Let $\Gamma$ be a finite multiplicative group of matrices with complex entries. Let $M$ denote the sum of the matrices in $\Gamma$. Prove that det $M$ is an integer.
Hint: square $M$ and use the distributivity of multiplication with respect to the sum of matrices.
I tried using the hint but I am not able to get to anything.
Best Answer
Let $G_1,G_2, \dots, G_k$ be the elements of $\Gamma$.
$$M^2 = (G_1+G_2+ \dots+ G_k)^2 = \sum_{i=1}^{k}G_i\left(\sum_{j=1}^k G_j\right)$$ $$= \sum_{i=1}^{k}G_i \left(\sum_{G \in \Gamma} G_i^{-1}G\right) = \sum_{G \in \Gamma} \sum_{i=1}^{k}G_i(G_i^{-1}G) = k\sum_{G \in \Gamma}G = kM$$ Taking determinants, we get that $(\det M)^2 = k^n \det M$. Hence either $\det M = 0$ or $\det M$ is equal to the order of $\Gamma$ raised to the $n$th power.