Projective special linear group

Question

What is it the minimum number of generators for $PSL(2,\, \mathbb{F}_q)$? Is it known? Is there some references I could see?

Answer 1

The group is generated by two elements. This was proved by L.E. Dickson, and that can be found in D. Gorensteins book on Finite Groups. Strictly, that proof does not cover the cases $q < 4$ and $q = 9.$ The case $q=2,3,4$ can be checked directly. Since ${\rm PSL}(2,9) \cong A_{6},$ this can be checked directly too.

Added later: Perhaps Dickson required the prime power $q$ to be odd, I can't remember. But in any case, when $q$ is a power of $2$ and $q >2,$ then $G = {\rm PSL}(2,q)$ can be generated by two elements, one of order $q-1$ and one of order $q+1.$ The first, $C$ say, can be taken to normalize a chosen Sylow $2$-subgroup, and permute its non-identity elements transitively by conjugation, and the second, $D$ say, permutes the Sylow $2$-subgroups transitively by conjugation. Now $\langle C,D \rangle$ must have odd order, for otherwise, it would contain a full Sylow $2$-subgroup, and hence have order divisible by $q(q-1)(q+1) = |{\rm PSL}(2,q)|$ in this case. Hence $CD = DC$ is a group of order $(q+1)(q-1).$ However, all odd order subgroups of ${\rm PSL}(2,q)$ are Abelian when $q$ is a power of $2$. There are several ways to finish-eg, N. Ito proved that a product of Abelian finite groups is metabelian and we have $G = S(CD)$ for $S \in {\rm Syl}_{2}(G)$ (also $S$ is Abelian), and ${\rm PSL}(2,q)$ is not metabelian for $q = 2^{n}$ when $n >1$- alternatively, one can just see that $G$ contains no element of order $(q-1)(q+1)$, and $CD$ would have to be cyclic.

(The $2$-generation of ${\rm PSL}(2,q)$ for $q >3$ also follows from a theorem of R. Steinberg).