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I have two matrices which are all permutations $C = \begin{pmatrix} 0 & 1 & 0 \\ 1& 0& 0\\0 & 0 &1\end{pmatrix}$and $ A= \begin{pmatrix} 0 & 0 & 1 \\ 0& 1& 0\\1 & 0 &0\end{pmatrix}$of the identity matrix. The question I want to answer is : Is it possible to create the remaining four matrices in this $S_3$ group. Also, can other permutations be generated from matrix products $\otimes$ of these two?

I have computed $C^2$ and I got a (321) permutation $C^3$ led to the identity matrix. So I concluded that no, it is impossible. $A^2 \rightarrow I_3$

On the matrix product part, I simply considered the fact that the product will be a 9 x 9 matrix and I said no.

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  • $\begingroup$ The matrix product of two $3$x$3$ matrices is still $3$x$3$. $\endgroup$
    – Harto Saarinen
    Feb 12, 2018 at 19:59
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    $\begingroup$ Welcome to stackexchange. You seem to know cycle notation for permutations (if not, look it up). Then $C$ corresponds to $(12)$ and $A$ to $(13)$. It's easy to show those two transpositions generate all of $S_3$. $\endgroup$
    – Ethan Bolker
    Feb 12, 2018 at 20:06
  • $\begingroup$ thank you Ethan and Harto $\endgroup$
    – TICHA
    Feb 12, 2018 at 20:18

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You can easily write down the group $S_3=\langle A,C\rangle$ with the usual matrix product, and not with the Kronecker product.

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  • $\begingroup$ you mean by multiplying the two? If so, must I then square , cube....the answer until I get to the identity matrix? Thank you for your hint. $\endgroup$
    – TICHA
    Feb 12, 2018 at 20:02
  • $\begingroup$ Yes. Note that $AC$ has order $3$, and $A$ and $C$ have order $2$, i.e., $A^2=C^2=I$, and $(AC)^3=(CA)^3=I$. $\endgroup$
    – Dietrich Burde
    Feb 12, 2018 at 20:04
  • $\begingroup$ Hah! Thanks a million @Dietrich $\endgroup$
    – TICHA
    Feb 12, 2018 at 20:08

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