MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $p$ be a prime number, and denote by $\mathbb{Z}_p$ the additive profinite group of p-adic integers. Let $G$ be a finitely generated profinite group of order coprime to $p$, and $V = \mathbb{Z}_p^{\mathbb{N}}$ a $G$-module equipped with the product topology ($V$ is a profinite $G$-module). My questions are:

  1. Must $V$ contain a closed, nontrivial, finitely generated $G$-submodule? (I want the submodule to be finitely generated as a profinite subgroup, in the topological sense, forgetting about the action of $G$).

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

I know that the answer to both questions is negative if the order of $G$ is not assumed to be coprime to $p$ as shown in: Action of a profinite group and in: Decomposing representations of finite groups. On the other hand, if $\mathbb{Z}_p$ is replaced by $\mathbb{F}_p$ (the field of cardinality $p$ with the discrete topology) both questions have positive answers (this can be shown by applying Pontryagin's duality and Maschke's theorem which give a direct product decomposition of $V$ into finite submodules).

In light of this, it seems possible that there is a proof using Pontryagin's duality, so I restate 2 in the dual language. Let $A = C_{p^{\infty}}$ be the multiplicative group of all complex roots of unity whose order is a finite power of $p$. Let $ U = \bigoplus_{n \in \mathbb{N}} A$ be the direct sum of countably many isomorphic copies of $A$ equipped with the discrete topology. Suppose that $U$ is a discrete $G$-module. Now 2 reads:

2.a. Must $U$ contain proper submodules $A,B \leq U$ satisfying $A + B = U$?

Yet another reformulation is:

2.b Does $U$ admit a decomposable (as a direct sum) quotient?

share|cite|improve this question

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.