x
∘
(
y
∘
z
)
=
(
x
∘
y
)
∘
z
. 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.