Action of one element of order $n$ and the $n$-th roots of unity Let $G$ be a finite group and $\rho : G\to GL(V)$ a representation of $G$ on the vector space $V$. If $g\in G$ is one element of order $n$ we have $g^n=1$, so that
$$\rho(g^n)=\rho(g)^n=I,$$
where $I$ is the identity linear transformation. Now, if we pick a basis we can convert $\rho(g)$ to a matrix. In particular, we can pick the Jordan basis of $\rho(g)$ so that the matrix $A$ of $\rho(g)$ takes the block-diagonal form
$$\begin{pmatrix}J_1 & 0 & \cdots & 0 \\0 & J_2 & \cdots  & 0 \\ \vdots & \vdots & \ddots & 0 \\ 0 & \cdots & \cdots & J_k \end{pmatrix}$$
where $J_i$ is a Jordan matrix, that is, for each $i\in \{1,\dots,k\}$ there is $\lambda_i$ such that $J_i$ is diagonal with $\lambda_i$ on the diagonal and with $1$'s right above.
I've heard that this implies that the action of $\rho(g)$ is diagonal with the $n$-th roots of unity on the diagonal.
I must confess that I'm not being able to understand this, though.


*

*How do we derive this result? I believe it has to do with the fact that the minimal polynomial of $\rho(g)$ divides $x^n-1$, but I don't know how to use this.

*For a given $n$ the number of $n$-th roots of unity is $n$. If I understood $A$ would have $n$ elements on the diagonal, implying that $A$ represents a linear map on one $n$-dimensional space. But $n$ is just the order of $g$, the representation $V$ is arbitrary! It could have dimension greater than $n$ or less than $n$. How can this be? 
So how to understand those two issues? How to show that $A$ is diagonal with the $n$-th roots of unity on the diagonal and how to reconcile this with the fact that $V$ can have any dimension?
 A: Let $ K $ be a field containing the $ k $th roots of unity such that its characteristic does not divide $ k $, and let $ A $ be a nilpotent matrix in $ GL_n(K) $ with $ A^k = I $. $ A $ is a root of $ x^k - 1 $, which splits into distinct linear factors in $ K[x] $. The minimal polynomial of $ A $ then also splits into distinct linear factors, and therefore we have that $ A $ is diagonalizable. Its eigenvalues are roots of its minimal polynomial which divides $ x^k - 1 $, therefore they are all $ k $th roots of unity.
A: Assume that $V$ is a vector space over a field $K$.
Supplementing Starfall's answer to your first question with the following facts about diagonalizability of $A$.
If $\operatorname{char} K=0$, then you can argue as follows. Any Jordan block $J$ necessarily also satisfies $J^n=I$. But if $J$ has more than one row (and hence at least a single $1$ adjacent to the diagonal), then an easy induction shows that
the diagonal entries of $J^m$ are all equal to $\lambda^m$, and the adjacent entries are all equal to $m\lambda^{m-1}$. So when $\lambda$ is a root of unity, and $m\cdot 1_K\neq0_K$ those non-diagonal entries are non-zero. Therefore all the Jordan blocks have size one and we are done.
On the other hand, if $\operatorname{char} K=p>0$, then the above argument fails. Actually the claim is also false. An element $g$ of order $p$ may be represented by the matrix
$$
\rho(g)=\pmatrix{1&1\cr0&1\cr}
$$
when we actually have
$$
\rho(g)^p=\pmatrix{1&p\cr0&1\cr}=I_2.
$$

Re: Your second question. Different Jordan blocks can share the same $\lambda$, and it is also possible that some roots of unity do not show up at all in the Jordan decomposition of $A$. So the number of distinct $n$th roots of unity cannot be compared with the size of the matrix $A$ at all.
