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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.

Maurizio Barbato
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2 Answers2

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$\newcommand{\bZ}{\mathbf Z}\newcommand{\bR}{\mathbf R}$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.

tomasz
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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.

Cronus
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  • Dear, @Cronus your argument must be flawed, even though my meagre knowledge does not allow me to say where it is the flaw: see my new post R is not isomorphic to a proper subfield of itself. Moreover, also your argument about the fact $O_n(\mathbb{R})$ has no proper subgroup which is topologically isomorphic to it seems flawed too: why should $H$ be open? – Maurizio Barbato Aug 02 '20 at 09:51
  • $K(D)$ cannot be the real numbers: the real numbers are not a purely transcendental extension of the real algebraic numbers. For the same reason, it cannot be isomorphic to the real numbers. – tomasz Aug 02 '20 at 11:07
  • Thanks to you both, and I'm sorry, I was sure this argument works.@tomasz - how come it is not purely transcendental? I realize the real algebraic numbers are not algebraically closed, so it is possible for an extension not to be purely transcendental, but it seems to me we don't solve any new polynomials in this extension, so that it must be purely transcendental... Where is the mistake? – Cronus Aug 02 '20 at 11:27
  • @Cronus: You do solve new polynomial equations: for example, $x^2-\pi=0$. It has nothing to do with the real algebraic numbers not being algebraically closed, though. The complex numbers are not purely transcendental over the algebraic numbers, either. – tomasz Aug 02 '20 at 11:30
  • @MaurizioBarbato: The part about topological subgroups should be true. A submanifold of codimension zero (without boundary) should be open, although I do not see any obvious argument for non-smooth manifolds. – tomasz Aug 02 '20 at 11:39
  • @tomasz I see, I guess I forgot a lot of stuff about basic field theory without realizing it. Thank you. – Cronus Aug 02 '20 at 11:40
  • @tomasz PS, every closed subgroup of a Lie group is an embedded submanifold, this is Cartan's theorem. – Cronus Aug 02 '20 at 11:41
  • @Cronus: I know that. But I think there is likely a more elementary argument there. – tomasz Aug 02 '20 at 11:42
  • @Cronus: Of course, this is just the invariance of domain. – tomasz Aug 02 '20 at 11:43
  • @tomasz I used Cartan's theorem in order to avoid invariance of domain, which is significantly less elementary (in my opinion). – Cronus Aug 02 '20 at 11:44
  • @tomasz By the way, I don't think solving $x^2-\pi$ counts, since this is not a polynomial over the real algebraic numbers. (Even though my reasoning was wrong anyway). – Cronus Aug 02 '20 at 11:45
  • @Cronus: Okay. To be honest, I don't remember their proofs at all, but you might be right. Still, the invariance of domain at least has a more elementary statement. :) – tomasz Aug 02 '20 at 11:46
  • @tomasz Yes, this is true :) Honestly, I just remember having less trouble reading the proof of Cartan's theorem ;) – Cronus Aug 02 '20 at 11:47
  • @Cronus: It does count. More precisely, given any $t\in \mathbf R$, either $x^2-t=0$ or $x^2+t=0$ has a real solution, which is not true in a purely transcendental extension of any field. – tomasz Aug 02 '20 at 11:47
  • @tomasz I see. Interesting. It's not what I meant by "solving new polynomials", but clearly what I had in mind simply doesn't work (since ultimately I was wrong). Thanks for correcting me, anyway :) – Cronus Aug 02 '20 at 11:48
  • @Cronus: For sure you are right about the consideration of $O_2(\mathbb{R})$ as a topological group. The ingredient I missed is the invariance of domain theorem! Thank you very much, Cronus, for your interest in my question! – Maurizio Barbato Aug 02 '20 at 15:28
  • @MaurizioBarbato Sure thing... And I apologize for giving a wrong answer at first. I really thought I knew what I was talking about ;) – Cronus Aug 02 '20 at 19:01