A curious identity concerning a finite group of automorphisms Let $E$ be a finite dimensional vector space over $\mathbb{C}$ and let $G$ be a finite subgroup of $Gl(E)$. Let $F=\{x \in E | \forall g \in G, g(x)=x\}$. Prove that
$$|G| \dim F=\sum_{g \in G}tr(g)$$

I have a simple proof if $G$ is cyclic, and I would like to know if my solution can be completed to the general case:
Suppose $G=\{id_E,g...,g^{k-1}\}$ with $g^k=id_E$. As $X^k-1$ is the minimal polynomial of $g$, $g$ is diagonalisable and the set of eigenvectors of $g$ is the roots of this polynomial: $\{\lambda_1,...,\lambda_k\}$. The trace of an endomorphism is the sum of its eigenvectors, thus
$\sum_{g \in G}tr(g)=\sum_{i=0}^{k-1}\sum_{j=1}^k\lambda_j^i=\sum_{j=1}^k\sum_{i=0}^{k-1}\lambda_j^i$
Since for each $j$ we have $\lambda_j^k=1$, by a geometric series the inner sum evaluates to $0$ if $\lambda_j \neq 1$, and $k$ otherwise. Hence $\sum_{g \in G}tr(g)$ is simply $k$ times the multiplicity of the eigenvalue $1$. Since $g$ is diagonalisable, the multiplicity is equal to the dimension of eigenspace associated to the eigenvalue $1$, i.e $\dim (\ker(g-id_E))=\dim(F)$. 
 A: (This is a proof gotten from expanding a fairly well-known proof in representation theory, applied to this special case.)
Note that the map $E \to E$, $x \mapsto \frac{1}{|G|} \sum_{g \in G} g(x)$, is a projection with image $F$.  Therefore, this map has trace $\dim F$; but by linearity of the trace, it is also equal to $\frac{1}{|G|} \sum_{g\in G} \operatorname{tr}(g)$.
A: Well, we can divide the proof in two parts: $(1)$ The assert is true for $F=E$ and $(2)$ It is true for $F=\{0\}$.
Given $(1)-(2)$ we can put them together this way: take $M\leq E$ a $G$-complement to $F$ (Maschke's theorem, as we are talking about a $\mathbb{C}G$-module, actually I think we could work on quotients as well) then we have that $tr(g)=tr_F(g)+tr_M(g)$, hence $\sum_{g\in G}tr(g)=\sum_{g\in G}tr_F(g)+\sum_{g\in G}tr_M(g)=|G|\dim{F}+0$ (as the fix space on $M$ is $\{0\}$). So let's prove $(1)$: We have that each $g\in G$ acts identically on $F=E$, hence summing them all we get $|G|$ times $I_{\dim{F}}=I$ from this we have the assert. Now part $(2)$: Let's suppose that no non-zero vector is fixed by all $G$, i.e. $F=\{0\}$ and consider $\phi=\sum_{g\in G}g$. Now note that if $v\in E$ then for each $h\in G$ we have $h(\phi(v))=(h·\phi)(v)=\phi(v)$. So $\phi(E)$ is point-fixed by every element of $G$, hence by assumption $\phi(E)\leq\{0\}$ and $\phi=0$. It then follows that $\sum_{g\in G}tr(g)=tr(\sum_{g\in G}g)=tr(\phi)=0$
