I'm studying Chevalley-Shephard-Todd theorem, in the version that states : let $G \subset GL(V)$ a finite group, where $V$ is a finite dimensional complex space. Let $S=S(V^*)$ indicates the symmetric algebra over the dual vector space and $R=S^G$ the ring of invariants.
These facts are equivalent:
1) $S$ is a free $R$ module
2) $R$ is a polynomial algebra
3) $G$ is generated by pseudoreflections
The implication I'm missing is $2)$ implies $3)$. To do that, one can take $H \subset G$ the subgroup generated by pseudoreflections contained in $G$.
Because of $3)$ implies $2)$ one has $S^H=\mathbb{C}[g_1,\cdots g_n]$ and during the proof of $3) \rightarrow 2)$ one can show that the $g_i$ can be taken to be homogeneous. Now, the problem is $S^G=\mathbb{C}[f_1 \cdots f_n]$ and to complete the proof one has to assume that these $f_i$ can be taken to be homogeneous, but I can't see why this should be true.