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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 $|X|=n$.

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    When referring to the $n$-th dihedral group -- despite the fact that its order is indeed $2n$ -- it is a very poor choice of syntax to refer to it as $D_{2n}$. $\mathrm{D}_n$ is a far better choice. Representing an arbitrary group $G$ as a subgroup of the symmetric group $\Sigma(A)$ on a certain set $A$ is equivalent to exhibiting a faithful action of $G$ on $A$. In your case since any of the subgroups of order $2$ of $\mathrm{D}_n$ is of index $n$ and of trivial normal core, any of them suffices to induce an action of the dihedral group on its left quotient w. rsp. to that subgroup. – ΑΘΩ Feb 18 '21 at 12:17
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    This probably helps: https://math.stackexchange.com/questions/2470880/defining-dihedral-groups-sigma-in-s-n-something – Nicky Hekster Feb 18 '21 at 12:32
  • Hint: $D_{2n}$ acts faithfully on the set of vertices of the regular $n$-gon. – user1892304 Feb 18 '21 at 12:34
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    Thank you, @user1892304. Yes, your hint is the classical argument when dealing with this result. However, here my aim is to prove it without any appeal to the geometry, i.e. by assuming the $n$-th dihedral group to be defined by the presentation $\langle r,s\mid r^n=s^2=(sr)^2=1\rangle$. –  Feb 18 '21 at 14:03

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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.

Mike Pierce
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