Dihedral groups

Dihedral groups are groups of symmetries of regular ngons (meaning regular polygons with n sides, for a fixed n). By a symmetry we mean a rigid transformation of these shapes, meaning that if we were to number the vertices, their order would always remain the same. In particular we have to send vertices to vertices and edges to edges. If you’re like me and have an incredibly hard time visualizing geometric shapes or abstract geometry related problems you might find a picture to be helpful For the rest of the blog post we’ll denote the rotations by r and powers of r, and the reflections by s (in more precise mathematical terms, r is the rotation of $\frac{2\pi}{n}$ radians and s is a symmetry along an axis)

This is a group by composition and it is usually denoted $D_{2n}$ and it is usually denoted as such because its order is 2n, which means the group has 2n elements (but some people are mean and denote it $D_n$ so be mindful of nonstandard notation).

On that note, let’s go over the meaning of order. But first a note on notation: from this point on 1 might denote the identity element in a group (unless it is ambiguous due to context, notation sucks sometimes)

The minimal $n \in \mathbb{Z}^+$ such that $x^n = 1$ is called the order of x when it exists. Otherwise we say that x has infinite order (here $x^n$ denotes $x \cdot x \cdot x ... \cdot x$ (n times) if n is positive, if n is negative then $x^n = (x^{-1})^{-n}$). The only element of order 1 is the identity. Remark: if G is a finite group, the cardinality of G is often referred to as its order. And now a proposition!

Proposition: let G be a group and $x \in G$, the order of x is exactly the size of $\{x^n, n \in \mathbb{Z}\}$

Proof: Assume that x has finite order (n). Pick t < s < n. Assume $x^s = x^t$ then $x^s(x^t)^{-1} = 1 \rightarrow x^s \cdot x^{-t} = x^{s-t}$. Therefore 0 < s-t < n, which is a contradiction since n is the minimal element. It follows that $x^t \neq x^s$. Let $s \geq n$ or s < 0, and s = qn+r $0 \leq r < n, x^s = x^{qn+r} = (x^n)^q\cdot x^r = 1^q\cdot x^r = x^r$ so $\{ x^s, s \in \mathbb{Z} \} = \{x^s : 0 \leq s < n \}$ and the size of that set is n. Assume x has infinite order and $s < t \in \mathbb{Z}$. If $x^s = x^t$ then $x^{t-s} = 1, t-s > 0$ which is a contradiction since x has infinite order. So $x^s \neq x^t$ and $\{x^s, s \in \mathbb{Z}\}$ has infinite elements.

To end this post I’ll go over some basic properties of $D_{2n}$

1. $1, r, r^2, ..., r^{n-1}$ are all distinct and $r^n = 1$ so the order of r is n
2. The order of s is 2
3. $s \neq r^i$ for any i
4. $sr^i \neq sr^j$ for all $0 \leq i, j \leq n-1$ with $i \neq j$ so $D_{2n} = \{1, r, r^2, ..., r^{n-1}, s, sr, sr^2, ..., sr^{n-1}\}$ ie each element can be written uniquely as $s^kr^i$ for some k=0 or 1 and $0 \leq i \leq n-1$
5. $rs = sr^{-1}$. In particular, r and s do not commute so $D_{2n}$ is not abelian
6. $r^is = sr^{-i}$ for all $0 \leq i \leq n$

Given these properties we can see that $D_{2n}$ is completely described by the following

• $D_{2n}$ is generated by r and s, meaning that all elements of $D_{2n}$ can be obtained by combining r and s a number of times.
• The following equations hold $r^n = 1, s^2 = 1, sr = r^{-1}s$