Q:Prove that the symmetry group of the circle is $O(2, \mathbb{R})$. What is its order?
I'm confused about this question. Here is what I have thought: I need to figure out a matrix that can represent the circle, and proving that this matrix is sufficient to the condition of $O(2, \mathbb{R})$. Is this right? But I don't know how to get the matrix. Any hint for this? Thank you!
If $M\in O(2,\mathbb{R})$ and $v\in S^1$, then $M.v\in S^1$. Furthermore, if $w\in S^1$, then $\bigl\|M.w-M.v\bigr\|=\|v-w\|$. Therefore, every element of $O(2,\mathbb{R})$ is an isometry of $S^1$. On the other hand, every isometry of $S^1$ is of this type.
Of course, $O(2,\mathbb{R})$ is an infinite group, since it contains all matrices for the type$$\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}$$($\theta\in\mathbb R$).