$D_{2n}$ Acting on the set consisting of Opposite Pairs of Vertices of a regular n-gon I have been asked to show that $D_{2n}$ acts on the set consisting of pairs of opposite vertices of a regular $n$-gon, where $n$ is taken to be a positive even integer. 
The first thing that came to my mind was to represent $r$ and $s$ in cycle notation, i.e., to show that $D_{2n}$ is isomorphic to a certain subgroup of $S_n$, and then I can show that permutations acting the two element sets of opposite vertices is a group action of $D_{2n}$ on them. More specifically, if $\sigma$ is a permutation and $\{v_1,v_2\}$ is a set consisting of opposite vertices, then I would show $\sigma\cdot \{v_1,v_2\} := \{\sigma(v_1), \sigma(v_2) \}$ is a group action. 
So, $r \mapsto (1~2~...~n)$ and $s \mapsto (2~n)(3~(n-1))~...?$.
There are a few things I am having difficulty with: (1) will this argument succeed; (2) how do I make the arguments more rigorous; (3) how do I determine exactly what $s$ gets mapped to; and (4), given an arbitrary vertex, what will its opposite vertex look like (in other words, what does the set $\{v_1,v_2\}$ look like)? 
Note, more questions may come. 
 A: Define the set consisting of pairs of opposite vertices of a regular n-gon as $A$. By the
definition of group action, we need to show any $h, g \in D_{2n}$,$ a \in A, g · a \in A, 1 · a = a,$
$h · (g · a) = (gh) · a.$
Since $1$ doesn’t make any change to the n-gon, thus $1 · a = a,$ for any $a \in A$.
And any element of $A$ is a combination of $s, r,$ thus in order to show $g · a \in A,$ we just need to show $s · a \in A, r · a \in A.$ We may define the vertices clockwise as $0, 1, 2, ..., n − 1$
Let the $\frac{n}{2}$ pairs of opposite vertices defined to be $a_i:= \{i, i + n/2\}, 0 ≤ i < n/2$.
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 $g \in D_{2n}, a \in A.$
Since any $g \in D_{2n}$ is in the form of $r^is^j$
, we just need to show $(r^as^b) · ((r^xs^y) · ai) =((r^as^b)(r^xs^y)) · a_i$
. Indeed, by computation, we can get
$((r^as^b)(r^xs^y)).a_i=(r^{a+(-1)^bx}s^{b+y}).a_i= a_{-1^{y+b}i + (-1)^{b+1} x-a \mod n/2}=r^as^b.a_{-1^yi-x \mod n/2}$ .Thus it is an action
since $r^xs^y.a_i= a_{-1^yi-x \mod n/2}$
Thus if $r^xs^y.a_i=a_i$ then$ x= n/2 ,0.y=0 $,
if $y = 1$ then $x$ depend on $i $, thus if $ r^xs^y.a_i=a_i $ for any$ i$ ,then$ y=0 , x= n/2, 0$.so kernel is $\{r^0=1, r^{n/2}\}$
