Here is the question that I am working on:
Let $\sigma$ be the permutation of the numbers $1,2,…,n$ which reverses their order completely. That is,
$$\sigma=\begin{pmatrix} 1 & 2 & 3 &…&n \\ n & n-1 & n-2&…&1 \end{pmatrix}$$
Is $\sigma$ even or odd?
Here is what I noticed. In general, if I want to find whether a permutation is even or odd, I can write down the permutation in disjoint cycle form and then express that as a composition of transpositions. So, for example, $(123)$ would be even because $(123)$ = $(13)(12)$. The problem is that I'm not sure if this approach can apply to my original question since the permutation $\sigma$ depends on the number $n$. Any more insight on this question would be helpful.
EDIT As explained by the users below, I initially misinterpreted the question, so disregard my first comments in the chat below.
Best Answer
It is hard to write the permutations neatly so I will use words instead.
Note that the first and last elements, $1$ and $n$, are just interchanged. Similarly, the second and second to last, $2$ and $n - 1$, are interchanged, $3$ and $n - 2$, etc.
If $n$ is even then every element is swapped and there are $\frac{n}{2}$ 2-cycles. So, if $\frac{n}{2}$ is even then the permutation is even and if $\frac{n}{2}$ is odd then the permutation is odd.
If $n$ is odd then the element in the middle, $\frac{n+1}{2}$ will be fixed. The remaining $n-1$ elements will be swapped by $\frac{n-1}{2}$ 2-cycles.
So summarising:
$n = 0 \mod 4$ No fixed element and the permutation is even.
$n = 1 \mod 4$ Middle element is fixed and the permutation is even.
$n = 2 \mod 4$ No fixed element and the permutation is odd.
$n = 3 \mod 4$ Middle element is fixed and the permutation is odd.