Here is the question I want to understand:
Assume $n$ is an even positive integer and show that $D_{2n}$ acts on the set consisting of pairs of opposite vertices of a regular $n$-gon. Find the kernel of this action (label vertices as usual).
Here is a part of the solution I found online:
" Let us assume that $n$ is even, and label the vertices of the $n$-gon clockwise as: $0,1,2, \dots, n-1.$ Now by using the set of ordered pairs defined by $\{a_i\},$ we can represent the $n/2$ pairs of opposite vertices as: $$\{a_i\} = \Big\{\left( i, \frac{n}{2} + i\right) : 0 \leq i \leq \frac{n}{2} \Big\}$$
Now on the elements $\{a_i\}$ define the action of $D_{2n},$ so that the elements of $D_{2n}$ permute the set as follows:
$r^k a_i = a_{{i-k}\mod n}$ and $s^k a_i = a_{{i-kn/2}\mod n}$ "
But I do not understand why the actions of $r^k$ and $s^k$ on $a_i$ looks like this. Could anyone clarify this to me please?