Let $G$ be a finite irreducible unitary reflection group (i.e. without G-invariant subspaces). Given orthonomal basis, we have that $g_1 \in GL(V)$ commutes with every element of $G$. It is said that in this case $g_1$ necessarily be a scalar matrix as a consequence of Schur's lemma which (in particular) states that two G-invariant hermitian forms are similar up to coefficient $\lambda >0$. But I'm confused in binding these two facts together.
Any suggestions will be helpful.
Suppose that the field is algebraically closed. Let $c$ be an eigenvalue of $g_1$, the eigenspace $V_c$ is stable by $G$, thus it is $V$ since the action is irreducible.