Is every subgroup of $\mathrm{Isom}(\mathbb{R}^n)$ a semidirect product $T \rtimes Q$?

Question

It is well known that the group of isometries of Euclidean space $\mathrm{Isom}(\mathbb{R}^n)$ splits as the semidirect product $\mathbb{R}^n \rtimes O(n)$. However, is it also true that every subgroup $G \le \mathrm{Isom}(\mathbb{R}^n)$ is isomorphic to the semidirect product $T \rtimes Q$, where $T = G \cap \mathbb{R}^n$ (this is a slight abuse of notation) and $Q$ is the projection of $G$ onto $O(n)$?

My initial instinct is that this is false. Indeed, suppose that we have a subgroup $G$ satisfying the condition that: for every $\psi \in G$, $\psi(x) = Ax + b$, where $A \in O(n)$ and $b \in \mathbb{R}^n$, if $A$ is not the identity, then $b \ne 0$. That is, in $G$ there are no isometries of the form $x \mapsto Ax$ where $A$ is not the identity. Suppose further that $G$ is not comprised solely of translations. Then $G$ cannot be isomorphic to the semidirect product, because $\{0\} \times Q \ne \{0\} \times \{I\}$ is a subgroup of $T \rtimes Q$. To be explicit, no map $\phi : T \rtimes Q \rightarrow G$ can be injective, since $G$ is in bijective correspondence with the proper subset $G' \subsetneq T \rtimes Q$ where $G' = \{(b,A) \in T \rtimes Q \, | \,\exists \psi \in G, \psi(x)=Ax+b\}$.

The only problem with this approach is that it assumes the existence of such a $G$, and thus far I have failed to produce an example. I would very much appreciate some help with this. Am I wrong; does the result in fact hold? Otherwise, is there a better way to prove its falsity? I feel like there must be a better way to do this.

Answer 1

This is not a proper answer but a proposal for where to potentially find simple counter examples.

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Consider a one parameter subgroup $\gamma:\Bbb{R}\to\mathrm{Isom}(\Bbb{R}^2)$ that

• actually rotates (e.g. $p(\gamma(t))$ is the rotation of angle $t$) and
• actually translates ($\gamma(2\pi)$ is a non-identity translation).

I expect under these circumstances $\gamma$ induces a homeomorphism onto its image. In any case, the image $G=\gamma(\Bbb{R})$ is connected.

The translations in the image, i.e. $T=G\cap\Bbb{R}^2$, form a discrete group and the projection $Q$ of $G$ is $\mathrm{SO}(2)$. Whatever semi-direct product structure you would consider involving $T$ and $Q$ it would produce group with infinitely many connected components and couldn't be isomorphic to the connected group $G$.

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I imagine it should be easier to do in $\Bbb{R}^3$ taking something like $$\gamma(t)\big(V\big)=R_t V + te_z$$ where $R_t$ is the rotation of angle $t$ around the axis $Oe_z$.