How to "show" a finite group representation $\rho(g)$ is diagonalizable over $\Bbb{C}$ for a given $g: V\to GL(V)$ and $\dim V=n$? And after that, how to show its eigenvalues are the unit roots?
As a hint I am given that if $\rho ^n(g)=1$ than it is diagonalizable, and that I could show it using, say, Jordan Diagonalization.
In an answer to a similar question, it is said that a Jordan form is idempotent if its dimension is 1, but what if there are only $k\le n$ $\lambda _i$ eigenvalues?
I would appreciate the help of telling me clearly whether or not there are $n$ eigenvalues. I think I am expected to arrive at $\lambda_1,…,\lambda_k, k\le n$ for a given $g\in G$ with $\rho (g) v= Av=\lambda_i v$ satisfying $\rho^n (g)=\rho (g^n)=Idv=A^nv=\lambda^n_i v$ and so over the complex field, $\lambda_i\in S^1$. But to what extent should I argue my steps? Should I just write a couple of sentences relying on Linear Algebra arguments under the assumption that I am already completely familiar with them, or should I actually find or describe a general diagonalization or a Jordan form of a given $\rho (g)$? I am really not sure what explanation I am required to give specifically, which demonstrates my understanding of the subject.
I've been studying math on and off and took Group Theory at least three years ago and linear algebra long before, and although I can still recall the main general elements of those, nothing is obvious to me in Group Representation Theory. This course is a bit advanced for me conceptually and is taking up most of my time while I should be practicing other courses as well. I could really use some guidance, please.
Best Answer
Just to follow up the hint you've been given, if $g$ has finite order $n$, then $$ 1 = \rho(1) = \rho(g^{n}) = \rho(g)^{n}. $$ So the minimal polynomial $f$ of $\rho(g)$ over $\mathbb{C}$ divides $x^{n} - 1$. The latter polynomial has $n$ distinct roots, the $n$-th roots of unity. Hence $f$, too, has distinct roots. Now linear algebra tells us that this implies that $\rho(g)$ is diagonalisable.