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Let $I$ be a set and suppose we have a partially defined multiplication on $I\times I$ satisfying "the cocycle condition" $$(i,j)(j,k)=(i,k)$$ for all $i,j,k\in I.$ Now I wonder what kind of algebraic structure (group, semigroup, ... etc) does $I\times I$ carry with it.

So far only thing I could extract from the cocycle condition is, for a given $(i,j)$ we have a left identity $(i,i)$ and a right identity $(j,j).$ Also if we define $$(i,j)^{-1}=(j,i),$$ then we can derive

  • $(i,j)(i,j)^{-1}=(i,i)$
  • $(i,j)^{-1}(i,j)=(j,j)$
  • $((i,j)^{-1})^{-1}=(i,j).$

But these properties are not enough to determine the structure completely. Also since the multiplication isn't defined for all ordered pairs, I can't see how we can understand the algebraic structure completely. Thank you for your help, in advance.

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The part of such a structure only involving multiplications that appear in your axiom is just a trivial groupoid, where $(i,j)$ represents the unique morphism $i\to j$. The rest of the structure is totally unaffected by the axiom, so there's nothing to say.

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