$\DeclareMathOperator\SL{SL}$Let $p$ be an odd prime and let $ \SL^1\_2(\mathbb{Z}\_p)$ denote the kernel of the natrual surjective morphism $\SL\_2(\mathbb{Z}\_p)\rightarrow \SL\_2(\mathbb{Z}\_p/p\mathbb{Z}\_p)$. Then the abelianization of $\SL^1\_2(\mathbb{Z}\_p)$ is $(\mathbb{Z}/p\mathbb{Z})^3$. Recall that there are only two nonabelian groups with order $p^3$ up to isomorphism as follows:


1. $G\_1=\langle x,y\mid x^{p^2}=y^p=1,~y^{-1}xy=x^{1+p}\rangle$.
2. $G\_2=\langle x,y,z\mid x^p=y^p=z^p=1,~xz=zx,~zy=yz,~xy=yxz\rangle$.


**Question:** Is there a continuous surjective morphism $\SL^1\_2(\mathbb{Z}\_p)\to G\_i$ for $i=1,2$?



Finite quotients of $p$-adic congruence subgroups of $\operatorname{SL}_2$
$\DeclareMathOperator\SL{SL}\newcommand{\Z}{\mathbf{Z}}$Indeed, $G\_1$ is a quotient, but not $G\_2$.


First, one checks that the third term in the lower central series is the kernel of $\SL\_2^1(\Z\_p)\to\SL\_2^1(\Z/p^3\Z)$. So this converts this into a question about quotients of a group of order $p^6$, namely $\SL\_2^1(\Z/p^3\Z)$.


For this group, the derived subgroup is central and generated by $p$-powers. This already implies that every quotient of $\SL\_2^1(\Z/p^3\Z)$ of exponent $p$ is abelian, and hence $G\_2$ is not a quotient.


One gets $G\_1$ by modding out by the normal subgroup of elements of the form $\begin{pmatrix}1+p^2a & 0\\ pc & 1-p^2a\end{pmatrix}$ for $a\in\Z/p\Z$, $c\in\Z/p^2\Z$.


