In set $S$, there is an operation $'' \circ ''$ such that $\forall a,b \in S$, a unique $a \circ b \in S$ exists. And (i) $\forall a,b,c \in S$, $(a \circ b) \circ c = a \circ (b \circ c)$. (ii) $a \circ b \neq b \circ a$ when $a \neq b$. Prove that: a.) $\forall a,b,c \in S$, $(a \circ b) \circ c = a \circ c$. b.) If $S = \{1,2, \ldots, 1990\}$, try to define an operation $'' \circ ''$ in $S$ with the above properties.
Problem
Source: China TST 1990, problem 3
Tags: function, algebra unsolved, algebra
28.06.2005 16:14
Observe first that $(a\circ a)\circ a=a\circ (a\circ a)$ implies $a\circ a=a$, for all $a$. Now, we have \[ a\circ\left( a\circ\left( b\circ a\right) \right) =\left( a\circ a\right) \circ\left( b\circ a\right) =a\circ\left( b\circ a\right) \] and% \[ \left( a\circ\left( b\circ a\right) \right) \circ a=\left( a\circ b\right) \circ\left( a\circ a\right) =\left( a\circ b\right) \circ a=a\circ\left( b\circ a\right) \] Again, we deduce that $a\circ\left( b\circ a\right) =a,$ for all $a,b.$ Finally,% \[ \left( a\circ c\right) \circ\left( a\circ\left( b\circ c\right) \right) =\left( a\circ\left( c\circ a\right) \right) \circ\left( b\circ c\right) =a\circ\left( b\circ c\right) \] and% \[ \left( a\circ\left( b\circ c\right) \right) \circ\left( a\circ c\right) =\left( a\circ b\right) \circ\left( c\circ\left( a\circ c\right) \right) =\left( a\circ b\right) \circ c, \] therefore% \[ a\circ\left( b\circ c\right) =a\circ c \] For the second part, a trivial example is $a\circ b=a,$ for all $a,b$ in $S.$ A less trivial example can be obtained by considering a set $S^{\prime }=\left\{ \left( x,y\right) |1\leq x\leq n,1\leq y\leq m\right\} ,$ with $nm=1990,$ defining a one-to-one function $f: S\rightarrow S^{\prime}$ and transporting through $f$ the structure defined on $S^{\prime}$ as follows: $\left( x,y\right) \circ\left( x^{\prime},y^{\prime}\right) =\left( x,y^{\prime}\right) ,$ which clearly satisfy the requested conditions. The previous trivial example is obtained for $n=1990$ and $m=1$