$D_{2n}$ is isomorphic to a subgroup of $S_n$ (for $n>2$).

Question

All the proofs I've come across of the fact in the title call into play the action of the group on the vertices of the regular $n$-gon, i.e. they rely on the geometrical definition of $D_{n}$. Am I wrong if I say that this result just follows from being $\{1,s\}\cap\{1,r^2s\}=\{1\}$, for $n>2$? In fact, this suffices to get that the action of $D_{n}$ by left multiplication on $X:=\{gH, g\in D_{n}\}$ is faithful, where $H:=\{1,s\}\le D_{n}$ and $|X|=n$.

Answer 1

You can show $D_{2n}$ is a subgroup of $S_n$ directly. Thinking about the elements $S_n$ in cycle notation, $D_{2n} = \langle r,s\rangle$ where $r = (1\,2\,3\,\dotsc\,n)$ and $$s = \begin{cases}(1\,n) \dotsb (\lfloor{n \over 2}\rfloor\,\lceil{n \over 2}\rceil)\;\text{if $n$ is odd}\\(2\,n)\dotsb({n \over 2}\,{n\over 2} -1)\text{if $n$ is even}\end{cases}$$ noting that this $s$ and $r$ satisfy your relations.