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For a symmetric group $S_n$. Is the decomposition starting from the permutation $(1,2,3,...,n)$? Also, how could I prove it if this is the case?

For a permutation $\sigma \in S_n$, it is even if $$\mathrm{sgn}(\sigma) = \prod_{1\leq i < j \leq n} \frac{\sigma(j)-\sigma(i)}{j-i} = 1$$

For example, $S_2$ would be the set of the following permutations

$$\begin{pmatrix} 1 & 2 \\ 1 & 2\end{pmatrix}, \begin{pmatrix} 1 & 2 \\ 2 & 1\end{pmatrix}$$

Where these are not matrices. the top row represents the domain and the bottom row represents the image of $\sigma$

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    $\begingroup$ Usually this is taken to be the definition of an even permutation. In your case, could you clarify what your definition is? $\endgroup$ – Misha Lavrov Apr 9 '17 at 16:46
  • $\begingroup$ I updated it with a few more things. Is the decomposition starting from the permutation $(1,2,...,n)$? $\endgroup$ – The Bosco Apr 9 '17 at 17:38
  • $\begingroup$ Yes, assuming that by this you mean the identity permutation. $\endgroup$ – Misha Lavrov Apr 9 '17 at 17:42
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An $r$-cycle $(i_1\ i_2\ldots i_r)$ can be decomposed into a product of transpositions $(i_1\ i_r)\ldots(i_1\ i_3)(i_1\ i_2)$, every permutation $\sigma \in S_n$ can be decomposed into a product of cycles and transpositions are odd.

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