Why is the dihedral group closed under composition?!

Question

I've been obsessing over this all day now. I understand associativity, presence of inverse elements and identity, but I don't get why a composition of a reflection with a rotation or other reflexions must be equal to a single reflection or rotation... I would really like to have an intuition of why it works, and maybe understand the proof!

Answer 1

One way of understanding why dihedral groups are closed under composition is to look at the matrix representations of rotations and reflections, which are
$$\begin{pmatrix} \cos\theta & - \sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} \text{ and } \begin{pmatrix} \cos\theta & \sin\theta \\ \sin\theta & -\cos\theta \end{pmatrix} \text{ for some } \theta$$ respectively. It's not too long/hard of an exercise to show that these are the general forms of $2$-by-$2$ orthogonal matrices (these are matrices with $AA^T = A^TA = I$) which form a group (using the definition is much easier than using the general forms!).

As the former set of the matrices have determinant 1 (corresponding to rotations), and the latter having determinant $-1$ (reflections), by the multiplicative property of the determinant, it is then straightforward to see that, for example, two reflections will give a rotation.

While this doesn't show that the dihedral group is closed under composition, it isn't too much work to do this then by noting that the group consists of transformations of the form $A^cB^d$, where $c \in \{0,1\}$, $d \in \{0, 1, \ldots, n-1 \}$, and $A/B$ are the reflections/rotations $$A = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \text{ and } B = \begin{pmatrix} \cos(\tfrac{2\pi}{n}) & - \sin(\tfrac{2\pi}{n}) \\ \sin(\tfrac{2\pi}{n}) & \cos(\tfrac{2\pi}{n}) \end{pmatrix}.$$

Admittedly, this is quite a tiresome and long way of thinking about it...