Can the difference of non-conjugate pseudoreflections lie in the commutator subgroup? Let $G$ be a finite group acting on a complex vector space $V$ by pseudoreflections (i.e. every element of $G$ is a product of elements which fix hyperplanes in $V$). I would like to understand the abelianization of $G$.
If $r_1$ and $r_2 = gr_1g^{-1}$ are conjugate pseudoreflections, then $r_1r_2^{-1} = r_1gr_1^{-1}g^{-1}$ is a commutator, so $r_1$ and $r_2$ are identified in the abelianization of $G$. Is the converse true?

If two pseudoreflections $r_1, r_2 \in G$ are identified in the abelianization of $G$, must they be conjugate?

In general, the product of commutators is not a commutator, so the answer is a resounding "no" if for arbitrary elements of an arbitrary finite group, but reflection groups are pretty special.
 A: For a Coxeter group, yes. Remember that in a Coxeter group, all the relations are of the form $s_i^2=1$ and $(s_i s_j)^{m_{ij}}=1$. So the corresponding relations in the abelianization are $2 \bar{s}_i=0$ and $\bar{s}_i = \bar{s}_j$ if $m_{ij}$ is odd. Letting $S$ index the simple generators and defining an equivalence relation $\sim$ on $S$ to be the transitive closure of $s_i \sim s_j$ if $m_{ij}$ is odd, we see that the abelianization is $(\mathbb{Z}/2)^{S/\sim}$ and that two elements of $S$ define the same class in the abelianization if and only if they are conjugate. Every reflection in a Coxeter group is conjugate to a reflection in $S$.
A: Sure: a slight enhancement of Geoff Robinson's comment will work. Let $s$ and $t$ be reflections (I'm going to write reflections instead of pseudoreflections) with the same image in the abelianization. Then for every one dimensional character $\chi$, we have $\chi(s)=\chi(t)$. Given a $G$-orbit of reflecting hyperplanes, there is a linear character $\chi$ defined as follows: take a linear form $\alpha_H$ defining each hyperplane $H$, and let $f$ be the product over the orbit of all these forms (in other words, a generator for the ideal of polynomial functions vanishing on the union of the hyperplanes in the orbit). Then $g(f)=\chi(g) f$ for all $g \in G$.
This character has the property that $\chi(g)=1$ if $g$ is a reflection in a hyperplane not belonging to the orbit, while $\chi(g)=\mathrm{det}(g)^{-1}$ for $g$ a reflection in a hyperplane in the orbit. So $s$ and $t$ must be reflections in hyperplanes in the same $G$-orbit. Supposing that $s$ is a reflection in a hyperplane $H$ and $t$ is a reflection in a hyperplane $H'$, let $s_{H'}$ and $s_H$ be generators for the pointwise fixers of $H$ and $H'$ in $G$---we may and will assume that $s_{H'}$ and $s_H$ are $G$-conjugates. Now $s=s_H^\ell$ and $t=s_{H'}^k$ for integers $\ell$ and $k$; these integers are determined by the determinants of $s$ and $t$, so by assumption $k=\ell$. Since $s_H$ and $s_{H'}$ are conjugate, this does it.
Here is a way to codify what we have done: write $\mathcal{A}$ for the set of reflecting hyperplanes of $G$ and for such a hyperplane write $G_H$ for its pointwise fixer. There is a map
$$\{\text{group homomorphisms} \ G \rightarrow \mathbb{C}^\times \} \rightarrow \{\text{group homomorphisms} \ \prod_{H \in \mathcal{A}} G_H \rightarrow \mathbb{C}^\times \}^G$$ given by restriction (I hope the notation on the target here is self-explanatory). The previous two paragraphs amount to showing that for any finite linear group $G$ this map is surjective; this doesn't really depend on $G$ being a reflection group. That it is also injective follows if the subgroups $G_H$ generate $G$, or in other words if $G$ is a reflection group.
