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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.

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    $\begingroup$ There are many ways to regard dihedral groups in terms other than reflections and rotations of the plane, so it's probably prudent not to refer to elements of an abstract dihedral group as rotations and reflections in their own right, without establishing a bit of geometric context first. $\endgroup$ – Travis Willse Jun 3 '15 at 9:48
  • $\begingroup$ I found a nice answer to this in an expository paper by Keith Conrad. He first points out that every point on a regular polygon is distinguished among all other points on the polygon by its distances from two adjacent vertices on the polygon. Using this, it is established that 2n is an upper bound on the number of elements in the group. Finally, it is simple to show that there are n unique pure rotations, and n unique pure reflections, and clearly no pure reflection equals a pure rotation. Thus there are exactly 2n elements, and these are all accounted for by the pure reflections and rotations $\endgroup$ – nilcit Jun 4 '15 at 10:31
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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.

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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.

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