Last night I taught an algebra tutorial, and while writing out the multiplication table for the units of $\mathbb{Z}/5\mathbb{Z}$, a student remarked that it looked like a sudoku puzzle. I noted that it was similar, as the rows and columns all satisfy the sudoku condition, however the 2 by 2 subsquares do not. The reason I gave is that multiplication is commutative in this group, and so the squares along the diagonal are symmetric. Afterwards I realized that there is an even simpler reason which excludes the possibility of nonabelian groups having multiplication tables which are sudoku squares: the identity element always messes up one of the subsquares along the diagonal. This leads to the natural question: do there exist associative sudoku squares of side length $n^2$? (Meaning interpret the sudoku square as defining a binary operation on the integers $1$ through $n^2$. Is this binary operation ever associative? Of course I care about when $n > 1$ as $n = 1$ is clear.)
If we are not obligated to keep the same order in rows and columns of the table, the answer is yes, and the group can be chosen to be commutative. For example
the group $G=\left(\mathbb{Z}/(n\mathbb{Z})\right) ^{\times2}$, Otherwise, the answer is no for any $n>1$, because in any table of this type both $a\cdot 1$ and $1\cdot a$ are in the same $n\times n$ square for some $a\in G$. Edit: If this operation doesn't need to define a group, then the answer is no too. Here is the proof. 

