# Can the permutation group of a matrix be generated by the subset of row permutation and column permutation?

Consider a 3*3 matrix $$\left( \begin{array}{*{20}{c}} {{m_{11}}}&{{m_{12}}}&{{m_{13}}}\\ {{m_{21}}}&{{m_{22}}}&{{m_{23}}}\\ {{m_{31}}}&{{m_{32}}}&{{m_{33}}} \end{array} \right)$$ and its permutation group, which consists of all the possible permutation over its element such as $$\left( {{m_{22}},{m_{33}}} \right) \cdot \left( {{m_{11}},{m_{21}},{m_{32}}} \right)$$. I wonder whether all these permutations can be expressed as a composition of several row permutation and column permutation? If so, does it still hold when we consider a non-square matrix?

• If I understand your question correctly, the answer is no. As you have defined it, there are $9!$ elements in your permutation group, but there are only $6$ permutation matrices. Apr 23, 2019 at 9:50

Of course not, for the following reason. When applying either a row or a column permutation, then the set of products $$m_{1\sigma(1)}\cdots m_{n\sigma(n)},\qquad\sigma\in\frak{S}_n$$ is unchanged. On the contrary, if you apply another permutation of the entries, this becomes false. For instance, if you apply the transposition $$(m_{22},m_{33})$$, then the product $$m_{12}m_{21}m_{33}$$ becomes $$m_{12}m_{21}m_{22}$$.