# Existence of a regular semisimple element over $\mathbb{F}_{q}$

This is probably old, a Chevalley level of old, but I'm not at all an expert in this field so I need help.

Let $$G$$ be a simply connected (almost) simple linear algebraic group defined over $$K=\mathbb{F}_{q}$$. Is there a clean reference in the literature to the fact that there must be at least one regular semisimple element in $$G(K)$$?

I know from the literature that regular semisimple elements are dense in $$G$$ and in every maximal torus $$T$$ (for example §2.3 and §2.5 in Humphreys's Conjugacy classes in semisimple algebraic groups), but this ensures only that there are many elements over the algebraic closure $$\overline{K}$$, not specifically over $$K$$. I know that this is true for $$\mathrm{char}(K)=0$$, using unirationality and the fact that $$K$$ is infinite, as in this question. I even know how to presumably shoot my question with a cannon in a few different ways, for instance:

1. Simply connected (almost) simple groups are in 1-1 correspondence with Dynkin diagrams (see for instance §32 in Humphreys's Linear algebraic groups), so I could manually write down one regular semisimple element over $$\mathbb{F}_{q}$$ for each such $$G$$. [EDIT: as pointed out in the comments below by LSpice, since the field in not algebraically closed there's no 1-1 correspondence as stated. One needs to consider automorphisms of the Dynkin diagrams as well.]
2. Fleischmann and Janiszczak (1993) have formulas for the number of regular semisimple elements over $$\mathbb{F}_{q}$$ at least for the classical $$G$$, i.e. of type $$A,B,C,D$$; they even announce on p. 484 that they obtained the number of semisimple conjugacy classes for $$E$$, so possibly they cover those cases as well. In any case, if there is a formula and its result is $$>0$$, then in particular there is one such element.
3. Lehrer (1992) showed among other things the "curious" result that, independently of characteristic, any semisimple simply connected $$G$$ has an odd number of regular semisimple conjugacy classes in $$G(K)$$ (see Corollary 3.5). Again, in particular if the number is odd then it is not $$0$$.

I would classify each of the routes above as "too strong for my question". I feel there must be some chapter-1-of-a-textbook kind of reference for "there exists a regular semisimple element over $$\mathbb{F}_{q}$$", but I could not find one.

Do you know where to find it? Also, bonus content: is the fact true and referenced if I relax the hypotheses (connected, or semisimple, or reductive...)?

• I think that this fact is probably less elementary than you expect, perhaps only for $q = 2$ and maybe $q = 3$. (As an example, I was sure that $\operatorname{SL}_2$ when $q = 2$ was a counterexample, but there's that elliptic torus.) In terms of hypotheses, once you can handle sc, reductive is enough: any regular element in the sc cover maps to a regular element; an sc group is a product of its simple components; and $(\operatorname{Res}_{E/K}G)(K)$ equals $G(E)$, so that it even suffices to assume $G$ absolutely simple. Commented Aug 7, 2023 at 0:12
• Steinberg - Regular elements of semisimple groups is the first place I would (and did!) look. I didn't find it there, but Theorem 1.7 is tantalising: it says that, if you can find just a rational regular semisimple class (i.e., the elements of the class might be permuted, but the class itself is fixed, by Galois), then you can find a regular semisimple element. Commented Aug 7, 2023 at 0:15
• Sorry for all the comments: I think that the classification in (1) is not true, and that you also need to account for the possibility of a non-trivial action of Galois on the Dynkin diagram. (Since the absolute Galois group is pro-cyclic, there's no triality $\mathsf D_4$ problem, which at least removes the nastiest kind of action.) Do you mean to assume that your group is split? Commented Aug 7, 2023 at 22:12
• Re, I believe that the assumption from p. 4 that the ground field $\mathsf K$ is algebraically closed stands throughout the paper; otherwise this is not even close to true. (Consider the special linear, and the special unitary, groups in the same number of variables over $\mathbb R$.) If your ground field is not algebraically closed, as in your example, then you need to impose the additional hypothesis that the groups are split to obtain the desired parameterisation. Commented Aug 12, 2023 at 18:02
• (E.g., there are still unitary groups over finite fields.) But the quasisplitness of groups over finite fields means that, in your setting, you need only a little more, namely an automorphism of the Dynkin diagram. Commented Aug 12, 2023 at 18:06

See Proposition 7.1.4 in Dat, Orlik, and Rapoport, Period domains over finite and $$p$$-adic fields, Cambridge tracts in Mathematics, vol. 183. While I don't have this book in front of me, I'm pretty sure that the result asserts the existence of elliptic semisimple elements in all finite reductive groups. Such elements are necessarily regular. Most groups are handled via a uniform argument, but several are treated as special cases. Technically, the proof contains a gap, as the case of $$G_2(\mathbb{F}_2)$$ was inadvertently omitted. Nonetheless, the result remains true in this case, where one can find a regular element by looking at the Coxeter torus $$T$$ of $$\operatorname{SL}_3$$, which we can consider as a subgroup of $$G_2$$. Every nontrivial element of $$T(\mathbb{F}_2)$$ is regular in $$G_2$$.
Let $$\sigma: G \rightarrow G$$ be the Frobenius map corresponding to the field automorphism $$x \mapsto x^q$$.
To show that there exists a regular semisimple element in $$G_{\sigma} = G(q)$$, by the Lang-Steinberg theorem it would be enough to show that there is a $$\sigma$$-invariant class of regular semisimple elements. See 2.7(a) in "Conjugacy Classes" by Springer and Steinberg (in Lecture Notes in Math 131).
• As far as I can see, 2.7(a) requires a homogeneous space $M$ for $G$. What $M$ are you using? (If you're letting $M$ be the variety of regular semisimple elements, then it is usually not a homogeneous space for $G$.) Commented Aug 11, 2023 at 4:28
• @LSpice: Yes you are right, to use 2.7(a) we need to assume that $G$ acts transitively on $M$. So this argument with Lang-Steinberg just shows that if there exists a $\sigma$-invariant conjugacy class $C$, then the fixed point set $C_{\sigma}$ is nonempty. Commented Aug 11, 2023 at 4:51