How can I prove that the sign of the permutation below is $(-1)^n$
$k↦n+1−k$
(A permutation with reflection property).
I tried using induction but it did not help ( For $n=1$ I get that the sign is $1$ (even permutation) but $(-1)^1=-1$
Thank you for your help!
You are correctly deducing that the sign for $n = 1$ is $1$. The sign of this permutation is simply not $(-1)^n$. We can see that when $n$ is even, we can write the given permutation down as product of transpositions in the following way (using the cycle notation):
$(1,n)(2,n-1)...(\frac{n}{2}-1,\frac{n}{2}+1)$
So if n is even the sign of the permutation is $(-1)^\frac{n}{2}$ since there are $\frac{n}{2}$ factors in the product above. If, on the other hand, n is odd then the product has the following form:
$(1,n)(2,n-1)...(\frac{n-1}{2}, \frac{n+1}{2})$
This time, the product consists of $\frac{n-1}{2}$ transpositions, so the solution in this case would be $(-1)^\frac{n-1}{2}$.
It not difficult to show that that in the general case ($n \in \mathbb{N}$) the solution is $(-1)^{\left \lfloor{\frac{n}{2}}\right \rfloor}$.