Diagonalizability of elements of finite subgroups of general linear group over an algebraically closed field How to show that every element of $G$, where $G$ is a finite subgroup of $GL_n(\mathbb{k})$, the general linear group of square matrices of order $n$ over some algebraic closed field $\mathbb{k}$, is diagnonalizable if $\mathbb{k}$ is an algebraic closure of $\mathbb{Q}$?
I know that a matrix is diagonalizable if its minimal polynomial is seperable in the field on which the matrix is defined. Now, what if the minimal polynomial has repeated roots? How do we ensure the diginalizability? Any hints. Thanks beforehand.
 A: Let $K$ be a field.
As you have already mentioned, a matrix $A \in \operatorname{M}_n(K)$ is diagonalizable (over $K$) if and only if there exists a polynomial $f(t) \in K[t]$ with $f(A) = 0$ such that $f$ decomposes into pairwise different linear factors over $K$.
For $A \in G$ and $n := |G|$ we have that $A^n = I$, so that $A$ satisfies the polynomial $f(t) := t^n - 1 \in \mathbb{k}[t]$.
The polynomial $f(t)$ decomposes into linear factors because $\mathbb{k}$ is algebraically closed.
It follows from $\operatorname{char}(\mathbb{k}) = 0$ that the polynomial $f$ is seperable (because $f(t) = t^n - 1$ and $f'(t) = n t^{n-1}$ are coprime), so that $f(t)$ decomposes into pairwise different linear factors.
Thus $A$ is diagonalizable.
(To see that $f(t)$ is seperable one can also embed $\mathbb{k}$ into $\mathbb{C}$ because $\mathbb{k}$ is an algebraic closure of $\mathbb{Q} \subseteq \mathbb{C}$.
As the roots of unity in $\mathbb{C}$ are pairwise different, the same goes for $\mathbb{k}$.)
A: Let $A\in G$ be a matrix. After changing bases, $A$ is in JNF. Let us write $A=D+N$ with $D$ diagonal and $N$ the nilpotent part. Now since $G$ is finite, there is some $m\in\Bbb N$ with $A^m=I$, the identity matrix. Since $DN=ND$, we have 
$$I = A^m = (D+N)^m = \sum_{k=0}^m \binom mk N^k D^{m-k} = D^m + \sum_{k=1}^m \binom mk N^k D^{m-k} =: D^m+\tilde N.
$$
Note that $\tilde N$ is a nilpotent, strictly upper triangular matrix. It follows that $\tilde N=0$ and $D^m=I$. Hence,
$$
0 = \tilde N = N\cdot\left(mD^{m-1}+\sum_{k=2}^{l} \binom mk N^{k-1} D^{m-k}\right)
$$
and as the second factor is an invertible upper trianglar matrix (because we are in characteristic zero and $m$ is invertible), so we must have $N=0$. In other words, $A=D$ is already diagonal. In other words, $A$ is diagonal up to changing bases, which means that it is diagonalizable.
