I was trying to understand the action of $D_{2n}$ on $A$ given here $D_{2n}$ Acting on the set consisting of Opposite Pairs of Vertices of a regular n-gon , specifically I was trying to understand this part of the solution:
` If we define $r(j) = j − 1\mod n$ with $0 ≤ j ≤ n − 1$, then we get $r(i) = i − 1,$ if $i ≥ 1 $and $r(0) = n − 1$.Then to any pair $(i, i + n/2)$ with $0 < i < n/2, r(i) = i − 1, r(i + n/2) =i − 1 + n/2, $i.e. $r(i, i + n/2) = (i − 1, i − 1 + n/2) = a_{i−1} \in A$
Thus we can conclude that $r(a_i) = a_{i−1} \mod \frac{n}{2}$
And we define the reflection is about the line connecting vertices $0, 3,$ thus $s(j) = −j \mod n,$ thus to any $a_i, s(i) = −i \mod n = n − i, s(i + n/2) = n/2 − i,$thus $s(a_i) = (n/2 − i, n − i) = a−i \mod n/2$. Thus we have proved s · $a \in A$, $r · a \in A$ for any $a \in A$, therefore $g · a \in A$, for any $g \in D_{2n}, a \in A.$`
But I have the following questions about this part of the solution:
1- why we need "mod n" in the definition of $r(j)$ and $s(j)$? what confuses me is that $0 ≤ j ≤ n − 1$ so I guess the value of $r(j)$ will never exceeds $n$ and similarlyy $s(j)$ I guess.
2- why $r(0) = n-1$ and not $n$?
3- why the value of $r(j)$ implies the given value of $r(i)$?
4- why $i \geq 1$ in a part and $0 < i < n/2$ in another part? are they different $i$? and shouldn 't it be $0 \leq i < n/2$ instead?
5- Why $r(i + n/2) =i − 1 + n/2$ and why $s(i + n/2) = n/2 − i$?
Could anyone help me understand the answers to those problems please?