Let $p$ be a prime number. A pro-$p$ group is the inverse limit of an inverse system of discrete finite $p$-groups. (I've just read the definition).

I have two questions:

  1. Is it true that a the $p$-part of an abelian profinite group is a pro-$p$ group? (I think so)
  2. How is an abelian pro-$p$ group a $\mathbb{Z}_p$-module?
  • 1
    $\begingroup$ A pro-$p$-group need not to be commutative: in such a case, it is not a $\Bbb{Z}_p$-module. $\endgroup$ – Crostul Aug 28 '16 at 18:29
  • $\begingroup$ @Crostul: ok, thanks. $\endgroup$ – user72870 Aug 28 '16 at 18:36
  • 4
    $\begingroup$ You can see an inverse limit as a subgroup of the product of your inverse system (tell me if you need some reference for this). In this case, operations are defined componentwise. If all the groups are abelian, their profinite completion is abelian too. Since the elements of a $p$-group have order a power of $p$, I believe that finite abelian $p$-groups are naturally $\mathbb{Z}_p$-modules (can you guess the action?), so that their profinite completion should be a $\mathbb{Z}_p$-module too, again with componentwise operations. $\endgroup$ – 57Jimmy Aug 28 '16 at 20:09
  • $\begingroup$ @57Jimmy: If a finite abelian group $A$ is of order $p^m$, then $\mathbb{Z}/p^m\mathbb{Z}$ acts on $A$ trought the rule $(k+p^m\mathbb{Z})a=ka$. And since $\mathbb{Z}_p=\varprojlim \mathbb{Z}/p^m\mathbb{Z}$ this defines also an action of $\mathbb{Z}_p$ on $A$. Right? $\endgroup$ – user72870 Aug 29 '16 at 9:19
  • 1
    $\begingroup$ 1. How do you define "the p-part of a pro-p-group" ? Already for a finite group, you can only define its p-Sylow-subgroups. $\endgroup$ – nguyen quang do Aug 29 '16 at 12:33

EDITED: (I have inserted here at the top an answer to your first question.)

There is a $p$-part of an abelian profinite group $G$, but it’s not what you wanted to define it as. You must define it as $$ G_p = \left\lbrace g\in G:\lim_{n\to\infty}g^{p^n}=e_G\right\rbrace\,. $$ You easily check that this $G_p$ is a subgroup of your abelian profinite group. Verification that $G_p$ is closed in $G$, thus compact, thus a profinite group, requires a few more words.

I’ll use additive notation in $G$, and use the letter $U$ for the open subgroups that define the topology of $G$. What is the condition that $g\in G_p$? \begin{align} g\in G_p&\Longleftrightarrow\forall U,\exists n_0\text{ such that }\forall n\ge n_0, p^ng\in U\\ &\Longleftrightarrow\forall U,\exists n\text{ such that }p^ng\in U\\ &\Longleftrightarrow\forall U,\exists n\text{ such that }g\in p^{-n}U\\ &\Longleftrightarrow g\in\bigcap_U\bigcup_np^{-n}U\, \end{align} Now, for a given U, this group $\bigcup_np^{-n}U$ is the total union of an ascending chain of opens, call it $U'$. It’s open, so closed, and we’re taking an intersection of closed subgroups $U'$, and that’s closed. Therefore compact, so a profinite group, and clearly a pro-$p$-group.

Here’s how an abelian pro-$p$-group $G$ is a $\Bbb Z_p$-module:

Let $z\in\Bbb Z_p$, and exhibit $z$ as a $p$-adically convergent sequence of positive integers, $z=\lim_in_i$. (If you like, you can take the $n_i$’s to be the partial sums in the standard representation of $z$ as a “power series in $p$”.) Now, for $g\in G$, define $g^z=\lim_ig^{n_i}$. You need to prove a few things, but they’re easy enough.

Note that the definition of this operation had nothing to do with abelianness of $G$: it’s always defined on a pro-$p$-group. It’s just that you don’t get a module structure without the abelian condition.

  • $\begingroup$ Thank you! Do you think is it positive the answer to the question (1)? So I can accept your answer. $\endgroup$ – user72870 Sep 1 '16 at 23:17
  • $\begingroup$ How do you define the $p$-part? $\endgroup$ – Lubin Sep 2 '16 at 2:34
  • $\begingroup$ If $G$ is an abelian group, I define the $p-$part of $G$ as $\{x\in G\mid \exists n\in \mathbb{N}, \text{ord}(x)=p^n\}$ $\endgroup$ – user72870 Sep 2 '16 at 8:23
  • $\begingroup$ But in a profinite group, most elements are not torsion. Consider $\widehat{\Bbb Z}$, where there is no torsion but your $p$-part is a copy of $\Bbb Z_p$. $\endgroup$ – Lubin Sep 2 '16 at 13:01
  • $\begingroup$ Are you saying that it makes much more sense to define the $p$−part of a profinite abelian group $G=\varprojlim G_i$ as the inverse limit of the $p-$Sylow of $G_i$? (I know that one should prove the existence of such inverse limit) $\endgroup$ – user72870 Sep 2 '16 at 18:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.