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 $\sum \limits_{g \in \mathcal{G}} g =0$ .

Solution

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

Finite matrix group

They might interest you:

Telescopic Fibonacci product

Double inequality involving matrix

On the zero function

Pierre Mounir series

Existence of constant (3)

Similarity implies equality?

Circumscribed and inscribed circle

Roots of equation

Trigonometric identity

Leave a Reply Cancel reply