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