Let $\sigma \in S_{2n}$ be the permutation defined by $$\sigma := \begin{pmatrix} 1 & 2 & 3 & \ldots & n & n+1 & n+2 & \ldots & 2n \\ 1 & 3 & 5 & \ldots & 2n-1 & 2 & 4 & \ldots & 2n \end{pmatrix} $$ What is the sign of $\sigma$?
I wanted to rewrite $\sigma$ as product of cycles but without success.
Any suggestions? Thanks in advance!
On
For this particular problem, I would use a different approach to sign. Recall that if a permutation $\sigma$ can be written as a product of $k$ transpositions, then it has sign $(-1)^k$. Another way to say this, is if we write out the permutation as a sequence $$\sigma(1)\,\dots\,\sigma(2n),$$ then we can swap adjacent entries $k$ times to get to $12\dots 2n$. For example, for $1324 \mapsto 1234$, so $1324$ is odd. For the next example, we have $$135246\mapsto 132546 \mapsto 123546 \mapsto 123456,$$ so this one is also odd. If you want to work out a pattern by observation, you will probably have to try up to at least $n=8.$ But if you find a formula for the number of swaps, it should not be too difficult to figure out the parity from that.
Hint: Count the number of swaps to get to the identity. First move $2$ up to second position, which takes $n-1$ swaps. Then move $4$ up to fourth position, which takes (how many?) swaps. Continue up to $2n-2$, which just has to swap once. Add them all up and see if the sum is even or odd.