Element of finite order is semisimple and eigenvalues are roots of unity

This page describes a useful fact in character theory/linear representation theory arising from rudimentary linear algebra
View other such facts OR View all facts related to linear representation theory

Statement

Suppose $k$ is a field, $V$ is a finite-dimensional vector space over $k$ and $g$ is an element in $GL(V)$ such that there exists $n$ with $g^n$ equal to the identity matrix. Then:

• All the eigenvalues of $g$ over the algebraic closure of $k$ are $n^{th}$ roots of unity.
• Suppose that either $k$ has characteristic zero or $n$ is relatively prime to the characteristic of $k$. $g$ is semisimple, i.e. it is diagonalizable over the algebraic closure of $k$.

Applications

• Trace of inverse is complex conjugate of trace
• Characters are cyclotomic integers

Proof

$g$ satisfies the polynomial $x^n - 1$, hence the minimal polynomial of $g$ must divide this polynomial. So, every eigenvalue of $g$ must satisfy the polynomial $x^n - 1$, hence must be a root of unity.

If $n$ is relatively prime to the characteristic, then the polynomial $x^n - 1$ has no repeated roots, hence the minimal polynomial of $g$ has no repeated roots. So, $g$ is semisimple, i.e. it is diagonalizable over the algebraic closure.