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I need some help solving the next permutations group equation: $$ x^{20} = \sigma $$ where

$$ \sigma = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 \\ 7 & 4 & 5 & 6 & 1 & 8 & 3 & 10 & 12 & 2 & 13 & 9 & 11 \end{pmatrix}. $$

I've already found out that the product of disjoint cycles is $(1 \ 7 \ 3 \ 5)(2 \ 4 \ 6 \ 8 \ 10)(9 \ 12)(11 \ 13)$, the order is $order(\sigma)=20$ and the sign is $sgn(\sigma)=1$, but I don't know how to use all these to solve the equation above.

Thomas Lesgourgues
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Bianca Pascu
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    The arguments I explained here apply to this problem as well. Solution #1: Because $20$ is a factor of $\operatorname{ord}(\sigma)$ it follows that $x$ must be of order $400$. There are no permutations of that order in the group $S_{13}$. Solution #2: If a fifth power of a permutations has a $5$-cycle, then it must have five of them. Here $\sigma=(x^4)^5$ should be a fifth power, but it only has a single 5-cycle. Of course, Vladimir's simple argument is a more delightful way of settling this! – Jyrki Lahtonen Jan 31 '19 at 11:20

2 Answers2

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Well, cycles of odd length are even permutations, and cycles of even length are odd permutations. Here we have three cycles of even length and one cycle of odd length, and so $\operatorname{sign}(\sigma)=(-1)^3\cdot1=-1$. On the other hand, $\operatorname{sign}(x^{20})=(\operatorname{sign}(x))^{20}=1$, and so the equation has no solutions at all.

Vladimir
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There are no solutions:

If $n=\mid x\mid$, then $\frac n{\operatorname{gcd}(n,20)}=20\implies n=400$.

But there are no elements of order $400$ in $S_{13}$. The smallest symmetric group with an element of order $400$ is $S_{33}$, where there is room for a permutation of cycle type $(8,25)$.