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

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

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.

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.