Page 102 - DMTH403_ABSTRACT_ALGEBRA

Unit 8: Permutation Groups

Notes


=   3 – 2  1 2  1 3    1.


 1  2  1 
Similarly, iff = (1 2)  S , then
3
f(2) – f(1) f(3) – f(1) f(3) – f(2)
sign f = . .
2 – 1 3 – 1 3 – 2


=   1 2  3 – 2  3 1     1.


 1  2  1 
Henceforth, whenever we talk of sign f, we shall assume that f  S for some n  2.
n
Theorem 3: Let f, g  S,. Then sign (f o g) = (sign f) (sign g).
Proof: By definition,

n f(g(j)) – f(g(i))
sign fog  
i,j 1 j – i

i j 
f(g(j)) f(g(i)) g(j) g(i)


  . 


i,j g(j) g(i) i,j j i
Now, as i and j take all possible pairs of distinct values from 1 to n, so do g(i) and g(j), since g is
a bijection.
f(g(j)) f(g(i))

   sign f.

i j  g(j) g(i)
 sign (fog) = (sign f) (sign g).
Now we will show that Im (sign) = (1, – 1}.
Theorem 4: (a) If t  S, is a transposition, then sign t = – 1.
(b) sign f = 1 or - 1  f  S,.
(c) Im (sign) = (1, –1).

Proof: (a) Let t = (p q), where p < q.
Now, only one factor of sign t involves both p and q, namely,

t (q) – t = p – 1 1.
q – p q - p 

Every factor of sign t that doesn’t contain p or q equals 1, since

t(i) t(j)  i  j  1, if i, j  p, q.

i  j i  j
The remaining factors contain either p or q, but not both. These can be paired together to form
one of the following products.
t(i) – t(p) t(i) – t(q) = i – q i – p l, if i > q,
.
.
i – p i – q i – p i – q 

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