This is a refinement of a [question asked on MSE](https://math.stackexchange.com/questions/1703749/can-the-quotient-by-a-nonabelian-group-yield-an-abelian-singularity).


Let $G$ be a finite group and let $V$ be a finite-dimensional faithful complex representation of $G$. Consider $V$ as an affine complex variety. In general (though not always), the image of the origin in the quotient variety $V/G$ is singular. By the "singularity type" of the pair $G,V$ I refer to the isomorphism class of the ring $\widehat{\mathcal{O}\_{V/G,0}}$, the completion of the local ring at this point. 


At broadest (soft-question) level, what I want to know is, "how fine an invariant of the pair $G,V$ is the singularity type?" But let me narrow the scope in order to be able to ask something precise:



> 
> Suppose $G$ is nonabelian, $V$ has no one-dimensional subrepresentations, and $V/G$ is singular at the image of the origin. Further suppose $A$ is a finite abelian group and $W$ is a faithful representation of $A$. Is it possible for $G,V$ to have the same singularity type as $A,W$?
> 
> 
> 


Comments: (1) The answer is certainly "yes" without the stipulation that $V/G$ be singular at the origin; both abelian and nonabelian groups can have smooth quotients, by Chevalley-Shephard-Todd since both abelian and nonabelian groups can be reflection groups. (2) The stipulation "$V$ has no one-dimensional subrepresentations" is there to rule out a match resulting from $G,V$'s singularity "actually" coming from $G$'s abelianization. This can happen for example if $V$ decomposes as a sum $W\oplus L$ with $W$ such that $W/G$ is smooth and $L$ is a sum of one-dimensional representations.



How fine an invariant of a representation is its quotient singularity?
To cherry-pick the answer from Jason Starr's comment above: the condition you want is that $G$ acts on $V$ without any element giving a pseudo-reflection. Chevalley-Shephard-Todd tells you that you must lose all information about pseudo-reflections, but you don't lose anything else. When there are no pseudo-reflections, the singular set of $V/G$ is exactly the image of the elements in $V$ with non-trivial stabilizer, so the smooth locus of $V/G$ is a quotient of a simply connected space by $G$, and thus its $\pi\_1$ is $G$. This should be reconstructable from the singularity type using the Galois theory of the fraction field of the completion at the origin (looking at the Galois group of the largest Galois extension with no ramification at smooth points, I think). 


