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?



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> If two pseudoreflections $r\_1, r\_2 \in G$ are identified in the abelianization of $G$, must they be conjugate?
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In general, the [product of commutators is not a commutator](https://math.stackexchange.com/questions/7811/derived-subgroup-where-not-every-element-is-a-commutator), so the answer is a resounding "no" if for arbitrary elements of an arbitrary finite group, but reflection groups are pretty special.



Can the difference of non-conjugate pseudoreflections lie in the commutator subgroup?
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.


