How to solve the equation:
$\sigma ^2 =\left({\begin{array}{*{20}c}1 & 2 & 3 & 4 & 5\\ 1 & 4 & 2 & 3 & 5\end{array}}\right)\ $ where $\sigma \in S_5$.
Is there a general method?
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Have you tried to find a cycle decomposition? – Neal Aug 03 '15 at 11:13
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I think a general method might be hard, since $\sigma$ does not have to be unique, e.g. repeating the swapping of two elements is equivalent to doing nothing. – Kenneth Goodenough Aug 03 '15 at 11:18
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I think this answer your question. http://math.stackexchange.com/questions/266569/how-to-find-the-root-of-permutation – Aymane Fihadi Aug 03 '15 at 11:19
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In your case you can write $\sigma^2=(243)$. Thus, being $\sigma\in S_5$, the only possibilities are $$ \sigma=(234) $$ and $$ \sigma=(15)(234)\;\,. $$
Joe
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Hints (corrected): Note that your permutation $\sigma^{2}$ has odd order. In general, a permutation $\tau$ of odd order $n$ has a square root which is $\tau^{\frac{n+1}{2}}$, though this is not usually unique. However, you should check out for yourself that you understand why this is so, and not just take my word for it.
Geoff Robinson
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You may wonder what happens to a single permutation $\sigma\in S_n$, represented as a product of disjoint cycles, when we square it: any cycle of odd length gives a cycle of the same length, any cycle of length $2m$ breaks as the product of two cycles of length $m$. By reversing the process, we may see that an element in $S_n$ is a square iff its cycles of even length come in pairs, and $$ (1)(5)(2\,4\,3)\in S_5 $$ is the square of $(1\,5)(2\,3\,4)$ or the square of $(1)(5)(2\,3\,4)$. In general, the number of square roots, provided that one exists, is $2^m$, where $m$ is the number of couples of cycles of the same length.
Jack D'Aurizio
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