The Sign |
Another important fact exhibits a normal subgroup An of Sn . In order to show this we introduce the sign e( p) as follows:
e( p):= Õ1 <= i<j <= n( pj- pi)/(j-i) Î Z, if n >= 2, while e(1S 0):= e(1S 1):=1 Z.
As i not =j implies pi not = pj, we have e( p) not =0. Moreover, the following sets of pairs are equal:
{ {i,j } | 1 £i<j £n }= { { pi, pj } | 1 £i<j £n },
and so we have e( p) = ±1 Z. Furthermore e is a homomorphism of Sn into {1,-1 }:
e( pr)= Õi<j( prj- pri)/(j-i)= Õi<j ( prj- pri)/( rj- ri) Õi<j( rj- ri)/(j-i)= e( p) e( r).
This proves
Corollary: The sign mape:Sn -> {1,-1 } :p -> e( p)is a homomorphism which is surjective for each n ³2. Hence its kernelAn := kere= { pÎSn | e( p)=1 }is a normal subgroup of Sn :An lefttriangleeq Sn , | An | = | Sn | /2= n!/2, if n ³2.
The elements of An are called even
permutations, while the elements of Sn \An are called odd
permutations. Correspondingly, an r-cycle is even if and only if r is odd. There is a program to compute the sign of various permutations.
harald.fripertinger "at" uni-graz.at | http://www-ang.kfunigraz.ac.at/~fripert/ | UNI-Graz | Institut für Mathematik | UNI-Bayreuth | Lehrstuhl II für Mathematik |
The Sign |