Given a representation $\rho$ of a finite group, why is $\rho(g)$ always diagonalisable?

Question

Why if we have representation $\rho:G \to GL(V)$ of finite group $G$ then $\rho(g)$ is diagonalisable matrix?

I read that it's because $x^{o(g)} -1$ splits, but I don't understand how this fact is helpful.

Answer 1

Over the complex numbers, every matrix of finite multiplicative order $d$ is diagonalisable (with $d$-th roots of unity as eigenvalues). This is because the polynomial $X^d-1$ that annihilates the matrix is split (because it is over the complex numbers) into distinct monic factors $X-\lambda$. Indeed $$X^d-1=\prod_{k=0}^{d-1}(X-\exp(k\frac{2\pi\mathbf i}d)).$$

This uses the theorem that any endomorphism annihilated by a product $\prod_k(X-a_k)$ with all $a_k$ distinct is diagonalisable.