I'm reading "Camila Jordan, David Jordan Groups – Modular Mathematics series", section 1.2 Symmetries of a circle.
In this section we consider the unit circle. A typical point on the unit circle has coordinates $(\cos\theta, \sin\theta)$ for some $\theta$. In contrast to the square, this circle has infinitely many symmetries. They are all the rotations $rot_{\theta}$ together with all the reflections $ref_{\theta}$. As for the square, these symmetries form a group, the group of symmetries of the circle, but this time the group is infinite. The neutral element is $rot_0$. Each rotation $rot_{\theta}$ has the rotation $rot_{-\theta}$ as inverse while each reflection $ref_{\theta}$ is its own inverse. This group is known as orthogonal group $O_2$. The group $O_2$ acts on the circle by rotations and reflections. Now let us consider the stabilizer of an arbitrary point $P=(\cos\theta, \sin\theta)$ on the circle. Which elements of the group $O_2$ send $P$ to itself? There are just two, namely $rot_0$ and $ref_{2\theta}$
My question is: why does this $O_2$ consist of both rotations and reflections? As if we take any arc on a circle, we can cover all the circle by rotating this arc. Why we would need reflections then?
Best Answer
Think of the circle as being oriented, say counterclockwise. Then rotations preserve this orientation, while reflections change it. Hence there are at least two kind of symmetries...
Note however that rotations can be obtained by composing two reflections about different axes. So in some sense reflections alone do suffice.