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Let \(\displaystyle{G}\) be a finite subgroup of \(\displaystyle{\left(GL_{n}(\mathbb{C}),\cdot\right)}\).

If \(\displaystyle{\sum_{g\in G}Tr(g)=0}\), then \(\displaystyle{\sum_{g\in G}g=\mathbb{O}}\).

Let us suppose that $|\mathcal{G}| = \kappa$ and $x= \frac{1}{\kappa} \sum \limits_{g \in \mathcal{G}} g $. We note that for every $h \in \mathcal{G}$ the depiction $\varphi: \mathcal{G} \rightarrow \mathcal{G}$ such that $\varphi(g)=h g $ is $1-1$ and onto. Thus:

\begin{align*}
x^2 &=\left ( \frac{1}{\kappa} \sum_{g \in \mathcal{G}} g \right )^2 \\
&= \frac{1}{\kappa^2} \sum_{g \in \mathcal{G}} \sum_{h \in \mathcal{G}} gh\\ &= \frac{1}{\kappa^2} \sum_{g \in \mathcal{G}} \sum_{h \in \mathcal{G}} g\\
&= \frac{1}{\kappa} \sum_{h \in \mathcal{G}} \left (\frac{1}{\kappa} \sum_{g \in \mathcal{G}} g \right ) \\
&= \frac{1}{\kappa} \sum_{h \in \mathcal{G}} x\\
&= \frac{1}{\kappa} \kappa x \\
&=x
\end{align*}

Thus the matrix $x$ is idempotent; its trace equals to its class. (since we are over $\mathbb{C}$ which is a field of zero characteristic.) Hence

$${\rm rank} \;(x) = {\rm trace} \;(x) = \frac{1}{\kappa} \sum_{g \in \mathcal{G}} {\rm trace} \; (g) =0$$

This implies that $x=0$ hence $\sum \limits_{g \in \mathcal{G}} g =0$.