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$\DeclareMathOperator\SL{SL}$The Cartan–Dieudonné theorem states that each element $g \in \operatorname{O}(V)$, where $V$ is a quadratic space of dimension $n$ over a field of characteristic $\neq 2$, can be written as a product of $\leq n$ reflections.

Something similar is true for $\SL_n(k)$ for $\operatorname{char}k \neq 2$: each element $g$ can be written as a product of elements of order $4.$ Indeed, it suffices to prove this for $n=2$. Then $s_t:=\begin{pmatrix} 0 & t \\ -\frac 1t & 0 \end{pmatrix}$ is of order $4$ . Let $h(t):=s_ts_{-1}=\begin{pmatrix} t & 0 \\ 0 & \frac 1t \end{pmatrix}$, then each element in $U$, the group of upper triangular matrices, is a product of two conjugates of elements $h(t),h(t')$ (provided $|k|\geq 4$). Similarly for the lower triangular matrices $V$, and then $\SL_2$ is generated by these subgroups.

Since a simply connected semisimple $k$-split group is generated by $\SL_2$'s, the same argument applies.

What can we say about other algebraic groups? Clearly unipotent groups (in characteristic 0) don't have elements of finite order, so nothing there.

What about (the $k$-rational points of) anisotropic groups?

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    $\begingroup$ The question needs better focus, I think. (More context and motivation would also help.) There is no obvious connection here with reflections or with Dieudonne's theorem: the square of your $s_t$ is not a "reflection" and no bounds on number of generators are implicit here. The type of field (finite, etc.) plays a role, along with the $k$-structure. Generation of semisimple groups comes up in work of Steinberg, Tits, and others; for more general groups you want to distinguish cases where a Levi decomposition exists and those in characteristic $p$ where it doesn't, etc. $\endgroup$ Apr 15, 2010 at 16:29
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    $\begingroup$ I think I remember there are (anisotropic) $\mathbf{Q}$-forms $G$ of $\mathrm{SL}_3$ for which $G(\mathbf{Q})$ is torsion-free. $\endgroup$
    – YCor
    Dec 19, 2023 at 22:47
  • $\begingroup$ @YCor, do you recall whether those were made from division algebras, or, rather, as (rationally anisotropic) unitary groups? $\endgroup$ Dec 20, 2023 at 3:03
  • $\begingroup$ @paulgarrett Division algebras. I (hopefully) retrieved the argument and posted it as an answer. $\endgroup$
    – YCor
    Dec 20, 2023 at 11:34

1 Answer 1

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Let $M$ be a division algebra of degree 3 (i.e., dimension 9) over $\mathbf{Q}$ that splits over $\mathbf{R}$, and $M_1$ its norm 1 subgroup. So $M_1$ is a $\mathbf{Q}$-anisotropic simple algebraic group, isomorphic to $\operatorname{SL}_3$ over $\mathbf{R}$.

I claim that $M_1(\mathbf{Q})$ is torsion-free.

Indeed, let us identify $M(\mathbf{R})$ with $\operatorname{Mat}_3(\mathbf{R})$. Then using that every element in $\operatorname{SO}(3)$ has the eigenvalue 1, we see that every finite order element $\theta$ in $M_1(\mathbf{R})$ has $\theta-1$ non-invertible. Let now $\theta$ be a finite order element in $M_1(\mathbf{Q})$. Then $\theta-1$ is non-invertible. Since $M(\mathbf{Q})$ is a division algebra, we deduce $\theta-1=0$. So $\theta=1$, proving that $M_1(\mathbf{Q})$ is torsion-free.

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  • $\begingroup$ Very nice! Persuasive, too! :) $\endgroup$ Dec 20, 2023 at 16:49
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    $\begingroup$ If I'm correct, it works in the same way in every odd degree, providing $\mathbf{Q}$-forms $G$ of $\mathrm{SL}_n$ for each odd $n\ge 3$ with $G(\mathbf{Q})$ torsion-free. $\endgroup$
    – YCor
    Dec 20, 2023 at 16:53
  • $\begingroup$ Yes, that's what I was thinking... maybe over number fields with at least one real place??? :) $\endgroup$ Dec 20, 2023 at 17:04
  • $\begingroup$ Once you'd shown what is true, it occurred to me that we can also give a "purely algebraic" proof, from the fact that the degrees of cyclotomic polynomials (apart from those with zeros $\pm 1$) are always even... so cannot be subfields of odd-dimensional division rings. :) $\endgroup$ Jan 29 at 21:13
  • $\begingroup$ The "S" requirement exactly-and-only excludes $-1$, ... :) $\endgroup$ Jan 29 at 21:22

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