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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?

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  • $\begingroup$ 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. $\endgroup$
    – Tony Huynh
    Apr 23, 2019 at 9:50

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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 $m_{12}m_{21}m_{22}$.

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