2D Rubik's cube? There is a $3\times3$ matrix filled by numbers 1~9 that might look like this
$$\begin{bmatrix}3 & 8 & 2 \\ 4 & 1 & 6 \\ 7 & 5 & 9\end{bmatrix}$$
All its rows and columns can be "rolled forwards and backwards" (like permutation acting on a single row/column)
Roll the second column upwards:
$$\begin{bmatrix}3 & 1 & 2 \\ 4 & 5 & 6 \\ 7 & 8 & 9\end{bmatrix}$$
And roll the first row to the left, we get
$$A=\begin{bmatrix}1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9\end{bmatrix}$$
Now, if given an arbitrary matrix like this, how can I tell if it can be restored back to $A$? If it always could, why?
 A: If we start with a matrix 
$$\begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i\end{bmatrix}$$
then the map $1 \mapsto a$, $2 \mapsto b$, etc, defines a permutation $\sigma$ of $\{1, 2, ..., 9\}$, i.e. an element of $S_9$. The valid "moves" consist of composing this permutation with one of the elements $(123), (456), (789), (147), (258), (369)$ or their inverses.
So the question becomes: what is the subgroup $G$ of $S_9$ generated by $\{(123), (456), (789), (147), (258), (369)\}$? For starters, these are all even permutations, so we're in fact looking for a subgroup of $A_9$. 
Next we notice that we can get from $A$ to 
$$\begin{bmatrix}2 & 3 & 1 \\ 5 & 6 & 4 \\ 8 & 9 & 7\end{bmatrix}$$
and then to
$$\begin{bmatrix}2 & 3 & 4 \\ 5 & 6 & 7 \\ 8 & 9 & 1\end{bmatrix}$$
so $(123456789) \in G$. And $(123456789)$ and $(123)$ together generate $A_9$, so $G = A_9$. 
So the answer to the first question: you can get from a matrix to $A$ if and only if the permutation $\sigma$ defined above is an even permutation.
