Permutation on an infinite set For $n \ge 3$, the dihedral group is $$D_{2n}=\langle r,s: s^2=e,r^n=e \text{ and } rs=sr^{-1} \rangle$$
and I know that $D_{2n}$ is isomorphic to the group of permutations on a regular $n$-gon. 
The infinite dihedral group is defined as $$D_{\infty}=\langle r,s: s^2=e\text{ and } rs=sr^{-1} \rangle$$
My Question: Can we show that $D_{\infty}$ is isomorphic to the group of permutations on an infinite set, for some well defined infinite set?
Thanks for the help!!
 A: Not a permutation representation per se, but here's where your countable $D_\infty$ shows up in something permutation-ish.  (You can define $D(A)$ for any abelian group $A$, but we are talking about $D_\infty= D(\mathbb{Z})$ in particular.  The classical dihedral group $D_n$ of order $2n$ shows up this way as $D(\mathbb{Z}_n)$.)
Give the set $\mathbb{Z}$ the topology generated by basic open sets 
$B(n) = \begin{cases} \{n\}, & n \text{ odd}\\ \{n-1, n, n+1\}, & n \text{  even.}\end{cases}$
It's not hard to check that these generate a legit topology on $\mathbb{Z}$.   This space is usually called the digital line, so let's write it as $\mathbb{D}$.  
As with any space $X$, you can ask to identify/compute its automorphism group $\mathrm{Aut}(X)$, which is the group of all self-homeomorphisms $f: X \to X$ under function composition.  
With some work, it turns out that $\mathrm{Aut}(\mathbb{D}) \cong D_\infty$ in a natural way.  You do this in a couple of stages.  First one proves that any automorphism of $\mathbb{D}$ is either a translation (necessarily by an even shift!) or a reflection.   Next, you can show that translation by $2$ and reflection about $0$ generate the entire automorphism group.  You can then check that you have an internal semi-direct decomposition 
$\mathrm{Aut}(\mathbb{D}) = \langle \rho_0 \rangle \ltimes T(\mathbb{D})$
where $\rho_0$ is the reflection about $0$ and $T(\mathbb{D})$ is the normal subgroup of translations.  This $\rho_0$ is your $s$ and the elements of $T(\mathbb{D})$ are your $r$'s.  This gives 
$\mathrm{Aut}(\mathbb{D}) \cong D(\mathbb{Z}) = D_\infty$.
So, your infinite dihedral group is the group of "continuous permutations" on $\mathbb{D}$. This is extra cute because $D(\mathbb{R})$ also gives the isometry group of the real line, and $D(S^1)$ gives $SO(2)$, the linear isometry group of the plane. 
