Solutions To Permutation Group Problems

1) Let A = {1, 2, 3, 4, 5} and s  be the result (st)  

 Find: 1 t, 2t , 3t , 4t, and 5t

Compare to:  1 st, 2st , 3st , 4st, and 5st

Soln.:

Let:   t =

 [1 2 3 4 5]

 [3 5 4 2 1]

1(t)  = 2,  2(t)  = 5, 3(t)  = 4, 4(t)  = 2,  5(t)  = 1

From def. illustration:  s = st  =  

[1 2 3 4 5]

[4 2 5 3 1] 

Then:

1(st)  = 4,  2(st)  = 2 , 3(st)  = 5, 4(st)  = 3,  5(st)  = 1

2) Express p =

 [1 2 3 4]

 [2 3 1 4] 

 as the product of transpositions, and determine the sign (+1 or -1) of the resulting end permutation. 

Solution:

Let T1 be the transposition 2 <-> 1 leaving 3, 4 fixed, so:

T1 p =

[1 2 3 4]
[1 3 2 4]


Let T2 be the transposition 2 <-> 3 leaving 1, 4 fixed, so:

T2 T1 p =

[1 2 3 4]
[1 2 3 4]

Then:

T2 T1 p = I (identity)

TWO transpositions (T1, T2) operated on p, so that the sign of the resulting permutation (to reach identity) is +1.  The permutation is therefore even.