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Let the permutation $$\sigma=\bigl(\begin{smallmatrix} 1 & 2 &3 &4 &5 &6 & 7 &8 & 9 & 10\\ 3& 4 & 5 &6 & 7 & 2 & 1 & 10 & 9 & 8 \end{smallmatrix}\bigr)$$

Analyze $\sigma$ as a product of disjoint cycles and calculate its order in $S_{10}$. Is $\sigma$ even or odd?

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I have done the following:

$$\sigma=(1 \ \ 3 \ \ 5 \ \ 7) \ (2 \ \ 4 \ \ 6) \ (8 \ \ 10)$$ $$ord{(\sigma)}=LCM(4, 3, 2)=12$$ $$\text{ Since } \sigma \text{ is written as a product of } 3 \text{ circles, }\sigma \text{ is odd. }$$

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Could you tell me if this is correct??

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    $\begingroup$ Do you mean disjoint cycle instead of foreign circle? $\endgroup$ – Hubble Jun 2 '14 at 23:20
  • $\begingroup$ Oh yes...I will edit my post... $\endgroup$ – Mary Star Jun 2 '14 at 23:21
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    $\begingroup$ Then you are entirely correct. $\endgroup$ – Hubble Jun 2 '14 at 23:21
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    $\begingroup$ Well, no @MaryStar: your permutation is not odd but even, since it is a product of two odd cycles (the 2 and 4 cycles) and one even cycle (the 3 cycle)... $\endgroup$ – DonAntonio Jun 3 '14 at 3:49
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    $\begingroup$ Hi @MaryStar. First, stop writing "circles": it is cycles . Second, yes: you must check each cycle, remembering that a cycle is an even permutation iff its length is odd, and the other way around. Thus, the cycle $\;(1\;2\;3\;4)\;$ is odd since its length (4) is even, and thus your whole permutation is odd.even.odd = even. $\endgroup$ – DonAntonio Jun 3 '14 at 13:10
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One way to prove that a cycle is even iff its lenght is odd is to write it as a product of transpositions, knowing that a transposition is odd ( always ):

$$\sigma:=(i_1\;i_2\;...i_k)=(i_1\;i_2)(i_2\;i_3)\cdot\ldots\cdot(i_{k-1}\;i_k)$$

As you can see, the cycle $\;\sigma\;$ is the product of $\;k-1\;$ transpositions, and thus (Syg:= sygnum = sign) :

$$Syg(\sigma)=(-1)^{k-1}=\begin{cases}\;\;\,1&,\;\;\;k\;\;\;\text{is odd}\\{}\\-1&,\;\;\;k\;\;\;\text{is even}\end{cases}$$

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