Permutation problem.

**PJani** · February 11th 2010, 07:10 AM

$\alpha=\left( {\begin{array}{cccccccccccc} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\ 5 & 6 & 7 & 8 & 3 & 11 & 9 & 12 & 1 & 10 & 2 & 4 \\ \end{array} } \right)$

$\beta=\left( {\begin{array}{cccccccccccc} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\ 6 & 7 & 4 & 5 & 12 & 1 & 11 & 9 & 2 & 3 & 10 & 8 \\ \end{array} } \right)$

$\beta^{-1}*\pi^{2006}*\beta=\alpha$

I dont know how to get cyclic structure of $\pi$ permutation and how to solve upper equation.

Any help would be apriciated.

**alunw** · February 11th 2010, 08:39 AM

You can easily find $\pi^{ 2006}$ , and see what its cycle structure is. Then you can think about the cycle structures that are possible for permutations of 12 symbols and work out what orders these might have. You certainly won't find 2006 in there since that is 2*17*59. And you can easily find what the order of $\pi^{2006}$ is. With those two pieces of information you should be able to narrow down the possibilities for $\pi$ itself a lot.
That's how I'd do it, but maybe there are more elegant methods I don't know about.

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