Are the integer matrices in SO(3,2) "boundedly generated"? 
Let $G$ be the subgroup of integer matrices in $\mathrm{SO}(3,2)$.

(The invertible linear maps from a $5$ dimensional real vector space to itself which leave invariant a nondegenerate symmetric billinear form of signature $(3,2)$, and have determinant $1$.)

$G$ is a group of real rank $2$, so conjecturally, it should be
  boundedly generated.

(There exists an $m \in \mathbb{N}$, and cyclic subgroups $C_1, \dots, C_m \leq G$ such that $G = C_1 \cdots C_m$.)
I would like to know whether the following much weaker statement holds:

There exists a positive integer $m$, and finitely generated subgroups
  $H_1, \dots, H_m \leq G$ of infinite index, such that $G = H_1 H_2
 \cdots H_m$.

 A: $\mathrm{SO}(3,2)_{\mathbb{Z}}$ is an orthogonal group of $\mathbb{Q}$-rank 2, so it is boundedly generated. This was proved by Tavgen [Math. USSR. Izvestiya 36 (1991) 101-127; MR1044049] for the quasi-split case, which includes $\mathrm{SO}(3,2)_{\mathbb{Z}}$, but the general case is due to Erovenko and Rapinchuk [J. Number Theory 119 (2006) 28-48; MR2228948]. For most arithmetic groups of real rank $\ge 2$, it is not known whether they are boundedly generated.
The Erovenko-Rapinchuk proof is by induction, and the induction step writes the group as a product of finitely many subgroups $H_i$, as in the question's weaker statement. (Each $H_i$ is the stabilizer of a vector.)  Although it does not seem to be spelled out in their paper, I have heard from the authors (in talks and elsewhere) that their methods to write the group as a product also work for unitary groups (and maybe others?), not just orthogonal groups. So the methods may suffice to establish the weaker statement for isotropic arithmetic groups of classical type. Or perhaps the weaker statement is fairly easy for isotropic groups?
