Is it true for any associative groupoid which elements are $a,b,c$, that $(a\ \circ \ b) \circ c = (b \ \circ c) \circ a $? As I know an associative groupoid is a semigroup. I'm assuming that it can be true, but I'm not sure, how can I give a proof for it?
 A: Let $G = (\{a,b,c\},\circ)$ be the groupoid where $\circ$ obeys the following table:
$$
\begin{array}{c|ccc} \circ & a & b & c \\ \hline a & a & a & a \\ b & b & b & b \\ c & a & a & c \end{array}
$$
One can check that $\circ$ is an associative operation. However
$$(a\circ b)\circ c = a \circ c = a \neq b = b\circ a = (b\circ c)\circ a.$$
My code found $7$ such groupoids (that is, up to isomorphism). Other than the one above, these are:
$$
\begin{array}{lcccr}
\begin{array}{c|ccc} \circ & a & b & c \\ \hline a & a & a & a \\ b & b & b & b \\ c & b & b & b \end{array} & &
\begin{array}{c|ccc} \circ & a & b & c \\ \hline a & a & a & a \\ b & b & b & b \\ c & c & c & c \end{array} & &
\begin{array}{c|ccc} \circ & a & b & c \\ \hline a & a & a & c \\ b & a & b & c \\ c & a & a & c \end{array} \\ \\
\begin{array}{c|ccc} \circ & a & b & c \\ \hline a & a & b & a \\ b & a & b & b \\ c & a & b & c \end{array} & &
\begin{array}{c|ccc} \circ & a & b & c \\ \hline a & a & c & c \\ b & a & b & c \\ c & a & c & c \end{array} & &
\begin{array}{c|ccc} \circ & a & b & c \\ \hline a & b & b & c \\ b & b & b & c \\ c & b & b & c \end{array} 
\end{array}
$$
If my code is correct this are the only $3$-element associative groupoids (up to isomorphism) that don't satisfy the identity in the stated problem.
