Proper Subgroup of $O_2(\mathbb{R})$ Isomorphic to $O_2(\mathbb{R})$ in his very beautiful answer to my post Subgroup of Plane Isometries Isomorphic to $O_2(\mathbb{R})$, Angina Seng stated without proof that there exists a proper subgroup of $O_2(\mathbb{R})$ which is isomorphic to $O_2(\mathbb{R})$.
He suggested to use the usual "Zorn's Lemma/Hamel basis" argument, but I cannot see what he really meant, since $O_2(\mathbb{R})$ is not even a vector space.
Could anyone help me please?
Thank you very much for your help in advance.
 A: $\newcommand{\bZ}{\mathbf Z}\newcommand{\bR}{\mathbf R}$Note that $O(2)\cong (\bR/\bZ)\rtimes (\bZ/2\bZ)$. Thus, it suffices to find a proper subgroup of the torus $\bR/\bZ$ (isomorphic to the torus).
To do this, just fix a basis $B$ of the reals containing $1$ (over the rationals). Then for every $B'\subseteq B$ containing $1$, having the cardinality of the continuum, you have $\operatorname{span}(B')/\bZ\cong \bR/\bZ$: just fix a bijection $B'\to B$ (fixing $1$), extend it to a linear isomorphism $\operatorname{span}(B')\to \mathbf R$ and note that it induces the desired isomorphism.
A: Let $K=\Bbb{R}\cap\bar {\Bbb{Q}}$, the algebraic closure of $\Bbb{Q}$ inside $\Bbb{R}$, and let $D\subset\Bbb{R}$ be a set of cardinality the continuum of algebraically independent elements. Assume moreover that $K(D)\neq \Bbb{R}$ (this is not a problem: if $K(D)=\Bbb{R}$, just replace $D$ with $D\setminus\{d\}$ for some $d\in D$). Then $K(D)$ is a proper subfield of $\Bbb{R}$ which is isomorphic to $\Bbb{R}$, and hence $\mathrm{O}_n(K(D))$ is a proper subgroup of $\mathrm{O}_n(\Bbb{R})$ which is isomorphic to $\mathrm{O}_n(\Bbb{R})$ (for every $n$).
This is more or less the answer given by Dietrich Burde (I just added the explanation of how to find a proper subfield of $\Bbb{R}$ which is isomorphic to $\Bbb{R}$), but what I would like to point out is the following: $K(D)$ is isomorphic to $\Bbb{R}$ only as an abstract field, and $\mathrm{O}_n(K(D))$ is isomorphic to $\mathrm{O}_n(\Bbb{R})$ only as an abstract group. It is not isomorphic to it as a topological group.
Here's a proof this can't happen when we consider $\mathrm{O}_2(\Bbb{R})$ as a topological group (rather than an abstract group). If $H$ is a subgroup of $\mathrm{O}_n(\Bbb{R})$ which is topologically isomorphic to it, then it is compact, and hence closed. In fact, it is also of the same dimension, so it is even open. Therefore it is either $\mathrm{O}_n(\Bbb{R})$ or $\mathrm{SO}_n(\Bbb{R})$. But $\mathrm{SO}_n(\Bbb{R})$ is not isomorphic to $\mathrm{O}_n(\Bbb{R})$, so $H$ must be $\mathrm{O}_n(\Bbb{R})$, so it is not a proper subgroup. So there are no proper subgroups of $\mathrm{O}_n(\Bbb{R})$ which are topologically isomorphic to it.
