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I was checking multiplication tables for groups with 4 elements, to see which tables "passed" the group axioms of closure, associativity, identity and inverses. But then I had a question, so hopefully someone can help me with this basic group theory question.

The proof that inverses are unique in a group depends upon the associativity axiom. Let a be an element that has two inverses b and c. Then b = be = b(ac) = (ba)c = ec = c. Thus the inverse is unique since b must equal c.

So my question is this. If we see a multiplication table for a finite group, and we can easily check closure, existence of identity, and existence of inverses, and further we can see that inverses are all unique, does this necessarily imply that the last axiom, associativity, holds? If yes, how?

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Not by far. Just try drawing up a multiplication that satisfies identity and existence and uniqueness of inverses. These laws put very few constraints on the multiplication table -- the remaining cells can be filled in at random, and it will be very unlikely that you happen to end up with a group.

Consider, for example, the structure with three elements where each is its own inverse and the inverses are clearly unique:

  |  e   A   B
--+-------------
e |  e   A   B
A |  A   e   B
B |  B   B   e

This is not a group, since $Be=BA$ but we cannot cancel the $B$ to conclude $e=A$.

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No. If an operation on a set has an identity and unique inverses (but not associativity) then this set is called a quasigroup.

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