4
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$\newcommand{\lcm}{\operatorname{lcm}}$ I was hoping to get some feedback.

Consider the permutation $$\sigma = \begin{pmatrix} 1&2&3&4&5&6&7&8\\ 5&8&3&6&7&4&1&2 \end{pmatrix}.$$ (a) Describe the orbits of $\sigma$.
(b) Express $\sigma$ as a product of disjoint cycles, and then as a product of transpositions.
(c) What is the order of $\sigma$? Explain.


(a) The orbits of $\sigma$ are $\{1,5,7\}, \{2,8\},\{3\},\{4,6\}$.

(b) As a product of disjoint cycles: $\sigma = (1\;5\;7)(2\;8)(3)(4\;6)$
As a product of transpositions: $\sigma = (1\;7)(1\;5)(2\;8)\boxed{(3)\,}(4\,6)$
I am not sure if I should include $(3)$ since transpositions are cycles of length $2$.

(c) I don't know what the order of a permutation is. My first guess was that it is the number of orbits of $\sigma$. So I would say $4$. But then I saw a post that used the $\lcm$. I infer that it is the $\lcm$ of the lengths of the orbits of $\sigma$; hence $\lcm(2,3) = 6$. Is this correct? If so, why is it the $\lcm$?

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    $\begingroup$ the order is least common multiple of $3,2,1,2$ corresponding with the lengths of the disjoint cycles in the product. So the order is $6$. It is correct to leave $(3)$ out of the product of transpositions. $\endgroup$ – drhab Feb 25 '16 at 12:26
  • $\begingroup$ Is that under the same topic as this ? $\endgroup$ – BCLC Feb 25 '16 at 12:37
  • $\begingroup$ @BCLC Possibly? Haha. $\endgroup$ – David Feb 25 '16 at 12:44
  • $\begingroup$ When you write a permutation as a product of cycles - i.e. like this $(1,2)(3,4,2)$ - it is generally agreed that all the elements that do not appear are left fixed. So, you can write the $1$-cycles (which are the fixed points), but you do not need to. If you want, you may as well notice that $(3)=id$ and eliminate it. $\endgroup$ – user228113 Feb 25 '16 at 13:25
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(a) -- Correct.

(b) -- No, you don't need to include $(3)$.

(c) -- The order of a permutation $\sigma$ is the smallest positive integer $\ell$ so that $\sigma^\ell=1$. Since the disjoints cycles of $\sigma$ are mutually commutative, $\ell$ is divisible by the order of each cycle. Therefore, $\ell$ is the l.c.m of those orders, which are exactly the lengths of the cycles.

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  • $\begingroup$ I understand what $\sigma^\ell(a)$ means, but what does $\sigma^\ell = 1$ mean? $\endgroup$ – David Feb 25 '16 at 12:33
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    $\begingroup$ $1$ means the identity permutation, which fixes all the letters $1,2,\dots,n$. $\endgroup$ – Quang Hoang Feb 25 '16 at 12:35

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