Is every finite group of isometries in $d$-dimensional Euclidean space a subgroup of a finite group generated by reflections?
By "reflection" I mean reflection in a hyperplane: the isometry fixing a hyperplane and moving every other point along the orthogonal line joining it to the hyperplane to the same distance on the other side.
Every isometry is a product of reflections; in $d$-space, at most $d+1$ reflections are needed.
So, every finite group of isometries is a subgroup of a group generated by reflections; just choose some reflections that yield each isometry, and take the group generated by those. The question is whether the reflections can always be chosen so that the resulting group is still finite, or even discrete.
Every finite group $G$ of isometries has a fixed point. If we regard this fixed point as the origin, then $G$ is a finite subgroup of the orthogonal group $O(d)$. Orthogonal transformations are the product of at most $d$ reflections in hyperplanes through the origin.
I have found some claims that the answer is "Yes", but it's not clear if they mean for arbitrary dimension, or just 3-space. Also, no proofs are provided.
This monograph "Symmetry Groups" claims "All the finite groups are reflection groups or subgroups thereof" (p.11).
This webpage says
"All symmetry groups will be subgroups of groups generated by reflections."
Best Answer
Here's a simple example for $n=4$, namely a finite subgroup of $\mathrm{GL}_4(\mathbf{R})$ not contained in any finite subgroup of $\mathrm{GL}_4(\mathbf{R})$ generated by reflections.
Let $p$ be a prime $\ge 7$, and let $M$ be the $2\times 2$ matrix of order $p$ given by rotation by $2\pi/p$. Consider the group $G\subset\mathrm{GL}_4(\mathbf{R})$ generated by $M_1=\begin{pmatrix}M &0\\ 0 & I_2\end{pmatrix}$, $M_2=\begin{pmatrix}I_2 &0\\ 0 & M\end{pmatrix}$, $s=\begin{pmatrix}0 & I_2\\ I_2 & 0\end{pmatrix}$. (It has order $2p^2$, with an abelian subgroup of index 2 $H$ generated by $M_1,M_2$, and is isomorphic to the wreath product $C_p\wr C_2$ where $C_n$ is cyclic of order $n$.)
I claim that $G$ is not contained in a finite subgroup of $\mathrm{GL}_4(\mathbf{R})$ generated by reflections. There are two steps.
1) $G$ is not contained in a finite subgroup of $\mathrm{GL}_4(\mathbf{R})$ generated by reflexions and normalizing $H$. Indeed, $H\simeq C_p\times C_p$ fixes exactly 2 elements in the 2-Grassmanian of $\mathbf{R}^4$, namely the planes $x_3=x_4=0$ and $x_1=x_2=0$. Hence the normalizer $L$ of $H$ preserves this unordered pair of planes, and hence has a subgroup $L'$ of index (at most) 2 preserving these two subspaces; this index is actually 2 since $s$ belongs to this normalizer but exchanges these 2 subspaces. But $L\smallsetminus L'$ consists of elements of trace zero and hence contains no reflection. Therefore any subgroup of $L$ containing $G$ fails to be generated by reflections.
2) any finite subgroup of $\mathrm{GL}_4(\mathbf{R})$ containing $H$ normalizes $H$. This follows from the classification of subgroups of $\mathrm{SO}(4)$. Indeed, first observe that it follows from the classification of finite subgroups of $\mathrm{SO}(3)$ that in such a finite subgroup, any cyclic subgroup of order $p$ is normal (we use here that $p\neq 2,3,5$), since finite subgroups of $\mathrm{SO}(3)$ with an element of order $p$ are cyclic or dihedral. That cyclic subgroups of order $p$ are normal and $p$-Sylow in all finite subgroups containing them therefore remains true in the double cover $\mathrm{SU}(2)$. It follows that in any finite subgroup $F$ of $\mathrm{SU}(2)\times\mathrm{SU}(2)$, any subgroup of $F$ isomorphic to $C_p\times C_p$ is normal and $p$-Sylow in $F$; and in turn the same conclusion holds in $\mathrm{SO}(4)=(\mathrm{SU}(2)\times\mathrm{SU}(2))/C_2$, and in turn in $\mathrm{O(4)}$, and in turn in $\mathrm{GL}_4(\mathbf{R})$.
The conclusion follows from 1) and 2).