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One of the questions as part of my college assignment asks us to give an example of a permutation in $\Bbb S_{10}$ based on a given order.

  1. For a permutation in $\Bbb S_{10}$ whose order is 6:

$o(\pi) = 6 = lcm(2,3)$ hence something like $(2,3)(4,5,6)$ would satifsy this, giving the permutation:

$\pi=\begin{bmatrix}1&2&3&4&5&6&7&8&9&10\\1 &3&2&5&6&4&7&8&9&10\end{bmatrix}$

  1. For a permutation in $\Bbb S_{10}$ whose order is 15:

$o(\beta) = 15 = lcm(3,5)$ hence something like $(2,3,4)(5,6,7,8,9)$ would satifsy this, giving the permutation:

$\beta=\begin{bmatrix}1&2&3&4&5&6&7&8&9&10\\1&3&4&2&6&7&8&9&5&10\end{bmatrix}$

  1. For a permutation in $\Bbb S_{10}$ whose order is 15:

$o(\gamma) = 30 = lcm(2,3,5)$ hence something like $(1,2)(3,4,5)(6,7,8,9,10)$ would satifsy this, giving the permutation:

$\gamma=\begin{bmatrix}1&2&3&4&5&6&7&8&9&10\\2&1&4&5&3&7&8&9&10&6\end{bmatrix}$

Is this correct or am I making a mistake? Is there a better way to do this?

erykkk
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1 Answers1

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Yes, this is correct. For 1., you also can just take $\sigma=(123456)\in S_{10}$. This means, $7,8,9,10$ are fixed by $\sigma$. Of course, the order of $\sigma$ in $S_{10}$ is $6$. In general, the order is given as follows:

Prove that the order of an element in $S_n$ equals the least common multiple of the lengths of the cycles in its cycle decomposition.

Dietrich Burde
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