The Cartan-Dieudonne theorem states that each element $g \in O(V)$, 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 $char\ k \neq 2$: 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?

  • $\begingroup$ 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. $\endgroup$ – Jim Humphreys Apr 15 '10 at 16:29

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