Let $G$ be a finite group, $p$ a prime number. We denote by $\mathbb{F}_p$ the field of cardinality $p$. Let $V$ be an infinite dimensional representation of $G$ over $\mathbb{F}_p$.

Must there be $G$-invariant, proper subspaces $U,W \leq V$ such that $U + W = V$?

I do not require the sum to be direct. The question should be equivalent to asking:

Must $V$ have a nontrivial decomposable image?

I am able to prove this if $p$ does not divide $|G|$ by applying Maschke's theorem to decompose $V$ into a direct sum of finite-dimensional irreducible subrepresentations. Even if in general the answer is negative, I would like to know about additional cases in which the conclusion holds, to say:

Under what conditions on $G$ can we find such subspaces?

Interesting cases can be abelian, solvable or any other "nice" classes of groups.


I am not sure that I follow Rickard's argument, but here is a direct proof. Given an infinite dimensional module $V$ (over an arbitrary field.) for a finite group $G$, I want to argue first that there exists a proper $G$-submodule $U$ with finite codimension. Let $X < V$ be an arbitrary proper subspace with finite codimension. Then $U = \bigcap_{g \in G} X^g$ is $G$-invariant and is an intersection of finitely many proper subspaces with finite codimension, so is proper and has finite codimension. Next, let $Y \subseteq V$ be a finite dimensional subspace such that $U + Y = V$. Let $W = \sum_{g \in G} Y^g$. Then $W$ is $G$-invariant and finite dimensional, so $W < V$. Also, $U + W = V$, as wanted.


Yes. In fact, $V$ has an infinite dimensional semisimple quotient, which is decomposable since any simple $\mathbb{F}_pG$-module is a quotient of $\mathbb{F}_pG$ and so is finite-dimensional.

Let $V'=\operatorname{rad}(V)=V.\operatorname{rad}(\mathbb{F}_pG)$. Then $V/V'$ is infinite dimensional if $V$ is, and is semisimple. (It's infinite dimensional since any epimorphism from a finite dimensional projective module to $V/V'$ would lift to an epimorphism to $V$.)

For finitely generated profinite groups, as asked about in comments, there are counterexamples. I asked a former colleague, John MacQuarrie, who's worked on modular representations of profinite groups, and the following example is based on his answer (although any errors introduced in translating it into terms I understand are my own work).

Let $V$ be a vector space over $\mathbb{F}_p$ with countable basis $\{e_1,e_2,\dots\}$, and let $\mathbb{Z}$ act on $V$ by letting a generator send $e_i$ to $e_{i-1}+e_i$ for $i>1$ and $e_1$ to $e_1$. If $q$ is a power of $p$ with $q>i$, then $q$ fixes $e_i$, so the action extends to a discrete action of the $p$-adic integers $\mathbb{Z}_p$. It easy to check that the span $V_n=\langle e_1,e_2,\dots,e_n\rangle$ is a submodule for any $0\leq n$, and that these are the only proper submodules. This module can be more naturally described as the Pontryagin dual of the regular representation of the profinite group algebra $\mathbb{F}_p[[\mathbb{Z}_p]]$.

  • $\begingroup$ Thanks a lot! Can this be generalized to the case of profinite $G$ somehow (In this case I assume that $V$ is a discrete $G$-module)? I think that I can construct some counterexample in the general case but what if, say, $G$ is finitely generated? Maybe more restrictions should be put on $G$ in order to make this true in the profinite case? $\endgroup$ – Pablo Apr 29 '14 at 16:03
  • $\begingroup$ I don't follow this. What does "and is semisimple" at the end mean? Projectives need not be semisimple. $\endgroup$ – Benjamin Steinberg Apr 29 '14 at 18:42
  • $\begingroup$ @BenjaminSteinberg: Sorry, I edited the sentence by putting in an explanation in the middle, making it confusing. I meant $V/V'$ is infinite dimensional and semisimple (with an explanation of why it's infinite dimensional in the middle). I'll edit to clarify. $\endgroup$ – Jeremy Rickard Apr 29 '14 at 18:47
  • $\begingroup$ @JeremyRickard, why must an epimorphism from a projective module lift to an epimorphism? $\endgroup$ – Benjamin Steinberg Apr 29 '14 at 19:29
  • $\begingroup$ @BenjaminSteinberg: By projectivity it lifts to a map to $V$. If that map is not an epimorphism and has image $V''$, then $V/V''$ is non-zero and so has a simple quotient, but every map from $V$ to a simple module factors through $V/\operatorname{rad}(V)$. $\endgroup$ – Jeremy Rickard Apr 29 '14 at 20:15

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.