Page 103 - DMTH403_ABSTRACT_ALGEBRA
P. 103
Abstract Algebra
Notes t(i) t(p) t(q) t(i) i q p i
.
i p q i = i p q i l, if q > i > p,
t(p) t(i) t(q) t(i) = q i p i l, if i > p,
.
p i q i p i q i
Taking the values of all the factors of sign t, we see that sign t = 1.
(b) Let f S,. By Theorem 3 we know that f = t ,t .... t, for some transpositions t , ..... t in S .
1
r
1
2
n
sign f = sign (t t . . . . t,)
2
1
= (sign t ) (sign t ) . . . . . sign (t ), by Theorem 3.
1
2
r
= (-1) , by (a) above.
r
sign f = 1 or 1.
(c) We know that Im (sign) {1, 1}.
We also know that sign t = 1, for any transposition t; and sign I = 1.
{1, 1} Im {sign}
Im (sign) = {1, 1}.
Now, we are in a position to prove what we said at the beginning of this section.
Theorem 5: Let f S, and let
f = t t ..... t = t t ..... t
4
1 2
r
1 2
be two factorisations of f into a product of transpositions. Then either both r and s are even
integers, or both are odd integers.
Proof: We apply the function sign: S {1, 1} to f = t t . . . . t .
r
1 2
n
By Theorem 4 we see that
sign f = (sign t ) (sign t ) ...... (sign t ) = (- 1).
2
r
1
sign (t t . . . t ) = (1) substituting t t . . . ts for f.
2
1
1
2
s
that is, ( -1) = (1) . r
s
This can only happen if both s and r are even, or both are odd.
So, we have shown that for f S, the number of factors occurring in any factorisation of f into
transposition is always even or always odd. Therefore, the following definition is meaningful.
Definition: A permutation f S , is called even if it can be written as a product of an even sign
n
number of transposition. f is called odd if it can be represented as a product of an odd number of
transpositions.
For example, (1 2) S is an odd permutation. In fact, any transposition is an odd permutation.
3
On the other hand, any 3-cycle is an even permutation, since
(i j k) = (i k) (i j)
Now, we define an important subset of S , namely,
n
A, = (f Sn, | f is even).
n!
Well show that A,, S , and that o(A ) = 2 , for n 2.
n
n
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