# 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.

1. 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.

2. 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?

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.

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.

• Of course, in your counterexample the exponent is divisible by the characteristic of your field; which makes it so that the minimal polynomial actually becomes inseparable. Still, it is interesting to note that such a counterexample is possible precisely when the characteristic of the field divides the order of $G$. – Starfall Jun 7 '16 at 7:55
• @Starfall: That's right! A related fact is that in characteristic $p$ there are no roots of unity of order $p$. So by requiring that the field $K$ contains the necessary roots of unity we can actually exclude this possibility :-) – Jyrki Lahtonen Jun 7 '16 at 8:08