# Find permutation that solves $\;\tau \circ X = \sigma$

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 $X$?

Hint: find the inverse of $\tau$ and left "multiply" each side of the equation by $\tau^{-1}$ to solve for $X$.
• $\tau^{-1} = (3,1,5)(4,2)$ so now I add $\sigma(3) = 1$ etc? – Chris Feb 24 '14 at 21:41
• No, you want to left-multiply each side of the equation to isolate $X$:$$\tau^{-1}\circ \tau\circ X = \tau^{-1}\circ\sigma\iff X = \tau^{-1}\circ\sigma$$ – Namaste Feb 24 '14 at 21:42
• $X = (1)(2,4,3,5)$ – Chris Feb 24 '14 at 21:47
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 $$\sigma = \begin{bmatrix} 1 & 2 & 3 & 4 & 5\\ 3 & 2 & 1 & 5 & 4 \end{bmatrix} = (1,3)(2)(4,5)$$