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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!!

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  • $\begingroup$ I'm no sure if i remember it correctly, but isn't $D_\infty$ the Symmetry group of the real numberline? $\endgroup$ – Verdruss Jul 28 '17 at 17:26
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    $\begingroup$ Sorry, not the real numberline but the numberline with the integers marked on it. $\endgroup$ – Verdruss Jul 28 '17 at 17:39
  • $\begingroup$ The set of all permutations of an infinite set $S$ is the set of all bijections $f:S\to S ,$ which is uncountable. What you mean is whether $D_{\infty}$ is the permutation group of some infinite $structure$, in the sense that $D_8$ is the permutation group of a square, not the permutation group of a $4$-member set. $\endgroup$ – DanielWainfleet Jul 28 '17 at 18:22
  • $\begingroup$ Isn’t $D_\infty$ the group of all order-preserving (or reversing) maps of $\Bbb Z$ onto $\Bbb Z$? Every such order-reversing map is an involution (square is the identity). $\endgroup$ – Lubin Jul 28 '17 at 21:59
  • $\begingroup$ @Lubin: I believe this is the cleaner, more succinct way of expressing my suggested answer below. Nice way to think about it. $\endgroup$ – Randall Jul 28 '17 at 22:00
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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.

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  • $\begingroup$ Minor mistake above: $D(S^1)$ gives $O(2)$, not $SO(2)$. $\endgroup$ – Randall Jul 28 '17 at 22:48

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