$G$ is a groupoid but not semigroup. On $G$ we can define so many binary operations. Is there any way we can show that at least two binary operations are the same?
I think its not possible, since it has only binary operation, but no other structure!!!
A groupoid G consists of a set, say G and a binary operation * on G, i.e.
* : G x G ---> G. So to check if two groupoids G = (G,*) and G' = (G,o) are the same,
we have to check whether two functions * and o are equal. They have the same domains, therefore the requirement reduces to check if x*y = xoy for all pairs x,y
from G. In case of finite G, this is the same as having equal Cayley tables.
From the mathematical point of view it is more interesting when they are the same up to the structure. In that case they are 'the same' if there is a renaming (an isomorphism) which makes them behave identically. Formally, an isomorphism is a bijection ('renaming') f : G ---> G such that f(x.y) = f(x) o f(y) for all pairs of x,y.
Sorry, the last formula in my previous post should read f(x*y) = f(x) o f(y).
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