Let $S^1 = \{z \in \mathbb{C} | |z| = 1\}$ and $O_2(\mathbb{R}) = \{A \in GL_2(\mathbb{R})|A^TA = I_2 \}$. Is the group $(S^1, \cdot)$ isomorphic to $(O_2(\mathbb{R}), \cdot)$? Prove it.
This is a question from "Prova Extramuros (2013)", a exam from Brazil. This question asks which of the options are isomorphisms. I easily found the correct option, but in this one I had a little trouble in proving that there was no isomorphism. I interpreted $(S^1, \cdot)$ as a rotation group and $(O_2(\mathbb{R}),\cdot)$ as a rotation and reflection group, so they could not be isomorphic. Is this reasoning correct? Is there a way to prove it more rigorously?
$S^1$ has exactly one element of order $2$: namely $-1$.
$O_2(\Bbb R)$ has at least three: $-I$ and two others formed by changing the sign of exactly one diagonal entry.
Since an isomorphism preserves orders, the two cannot be isomorphic.