# How boundedly generated is $SL_3(\mathbb{Z})$?

The group $G = \mathrm{SL}_3(\mathbb{Z})$ is known to be boundedly generated, that is, there exists some $m \in \mathbb{N}$, and $g_1, \dots, g_m \in G$ such that we have the following equality of sets: $$G = \langle g_1 \rangle \cdots \langle g_m \rangle.$$

Is there an elementary/accessible proof of this fact? I would like to have a reference to a proof.

How small can we take $m$ to be?

Is it possible to take $m = 8$?

• The Wikipedia article cites a 1984 paper by Carter and Keller, Elementary expressions for unimodular matrices. They determine that for $G = \mathrm{SL}_n(\mathbb Z)$ ($n > 2$) you can take elementary matrices for all the $g_i$ and get an upper bound $m \le \frac12(3n^2-n) + 36$. – algorithmshark Sep 11 '15 at 16:49
• Section 4.1 gives the proof. It uses Dirichlet's theorem on primes in arithmetic progressions. perso.univ-rennes1.fr/bachir.bekka/KazhdanTotal.pdf – Ian Agol Sep 11 '15 at 20:14
• Another exposition of the elementary proof can be found in a paper by Adian and Mennicke worldscientific.com/doi/abs/10.1142/s0218196792000220 – Andrei Smolensky Sep 12 '15 at 11:59
• $m=2$ is clearly impossible: you can't even generate the invertible matrices mod 2 this way, because each $\left|\langle g_i\rangle\right| \leq 7$ and $7 \times 7 < 168$. – Noam D. Elkies Sep 12 '15 at 19:48
• You're welcome. I see that you've now changed "$m=2$" to "$m=5$". In fact I think any $m<8$ can be excluded by a $p$-adic dimension argument: for any $p$, each $\langle g_i \rangle$ is contained in the union of finitely many one-dimensional $p$-adic patches, so their product has dimension at most $m$, and thus cannot exhaust ${\rm SL}_3({\bf Z})$ beause this group's closure in ${\rm SL}_3({\bf Z}_p)$ has dimension $8$. – Noam D. Elkies Sep 12 '15 at 20:36