I came across this statement which says that "A finite subgroup $G$ of $GL(n,\mathbb{C})$ is conjugate to a subgroup of $U(n)$". As trying to understand this, the following question came to me:
Assuming I have a finite subgroup $G$ of $GL(n,\mathbb{C})$ together with an $G$-invariant, non-degenerate hermitian Form $F$ on $\mathbb{C}^n$, then i get that $G \subset U_F(\mathbb{C}) =\{ X \in GL(n,\mathbb{C})\mid X^tF\overline{X}=F \}$. But how does the finitenes of $G$ imply, that $F$ has to be definite, i.e. that up to conjugation $G \subset U(n,\mathbb{C})$?
