# 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 $n>2$?

• Elements of $D_n$ are faithfully represented by their action on the vertices of the regular $n$-gon, so... Yes! Commented Aug 31, 2014 at 20:25
• @JyrkiLahtonen, though some years late... may I ask you how is formally defined such an action? I mean something like "$\alpha\cdot i:=\dots$". I ask this to get where to start from to prove that this is really an action (or is this just "geometrically evident"?).
– user943729
Commented Oct 4, 2021 at 10:44
• @CAB To me the action is a part of the definition of a dihedral group. That is, after you justify, why a symmetry of the regular $n$-gon must permute the vertices. And, why any such symmetry is uniquely determined by where it sends the vertices. But, the dihedral groups can be defined in various ways. If you are using a definition by generators and relations, then you need to use those instead. And the general result of how you justify the existence of a homomorphism from a group defined by generators and relations. If you use yet a different definition, you need to adjust. Commented Oct 4, 2021 at 11:43
• @JyrkiLahtonen, ok - thanks.
– user943729
Commented Oct 4, 2021 at 11:50

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 $$S_n$$
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 $$|X|=n$$.