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[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!

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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.

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  • $\begingroup$ Many thanks for pointing this out! Embarrassing that I forgot this simple Galois theory trick. $\endgroup$ – Cheng-Chiang Tsai Mar 24 '15 at 18:24

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