Elements of Dihedral Groups

Question

Is there an elegant way of showing that the elements of a dihedral group are only rotations and reflections? Specifically, I'm having trouble convincing myself that a composition of a rotation and a reflection must always result in either a rotation or reflection.

Answer 1

One can show that for integers $n > 1$ the (matrix) subgroup $G \leq GL(2, \Bbb R)$ with elements $$\pmatrix{\frac{\cos 2 \pi k}{n} & \mp\frac{\sin 2 \pi k}{n} \\ \frac{\sin 2 \pi k}{n} & \pm\frac{\cos 2 \pi k}{n}}$$ is isomorphic to the dihedral group $D_{2n}$ of order $2n$. (Hence the map $D_{2 n} \to GL(2, \Bbb R)$ is a faithful representation of $D_{2n}$.) In particular, this is actually a subgroup of the orthogonal group $O(2, \Bbb R)$, and so the elements of this group can be identified by with the linear transformations of the plane that preserve the standard inner product; these are precisely the rotations (given by taking $\pm$ to be $+$ in the above) and the reflections ($-$).

With this representation in hand, it's easy to check (either directly, by multiplying generic elements of $G$, or using that $\det G \to \Bbb R^*$ is a group homomorphism) that the product of a rotation and a reflection (in either order) is a reflection.

Answer 2

Here is a somewhat more synthetic answer to the question.

Given a rotation $f \in \text{Isom}(\mathbb{E}^2)$ and a reflection $g \in \text{Isom}(\mathbb{E}^2)$, it is not generally true that the composition $g \circ f$ is a pure reflection. That composition is either a pure reflection or a glide reflection, and the latter occurs if and only if $f$ is a rotation through angle $\pi$ and its center of rotation does not lie on the reflection line of $g$. But that cannot happen if $f,g$ are elements of a standard dihedral subgroup of $\text{Isom}(\mathbb{E}^2)$, because all elements of the subgroup share a common fixed point.