Subgroups of Symmetric groups isomorphic to dihedral group

Question

Is $D_n$, the dihedral group of order $2n$, isomorphic to a subgroup of $S_n$ ( symmetric group of $n$ letters) for all $n>2$?

Answer 1

Let $\left\{a_1,a_2,\ldots,a_n\right\}$ denote the vertices of an $n$-gon. The cyclic permutation $\alpha=(a_1,a_2,\ldots,a_n) \in S_n$. Choose any vertex, e.g. $a_1$ and consider it the fixed point of a reflexion of the plane (i.e the permutation $\beta=(a_2,a_n)(a_3,a_{n-1})\ldots$ of order $2$). This permutation also $\in S_n$. Now $D_n$ is generated by $\alpha, \beta \in S_n$ which shows that the $D_n$ thus defined is a subgroup of $S_n$

Answer 2

Yes, with the standard presentation of $D_n$ the claim follows by just noting that $r\{1,s\}r^{-1}=\{1,r^2s\}$, and $\{1,s\}\cap\{1,r^2s\}=\{1\}$ for $n>2$: 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 $|X|=n$.