For $n≥3$, the Dihedral groups may be defined as a collection of rotational isometries $r, r^2, \ldots, r^n=e$ and reflection isometries $s, sr, sr^2, \ldots, sr^{n-1}$ satisfying $(sr)^2 =e$ of $n-$gon.
But how does one make sense of smaller dihedral groups $D_1$ and $D_2$, in that what is a $1$-gon or $2$-gon.
Here is a related question but I don't understand how that $1$-gon or $2$-gon is drawn. As I understand it, a polygon is defined as closed plane figure with "edges" and "vertices". What are the edges and vertices in this case? And hence what would a rotation or reflection look like here? Is there any intuition to it?
I'm new to Dihedral groups so I'd really appreciate an answer in simple English.
Best Answer
The case of $D_2$ can be reasonably done "geometrically" following the usual description of $D_n$ as the "rigid motions of the regular $n$-gon", provided you are okay with degenerate geometrical objects.
There regular $n$-gon consists of $n$ vertices, numbered $1$ through $n$, with edges going from $v_1$ to $v_2$, from $v_2$ to $v_3$, and so on, until we get to an edge from $v_{n-1}$ to $v_n$, and one from $v_n$ to $v_1$. Thus, among other things the regular $n$-gon has $n$ vertices and $n$ edges. We may want to think of the edges as having "direction", which rigid motions may preserve (when they are rotations) or reverse (when they are reflections)
What would be the "regular $2$-gon", then? Well, it consists of two vertices, $v_1$ and $v_2$, an edge from $v_1$ to $v_2$, and an edge from $v_2$ to $v_1$. It's important to note, however, that they are not the same edge: they don't have the same direction, and we need to consider them as two different edges, so that we keep the same pattern of $2$ vertices and $2$ edges. You could think of them as two edges superimposed on one another, or as two edges infinitesimally close to one another. The picture in your link shows them as "curved" segments to emphasize that they are two different edges.
What are the rigid motions of this geometric object? You can fix the two vertices and exchange the two edges (a "reflection"); or you can exchange the two vertices and fix the two edges (a "reflection" again); or you can "rotate" by exchanging the two vertices and the two edges. A little bit of work will show that you get the group $\langle \rho,\sigma\mid \rho^2=\sigma^2=1,\ \sigma\rho=\rho\sigma\rangle$, which fits perfectly into the dihedral framework once you note that $\rho\sigma$ is the same as $\rho^{-1}\sigma$, in the presence of the relation $\rho^2=1$. Here, $\sigma$ is the first reflection I described, and $\rho$ is the rotation. This presentation gives the Klein $4$-group, as expected.
Following the same idea, what would be a regular $1$-gon? It would consists of exactly one vertex $v_1$, and exactly one edge going from $v_1$ to $v_1$; pictorially that would just look like a point, perhaps with an infinitesimally small "loop" around it. That's the picture you have in your link, only greatly magnified to show the "loop".
What are the rigid motions here? Well, $v_1$ must go to $v_1$. But if you think of the edge as having a "direction", then you can either keep its direction (the "rotation") or you can "reverse" it (the "reflection"). It's a conceit or fiction, but one that fits into the general dihedral framework again, so it is worth being fictional so that you get the group $$D_1 = \langle \rho,\sigma\mid \rho^1=\sigma^2=1,\quad \sigma\rho=\rho^{-1}\sigma\rangle,$$ which is a somewhat over-complicated way of presenting the cyclic group of order $2$.