# Proving that the dihedral group $D_n$ has $2n$ elements

I am trying to prove that the dihedral group $D_n$ has $2n$ elements by using the theory of group actions. Specifically I want to use the orbit stabilizer theorem. So I need $D_n$ to act on a specific set $X$ and then compute the order of the stabilizer and the orbit for some $x\in X$.

My question is presuming $X=\{1,2,\cdots,n\}$ and if $D_n$ acts in the canonical way on $X$ what will be the orbit and stabilizer of $x\in X$?

By the dihedral group $D_n$ I understand the subgroup of $S_n$ generated by the two permutations $\sigma=\left(\begin{smallmatrix}1&2&\cdots&n\end{smallmatrix}\right)$ and $\tau=\left(\begin{smallmatrix}2&3&\cdots&n-1&n\\n&n-1&\cdots&3&2\end{smallmatrix}\right)$.

Update: Here is my attempt: Let $D_n$ act on $X=\{1,2,\ldots,n\}$ in the canonical way, that is by $\rho\cdot x=\rho(x)$. Then $\mathcal{O}_1=X$ as $X=\{\sigma^i(1):1\le i\le n\}\subset\mathcal{O}_1$. So the action is transitive. Furthermore the stabilizer of $1$ is $\{\rho:\rho(1)=1\}$, i.e. all those permutations which fix $1$. Clearly $Id,\tau$ are such permutations.

At this point I wish to prove that there are no other permutations which fix $1$. How do I do that? I am in particular bothered by permutations like $\sigma^{n-2}\tau\sigma^{n-2}\tau\sigma^{4}$? They fix $1$ but how do I show them to be the identity/$\tau$?

(I do not wish to use the fact that $D_n=\{\sigma^i\circ\tau^j:0\le i<n,j=0,1\}$.)

The orbit is the whole set $X$, since we can rotate the polygon with vertices $X$ to move $x$ to any other vertex. The stabilizer is the identity and the reflection about a line through the origin and vertex $x$. So the group has order $n\cdot 2= 2n$.
• Geometrically it is clear, but what will be the stabilizer in terms of the dihedral group as I have defined it? Will it be $\{Id,\tau\sigma^x\}$? – Shahab Apr 2 '15 at 12:43
• Or rather $\{Id,\tau\sigma^{x-1}\}$? – Shahab Apr 2 '15 at 13:06
• $\sigma ^{2x-2}\tau (x)=x$ for $1\leq x\leq n$ – Lozenges Apr 2 '15 at 17:37