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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.

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up vote 14 down vote accepted

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.

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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 $\mathbb{F}_p[[\mathbb{Z}_p]]$.

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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? –  Pablo Apr 29 at 16:03
    
I don't follow this. What does "and is semisimple" at the end mean? Projectives need not be semisimple. –  Benjamin Steinberg Apr 29 at 18:42
    
@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. –  Jeremy Rickard Apr 29 at 18:47
    
@JeremyRickard, why must an epimorphism from a projective module lift to an epimorphism? –  Benjamin Steinberg Apr 29 at 19:29
    
@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 $V/\operatorname{rad}(V)$. –  Jeremy Rickard Apr 29 at 20:15
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