Show that in $D_{2n}$, $rs = sr^{-1}$ [First work out what permutation $s$ effects on $\{1, 2, ... , n\}$ and then work out separately what each side in this equation does to vertices $1$ and $2$.] This shows in particular that $r$ and $s$ do not commute so that $D_{2n}$ is non-abelian.
Now consider $D_{2n}$ for $n\geq 4$. Let $S$ be the refelction about the line passing through vertex $2$ and center of regular n gon. Now for this reflection we have $2 \to 2$, and after analyzing other Dihedral groups for $n=5,6,7,8,9$ I have found a pattern that always $1 \to 3, 3 \to 1, n\to 4$. Now if we take $r$ be the rotation $i \to i+1$, then we can show $rs=sr^{-1}$ as $1 \to 4, 2 \to 3$ for both $rs,sr^{-1}$. As $1,2$ to matches so we have equality.
But I have tried analyzing the refection about vertex $n=5$ for Dihedral group $n=5,6,7,8,9$ ,there I got the pattern $4 \to 6, 6 \to 4$. So I think the pattern for refelction theorugh vertex $n=i$ is $i-1 \to i+1, i+1 \to i-1$. But the pattern $n \to 4$ or something that is fixed other than $i,i+1,i-1$, is vanishes.
Why this is happening or am I mistaken something? Or is completely normal?