No cycles in permutation N how to calculate sgn(N^2)?

[ilyas.h]

N is a 2 x n matrix:

N =

1 2 3 4 ... n-1 n
n n-1 ... 4 3 2 1

then N^2 =

1 2 3 4 ... n-1 n
1 2 3 4 ... n-1 nYou COULD use the theorem: sgn(N^2) = sgn(N)sgn(N)

however, I am asked to find sgn(N^2) by the traditional method: sgn(N^2) = (-1)^([L1 - 1] + [L2 - 1]...) where L represents the respective lengths of each cycle. However, N^2 has no cycles, so I am confused.

How would I go about this? thanks.

 

[ilyas.h]

I think I know:

since there are no cycles, you just have sgn(N^2) = (-1)^0 = 1.

can anyone confirm if my method is correct? thanks

 

[ChrisVer]

If that's the N^2 you have, the cycles are 1-cycles... So their parity is + ...
In particular you have the identity element of Sn:
$e=(1)(2)...(n)$
and $W(e)= (1-1)_{first} + (1-1)_{second} + ... + (1-1)_{n-th} =0$ and so the parity:

$\delta_P(e) = (-1)^{W(e)} = +1$