# Action of a profinite group

Let $G$ be a finitely generated profinite group, $p$ a prime number. Put $$V = \prod_{i \in I} \mathbb{Z}_p$$ a (profinite) group equipped with the product topology (for convenience, $I$ may be assumed to be countable). Suppose that $G$ acts by continuous automorphisms on $V$ (this means that $G$ acts continuously on $V$ respecting its group law, i.e $V$ is a profinte $G$-module. I am mainly interested in actions coming from extensions).

1. Must $V$ contain a closed, nontrivial, topologically finitely generated subgroup invariant under the action of $G$?

2. Must $V$ contain a nontrivial pair of trivially intersecting $G$-submodules?

3. Is there a way to decompose $V$ into a nontrivial direct product of $G$-submodules?

I am equally interested in the case of $\mathbb{F}_p$ (the field of cardinality $p$) in place of $\mathbb{Z}_p$ (the $p$-adic integers).

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Does $\mathbb{Z}_p$ denotes the $p$-adics? Do you mean $p\in I$ instead of $i\in I$ (otherwise the product is meaningless). What is $\mathbb{F}_p$? Also "i.e. $V$ is a profinite $G$-module" is not just a restatement: there is something (not hard) to check, which fails when $G$ is not assumed compact. –  YCor Apr 22 '14 at 17:05
$p$ is a fixed prime not to be changed. $\mathbb{F}_p$ is a field with $p$ elements. The infinite product means that we take a power of disjoint copies of the same group, to say, vectors with coordinates in $\mathbb{Z}_p$ (only one group - not changing the prime). $G$ is assumed to be a finitely generated profinite group so compactness is not an issue. –  Pablo Apr 22 '14 at 17:12
Ah OK you just mean $\mathbb{Z}_p^I$... Question 1 looks weird: $\{0\}$ answers positively the question. Also the closed submodule generated by any element is t.f.g. as a submodule... What do you mean? A nonzero submodule that is t.f.g. as topological group? –  YCor Apr 22 '14 at 17:15
Yes, I mean that the submodule will be nonntrivial, and generated as a group (forgetting the action of $G$) by finitely many elements. –  Pablo Apr 22 '14 at 17:21
I suspect that I don't understand the question, but if you take $G$ finite of order not divisible by $p$, $M$ any non-trivial irreducible $\mathbb{Q}_p[G]$-module, and $V$ a $\mathbb{Z}_p[G]$-lattice in $M$, then isn't the answer to all your questions trivially (no pun intended) "no"? –  Alex B. Apr 22 '14 at 20:24

As with my answer to Pablo's Pontryagin dual version of this question in the other thread, the following emerged from discussions with John MacQuarrie, who knows much more about this stuff than I do.

Let $G=\mathbb{Z}_p$. Then the completed group algebra $\mathbb{Z}_p[[G]]$ is isomorphic to the power series algebra $\mathbb{Z}_p[[T]]$, where a generator of $G$ corresponds to $1+T$ (see, for example, Theorem 7.3.3 in John Wilson's book "Profinite Groups").

Let $V$ be the regular representation of $\mathbb{Z}_p[[G]]$, so as a $\mathbb{Z}_p$-module it is a countable direct product of copies of $\mathbb{Z}_p$ as required

In fact, this is precisely the same module that Julian Rosen used to give an answer to question 1.

Suppose $M$ and $N$ are closed non-trivial submodules, and let $m$ and $n$ be non-zero elements. Then $mn\in U\cap V$, and $mn\neq0$ since the power series ring has no zero divisors. So the intersection of two non-trivial submodules can never be trivial, answering question 2 (assuming the submodules are supposed to be closed).

Also, of course this also shows that the answer to question 3 is "no" for this module.

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Is $V$ finitely generated as a $G$-module in this case? –  Pablo Jun 4 '14 at 7:08
@Pablo: It's certainly finitely generated in the topological sense (the smallest closed submodule containing $1$ is the whole of $V$). Is that what you mean? –  Jeremy Rickard Jun 4 '14 at 10:00
Yes I think so. This means that if I take the semidirect product of $G$ by $V$ with this action I get a finitely generated profinite group, right? –  Pablo Jun 4 '14 at 13:16

The answer to the first question is no. Take as index set $I=\mathbb{N}$. Let $\varphi$ be the continuous automorphism of $V=\mathbb{Z}_p^{\mathbb{N}}$ given by $$\varphi:(x_1,x_2,\ldots)\mapsto (x_1,x_2+x_1,x_3+x_2,\ldots,x_n+x_{n-1},\ldots).$$ This gives an action of $\mathbb{Z}$ on $V$, letting $1$ act by $\varphi$. For $v\in V$, it can be checked that $\varphi^{p^n}(v)\to v$ as $n\to\infty$ uniformly in $v$, so our action extends to an action of $\mathbb{Z}_p$ on $V$. For non-zero $v\in V$, the elements $(\varphi-\mathrm{Id}_V)^n(v)$, for $n=1,2,\ldots$, are linearly independent over $\mathbb{Z}_p$ (their first non-zero coordinates are all in different places). Every non-zero $G$-invariant subgroup contains the orbit of a non-zero element, so cannot be topologically finitely generated.

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This works also for $\mathbb{F}_p$ instead of $\mathbb{Z}_p$? meaning that $V$ is changed accordingly and the action of $G$ is the same. –  Pablo Apr 22 '14 at 21:37
Yes, I think so. –  Julian Rosen Apr 22 '14 at 21:38
Is $V$ finitely generated as a $G$-module in this case? –  Pablo Jun 4 '14 at 7:08

By applying Pontryagin's duality, $V^*$ becomes a discrete $G$-module. The appropriate reformulation of question 2 to this case is treated in Decomposing representations of finite groups

An example of an action is given there so the answer to the question is "no", for $\mathbb{F}_p$ at least.

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I hadn't noticed this question before. You probably realize this, but Julian's example, done over $\mathbb{F}_p$ instead of $\mathbb{Z}_p$, is precisely the Pontryagin dual of the example I gave in the other thread. Unfortunately, over $\mathbb{Z}_p$ it doesn't give an answer to question 2. –  Jeremy Rickard Apr 30 '14 at 16:12
Yes, you are right. I am still curious about the case of $\mathbb{Z}_p$. What happens there? –  Pablo Apr 30 '14 at 17:15
In Julian's example, I think that $(p^2,1,0,0,\dots)$ and $(p^2,p,0,0,\dots)$ generate closed submodules with trivial intersection, so the counterexample that works over $\mathbb{F}_p$ doesn't seem to generalize in a straightforward way to $\mathbb{Z}_p$. –  Jeremy Rickard Apr 30 '14 at 18:46
What if $G$ is of order coprime to $p$? Can we say something in the $\mathbb{Z}_p$ case then? –  Pablo Apr 30 '14 at 19:10
Ignore my previous comment: I think I was confused, and Julian's example does answers question in the negative, even over $\mathbb{Z}_p$. –  Jeremy Rickard May 1 '14 at 9:38