Is every finitely generated reflection group a Coxeter group? Suppose $G$ is a subgroup of $O_n(\mathbb{R})$ generated by finitely many
reflections, where a reflection is defined to be a linear
transformation fixing a hyperplane and sending a normal vector to the
hyperplane to its negative. Suppose the normal vectors corresponding
to these hyperplanes are linearly independent. Is G necessarily
isomorphic to a Coxeter group?
This problem arose upon considering the fact that every Coxeter group
is isomorphic to its Tits reflection representation, and I was
wondering if there was a converse statement. Looking at the
literature, a common statement is that this is true for finite
reflection groups, but I could not find a statement for finitely
generated reflection groups.
I might be missing some easy counterexample here as it seems like it would be a standard result if it were true.
Thanks!
 A: No. For a specific counterexample, let $p$ be the root of 
$$p^3-3p^2+1$$
which is roughly $0.652704$ and chose three vectors in $\mathbb{R}^3$ at angles $\cos^{-1}(p) \approx 0.273635 \pi$ with respect to each other. (Since $0.273635<1/2$, the angle is acute and this is clearly possible.) Let $a$, $b$ and $c$ be the reflections in these vectors. Then $\langle a,b \rangle$, $\langle a,c \rangle$ and $\langle b,c \rangle$ are infinite dihedral groups, but $(abc)^6=\mathrm{Id}$. So this is not a Coxeter group with those generators, and I highly doubt it is a Coxeter group at all.

OK, so how did I find this? I started out looking for reflections $a$, $b$ and $c$ such that the dihedral groups $\langle a,b \rangle$, $\langle a,c \rangle$, $\langle b,c \rangle$ are infinite but $abc$ had finite order. I knew that a good way to compute with vectors in $\mathbb{R}^3$ was in terms of their matrix of inner products. Let $p = \langle v_a, v_b \rangle$, $q = \langle v_a, v_c \rangle$, $r = \langle v_b, v_c \rangle$ where $s$ is the reflection in unit vector $v_s$. I claim that $a$, $b$, $c$ will be a counter-example as long as $(p,q,r)$ are real numbers in the interval $(-1,1)$ such that
(1) $\frac{1}{\pi} \cos^{-1} p$, $\frac{1}{\pi} \cos^{-1} q$, $\frac{1}{\pi} \cos^{-1} r$ are irrational
(2) $1+2pqr > p^2+q^2+r^2$ and
(3) $\frac{1}{\pi} \cos^{-1} \left( \frac{p^2+q^2+r^2-pqr-2}{2} \right)$ is rational.  
Proof: First we show that the vectors exist. There is a collection of matrices with the given inner products if and only if the Gram matrix 
$$\begin{pmatrix} 
1 & p & q \\ 
p & 1 & r \\
q & r & 1 \\
\end{pmatrix}$$
is positive definite. By Sylvester's criterion (since $p \in (-1,1)$), this will occur if and only if the determinant of the Gram matrix is positive. That determinant is $1+2pqr - p^2-q^2-r^2$, explaining condition (2).
The pairs $(a,b)$, $(a,c)$ and $(b,c)$ will generate infinite dihedral groups as long as $\frac{1}{\pi} \cos^{-1}(p)$, $\frac{1}{\pi} \cos^{-1}(q)$ and $\frac{1}{\pi} \cos^{-1}(r)$ are all irrational. That explains condition (1).
I had Mathematica compute the characteristic polynomial of $abc$. It is 
$$(x+1) (x^2 - (p^2+q^2+r^2 - pqr -2) x + 1).$$
(We knew that the $-1$ root would be there, since it $abc$ is an orientation reversing orthogonal map.) 
So $abc$ will have finite order if and only if $\frac{1}{\pi} \cos^{-1} \left( \frac{p^2+q^2+r^2-pqr-2}{2} \right)$ is rational, explaining condition (3). 

At this point, you should have the strong feeling that you are already done. Fixing some candidate value $2 \cos (\frac{m}{n} \pi)$ for $p^2+q^2+r^2-pqr-2$, there should be a whole surface of solutions to (3). Then (2) will cut out an open subset of that surface, and (1) will cut out a dense subset of that. So there should be such points. 
Looking for an explicit solution to give you, I decided to set $p=q=r$. (This was motivated by ease of computation, not by any deep theory.) Condition (2) then becomes 
$p \in (-1/2, 1)$. Numeric computation shows that $p^2+q^2+r^2-pqr-2$ is then in $[-2,0)$; I decided to set it equal to $-1 = 2 \cos (2 \pi/3)$ to enforce condition (3). So my $p$ is one of the roots of 
$$3 p^2 - p^3 - 2 = -1$$
which lies in the desired range.
The roots of this equation are $(-0.532089, 0.652704, 2.87939)$, so the root at $0.652704$ is valid for my purposes. Note that these roots provide an immediate proof that $\frac{1}{\pi} \cos^{-1}(p)$ is not rational: If $p$ were in $\cos (\pi \mathbb{Q})$, then so would all its Galois conjugates be, and this polynomial would not have a root outside $(-1,1)$.
A: In a different direction, you might be interested in the following result of Deodhar (see also Dyer). Let $W$ be any Coxeter group, possibly infinite, or even of infinite rank. (Rank is the number of Coxeter generators.) Let $W'$ be any subgroup of $W$ generated by some set of reflections of $W$. Then $W'$ is a Coxeter group. 
Warning 1: It can happen that the rank of $W'$ is larger than that of $W$, and even that $W$ has finite rank and $W'$ has infinite rank. 
Warning 2: If $V$ and $V'$ are the reflection representations of $W$ and $W'$, then there is a natural $W'$-equivariant map $V' \to V$, but this map need be neither injective nor surjective.
