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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 $K$.

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  • $\begingroup$ This could depend in a complicated manner on the $p_i$. The answer will be something between $2r$ and $3r$. If all $p_i \equiv 1 \bmod 4$, then there is a $2$-subgroup of rank $3r$, It might be hard to calculate the exact answer in other cases. $\endgroup$
    – Derek Holt
    Commented Sep 29, 2015 at 9:28
  • $\begingroup$ @DerekHolt you may choose the primes $p_1, \dots, p_r$ as you wish as long as they are distinct (if it makes life easier to assume that they lie in some arithmetic progression - go ahead!). Can you get something better than $3r$? Or at least get something linear in $r$ but with index in $G$ which is not so big? $\endgroup$
    – Pablo
    Commented Sep 29, 2015 at 9:39
  • $\begingroup$ I sketched a proof in answer to your previous question that $3$ is the highest rank of a subgroup of each ${\rm SL}(3,p_i)$. So $3r$ is maximal, and that can be achieved by choosing all $p_i \equiv 1 \bmod 4$. $\endgroup$
    – Derek Holt
    Commented Sep 29, 2015 at 10:12
  • $\begingroup$ @DerekHolt is it obvious that it is $3r$ from the fact that it is $3$ for each prime? A subgroup of a direct product does not have to be a direct product itself... $\endgroup$
    – Pablo
    Commented Sep 29, 2015 at 10:23
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    $\begingroup$ Yes I think it is obvious! There is a normal series of length $r$ in which the factors are subgroups of ${\rm SL}(3,p_i)$. $\endgroup$
    – Derek Holt
    Commented Sep 29, 2015 at 11:55

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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 ${\rm SL}(3,p_i)$

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