13
$\begingroup$

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."

$\endgroup$
6
  • $\begingroup$ You might be interested in this : en.wikipedia.org/wiki/… $\endgroup$
    – Patrick Da Silva
    Aug 29, 2014 at 22:22
  • 1
    $\begingroup$ $S_n$ is a finite reflection group and every finite group occurs as a subgroup of some $S_n$ by Cayley's theorem. This is a weaker statement than the claim that every finite subgroup of $\text{O}(n)$ is contained in a finite subgroup of $\text{O}(n)$ generated by reflections, though. $\endgroup$
    – Qiaochu Yuan
    Aug 30, 2014 at 0:28
  • $\begingroup$ I think that $SL_2(7)$ occurs as a maximal finite subgroup of $SL_4(C)$, since $PSL_2(7)$ is not a Jordan-Holder factor of any finite Coxeter group, I think this implies that this $SL_2(7)$ is not contained in a subgroup of $GL_4(C)$ generated by reflections. $\endgroup$
    – YCor
    Aug 31, 2014 at 22:48
  • $\begingroup$ @YCor: Would this also mean $SL_2(7)$ occurs as a maximal finite subgroup of $GL_8(\mathbb{R})$? Can you tell what its generating isometries would be? $\endgroup$
    – Nick Matteo
    Sep 1, 2014 at 17:45
  • $\begingroup$ No I don't claim this. Actually I answered in the complex case instead of real, and $SL_2(7)$ doesn't occur in $GL_4(\mathbf{R})$. So another example should be found, but I don't find any data about maximal finite subgroups of $SO(n)$ for small values of $n$. $\endgroup$
    – YCor
    Sep 1, 2014 at 17:56

1 Answer 1

Reset to default
10
$\begingroup$

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).

$\endgroup$
7
  • $\begingroup$ I am pretty sure that the group $G$ is a subgroup of the symmetry group of the duoprism $\{p\}\times\{p\}$, that is, the Cartesian product of two regular $p$-gons. The vertices are $(\cos(2j\pi/p),\sin(2j\pi/p),\cos(2k\pi/p),\sin(2k\pi/p))$ for $0 \leq j,k \leq (p-1)$, and $M_1$, $M_2$, and $s$ all seem to carry these onto themselves. So $G$ must be a subgroup of its symmetry group (of order $8p^2$), which is a finite reflection group. $\endgroup$
    – Nick Matteo
    Sep 1, 2014 at 23:53
  • $\begingroup$ I agree except in the last line: this symmetry group has order $8p^2$, is generated by its elements of order 2, but its reflections only generate a subgroup of index 2 (which is the direct product of 2 dihedral groups of order $2p$). $\endgroup$
    – YCor
    Sep 2, 2014 at 8:50
  • $\begingroup$ You're right; I was mistaken in thinking it was a reflection group. Thank you! $\endgroup$
    – Nick Matteo
    Sep 3, 2014 at 13:32
  • $\begingroup$ @YCor - I am having trouble following the logic in (2). I agree that if a finite subgroup of $SO(3)$ contains a subgroup of order $p$, it is normal and $p$-sylow. I can reconstruct the same claim for $SU(2)$ as follows: let $H$ be a subgroup of $SU(2)$ containing a group $P$ of order $p$. Then $H$'s image $\pi H$ in $SO(3)$ also contains (an isomorphic copy of) $P$, normally; its pullback $\pi^{-1}\pi H$ in $SU(2)$ thus normally contains $P\times \{\pm I\}$, which is cyclic, thus $P$ is characteristic in it, so normal in $\pi^{-1}\pi H$ and thus in $H$ (as $H\subset \pi^{-1}\pi H$). Cont'd... $\endgroup$
    – Ben Blum-Smith
    Jun 18, 2016 at 0:15
  • 1
    $\begingroup$ what next? consider the projections $F_1,F_2,P_1,P_2$ on the two $SU(2)$ factors. Then $P_i\subset F_i$. Also, $P_i$ can't be trivial because otherwise the other $P_j$ would be of order $p^2$ in a finite subgroup of $SU(2)$. So both $P_i$ are cyclic of order $p$, so $P_i$ is normal and Sylow in $F_i$, hence $P=P_1\times P_2$ is normal and Sylow in $F_1\times F_2$, and in particular in its subgroup $F$. $\endgroup$
    – YCor
    Jun 18, 2016 at 7:05

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .