Number of involutions in finite reductive groups Let $G$ be a connected split reductive group over $\mathbb{Z}$. Let $n$ be a positive integer. Let $i_n(q)$ be the number of elements of $G(\mathbb{F}_q)$ satisfying $x^n=1$.
Question: Is there a "nice" case-free formula for $i_n(q)$? Is it always a polynomial in $q$?
The "best hope" is a formula akin to Steinberg's formula for the number of elements of $G(\mathbb{F}_q)$ (cf. Section 7.3 of these notes). Can this best hope be realised?
For instance, suppose $n=2$. Then we are considering the number of involutions in finite reductive groups. A case by case analysis shows that there exists polynomials $P, R\in \mathbb{Z}[t]$ such that $i_2(q)=P(q)$ for all even $q$ and $i_2(q)=R(q)$ for all odd $q$, cf. this paper. However, I do not know a case-free proof of this fact. Nor do I know a case free expression for these polynomials.
Remark 1. If it helps, we can assume that the characteristic is large enough.
Remark 2. Aforementioned result of Steinberg amounts to giving an expression for $|\mathrm{Hom}(\mathbb{Z}, G(\mathbb{F}_q)|$. The above question asks for an expression for $|\mathrm{Hom}(\mathbb{Z}/n, G(\mathbb{F}_q)|$.
 A: Let's assume $q$ is odd. If $G={\rm GL}_n({\mathbb F}_q)$, then any involution in $G$ has minimal polynomial $t^2-1$, so is diagonalizable over ${\mathbb F}_q$ and is therefore conjugate to (exactly) one of the matrices of the form $$D_{r,n-r}=\begin{pmatrix} I_r & 0 \\ 0 & -I_{n-r} \end{pmatrix}.$$
Now it's easy to determine the order of the conjugacy class of $D_{r,n-r}$: it's $\frac{p_n(q)}{p_r(q)p_{n-r}(q)}$ where $p_n(q)=|{\rm GL}_n({\mathbb F}_q)|$. This is obviously a rational function; to show it's a polynomial you need to know that $(q-1)\cdots (q^r-1)$ divides $(q^{n-r+1}-1)\cdots (q^n-1)$.
Summing up the orders of all these conjugacy classes will give you the number of elements of order 2 (or order $\leq 2$ if you add 1 for the identity). You can certainly modify this argument for ${\rm SL}_n({\mathbb F}_q)$. A similar argument probably also works for the other split simple algebraic groups, e.g. in ${\rm Sp}_{4}({\mathbb F}_q)$ there are two classes of involutions, with characteristic polynomials $(t+1)^4$ and $(t-1)^2(t+1)^2$. The former has just one element, while the latter has $|{\rm Sp}_4({\mathbb F}_q)|/|{\rm Sp}_2({\mathbb F}_q)|^2 = q^4+q^2$ elements. So we get $q^4+q^2+2$ elements of ${\rm Sp}_4({\mathbb F}_q)$ satisfying $x^2=I$.
For the case $q$ even, you aren't dealing with diagonalizable elements any more. These are (a subset of the) unipotent elements. This is more complicated because (a) while the unipotent conjugacy classes over algebraically closed fields are known, I don't think the intersection of any such conjugacy class with $G({\mathbb F}_q)$ has to be a single conjugacy class; (b) 2 is a bad prime for simple groups other than ${\rm SL}_n$ (and the classification of unipotent orbits in $G(\overline{{\mathbb F}_q})$ is rather a niche topic). But in principal I think you could find a conceptual argument in this case too.
A: The set of involutions in $Sp(n,\mathbb F_q)$ is determined in Pantoja, Soto Andrade, Vargas, Journal of Lie Theory  Vol 25 (2015) page 1069. We also determine the set of anti-involutions.
