Symmetry group of the unit circle in $\mathbb{R}^2$ versus $\mathbb{R}/\mathbb{Z}$.

Question

I am studying a set of lecture notes on group theory, and I don't think I understand a point the author makes about the unit circle and its symmetry group in relation to $\mathbb{R}/\mathbb{Z}$.

Let $S^1$ denote the set of $(x,y) \in \mathbb{R}^2$ with $x^2 + y^2 = 1$. (This could be complex, I believe, and we'd arrive at the same conclusion. That could be the missing link my understanding.) The symmetry group of $S^1$ consists of rotations and reflections, which is infinite. I cannot fully visualize this, but my mental picture is that it takes a point $(x,y)$ on the circle and either moves it counterclockwise along the unit circle (in polar coordinates, the value of $\theta$ increases mod $2\pi$) or reflects it in any line through the origin.

The author claims that the rotation of $S^1$ is isomorphic to $\mathbb{R}/\mathbb{Z}$. I'm not certain I fully understand this. I believe the complex unit circle is isomorphic to $\mathbb{R}/\mathbb{Z}$ with isomorphism given by (denoting the complex unit circle by $C^1$) $$ f: \mathbb{R}/\mathbb{Z} \to C^1, \; t \mapsto \exp(2\pi t i). $$ Here I am thinking of $\mathbb{R}/\mathbb{Z}$ as the set $[0,1)$, so varying $t$ amounts to varying the polar angle in the complex plane. Since $\mathbb{C} \cong \mathbb{R}^2$ as a vector space, it seems to me that this conclusion should be unchanged. But then $\mathbb{R}/\mathbb{Z}$ is isomorphic to both the unit circle in $\mathbb{R}^2$ and the group of rotations of the unit circle, so the unit circle is isomorphic to its rotation group. This doesn't make sense to me, so something must be wrong with my understanding.

Answer 1

There's only one thing keeping the unit circle from being isomorphic to its rotation group, which is that the unit circle is not technically a group - just a set. There's no intrinsic binary operator on that set of points.

But you can quite easily give it a reasonable group operation - identify the circle with the complex unit circle as you do in the question, and use complex multiplication as your operation. Under this operation, the circle is indeed isomorphic to its set of rotations, and you can easily construct the isomorphism: Just map every point $z$ to the unique rotation that takes the complex number $1$ to $z$.