Timeline for Decomposing representations of finite groups
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| Apr 30, 2014 at 11:22 | comment | added | Pablo | Great! It answers my question on the profinite case. Maybe you know what happens if you replace $\mathbb{F}_p$ by $\mathbb{Z}_p$ before applying Pontryagin's duality? What I mean is that I take a finitely generated profinite group $G$, and a profinite $G$-module $V$, which is a power of $\mathbb{Z}_p$. Must there be two nontrivial disjoint $G$-submodules? Your example answers this question in the negative in the $\mathbb{F}_p$ case. Dose the analogous example works for the $\mathbb{Z}_p$ case? Maybe there is another counterexample instead? | |
| Apr 30, 2014 at 9:50 | comment | added | Jeremy Rickard | @Pablo: I've edited my answer to add an answer to your question about profinite groups. | |
| Apr 30, 2014 at 9:40 | history | edited | Jeremy Rickard | CC BY-SA 3.0 |
added counterexample for profinite groups
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| Apr 29, 2014 at 22:08 | comment | added | Marty Isaacs | @JeremyRickard Right. Every module over a semisimple algebra is semisimple by a Zorn argument, and that implies that over a semisimple algebra, every nonzero module has a simple quotient. I see now that that does answer the original question using the fact that V/V' is not simple because it is infinite dimensional. | |
| Apr 29, 2014 at 21:08 | comment | added | Jeremy Rickard | $V/V'$ is a module for $\mathbb{F}_pG/\operatorname{rad}(\mathbb{F}_pG)$, which is a semisimple algebra, and all modules (even infinitely generated) for semisimple algebras are direct sums of simple modules. | |
| Apr 29, 2014 at 20:59 | comment | added | Benjamin Steinberg | Fg is not needed to give that modules over a semisimple ring are direct sums of simples. But I think fg is need to guarantee a maximal proper submodule. | |
| Apr 29, 2014 at 20:52 | comment | added | Marty Isaacs | @JeremyRickard I suppose that your assertion that V/V' is semisimple means that it is a direct sum of simples, but I don't see why that holds, even if the radical is trivial. Doesn't proof of semisimplicty require assuming that the module is finitely generated. I don't see why your argument shows that V has a simple quotient or why it answers the original question. Can you explain? | |
| Apr 29, 2014 at 20:15 | comment | added | Jeremy Rickard | @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 $V/\operatorname{rad}(V)$. | |
| Apr 29, 2014 at 19:29 | comment | added | Benjamin Steinberg | @JeremyRickard, why must an epimorphism from a projective module lift to an epimorphism? | |
| Apr 29, 2014 at 18:49 | history | edited | Jeremy Rickard | CC BY-SA 3.0 |
Tried to make the argument clearer.
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| Apr 29, 2014 at 18:47 | comment | added | Jeremy Rickard | @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. | |
| Apr 29, 2014 at 18:42 | comment | added | Benjamin Steinberg | I don't follow this. What does "and is semisimple" at the end mean? Projectives need not be semisimple. | |
| Apr 29, 2014 at 17:06 | vote | accept | Pablo | ||
| Apr 29, 2014 at 20:48 | |||||
| Apr 29, 2014 at 16:03 | comment | added | Pablo | 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? | |
| Apr 29, 2014 at 15:50 | history | answered | Jeremy Rickard | CC BY-SA 3.0 |