Action of a profinite group

Question

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).

Answer 1

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.

Answer 2

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.

Answer 3

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.