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Finite matrix group

Let \mathcal{G} be a finite subgroup of {\rm GL}_n(\mathbb{C})  this is the group of the  n \times n invertible matrices over \mathbb{C}). If \sum \limits_{g \in \mathcal{G}} {\rm Tr}(g)=0 then prove that \sum \limits_{g \in \mathcal{G}} g =0.


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. thus its trace equals to its class. (since we are over \mathbb{C} which is a field of zero characteristic.) Hence

\displaystyle {\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.

The exercise can also be found at mathematica.gr

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