# Unipotent orbit in adjoint group over finite field

[Editted: The assertion is wrong; see Jay's answer]

My apology if this question is too simple. I am reading Deligne-Lusztig "Reductive groups over finite fields" and at the beginning of Chap. 4, there is a statement saying something about Green functions which seems to claim the following: Let $G$ be a adjoint semisimple group over a finite field $k$ and let $u\in G(k)$ be any unipotent element. Then $u$ is conjugate to $u^n$ in $G(k)$ for any $n$ prime to $p=\text{char}(k)$.

I know how to prove this when $p$ is good for $G$. It will be great to know how to do it in general. Thanks!

That's not what they're claiming and your statement is not true. Your claim is that every unipotent element is rational. However Lemma 5.6 of this article by Tiep and Zalesskii provides a counter example. Indeed, non-rational unipotent elements exist for instance in $\mathrm{F}_4(q)$ when $q$ is a power of 2.

The claim of Deligne-Lusztig is that, as the Green functions are integer valued and are the restriction of a character, we must have $Q_{T,G}(u) = Q_{T,G}(u^n)$ if $(n,p) = 1$. This is the reverse argument, which you can deduce from the argument given in Lemma 2.6(i) of the article from Tiep and Zalesskii.

Edit: Actually, I thought about this on the way home and I should clarify my last comment. The deduction can be seen as follows. Let $G$ be a finite group and $u \in G$ an element of order $p^a$ for $p$ a prime. Assume $\chi : G \to \mathbb{C}$ is a character of $G$ then we have $\chi(u) \in \mathbb{Q}(\zeta)$ where $\zeta \in \mathbb{C}$ is a primitive $p^a$th root of unity. If $\rho : G \to \mathrm{GL}_n(\mathbb{C})$ is the representation affording $\chi$ then we have

$$\chi(u) = \lambda_1 + \cdots + \lambda_k$$

where $\lambda_1,\dots,\lambda_k \in \mathbb{Q}(\zeta)$ are the eigenvalues of $\rho(u)$. Clearly we have

$$\chi(u^m) = \lambda_1^m + \cdots + \lambda_k^m$$

for any $m$. Now assume $m$ is coprime to $p$ then the map $\gamma : \mathbb{Q}(\zeta) \to \mathbb{Q}(\zeta)$ defined by the $\mathbb{Q}$-linear extension of $\zeta \mapsto \zeta^m$ is a field automorphism fixing $\mathbb{Q}$. As $\chi(u) \in \mathbb{Z}$ is integral we have

$$\chi(u) = \gamma(\chi(u)) = \chi(u^m)$$

as desired.

• Many thanks for pointing this out! Embarrassing that I forgot this simple Galois theory trick. Mar 24, 2015 at 18:24