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According to Wikipedia, "the infinite dihedral group Dih∞ is an infinite group with properties analogous to those of the finite dihedral groups." However, it doesn't appear that this has anything to do with the symmetries of the circle, which surprised me, because that seems like the most natural generalization of the finite-order dihedral groups, which are sets of symmetries of regular polygons.

Put another way, since regular polygons "approach" becoming a circle as the number of vertices, $n\to\infty$, it seems like $D_\infty$ should be the symmetries of the circle.

So what kinds of symmetries does $D_\infty$ represent, and why was this group chosen for extension of the finite dihedral groups over the symmetries of the circle? (Or are they related in some way I'm just not grasping?)

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    $\begingroup$ There is some explanation later in the Wikipedia page, particularly in the "Definition" section. $\endgroup$ – Eric Wofsey Nov 2 '16 at 5:24
  • $\begingroup$ Taking the limit of symmetry groups is not going to work... $\endgroup$ – Mariano Suárez-Álvarez Nov 2 '16 at 5:29
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    $\begingroup$ Some properties are not preserved by limits (even when there is a sensible way to take the limit of the property). eg an $n$-gon has $n$ vertices, but a circle has none (suppose we interpret vertex as point where the curve isn't differentiable). $\endgroup$ – stewbasic Nov 2 '16 at 5:39
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    $\begingroup$ I realize you can't actually take a limit, I just provided that description as a sort of intuitive feel for where my thoughts are going and why it seems to me like the symmetries of a circle seem like the more natural infinite generalization of the (finite) dihedral groups. $\endgroup$ – MightyTyGuy Nov 2 '16 at 6:39
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Although polygons "approach" the circle, the circle has quite a few more points than just the union of the vertices of all the $n$-gons inscribing it, and consequently it "should" have more symmetries. The symmetry group (rotations and reflections) of the circle is known as $\text{O}(2, \mathbb{R})$ or sometimes just $\text{O}(2)$, which can be seen as the set of all orthogonal ($A^t = A^{-1}$) 2-by-2 matrices with real entries, under multiplication.

Instead, you could think of $D_\infty$ as being the "limit" of the presentations for the finite dihedral groups, $$D_n = \langle r, s \mid r^n = s^2 = 1,\, srs = r^{-1} \rangle.$$ As $n$ goes to infinity, the statement that $r^n = 1$ simply drops out, because what "$r^n = 1$" is really saying is "$\text{ord}(r) = n$", and we write $\text{ord}(r) = \infty$ to mean $r$ has infinite order. So $$D_\infty = \langle r, s \mid s^2 = 1,\, srs = r^{-1}\rangle$$ which some authors take as the definition of $D_\infty$. Others define the infinite dihedral group as the isometries of $\mathbb{Z}$ as a subset of $\mathbb{R}$, in which case $x \mapsto x + 1$ can be thought of as $r$ (a "rotation") and $x \mapsto -x$ can be thought of as $s$ (a "reflection").

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What you think of when taking $n$ to $\infty$ is this:

enter image description here

That is, you've got polygons inscribed in a common circle. Note however that even in that model, the limit to infinity would not get the full circle, because only rational multiples of $2\pi$ will appear.

But the limit that leads to $D_\infty$ is more like this:

enter image description here

And if you continue that to infinity, it's easy to see that the the sides will approach a straight line with equidistant points, and a rotation will end up as a translation on that line.

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    $\begingroup$ This is really clever and insightful. I hadn't ever really thought about the problem of only rational multiples of $2\pi$ being included - that works well with Kaj Hansen's answer above - that the cardinalities don't match up, since the model $D_\infty$ follows has the same cardinality as the rational numbers, whereas the symmetries of a circle are uncountable. $\endgroup$ – MightyTyGuy Nov 3 '16 at 2:40
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Finite dihedral groups, in general, are defined as particular quotients of the free group on two letters, with the associated normal subgroup being generated by the relations $r^n = f^2 = (rf)^2 = 1$. The natural extension of this idea to the infinite case is to let $\langle r \rangle \cong \mathbb{Z}$ since the rotation subgroup is always cyclic, and the size of the rotation subgroup is the only thing differentiating all the various finite dihedral groups, structurally speaking.

One problem with thinking about $D_{\infty}$ as the symmetry group of a circle is that the cardinalities simply don't match up. As a quotient of a free group on two letters, $D_{\infty}$ is countably infinite. On the other hand, the rotation subgroup of the group of symmetries of a circle is uncountably infinite as we can rotate $S^1$ by any angle $\theta \in [0, 2\pi)$.

Note that we can realize $D_{\infty}$ as the symmetry group of the set of integer points on a real number line: the rotation can be thought of as the translation $x \mapsto x \! + \! 1$ and the reflection as reflection across the origin. And this kind of makes sense given that the internal angle of a regular $n$-gon measures $\displaystyle 180- \frac{360}{n}$ degrees, and this approaches $180^{\circ}$ as $n \to \infty$.

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  • $\begingroup$ I'm not sure how sequential compactness comes into play here; AFAIK $D_\infty$ is not a topological group. Indeed, you can embed it into $O(2)$ by mapping the operation $x\mapsto x+1$ to the rotation by the angle $1$ (which is incommensurable with $2\pi$, so all integers will lead to different rotations) and the reflection $x\mapsto -x$ to mirroring on the horizontal axis. $\endgroup$ – celtschk Jun 26 at 4:20
  • $\begingroup$ Wow, that's such a beautiful idea that I wasn't aware of; thanks for commenting @celtschk. I assumed that the order of the points around the circle would need to be preserved by the rotation, which would require each point to be equidistant from its immediate neighbors, and I was claiming that sequential compactness makes this impossible. Or so I think. $\endgroup$ – Kaj Hansen Jun 26 at 9:46
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$D_\infty$ is the group of symmetries of an $\infty$-agon, of which a simple model is the set $\mathbb Z$ inside of $\mathbb R$.

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