3 Clarified about fixed point sets

There are indeed many presentations (if I remember correctly Bourbaki has it) but the proof is very elegant and short so that I find it hard to refrain from giving it. Let $H$ be the normal subgroup of the finite $G\subset \mathrm{GL}_n$ generated by the pseudo-reflections. By the other direction $X:=\mathbb{A}^n/H$ is again affine space and in particular is smooth. We have an action of $G/H$ on $X$ and a moment's thought reveals that it acts freely on the hypersurfaces of in codimension $X$ 1$(as a hypersurface point fixed by a non-identity element would lie belowq below a reflection hyperplane of$\mathbb{A}^n$and the fixing element below a pseudo-reflection). Hence$X \to X/(G/H)=\mathbb{A}^n/G$is étale in codimension$1$. If$\mathbb{A}^n/G$were smooth, purity of the branch locus would imply that the map were étale. However, that forces$G/H$to act freely on$X$but the image of the origin is fixed by all of$G/H$and therefore$G=H$. 2 deleted 1 characters in body There are indeed many presentations (if I remember correctly Bourbaki has it) but the proof is very elegant and short so that I find it hard toq to refrain from giving it. Let$H$be the normal subgroup of the finite$G\subset \mathrm{GL}_n$generated by the pseudo-reflections. By the other direction$X:=\mathbb{A}^n/H$is again affine space and in particular is smooth. We have an action of$G/H$on$X$and a moment's thought reveals that it acts freely on the hypersurfaces of$X$(as a hypersurface fixed by a non-identity element would lie belowq a reflection hyperplane of$\mathbb{A}^n$and the fixing element below a pseudo-reflection). Hence$X \to X/(G/H)=\mathbb{A}^n/G$is étale in codimension$1$. If$\mathbb{A}^n/G$were smooth, purity of the branch locus would imply that the map were étale. However, that forces$G/H$to act freely on$X$but the image of the origin is fixed by all of$G/H$and therefore$G=H$. 1 There are indeed many presentations (if I remember correctly Bourbaki has it) but the proof is very elegant and short so that I find it hard toq refrain from giving it. Let$H$be the normal subgroup of the finite$G\subset \mathrm{GL}_n$generated by the pseudo-reflections. By the other direction$X:=\mathbb{A}^n/H$is again affine space and in particular is smooth. We have an action of$G/H$on$X$and a moment's thought reveals that it acts freely on the hypersurfaces of$X$(as a hypersurface fixed by a non-identity element would lie belowq a reflection hyperplane of$\mathbb{A}^n$and the fixing element below a pseudo-reflection). Hence$X \to X/(G/H)=\mathbb{A}^n/G$is étale in codimension$1$. If$\mathbb{A}^n/G$were smooth, purity of the branch locus would imply that the map were étale. However, that forces$G/H$to act freely on$X$but the image of the origin is fixed by all of$G/H$and therefore$G=H\$.