Let $X$ be a hypersurfaces in $\mathbb C^3$ defined by $f(x,y,z)=0$ with an ADE type singularity at $0$. Denote $\mu$ the Milnor number of the singularity.
On the one hand, we can fit $(X,0)$ into the semi-universal deformation parametrized by $\mathbb C^{\mu}\cong \mathbb C[x,y,z]/\langle\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z}\rangle$, and the monodromy group on the general fiber should be the Weyl group $W$ of the Lie algebra corresponding to the singularity.
On the other hand, we have Milnor fiber theory for the hypersurface singularity: There is a perturbation $\tilde{f}$ of $f$ such that $\tilde{f}$ is Morse in the sense that all critical points are non-degenerate, and all critical values $t_1,...,t_{\mu}$ are distinct and are contained in a small neighborhood $D$ of $0$. Milnor showed the existence of such $\tilde{f}$ by adding a general linear functional $l\in (\mathbb C^3)^*$ to $f$ (see Milnor's book Singular points of complex Hypersurfaces, Lemma B.3 in page 113).
Definition: Let $F$ denote the Milnor fiber, then the monodromy group of the Milnor fiber is defined as the image of $$\rho: \pi_1(D\setminus\{t_1,...,t_{\mu}\},*)\mapsto \text{Aut}~H^2(F,\mathbb Z).$$ Namely, it is the group generated by the loops around $t_i$, $1\le i\le \mu$, which correspond to reflections $$T_i(\alpha)=\alpha+(\delta_i,\alpha)\delta_i$$ by Picard-Lefschetz formula. These reflections correspond to a reflections on the root system, therefore generate a subgroup of $W$, but I would like to know the converse:
Question: Is $\text{Im}(\rho)=W$?
I guess the answer should be yes and it should be classical, but I can't find such a result. Could someone help me?
