# Equation in Permutations

Be on Lookout

Solve in permutations

$\begin{align} xyz^{2} &= \left(\begin{matrix}1 & 2 & 3 & 4 & 5\\1 & 2 & 4 & 3 & 5\end{matrix}\right)z^{4},\\ xy^{2}z &= \left(\begin{matrix}1 & 2 & 3 & 4 & 5\\1 & 3 & 2 & 4 & 5\end{matrix}\right)z^{2},\\ x^{2}yz &= \left(\begin{matrix}1 & 2 & 3 & 4 & 5\\5 & 3 & 2 & 4 & 1\end{matrix}\right)z^{2}. \end{align}$

### Solution

Let's denote the signature of permutation $p$ as $\chi (p).$ Denote also the three given permutations on the right side of the equations $u,$ $v,$ and $w.$ Using the multiplicative property of the signature, Having the three identities would imply

$ \chi(x)\chi(y)\chi(z)^{2} = \chi(u)\chi(z)^{4},\\ \chi(x)\chi(y)^{2}\chi(z) = \chi(v)\chi(z)^{2},\\ \chi(x)^{2}\chi(y)\chi(z) = \chi(w)\chi(z)^{2}.$

The product of the three gives

$\chi(x)^{4}\chi(y)^{4}\chi(z)^{4} = \chi(u)\chi(v)\chi(w)\chi(z)^{8}.$

This would imply that $\chi(u)\chi(v)\chi(w)=1.$ However, note that each of the three given permutations $u,$ $v,$ or $w$ may be split into $3$ transpositions, such that, for each, the signature is $-1,$ which leads to a contradiction. Thus the system has no solution.

### Acknowledgment

The problem is from Dan Sitaru's book *Ice Math*. The solution is by Leo Giugiuc.

### Permutations

- Various ways to define a permutation
- Counting and listing all permutations
- Johnson-Trotter Algorithm: Listing All Permutations

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