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An infinite profinite group $G$ is called just-infinite if all non-trivial closed normal subgroups of $G$ have finite index. A profinite group is called hereditarily just-infinite if every open subgroup is just-infinite. If $p$ is an odd prime, then it's known that the pro-$p$ group ${\rm SL}_{n}^{1}(\mathbb{Z}_{p}):=\ker({\rm SL}_{n}(\mathbb{Z}_{p})\to {\rm SL}_{n}(\mathbb{F}_{p}))$ is hereditarily just-infinite where $\mathbb{Z}_p$ is the ring of $p$-adic integers. However, it is obvious that ${\rm SL}_{2}^{1}(\mathbb{Z}_{2}):=\ker({\rm SL}_{2}(\mathbb{Z}_{2})\to {\rm SL}_{2}(\mathbb{F}_{2}))$ is not just-infinite since $\{\pm I_2\}$ is a closed normal subgroup of ${\rm SL}_{2}^{1}(\mathbb{Z}_{2})$. This leads to the following questions:

  1. Is ${\rm SL}_{2}^{1}(\mathbb{Z}_{2})/\{\pm I_2\}$ hereditarily just-infinite?
  2. Let ${\rm SL}_{2}^{2}(\mathbb{Z}_{2}):=\ker({\rm SL}_{2}(\mathbb{Z}_{2})\twoheadrightarrow {\rm SL}_{2}(\mathbb{Z}_{2}/4\mathbb{Z}_{2})$. Is ${\rm SL}_{2}^{2}(\mathbb{Z}_{2})$ hereditarily just-infinite?
  3. Is the projective special linear group ${\rm PSL}_{2}(\mathbb{Z}_{2})$ over $\mathbb{Z}_2$ hereditarily just-infinite?

Any references will be appreciated.

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    $\begingroup$ I think the answer is positive for all three questions. For some of them it should follow from Margulis Normal Subgroup Theorem. Otherwise, one can prove it by looking at the Lie algebra given by the lower central series and prove that that is simple. $\endgroup$
    – Matteo Vannacci
    Nov 30 at 8:56

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