MHB Challenge Problem #9 [Olinguito]

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The discussion addresses two assertions regarding permutations in the group S_n. The first assertion is confirmed as true, as the sum of the differences between the permutation and the identity is zero. The second assertion posits that the sum of absolute differences, σ_π, is even, which is also validated based on the first assertion. A bonus challenge involves determining the maximum value of σ_π, with preliminary findings suggesting it may equal ⌊n²/2⌋, though a formal proof is still needed. The exploration of these properties highlights interesting characteristics of permutations and their behavior in relation to their indices.
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Let $S_n$ be the group of all permutations of the set $\{1,\ldots,n\}$. Determine whether the following assertions are true or false.

1. For each $\pi\in S_n$,
$$\sum_{i=1}^n\,(\pi(i)-i)\ =\ 0.$$

2. If
$$\sigma_\pi\ =\ \sum_{i=1}^n\,\left|\pi(i)-i\right|$$
for each $\pi\in S_n$, then $\sigma_\pi$ is an even number.

Bonus challenge: Find $\displaystyle\max_{\pi\in S_n}\,\sigma_\pi$.
 
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I have solved the first part of the problem.

$$\sum_{i=1}^n\,(\pi(i)-i))\ =\ 0$$
follows from the fact that
$$\sum_{i=1}^n\,\pi(i)\ =\ \sum_{i=1}^n\,i$$
(the terms of the LHS sum are precisely those of the RHS sum in a different order).
 
[sp]For the second part, notice that by the first part, the sum of the negative terms in $\sum(\pi(i) - i)$ is the negative of the sum of the positive terms. So when you replace the negative terms by their absolute values, the resulting sum is twice the sum of the positive terms. In other words, it must be an even number.
[/sp]
 
Olinguito said:
Bonus challenge: Find $\displaystyle\max_{\pi\in S_n}\,\sigma_\pi$.
[sp]That part looks harder. After experimenting with small values of $n$, I believe that $$\max_{\pi\in S_n}\sigma_\pi = \bigl\lfloor\tfrac{n^2}2\bigr\rfloor$$. But I don't have a proof of that.
[/sp]
 
Opalg said:
[sp]That part looks harder. After experimenting with small values of $n$, I believe that $$\max_{\pi\in S_n}\sigma_\pi = \bigl\lfloor\tfrac{n^2}2\bigr\rfloor$$. But I don't have a proof of that.
[/sp]
It would be $\dfrac{n^2}2$ if $n$ is even and $\dfrac{n^2-1}2$ if $n$ is odd – a simple induction clinches the proof.

Thanks for helping me with the challenge! (Clapping)
 
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