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I only know a few examples of pro-$p$ groups.

  • Of course the $p$-adics $\mathbf{Z}_p$, and any finite $p$-group.
  • Congruence subgroups of $\text{GL}_n(\mathbf{Z}_p)$: e.g. $\Gamma_1:=\{g\in \text{SL}_n(\mathbf{Z}_p) : g\equiv \text{id} \, (\text{mod } p)\}$.
  • The pro-$p$ completion of any topological group, for example the free pro-$p$ group $\widehat{F}_p(X)$ on a set $X$ is the pro-$p$ completion of the discrete free group on $X$.

I'm not quite convinced that pro-$p$ groups are ubiquitous. For example when you first learn about solvable groups, it might not be clear that they are common. But if you view them as groups which can be built up as extension by abelian groups then suddenly you nod and say "Yeah I guess I could make one if I wanted to."

(I suppose you could say the same thing about pro-$p$ groups: if you really wanted to, you could sit down and make an inverse system of finite $p$-groups.)

On the other hand I don't feel that the definition of a pro-$p$ group is contrived, but I can't justify this without having accessible examples. I have no idea how to construct counterexamples. My main issues are here:

  • Find non-powerful pro-$p$ groups, one finite and one infinite.
  • Find a non-nilpotent pro-$p$ group.
  • Find a (finitely-generated) pro-$p$ group of infinite rank.
  • Find a finite rank pro-$p$ group with an infinitely-generated subgroup.

Do people usually use generators and relations to construct such groups?

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    $\begingroup$ In my not so humble option, you need to be more conservative with your use of the term "trivial". Trivial does not mean "well know". $\endgroup$
    – RghtHndSd
    Dec 1, 2014 at 2:23
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    $\begingroup$ Isn't $\text{GL}_2(\mathbb{Z}_p)$ already an example for the first two? And for the fourth, if "rank" means Prüfer rank then doesn't finite rank imply that every subgroup is finitely generated or am I misreading the Wikipedia article? $\endgroup$ Dec 1, 2014 at 10:21
  • $\begingroup$ $\text{GL}_2(\mathbf{Z}_p)$ is certainly a $p$-adic analytic group, I don't know if it's pro-$p$ though. And indeed you're right for the second part, I should have said find a finitely-generated pro-$p$ group with infinitely-generated subgroup. For powerful groups it's known that if there are $d$ generators then every subgroup has $\leq d$ generators, i.e. the rank is equal to the minimal number of generators. $\endgroup$
    – Ehsaan
    Dec 2, 2014 at 15:42

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To answer some of the questions you can consider $\mathbb{Z}_p \wr \mathbb{Z}_p$. It is a finitely generated pro-$p$ group of infinite rank. Also it is solvable but not nilpotent.

Finally, in the fourth example, a pro-$p$ group of finite rank cannot have a (topologically) infinitely generated (closed) subgroup by definition. Did you mean something different?

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  • $\begingroup$ Could you please clarify what do you mean with $\mathbb{Z}_p \wr \mathbb{Z}_p$? $\endgroup$
    – N.B.
    Nov 8, 2019 at 15:26
  • $\begingroup$ @N.B. It's a wreath product :) $\endgroup$
    – Ehsaan
    Nov 8, 2019 at 16:50
  • $\begingroup$ Can you define it explicitly? Is it $\mathbb{Z}_p^{\mathbb{Z}_p}\rtimes \mathbb{Z}_p$ with action of the second copy of $\mathbb{Z_p}$ on the coordinates of $\mathbb{Z}_p^{\mathbb{Z}_p}$? $\endgroup$
    – N.B.
    Nov 11, 2019 at 14:58
  • $\begingroup$ @N.B. Yes indeed $\endgroup$ Dec 11, 2019 at 11:28
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    $\begingroup$ @debanjana yes, it is true. It follows from the same fact for Lie algebras $\endgroup$ Mar 10, 2020 at 9:17

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