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?

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- Oct 9th 2010, 07:22 PMjzelltPermutations
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? - Oct 9th 2010, 09:20 PMtonio
- Oct 9th 2010, 09:32 PMjzellt
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? - Oct 9th 2010, 10:24 PMRoam
Yes, that is fine. You could also write your group in the cycle notation as a product of disjoint cycles $\displaystyle f=(1,6,2)(4,5)$. In this notation it's preferred not to write cycles that have only one entry. Any missing element is mapped to itself.

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

which is the same as your answer in array notation.