### The Sign

Another important fact exhibits a normal subgroup * A*_{n} of
* S*_{n} . In order to show this we introduce the *sign*
* e( p)*
as follows:
* e( p):= Õ*_{1 <= i<j <= n}**(** pj- pi**)/(**j-i**)** Î **Z**,
if
n >= 2, while e(1_{S 0}):= e(1_{S
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 * S*_{n} into
* {1,-1 }*:
* e( pr)= Õ*_{i<j}**(** prj- pri**)/(**j-i**)**= Õ_{i<j}
**(** prj- pri**)/(** rj- ri**)** Õ_{i<j}**(** rj- ri**)/(**j-i**)**= e( p) e( r).

This proves
**Corollary: **
*
The sign map
** e:S*_{n} -> {1,-1 } :p -> e( p)

is a homomorphism which is surjective for each *n ³2*.
Hence its kernel
* A*_{n} := kere= { pÎS_{n} | e( p)=1 }

is a normal subgroup of * S*_{n} :
* A*_{n} lefttriangleeq S_{n} , | A_{n} | = | S_{n} | /2= n!/2, if n ³2.

The elements of * A*_{n} are called *even*
permutations, while the elements of * S*_{n} \A_{n} 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@kfunigraz.ac.at,

last changed: August 28, 2001