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I'm currently doing work on discrete mathematics in my free time and am having some difficulties with understanding some questions pertaining to Relations and Functions. To be specific, I'm stuck on the following practice question:

Let P1 = (3 4 1 2 5), P2 = (3 5 1 2 4) and P3 = (5 1 4 2 3) be three
permutations

1.) Find the permutations below:
    a.) P3^-1
    b.) (P3 ◦ P1)^-1
    c.) P1^-1 ◦ P2

2.) Show that P1 ◦ P2 != P2 ◦ P1

Starting with the easy stuff I know P3^-1 would be (2 4 5 3 1). But after that point I'm not sure what to do to determine the other answers. Any help is appreciated! Thank you.

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    $\begingroup$ Why do you think P3^-1 would be (2 4 5 3 1)? $\endgroup$
    – fleablood
    Oct 6, 2015 at 19:54

1 Answer 1

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To find the inverse of a permutation that is a cycle all we have to do is write the elements of the cycle in reverse order. Thus the inverse of (1 2 3 4) is (4 3 2 1). Since a cycle can be written with any of its elements as the first term we can also write this inverse as (1 4 3 2). This gives an alternative way to write down the inverse of a cycle. Fix the first element in the cycle and write the remaining elements in reverse order. Thus, the inverse of (1 2 3 4 5) is (1 5 4 3 2).

so $P_3^{-1}=(3,2,4,1,5)=(1,5,3,2,4)$

$ (P_3 ◦ P_1)=(5 1 4 2 3)◦(3 4 1 2 5)=(1,3,2)\Rightarrow (P_3 ◦ P_1)^{-1}=(1,2,3)=(2,3,1) $....

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