Subgroups of Symmetric groups isomorphic to dihedral group Is $D_n$, the dihedral group of order $2n$, isomorphic to a subgroup of $S_n$ ( symmetric group of $n$ letters) for all $n&gt;2$?
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 $S_n$
If you can build up a faithful action of $D_{2n}$ on a set of $n$ elements, then you are done. I'll adopt this definition of the dihedral group of order $2n$:
$$D_{2n}:=\langle r,s\mid r^n=s^2=(sr)^2=1\rangle \tag 1$$
Let's consider the action of $D_{2n}$ by left multiplication on the set of left cosets of some subgroup $H$ of index $n$, i.e. of order $2$; such an action, say $\varphi$, has kernel $\operatorname{ker}\varphi=\bigcap_{g\in D_{2n}}gHg^{-1}$, and hence if there is a $\tilde g\in D_{2n}$ such that $\tilde gH\tilde g^{-1}\cap H=\{1\}$, then the action is faithful. Now, take $H=\{1,s\}$ and $\tilde g=r$; then $\tilde gH\tilde g^{-1}\cap H=\{1,r^2s\}\cap\{1,s\}=\{1\}$ as soon as $n&gt;2$. Therefore, indeed $D_{2n}\hookrightarrow S_n$ for $n&gt;2$.