The maximal possible rank of a subgroup of a product of special linear groups In this question I ask for a generalization of What is the maximal possible rank of a subgroup of a special linear group mod a prime?
Let $p_1, \dots, p_r$ be $r$ distinct odd primes.
Set $$G = \prod_{i=1}^r \mathrm{SL}_3(\mathbb{F}_{p_i})$$ 
What is $\max \{d(H) : H \leq G\}$?
Here we denote by $d(K)$ the smallest cardinality of a generating set of a group $K$. 
 A: I should answer the question! In my answer to the earlier question I sketched a proof that the maximal rank of a subgroup of ${\rm SL}(3,p)$ with $p$ prime is $3$, except for when $p=2$, when it is $2$. So $3r$ is an upper bound on ranks of subgroups of the group $G$ in the question.
Now, when $p \equiv 1 \bmod 4$, ${\rm GL}(2,p)$ has an irreducible subgroup of rank $3$ equal to a central product of $Q_8$ and a cyclic scalar subgroup of order $4$. Hence so does ${\rm SL}(3,p)$.
So, if we choose $p_i \equiv 1 \bmod 4$ for $1 \le i \le r$, then the direct product of these $r$ rank $3$ subgroups is a $2$-subgroup of $G$ of rank $3r$.
Note that the largest proper subgroup of ${\rm SL}(3,p)$ is the parabolic subgroup $p^2:{\rm GL}(2,p)$, which has index $p^2+p+1$. Since this has a quotient of order $2$, the direct product of these subgroups is a subgroup of $G$ of rank (at least, and probably exactly) $r$ and index $\prod_{i=1}^r (p_i^2+p_i+1)$.
I would guess that no subgroup of larger order has rank at least $r$, since any such subgroup would have to contain at least one of the quasisimple direct factors ${\rm SL}(3,p_i)$
