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$\DeclareMathOperator\GL{GL}\DeclareMathOperator\SL{SL}$Let $p$ be a odd prime. We say that a pro-$p$ group has finite rank if it is isomorphic to a closed subgroup of $\GL_d(\mathbb{Z}_p)$ for some integer $d$. A pro-$p$ group $G$ is called FAb if $U^{\text{ab}}$ is finite for all open subgroups $U$ of $G$ where $U^{\text{ab}}$ denotes the abelianization of $U$. For example, any open pro-$p$ subgroup of $\SL_n(\mathbb{Z}_p)$ is FAb. We will denote the first congruence subgroup of $\GL_n(\mathbb{Z}_p)$ by $\GL_{n}^{1}(\mathbb{Z}_p)$, i.e. $$\GL_{n}^{1}(\mathbb{Z}_p):=\ker(\GL_n(\mathbb{Z}_p)\to \GL_n(\mathbb{F}_p)).$$

Question: Suppose that $G$ is a torsion-free pro-$p$ group of finite rank.

  1. Is there a positive integer $N$ such that $G$ admits an injective homomorphism $G\to \GL_{N}^{1}(\mathbb{Z}_p)$ ?
  2. Assume further that $G$ is FAb. Is there an affirmative answer to Question 1? Note that we may assume that $G$ is non-solvable in this case. (Otherwise, it must be trivial.)
  3. If the answer to the Question 1 is no, then is it possible to give some sufficient condtions?
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  • $\begingroup$ Oops, sorry. I only arranged that the determinant of the reduction mod $p$ is 1. I erased my comment. $\endgroup$
    – YCor
    Nov 18, 2022 at 14:11
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    $\begingroup$ I think that the most important part of this question is that you have a chance to incorporate base change somehow and ask whether a group is absolutely FAb. \\ Seriously, this is an interesting question. Why do you exclude $p = 2$? Do you know counterexamples, or is it just the usual feeling that something will go wrong for $p = 2$? $\endgroup$
    – LSpice
    Nov 18, 2022 at 14:28
  • $\begingroup$ @LSpice Thanks for your comment. I do not know a counterexample to the case of $p=2$, and I just want to keep it simple in some sense. $\endgroup$
    – stupid boy
    Nov 18, 2022 at 15:12
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    $\begingroup$ @LSpice +1 for the sitcom reference $\endgroup$
    – krl
    Nov 18, 2022 at 16:03

1 Answer 1

3
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The answer to 1 is No. By Corollary 4.5 and Example 4.2 in Analytic pro-p groups of small dimensions, we see that

Every $p$-adic analytic pro-$p$ group of dimension $p$ which embeds into a saturable pro-$p$ group is itself saturable. On the other hand, there exists a torsion-free $p$-adic analytic pro-$p$ group of dimension $p$ which is not saturable.

In other words, any torsion-free $p$-adic analytic pro-$p$ group of dimension $p$ which is not saturable cannot be embeded into a saturable pro-$p$ group, and there really exist torsion-free $p$-adic analytic pro-$p$ group $G$ of dimension $p$ which is not saturable. Since ${\rm GL}_{n}^{1}(\mathbb{Z}_p)$ is saturable, this gives a counterexample to Question 1. However, the group $G$ in the above example is not FAb. So I still don't know the answer to Question 2.

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  • $\begingroup$ For those without access to the article, can you describe the non-saturable group? $\endgroup$ Feb 4, 2023 at 5:22
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    $\begingroup$ @TorstenSchoeneberg arxiv.org/abs/0806.2968 $\endgroup$
    – stupid boy
    Feb 4, 2023 at 14:19

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