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This question already has an answer here:

A permutation can't be both even and Odd. How?? Is their any proof?? Kindly tell me.!

Thanks beforehand

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marked as duplicate by Martin R, Zain Patel, JMP, Namaste abstract-algebra May 2 '17 at 12:53

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  • $\begingroup$ See also here or here. $\endgroup$ – user228113 May 2 '17 at 11:50
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Given a permutation $\pi$ we define $$\sigma(\pi) = |\{(i, j) : i < j \textrm{ and } \pi(i) > \pi(j)\}| \mod 2$$

It is easy to show inductively that $\sigma$ is $1\mod 2$ iff $\pi$ can be expressed as the product of a odd number of transpositions. By showing every transposition changes $\sigma$ by one. Once you have convinced yourself of this it follows that the parity of permutation is well defined.

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