Dummit & Foote 1.7.12 - finding a nicer expression for $r^xs^y\cdot a$ for $a\in A$ 
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.

Here's my solution to the first part, showing that this is a group action:

Let $$A = \left\{\left\{i, i+\frac{n}{2}\right\}: 1\le i\le \frac{n}{2}\right\}$$
be the set of pairs of opposite vertices, labeled from $1$ to $n$. In words, we are required to first show that two opposite vertices remain opposite after rotations and reflections of the $n$-gon, as dictated by elements of $D_{2n}$. Given this, we can define a map $D_{2n}\times A \to A$, and check that it is a group action. First of all, to ensure that $g\cdot a\in A, \forall g\in D_{2n},\forall a\in A$, we need to check that $r\cdot a, s\cdot a \in A$ for all $a\in A$ (because any $g\in G$ is just $g = r^is^j$).
Consider arbitrary $a\in A$ given by $\{i,i+\frac{n}{2}\}$ for $1\le i\le \frac{n}{2}$. $r\cdot a = \{i+1 \bmod n, i+\frac{n}{2}+1\bmod n\} \in A$. $s\cdot a = \{n+2-i, \frac{n}{2}+2-i\} \in A$.
Also, $1_G\cdot a = a, \forall a\in A$.
Let $g_1 = r^cs^d, g_2 = r^es^f \in D_{2n}$. Consider arbitrary $a\in A$ given by $\{i,i+\frac{n}{2}\}$ for $1\le i\le \frac{n}{2}$. Note that $r^c\cdot a = \{i+c \bmod n, i+\frac{n}{2}+c\bmod n\}$. Also, $s^d\cdot a = a$ if $d$ is even, and $s^d\cdot a = s\cdot a$ if $d$ is odd. Verify $g_1\cdot (g_2\cdot a) = (g_1g_2)\cdot a$. Hence, this is a group action.

The last part in bold is where I need help. I know $r^c\cdot a$ and $s\cdot a$ but to verify $g_1\cdot (g_2\cdot a) = (g_1g_2)\cdot a$ I would need a nice expression for $r^cs^d\cdot a$. Otherwise, it's too much computation that's involved, i.e. assuming $d=0,1$ and going ahead making cases. Is there an easier way to do this?
Also, I can see intuitively see that the $r^{n/2}$ and the identity element of $D_{2n}$ are in the kernel of this action. I will need an expression for $r^xs^y\cdot a$ for $a\in A$ to find other elements in the kernel, or to show that there are none.
Thank you!
 A: The trick is to note that $r^{n/2}$ is in the center, which spares you all the symbolic work. For the Kernel, we use orbit stabilizer to find an upper bound of two possible the elements, which we already know.
I will be overly verbose here, but you can condense the argument a lot.
Recall that the dihedral group $D_{2n}$ has a natural action on the set $V := \{1,\cdots,n\}$ by defining it to be generated by the permutation action $r := (1,\cdots,n)$ and $s(i) := ((i+n/2 -1) \operatorname{mod}\: n) +1$.  Also, $r^n=s^2=(rs)^2=e$.  A simple inductive argument shows that $r^j s = sr^{-j}$ for all integers $j$, which in particular implies that $r^{n/2}$ commutes with both $r$ and $s$, and hence is in the center.
We will denote the aforementioned action with a dot, i.e. by $r.i := r(i)$.

Now, we will define $\overline V$ to be $V$ modulo the antipodal equivalence relation you established, and writing $[i]$ for the equivalence class of $i\in V$, we define a new map $$
  D_{2n}\times \overline V\to \overline V,\qquad g.[i]:= [g.i],
$$ And show that this is well-defined.
Indeed, let $i\neq i'\in [i]$. Since $i'$ is the antipode, we can represent it as $r^{n/2}.i$, and since $r^{n/2}$ is in the center, we have $$
g.[i'] = [g.i'] = [g.(r^{n/2}.i)] = [(gr^{n/2}).i] = [(r^{n/2}g).i] =[r^{n/2}.(g.i)] = [g.i] = g.[i],
$$ where the second to last equality follows from the fact that application of $r^{n/2}$ does not change the equivalence class.
The fact that this does constitute an action follows then directly from the fact that $-.-\colon D_{2n}\times V\to V$ is an action.

To find the Kernel of that action, recall that the orbit stabilizer theorem applied to $x := [1]\in \overline V$ gives us $$
|D_{2n}| = |D_{2n}.x| \cdot |(D_{2n})_x|,
$$ and since the action is transitive (powers of $r$ can map $x=[1]$ to an arbitrary element), the cardinality $|D_{2n}.x|$ of the stabilizer is $|\overline V|=n$, implying the stabilizer has exactly two elements.
Furthermore, we know that the kernel $K$ of the action contains group elements that stabilize all of $\overline V$, implying that those in particular stabilize $x$.  So we have $K\subseteq (D_{2n})_x$.
On the other hand, at least the identity and $r^{n/2}$ do stabilize everything in $\overline V$, which means we have $$
  \{1, r^{n/2}\} \subseteq K = (D_{2n})_x,
$$ and since the right hand side has cardinality two, we have equality.
