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I know that if you are given the matrix itself or the disjoint cycles, you can easily express the permutation as a product of transpositions, but if you are only given the transpositions, can you go backwards and find the permutation matrix? If so, how?

For example if you had $(1,6)(2,3)(2,5) \in S_6$, how would you reconstruct the permutation in matrix form?

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    $\begingroup$ Obviously, you need to know the order of the transpositions. If you have the order, then you can just multiply the matrices, can't you? $\endgroup$ – Thomas Andrews Dec 18 '14 at 3:58
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First the notation is (i j) and not (i,j). Now if you want to get the matrix form of a permutation, you will need to know how it works in every $i=1,...,n$. For example, in your specific example $\sigma =(1 6)(2 3)(2 5)$ send 1 to 6, since 1 is fixed by (2 3) and (2 5) and $(1 6)$ sends 1 to 6. Then $\sigma(1)=6$. Now you can apply the same idea to get the value of $\sigma(2)=5$, $\sigma(3)=2$, $\sigma(4)=4$, $\sigma(5)=3$ and $\sigma(6)=1$, the the matrix form of your specific permutation is: $$\sigma= \left( \begin{array}{ccc} 1 & 2 & 3 & 4 & 5 & 6\\ 6 & 5 & 2 & 4 & 3 & 1\end{array} \right)$$

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