# Subgroups of $SL_2(F)$ generated by unipotent elements

I am interested in the following problem : given a finite field $F$ and two unipotent elements $g_1,g_2\in\mathrm{SL}_2(F)$ which do not commute, what can we say about the subgroup they generate? More specifically, is there an additional condition under which it is known that they generate all of $\mathrm{SL}_2(F)$?

For $F=\mathbf F_p$ it is easily seen that this is always the case. For $F=\mathbf{F}_{p^2}$ I convinced myself a while ago (using lenghty computations of products of $g_1,g_2$) that they generate either the whole of $\mathrm{SL}_2(F)$ or a subgroup conjugated to $\mathrm{SL}_2(\mathbf{F}_p)$. I would be glad to see a more enlightening proof of this fact, and even more if it pointed at generalizations to other algebraic groups than $\mathrm{SL}_2$.

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You should look up Dickson's Theorem. It deals pretty comprehensively with the two dimensional case when $F$ is finite, at least in odd characteristic. The proof may be found in Gorenstein's 1968 book "Finite Groups". I don't remember the statement precisely, but I think it's that $\langle \left( \begin{array}{clcr} 1 & 0\\1 &1 \end{array} \right),\left( \begin{array}{clcr} 1 & \lambda\\0 &1 \end{array} \right) \rangle$ is all of ${\rm SL}(2,K)$, where $K$ is the subfield of $F$ generated by $\lambda \neq 0,$ except when $|F| =9,$ in which case ${\rm SL}(2,5)$ may occur.

There is also the work by J.G. Thompson on quadratic pairs, which may be seen as a massive generalization of this result. Going back further historically than that, there is the Hall-Higman theorem, and more recently there has been extensive work on so-called $2F$-modules, which is relevant to the classification of finite simple groups.

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Thanks! I'll look up Dickson's theorem first. – Jean Raimbault Mar 14 '13 at 20:49
Maybe one ought to remark here that any two non-commuting unipotent elements are conjugate to a pair of this type. A unipotent element fixes a unique point of $F\mathbb{P}^1$. The action of $SL(2,F)$ on $F\mathbb{P}^1$ is transitive on pairs of points in $F\mathbb{P}^1$, so one may assume the fixed point of one unipotent is $\infty$, and the other is $0$, giving the desired normalization (but where the lower left entry of the first matrix might not be $1$). – Ian Agol Mar 14 '13 at 21:41
@Agol: A valid point; there are many ways to get there. – Geoff Robinson Mar 14 '13 at 22:02

A few complements to Geoff's answer and Agol's comment:

1) Dickson's theorem is worked out in detail (somewhat lengthy) in Gorenstein's Chapter 2, page 44+, as Theorem 8.4. His section 8 is devoted to a study of this matrix group and its projective quotient. The treatment is elementary but not conceptual. In the exceptional case of the field of 9 elements, the proper subgroup you might generate is of order 120 and has the simple group of order 60 as a quotient.

2) Larger matrix groups are definitely more complicated, though the theme of generation by unipotent matrices (geometrically, transvections) is in any case prominent in the study of such classical groups. For instance, when you look at $3 \times 3$ matrices over a finite field, you easily find two unipotent Jordan blocks (with a single block just above the diagonal) which fail to commute and only manage to generate a subgroup of upper triangular matrices.

3) As Agol comments, you still need to get your arbitrary non-commuting unipotent matrices to be conjugate to a pair treated in Dickson's theorem. What Agol suggests is in the more sophisticated but also more conceptual framework of Chevalley groups (as in Steinberg's old Yale lectures), or more generally semisimple linear algebraic groups. Looking at $\mathrm{SL}_2$ in this spirit, there are two Borel subgroups containing a typical maximal torus such as the group of diagonal matrices (and any two distinct Borels intersect in such a torus). In turn, the unipotent subgroups (say upper and lower unitriangular matrices) suffice to generate the entire group. Such a unipotent group is cyclic over the prime field, but otherwise elementary abelian (which creates complications for applying Dickson's theorem). Moreover, the quotient of the full group by a Borel subgroup is a projective line, on which the torus has precisely two fixed points, etc. This is where Agol's comment comes from.

4) Most of this Chevalley group formalism is well encoded in the more elementary theory of split BN-pairs, where it's still fairly easy to see conceptually how the generation by unipotents works. See Steinberg's lecture notes (now online at his UCLA homepage) or Carter's 1968 book for results in this direction.

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This is a much more comprehensive and conceptual answer than mine! – Geoff Robinson Mar 15 '13 at 19:30