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Anton Geraschenko
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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.

Anton Geraschenko
  • 24k
  • 17
  • 127
  • 180