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A theorem of Chevalley, Shepard, and Todd states that if $G$ is a finite group and $\rho: G \rightarrow GL_n(\mathbb{C})$ a representation so that $\rho(G)$ is generated by pseudoreflections, then $\mathbb{C}[z_1,\dots,z_n]^G$ (the subring of $G$-invariant polynomials) is again a polynomial ring.

From my understanding, a pseudoreflection is diagonalizable with diagonal form $\text{diag}(1,\dots,1,\zeta)$ where $\zeta$ is a root of unity.

My question is whether the property of being generated by pseudoreflections is a property of the representation or of the group itself, and if so what is the equivalent abstract group-theoretic property. If it is not the case, could someone point me towards a group which furnishes a counterexample?

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up vote 6 down vote accepted

It's a property of the representation. For example, under the 1-dimensional representation of $\newcommand{\Z}{\mathbb Z} \Z/2$ given by $x\mapsto -x$, $\Z/2$ is generated by pseudoreflections, but under the 2-dimensional representation $(x,y)\mapsto(-x,-y)$, it's not.

It makes sense to ask if there is an abstract finite group $G$ which cannot be generated by pseudoreflections (i.e. there is no faithful representation under which it is generated by pseudoreflections). I feel like I should know the answer (I even feel like I've thought about this question before), but I don't.

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Thanks! This is a very simple example of why it cannot be an intrinsic property of the group. So maybe a better question, more in line with your second statement, is whether restricting to a simple representation turns it into a property of the group. I will update the question. – Merrick Brown Jan 19 '11 at 3:03
@Merrick, not really: it may well be the case there non of the pseudoreflections of $S_n$ in that representation do not belong to your subgroup! – Mariano Suárez-Alvarez Jan 19 '11 at 3:17

Regarding a comment on Anton's answer:

Todd and Shephard classified in [Shephard, G. C.; Todd, J. A. Finite unitary reflection groups. Canadian J. Math. 6, (1954). 274--304. MR0059914 (15,600b)] all finite groups generated by pseudo-reflections. Not all groups are isomorphic to one in their list.

For example, the center of a group generated by pseudo-reflections which is irreducible (i.e., not a direct product of two smaller groups of the same kind) is cyclic. So to find an example, it is enough to find a group which is not a direct product and which has non-cyclic center.

Restricting to $p$-groups with even prime $p$, GAP tells me there is such a group of order $2^4$, the semi-direct product of $C_4$ by itself, and another, a semi-direct product of $C_4\times C_2$ by $C_2$, both with center isomorphic to $C_2\times C_2$.

Another source of examples: if $G$ is a finite group with a faithful representation generated by pseudo-reflections, then the corresponding determinant representation is non-trivial (as the determinant of a pseudo-reflection is not $1$) It follows that $G$ is not simple.

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+1. Your second source of examples is beautiful (because the explanation is so self-contained): any non-cyclic simple group cannot be generated by pseudoreflections. – Anton Geraschenko Jan 19 '11 at 4:21

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