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I'm struggling a lot with understanding the steps to find the inverse of a permutation. I'm hoping I can get an extremely simple explanation outlining the process. Let's use an easy example,

Let $P_3 = (5, 1, 4, 2, 3)$

Find $P_3^{-1}$

I know that if we can define each element of $P$ based on its $n$-th position. That is, $P(1)=5, P(2)=1,$ and so on. So, obviously, we can define each individual inverse as $P^{-1}(5)=1,P^{-1}(1)=2,$ and so on. What would be the next step to finding the inverse set? I think my confusions occurs with what exactly the inverse of a permutation set is supposed to be.

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  • $\begingroup$ You have found the inverse - just finish the "and so on" and write down the result. (Calculating the inverse is much easier if the permutation is written in cycle notation: a search for cycle notation permutation will find lots of explanations. $\endgroup$ Dec 30, 2020 at 1:12
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    $\begingroup$ the permutation sends P(5)->1, P(1)->4, P(4)->2, P(2)->3, P(3)->5 $\endgroup$
    – yugikaiba
    Dec 30, 2020 at 1:26
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    $\begingroup$ P(1)=4 (not 5), P(2)=3 (not 1) $\endgroup$
    – yugikaiba
    Dec 30, 2020 at 1:27
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    $\begingroup$ Also note that (5, 1, 4, 2, 3) = (1, 4, 2, 3, 5) = (4, 2, 3, 5, 1) = (2, 3, 5, 1, 4) = (3, 5, 1, 4, 2) $\endgroup$
    – yugikaiba
    Dec 30, 2020 at 1:29

2 Answers 2

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Well if $P=(5,1,4,2,3)$ then $P^{-1}=(2,4,5,3,1)$ so that $P(P^{-1}(x))=P^{-1}(P(x))=x$ for each $x\in 1,2,3,4, 5$.

Inverses are always defined in the sense that if you apply a procedure or function to a given element and then the inverse function or procedure to that result, it equates to applying the identity operation and you get the same element you started with.


Method: Begin with $$\bigg(\begin{matrix} 1 & 2 & 3 & 4 & 5 \\ 5 & 1 & 4 & 2 & 3\end{matrix}\bigg)$$

Flip it over: $$\bigg(\begin{matrix} 5 & 1 & 4 & 2 & 3 \\1 & 2 & 3 & 4 & 5\end{matrix}\bigg)$$ and rearrange so that the top row is ordered again: $$\bigg(\begin{matrix} 1&2&3&4&5 \\2 & 4 & 5 & 3 & 1\end{matrix}\bigg)$$

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  • $\begingroup$ Not quite, $P^{-1}=(2,4,5,3,1)$ but $P^{-1}=(2, 4, 1, 5, 3)$ instead $\endgroup$
    – yugikaiba
    Dec 30, 2020 at 1:31
  • $\begingroup$ What exact arithmetic steps are taken to go from $P=(5,1,4,2,3)$ to $P^{-1}=(2,4,5,3,1)$? I don't understand how I'm supposed to figure out $P^{-1}$. $\endgroup$
    – Bobby B
    Dec 30, 2020 at 1:34
  • $\begingroup$ Just draw 5 points 1 2 3 4 5 on the left, then draw them again on the right side, then connect them 5->1, 1->4, 4->2, 2->3, 3->5 and just observe that the inverse 1->5, 4->1, 2->4, 3->2, 5->3. $\endgroup$
    – yugikaiba
    Dec 30, 2020 at 1:39
  • $\begingroup$ @RhysHughes could you offer a step by step process for how you went from $P=(5,1,4,2,3)$ to $P^{-1}=(2,4,5,3,1)$? I don't understand the arithmetic logic for how I'm supposed to go from $P$ to $P^{-1}$ $\endgroup$
    – Bobby B
    Dec 30, 2020 at 1:59
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    $\begingroup$ @Dylan added to my answer for that purpose $\endgroup$ Dec 30, 2020 at 2:47
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That is a cycle permutation, to find the inverse of a cycle, just run the cycle backwards.

i.e. $P=(5,1,4,2,3)$ $\rightarrow$ $P^{-1}=(3,2,4,1,5)$.

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  • $\begingroup$ Could you offer a step by step explanation for how you arrived at $P^{-1}$? I know you simply reversed the order of the set, but why is this the method to find the inverse? $\endgroup$
    – Bobby B
    Dec 30, 2020 at 1:38
  • $\begingroup$ Just draw 5 points 1 2 3 4 5 on the left, then draw them again on the right side, then connect them 5->1, 1->4, 4->2, 2->3, 3->5 and just observe that the inverse 1->5, 4->1, 2->4, 3->2, 5->3. $\endgroup$
    – yugikaiba
    Dec 30, 2020 at 1:40
  • $\begingroup$ So does $5 \rightarrow 1$ as $5$ is the first number in the set? But why does $1 \rightarrow 4$? Same with the rest, I don't understand the logic for how I'm supposed to know which number goes to which. $\endgroup$
    – Bobby B
    Dec 30, 2020 at 1:45
  • $\begingroup$ Every permutation can be written as a disjoint product of cycle decompositions, what you are using is the notation of cycle decompositions. And the rule is (a,b,c,d,e) is for a to go to b, b to go to c, c to go to d, d to go to e, e to go to a. $\endgroup$
    – yugikaiba
    Dec 30, 2020 at 1:48

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