Consider the binary operation * and 0 defined by the following tables on set S = {a, b, c, d}.

We observe the following:



a * b = b * a = a a * a = a


c * a = a * c = a b * b = b


a * d = d * a = a c * c = d


b * c = c * b = c d * d = c


b * d = d * b = d


c * d = d * c = b


There ‘ * ’ is commutative.


Also,


a * (b * c) = a * (c) = a (From above)


(a * b) * c = (a) * c = a (Also from above)


Hence, ‘ * ’ is associative too.


Therefore, to find the identity element, e for e belong to S, we need:


a * e = e * a = a, a belong to S.


Therefore, a * e = a


We find that there is no unique element e which satisfies the condition.


e = a or b or c or d for a.


Since the identity is not unique, the inverse will also be not unique.


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