Let $p$ be a prime number, and let $\mathbb{F}\_p$ be the unique field of cardinality $p$. 


What is $\max \{d(H) : H \leq \mathrm{SL}\_3(\mathbb{F}\_p)\}$?


Here we denote by $d(G)$ the smallest cardinality of a generating set of a group $G$. 



What is the maximal possible rank of a subgroup of a special linear group mod a prime?
I am very confident that, for $q=p^e$ with $p$ prime, the answer is $2e$ when $q$ is even, and $2e+1$ when $q$ is odd. There is an elementary abelian subgroup $P$ of rank $2e$ with
$$P=\left( \begin{array}{ccc}1&a&b\\0&1&0\\0&0&1\end{array} \right).$$
When $q$ is odd, all elements of $P$ are inverted by the diagonal matrix $t$ with entries $(1,-1,-1)$, and $\langle P,t \rangle$ has rank $2e+1$. Note that this maximal rank is at least $3$ except in the case $q=2$, when you can check directly that it is $2$.


I have not attempted to write down a complete formal proof, and I don't plan to. If you need a formal proof, then I would happy to help! Here is a sketch proof.


Let $G \le {\rm SL}(3,q)$. I proved a result in a recent paper with Roney-Dougal that any finite primitive irreducible subgroup of ${\rm GL}(n,K)$ for any field $K$ has rank at most $1 + \lceil(2\log\_3 2)\log\_2 n \rceil$, which gives rank at most $3$ for $n=3$.


An imprimitive irreducible subgroup is a subgroup of $(q-1)^2:S\_3$ and it is not hard to show that any such group has rank at most $3$.


So suppose that $G$ is reducible. By replacing $G$ by its dual if necessary, we can assume that it fixes a $2$-dimensional subspace, and so it is a subgroup of the maximal parabolic subgroup $N:{\rm GL}(2,q)$, where $N$ is elementary abelian of order $q^2$ with the natural induced action of ${\rm GL}(2,q)$ on $N$. Let $H$ be the image of the projection of $G$ onto ${\rm GL}(2,q)$.


If $H$ acts irreducibly on $N$, then the rank of $G$ is equal to the rank of $H$ (that is a known result). In a 1991 paper, Kovacs and Robinson proved that any finite completely reducible subgroup of ${\rm GL}(n,K)$ (for any field $K$) has rank at most $3n/2$. So $H$ and hence $G$ has rank at most $3$ in this case.


So we are left with the situation when $H$ does not act irreducibly on $N$, when $G$ is conjugate to a group $U$ of upper triangular matrices. Then $U$ has the strucure $Q:D$, where $Q$ is a $p$-group of class $2$ and rank $2e$, and $D \cong C\_{q-1}^2$. I haven't tried to write out a formal proof in this case. The maximal ranks of subgroups of $Q$ and $D$ are $2e$ and 2, so we easily get a bound of $2e+2$. But in fact generators projecting nontrivially onto $D$ generally reduce the numbers of generators in $Q$ that you need. The only exception is the involution $t$ above, giving the maximal rank $2e+1$ for odd $q$.


