closed non associative binary operation I'm Trying to show that a binary operation does not have to be associative in order to maintain closure, identity element and inverse element - on a 5 elements set.
Any ideas for a set + binary operation that shows that?
I can't find a way to lose associativity without throwing out closure as well.
 A: Consider the set $\{0,1,2,3,4\}$ and the opertion $+$ thus defined:


*

*$n+n=n$;

*$0+n=n+0=n$;

*otherwise, $m+n=0$.


It has an identity element ($0$) and each element has an inverse (several, indeed, except for $0$). But $(2+1)+1=0+1=1$, whereas $2+(1+1)=2+1=0$.
A: The example given in the accepted answer is brilliant but, there, the inverse element is not unique for any given element of the set. Otherwise, in a group context, inverse must be unique.
For $n=5$, we have six different isotopy classes of loops, only one being a group (integers mod $5$). Two of the other five:
$$
\begin{array}{c|ccccc}
\ast & 0 & 1 & 2 & 3 & 4\\
\hline
0 & 0 & 1 & 2 & 3 & 4\\
1 & 1 & 0 & 3 & 4 & 2\\
2 & 2 & 4 & 0 & 1 & 3\\
3 & 3 & 2 & 4 & 0 & 1\\
4 & 4 & 3 & 1 & 2 & 0
\end{array}\qquad\begin{array}{c|ccccc}
\bullet & 0 & 1 & 2 & 3 & 4\\
\hline
0 & 0 & 1 & 2 & 3 & 4\\
1 & 1 & 0 & 4 & 2 & 3\\
2 & 2 & 3 & 0 & 4 & 1\\
3 & 3 & 4 & 1 & 0 & 2\\
4 & 4 & 2 & 3 & 1 & 0
\end{array}
$$
The set $A=\{0,1,2,3,4\}$ is closed under both $\ast$ and $\bullet$.
None of the two is commutative since
$1\ast 2\ne 2\ast 1$ and $1\bullet 2\ne 2\bullet 1$.
None of the two is associative since
$$
(2\ast 3)\ast 4\ne 2\ast (3\ast 4)
$$
and
$$
(1\bullet 3)\bullet 4\ne 1\bullet (3\bullet 4).
$$
Finally, for both the above operations, $0$ is the identity element and each element of $A$ is the inverse of itself (unique).
