Consider permutations $\pi$ of the set $\{1,\dots,n\}$ having the symmetry property $\pi \pi^* \pi = \pi^*$, where $\pi^*$ is the "reflection" $k \mapsto n+1-k$. Are there references or other information about such a $\pi$?
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asked Sep 12, 2018 at 18:28
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$\begingroup$ You really do mean $\pi\pi^*\pi$, not $\pi\pi^*\pi^{-1}$? $\endgroup$– LSpiceCommented Sep 12, 2018 at 19:09
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2$\begingroup$ Also, is $\pi^*$ meant to depend on $\pi$, say as $k \mapsto n + 1 - \pi(k)$, or is it really a fixed permutation independent of $\pi$? If the latter, then you are just looking for permutations conjugate by a fixed involution to their inverses. $\endgroup$– LSpiceCommented Sep 12, 2018 at 19:10
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4$\begingroup$ In other words, $\pi \pi^{*} $ is involution. It should reduce the questions about your permutations to the questions about involutions. $\endgroup$– Fedor PetrovCommented Sep 12, 2018 at 19:11
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$\begingroup$ @FedorPetrov's re-phrasing is better than mine. $\endgroup$– LSpiceCommented Sep 12, 2018 at 19:11
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2$\begingroup$ Why don't you use a notation without the letter $\pi$ to denote the reflection? $\endgroup$– YCorCommented Sep 15, 2018 at 16:59
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1 Answer
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The problem boils down to finding permutations $\pi$ such that $\pi(n+1-x)=y\iff\pi(n+1-y)=x$. Let $X\subset\{1,...,n\}$ be such that $n-\lvert X\rvert$ is even and let $P$ be a partition of $\{1,...,n\}\backslash X$ into pairs and denote, for every element $k$ of $\{1,...,n\}\backslash X$, its companion by $P(k)$. Then, the permutation defined by $\pi(n+1-k)=P(k)$ for every $k\in\{1,...,n\}\backslash X $ and $\pi(n+1-x)=x$ for every $x\in X$ satisfies the required condition. It is clear that appropriate permutations are in a one-to-one correspondence with $(X,P)$ couples.
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$\begingroup$ Isn't this just a complicated way of saying "$\pi$ is of the form $\sigma\pi^*$, where $\sigma$ is an involution" (which is what @FedorPetrov's comment's that you quote says)? I guess it depends on the poster's needs which form of the statement is more convenient …. $\endgroup$– LSpiceCommented Sep 15, 2018 at 12:49
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$\begingroup$ I thought a constructive/descriptive approach might be helpful depending on the OP's needs indeed. I just lightened my response a bit. $\endgroup$ Commented Sep 15, 2018 at 16:05
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$\begingroup$ By the way, as a TeX matter,
\midis only intended as a binary operator (like2 \mid 6: $2 \mid 6$). For delimiters, you want\lvertand\rvert(like\lvert2\rvert: $\lvert2\rvert$). I made the edit here. $\endgroup$– LSpiceCommented Sep 15, 2018 at 17:41