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OpenStudy (anonymous):

Question about permutations.

OpenStudy (anonymous):

Prove that: \[sgn(\sigma)=sgn(\sigma^{-1})\forall \sigma \in S_n\]

OpenStudy (anonymous):

what's \(S_n\)?

OpenStudy (jamesj):

\[ 1 = sgn(e) \ \ \ \ \ \ \ \ \ \ \ \ \ \hbox{ where e is the identity permutation} \] \[ = sgn(\sigma \sigma^{-1}) \] \[ = sgn(\sigma).sgn(\sigma^{-1}) \] And now you have it.

OpenStudy (anonymous):

Indeed! I figured it was a simple one but I wanted to post something xD

OpenStudy (anonymous):

who can answer my question? or it's too hard?

OpenStudy (jamesj):

oh, S_n is the group of permutation on n objects.

OpenStudy (anonymous):

are you speaking about same permutation as in combinatorics \[nPk=\frac{n!}{(n-k)!}\] because I hardly understand it here

OpenStudy (jamesj):

No, I'm talking about this: http://en.wikipedia.org/wiki/Permutation_group

OpenStudy (anonymous):

The symmetric group right?

OpenStudy (jamesj):

Yes

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