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I need to find a permutation $X$ that solves $\;\tau \circ X = \sigma,\;$ given $$\tau = \begin{bmatrix} 1 & 2 & 3 & 4 & 5\\ 3 & 4 & 5 & 2 & 1 \end{bmatrix} = (1,3,5)(2,4)$$ $$\sigma = \begin{bmatrix} 1 & 2 & 3 & 4 & 5\\ 3 & 2 & 1 & 5 & 4 \end{bmatrix} = (1,3)(2)(4,5)$$

Is there a trick on how to solve for $X$?

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Hint: find the inverse of $\tau$ and left "multiply" each side of the equation by $\tau^{-1}$ to solve for $X$.

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  • $\begingroup$ $\tau^{-1} = (3,1,5)(4,2)$ so now I add $\sigma(3) = 1$ etc? $\endgroup$ – Chris Feb 24 '14 at 21:41
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    $\begingroup$ No, you want to left-multiply each side of the equation to isolate $X$:$$\tau^{-1}\circ \tau\circ X = \tau^{-1}\circ\sigma\iff X = \tau^{-1}\circ\sigma$$ $\endgroup$ – Namaste Feb 24 '14 at 21:42
  • $\begingroup$ $X = (1)(2,4,3,5)$ $\endgroup$ – Chris Feb 24 '14 at 21:47
  • $\begingroup$ Yes, Chris, you've got it! $\endgroup$ – Namaste Feb 24 '14 at 21:48
  • $\begingroup$ thanks alot. Sounds logic now. $\endgroup$ – Chris Feb 24 '14 at 21:49
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Assuming you compose left to right: $$\tau = \begin{bmatrix} 1 & 2 & 3 & 4 & 5\\ 3 & 4 & 5 & 2 & 1 \end{bmatrix} = (1,3,5)(2,4)$$ times $$X = \begin{bmatrix} 3 & 4 & 5 & 2 & 1\\ 3 & 2 & 1 & 5 & 4 \end{bmatrix} = (1,4,2,5)(3)$$ equals $$\sigma = \begin{bmatrix} 1 & 2 & 3 & 4 & 5\\ 3 & 2 & 1 & 5 & 4 \end{bmatrix} = (1,3)(2)(4,5)$$

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