# Permutations

• October 9th 2010, 08:22 PM
jzellt
Permutations
Consider the following permutation f in S6:

f = 1 2 3 4 5 6
.....6 1 3 5 4 2

(The above is one set all in parentheses...)

Calculate f^-1

How do I do this?
• October 9th 2010, 10:20 PM
tonio
Quote:

Originally Posted by jzellt
Consider the following permutation f in S6:

f = 1 2 3 4 5 6
.....6 1 3 5 4 2

(The above is one set all in parentheses...)

Calculate f^-1

How do I do this?

Thus, $1 |\rightarrow 2\,,\,2|\rightarrow 6$ and etc.

Tonio
• October 9th 2010, 10:32 PM
jzellt
So, if I understand you correctly, this would be f^-1:

f^-1 = 1 2 3 4 5 6
..........2 6 3 5 4 1

Is this correct?
• October 9th 2010, 11:24 PM
Roam
Yes, that is fine. You could also write your group in the cycle notation as a product of disjoint cycles $f=(1,6,2)(4,5)$. In this notation it's preferred not to write cycles that have only one entry. Any missing element is mapped to itself.

$f^{-1}=(1,6,2)^{-1}(4,5)^{-1}=(2,6,1)(5,4)$.