I'm looking for a reference (or quick proof) of the following fact. Fix some $n \geq 3$ and some $\ell \geq 2$. Set $\Gamma\_n(\ell) = \text{ker}(\text{SL}\_n(\mathbb{Z}) \rightarrow \text{SL}\_n(\mathbb{Z}/\ell \mathbb{Z}))$. Next, set


$$\Gamma\_n'(\ell) = \{\text{$A \in \Gamma\_n(\ell)$ $|$ for all diagonal entries $a\_{ii}$ of $A$, we have $\ell^2 | (a\_{ii}-1)$}\}.$$


It is an easy exercise to show that $\Gamma\_n'(\ell)$ is a subgroup of $\Gamma\_n(\ell)$. Finally, for all $1 \leq i,j \leq n$ with $i \neq j$, let $e\_{ij}$ be the elementary matrix obtained by taking the identity matrix and placing a $1$ at position $(i,j)$.


Observe that $e\_{ij}^{\ell} \in \Gamma\_n'(\ell)$. The fact I'm interested in is the fact that $\Gamma\_n'(\ell)$ is generated by the set of all the $e\_{ij}^{\ell}$. I've seen this asserted without proof in print before (though I don't recall exactly where), but I've been unable to work out a proof myself.



Generators for a certain congruence subgroup of SL(n,Z)
There is a Theorem of Tits which says that the group generated by $e\_{ij}^l$ has finite index in $SL\_n({\mathbb Z})$ if $n\geq 3$. The paper is a Comptes rendus announcement (generating systems of congruence subgroups, CR. ACad. Sci 283 (1976), no. 9 A693-A695 (see the following review <http://www.ams.org/mathscinet/search/publdoc.html?amp=&loc=revcit&revcit=244257&vfpref=html&r=16&mx-pid=424966>). By a result of Mennicke, finite index subgroups of $SL\_n({\mathbb Z})$ are congruence subgroups ($n\geq 3$). So, you need only check that the congruence closure of the group generated by $e\_{ij}^l$ in $SL\_n({\mathbb Z})$ is of the desired type. This is a purely local checking (prime by prime, for primes dividing $l$) and can be done easily. 


ADDED: what you are asking for is (seemingly) stronger than Tits' theorem: not only do the $e\_{ij}^l$ generate a finite index subgroup, they generate a specific congruence subgroup. But, as the previous paragraph shows, this is $equivalent$ to the result of Tits. 


