Return to Answer

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 Group $G=\left(\mathbb{Z}/(n\mathbb{Z})\right) ^{\times2}$,
the order of columns: $(1,1), (1,2), \dots, (1,n), (2,1), \dots, (n,n)$,
the order of rows: $(1,1), (2,1), \dots, (n,1), (1,2), \dots, (n,n)$.

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.
Suppose $G$ is $\{1,\dots,n^2\}$ endowed with some associative operation and suppose, that the table of this operation is a sudoku table. Take any $x\in G$. In $x$-th line of the table we can find element $x$. Therefore $xy=x$ for some $y\in G$. Then take $a\in G$, such that $a\neq y$ and $ay$ and $ya$ are in the same square of the table. We have $(xy)a=x(ya)$ and, therefore $xa=x(ya)$. Using, that there is only one place in $x$-th line of the table, where we can find $xa$, we see that $a=ya$. From this we have $a(ya)=(ay)a$, $aa=(ay)a$ and $a=ay$. Therefore $ay=a=ya$ and the table is not a sudoku table.

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 $G=\left(\mathbb{Z}/(n\mathbb{Z})\right) ^{\times2}$,
the order of columns: $(1,1), (1,2), \dots, (1,n), (2,1), \dots, (n,n)$,
the order of rows: $(1,1), (2,1), \dots, (n,1), (1,2), \dots, (n,n)$.

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.
Suppose $G$ is $\{1,\dots,n^2\}$ endowed with some associative operation and suppose, that the table of this operation is a sudoku table. Take any $x\in G$. In $x$-th line of the table we can find element $x$. Therefore $xy=x$ for some $y\in G$. Then take $a\in G$, such that $a\neq y$ and $ay$ and $ya$ are in the same square of the table. We have $(xy)a=x(ya)$ and, therefore $xa=x(ya)$. Using, that there is only one place in $x$-th line of the table, where we can find $xa$, we see that $a=ya$. From this we have $a(ya)=(ay)a$, $aa=(ay)a$ and $a=ay$. Therefore $ay=a=ya$ and the table is not a sudoku table.

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 Group $G=\left(\mathbb{Z}/(n\mathbb{Z})\right) ^{\times2}$,
the order of columns: $(1,1), (1,2), \dots, (1,n), (2,1), \dots, (n,n)$,
the order of rows: $(1,1), (2,1), \dots, (n,1), (1,2), \dots, (n,n)$.

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 $a\in G$.

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 Group $G=\left(\mathbb{Z}/(n\mathbb{Z})\right) ^{\times2}$,
the order of columns: $(1,1), (1,2), \dots, (1,n), (2,1), \dots, (n,n)$,
the order of rows: $(1,1), (2,1), \dots, (n,1), (1,2), \dots, (n,n)$.

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.
Suppose $G$ is $\{1,\dots,n^2\}$ endowed with some associative operation and suppose, that the table of this operation is a sudoku table. Take any $x\in G$. In $x$-th line of the table we can find element $x$. Therefore $xy=x$ for some $y\in G$. Then take $a\in G$, such that $a\neq y$ and $ay$ and $ya$ are in the same square of the table. We have $(xy)a=x(ya)$ and, therefore $xa=x(ya)$. Using, that there is only one place in $x$-th line of the table, where we can find $xa$, we see that $a=ya$. From this we have $a(ya)=(ay)a$, $aa=(ay)a$ and $a=ay$. Therefore $ay=a=ya$ and the table is not a sudoku table.

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 Group $G=\left(\mathbb{Z}/(n\mathbb{Z})\right) ^{\times2}$,
the order of columns: $(1,1), (1,2), \dots, (1,n), (2,1), \dots, (n,n)$,
the order of rows: $(1,1), (2,1), \dots, (n,1), (1,2), \dots, (n,n)$.

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 Group $G=\left(\mathbb{Z}/(n\mathbb{Z})\right) ^{\times2}$,
the order of columns: $(1,1), (1,2), \dots, (1,n), (2,1), \dots, (n,n)$,
the order of rows: $(1,1), (2,1), \dots, (n,1), (1,2), \dots, (n,n)$.

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 $a\in G$.