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Definition: A transposition is a 2-cycle permutation.

Definition: A permutation $\sigma$ is called even if its decomposition into transpositions has even number of transpositions; analogously for odd permutations.

Now, coming to the question, how do I prove using this definition that the sign map $\operatorname{sgn}\colon S_n\to\{\pm 1\}$ assigning a permutation in the symmetric group $S_n$ to its sign (+1 if even, -1 if odd) is a homomorphism?

I know how to prove it using the definition of the sign map being the determinant of the $n\times n$ permutation matrix associated to a permutation (in which case it is obvious since $\operatorname{det}$ is multiplicative).

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If it enough to know that the parity of the number of transpositions which appear in a given decomposition does not depend of the specific decomposition.

Then if $\sigma=t_1...t_n, sign(\sigma)=(-1)^n$ this is well defined since the parity of $n$ does not depend of the composition. If $\sigma'=t'_1....t'_{n'}, sign(\sigma')=(-1)^{n'}$ and as $\sigma\sigma'=t_1....t_nt'_1...t'_{n'}, sign(\sigma\sigma')=(-1)^{n+n'}$.

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  • $\begingroup$ I know that; I don't see how that answers my question. $\endgroup$ – learner Dec 26 '18 at 17:28
  • $\begingroup$ you see now how it answer ? $\endgroup$ – Tsemo Aristide Dec 26 '18 at 17:30
  • $\begingroup$ Yes, I do. And I feel quite silly that I didn't think of it myself. Thanks. $\endgroup$ – learner Dec 26 '18 at 17:32

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