Subgroup of Plane Isometries Isomorphic to $O_2(\mathbb{R})$

Question

Let $\mathcal{Isom}(\mathbf{E})$ be the group of the isometries of the euclidean plane $\mathbf{E}$, and for every point $P \in \mathbf{E}$, let $\mathcal{Isom}_P$ be the subgroup of all isometries of $\mathbf{E}$ that fix $P$. It is well known that $\mathcal{Isom}_P \simeq O_2(\mathbb{R})$. Is the converse true? That is, if $H$ is a subgroup of $\mathcal{Isom}(\mathbf{E})$, and $H \simeq O_2(\mathbb{R})$, is is true that there exists some point $P \in \mathbf{E}$ such that $H = \mathcal{Isom}_P$?

Even though it seems quite a reasonable statement, every attempt that I made to prove it has failed up to now, so any help is welcome in answering this question. Thank you very much in dvance for your kind attention.

Answer 1

$H$ has a subgroup $K$ of index $2$ isomorphic to $SO_2$ and so commutative. Then $K$ has an element $g$ of order $3$. The isometries of the plane of order $3$ are rotations through angle $\pm2\pi/3$. So $g$ is a rotation centred at a some point $P$. The centralizer of $g$ in the isometry group of the plane is the set of all rotations with centre $P$. So $K\subseteq\text{Rotations}_P$

The elements of $H-K$ don't centralise $g$ but normalise the subgroup it generates. The isometries which do this are the reflections fixing $P$. Therefore $H$ is a subgroup of the stabiliser of $P$.

But there are proper subgroups of $O_2$ which are isomorphic to $O_2$ so we cannot say that $H$ is the stabiliser of $P$, but if say $H$ is a closed subgroup of the isometry group, then it is.

Answer 2

This is not a separate answer, but simply a long comment to Angina Seng's answer which has the purpose of fill in the missing details. First of all, let us recall Chasles classification theorem for plane isometries: an isometry of the plane is a translation, a rotation, a reflection or a glide reflection.

Now let $g \in K$ be an element of order 3. Since translations and glide reflections have infinite order, while reflections have order 2, $g$ must be a rotation around some point $P$, and since $g$ has order 3, $g$ must be a reflection of $\frac{2 \pi}{3}$ or of $- \frac{2 \pi}{3}$. Now consider the centralizer $\mathcal{C}(g)= \{ f \in \mathcal{Isom}(\mathbf{E}): gf = gf \}$. Clearly every rotation around $P$ belongs to $\mathcal{C}(g)$. Conversely, let $f \in \mathcal{C}(g)$, and let $f(P)=Q$. Then $(gf)(P)=g(f(P))=g(Q)$, while $(fg)(P)=f(g(P))=Q$. So $g(Q)=Q$, and since $g$ fixes only $P$, we must have $P=Q$, that is $f$ fixes $P$. So $f$ must be a rotation around $P$ or a reflection with respect to a straight line $r$ passing through $P$. In this latter case we would have (remember that if $s$ is a reflection with respect to a straight line passing through $P$ and $r$ is a rotation around $P$ we have $rs=sr^{-1}$): $gf=fg^{-1} \neq fg$, since $g \neq g^{-1}$. So $f$ must be a rotation around $P$. We conclude that

$\mathcal{C}(g)$ coincides with the set of all rotations through $P$.

Since $K$ is abelian, we have $K \subset \mathcal{C}(g)$, and so $K$ is a subset of the set of all rotations around $P$.

Now consider $H \backslash K$. Since $H \simeq O_2$ and $K \simeq SO_2$, the elements of $H \backslash K$ correspond to those of $O_2 \backslash SO_2$, and so they have the same algebraic properties. In particular, if we denote by $G$ the subgroup generated by $g$ (that is $G=\{id, g, g^2 \}$), we have that each element $h \in H \backslash K$ satisfies the two properties:

(i) $h \in N(G)=\{ f \in \mathcal{Isom}(\mathbf{E}): fGf^{-1}=G \}$;

(ii) $h \notin \mathcal{C}(g)$.

Now, consider the set $N(G) \cap (\mathcal{Isom}(\mathbf{E}) \backslash \mathcal{C}(g))$, and let $f \in N(G) \cap (\mathcal{Isom}(\mathbf{E}) \backslash \mathcal{C}(g))$. Let $f(P)=Q$. Since $fgf^{-1} \in G$ and $fgf^{-1} \neq id$ (since the only conjugate of the identity is the identity itself), we must have $fgf^{-1}=g^2$. So $fgf^{-1}=g^2$ fixes only $P$. But we have $(fgf^{-1})(Q)=Q$, and we conclude that $P=Q$, that is $f$ fixes $P$, so $f$ is a rotation around $P$ or it is a reflection with respect to a straight line passing through $P$. But since $f \notin \mathcal{C}(g)$, the first case is not possible, and we conclude that $f$ must be a reflection with respect to a straight line passing through $P$. On the other hand, every reflection $f$ with respect to a straight line passing through $P$ is in $N(G) \cap (\mathcal{Isom}(\mathbf{E}) \backslash \mathcal{C}(g))$. Indeed, as seen above $f \notin in \mathcal{C}(g)$, while, since we have $fgf^{-1}=g^{-1}=g^{2}$ and $fg^{2}f^{-1}=g^{-2}=g$, we have $f \in N(G)$. We conclude that

$N(G) \cap (\mathcal{Isom}(\mathbf{E}) \backslash \mathcal{C}(g))$ is the set of all reflections with respect to straight lines passing through $P$.

In particular $H \backslash K$ is a subset of the set all reflections with respect to straight lines passing through $P$.

So $H$ is a subgroup of $\mathcal{Isom}_P$.

Now we shall prove that with the additional assumption that $H$ is a closed subgroup of $\mathcal{Isom}(\mathbf{E})$, then $H= \mathcal{Isom}_P$. First, let us note that if $r$ is an element of $K$ of order $m$, where $m$ is a positive integer, then $r$ is a rotation around $P$ of order $m$, so that the subgroup $\langle r \rangle$ of $H$ generated by $r$ coincides with the subgroup $\langle s \rangle$ of $\mathcal{Isom}_P$ generated by the rotation $s$ of $\frac{2 \pi}{m}$. So the set \begin{equation} S= \{ r \in K : \text{r has finite order} \}, \end{equation} is dense in the subgroup $\mathcal{Rot}_P$ of all rotations around $P$. On the other hand, since for any reflection $\rho$ with respect to a straight line $r$ passing through $P$, and any rotation $R \in \mathcal{Rot}_P$ of angle $\theta$, $R \rho$ is the reflection with respect to the straight line $s$ passing through $P$ and rotated by an angle $\theta/2$ with respect to $r$, we see that, chosen any element $\rho \in H \backslash K$, the subset of $H$ defined by \begin{equation} \rho S = \{ \rho R : R \in S \}, \end{equation} is dense in $\mathcal{Isom}_P \backslash \mathcal{Rot}_P$. We conclude then that $H$ is dense in $\mathcal{Isom}_P$, and so $H= \mathcal{Isom}_P$.

The final issue to settle is to show that, without any topological hypothesis on $H$, we can have that $H$ is a proper subgroup of $\mathcal{Isom}_P$, which is equivalent to say that $O_2(\mathbb{R})$ admits some proper subgroup which is isomorphic to $O_2(\mathbb{R})$ itself. This has been proved by tomasz in his very beautiful answer to my post Proper Subgroup of $O_2(\mathbb{R})$ Isomorphic to $O_2(\mathbb{R})$.